Discrete Mathematical Structures, Sixth Edition, offers a clear and concise presentation of the fundamental concepts of

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*Table of contents : CoverTable of Contents1. Fundamentals2. Logic3. Counting4. Relations and Digraphs5. Functions6. Order Relations and Structures7. Trees8. Topics in Graph Theory9. Semigroups and Groups10. Groups and Coding11. Languages and Finite-State MachinesAppendix: Algorithms and PseudocodeAppendix: Additional Experiments in Discrete MathematicsGlossary A B C D E F G I L M O P R S T U WIndex A B C D E F G H I L M N O P Q R S T U V W Y Z*

Discrete Mathematical Structures B. Kolman R. Busby S. Ross Sixth Edition

Pearson Education Limited Edinburgh Gate Harlow Essex CM20 2JE England and Associated Companies throughout the world Visit us on the World Wide Web at: www.pearsoned.co.uk © Pearson Education Limited 2014 All rights reserved. No part of this publication may be reproduced, stored in a retrieval system, or transmitted in any form or by any means, electronic, mechanical, photocopying, recording or otherwise, without either the prior written permission of the publisher or a licence permitting restricted copying in the United Kingdom issued by the Copyright Licensing Agency Ltd, Saffron House, 6–10 Kirby Street, London EC1N 8TS. All trademarks used herein are the property of their respective owners. The use of any trademark in this text does not vest in the author or publisher any trademark ownership rights in such trademarks, nor does the use of such trademarks imply any afﬁliation with or endorsement of this book by such owners.

ISBN 10: 1-292-02484-4 ISBN 13: 978-1-292-02484-4

British Library Cataloguing-in-Publication Data A catalogue record for this book is available from the British Library Printed in the United States of America

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Table of Contents 1. Fundamentals Bernard Kolman/Robert C. Busby/Sharon Cutler Ross

1

2. Logic Bernard Kolman/Robert C. Busby/Sharon Cutler Ross

57

3. Counting Bernard Kolman/Robert C. Busby/Sharon Cutler Ross

105

4. Relations and Digraphs Bernard Kolman/Robert C. Busby/Sharon Cutler Ross

139

5. Functions Bernard Kolman/Robert C. Busby/Sharon Cutler Ross

205

6. Order Relations and Structures Bernard Kolman/Robert C. Busby/Sharon Cutler Ross

247

7. Trees Bernard Kolman/Robert C. Busby/Sharon Cutler Ross

307

8. Topics in Graph Theory Bernard Kolman/Robert C. Busby/Sharon Cutler Ross

349

9. Semigroups and Groups Bernard Kolman/Robert C. Busby/Sharon Cutler Ross

393

10. Groups and Coding Bernard Kolman/Robert C. Busby/Sharon Cutler Ross

441

11. Languages and Finite-State Machines Bernard Kolman/Robert C. Busby/Sharon Cutler Ross

469

Appendix: Algorithms and Pseudocode Bernard Kolman/Robert C. Busby/Sharon Cutler Ross

519

Appendix: Additional Experiments in Discrete Mathematics Bernard Kolman/Robert C. Busby/Sharon Cutler Ross

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I

II

Glossary Bernard Kolman/Robert C. Busby/Sharon Cutler Ross

535

Index

537

Fundamentals Prerequisites: There are no formal prerequisites for this chapter; the reader is encouraged to read carefully and work through all examples. In this chapter we introduce some of the basic tools of discrete mathematics. We begin with sets, subsets, and their operations, notions with which you may already be familiar. Next we deal with sequences, using both explicit and recursive patterns. Then we review some of the basic properties of the integers. Finally we introduce matrices and matrix operations. This gives us the background needed to begin our exploration of mathematical structures.

Looking Back: Matrices The origin of matrices goes back to approximately 200 b.c.e., when they were used by the Chinese to solve linear systems of equations. After being in the shadows for nearly two thousand years, matrices came back into mathematics toward the end of the seventeenth century and from then research in this area proceeded at a rapid pace. The term “matrix” (the singular of “matrices”) was coined in 1850 by James Joseph Sylvester

(1814–1897), a British mathematician and lawyer. In 1851, Sylvester met Arthur Cayley (1821–1895), also a British lawyer with a strong interest in mathematics. Cayley quickly realized the importance of the notion of a matrix and in 1858 published a book showing the basic operations on matrices. He also discovered a number of important results in matrix theory.

James Joseph Sylvester Library of Congress

Arthur Cayley CORBIS BETTMANN

From Discrete Mathematical Structures, Sixth Edition, Bernard Kolman, Robert C. Busby, Sharon Cutler Ross. Copyright © 2009 by Pearson Education, Inc. Published by Prentice Hall. All rights reserved.

1

Fundamentals

1

Sets and Subsets Sets A set is any well-defined collection of objects called the elements or members of the set. For example, the collection of all wooden chairs, the collection of all one-legged black birds, or the collection of real numbers between zero and one are all sets. Well-defined just means that it is possible to decide if a given object belongs to the collection or not. Almost all mathematical objects are first of all sets, regardless of any additional properties they may possess. Thus set theory is, in a sense, the foundation on which virtually all of mathematics is constructed. In spite of this, set theory (at least the informal brand we need) is quite easy to learn and use. One way of describing a set that has a finite number of elements is by listing the elements of the set between braces. Thus the set of all positive integers that are less than 4 can be written as {1, 2, 3}.

(1)

The order in which the elements of a set are listed is not important. Thus {1, 3, 2}, {3, 2, 1}, {3, 1, 2}, {2, 1, 3}, and {2, 3, 1} are all representations of the set given in (1). Moreover, repeated elements in the listing of the elements of a set can be ignored. Thus, {1, 3, 2, 3, 1} is another representation of the set given in (1). We use uppercase letters such as A, B, C to denote sets, and lowercase letters such as a, b, c, x, y, z, t to denote the members (or elements) of sets. We indicate the fact that x is an element of the set A by writing x ∈ A, and we indicate the fact that x is not an element of A by writing x ∈ / A. Example 1

Let A = {1, 3, 5, 7}. Then 1 ∈ A, 3 ∈ A, but 2 ∈ / A.

◆

Sometimes it is inconvenient or impossible to describe a set by listing all of its elements. Another useful way to define a set is by specifying a property that the elements of the set have in common. We use the notation P(x) to denote a sentence or statement P concerning the variable object x. The set defined by P(x), written {x | P(x)}, is just the collection of all objects for which P is sensible and true. {x | P(x)} is read, “the set of all x such that P(x).” For example, {x | x is a positive integer less than 4} is the set {1, 2, 3} described in (1) by listing its elements. Example 2

The set consisting of all the letters in the word “byte” can be denoted by {b, y, t, e} ◆ or by {x | x is a letter in the word “byte”}.

Example 3

We introduce here several sets and their notations. (a) Z+ = {x | x is a positive integer}. Thus Z+ consists of the numbers used for counting: 1, 2, 3 . . . . (b) N = {x | x is a positive integer or zero} = {x | x is a natural number}. Thus N consists of the positive integers and zero: 0, 1, 2, . . . . (c) Z = {x | x is an integer}. Thus Z consists of all the integers: . . . , −3, −2, −1, 0, 1, 2, 3, . . . . (d) Q = {x | x is a rational number}. a Thus Q consists of numbers that can be written as , where a and b are integers b and b is not 0.

2

Fundamentals (e) R = {x | x is a real number}. (f) The set that has no elements in it is denoted either by { } or the symbol ∅ and is called the empty set. ◆ Example 4

Since the square of a real number is always nonnegative, {x | x is a real number and x2 = −1} = ∅.

◆

Sets are completely known when their members are all known. Thus we say two sets A and B are equal if they have the same elements, and we write A = B. Example 5

If A = {1, 2, 3} and B = {x | x is a positive integer and x2 < 12}, then A = B. ◆

Example 6

If A = {JAVA, PASCAL, C++} and B = {C++, JAVA, PASCAL}, then A = B. ◆

B

Subsets

B

A

A

A⊆B

AB

Figure 1

If every element of A is also an element of B, that is, if whenever x ∈ A then x ∈ B, we say that A is a subset of B or that A is contained in B, and we write A ⊆ B. If A is not a subset of B, we write A B. (See Figure 1.) Diagrams, such as those in Figure 1, which are used to show relationships between sets, are called Venn diagrams after the British logician John Venn. Venn diagrams will be used extensively in Section 2.

Example 7

We have Z+ ⊆ Z. Moreover, if Q denotes the set of rational numbers, then Z ⊆ Q. ◆

Example 8

Let A = {1, 2, 3, 4, 5, 6}, B = {2, 4, 5}, and C = {1, 2, 3, 4, 5}. Then B ⊆ A, ◆ B ⊆ C, and C ⊆ A. However, A B, A C, and C B.

Example 9

If A is any set, then A ⊆ A. That is, every set is a subset of itself.

Example 10

◆

Let A be a set and let B = {A, {A}}. Then, since A and {A} are elements of B, we have A ∈ B and {A} ∈ B. It follows that {A} ⊆ B and {{A}} ⊆ B. However, it is ◆ not true that A ⊆ B. For any set A, since there are no elements of ∅ that are not in A, we have

∅ ⊆ A.

U

A

Figure 2

It is easy to see that A = B if and only if A ⊆ B and B ⊆ A. This simple statement is the basis for proofs of many statements about sets. The collection of everything, it turns out, cannot be considered a set without presenting serious logical difficulties. To avoid this and other problems, which need not concern us here, we will assume that for each discussion there is a “universal set” U (which will vary with the discussion) containing all objects for which the discussion is meaningful. Any other set mentioned in the discussion will automatically be assumed to be a subset of U. Thus, if we are discussing real numbers and we mention sets A and B, then A and B must (we assume) be sets of real numbers, not matrices, electronic circuits, or rhesus monkeys. In most problems, a universal set will be apparent from the setting of the problem. In Venn diagrams, the universal set U will be denoted by a rectangle, while sets within U will be denoted by circles as shown in Figure 2.

3

Fundamentals A set A is called finite if it has n distinct elements, where n ∈ N. In this case, n is called the cardinality of A and is denoted by |A|. Thus, the sets of Examples 1, 2, 4, 5, and 6 are finite. A set that is not finite is called infinite. The sets introduced in Example 3 (except ∅) are infinite sets. If A is a set, then the set of all subsets of A is called the power set of A and is denoted by P(A). (Be sure to distinguish between P(A), a statement about A, and P(A), the power set of A.) Example 11

Let A = {1, 2, 3}. Then P(A) consists of the following subsets of A: { }, {1}, {2}, {3}, {1, 2}, {1, 3}, {2, 3}, and {1, 2, 3} (or A). In a later section, we will count the ◆ number of subsets that a set can have.

1 Exercises 1. Let A = {1, 2, 4, a, b, c}. Identify each of the following as true or false. (b) 3 ∈ A (c) c ∈ /A (a) 2 ∈ A (e) { } ∈ /A (f) A ∈ A (d) ∅ ∈ A 2. Let A = {x | x is a real number and x < 6}. Identify each of the following as true or false. (b) 6 ∈ A (c) 5 ∈ /A (a) 3 ∈ A /A (e) −8 ∈ A (f) 3.4 ∈ /A (d) 8 ∈ 3. In each part, give the set of letters in each word by listing the elements of the set. (b) BOOK (a) AARDVARK (c) MISSISSIPPI 4. Give the set by listing its elements. (a) The set of all positive integers that are less than ten. (b) {x | x ∈ Z and x2 < 12} 5. Let A = {1, {2, 3}, 4}. Identify each of the following as true or false. (b) {1, 4} ⊆ A (c) {2, 3} ⊆ A (a) 3 ∈ A (d) {2, 3} ∈ A

(e) {4} ∈ A

(f) {1, 2, 3} ⊆ A

In Exercises 6 through 9, write the set in the form {x | P(x)}, where P(x) is a property that describes the elements of the set. 6. {2, 4, 6, 8, 10}

7. {a, e, i, o, u}

8. {1, 8, 27, 64, 125}

9. {−2, −1, 0, 1, 2}

10. Let A = {1, 2, 3, 4, 5}. Which of the following sets are equal to A? (a) {4, 1, 2, 3, 5} (b) {2, 3, 4} (c) {1, 2, 3, 4, 5, 6} (d) {x | x is an integer and x2 ≤ 25} (e) {x | x is a positive integer and x ≤ 5} (f) {x | x is a positive rational number and x ≤ 5} 11. Which of the following sets are the empty set? (a) {x | x is a real number and x2 − 1 = 0} (b) {x | x is a real number and x2 + 1 = 0} (c) {x | x is a real number and x2 = −9}

4

(d) {x | x is a real number and x = 2x + 1} (e) {x | x is a real number and x = x + 1} 12. List all the subsets of {a, b}. 13. List all the subsets of {JAVA, PASCAL, C++}. 14. List all the subsets of { }. 15. Let A = {1, 2, 5, 8, 11}. Identify each of the following as true or false. (b) {8, 1} ∈ A (a) {5, 1} ⊆ A (c) {1, 8, 2, 11, 5} A (d) ∅ ⊆ A (f) {2} ⊆ A (e) {1, 6} A (g) {3} ∈ /A (h) A ⊆ {11, 2, 5, 1, 8, 4} 16. Let A = {x | x is an integer and x2 < 16}. Identify each of the following as true or false. (b) {−3, −2, −1} ⊆ A (a) {0, 1, 2, 3} ⊆ A (c) { } ⊆ A (d) {x | x is an integer and |x| < 4} ⊆ A (e) A ⊆ {−3, −2, −1, 0, 1, 2, 3} 17. Let A = {1}, B = {1, a, 2, b, c}, C = {b, c}, D = {a, b}, and E = {1, a, 2, b, c, d}. For each part, replace the symbol 2 with either ⊆ or to give a true statement. (b) ∅ 2 A (c) B 2 C (a) A 2 B (e) D 2 C (f) B 2 E (d) C 2 E In Exercises 18 through 20, find the set of smallest cardinality that contains the given sets as subsets. 18. {a, b, c}, {a, d, e, f }, {b, c, e, g} 19. {1, 2}, {1, 3}, ∅ 20. {2, 4, 6, . . . , 20}, {3, 6, 9, . . . , 21} 21. Is it possible to have two different (appropriate) universal sets for a collection of sets? Would having different universal sets create any problems? Explain. 22. Use the Venn diagram in Figure 3 to identify each of the following as true or false. (b) B ⊆ A (c) C ⊆ B (a) A ⊆ B (e) x ∈ A (f) y ∈ B (d) x ∈ B

Fundamentals U y A

B x

25. Complete the following statement. A generic Venn diaregions. Describe them in gram for three sets has words. 26. (a) If A = {3, 7}, find P(A).

C

(b) What is |A|?

(c) What is |P(A)|?

27. If P(B) = {{ }, {m}, {n}, {m, n}}, then find B. 28. (a) If A = {3, 7, 2}, find P(A). (b) What is |A|?

Figure 3 23. Use the Venn diagram in Figure 4 to identify each of the following as true or false. (b) A ⊆ C (c) C ⊆ B (a) B ⊆ A (e) t ∈ A (f) w ∈ B (d) w ∈ A U A

B

t

29. If P(B) = {{a}, { }, {c}, {b, c}, {a, b}, . . . } and |P(B)| = 8, then B = . In Exercises 30 through 32, draw a Venn diagram that represents these relationships. 30. A ⊆ B, A ⊆ C, B C, and C B 31. x ∈ A, x ∈ B, x ∈ / C, y ∈ B, y ∈ C, and y ∈ /A 32. A ⊆ B, x ∈ / A, x ∈ B, A C, y ∈ B, y ∈ C 33. Describe all the subset relationships that hold for the sets given in Example 3.

w

34. Show that if A ⊆ B and B ⊆ C, then A ⊆ C. 35. The statement about sets in Exercise 34 can be restated as “Any subset of is also a subset of any set that contains .”

C

Figure 4 24. (a) Complete the following statement. A generic Venn diagram for a single set has regions. Describe them in words. (b) Complete the following statement. A generic Venn regions. Describe diagram for two sets has them in words.

2

(c) What is |P(A)|?

36. Suppose we know that set A has n subsets, S1 , S2 , . . . , Sn . If set B consists of the elements of A and one more element so |B| = |A| + 1, show that B must have 2n subsets. 37. Compare the results of Exercises 12, 13, 26, and 28 and complete the following: Any set with two elements has subsets. Any set with three elements has subsets.

Operations on Sets In this section we will discuss several operations that will combine given sets to yield new sets. These operations, which are analogous to the familiar operations on the real numbers, will play a key role in the many applications and ideas that follow. If A and B are sets, we define their union as the set consisting of all elements that belong to A or B and denote it by A ∪ B. Thus A ∪ B = {x | x ∈ A or x ∈ B}. Observe that x ∈ A ∪ B if x ∈ A or x ∈ B or x belongs to both A and B.

Example 1

Let A = {a, b, c, e, f } and B = {b, d, r, s}. Find A ∪ B. Solution

Since A ∪ B consists of all the elements that belong to either A or B, A ∪ B = ◆ {a, b, c, d, e, f, r, s}. We can illustrate the union of two sets with a Venn diagram as follows. If A and B are the sets in Figure 5(a), then A ∪ B is the set represented by the shaded region in Figure 5(b).

5

Fundamentals A

A B

B

U

U (b) A ∪ B

(a)

Figure 5

If A and B are sets, we define their intersection as the set consisting of all elements that belong to both A and B and denote it by A ∩ B. Thus A ∩ B = {x | x ∈ A and x ∈ B}. Let A = {a, b, c, e, f }, B = {b, e, f, r, s}, and C = {a, t, u, v}. Find A ∩ B, A ∩ C, and B ∩ C.

Example 2

Solution

The elements b, e, and f are the only ones that belong to both A and B, so A ∩ B = {b, e, f }. Similarly, A ∩ C = {a}. There are no elements that belong to both B and ◆ C, so B ∩ C = { }. Two sets that have no common elements, such as B and C in Example 2, are called disjoint sets. We can illustrate the intersection of two sets by a Venn diagram as follows. If A and B are the sets given in Figure 6(a), then A ∩ B is the set represented by the shaded region in Figure 6(b). Figure 7 illustrates a Venn diagram for two disjoint sets. A

A

U

B

B A

U (a)

Figure 6

B

U (b) A ∩ B

Figure 7

The operations of union and intersection can be defined for three or more sets in an obvious manner: A ∪ B ∪ C = {x | x ∈ A or x ∈ B or x ∈ C} and

A ∩ B ∩ C = {x | x ∈ A and x ∈ B and x ∈ C}.

The shaded region in Figure 8(b) is the union of the sets A, B, and C shown in Figure 8(a), and the shaded region in Figure 8(c) is the intersection of the sets A, B, and C. Note that Figure 8(a) says nothing about possible relationships between the sets, but allows for all possible relationships. In general, if A1 , A2 , . . . , An are n subsets of U, then A1 ∪ A2 ∪ · · · ∪ An will be denoted by Ak , read as, “the union k=1

6

Fundamentals U

U

U

B

B

A

B

A

A

C

C

C (c) A ∩ B ∩ C

(b) A ∪ B ∪ C

(a)

Figure 8 n from 1 to n of A sub k,” and A1 ∩ A2 ∩ · · · ∩ An will be denoted by Ak , read as, k=1 “the intersection from 1 to n of A sub k.”

Example 3

Let A = {1, 2, 3, 4, 5, 7}, B = {1, 3, 8, 9}, and C = {1, 3, 6, 8}. Then A ∩ B ∩ C ◆ is the set of elements that belong to A, B, and C. Thus A ∩ B ∩ C = {1, 3}. If A and B are two sets, we define the complement of B with respect to A as the set of all elements that belong to A but not to B, and we denote it by A − B. Thus A − B = {x | x ∈ A and x ∈ / B}.

Example 4

Let A = {a, b, c} and B = {b, c, d, e}. Then A − B = {a} and B − A = {d, e}. ◆ If A and B are the sets in Figure 9(a), then A − B and B − A are represented by the shaded regions in Figures 9(b) and 9(c), respectively. U

A

B

(a)

U

U

A

B

(b) A − B

A

B

(c) B − A

Figure 9

If U is a universal set containing A, then U − A is called the complement of / A}. A and is denoted by A. Thus A = {x | x ∈ Example 5

Let A = {x | x is an integer and x ≤ 4} and U = Z. Then A = {x | x is an integer ◆ and x > 4}. If A is the set in Figure 10, its complement is the shaded region in that figure. If A and B are two sets, we define their symmetric difference as the set of all elements that belong to A or to B, but not to both A and B, and we denote it by A ⊕ B. Thus A ⊕ B = {x | (x ∈ A and x ∈ / B) or (x ∈ B and x ∈ / A)}.

Example 6

Let A = {a, b, c, d} and B = {a, c, e, f, g}. Then A ⊕ B = {b, d, e, f, g}.

◆

7

Fundamentals U

A

Figure 10

If A and B are as indicated in Figure 11(a), their symmetric difference is the shaded region shown in Figure 11(b). It is easy to see that A ⊕ B = (A − B) ∪ (B − A).

U A

B

U A

(a)

B

(b) A ∩ B

Figure 11

Algebraic Properties of Set Operations The operations on sets that we have just defined satisfy many algebraic properties, some of which resemble the algebraic properties satisfied by the real numbers and their operations. All the principal properties listed here can be proved using the definitions given and the rules of logic. We shall prove only one of the properties and leave proofs of the remaining ones as exercises for the reader. Proofs are fundamental to mathematics. Some simple proofs are required in the exercises. Venn diagrams are often useful to suggest or justify the method of proof. THEOREM 1

The operations defined on sets satisfy the following properties: Commutative Properties 1. A ∪ B = B ∪ A 2. A ∩ B = B ∩ A Associative Properties 3. A ∪ (B ∪ C) = (A ∪ B) ∪ C 4. A ∩ (B ∩ C) = (A ∩ B) ∩ C Distributive Properties 5. A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C) 6. A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C)

8

Fundamentals Idempotent Properties 7. A ∪ A = A 8. A ∩ A = A Properties of the Complement 9. 10. 11. 12. 13. 14. 15.

(A) = A A∪A=U A∩A=∅ ∅=U U=∅ A∪B =A∩B A∩B =A∪B

Properties 14 and 15 are known as De Morgan’s laws.

Properties of a Universal Set 16. A ∪ U = U 17. A ∩ U = A Properties of the Empty Set 18. A ∪ ∅ = A or A ∪ { } = A 19. A ∩ ∅ = ∅ or A ∩ { } = { } Proof We prove Property 14 here and leave proofs of the remaining properties as exercises for the reader. A common style of proof for statements about sets is to choose an element in one of the sets and see what we know about it. Suppose that x ∈ A ∪ B. Then we know that x ∈ / A ∪ B, so x ∈ / A and x ∈ / B. (Why?) This means x ∈ A ∩ B (why?), so each element of A ∪ B belongs to A ∩ B. Thus A ∪ B ⊆ A ∩ B. / A and x ∈ / B (why?), so x ∈ / A ∪ B, Conversely, suppose that x ∈ A ∩ B. Then x ∈ which means that x ∈ A ∪ B. Thus each element of A ∩ B also belongs to A ∪ B, ■ and A ∩ B ⊆ A ∪ B. Now we see that A ∪ B = A ∩ B. A∩B

A

The Addition Principle B

Figure 12

THEOREM 2 Example 7

Suppose now that A and B are finite subsets of a universal set U. It is frequently useful to have a formula for |A ∪ B|, the cardinality of the union. If A and B are disjoint sets, that is, if A∩B = ∅, then each element of A∪B appears in either A or B, but not in both; therefore, |A ∪ B| = |A| + |B|. If A and B overlap, as shown in Figure 12, then elements in A ∩ B belong to both sets, and the sum |A| + |B| counts these elements twice. To correct for this double counting, we subtract |A ∩ B|. Thus we have the following theorem, sometimes called the addition principle. Because of Figure 12, this is also called the inclusion-exclusion principle. If A and B are finite sets, then |A ∪ B| = |A| + |B| − |A ∩ B|.

■

Let A = {a, b, c, d, e} and B = {c, e, f, h, k, m}. Verify Theorem 2. Solution

We have A ∪ B = {a, b, c, d, e, f, h, k, m} and A ∩ B = {c, e}. Also, |A| = 5, |B| = 6, |A ∪ B| = 9, and |A ∩ B| = 2. Then |A| + |B| − |A ∩ B| = 5 + 6 − 2 or ◆ 9 and Theorem 2 is verified.

9

Fundamentals A∩B

B

A

A∩B∩C

A∩C

If A and B are disjoint sets, A ∩ B = ∅ and |A ∩ B| = 0, so the formula in Theorem 2 now becomes |A ∪ B| = |A| + |B|. This special case can be stated in a way that is useful in a variety of counting situations. If a task T1 can be performed in exactly n ways, and a different task T2 can be performed in exactly m ways, then the number of ways of performing task T1 or task T2 is n + m.

C B∩C

Figure 13

THEOREM 3

The situation for three sets is shown in Figure 13. We state the three-set addition principle without discussion. Let A, B, and C be finite sets. Then |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − ■ |B ∩ C| − |A ∩ C| + |A ∩ B ∩ C|. Theorem 3 can be generalized for more than three sets. This is done in Exercises 50 and 51.

Example 8

Let A = {a, b, c, d, e}, B = {a, b, e, g, h}, and C = {b, d, e, g, h, k, m, n}. Verify Theorem 3. Solution

We have A ∪ B ∪ C = {a, b, c, d, e, g, h, k, m, n}, A ∩ B = {a, b, e}, A ∩ C = {b, d, e}, B∩C = {b, e, g, h}, and A∩B∩C = {b, e}, so |A| = 5, |B| = 5, |C| = 8, |A∪B∪C| = 10, |A∩B| = 3, |A∩C| = 3, |B∩C| = 4, and |A∩B∩C| = 2. Thus |A|+|B|+|C|−|A∩B|−|B∩C|−|A∩C|+|A∩B∩C| = 5+5+8−3−3−4+2 ◆ or 10, and Theorem 3 is verified. Example 9

Acomputer company wants to hire 25 programmers to handle systems programming jobs and 40 programmers for applications programming. Of those hired, ten will be expected to perform jobs of both types. How many programmers must be hired? Solution

Let A be the set of systems programmers hired and B be the set of applications programmers hired. The company must have |A| = 25 and |B| = 40, and |A∩B| = 10. The number of programmers that must be hired is |A ∪ B|, but |A ∪ B| = |A| + |B| − |A ∩ B|. So the company must hire 25 + 40 − 10 or 55 programmers. ◆ Example 10

Asurvey has been taken on methods of commuter travel. Each respondent was asked to check BUS, TRAIN, or AUTOMOBILE as a major method of traveling to work. More than one answer was permitted. The results reported were as follows: BUS, 30 people; TRAIN, 35 people; AUTOMOBILE, 100 people; BUS and TRAIN, 15 people; BUS and AUTOMOBILE, 15 people; TRAIN and AUTOMOBILE, 20 people; and all three methods, 5 people. How many people completed a survey form? Solution

Let B, T , and A be the sets of people who checked BUS, TRAIN, and AUTOMOBILE, respectively. We know |B| = 30, |T | = 35, |A| = 100, |B ∩ T | = 15, |B ∩ A| = 15, |T ∩ A| = 20, and |B ∩ T ∩ A| = 5. So |B| + |T | + |A| − |B ∩ T | − |B ∩ A| − |T ∩ A| + |B ∩ T ∩ A| = 30 + 35 + 100 − 15 − 15 − 20 + 5 or ◆ 120 is |A ∪ B ∪ C|, the number of people who responded.

10

Fundamentals

2 Exercises In Exercises 1 through 4, let U = {a, b, c, d, e, f, g, h, k}, A = {a, b, c, g}, B = {d, e, f, g}, C = {a, c, f }, and D = {f, h, k}. 1. Compute (a) A ∪ B

(b) B ∪ C

(c) A ∩ C

(e) (A ∪ B) − C (f) A − B (d) B ∩ D (g) A (h) A ⊕ B (i) A ⊕ C (j) (A ∩ B) − C 2. Compute (b) B ∪ D (c) C ∩ D (a) A ∪ D (e) (A ∪ B) − (C ∪ B) (d) A ∩ D (h) C − B (g) B (f) B − C (j) (A ∩ B) − (B ∩ D) (i) C ⊕ D 3. Compute (b) A ∩ B ∩ C (a) A ∪ B ∪ C (d) (A ∪ B) ∩ C (c) A ∩ (B ∪ C) (f) A ∩ B (e) A ∪ B 4. Compute (b) A ∪ U (c) B ∪ B (a) A ∪ ∅ (d) C ∩ { }

(e) C ∪ D

(f) C ∩ D

In Exercises 5 through 8, let U = {1, 2, 3, 4, 5, 6, 7, 8, 9}, A = {1, 2, 4, 6, 8}, B = {2, 4, 5, 9}, C = {x | x is a positive integer and x2 ≤ 16}, and D = {7, 8}. 5. Compute (a) A ∪ B (d) B ∪ C (g) B ∩ C 6. Compute (a) A − B

(b) A ∪ C

(c) A ∪ D

(e) A ∩ C

(f)

A∩D

(c) A ∪ A

In Exercises 12 and 13, refer to Figure 14. U A B w

z

y

x

u v

C

Figure 14 12. Identify the following as true or false. (b) x ∈ B ∪ C (a) y ∈ A ∩ B (d) u ∈ /C (c) w ∈ B ∩ C 13. Identify the following as true or false. (b) y ∈ A ∪ B ∪ C (a) x ∈ A ∩ B ∩ C (d) v ∈ B ∩ C (c) z ∈ A ∩ C 14. Describe the shaded region shown in Figure 15 using unions and intersections of the sets A, B, and C. (Several descriptions are possible.) A

(b) B − A

(c) A ∩ (B ∪ C) (e) A ∪ B 8. Compute (a) B ∪ C ∪ D

B

(c) C − D

(d) (A ∪ B) ∩ D

C

Figure 15

A∩B

15. Let A, B, and C be finite sets with |A| = 6, |B| = 8, |C| = 6, |A ∪ B ∪ C| = 11, |A ∩ B| = 3, |A ∩ C| = 2, and |B ∩ C| = 5. Find |A ∩ B ∩ C|.

(b) B ∩ C ∩ D

In Exercises 16 through 18, verify Theorem 2 for the given sets.

(d) A ∩ A

16. (a) A = {1, 2, 3, 4}, B = {2, 3, 5, 6, 8}

(f)

(c) A ∪ A (e) A ∪ A

(f)

A ∩ (C ∪ D)

In Exercises 9 and 10, let U = {a, b, c, d, e, f, g, h}, A = {a, c, f, g}, B = {a, e}, and C = {b, h}.

(d) A ∩ B

(b) B ∪ C

(e) A ⊕ B (f) B ⊕ C (d) C ∩ C 11. Let U be the set of real numbers, A = {x | x is a solution of x2 − 1 = 0}, and B = {−1, 4}. Compute (b) B (c) A ∪ B (d) A ∩ B (a) A

(h) C ∩ D

(e) A (f) A ⊕ B (d) C (g) C ⊕ D (h) B ⊕ C 7. Compute (b) A ∩ B ∩ C (a) A ∪ B ∪ C

9. Compute (a) A

10. Compute (a) A ∩ B

(b) B

(c) A ∪ B

(e) U

(f) A − B

(b) A = {1, 2, 3, 4}, B = {5, 6, 7, 8, 9} 17. (a) A = {a, b, c, d, e, f }, B = {a, c, f, g, h, i, r} (b) A = {a, b, c, d, e}, B = {f, g, r, s, t, u} 18. (a) A = {x | x is a positive integer < 8}, B = {x | x is an integer such that 2 ≤ x ≤ 5} (b) A = {x | x is a positive integer and x2 ≤ 16}, B = {x | x is a negative integer and x2 ≤ 25}

11

Fundamentals 19. If A and B are disjoint sets such that |A ∪ B| = |A|, what must be true about B? 20. Write Property 14 of Theorem 1 in ordinary English. 21. Write Property 15 of Theorem 1 in ordinary English. In Exercises 22 through 24, verify Theorem 3 for the given sets. 22. A = {a, b, c, d, e}, B = {d, e, f, g, h, i, k}, C = {a, c, d, e, k, r, s, t} 23. A = {1, 2, 3, 4, 5, 6}, B = {2, 4, 7, 8, 9}, C = {1, 2, 4, 7, 10, 12} 24. A = {x | x is a positive integer < 8}, B = {x | x is an integer such that 2 ≤ x ≤ 4}, C = {x | x is an integer such that x2 < 16} 25. In a survey of 260 college students, the following data were obtained: 64 had taken a mathematics course, 94 had taken a computer science course, 58 had taken a business course, 28 had taken both a mathematics and a business course, 26 had taken both a mathematics and a computer science course, 22 had taken both a computer science and a business course, and 14 had taken all three types of courses. (a) How many students were surveyed who had taken none of the three types of courses? (b) Of the students surveyed, how many had taken only a computer science course? 26. A survey of 500 television watchers produced the following information: 285 watch football games, 195 watch hockey games, 115 watch basketball games, 45 watch football and basketball games, 70 watch football and hockey games, 50 watch hockey and basketball games, and 50 do not watch any of the three kinds of games. (a) How many people in the survey watch all three kinds of games? (b) How many people watch exactly one of the sports? 27. The Journalism 101 class recently took a survey to determine where the city’s people obtained their news. Unfortunately, some of the reports were damaged. What we know is that 88 people said they obtained their news from television, 73 from the local paper, and 46 from a news magazine. Thirty-four people reported that they obtained news from television and the local paper, 16 said they obtained their news from television and a news magazine, and 12 obtained theirs from the local paper and a news magazine. A total of five people reported that they used all three media. If 166 people were surveyed, how many use none of the three media to obtain their news? How many obtain their news from a news magazine exclusively? 28. The college catering service must decide if the mix of food that is supplied for receptions is appropriate. Of 100

12

people questioned, 37 say they eat fruits, 33 say they eat vegetables, 9 say they eat cheese and fruits, 12 eat cheese and vegetables, 10 eat fruits and vegetables, 12 eat only cheese, and 3 report they eat all three offerings. How many people surveyed eat cheese? How many do not eat any of the offerings? 29. In a psychology experiment, the subjects under study were classified according to body type and gender as follows:

ENDOMORPH

ECTOMORPH

MESOMORPH

72 62

54 64

36 38

Male Female

(a) How many male subjects were there? (b) How many subjects were ectomorphs? (c) How many subjects were either female or endomorphs? (d) How many subjects were not male mesomorphs? (e) How many subjects were either male, ectomorph, or mesomorph? 30. The following table displays information about the sophomore, junior, and senior classes at Old U.

Class Sophomore (S) Junior (J) Senior (R)

Major Declared (D)

Major Undeclared (U)

143 245 392

289 158 36

For each of the following tell how many students are in the set and describe those students in words. (b) U ∪ R (c) (D ∪ S) ∩ R (a) D ∩ J 31. Create a Venn diagram that displays the information in the table in Exercise 30. 32. Complete the following proof that A ⊆ A ∪ B. Suppose x ∈ A. Then x ∈ A ∪ B, because . Thus by the definition of subset A ⊆ A ∪ B. In Exercises 33 through 38, classify each statement as true, false, or not possible to identify as true or false. 33. Choose x ∈ A ∩ B. (b) x ∈ B (c) x ∈ /A (d) x ∈ /B (a) x ∈ A 34. Choose y ∈ A ∪ B. (b) y ∈ B (c) y ∈ /A (a) y ∈ A /B (e) y ∈ A ∩ B (d) y ∈ 35. Choose z ∈ A ∪ (B ∩ C). (b) z ∈ B (a) z ∈ A (d) z ∈ B ∩ C

(e) z ∈ A

(f) y ∈ / A∩B (c) z ∈ C (f) z ∈ /C

Fundamentals 36. Choose w ∈ D ∩ (E ∪ F). (b) w ∈ E (c) w ∈ F (a) w ∈ D (e) w ∈ F ∪ E (d) w ∈ D (f) w ∈ (D ∩ E) ∪ (D ∩ F) 37. Choose t ∈ D ∩ E. (b) t ∈ E (c) t ∈ D (a) t ∈ D (e) t ∈ D ∪ E (d) t ∈ E 38. Choose x ∈ A ∪ (B ∩ C). (b) x ∈ B (c) x ∈ C (a) x ∈ A (e) x ∈ (A ∪ B) ∩ (A ∪ C) (d) x ∈ A ∪ B 39. Complete the following proof that A ∩ B ⊆ A. Suppose x ∈ A ∩ B. Then x belongs to . Thus A ∩ B ⊆ A. 40. (a) Draw a Venn diagram to represent the situation C ⊆ A and C ⊆ B. (b) To prove C ⊆ A ∪ B, we should choose an element from which set? (c) Prove that if C ⊆ A and C ⊆ B, then C ⊆ A ∪ B. 41. (a) Draw a Venn diagram to represent the situation A ⊆ C and B ⊆ C.

3

(b) To prove A ∪ B ⊆ C, we should choose an element from which set? (c) Prove that if A ⊆ C and B ⊆ C, then A ∪ B ⊆ C. 42. Prove that A − (A − B) ⊆ B. 43. Suppose that A ⊕ B = A ⊕ C. Does this guarantee that B = C? Justify your conclusion. 44. Prove that A − B = A ∩ B. 45. If A ∪ B = A ∪ C, must B = C? Explain. 46. If A ∩ B = A ∩ C, must B = C? Explain. 47. Prove that if A ⊆ B and C ⊆ D, then A ∪ C ⊆ B ∪ D and A ∩ C ⊆ B ∩ D. 48. When is A − B = B − A? Explain. 49. Explain the last term in the sum in Theorem 3. Why is |A ∩ B ∩ C| added and |B ∩ C| subtracted? 50. Write the four-set version of Theorem 3; that is, |A ∪ B ∪ C ∪ D| = · · · . 51. Describe in words the n-set version of Theorem 3.

Sequences Some of the most important sets arise in connection with sequences. A sequence is simply a list of objects arranged in a definite order; a first element, second element, third element, and so on. The list may stop after n steps, n ∈ N, or it may go on forever. In the first case we say that the sequence is finite, and in the second case we say that it is infinite. The elements may all be different, or some may be repeated.

Example 1

The sequence 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 1 is a finite sequence with repeated items. The digit zero, for example, occurs as the second, third, fifth, seventh, and eighth ◆ elements of the sequence.

Example 2

The list 3, 8, 13, 18, 23, . . . is an infinite sequence. The three dots in the expression mean “and so on,” that is, continue the pattern established by the first few elements. ◆

Example 3

Another infinite sequence is 1, 4, 9, 16, 25, . . . , the list of the squares of all positive ◆ integers. It may happen that how a sequence is to continue is not clear from the first few terms. Also, it may be useful to have a compact notation to describe a sequence. Two kinds of formulas are commonly used to describe sequences. In Example 2, a natural description of the sequence is that successive terms are produced by adding 5 to the previous term. If we use a subscript to indicate a term’s position in the sequence, we can describe the sequence in Example 2 as a1 = 3, an = an−1 + 5, 2 ≤ n. A formula, like this one, that refers to previous terms to define the next term is called recursive. Every recursive formula must include a starting place. On the other hand, in Example 3 it is easy to describe a term using only its position number. In the nth position is the square of n; bn = n2 , 1 ≤ n. This type of formula is called explicit, because it tells us exactly what value any particular term has.

13

Fundamentals Example 4

The recursive formula c1 = 5, cn = 2cn−1 , 2 ≤ n ≤ 6, defines the finite sequence ◆ 5, 10, 20, 40, 80, 160.

Example 5

The infinite sequence 3, 7, 11, 15, 19, 23, . . . can be defined by the recursive ◆ formula d1 = 3, dn = dn−1 + 4.

Example 6

The explicit formula sn = (−4)n , 1 ≤ n, describes the infinite sequence ◆ −4, 16, −64, 256, . . . .

Example 7

The finite sequence 87, 82, 77, 72, 67 can be defined by the explicit formula ◆ tn = 92 − 5n, 1 ≤ n ≤ 5.

Example 8

An ordinary English word such as “sturdy” can be viewed as the finite sequence s, t, u, r, d, y composed of letters from the ordinary English alphabet.

◆

In examples such as Example 8, it is common to omit the commas and write the word in the usual way, if no confusion results. Similarly, even a meaningless word such as “abacabcd” may be regarded as a finite sequence of length 8. Sequences of letters or other symbols, written without the commas, are also referred to as strings. Example 9

An infinite string such as abababab. . . may be regarded as the infinite sequence a, ◆ b, a, b, a, b, . . . .

Example 10

The sentence “now is the time for the test” can be regarded as a finite sequence of English words: now, is, the, time, for, the, test. Here the elements of the sequence are themselves words of varying length, so we would not be able simply to omit ◆ the commas. The custom is to use spaces instead of commas in this case. The set corresponding to a sequence is simply the set of all distinct elements in the sequence. Note that an essential feature of a sequence is the order in which the elements are listed. However, the order in which the elements of a set are listed is of no significance at all.

Example 11

(a) The set corresponding to the sequence in Example 3 is {1,4,9,16,25, . . .}. ◆ (b) The set corresponding to the sequence in Example 9 is simply {a, b}. The idea of a sequence is important in computer science, where a sequence is sometimes called a linear array or list. We will make a slight but useful distinction between a sequence and an array, and use a slightly different notation. If we have a sequence S : s1 , s2 , s3 , . . . , we think of all the elements of S as completely determined. The element s4 , for example, is some fixed element of S, located in position four. Moreover, if we change any of the elements, we have a new sequence and will probably name it something other than S. Thus if we begin with the finite sequence S : 0, 1, 2, 3, 2, 1, 1 and we change the 3 to a 4, getting 0, 1, 2, 4, 2, 1, 1, we would think of this as a different sequence, say S .

14

Fundamentals S[1]

S[2]

S[3] ···

Array S: 1

2

3

Figure 16

An array, on the other hand, may be viewed as a sequence of positions, which we represent in Figure 16 as boxes. The positions form a finite or infinite list, depending on the desired size of the array. Elements from some set may be assigned to the positions of the array S. The element assigned to position n will be denoted by S[n], and the sequence S[1], S[2], S[3], . . . will be called the sequence of values of the array S. The point is that S is considered to be a well-defined object, even if some of the positions have not been assigned values, or if some values are changed during the discussion. The following shows one use of arrays.

Characteristic Functions A very useful concept for sets is the characteristic function. Think of a function on a set as a rule that assigns some “value” to each element of the set. If A is a subset of a universal set U, the characteristic function fA of A is defined for each x ∈ U as follows: 1 if x ∈ A f A (x) = 0 if x ∈ / A. We may add and multiply characteristic functions, since their values are numbers, and these operations sometimes help us prove theorems about properties of subsets. THEOREM 1

Characteristic functions of subsets satisfy the following properties: (a) fA∩B = fA fB ; that is, fA∩B (x) = fA (x)fB (x) for all x. (b) fA∪B = fA + fB − fA fB ; that is, fA∪B (x) = fA (x) + fB (x) − fA (x)fB (x) for all x. (c) fA⊕B = fA +fB −2fA fB ; that is, fA⊕B (x) = fA (x)+fB (x)−2fA (x)fB (x) for all x. Proof (a) fA (x)fB (x) equals 1 if and only if both fA (x) and fB (x) are equal to 1, and this happens if and only if x is in A and x is in B, that is, x is in A ∩ B. Since fA fB is 1 on A ∩ B and 0 otherwise, it must be fA∩B . (b) If x ∈ A, then fA (x) = 1, so fA (x) + fB (x) − fA (x)fB (x) = 1 + fB (x) − fB (x) = 1. Similarly, when x ∈ B, fA (x) + fB (x) − fA (x)fB (x) = 1. If x is not in A or B, then fA (x) and fB (x) are 0, so fA (x) + fB (x) − fA (x)fB (x) = 0. Thus fA + fB − fA fB is 1 on A ∪ B and 0 otherwise, so it must be fA∪B . (c) We leave the proof of (c) as an exercise. ■ Note that the proof of Theorem 1 proceeds by direct application of the definition of the characteristic function.

15

Fundamentals

Computer Representation of Sets and Subsets Another use of characteristic functions is in representing sets in a computer. To represent a set in a computer, the elements of the set must be arranged in a sequence. The particular sequence selected is of no importance. When we list the set A = {a, b, c, . . . , r} we normally assume no particular ordering of the elements in A. Let us identify for now the set A with the sequence a, b, c, . . . , r. When a universal set U is finite, say U = {x1 , x2 , . . . , xn }, and A is a subset of U, then the characteristic function assigns 1 to an element that belongs to A and 0 to an element that does not belong to A. Thus fA can be represented by a sequence of 0’s and 1’s of length n. Example 12

Let U = {1, 2, 3, 4, 5, 6}, A = {1, 2}, B = {2, 4, 6}, and C = {4, 5, 6}. Then fA (x) has value 1 when x is 1 or 2, and otherwise is 0. Hence fA corresponds to the sequence 1, 1, 0, 0, 0, 0. In a similar way, the finite sequence 0, 1, 0, 1, 0, 1 ◆ represents fB and 0, 0, 0, 1, 1, 1 represents fC . Any set with n elements can be arranged in a sequence of length n, so each of its subsets corresponds to a sequence of zeros and ones of length n, representing the characteristic function of that subset. This fact allows us to represent a universal set in a computer as an array A of length n. Assignment of a zero or one to each location A[k] of the array specifies a unique subset of U.

Example 13

Let U = {a, b, e, g, h, r, s, w}. The array of length 8 shown in Figure 17 represents U, since A[k] = 1 for 1 ≤ k ≤ 8. 1

1

1

1

1

1

1

1

Figure 17

If S = {a, e, r, w}, then fS (x) =

1 0

if x = a, e, r, w if x = b, g, h, s.

Hence the array in Figure 18 represents the subset S. 1

0

1

Figure 18

0

0

1

0

1 ◆

A set is called countable if it is the set corresponding to some sequence. Informally, this means that the members of the set can be arranged in a list, with a first, second, third, . . . , element, and the set can therefore be “counted.” Not all infinite sets are countable. A set that is not countable is called uncountable. The most accessible example of an uncountable set is the set of all real numbers that can be represented by an infinite decimal of the form 0.a1 a2 a3 . . . , where ai is an integer and 0 ≤ ai ≤ 9. We shall now show that this set is uncountable. We will prove this result by contradiction; that is, we will show the countability of this set implies an impossible situation.

16

Fundamentals Assume that the set of all decimals 0.a1 a2 a3 . . . is countable. Then we could form the following list (sequence), containing all such decimals: d1 = 0.a1 a2 a3 . . . d2 = 0.b1 b2 b3 . . . d3 = 0.c1 c2 c3 . . . .. . Each of our infinite decimals must appear somewhere on this list. We shall establish a contradiction by constructing an infinite decimal of this type that is not on the list. Now construct a number x as follows: x = 0.x1 x2 x3 . . . , where x1 is 1 if a1 = 2, otherwise x1 is 2; x2 = 1 if b2 = 2, otherwise x2 = 2; x3 = 1 if c3 = 2, otherwise x3 = 2. This process can clearly be continued indefinitely. The resulting number is an infinite decimal consisting of 1’s and 2’s, but by its construction x differs from each number in the list at some position. Thus x is not on the list, a contradiction to our assumption. Hence no matter how the list is constructed, there is some real number of the form 0.x1 x2 x3 . . . that is not in the list. On the other hand, it can be shown that the set of rational numbers is countable.

Strings and Regular Expressions Given a set A, we can construct the set A∗ consisting of all finite sequences of elements of A. Often, the set A is not a set of numbers, but some set of symbols. In this case, A is called an alphabet, and the finite sequences in A∗ are called words from A, or sometimes strings from A. For this case in particular, the sequences in A∗ are not written with commas. We assume that A∗ contains the empty sequence or empty string, containing no symbols, and we denote this string by . Example 14

Let A = {a, b, c, . . . , z}, the usual English alphabet. Then A∗ consists of ordinary words, such as ape, sequence, antidisestablishmentarianism, and so on, as well as “words” such as yxaloble, zigadongdong, esy, and pqrst. All finite sequences from ◆ A are in A∗ , whether they have meaning or not. If w1 = s1 s2 s3 . . . sn and w2 = t1 t2 t3 . . . tk are elements of A∗ for some set A, we define the catenation of w1 and w2 as the sequence s1 s2 s3 . . . sn t1 t2 t3 . . . tk . The catenation of w1 with w2 is written as w1 · w2 or w1 w2 , and is another element of A∗ . Note that if w belongs to A∗ , then w · = w and · w = w. This property is convenient and is one of the main reasons for defining the empty string .

Example 15

Let A = {John, Sam, Jane, swims, runs, well, quickly, slowly}. Then A∗ contains real sentences such as “Jane swims quickly” and “Sam runs well,” as well as nonsense sentences such as “Well swims Jane slowly John.” Here we separate the elements in each sequence with spaces. This is often done when the elements of A ◆ are words. The idea of a recursive formula for a sequence is useful in more general settings as well. Regular expressions are defined recursively. A regular expression over A is a string constructed from the elements of A and the symbols (, ), ∨, ∗, , according to the following definition. RE1. The symbol is a regular expression.

17

Fundamentals RE2. RE3. RE4. RE5.

If x ∈ A, the symbol x is a regular expression. If α and β are regular expressions, then the expression αβ is regular. If α and β are regular expressions, then the expression (α∨β) is regular. If α is a regular expression, then the expression (α)∗ is regular.

Note here that RE1 and RE2 provide initial regular expressions. The other parts of the definition are used repetitively to define successively larger sets of regular expressions from those already defined. Thus the definition is recursive. By convention, if the regular expression α consists of a single symbol x, where x ∈ A, or if α begins and ends with parentheses, then we write (α)∗ simply as α∗ . When no confusion results, we will refer to a regular expression over A simply as a regular expression (omitting reference to A). Example 16

Let A = {0, 1}. Show that the following expressions are all regular expressions over A. (a) 0∗ (0 ∨ 1)∗

(b) 00∗ (0 ∨ 1)∗ 1

(c) (01)∗ (01 ∨ 1∗ )

Solution

(a) By RE2, 0 and 1 are regular expressions. Thus (0 ∨ 1) is regular by RE4, and so 0∗ and (0 ∨ 1)∗ are regular by RE5 (and the convention mentioned previously). Finally, we see that 0∗ (0 ∨ 1)∗ is regular by RE3. (b) We know that 0, 1, and 0∗ (0 ∨ 1)∗ are all regular. Thus, using RE3 twice, 00∗ (0 ∨ 1)∗ 1 must be regular. (c) By RE3, 01 is a regular expression. Since 1∗ is regular, (01 ∨ 1∗ ) is regular by RE4, and (01)∗ is regular by RE5. Then the regularity of (01)∗ (01∨1∗ ) ◆ follows from RE3. Associated with each regular expression over A, there is a corresponding subset of A∗ . Such sets are called regular subsets of A∗ or just regular sets if no reference to A is needed. To compute the regular set corresponding to a regular expression, we use the following correspondence rules. 1. The expression corresponds to the set {}, where is the empty string in A∗ . 2. If x ∈ A, then the regular expression x corresponds to the set {x}. 3. If α and β are regular expressions corresponding to the subsets M and N of A∗ , then αβ corresponds to M · N = {s · t | s ∈ M and t ∈ N}. Thus M · N is the set of all catenations of strings in M with strings in N. 4. If the regular expressions α and β correspond to the subsets M and N of A∗ , then (α ∨ β) corresponds to M ∪ N. 5. If the regular expression α corresponds to the subset M of A∗ , then (α)∗ corresponds to the set M ∗ . Note that M is a set of strings from A. Elements from M ∗ are finite sequences of such strings, and thus may themselves be interpreted as strings from A. Note also that we always have ∈ M ∗ . Example 17

18

Let A = {a, b, c}. Then the regular expression a∗ corresponds to the set of all finite sequences of a’s, such as aaa, aaaaaaa, and so on. The regular expression a(b ∨ c) corresponds to the set {ab, ac} ⊆ A∗ . Finally, the regular expression ab(bc)∗ corresponds to the set of all strings that begin with ab, and then repeat the symbols bc n times, where n ≥ 0. This set includes the strings ab, abbcbc, ◆ abbcbcbcbc, and so on.

Fundamentals Example 18

Let A = {0, 1}. Find regular sets corresponding to the three regular expressions in Example 16. Solution

(a) The set corresponding to 0∗ (0 ∨ 1)∗ consists of all sequences of 0’s and 1’s. Thus, the set is A∗ . (b) The expression 00∗ (0 ∨ 1)∗ 1 corresponds to the set of all sequences of 0’s and 1’s that begin with at least one 0 and end with at least one 1. (c) The expression (01)∗ (01 ∨ 1∗ ) corresponds to the set of all sequences of 0’s and 1’s that either repeat the string 01 a total of n ≥ 1 times, or begin with a total of n ≥ 0 repetitions of 01 and end with some number k ≥ 0 of 1’s. This set includes, for example, the strings 1111, 01, 010101, ◆ 0101010111111, and 011.

3 Exercises In Exercises 1 through 4, give the set corresponding to the sequence. 1. 1, 2, 1, 2, 1, 2, 1, 2, 1 2. 0, 2, 4, 6, 8, 10, . . . 3. aabbccddee. . . zz 4. abbcccdddd 5. Give three different sequences that have {x, y, z} as a corresponding set. 6. Give three different sequences that have {1, 2, 3, . . . } as a corresponding set. In Exercises 7 through 14, write out the first four terms (begin with n = 1) of the sequence whose general term is given. 7. an = 5

n

8. bn = 3n2 + 2n − 6 9. gn = 1 · 2 · 3 · · · · · n an − 1 10. hn = , a = 1 a−1 11. c1 = 2.5, cn = cn−1 + 1.5 12. d1 = −3, dn = −2dn−1 + 1

23. Let A = {x | x is a real number and 0 < x < 1}, B = {x | x is a real number and x2 + 1 = 0}, C = {x | x = 4m, m ∈ Z}, D = {(x, 3) | x is an English word whose length is 3}, and E = {x | x ∈ Z and x2 ≤ 100}. Identify each set as finite, countable, or uncountable. 24. Let A = W ∗ for W = {a, b}, B = {x | x ∈ R and x2 + 41x + 41 = 0}, C = {x | x = mn , m, n ∈ Z+ , n > 4}, D = {x | x ∈ R and x2 + 3x + 2 = 0}, and E = {(x, y, z) | x ∈ Z, y ∈ R+ , z ∈ Z+ }. Identify each set as finite, countable, or uncountable. 25. Let A = {ab, bc, ba}. In each part, tell whether the string belongs to A∗ . (b) abc (c) abba (a) ababab (e) bcabbab (f) abbbcba (d) abbcbaba 26. Let U = {arithmetic, algebra, calculus, geometry, trigonometry, analysis, topology, statistics}, B = {analysis, topology, calculus}, C = {algebra, calculus, trigonometry, analysis}, D = {arithmetic, algebra, calculus, topology, statistics}, E = {algebra, calculus, geometry, trigonometry, analysis}. In each of the following, represent the given set by an array of zeros and ones. (b) C ∩ D (a) B ∪ C (c) B ∩ (D ∩ E)

13. e1 = 0, en = en−1 − 2 14. f1 = 4, fn = n · fn−1 In Exercises 15 through 20, write a formula for the nth term of the sequence. Identify your formula as recursive or explicit. 15. 1, 3, 5, 7, . . .

16. 0, 3, 8, 15, 24, 35, . . .

17. 1, −1, 1, −1, 1, −1, . . .

18. 0, 2, 0, 2, 0, 2, . . .

19. 1, 4, 7, 10, 13, 16

20. 1, 12 , 41 , 18 ,

1 , 16

...

21. Write an explicit formula for the sequence 2, 5, 8, 11, 14, 17, . . . . 22. Write a recursive formula for the sequence 2, 5, 7, 12, 19, 31, . . . .

(d) B ∪ E

(e) C ∩ (B ∪ E) 27. Let U = {b, d, e, g, h, k, m, n}, B = {b}, C = {d, g, m, n}, and D = {d, k, n}. (b) What is fC (e)? (a) What is fB (b)? (c) Find the sequences of length 8 that correspond to fB , fC , and fD . (d) Represent B ∪ C, C ∪ D, and C ∩ D by arrays of zeros and ones. 28. Complete the proof that fA⊕B = fA + fB − 2fA fB [Theorem 1(c)]. Suppose x ∈ A and x ∈ / B. Then fA (x) = , fB (x) = , and fA (x)fB (x) = , so fA (x) + fB (x) − 2fA (x)fB (x) = .

19

Fundamentals Now suppose x ∈ / A and x ∈ B. Then fA (x) = , fB (x) = , and fA (x)fB (x) = , so fA (x) + fB (x) − 2fA (x)fB (x) = . The remaining case to check is x ∈ / A ⊕ B. If x ∈ / A ⊕ B, then x ∈ and fA (x) + fB (x) − 2fA (x)fB (x) = . Explain how these steps prove Theorem 1(c). 29. Using characteristic functions, prove that (A ⊕ B) ⊕ C = A ⊕ (B ⊕ C). 30. Let A = {+, ×, a, b}. Show that the following expressions are regular over A. (a) a + b(ab)∗ (a × b ∨ a) (b) a + b × (a∗ ∨ b) (c) (a∗ b ∨ +)∗ ∨ × b∗ In Exercises 31 and 32, let A = {a, b, c}. In each exercise a string in A∗ is listed and a regular expression over A is given. In each case, tell whether or not the string on the left belongs to the regular set corresponding to the regular expression on the right. 31. (a) ac a∗ b∗ c (b) abcc (abc ∨ c)∗ (c) aaabc ((a ∨ b) ∨ c)∗ 32. (a) ac (a∗ b ∨ c) (b) abab (ab)∗ c (c) aaccc (a∗ ∨ b)c∗ 33. Give three expressions that are not regular over the A given for Exercises 31 and 32. 34. Let A = {p, q, r}. Give the regular set corresponding to the regular expression given. (b) p(qq)∗ r (a) (p ∨ q)rq∗ 35. Let S = {0, 1}. Give the regular expression corresponding to the regular set given.

4

(a) {00, 010, 0110, 011110, . . . } (b) {0, 001, 000, 00001, 00000, 0000001, . . . } 36. We define T -numbers recursively as follows: 1. 0 is a T -number. 2. If X is a T -number, X + 3 is a T -number. Write a description of the set of T -numbers. 37. Define an S-number by 1. 8 is an S-number. 2. If X is an S-number and Y is a multiple of X, then Y is an S-number. 3. If X is an S-number and X is a multiple of Y , then Y is an S-number. Describe the set of S-numbers. 38. Let F be a function defined for all nonnegative integers by the following recursive definition. F(0) = 0, F(1) = 1 F(n + 2) = 2F(n) + F(n + 1), n ≥ 0 Compute the first six values of F ; that is, write the values of F(n) for n = 0, 1, 2, 3, 4, 5. 39. Let G be a function defined for all nonnegative integers by the following recursive definition. G(0) = 1,

G(1) = 2

G(n + 2) = G(n) + G(n + 1), 2

n≥0

Compute the first five values of G.

Properties of Integers We shall now discuss some results needed later about division and factoring in the integers. If m is an integer and n is a positive integer, we can plot the integer multiples of n on a line, and locate m as in Figure 19. If m is a multiple of n, say m = qn, then we can write m = qn + r, where r is 0. On the other hand (as shown in Figure 19), if m is not a multiple of n, we let qn be the first multiple of n lying to the left of m and let r be m − qn. Then r is the distance from qn to m, so clearly 0 < r < n, and again we have m = qn + r. We state these observations as a theorem. m

...

–2n

–n

0

n

2n

...

qn

(q + 1)n

...

Figure 19

THEOREM 1

20

If n and m are integers and n > 0, we can write m = qn + r for integers q and r ■ with 0 ≤ r < n. Moreover, there is just one way to do this.

Fundamentals Example 1

(a) If n is 3 and m is 16, then 16 = 5(3) + 1 so q is 5 and r is 1. (b) If n is 10 and m is 3, then 3 = 0(10) + 3 so q is 0 and r is 3. (c) If n is 5 and m is −11, then −11 = −3(5) + 4 so q is −3 and r is 4.

◆

If the r in Theorem 1 is zero, so that m is a multiple of n, we write n | m, which is read “n divides m.” If n | m, then m = qn and n ≤ |m|. If m is not a multiple of n, we write n m, which is read “n does not divide m.” We now prove some simple properties of divisibility. THEOREM 2

Let a, b, and c be integers. (a) If a | b and a | c, then a | (b + c). (b) If a | b and a | c, where b > c, then a | (b − c). (c) If a | b or a | c, then a | bc. (d) If a | b and b | c, then a | c. Proof (a) If a | b and a | c, then b = k1 a and c = k2 a for integers k1 and k2 . So b + c = (k1 + k2 )a and a | (b + c). (b) This can be proved in exactly the same way as (a). (c) As in (a), we have b = k1 a or c = k2 a. Then either bc = k1 ac or bc = k2 ab, so in either case bc is a multiple of a and a | bc. (d) If a | b and b | c, we have b = k1 a and c = k2 b, so c = k2 b = k2 (k1 a) = ■ (k2 k1 )a and hence a | c. Note that again we have a proof that proceeds directly from a definition by restating the original conditions. As a consequence of Theorem 2, we have that if a | b and a | c, then a | (mb + nc), for any integers m and n. (See Exercises 23 and 24.) A number p > 1 in Z+ is called prime if the only positive integers that divide p are p and 1.

Example 2

The numbers 2, 3, 5, 7, 11, and 13 are prime, while 4, 10, 16, and 21 are not prime. ◆ It is easy to write a set of steps, or an algorithm,∗ to determine if a positive integer n > 1 is a prime number. First we check to see if n is 2. If n > 2, we could divide by every integer from 2 to n − 1, and if none of these is a divisor of n, then n is prime. To make the process√more efficient, we note that if mk = n, then either m or k is less than or equal to n. This √ means that if n is not prime, it has a divisor k satisfying the inequality 1 < k ≤ n, so we need only test for divisors in this range. Also, if n has any even number as a divisor, it must have 2 as a divisor. Thus after checking for divisibility by 2, we may skip all even integers. Algorithm

To test whether an integer N > 1 is prime: Step 1 Check whether N is 2. If so, N is prime. If not, proceed to Step 2 Check whether 2 | N. If so, N is not √ prime; otherwise, proceed to Step 3 Compute the largest integer K ≤ N. Then ∗Algorithms

are discussed in Appendix A.

21

Fundamentals Step 4 Check whether D | N, where D is any odd number such that 1 < D ≤ K. If D | N, then N is not prime; otherwise, N is prime.

●

Testing whether an integer is prime is a common task for computers. The algorithm given here is too inefficient for testing very large numbers, but there are many other algorithms for testing whether an integer is prime. THEOREM 3

Example 3

k

k

Every positive integer n > 1 can be written uniquely as p11 p22 · · · pks s , where p1 < p2 < · · · < ps are distinct primes that divide n and the k’s are positive integers giving the number of times each prime occurs as a factor of n. ■ (a) 9 = 3 · 3 = 32 (b) 24 = 8 · 3 = 2 · 2 · 2 · 3 = 23 · 3 (c) 30 = 2 · 3 · 5

◆

Greatest Common Divisor If a, b, and k are in Z+ , and k | a and k | b, we say that k is a common divisor of a and b. If d is the largest such k, d is called the greatest common divisor, or GCD, of a and b, and we write d = GCD(a, b). This number has some interesting properties. It can be written as a combination of a and b, and it is not only larger than all the other common divisors, it is a multiple of each of them. THEOREM 4

If d is GCD(a, b), then (a) d = sa + tb for some integers s and t. (These are not necessarily positive.) (b) If c is any other common divisor of a and b, then c | d. Proof Let x be the smallest positive integer that can be written as sa + tb for some integers s and t, and let c be a common divisor of a and b. Since c | a and c | b, it follows from Theorem 2 that c | x, so c ≤ x. If we can show that x is a common divisor of a and b, it will then be the greatest common divisor of a and b and both parts of the theorem will have been proved. By Theorem 1, a = qx + r with 0 ≤ r < x. Solving for r, we have r = a − qx = a − q(sa + tb) = a − qsa − qtb = (1 − qs)a + (−qt)b. If r is not zero, then since r < x and r is the sum of a multiple of a and a multiple of b, we will have a contradiction to the fact that x is the smallest positive number that is a sum of multiples of a and b. Thus r must be 0 and x | a. In the same way ■ we can show that x | b, and this completes the proof. This proof is more complex than the earlier ones. At this stage you should focus on understanding the details of each step. From the definition of greatest common divisor and Theorem 4(b), we have the following result: Let a, b, and d be in Z+ . The integer d is the greatest common divisor of a and b if and only if (a) d | a and d | b. (b) Whenever c | a and c | b, then c | d.

22

Fundamentals Example 4

(a) The common divisors of 12 and 30 are 1, 2, 3, and 6, so that GCD(12, 30) = 6

and

6 = 1 · 30 + (−2) · 12.

(b) It is clear that GCD(17, 95) = 1 since 17 is prime and 17 95, and the reader ◆ may verify that 1 = 28 · 17 + (−5) · 95. If GCD(a, b) = 1, as in Example 4(b), we say a and b are relatively prime. One remaining question is that of how to compute the GCD conveniently in general. Repeated application of Theorem 1 provides the key to doing this. We now present a procedure, called the Euclidean algorithm, for finding GCD(a, b). Suppose that a > b > 0 (otherwise interchange a and b). Then by Theorem 1, we may write a = k1 b + r1 ,

where k1 is in Z+ and 0 ≤ r1 < b.

(1)

Now Theorem 2 tells us that if n divides a and b, then it must divide r1 , since r1 = a − k1 b. Similarly, if n divides b and r1 , then it must divide a. We see that the common divisors of a and b are the same as the common divisors of b and r1 , so GCD(a, b) = GCD(b, r1 ). We now continue using Theorem 1 as follows: divide b by r1 : divide r1 by r2 : divide r2 by r3 : .. . divide rn−2 by rn−1 : divide rn−1 by rn :

b = k2 r1 + r2 r1 = k3 r2 + r3 r2 = k4 r3 + r4 .. . rn−2 = kn rn−1 + rn rn−1 = kn+1 rn + rn+1

0 ≤ r2 < r1 0 ≤ r3 < r2 0 ≤ r4 < r3 .. .

(2)

0 ≤ rn < rn−1 0 ≤ rn+1 < rn .

Since a > b > r1 > r2 > r3 > r4 > · · · , the remainder will eventually become zero, so at some point we have rn+1 = 0. We now show that rn = GCD(a, b). We saw previously that GCD(a, b) = GCD(b, r1 ). Repeating this argument with b and r1 , we see that GCD(b, r1 ) = GCD(r1 , r2 ). Upon continuing, we have GCD(a, b) = GCD(b, r1 ) = GCD(r1 , r2 ) = · · · = GCD(rn−1 , rn ). Since rn−1 = kn+1 rn , we see that GCD(rn−1 , rn ) = rn . Hence rn = GCD(a, b). Example 5

Compute GCD(273, 98). Let a be 273 and b be 98. Using the Euclidean algorithm, divide 273 by 98: divide 98 by 77: divide 77 by 21: divide 21 by 14: divide 14 by 7:

273 = 2 · 98 + 77 98 = 1 · 77 + 21 77 = 3 · 21 + 14 21 = 1 · 14 + 7 14 = 2 · 7 + 0

so GCD(273, 98) = 7, the last of the nonzero divisors.

◆

23

Fundamentals In Theorem 4(a), we observed that if d = GCD(a, b), we can find integers s and t such that d = sa + tb. The integers s and t can be found as follows. Solve the next-to-last equation in (2) for rn : rn = rn−2 − kn rn−1 .

(3)

Now solve the second-to-last equation in (2), rn−3 = kn−1 rn−2 + rn−1 for rn−1 : rn−1 = rn−3 − kn−1 rn−2 and substitute this expression in (3): rn = rn−2 − kn [rn−3 − kn−1 rn−2 ]. Continue to work up through the equations in (2) and (1), replacing ri by an expression involving ri−1 and ri−2 , and finally arriving at an expression involving only a and b. Example 6

(a) Let a be 273 and b be 98, as in Example 5. Then GCD(273, 98) = 7 = 21 − 1 · 14 = 21 − 1(77 − 3 · 21) = 4 · 21 − 1 · 77 = 4(98 − 1 · 77) − 1 · 77 = 4 · 98 − 5 · 77 = 4 · 98 − 5(273 − 2 · 98) = 14(98) − 5(273)

14 = 77 − 3 · 21 21 = 98 − 1 · 77 77 = 273 − 2 · 98

Hence s = −5 and t = 14. Note that the key is to carry out the arithmetic only partially. (b) Let a = 108 and b = 60. Then GCD(108, 60) = 12 = 60 − 1(48) = 60 − 1[108 − 1(60)] = 2(60) − 108.

48 = 108 − 1 · 60

Hence s = −1 and t = 2.

THEOREM 5

◆

If a and b are in Z+ , b > a, then GCD(a, b) = GCD(b, b ± a). Proof If c divides a and b, it divides b ± a, by Theorem 2. Since a = b − (b − a) = −b + (b + a), we see, also by Theorem 2, that a common divisor of b and b ± a also divides a and b. Since a and b have the same common divisors as b and b ± a, ■ they must have the same greatest common divisor. This is another direct proof, but one that uses a previous theorem as well as definitions.

24

Fundamentals

Least Common Multiple If a, b, and k are in Z+ , and a | k, b | k, we say k is a common multiple of a and b. The smallest such k, call it c, is called the least common multiple, or LCM, of a and b, and we write c = LCM(a, b). The following result shows that we can obtain the least common multiple from the greatest common divisor, so we do not need a separate procedure for finding the least common multiple. THEOREM 6

If a and b are two positive integers, then GCD(a, b) · LCM(a, b) = ab. Proof Let p1 , p2 , . . . , pk be all the prime factors of either a or b. Then we can write a

a

a

a = p11 p22 · · · pkk

b

b

b

b = p11 p22 · · · pkk

and

where some of the ai and bi may be zero. It then follows that min(a1 ,b1 ) min(a2 ,b2 ) p2

GCD(a, b) = p1

min(ak ,bk )

· · · pk

and max(ak ,bk )

max(a1 ,b1 ) max(a2 ,b2 ) p2

· · · pk

a +b1

· · · pkk

LCM(a, b) = p1

.

Hence GCD(a, b) · LCM(a, b) = p11

a

a +b2

p22

a +bk

a

a

b

b

b

= (p11 p22 · · · pkk ) · (p11 p22 · · · pkk ) = ab. Example 7

■

Let a = 540 and b = 504. Factoring a and b into primes, we obtain a = 540 = 22 · 33 · 5

and

b = 504 = 23 · 32 · 7.

Thus all the prime numbers that are factors of either a or b are p1 = 2, p2 = 3, p3 = 5, and p4 = 7. Then a = 22 · 33 · 51 · 70 and b = 23 · 32 · 50 · 71 . We then have GCD(540, 504) = 2min(2,3) · 3min(3,2) · 5min(1,0) · 7min(0,1) = 22 · 32 · 50 · 70 = 22 · 32 or 36. Also, LCM(540, 504) = 2max(2,3) · 3max(3,2) · 5max(1,0) · 7max(0,1) = 23 · 33 · 51 · 71 or 7560. Then GCD(540, 504) · LCM(540, 504) = 36 · 7560 = 272,160 = 540 · 504. As a verification, we can also compute GCD(540, 504) by the Euclidean algorithm ◆ and obtain the same result. If n and m are integers and n > 1, Theorem 1 tells us we can write m = qn + r, 0 ≤ r < n. Sometimes the remainder r is more important than the quotient q.

25

Fundamentals Example 8

If the time is now 4 o’clock, what time will it be 101 hours from now? Solution

Let n = 12 and m = 4 + 101 or 105. Then we have 105 = 8 · 12 + 9. The remainder 9 answers the question. In 101 hours it will be 9 o’clock. ◆ For each n ∈ Z+ , we define a function fn , the mod-n function, as follows: If z is a nonnegative integer, fn (z) = r, the remainder when z is divided by n. (As in Section 3, we need only think of a function as a rule that assigns some “value” to each member of a set.) Example 9

(a) f3 (14) = 2, because 14 = 4 · 3 + 2. (b) f7 (153) = 6.

◆

Pseudocode Versions An alternative to expressing an algorithm in ordinary English as we did in this section is to express it in something like a computer language. Here we give pseudocode versions for an algorithm that determines if an integer is prime and for an algorithm that calculates the greatest common divisor of two integers. In the pseudocode for the algorithm to determine if an integer is prime, we assume the existence of√functions SQR and INT, where SQR(N) returns the greatest integer not exceeding N, and INT(X) returns the greatest integer not exceeding X. For example, SQR(10) = 3, SQR(25) = 5, INT(7.124) = 7, and INT(8) = 8. SUBROUTINE PRIME(N ) 1. IF (N = 2) THEN a. PRINT (’PRIME’) b. RETURN 2. ELSE a. IF (N/2 = INT(N/2)) THEN 1. PRINT (’NOT PRIME’) 2. RETURN b. ELSE 1. FOR D = 3 THRU SQR(N) BY 2 a. IF (N/D = INT(N/D)) THEN 1. PRINT (’NOT PRIME’) 2. RETURN 2. PRINT (’PRIME’) 3. RETURN END OF SUBROUTINE PRIME

The following gives a pseudocode program for finding the greatest common divisor of two positive integers. This procedure is different from the Euclidean algorithm. FUNCTION GCD(X,Y) 1. WHILE (X = Y) a. IF (X > Y) THEN 1. X ← X − Y

26

Fundamentals b. ELSE 1. Y ← Y − X 2. RETURN (X) END OF FUNCTION GCD

Example 10

Use the pseudocode for GCD to calculate the greatest common divisor of 190 and 34. Solution

The following table gives the values of X, Y , X − Y , or Y − X as we go through the program. X

Y

X−Y

190 156 122 88 54 20 20 6 6 6 4 2

34 34 34 34 34 34 14 14 8 2 2 2

156 122 88 54 20

Y −X

14 6 8 2 4 2

Since the last value of X is 2, the greatest common divisor of 190 and 34 is 2. ◆

Representations of Integers The decimal representation of an integer is so familiar that we sometimes regard it as the symbol, or name for that integer. For example, when we write the integer 3264, we are saying that the number is the result of adding 3 times 103 , 2 times 102 , 6 times 101 , and 4, or 4 times 100 . We say 3264 is the base 10 expansion of n or the decimal expansion of n; 10 is called the base of this expansion. There is nothing special about using the number 10 as a base, and it is likely the result of our having 10 fingers on which to count. Any positive integer b > 1 can be used in a similar way, and these expansions are often of much greater use than the standard decimal expansion. The bases 2, 8, and 16 are frequently used in computer science, and the base 26 is sometimes used in cryptology (the science of producing and deciphering secret codes). THEOREM 7

If b > 1 is an integer, then every positive integer n can be uniquely expressed in the form n = dk bk + dk−1 bk−1 + · · · + d1 b + d0 ,

(4)

where 0 ≤ di < b, i = 0, 1, . . . , k, and dk = 0. The sequence dk dk−1 . . . d1 d0 is called the base b expansion of n. If we need to explicitly indicate the base b, we will write the above sequence as (dk dk−1 . . . d1 d0 )b .

27

Fundamentals Proof Suppose that k is the largest nonnegative integer so that bk ≤ n (k could be 0). By Theorem 1 we can uniquely write n = qbk + r, where 0 ≤ r < bk . Let dk = q. If k = 0, then r = 0 and we are done (n = d0 ). Otherwise we have n = dk bk + r. Repeat this process, using r in place of n. Let s be the largest nonnegative integer so that bs ≤ r, write r = qbs + r1 , where 0 ≤ r1 < bs , and define ds to be q. If s < k, define ds+1 , . . . , dk−1 to be 0. Then n = dk bk + dk−1 bk−1 + · · · + ds bs + r1 . Continuing this process using r1 , r2 , . . . , we will eventually arrive at (4).

■

Now that we know that the base b expansion exists, we can find the digits dk , dk−1 , . . . , d1 , d0 by a more direct method, easily implemented on a computer. Note that n = (dk bk−1 + dk−1 bk−2 + · · · + d1 )b + d0 so that d0 is the remainder after dividing n by b, and the quotient is dk bk−1 + dk−1 bk−2 + · · · + d1 . Similarly, if this quotient is divided by b, the remainder is d1 . By repeatedly dividing the quotients by b and saving the remainders, we produce the digits of the base b representation of n from right to left. Example 11

Find the base 4 representation of 158. Solution We repeatedly divide by 4 and save the remainder:

4 158 4 39 4 9 4 2 0 Thus 158 = (2132)4 .

2 3 1 2 ◆

The following pseudocode algorithm returns the digits of the base b expansion of an integer. We use the expression m mod n to denote the mod-n function value for m, that is, the remainder after dividing m by n. These are the functions defined after Example 8. The mod-n functions are commonly implemented in most programming languages. SUBROUTINE EXPANSION(N ) 1. Q ← N 2. K ← 0 3. WHILE (Q = 0) a. Dk ← Q mod B b. Q ← INT(Q/B) c. K ← K + 1 4. RETURN END OF SUBROUTINE EXPANSION

When this subroutine ends, the base B expansion of N consists of the integers Di , which can then be further processed.

28

Fundamentals No matter what base is used to represent integers, the elementary rules of addition and multiplication are still valid. Only the appearance of the numbers changes. Example 12

Let m = (313)4 and n = (322)4 . Find the base 4 expansion of m + n. Solution

Adding the digits in the last column, we have 3 + 2 = (11)4 , so we record a 1 and carry a 1 to the next column. 1 313 + 322 1 Adding the digits in the second column, we have 1 + 1 + 2 = (10)4 , so we record a 0 and carry the 1. 11 313 + 322 01 Finally, adding the digits in the first column, we obtain 3 + 3 + 1 = (13)4 , so the ◆ answer is (1301)4 . The most common expansion used in computer work is the base 2 or binary expansion. Since the only remainders of division by 2 are 0 and 1, the binary expansion of every number needs only the digits 0 and 1 and so is easily implemented and manipulated by the on-off digital circuits of a computer. Example 13

(a) The binary expansion of 39 is (100111)2 . (b) (110101)2 = 32 + 16 + 4 + 1 = 53.

◆

Binary addition and multiplication are usually implemented in computer circuitry rather than by software. Another common base in computer science is base 16, or hexadecimal (or hex) representation. This representation requires six additional symbols to use with 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 to represent the digits usually written 10, 11, 12, 13, 14, 15. It is customary to choose the letters A, B, C, D, E, F for this purpose. Example 14

To use 26 as a base, we can use the letters A, B, . . . , Z of the English alphabet to represent the digits 0, 1, . . . , 25. In this way we can interpret any text as the base 26 representation of an integer. Thus we can interpret “TWO” as the base 26 representation of (T×262 )+(W×26)+O = (19×676)+(22×26)+14 = 13430. The ability to “add” words can be made the basis of cryptographic encoding. ◆

Example 15

As an example of cryptology, we consider a remarkable code due to Sir Francis Bacon, the English writer and philosopher. Suppose that we wish to encode a message, say FLEE NOW. We first write the base 2 representation of the position of each letter of the message in the English alphabet, starting with A in position 0, and ending with Z in position 25. Thus we have the following table. F 00101

L 01011

E 00100

E 00100

N 01101

O 01110

W 10110

29

Fundamentals Since F is in position 5, its binary representation is (00101)2 , and so on. Now choose an unrelated “dummy” message exactly five times as long (padded with a few extra letters if necessary). Place the dummy message in a third row. The letters of the dummy message correspond exactly to the string of 0’s and 1’s in the second row of the above table. We agree to write each letter in one font (say Times Roman) if it corresponds to a 0 in the table, and another font (say Times Roman Italic) if it corresponds to a 1 (Bacon used fonts that were even more similar). Thus if the dummy message is ONCE UPON A TIME IN THE WEST THERE WAS A TOWN, we would write that message as ONCE UPON A T IME IN THE WEST T HERE WAS A TOW N. Note that when the letters in this message are arranged in a third row of the table, the patterns of nonitalic for 0 and italic for 1 allow us to decode the message. F 00101 ONCEU

L 01011 PONAT

E 00100 IMEIN

E 00100 THEWE

N 01101 STT HE

O 01110 REWAS

W 10110 ATOW N ◆

Example 16

Suppose that we wish to decode the following dummy message, using the Bacon code NOW IS T HE T IME FOR ALL GOOD MEN TO AID THE COUNTRY Since there are 40 letters, the true message must have 8 letters. Arrange the dummy message, 5 letters at a time in the following table, and then list the corresponding binary digits, using 1 for italic and 0 for plain text. NOWIS T HET I MEFOR ALLGO ODMEN TOAID THECO UNTRY 10010 10011 00000 10001 10011 01001 01110 00001 The binary representations in the second row correspond respectively to the numbers 18, 19, 0, 17, 19, 9, 14, and 1, and therefore represent the letters STARTJOB. ◆ Thus the decoded message is START JOB. If the fonts used are quite similar, it will not be obvious that the text is a coded message. Thus this example also illustrates steganography, the science of concealment of coded information. Modern versions include hiding information in stray bits of a digital photograph. One use of this is to watermark copyrighted artistic material.

4 Exercises In Exercises 1 through 4, for the given integers m and n, write m as qn + r, with 0 ≤ r < n. 1. m = 20, n = 3

6. a = 60, b = 100

2. m = 64, n = 37

7. a = 45, b = 33

3. m = 3, n = 22

8. a = 34, b = 58

4. m = 48, n = 12

9. a = 77, b = 128

5. Write each integer as a product of powers of primes (as in Theorem 3). (b) 1666 (c) 1781 (a) 828 (d) 1125

30

In Exercises 6 through 9, find the greatest common divisor d of the integers a and b, and write d as sa + tb.

(e) 107

In Exercises 10 through 13, find the least common multiple of the integers. 10. 72, 108

11. 150, 70

12. 175, 245

13. 32, 27

Fundamentals 14. If f is the mod-7 function, compute each of the following. (b) f(48) (c) f(1207) (a) f(17) (d) f(130)

(e) f(93)

(f) f(169)

15. If f is the mod-11 function, compute each of the following. (b) f(386) (c) f(1232) (a) f(39) (d) f(573)

(e) 2f(87)

(f) f(175) + 4

16. If f is the mod-7 function, compute each of the following. (b) f(752) + f(793) (a) f(752 + 793) (c) f(3 · 1759)

(d) 3 · f(1759)

17. If f is the mod-12 function, compute each of the following. (b) f(1259) + f(743) (a) f(1259 + 743) (c) f(2 · 319)

(d) 2 · f(319)

18. Let f be the mod-n function for a fixed n. What do the results of Exercises 16 and 17 suggest about the relationship between k · f(a) and f(k · a)? 19. Let f be the mod-n function for a fixed n. Based on the results of Exercises 16 and 17, explain why f(a + b) does not always equal f(a) + f(b). 20. Let f be the mod-n function for a fixed n. Explain when f(a + b) = f(a) + f(b) is true. 21. If g is the mod-5 function, solve each of the following. (b) g(n) = 4 (a) g(n) = 2 22. If g is the mod-6 function, solve each of the following. (b) g(n) = 1 (a) g(n) = 3 23. Prove that if a | b, then a | mb, for any m ∈ Z. 24. Prove that if a | b and a | c, then a | mb + nc, for any m, n ∈ Z. 25. Complete the following proof. Let a and b be integers. If p is a prime and p | ab, then p | a or p | b. We need to show that if p a, then p must divide b. If p a, then GCD(a, p) = 1, because . By Theorem 4, we can write 1 = sa + tp for some integers s and t. Then b = sab + tpb. (Why?) Then p must divide sab + tpb, because . So p | b. (Why?) 26. Show that if GCD(a, c) = 1 and c | ab, then c | b. (Hint: Model the proof on the one in Exercise 25.) 27. Show that if GCD(a, c) = 1, a | m, and c | m, then ac | m. (Hint: Use Exercise 26.) 28. Show that if d = GCD(a, b), a | b, and c | b, then ac | bd. 29. Show that GCD(ca, cb) = c GCD(a, b). 30. Show that LCM(a, ab) = ab. 31. Show that if GCD(a, b) = 1, then LCM(a, b) = ab. 32. Let c = LCM(a, b). Show that if a | k and b | k, then c | k. 33. Prove that if a and b are positive integers such that a | b and b | a, then a = b.

34. Let a be an integer and let p be a positive integer. Prove that if p | a, then p = GCD(a, p). 35. Theorem 2(c) says that if a | b or a | c, then a | bc. Is the converse true; that is, if a | bc, then a | b or a | c? Justify your conclusion. 36. Prove that if m and n are relatively prime and mn is a perfect square, then m and n are each perfect squares. 37. Is the statement in Exercise 36 true for cubes? For any fixed power? Justify your conclusion. In Exercises 38 through 40, let U = {1, 2, 3, . . . , 1689}, A = {x | x ∈ U and 3 | x}, B = {y | y ∈ U and 5 | y}, and C = {z | z ∈ U and 11 | z}. Compute each of the following. 38. (a) |A| (b) |B| (c) |C| 39. (a) The number of elements in U that are divisible by 15 (b) The number of elements of U that are divisible by 165 (c) The number of elements of U that are divisible by 55 40. Use the results of Exercises 38 and 39 to compute each of the following. (b) |A ∪ B ∪ C| (a) |A ∪ B| 41. (a) Write the expansion in base 5 of each of the following numbers. (i) 29

(ii) 73

(iii) 215

(iv) 732

(b) Write the expansion in base 10 of each of the following numbers. (ii) (320)5 (i) (144)5 (iv) (11231)5 (1242) (iii) 5 42. (a) Write the expansion in base 7 of each of the following numbers. (i) 29

(ii) 73

(iii) 215

(iv) 732

(b) Write the expansion in base 10 of each of the following numbers. (ii) (161)7 (i) (102)7 (iv) (1613)7 (iii) (460)7 43. For each of the following, write the expansion in the specified base. (ii) 73 (iii) 215 (iv) 732 (i) 29 (b) 4 (c) 16 (a) 2 44. (a) How are the numbers 2, 4, and 16 related? (b) Because of the way 2, 4, and 16 are related, it is possible to change the expansion of a number relative to one of these numbers to the expansion relative to another directly without using the number’s base 10 expansion. Examine the results of Exercise 43 (and other examples, if needed) and describe how to change from (i) a base 2 expansion to a base 4 expansion. (ii) a base 16 expansion to a base 2 expansion. (iii) a base 4 expansion to a base 16 expansion.

31

Fundamentals 45. Use Bacon’s code as given in Example 15 (a) to create a dummy message for COME BACK (b) to decode W HEN THE MOON COMES OVER T HE MOUNTAIN, THE OWLS FLY HIGH. 46. (a) For Bacon’s code as given in Example 15, why should the dummy message be five times as long as the true message? (b) Modify Bacon’s code so that it handles the spaces

5

between words and the digits 0, 1, 2, . . . , 9. 47. If ONE and TWO are the base 26 representations of two integers, then what is the base 26 representation of the sum ONE + TWO? 48. Use Bacon’s code, as given in Example 15, to decode DO YOU KNOW T HAT THE NUMBER PI IS NOW KNOWN T O MORE THAN FOUR HUNDRED MILLION DECIMAL PLACES.

Matrices A matrix is a rectangular array of numbers arranged in m horizontal rows and n vertical columns: ⎤ ⎡ a11 a12 · · · a1n ⎢ a21 a22 · · · a2n ⎥ (1) A=⎢ .. .. ⎥ ⎣ ... . . ⎦

am1 am2 · · · amn · · · ain , 1 ≤ i ≤ m, and the jth column of A is

The ith row of A is ai1 ai2 ⎡ ⎤ a1j ⎢ a2j ⎥ ⎢ . ⎥, 1 ≤ j ≤ n. We say that A is m by n, written m × n. If m = n, we say ⎣ .. ⎦ amj A is a square matrix of order n and that the numbers a11 , a22 , . . . , ann form the main diagonal of A. We refer to the number aij , which is in the ith row and jth column of A as the i, jth element of A or as the (i, j) entry of A, and we often write (1) as A = aij . Note that first the row is named and then the column. Example 1

Let

A=

5 2 3 , B= , C = 1 −1 3 4 2 4 6 ⎡ ⎤ ⎡ ⎤ −1 1 0 −1 2 3 ⎦. D = ⎣ 2 ⎦ , and E = ⎣ −1 0 2 4 5

2 3 0 −1

Then A is 2 × 3 with a12 = 3 and a23 = 2, B is 2 × 2 with b21 = 4, C is 1 × 4, ◆ D is 3 × 1, and E is 3 × 3. A square matrix A = aij for which every entry off the main diagonal is zero, that is, aij = 0 for i = j, is called a diagonal matrix. Example 2

Each of the following is a diagonal matrix. ⎡ ⎤

2 0 0 4 0 0 ⎦, F= , G = ⎣ 0 −3 0 3 0 0 5

⎡

and

⎤ 0 0 0 H=⎣0 7 0⎦ 0 0 6

◆

Matrices are used in many applications in computer science. We present the following simple application showing how matrices can be used to display data in a tabular form.

32

Fundamentals Example 3

The following matrix gives the airline distances between the cities indicated. ⎡

London

0 London Madrid ⎢ 785 ⎣ 3469 New York Tokyo 5959

Madrid

New York

785 0 3593 6706

3469 3593 0 6757

Tokyo

⎤ 5959 6706 ⎥ 6757 ⎦ 0

◆

Two m × n matrices A = aij and B = bij are said to be equal if aij = bij , 1 ≤ i ≤ m, 1 ≤ j ≤ n; that is, if corresponding elements are the same. Notice how easy it is to state the definition using generic elements aij , bij . Example 4

Example 5

If

⎡

⎤ 2 −3 −1 5 2⎦ A=⎣0 4 −4 6

⎡

⎤ 2 x −1 5 2 ⎦, B=⎣y 4 −4 z

and

◆ then A = B if and only if x = −3, y = 0, and z = 6. b If A = aij and B = ij are m × n matrices, then the sum of A and B is the matrix C = cij defined by cij = aij + bij , 1 ≤ i ≤ m, 1 ≤ j ≤ n. That is, C is obtained by adding the corresponding elements of A and B. Once again the use of generic elements makes it easy to state the definition.

3 4 −1 4 5 3 Let A = and B = . Then 5 0 −2 0 −3 2

7 9 2 3+4 4+5 −1 + 3 A+B= = . 5 + 0 0 + (−3) −2 + 2 5 −3 0 ◆

Observe that the sum of the matrices A and B is defined only when A and B have the same number of rows and the same number of columns. We agree to write A + B only when the sum is defined. A matrix all of whose entries are zero is called a zero matrix and is denoted by 0. Example 6

Each of the following is a zero matrix.

0 0 0 0

0 0 0 0 0 0

⎡

⎤ 0 0 0 ⎣0 0 0⎦ 0 0 0

◆

The following theorem gives some basic properties of matrix addition; the proofs are omitted. THEOREM 1

(a) A + B = B + A (b) (A + B) + C = A + (B + C) (c) A + 0 = 0 + A = A ■ If A = aij is an m × p matrix and B = bij is a p × n matrix, then the product of A and B, denoted AB, is the m × n matrix C = cij defined by cij = ai1 b1j + ai2 b2j + · · · + aip bpj

1 ≤ i ≤ m, 1 ≤ j ≤ n.

(2)

33

Fundamentals ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

⎤

a11

a12

...

a1p

a21 .. . ai1 .. .

a22 .. . ai2 .. .

...

a2p .. . aip .. .

am1

am2

. . . amp

...

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎡ b11

⎢ ⎢ ⎢ b21 ⎢ ⎢ .. ⎢ . ⎣ bp1

⎤

b12

. . . b1j

...

b1n

b22 .. .

. . . b2j .. .

...

b2n .. .

bp2

. . . bpj

. . . bpn

ai1

b1j

ai2 .. .

b2j .. .

aip

bpj

⎡ c11

⎥ ⎢ ⎥ ⎢ ⎥ ⎢ c21 ⎥ =⎢ ⎥ ⎢ .. ⎥ ⎢ . ⎦ ⎣ cm1

⎤

c12

...

c1n

c22 .. .

...

c2n .. .

cm2

. . . cmn

cij

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎤ Multiply corresponding ⎢ elements together and ⎥ ⎦ ⎣ add the results to form cij. ⎡

Figure 20

Let us explain (2) in more detail. The elements ai1 , ai2 , . . . , aip form the ith row of A, and the elements b1j , b2j , . . . , bpj form the jth column of B. Then (2) states that for any i and j, the element cij of C = AB can be computed in the following way, illustrated in Figure 20. 1. Select row i of A and column j of B, and place them side by side. 2. Multiply corresponding entries and add all the products.

Example 7

Let A =

2 1

3 −4 2 3

⎡

⎤ 3 1 2 ⎦. Then and B = ⎣ −2 5 −3

(2)(3) + (3)(−2) + (−4)(5) (2)(1) + (3)(2) + (−4)(−3) AB = (1)(3) + (2)(−2) + (3)(5) (1)(1) + (2)(2) + (3)(−3)

−20 20 = . 14 −4

◆

An array of dimension two is a modification of the idea of a matrix, in the same way that a linear array is a modification of the idea of a sequence. By an m×n array A we will mean an m × n matrix A of mn positions. We may assign numbers to these positions later, make further changes in these assignments, and still refer to the array as A. This is a model for two-dimensional storage of information in a computer. The number assigned to row i and column j of an array A will be denoted A[i, j]. As we have seen, the properties of matrix addition resemble the familiar properties for the addition of real numbers. However, some of the properties of matrix multiplication do not resemble those of real number multiplication. First, observe that if A is an m × p matrix and B is a p × n matrix, then AB can be computed and is an m × n matrix. As for BA, we have the following four possibilities: 1. BA may not be defined; we may have n = m. 2. BA may be defined and then BA is p × p, while AB is m × m and p = m. Thus AB and BA are not equal. 3. AB and BA may both be the same size, but not be equal as matrices. 4. AB = BA.

34

Fundamentals

Example 8

We agree as before to write AB only when the product is defined.

4 −5 1 −1 2 1 and . Then AB = and B = Let A = −1 3 2 −3 3 −2

−1 3 BA = . ◆ −5 8 The basic properties of matrix multiplication are given by the following theorem.

THEOREM 2

(a) A(BC) = (AB)C (b) A(B + C) = AB + AC (c) (A + B)C = AC + BC

■

The n × n diagonal matrix ⎡

⎤ 1 0 ··· 0 ⎢ 0 1 ··· 0 ⎥ In = ⎢ , .. ⎥ ⎣ ... ... .⎦ 0 0 ··· 1 all of whose diagonal elements are 1, is called the identity matrix of order n. If A is an m × n matrix, it is easy to verify that Im A = AIn = A. If A is an n × n matrix and p is a positive integer, we define Ap = A · A· · · A

and

A0 = In .

p factors

If p and q are nonnegative integers, we can prove the following laws of exponents for matrices: Ap Aq = Ap+q and (Ap )q = Apq . Observe that the rule (AB)p = Ap Bp does not hold for all square matrices. Howp p p ever, if AB = then (AB) = A B . BA, If A = aij is an m × n matrix, then the n × m matrix AT = aijT , where aijT = aji , 1 ≤ i ≤ m, 1 ≤ j ≤ n, is called the transpose of A. Thus the transpose

Example 9

of A is obtained by interchanging the rows and columns of A. ⎡ ⎤

3 4 5 2 −3 5 0 ⎦. Then Let A = and B = ⎣ 2 −1 6 1 3 1 6 −2 ⎡ ⎡ ⎤ ⎤ 2 6 3 2 1 1 ⎦ and BT = ⎣ 4 −1 6 ⎦. AT = ⎣ −3 5 3 5 0 −2

◆

The following theorem summarizes the basic properties of the transpose operation. THEOREM 3

If A and B are matrices, then (a) (AT )T = A (b) (A + B)T = AT + BT (c) (AB)T = BT AT

■

35

Fundamentals

Example 10

Example 11

A matrix A = aij is called symmetric if AT = A. Thus, if A is symmetric, it must be a square matrix. It is easy to show that A is symmetric if and only if aij = aji . That is, A is symmetric if and only if the entries of A are symmetric with respect to the main diagonal of A. ⎡ ⎤ ⎡ ⎤ 1 2 −3 1 2 −3 4 5 ⎦ and B = ⎣ 2 4 0 ⎦, then A is symmetric and B If A = ⎣ 2 −3 5 6 3 2 1 ◆ is not symmetric. If x is a nonzero number, there is a number y such that xy = 1. The number y is called the multiplicative inverse of x. When we have a multiplication for objects other than numbers, it is reasonable to ask if a multiplicative inverse exists. One example is matrix multiplication, described in this section. If A and B are n × n matrices, then we say that B is an inverse of A if AB = In and BA = In , where In is the n × n identity matrix defined earlier. The identity matrix behaves like the number 1, in that In A = AIn = A for any n × n matrix A, so the matrix inverse is analogous to the reciprocal of a nonzero number. However, it is not clear how to construct inverses, or even when they exist. ⎡ ⎤ ⎡ ⎤ 1 3 0 −2 3 −3 1 ⎦. This can An inverse of the matrix ⎣ 2 2 1 ⎦ is the matrix ⎣ 1 −1 1 0 1 2 −3 4 be verified by checking that ⎡ ⎤⎡ ⎤ ⎡ ⎤ 1 3 0 −2 3 −3 1 0 0 ⎣ 2 2 1 ⎦ ⎣ 1 −1 1⎦=⎣0 1 0⎦ 1 0 1 2 −3 4 0 0 1 and ⎡

⎤⎡ ⎤ ⎡ ⎤ −2 3 −3 1 3 0 1 0 0 ⎣ 1 −1 1 ⎦⎣ 2 2 1 ⎦ = ⎣ 0 1 0 ⎦. 2 −3 4 1 0 1 0 0 1

◆

There are tests to determine if an n × n matrix has an inverse, and formulas for computing an inverse if it exists. Many of these are programmed into calculators and computer algebra systems. We will be content to give the results for the case of 2 × 2 matrices.

a b e f Suppose that the 2 × 2 matrix A = has an inverse B = . c d g h Then

a b e f 1 0 = c d g h 0 1 and we have two pairs of equations ae + bg = 1 ce + dg = 0

and

af + bh = 0 cf + dh = 1.

When we solve the first pair for e and g, we find that we must divide by ad − bc. This can only be done if ad − bc = 0. The same condition is needed when solving the second pair for f and h. The results of solving for e, f , g, and h are summarized in Theorem 4.

36

THEOREM 4

Fundamentals

a b A matrix A = has a unique inverse if and only if ad − bc = 0. In this c d case we have ⎤ ⎡ b d − ⎥ ⎢ ad − bc ⎥ ⎢ ad − bc −1 ⎥. ⎢ A =⎢ ⎥ a c ⎦ ⎣ − ad − bc ad − bc ■

Boolean Matrix Operations A Boolean matrix (also called a bit matrix) is an m × n matrix whose entries are either zero or one. We shall now define three operations on Boolean matrices. Let A = aij and B = bij be m × n Boolean matrices. We define A ∨ B = C = cij , the join of A and B, by 1 if aij = 1 or bij = 1 cij = 0 if aij and bij are both 0 and A ∧ B = D = dij , the meet of A and B, by 1 if aij and bij are both 1 dij = 0 if aij = 0 or bij = 0. Note that these operations are only possible when A and B have the same size, just as in the case of matrix addition. Instead of adding corresponding elements in A and B, to compute the entries of the result, we simply examine the corresponding elements for particular patterns. ⎡

Example 12

1 ⎢0 Let A = ⎣ 1 0

0 1 1 0

⎤ ⎡ 1 1 1⎥ ⎢1 and B = ⎣ 0⎦ 0 0 1

(a) Compute A ∨ B.

1 0 0 1

⎤ 0 1⎥ . 1⎦ 0

(b) Compute A ∧ B.

Solution

(a) Let A∨B = cij . Then, since a43 and b43 are both 0, we see that c43 = 0. In all other cases, either aij or bij is 1, so cij is also 1. Thus ⎡ ⎤ 1 1 1 ⎢1 1 1⎥ A∨B=⎣ . 1 1 1⎦ 1 1 0 (b) Let A ∧ B = dij . Then, since a11 and b11 are both 1, d11 = 1, and since a23 and b23 are both 1, d23 = 1. In all other cases, either aij or bij is 0, so dij = 0. Thus ⎡ ⎤ 1 0 0 ⎢0 0 1⎥ A∧B=⎣ . 0 0 0⎦ 0 0 0 ◆

37

Fundamentals Finally, suppose that A = aij is an m × p Boolean matrix and B = bij is a p × n Boolean matrix. Notice that the condition on the sizes of A and B is exactly the condition needed to form the matrix product AB. We now define another kind of product. The product of A and B, denoted A B, is the m × n Boolean matrix Boolean C = cij defined by cij =

if aik = 1 and bkj = 1 for some k, 1 ≤ k ≤ p otherwise.

1 0

This multiplication is similar to ordinary matrix multiplication. The preceding formula states that for any i and j the element cij of C = A B can be computed in the following way, as illustrated in Figure 21. (Compare this with Figure 20.) 1. Select row i of A and column j of B, and arrange them side by side. 2. Compare corresponding entries. If even a single pair of corresponding entries consists of two 1’s, then cij = 1. If this is not the case, then cij = 0. ⎡ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎢ ⎣

⎤

a11

a12

...

a1p

a21 .. . ai1 .. .

a22 .. . ai2 .. .

...

a2p .. . aip .. .

am1

am2

. . . amp

...

⎥ ⎡ ⎥ b ⎥ ⎥ ⎢ 11 ⎥ ⎢b ⎢ ⎥ ⎥ ⎢ .21 ⎥ ⎢ . ⎥ ⎢ . ⎥ ⎣ ⎥ bp1 ⎥ ⎦

⎤

b12

. . . b1j

...

b1n

b22 .. .

. . . b2j .. .

...

b2n .. .

bp2

. . . bpj

. . . bpn

ai1

b1j

ai2 .. .

b2j .. .

aip

bpj

⎡ c11

⎥ ⎢ ⎥ ⎢ ⎥ ⎢c ⎥ = ⎢ 21 ⎥ ⎢ .. ⎥ ⎢ . ⎦ ⎣ cm1

⎤

c12

...

c1n

c22 .. .

...

c2n .. .

cm2

. . . cmn

cij

⎥ ⎥ ⎥ ⎥ ⎥ ⎥ ⎦

⎡

⎤ If any corresponding pair of entries are both ⎢ ⎥ ⎣ ⎦ equal to 1, then cij = 1; otherwise cij = 0.

Figure 21

We can easily perform the indicated comparisons and checks for each position of the Boolean product. Thus, at least for human beings, the computation of elements in A B is considerably easier than the computation of elements in AB. ⎡

Example 13

1 ⎢0 Let A = ⎣ 1 0 Solution

1 1 1 0

⎤ ⎡ ⎤ 0 1 0 0 0 0⎥ and B = ⎣ 0 1 1 0 ⎦. Compute A B. 0⎦ 1 0 1 1 1

Let A B = eij . Then e11 = 1, since row 1 of A and column 1 of B each have a 1 as the first entry. Similarly, e12 = 1, since a12 = 1 and b22 = 1; that is, the first row of A and the second column of B have a 1 in the second position. In a similar way we see that e13 = 1. On the other hand, e14 = 0, since row 1 of A and column

38

Fundamentals 4 of B do not have common 1’s in any position. Proceeding in this way, we obtain ⎤ ⎡ 1 1 1 0 ⎢0 1 1 0⎥ . AB=⎣ 1 1 1 0⎦ 1 0 1 1 ◆ The following theorem, whose proof is left as an exercise, summarizes the basic properties of the Boolean matrix operations just defined. THEOREM 5

5 Exercises

1. Let A = ⎡

−2 1

3 4

2 C=⎣5 2

3 6 0

If A, B, and C are Boolean matrices of compatible sizes, then 1. (a) A ∨ B = B ∨ A (b) A ∧ B = B ∧ A 2. (a) (A ∨ B) ∨ C = A ∨ (B ∨ C) (b) (A ∧ B) ∧ C = A ∧ (B ∧ C) 3. (a) A ∧ (B ∨ C) = (A ∧ B) ∨ (A ∧ C) (b) A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C) 4. (A B) C = A (B C)

⎡ ⎤ 3 5 , B = ⎣ −2 ⎦, and 2 4 ⎤

4 −1 ⎦. 8

(b) What is b11 , b31 ? (c) What is c13 , c23 , c33 ? (d) List the elements on the main diagonal of C. 2. Which of the following are diagonal matrices? ⎡ ⎤

3 0 0 2 3 0⎦ (b) B = ⎣ 0 −2 (a) A = 0 0 0 0 5 ⎡ ⎤ 0 0 0 (c) C = ⎣ 0 0 0 ⎦ 0 0 0 ⎡ ⎤ 2 6 −2 0⎦ (d) D = ⎣ 0 −1 0 0 3 ⎡ ⎤ 4 0 0 (e) E = ⎣ 0 4 0 ⎦ 0 0 4

a+b c+d 4 6 3. If = , find a, b, c, and d. c−d a−b 10 2

4. If

a + 2b 2c + d

2a − b c − 2d

=

In Exercises 5 through 10, let

2 4

1 1

1 C=⎣4 3 ⎡ 3 E=⎣5 0

−2 2 1

A= ⎡

(a) What is a12 , a22 , a23 ?

4 4

−2 , find a, b, c, and d. −3

■

2 4 1

⎤ 0 1 3 , B = ⎣ 1 2 ⎦, −2 2 3 ⎤

3 −3 2 5 ⎦, D = , 4 1 2 ⎤

−1 −2 3 −3 ⎦ , F = . 4 5 2

⎡

5. If possible, compute each of the following. (b) AB (a) C + E (d) AB + DF (c) CB + F 6. If possible, compute each of the following. (a) A(BD) and (AB)D (b) A(C + E) and AC + AE (c) FD + AB 7. If possible, compute each of the following. (a) EB + FA (b) A(B + D) and AB + AD (d) AC + DE (c) (F + D)A 8. If possible, compute each of the following. (a) AT and (AT )T (b) (C + E)T and CT + ET (d) (BT C) + A (c) (AB)T and BT AT 9. If possible, compute each of the following. (b) (BC)T and CT BT (a) AT (D + F)

39

Fundamentals (c) (BT + A)C 10. Compute D3 .

(d) (DT + E)F

11. Let A be an m × n matrix. Show that Im A = AIn = A. (Hint: Choose a generic element of Im A.)

2 1 −1 2 12. Let A = and B = . Show that 3 −2 3 4 AB = BA. ⎡ ⎤ 3 0 0 0 ⎦. 13. Let A = ⎣ 0 −2 0 0 4 (b) What is Ak ? (a) Compute A3 . 14. Show that A0 = 0 for any matrix A. 15. Show that InT = In . 16. (a) Show that if A has a row of zeros, then AB has a corresponding row of zeros. (Hint: Use the generic element definition of AB given in this section.) (b) Show that if B has a column of zeros, then AB has a corresponding column of zeros. 17. Show that the jth column of the matrix product AB is equal to the matrix product ABj , where Bj is the jth column of B. 18. If 0 is the 2 × 2 zero matrix, find two 2 × 2 matrices A and B, with A = 0 and B = 0, such that AB = 0.

0 1 19. If A = , show that A2 = I2 . 1 0

0 a 20. Determine all 2 × 2 matrices A = such that b c A2 = I2 . 21. Let A and B be symmetric matrices. (a) Show that A + B is also symmetric. (b) Is AB also symmetric? 22. Let A be an n × n matrix. (a) Show that AAT and AT A are symmetric. (b) Show that A + AT is symmetric. 23. Prove Theorem 3. [Hint: For part (c), show that the i, jth element of (AB)T equals the i, jth element of BT AT .] 24. Let A be a symmetric 2 × 2 matrix that has an inverse. Must A−1 also be symmetric? Explain your reasoning. 25. Find the inverse of each matrix.

2 1 −3 (b) (a) 5 7 0

6 5 (c) 4 −2 26. Find the inverse of each matrix.

−8 3 3 (b) (a) 9 −2 −1

−2 −9 (c) −6 −4

40

4 9

4 10

For Exercises 27 and 28, let ⎡ ⎤ ⎡ ⎤ 1 0 1 0.2 0.4 0.2 0.2 0.6 ⎦ , A = ⎣ 0 0 1 ⎦ , B = ⎣ −0.4 1 1 1 0.2 −0.6 0.2 ⎡ ⎤ ⎡ ⎤ 2 −1 1 1 −1 0 0 −1 ⎦ . 0 1 ⎦ , and D = ⎣ 1 C = ⎣ −1 1 1 1 0 1 0 27. (a) Verify that C is an inverse of A. (b) Verify that D is an inverse of B. 28. Determine whether CD is the inverse of AB. Explain your reasoning. 29. Show that if A and B are n × n matrices and A−1 , B−1 both exist, then (AB)−1 = (B−1 A−1 ). In Exercises 30 and 31, compute A ∨ B, A ∧ B, and A B for the given matrices A and B.

1 0 1 1 30. (a) A = , B= 0 1 0 1

1 1 0 0 , B= (b) A = 0 1 1 1

1 1 0 0 , B= (c) A = 1 1 1 0 ⎡ ⎤ ⎡ ⎤ 1 0 0 1 1 1 31. (a) A = ⎣ 0 1 1 ⎦, B = ⎣ 0 0 1 ⎦ 1 0 0 1 0 1 ⎡ ⎤ ⎡ ⎤ 0 0 1 0 1 1 (b) A = ⎣ 1 1 0 ⎦, B = ⎣ 1 1 0 ⎦ 1 0 0 1 0 1 ⎡ ⎤ ⎡ ⎤ 1 0 0 1 1 1 (c) A = ⎣ 0 0 1 ⎦, B = ⎣ 1 1 1 ⎦ 1 0 1 1 0 0 In Exercises 32 and 33, let F = fij be a p × q matrix. 32. (a) (b) (c) (d) 33. (a) (b) (c) (d) (e) (f)

Give a generic element in row k of F. What is the largest value of j? Give a generic element in column l of F. What is the largest value of i? Give the element in the lower left corner of F. Give the element in the upper left corner of F. Give the element in the lower right corner of F. Give the element in the upper right corner of F. Give a generic element in row r of FT . Give a generic element on the diagonal of F. In Exercises 34 and 35, let M = mij , N = nij , and P = pij be square matrices of the same size. Use the entries of M, N, and P to represent the entry in the ij-position of each of the following. 34. (a) P + M + N (b) PT + M (c) NT (d) NM 35. (a) PM (b) MP (c) P(M + N) (d) MN + PN

Fundamentals In Exercises 36 and 37, let M = mij , N = nij , and P = pij be square Boolean matrices of the same size. Use the entries of M, N, and P to represent the entry in the ij-position of each of the following. 36. (a) M ∧ P

(b) P ∨ N

(c) (M∧N)∨(M∧P)

37. (a) M P

(b) P M

(c) (M N) P

38. Complete the following proofs. (a) A ∨ A = A. Proof: Let bij be an element of A ∨ A. If bij = 0, then aij = , because . If bij = 1, then aij = because . Hence bij = aij for each i, j pair. (b) A ∧ A = A. Proof: Let bij be an element of A ∧ A. If bij = 0, then . If bij = 1, then . (Explain.) Hence bij = aij for each i, j pair. 39. Show that A ∨ B = B ∨ A. 40. Show that A ∧ B = B ∧ A. 41. Show that A ∨ (B ∨ C) = (A ∨ B) ∨ C. 42. Show that A ∧ (B ∧ C) = (A ∧ B) ∧ C. 43. Show that A (B C) = (A B) C. 44. Show that A ∧ (B ∨ C) = (A ∧ B) ∨ (A ∧ C). 45. Show that A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C). 46. What fact does Example 8 illustrate? 47. Let A = aij and B = bij be two n × n matrices and let C = cij represent AB. Prove that if k is an integer and k | aij for all i, j, then k | cij for all i, j. 48. Let p be a prime number with p > 2, and let A and B be matrices all of whose entries are integers. Suppose that p divides all the entries of A + B and all the entries of A − B.

6

Prove that p divides all the entries of A and all the entries of B. Another operation on matrices is scalar multiplication. Let k be a real number and A = aij be an m×n matrix. The result of multiplying A by the scalar k is the matrix kA = kaij . For Exercises 49 through 53, use the definition of scalar multiplication and the matrices given. ⎡ ⎤ ⎡ ⎤ 2 −3 −1 2 4 −6 5 2 ⎦, B = ⎣ 4 0 9 ⎦, A=⎣0 4 −4 6 7 −1 3 and ⎡

4 C=⎣ 3 −2

⎤ 0 1⎦ 5

49. Compute each of the following. (b) 5B (c) −1C (a) 3A 50. Compute each of the following. (b) 3A + 3B (a) 3(A + B) (d) A(−2C) (c) −2(AC) 51. Show that scalar multiplication has the following property. k(A + B) = kA + kB 52. Show that scalar multiplication has the following property. k(AB) = (kA)B = A(kB) 53. Let A be an m × n matrix. Find a matrix K such that KA = kA, for a fixed k.

Mathematical Structures Several times in this chapter, we have defined a new kind of mathematical object; for example, a set or a matrix. Then notation was introduced for representing the new type of object and a way to determine whether two objects are the same was described. Next we classified objects of the new type; for example, finite or infinite for sets, and Boolean or symmetric for matrices. And then operations were defined for the objects and the properties of these operations were examined. Such a collection of objects with operations defined on them and the accompanying properties form a mathematical structure or system. Here we deal only with discrete mathematical structures.

Example 1

The collection of sets with the operations of union, intersection, and complement and their accompanying properties is a (discrete) mathematical structure. We denote ◆ this structure by (sets, ∪, ∩, ).

Example 2

The collection of 3 × 3 matrices with the operations of addition, multiplication, and transpose is a mathematical structure denoted by (3 × 3 matrices, +, ∗, T ). ◆

41

Fundamentals An important property we have not identified before is closure. A structure is closed with respect to an operation if that operation always produces another member of the collection of objects. Example 3

The structure (5 × 5 matrices, +, ∗, T ) is closed with respect to addition because ◆ the sum of two 5 × 5 matrices is another 5 × 5 matrix.

Example 4

The structure (odd integers, +, ∗) is not closed with respect to addition. The sum of two odd integers is an even integer. This structure does have the closure property ◆ for multiplication, since the product of two odd numbers is an odd number. An operation that combines two objects is a binary operation. An operation that requires only one object is a unary operation. Binary operations often have similar properties, as we have seen earlier.

Example 5

(a) Set intersection is a binary operation since it combines two sets to produce a new set. ◆ (b) Producing the transpose of a matrix is a unary operation. Common properties have been given names. For example, if the order of the objects does not affect the outcome of a binary operation, we say that the operation is commutative. That is, if x 2 y = y 2 x, where 2 is some binary operation, 2 is commutative.

Example 6

(a) Join and meet for Boolean matrices are commutative operations. A∨B=B∨A

and

A ∧ B = B ∧ A.

(b) Ordinary matrix multiplication is not a commutative operation. AB = BA. ◆ Note that when we say an operation has a property, this means that the statement of the property is true when the operation is used with any objects in the structure. If there is even one case when the statement is not true, the operation does not have that property. If 2 is a binary operation, then 2 is associative or has the associative property if (x 2 y) 2 z = x 2 (y 2 z). Example 7

Set union is an associative operation, since (A ∪ B) ∪ C = A ∪ (B ∪ C) is always true. ◆ If a mathematical structure has two binary operations, say 2 and , a distributive property has the following pattern: x 2 (y z) = (x 2 y) (x 2 z). We say that “2 distributes over .”

Example 8

(a) We are familiar with the distributive property for real numbers; if a, b, and c are real numbers, then a · (b + c) = a · b + a · c. Note that because we have an agreement about real number arithmetic to multiply before adding, parentheses are not needed on the right-hand side. (b) The structure (sets, ∪, ∩, ) has two distributive properties: A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C) and A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C).

42

◆

Fundamentals Several of the structures we have seen have a unary operation and two binary operations. For such structures we can ask whether De Morgan’s laws are properties of the system. If the unary operation is ◦ and the binary operations are 2 and , then De Morgan’s laws are (x 2 y)◦ = x◦ y◦ Example 9

(x y)◦ = x◦ 2 y◦ .

and

(a) As we saw in Section 2, sets satisfy De Morgan’s laws for union, intersection, and complement: (A ∪ B) = A ∩ B and (A ∩ B) = A ∪ B. √ (real +, ∗, ) does not satisfy De Morgan’s laws, (b) The structure √ numbers, √ √ ◆ since x + y = x ∗ y. A structure with a binary operation 2 may contain a distinguished object e, with the property x 2 e = e 2 x = x for all x in the collection. We call e an identity for 2. In fact, an identity for an operation must be unique.

THEOREM 1

If e is an identity for a binary operation 2, then e is unique. Proof Assume another object i also has the identity property, so x 2 i = i 2 x = x. Then e 2 i = e, but since e is an identity for 2, i 2 e = e 2 i = i. Thus, i = e. ■ Therefore there is at most one object with the identity property for 2. This is one of our first examples of a proof that does not proceed directly. We assumed that there were two identity elements and showed that they were in fact the same element.

Example 10

For (n × n matrices, +, ∗, T ), In is the identity for matrix multiplication and the ◆ n × n zero matrix is the identity for matrix addition. If a binary operation 2 has an identity e, we say y is a 2-inverse of x if x 2 y = y 2 x = e.

THEOREM 2

Example 11

Example 12

If 2 is an associative operation and x has a 2-inverse y, then y is unique. Proof Assume there is another 2-inverse for x, say z. Then (z 2 x) 2 y = e 2 y = y and z 2 (x 2 y) = z 2 e = z. Since 2 is associative, (z 2 x) 2 y = z 2 (x 2 y) ■ and so y = z. (a) In the structure (3 × 3 matrices, +, ∗, T ), each matrix A = aij has a +-inverse, or additive inverse, −A = −aij . (b) In the structure (integers, +, ∗), only the integers 1 and −1 have multiplicative ◆ inverses. Let 2, , and ◦ be defined for the set {0, 1} by the following tables. 2

0 1

0 0 1

1 1 0

0 1

0 0 0

1 0 1

x◦ 0 1

x 1 0

Thus 1 2 0 = 1, 0 1 = 0, and 1◦ = 0. Determine if each of the following is true for ({0, 1}, 2, , ◦). (a) 2 is commutative.

(b) is associative.

43

Fundamentals (c) De Morgan’s laws hold. (d) Two distributive properties hold for the structure. Solution

(a) The statement x 2 y = y 2 x must be true for all choices of x and y. Here there is only one case to check: Is 0 2 1 = 1 2 0 true? Since both 0 2 1 and 1 2 0 are 1, 2 is commutative. (b) The eight possible cases to be checked are left as an exercise. See Exercise 6(b). 0◦ 0◦ = 1 1 = 1 (c) (0 2 0)◦ = 0◦ = 1 ◦ ◦ 0◦ 1◦ = 1 0 = 0 (0 2 1) = 1 = 0 ◦ ◦ 1◦ 1◦ = 0 0 = 0 (1 2 1) = 0 = 1 The last pair shows that De Morgan’s laws do not hold in this structure. (d) One possible distributive property is x 2 (y z) = (x 2 y) (x 2 z). We must check all possible cases. One way to organize this is shown in a table. x y z 0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

y z x 2 (y z) x 2 y 0 0 0 1 0 0 0 1

0 0 0 1 1 1 1 0 (A)

0 0 1 1 1 1 0 0

x 2 z (x 2 y) (x 2 z) 0 1 0 1 1 0 1 0

0 0 0 1 1 0 0 0 (B)

Since columns (A) and (B) are not identical, this possible distributive property does not hold in this structure. The check for the other distributive ◆ property is Exercise 7. It is useful to consider mathematical structures themselves as objects and to classify them according to the properties associated with their operations.

6 Exercises In Exercises 1 and 2, tell whether the structure has the closure property with respect to the operation. 1. (a) (b) 2. (a) (b)

(sets, ∪, ∩, ) union complement (sets, ∪, ∩, ) multiplication (4 × 4 matrices, +, ∗, T ) T transpose (3 × 5 matrices, +, ∗, )

In Exercises 3 and 4, tell whether the structure has the closure property with respect to the operation. 3. (a) (integers, +, −, ∗, ÷) division catenation (b) (A∗ , catenation) 4. (a) (n × n Boolean matrices, ∨, ∧, T )

44

meet

5. 6. 7. 8.

addition (b) (prime numbers, +, ∗) Show that ⊕ is a commutative operation for sets. Using the definitions in Example 12, (a) show that 2 is associative. (b) Show that is associative. Using the definitions in Example 12, determine if the other possible distributive property holds. Give the identity element, if one exists, for each binary operation in the given structure. √ ) (a) (real numbers, +, ∗, (b) (sets, ∪, ∩, ) (c) ({0, 1}, 2, , ∗) as defined in Example 12 (d) (subsets of a finite set A, ⊕, )

Fundamentals 9. Give the identity element, if one exists, for each binary operation in the structure (5 × 5 Boolean matrices, ∨, ∧, ). In Exercises 10 through 16, use the structure S = (n × n diagonal matrices, +, ∗, T ).

29. Let R = (2 × 1 matrices, ), where

w x+w x = . z y+z+1 y

10. Show that S is closed with respect to addition. 11. Show that S is closed with respect to multiplication. 12. Show that S is closed with respect to the transpose operation.

30.

13. Does S have an identity for addition? If so, what is it? 14. Does S have an identity for multiplication? If so, what is it?

31.

15. Let A be an n × n diagonal matrix. Describe the additive inverse of A. 16. Let A be an n × n diagonal matrix. Describe the multiplicative inverse of A. In Exercises 17 through 23, use the structure R = T (M, +, ∗,

), where M is the set of matrices of the form a 0 , where a is a real number. 0 0

32.

17. Show that R is closed with respect to addition.

33.

18. Show that R is closed with respect to multiplication. 19. Show that R is closed with respect to the transpose operation. 20. Does R have an identity for addition? If so, what is it?

34.

21. Does R have an identity for multiplication? If so, what is it? 22. Let A be an element of M. Describe the additive inverse for A. 23. Let A be an element of M. Describe the multiplicative inverse for A. In Exercises 24 through 28, let R = (Q, 2), where x 2 y = x+y . Determine which of the following properties hold for 2 this structure: 24. Closure 25. Commutative

35.

Determine which of the following properties hold for this structure. (b) Commutative (a) Closure (c) Associative Let R be as in Exercise 29. Determine which of the following properties hold for this structure. (a) An identity element (b) An inverse for every element Let S = (1×2 matrices, 2), where x y 2 w z = y+z . Determine which of the following x+w 2 properties hold for this structure. (b) Commutative (a) Closure (c) Associative Let S be as in Exercise 31. Determine which of the following properties hold for this structure. (a) An identity element (b) An inverse for every element (a) Give a symbolic statement of the distributive property for scalar multiplication over as defined in Exercise 29. (b) Is the distributive property in part (a) a property of R? (a) Give a symbolic statement of the distributive property for scalar multiplication over 2 as defined in Exercise 31. (b) Is the distributive property in part (a) a property of S? For a Boolean matrix B, we define comp B to be the matrix formed by changing each 0 entry of B to 1 and each 1 entry of B to 0. Let R = (5 × 5 Boolean matrices, ∧, ∨, comp). Do De Morgan’s laws hold for R? Justify your answer.

The properties of a mathematical structure can be used to rewrite expressions just as is done in ordinary algebra. In Exercises 36 through 39, rewrite the given expression to produce the requested result.

27. An identity element

36. (A ∪ B) ∩ (A ∪ B) one set, no operations 37. (A ∩ B) ∩ A two sets, two operations 38. (A ∪ B) ∪ (A ∩ B) two sets, two operations

28. An inverse for every element

39. (A ∪ B) ∩ (A ∪ B)

26. Associative

one set, no operations

45

Fundamentals Tips for Proofs

Many exercises in this chapter ask that you show, prove, or verify a statement. To show or prove a statement means to give a written explanation demonstrating that the statement is always true. To verify a statement means to check its truth for a particular case; see, for example, Section 2, Exercises 16 and 24. Most proofs required in this chapter proceed directly from the given conditions using definitions and previously proven facts; an example is Section 4, Theorem 2. A powerful tool for constructing a proof is to choose a generic object of the type in the statement and to see what you know about this object. Remember that you must explain why the statement is always true, so choosing a specific object will only verify the statement for that object. The most common way to show that two sets are equal is to show each is a subset of the other (Section 2, Theorem 1). In proving statements about sets or matrix operations, try to work at the level of object names rather than at the element or entry-level. For example, Section 5, Exercise 22 is more easily proved by using the facts that if Ais symmetric, then AT = A and Theorem 3 rather than by using the fact that if A = aij is symmetric, then aij = aji for each i and j. One other style of direct proof is seen in Section 6, Example 12. Sometimes we show the statement is always true by examining all possible cases.

Key Ideas for Review • Set: a well-defined collection of objects • ∅ (empty set): the set with no elements

• Characteristic function of a set A: fA (x) =

1 0

if x ∈ A if x ∈ /A

• Equal sets: sets with the same elements

• Countable set: a set that corresponds to a sequence

• A ⊆ B (A is a subset of B): Every element of A is an element

• Word: finite sequence of elements of A

• • • • • • •

of B. |A| (cardinality of A): the number of elements of A Infinite set P(A) (power set of A): the set of all subsets of A A ∪ B (union of A and B): {x | x ∈ A or x ∈ B} A ∩ B (intersection of A and B):{x | x ∈ A and x ∈ B} Disjoint sets: two sets with no elements in common A − B (complement of B with respect to A): {x | x ∈ A and x ∈ / B}

• A (complement of A): {x | x ∈ / A} • Algebraic properties of set operations

• Regular expression • Theorem: If n and m are integers and n > 0, we can write

m = qn + r for integers q and r with 0 ≤ r < n. Moreover, there is just one way to do this.

• GCD(a, b): d = GCD(a, b) if d | a, d | b, and d is the

largest common divisor of a and b. • Theorem: If d is GCD(a, b), then

(a) d = sa + tb for some integers s and t. (b) If c | a and c | b, then c | d.

• Relatively prime: two integers a and b with GCD(a, b) = 1 • Euclidean algorithm: method used to find GCD(a, b)

• Theorem (the addition principle): If A and B are finite sets,

• LCM(a, b): c = LCM(a, b) if a | c, b | c, and c is the

• Theorem (the three-set addition principle): If A, B, and C

• Theorem: GCD(a, b) · LCM(a, b) = ab

then |A ∪ B| = |A| + |B| − |A ∩ B|.

are finite sets, then |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|.

• Inclusion-exclusion principle

smallest common multiple of a and b • Base b expansion of a number • Cryptology: the science of producing and deciphering secret

codes

• Sequence: list of objects arranged in a definite order

• Bacon’s code

• Recursive formula: formula that uses previously defined

• Steganography: the science of concealment of coded infor-

terms • Explicit formula: formula that does not use previously defined terms • Linear array

46

mation • mod-n function: fn (z) = r, where r is the remainder when

z is divided by n • Matrix: rectangular array of numbers

Fundamentals • Size of a matrix: A is m × n if it has m rows and n columns • Diagonal matrix: a square matrix with zero entries off the

main diagonal • Equal matrices: matrices of the same size whose correspond-

ing entries are equal • A + B: the matrix obtained by adding corresponding entries • • • • • • • •

of A and B Zero matrix: a matrix all of whose entries are zero AB In (identity matrix): a square matrix with ones on the diagonal and zeros elsewhere Array of dimension 2 AT : the matrix obtained from A by interchanging the rows and columns of A Symmetric matrix: AT = A Inverse of a matrix Boolean matrix: a matrix whose entries are either one or zero

• A∨B • A∧B • AB • Properties of Boolean matrix operations • Mathematical structure: a collection of objects with opera-

tions defined on them and the accompanying properties • Binary operation: an operation that combines two objects • Unary operation: an operation that requires only one object • Closure property: each application of the operation produces

another object in the collection • Associative property: (x 2 y) 2 z = x 2 (y 2 z) • Distributive property: x 2 (y z) = (x 2 y) (x 2 z) • De Morgan’s laws: (x 2 y)◦ = x◦ y◦ and

(x y)◦ = x◦ 2 y◦ • Identity for 2: an element e such that x 2 e = e 2 x = x for all x in the structure • 2-inverse for x: an element y such that x 2 y = y 2 x = e, where e is the identity for 2

Chapter Self-Test 1. What kind of mathematical object is P(A)? 2. What kind of mathematical object is |P(A)|? 3. What kind of mathematical object is LCM(a, b)? 4. What kind of mathematical object is kA? 5. What are the components of a mathematical structure? 6. Let A = {x | x is a real number and 0 < x < 1}, B = {x | x is a real number and x2 + 1 = 0}, C = {x | x = 4m, m ∈ Z}, D = {0, 2, 4, 6, . . . }, and E = {x | x ∈ Z and x2 ≤ 100}. (a) Identify the following as true or false. (i) C ⊆ D (iii) {4, 16} ⊆ E (v) B ⊆ ∅

(ii) {4, 16} ⊆ C (iv) D ⊆ D

(b) Identify the following as true or false. (i) (ii) (iv) (v)

C ∩ E (C ∪ E) ∅ ⊆ (A ∩ B) (iii) C ∩ D = D C∪E ⊆D A∩D⊆A∩C

7. Let A = {x | x = 2n, n ∈ Z+ }, B = {x | x = 2n+1, n ∈ Z+ }, C = {x | x = 4n, n ∈ Z+ }, and D = {x | x(x2 − 6x + 8) = 0, x ∈ Z}. Use Z as the universal set and find (b) A (a) A ∪ B (c) (A ∩ D) ⊕ (A ∩ B)

(d) A ∪ C

(e) A − C 8. Draw a Venn diagram to represent (a) A ∩ B and (b) A ∩ B.

9. Under what conditions will A ∩ B = A ∪ B? 10. Suppose that 109 of the 150 mathematics students at Verysmall College take at least one of the following computer languages: PASCAL, BASIC, C++ . Suppose 45 study BASIC, 61 study PASCAL, 53 study C++ , 18 study BASIC and PASCAL, 15 study BASIC and C++ , and 23 study PASCAL and C++ . (a) How many students study all three languages? (b) How many students study only BASIC? (c) How many students do not study any of the languages? 11. Define a sequence as follows: a0 = 0, a1 = 0, an = 1 − 3an−1 + 2an−2 . Compute the first six terms of this sequence. 12. Let U = {a, b, c, d, e, f, g, h, i, j}, A = {a, b, d, f }, B = {a, b, c, h, j}, C = {b, c, f, h, i}, and D = {g, h}. Represent each of the following sets by an array of zeros and ones. (b) A ∩ B (a) A ∪ B (d) (A ∩ B) ∪ D (c) A ∩ (B ∪ C) 13. Let I = {a, b, c}. In each part that follows is listed a string in I ∗ and a regular expression over I. For each, state whether the string belongs to the regular set corresponding to the expression. (b) acbb ((acb) ∨ b)∗ (a) ab a∗ bc∗ (d) abaca (ab)∗ ac (c) bc ((ab∗ ) ∨ c) 14. Use the Euclidean algorithm to compute GCD(4389, 7293) and write it as s(7293) + t(4389).

47

Fundamentals

2 6 4 2 0 and B = . Compute, −1 3 2 −3 1 if possible, each of the following. (b) BA (c) BT (a) AB

15. Let A =

(d) A + B (g) B−1 A

(e) AT B

(f) B−1

⎡

⎤ ⎡ ⎤ 1 0 1 1 1 0 16. Let C = ⎣ 1 1 0 ⎦ and D = ⎣ 0 1 0 ⎦. 0 1 1 1 1 0 Compute each of the following. (b) C ∨ D (c) C ∧ D (a) C D 17. Let S = (2 × 2 Boolean matrices, ∧, ∨, ) and A be a 2 × 2 Boolean matrix. Describe the ∧-inverse of A in S.

Experiment 1 In many voting procedures the rules are one person, one vote, and a simple majority is required to elect a candidate or to pass a motion. But it is not unusual to have a procedure where individual voters have more than one vote or where something other than a simple majority is required. An example of such a situation is when not all shareholders in a company own the same number of shares, and each shareholder has as many votes as shares. Does a shareholder with twice as many shares as another have twice as much control, or power, over the company? In this experiment you will investigate this question and some related ones. First, we begin with some definitions. The number of votes that a voter has is called the voter’s weight. Here only counting numbers can be weights. The total number of votes needed to elect a candidate or to pass a motion is the quota. The collection of the quota and the individual weights for all voters is called a weighted voting system. If the voters are designated v1 , v2 , . . . , vk with corresponding weights w1 , w2 , . . . , wk and q is the quota, then the weighted voting system may be conveniently represented by [q : w1 , w2 , . . . , wk ]. For ease of computations, the weights are usually listed from largest to smallest. 1. For the weighted voting system [9 : 9, 4, 2, 1], what is the quota? How many voters are there? What is the total number of votes available? 2. In a weighted voting system [q : w1 , w2 , . . . , wk ], what are the restrictions on the possible values of q? Explain each restriction. 3. For the weighted voting system [9 : 9, 4, 2, 1], describe how much power voter v1 has. Such a voter is called a dictator. Why is this appropriate? Could a system have two dictators? Explain why or why not. 4. For [8 : 5, 3, 2, 1], is v1 a dictator? Describe v1 ’s power relative to the other voters. More interesting cases arise when the power of each voter is not so obvious as in these first examples. One way to measure a voter’s power was developed by John Banzhaf in 1965. A coalition is a subset of the voters in a weighted voting system. If the total number of votes controlled by the members of the coalition equals or exceeds the quota, we call the coalition a winning coalition. If not, this is a losing coalition. 5. (a) List all the coalitions for [9 : 9, 4, 2, 1]. Which of these are winning coalitions? (b) List all the winning coalitions for [8 : 5, 3, 2, 1]. Banzhaf’s idea is to measure a voter’s power by examining how many times removal of that voter from a coalition would change the coalition from winning to losing. Consider the system [7 : 5, 4, 3]. The winning coalitions are {v1 , v2 }, {v1 , v3 }, {v2 , v3 }, and {v1 , v2 , v3 }. Each member of the first three coalitions has the power to change it from winning to losing, but none have this power in the last

48

Fundamentals coalition. All together there are six opportunities for change. Each of v1 , v2 , v3 has two of these opportunities. We record this information as the Banzhaf power distribution for the system: v1 : 26 , v2 : 26 , v3 : 26 . According to this analysis, all three voters have the same amount of power despite having different weights. The fraction of power assigned to a voter is the voter’s Banzhaf power index. 6. Here is a test for Banzhaf’s definition of power. Calculate the Banzhaf power distribution for [9 : 9, 4, 2, 1]. Explain how the results are consistent with the designation of v1 as a dictator. 7. Calculate the Banzhaf power distribution for [8 : 5, 3, 2, 1]. A voter like v1 that must belong to every winning coalition has veto power in the system. 8. Let [q : 6, 3, 1] be a weighted voting system. (a) Give values for q for which the system has a dictator and identify that voter. (b) Give values for q for which one or more voters have veto power and identify these voters. (c) Give values for q for which at least one player is powerless, but there is no dictator. Which player is powerless? Banzhaf’s idea is adaptable to cases where each voter has one vote, but special rules for voting apply. 9. The four partners in a company agree that each partner has one vote and a simple majority passes a motion. In the case of a tie the coalition containing the senior partner is the winning coalition. Give the Banzhaf power distribution for this system. Would the distribution change if the tie-breaking rule were changed to the coalition containing the most junior member is the losing coalition? Explain. 10. Suppose you are the voter with weight one in [8 : 5, 3, 2, 1]. (a) What is your Banzhaf power index? (b) Unhappy with this situation, you offer to buy a vote from one of the other voters. If each is willing to sell and each asks the same price, from whom should you buy a vote and why? Give the Banzhaf power distribution for this system for the resulting weighted voting system. 11. Here is another feature of Banzhaf’s way of measuring power. Let [q : w1 , w2 , . . . , wk ] be a weighted voting system and n be a positive integer. Prove that the Banzhaf power distributions for [q : w1 , w2 , . . . , wk ] and [nq : nw1 , nw2 , . . . , nwk ] are the same. 12. We now return to the original question about power. Suppose we have a weighted voting system in which v1 has weight w1 , v2 has weight w2 , and w1 = 2w2 . Construct such a system where the Banzhaf power index of v1 is (a) the same as that of v2 (b) twice that of v2 (c) more than twice that of v2 .

Coding Exercises In Exercises 1 through 3, assume that A and B are finite sets of integers. Write a subroutine to compute the specified set. 1.

A∪B

2. A ∩ B

3. A − B

49

Fundamentals 4.

Consider the sequence recursively defined by g(0) = 1, g(1) = −1, g(n) = 3g(n − 1) − 2g(n − 2). (a) Write a subroutine that will print the first 20 terms of the sequence. (b) Write a subroutine that will print the first n terms of the sequence. The user should be able to supply the value of n at runtime.

5. Write a subroutine to find the least common multiple of two positive integers.

Answers to Odd-Numbered Exercises Exercise Set 1

Exercise Set 2

1. (a) True.

(b) False.

(c) False.

(d) False.

(e) True.

(f) False.

3. (a) {A, R, D, V, K}.

(b) {B, O, K}.

(c) {M, I, S, P}. 5. (a) False.

(b) True.

(c) False.

(d) True.

(e) False.

(f) False.

7. {x | x is a vowel}. 9. {x | x ∈ Z and x2 < 5}.

(b) {a, c, d, e, f, g}.

(c) {a, c}.

(d) {f }.

(e) {b, g, d, e}. (g) {d, e, f, h, k}.

(f) {a, b, c}.

(i) {b, g, f }.

(j) {g}.

(h) {a, b, c, d, e, f }.

3. (a) {a, b, c, d, e, f, g}.

(b) { }.

(c) {a, c, g}.

(d) {a, c, f }.

(e) {h, k}.

(f) {a, b, c, d, e, f, h, k}.

5. (a) {1, 2, 4, 5, 6, 8, 9}.

(b) {1, 2, 3, 4, 6, 8}.

11. (b), (c), (e).

(c) {1, 2, 4, 6, 7, 8}.

(d) {1, 2, 3, 4, 5, 9}.

13. { }, {JAVA}, {PASCAL}, {C++}, {JAVA, PASCAL}, {JAVA, C++}, {PASCAL, C++}, {JAVA, PASCAL, C++}.

(e) {1, 2, 4}. (g) {2, 4}.

(f) {8}.

15. (a) True.

(b) False.

(c) False.

(d) True.

(e) True.

(f) True.

(g) True.

(h) True.

(h) { }.

7. (a) {1, 2, 3, 4, 5, 6, 8, 9}.

(b) {2, 4}.

(c) {1, 2, 4}.

(d) {8}.

(e) {3, 7}.

(f) {1, 3, 5, 6, 7, 8, 9}.

17. (a) ⊆.

(b) ⊆.

(c) .

9. (a) {b, d, e, h}.

(d) ⊆.

(e) .

(f) ⊆.

(c) {b, d, h}.

(d) {b, c, d, e, f, g, h}.

(e) { }.

(f) {c, f, g}.

19. {1, 2, 3} 21. Yes, Yes, the complement of a set would not be defined unambiguously. 23. (a) False.

(b) False.

(d) False.

(e) True.

(c) Insufficient information. (f) True.

25. Eight. There are three parts that represent what is left of each set when common parts are removed, three regions that each represent the part shared by one of the three pairs of sets, a region that represents what all three sets have in common, and a region outside all three sets. 27. B = {m, n}.

29. B = {a, b, c}.

31.

is one solution. A

x

B

(b) {b, c, d, f, g, h}.

11. (a) All real numbers except −1 and 1. (b) All real numbers except −1 and 4. (c) All real numbers except −1, 1, and 4. (d) All real numbers except −1. 13. (a) True.

(b) True.

(c) False.

(d) False.

15. 1. 17. (a) |A ∪ B| = 10, |A| = 6, |B| = 7, |A ∩ B| = 3. Hence |A ∪ B| = |A| + |B| − |A ∩ B|. (b) |A ∪ B| = 11, |A| = 5, |B| = 6, |A ∩ B| = 0. Hence |A ∪ B| = |A| + |B| − |A ∩ B|. 19. B must be the empty set. 21. The complement of the intersection of two sets is the union of the complements of the two sets.

y

33. ∅ ⊆ Z+ ⊆ N ⊆ Z ⊆ Q ⊆ R.

23. |A| = 6, |B| = 5, |C| = 6, |A ∩ B| = 2, |A ∩ C| = 3, |B ∩ C| = 3, |A ∩ B ∩ C| = 2, |A ∪ B ∪ C| = 11. Hence |A ∪ B ∪ C| = |A| + |B| + |C| − |A ∩ B| − |A ∩ C| − |B ∩ C| + |A ∩ B ∩ C|.

35. B; B.

25. (a) 106.

C

50

1. (a) {a, b, c, d, e, f, g}.

37. 4; 8.

(b) 60.

Fundamentals 51. The cardinality of the union of n sets is the sum of the cardinalities of each of the n sets minus the sum of the cardinalities of the n C2 different intersections of two of the sets plus the sum of the n C3 different intersections of three of the sets and so on, alternating plus and minus the sum of the n Ck k-set intersections, k = 4, . . . , n.

27. 16; 23. 29. (a) 162.

(b) 118.

(d) 290.

(e) 264.

(c) 236.

31. D

U

B

Exercise Set 3 392

36

245

158

143

289

1. {1, 2}.

J

3. {a, b, c, . . . , z}. 5. Possible answers include xyzxyz. . . , xxyyzzxxyyzz, and yzxyzx. . . .

S

7. 5, 25, 125, 625. 9. 1, 2, 6, 24. 11. 2.5, 4, 5.5, 7.

33. (a) True.

(b) True.

(c) False.

(d) False.

35. (a) Not possible to decide. (b) Not possible to decide. (c) Not possible to decide. (d) Not possible to decide. (e) Not possible to decide. (f) Not possible to decide. 37. (a) Not possible to decide. (b) Not possible to decide.

13. 0, −2, −4, −6. 15. an = an−1 + 2, a1 = 1, recursive; an = 2n − 1, explicit. 17. cn = (−1)n+1 , explicit. 19. en = en−1 + 3, e1 = 1, recursive.

(c) Not possible to decide. (d) Not possible to decide.

21. an = 2 + 3(n − 1).

(e) Not possible to decide.

23. A, uncountable; B, finite; C, countable; D, finite; E, finite.

39. A and to B. 41. (a)

C

(b) A ∪ B.

B A

(c) Let x ∈ A ∪ B. Then x ∈ A or x ∈ B. Since A ⊆ C and B ⊆ C, x ∈ A or x ∈ B means that x ∈ C. Hence A ∪ B ⊆ C. 43. Yes. Suppose x ∈ B. Either x ∈ A or x ∈ / A. If x ∈ A, then x∈ / A ⊕ B = A ⊕ C. But then x must be in C. If x ∈ / A, then x ∈ A ⊕ B = A ⊕ C, and again x must be in C. So B ⊆ C. A similar argument shows that C ⊆ B, so B = C. 45. No. Let A = {1, 2, 3}, B = {4}, and C = {3, 4}. Then A ∪ B = A ∪ C and B = C. 47. (a) Let x ∈ A ∪ C. Then x ∈ A or x ∈ C, so x ∈ B or x ∈ D and x ∈ B ∪ D. Hence A ∪ C ⊆ B ∪ D. (b) Let x ∈ A ∩ C. Then x ∈ A and x ∈ C, so x ∈ B and x ∈ D. Hence x ∈ B ∩ D. Thus A ∩ C ⊆ B ∩ D. 49. We must subtract |B ∩ C|, because each element in B ∩ C has been counted twice in |A| + |B| + |C|. But when we subtract both |B ∩ C| and |A ∩ C|, we have “uncounted” all the elements of C that also belong to B and A. These (and the similar elements of A and B) are counted again by adding |A ∩ B ∩ C|.

25. (a) Yes.

(b) No.

(c) Yes.

(d) Yes.

(e) No.

(f) No.

27. (a) 1.

(b) 0.

(c) fB : 10000000, fC : 01010011, fD : 01000101. (d) 11010011, 01010111, 01000001. 29. f(A⊕B)⊕C = fA⊕B + fC − 2fA⊕B fC by Theorem 4 = (fA + fB − 2fA fB ) + fC − 2(fA + fB − 2fA fB )fC = fA + (fB + fC − 2fB fC ) − 2fA (fB + fC − 2fB fC ) = fA + fB⊕C − 2fA fB⊕C = fA⊕(B⊕C) Since the characteristic functions are the same, the sets must be the same. 31. (a) Yes.

(b) Yes.

(c) Yes.

33. Possible answers include ∨ ac 35. (a) 01∗ 0.

a∨

ab(.

(b) 0(00)∗ ∨ (00)∗ 1.

37. By (1), 8 is an S-number. By (3), 1 is an S-number. By (2), all multiples of 1, that is, all integers are S-numbers. 39. 1, 2, 3, 7, 16.

51

Fundamentals Exercise Set 4

45. (a) Answers will vary, but the pattern of italicized and nonitalized letters should match AAAAAAAAAAAAAAAAAAAAAAAAAA AAAAAAAAA.

1. 20 = 6 · 3 + 2. 3. 3 = 0 · 22 + 3. 5. (a) 828 = 22 · 32 · 23. (c) 1781 = 13 · 137.

(b) 1666 = 2 · 72 · 17.

(b) STUDY WELL.

(d) 1125 = 32 · 53 .

47. BIJS.

(e) 107.

Exercise Set 5

7. d = 3; 3 = 3 · 45 − 4 · 33.

1. (a) −2, 1, 2.

9. d = 1; 1 = 5 · 77 − 3 · 128. 11. 1050. 15. (a) 6. (d) 1. 17. (a) 10.

(b) 1.

(c) 0.

(e) 20.

(f) 14.

(b) 22.

(c) 2.

(d) 14.

19. f(a) + f(b) may be greater than n. (b) {1, 6, 11, 16, . . . }.

23. If a | b, then b = ka, for some k ∈ Z. Thus, mb = m(ka) = (mk)a and mb is a multiple of a. 25. The only divisors of p are ±p and ±1, but p does not divide a. (Multiply both sides by b.) p | sab and p | tpb. If p divides the right side of the equation, then it must divide the left side also. 27. Because a | m, ac | mc and because c | m, ac | am. If GCD(a, c) = 1, there are integers s, t such that 1 = sa + tc. Thus m = sam + tcm. But ac divides each term on the left so ac | m. 29. Let d = GCD(a, b). Then cd | ca and cd | cb; that is, cd is a common divisor of ca and cb. Let e = GCD(ca, cd). Then cd | e and e = cdk is a divisor of ca and cb. But then dk | a and dk | b. Because d is the greatest common divisor of a and b, k must be 1 and e = cd. 31. By Theorem 6, GCD(a, b) · LCM(a, b) = ab. GCD(a, b) = 1, we have LCM(a, b) = ab.

Since

33. a | b means b = am. b | a means a = bn. Thus b = am = bnm. Hence nm = 1 and n = m = 1, because a and b are positive. 35. No; consider a = 6, b = 4, c = 3. 37. Yes. Using the same reasoning as in Exercise 36, m and n share no prime factors, so for mn to be a perfect tth power, each prime in the factorizations of m and n must appear a multiple of t times. But this means each of m and n are also perfect tth powers. 39. (a) 112.

(b) 10.

(c) 30.

41. (a) (i) (104)5 . (ii) (243)5 . (iii) (1330)5 . (iv) (10412)5 . (b) (i) 49.

(ii) 85.

(iii) 197.

(iv) 816.

43. (a) (11101)2 . (1001001)2 . (11010111)2 . (1011011100)2 . (b) (131)4 . (1021)4 . (3113)4 . (23130)4 . (c) (1D)16 . (49)16 . (D7)16 . (2DC)16 .

52

(c) 4, −1, 8.

(d) 2, 6, 8.

13. 864.

21. (a) {2, 7, 12, 17, . . . }.

(b) 3, 4.

3. a is 3, b is 1, c is 8, and d is −2. ⎡ ⎤ 4 0 2 5. (a) ⎣ 9 6 2 ⎦. 3 2 4

7 13 . (b) AB = −3 0

21 (d) (c) Not possible. −7

14 . 17

7. (a) EB is 3 × 2 and FA is 2 × 3; the sum is undefined. (b) B + D does not exist.

10 0 −25 . (d) DE does not exist. (c) 40 14 12 ⎡ ⎤ 22 34 3 11 ⎦. 9. (a) ⎣ (b) BC is not defined. −31 3

25 5 26 . (c) 20 −3 32 (d) DT + E is not defined. m 11. Let B = bjk = Im A. Then bjk = l=1 ijl alk , for 1 ≤ j ≤ m and 1 ≤ k ≤ n. But ijj = 1 and ijl = 0 if j = l. Hence bjk = ijj ajk , 1 ≤ j ≤ m, 1 ≤ k ≤ n. This means B = Im A = A. Similarly, if C = AIn = cjk , n cjk = l=1 ajl ilk = ajk ikk = ajk for 1 ≤ j ≤ m, 1 ≤ k ≤ n. ⎤ ⎡ ⎤ ⎡ 3 0 0 3 27 0 0 0 ⎦ or ⎣ 0 (−2)3 0 ⎦, Ak = 13. A3 = ⎣ 0 −8 0 0 64 0 0 43 ⎡ k ⎤ 3 0 0 ⎣ 0 (−2)k 0 ⎦. 0 0 4k 15. The entries of InT satisfy i kj = ijk . But ijk = 1 if j = k and is 0 otherwise. Thus i kj = 1 if k = j and is 0 if k = j for 1 ≤ j ≤ n, 1 ≤ k ≤ n. 17. The jth of AB has entries cij = nk=1 aik bkj . Let column dij = ABj , where Bj is the jth column of B. Then D= dij = nm=1 aim bmj = cij .

1 0 19. A2 = = I2 . 0 1 21. (a) (A + B)T = AT + BT by Theorem 3. Since A and B are symmetric, AT + BT = A + B and A + B is also symmetric.

Fundamentals (b) (AB)T = BT AT = BA, but this may not

be AB, so 1 2 AB may not be symmetric. Let A = and 2 4

3 −1 B= . Then AB is not symmetric. −1 2 23. (a) The i, jth element of (AT )T is the j, ith element of AT . But the j, ith element of AT is the i, jth element of A. Thus (AT )T = A. (b) The i, jth element of (A + B)T is the j, ith element of A + B, aji + bji . But this is the sum of the j, ith entry of A and the j, ith entry of B. It is also the sum of the i, jth entry of AT and the i, jth entry of BT . Thus (A + B)T = AT + BT . cij = nk=1 ajk bki , the (c) Let C = cij = (AB)T . Then j, ith entry of AB. Let D = dij = BT AT , then dij =

n k=1

bik akj =

n

bki ajk =

k=1

Hence (AB)T = BT AT . ⎤ ⎡ ⎡ 7 1 − −9 9 ⎦ . (b) ⎣ 27 25. (a) ⎣ 9 0 2 − 59 9 27

n

ajk bki = cij .

k=1

⎤ 4 27 3 27

⎡

⎦. (c) ⎣

2 32 4 32

5 32 − 326

1 0 1 (b) 0 1 (c) 0

37. (a)

if mik = pkj = 1 for some k otherwise. if pik = mkj = 1 for some k otherwise.

if mik = nkl = plj = 1 for some k, l otherwise. 39. Let C = cij = A ∨ B and D = dij = B ∨ A. 1 if aij = 1 or bij = 1 = dij cij = 0 if aij = 0 = bij

Hence C = D. 41. Let dij = B ∨ C, eij = A ∨ (B ∨ C), fij = A ∨ B, and gij = (A ∨ B) ∨ C. Then 1 if bij = 1 or cij = 1 dij = 0 otherwise ⎤ 1 if aij = 1 or dij = 1 eij = ⎦. 0 otherwise.

27. (a) C = A−1 since CA = I3 . (b) D = B−1 since DB = I3 . 29. Since (B−1 A−1 )(AB) = B−1 (A−1 A)B = B−1 In B = B−1 B = In , (B−1 A−1 ) = (AB)−1 . ⎡ ⎤ ⎡ ⎤ 1 1 1 1 0 0 31. (a) A ∨ B = ⎣ 0 1 1 ⎦; A ∧ B = ⎣ 0 0 1 ⎦; 1 0 1 1 0 0 ⎡ ⎤ 1 1 1 A B = ⎣ 0 0 1 ⎦. 1 1 1 ⎡ ⎤ ⎡ ⎤ 0 1 1 0 0 1 (b) A ∨ B = ⎣ 1 1 0 ⎦; A ∧ B = ⎣ 1 1 0 ⎦; 1 0 1 1 0 0 ⎡ ⎤ 1 0 1 A B = ⎣ 1 1 1 ⎦. 0 1 1 ⎡ ⎤ ⎡ ⎤ 1 1 1 1 0 0 (c) A ∨ B = ⎣ 1 1 1 ⎦; A ∧ B = ⎣ 0 0 1 ⎦; 1 0 1 1 0 0 ⎡ ⎤ 1 1 1 A B = ⎣ 1 0 0 ⎦. 1 1 1 33. (a) (d) 35. (a) (b) (c) (d)

(b) f11 . (c) fpq . fp1 . (e) fir . (f) fii if p = q. f1q . pi1 m1j + pi2 m2j + · · · + pit mtj . mi1 p1j + mi2 p2j + · · · + mit ptj . pi1 (m1j + n1j ) + pi2 (m2j + n2j ) + · · · + pit (mtj + ntj ). (mi1 n1j + mi2 n2j + · · · + mit ntj ) + (pi1 n1j + pi2 n2j + · · · + pit ntj ).

But this means dij = 1 if aij = 1 or bij = 1 or cij = 1 and dij = 0 otherwise. 1 if aij = 1 or bij = 1 fij = 0 otherwise 1 if fij = 1 or cij = 1 gij = 0 otherwise. But this means gij = 1 if aij = 1 or bij = 1 or cij = 1 and gij = 0 otherwise. Hence A ∨ (B ∨ C) = (A ∨ B) ∨ C. 43. Let dij = B C, eij = A (B C), fij = A B, and gij = (A B) C. Then 1 if bik = ckj for some k dij = 0 otherwise and eij =

1 if ail = 1 = dlj for some l 0 otherwise.

But this means 1 if ail = 1 = blk = ckj for some k, l eij = 0 otherwise. 1 if aik = 1 = bkj for some k fij = 0 otherwise and gij =

1 if fil = 1 = clj for some l 0 otherwise.

53

Fundamentals But then gij =

1 if aik = 1 = bkl = clj for some k, l 0 otherwise

and A (B C) = (A B) C. 45. An argument similar to that in Exercise 44 shows that A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C). n 47. Since cij = ait btj and k | ait for any i and t, k divides each t=1

term in cij , and thus k | cij for all i and j. ⎡ ⎤ 6 −9 −3 15 6 ⎦. 49. (a) ⎣ 0 12 −12 18 ⎡ ⎤ 10 20 −30 0 45 ⎦. (b) ⎣ 20 35 −5 15 ⎡ ⎤ −4 0 (c) ⎣ −3 −1 ⎦. 2 −5 51. Let A = aij and B = bij be two m × n matri ces. Then k(A + B) = k aij + bij = k(aij + bij ) = kaij + kbij = kaij + kbij = kA + kB. 53. Let K be the m × m diagonal matrix with each diagonal entry equal to k. Then KA = kA. Exercise Set 6 1. (a) Yes. (b) Yes. 3. (a) No. (b) Yes. 5. A ⊕ B = {x | (x ∈ A ∪ B) and (x ∈ / A ∩ B)} = {x | (x ∈ B ∪ A) and (x ∈ / B ∩ A)} = B ⊕ A. 7. x y z y 2 z x (y 2 z) x y x z (x y) 2 (x z) 0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

0 1 1 0 0 1 1 0

0 0 0 0 0 1 1 0

0 0 0 0 0 0 1 1

0 0 0 0 0 1 0 1

(A)

0 0 0 0 0 1 1 0 (B)

Since columns (A) and (B) are identical, the distributive property x (y 2 z) = (x y) 2 (x z) holds. 9. 5 × 5 zero matrix for ∨; 5 × 5 matrix of 1’s for ∧; I5 for . 11. Let A, B be n×n diagonal matrices. Let cij = AB. Then cij = nk=1 aik bkj , but aik = 0 if i = k. Hence cij = aii bij . But bij = 0 if i = j. Thus cij = 0 if i = j and AB is an n × n diagonal matrix. 13. Yes, the n × n zero matrix, which is a diagonal matrix. 15. −A is the diagonal matrix with i, ith entry −aii .

a 0 b 0 a+b 0 17. + or belongs to M. 0 0 0 0 0 0

T a 0 a 0 or belongs to M. 19. 0 0 0 0

1 0 21. Yes, . 0 0 1 0 −1 a . 23. If a = 0, then A = 0 0 25. Yes.

27. No.

29. (a) Yes.

(b) Yes.

0 . (d) Yes, −1

(c) Yes. (e) Yes,

−x . −y − 2

31. (a) Yes. (b) Yes. (c) No.

x w x w 33. (a) k = k k . y z y z (b) No.

35. Let C = cij = comp(A ∨ B) and D = dij = comp(A) ∧ comp(B). Then 0 if aij = 1 or bij = 1 cij = 1 if aij = 0 = bij

and dij =

0 if aij = 1 or bij = 1 1 if aij = 0 = bij .

Hence, C = D. Similarly, we can show that comp(A∧B) = comp(A) ∨ comp(B). 37. A ∩ B.

39. { }.

Answers to Chapter Self-Tests 1. P(A) is a set of sets. 2. |P(A)| is a counting number or infinity. 3. LCM(a, b) is a positive integer. 4. kA is a matrix of the same size as A. 5. A mathematical structure consists of a set of objects, operations on those objects, and the properties of those operations.

54

6. (a) (i) False. (iv) True.

(ii) True. (v) True.

(iii) False.

(b) (i) False. (iv) False.

(ii) True. (v) True.

(iii) False.

7. (a) {1, 2, 3, 4, . . . , }. (b) {. . . , −3, −2, −1, 0} ∪ B. (c) {2, 4}.

(d) A.

Fundamentals (e) {2, 6, 10, 14, . . . }. 8. (a)

(b)

B

A

A

B U

U

9. A ∩ B is always a subset of A ∪ B. A ∪ B ⊆ A ∩ B if and only if A = B. 10. (a) 6. (b) 18. (c) 41. 11. 0, 0, 1, −2, 9, −30. 12. (a) 1 1 1 1 0 1 0 1 0 1 (b)

1

1

0

0

0

0

0

0

0

0

(c)

1

1

0

0

0

1

0

0

0

0

(d) 0 0 1 0 0 0 1 13. (a) Yes. (b) Yes. (c) No. 14. 33 = 65(7293) − 108(4389). 15. (a) AB does not exist.

1

0 1 (d) No.

4 12 (b) BA = −7 −15

2 −3 . (c) 0 1 ⎡ ⎤ 7 −1 3 ⎦. (e) ⎣ 3 2 2

1 3 2 (g) . 2 12 8 ⎡ ⎤ 1 1 0 16. (a) ⎣ 1 1 0 ⎦. 1 1 0 ⎡ ⎤ 1 0 0 (c) ⎣ 0 1 0 ⎦. 0 1 0

8 . −10 (d) A + B does not exist. ⎤ ⎡ 1 0 ⎦. (f) ⎣ 2 3 1 2 ⎡

1 (b) ⎣ 1 1

1 1 1

17. A has a ∧-inverse if and only if A =

⎤ 1 0 ⎦. 1

1 1

1 , the ∧1

identity.

55

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Logic

From Discrete Mathematical Structures, Sixth Edition, Bernard Kolman, Robert C. Busby, Sharon Cutler Ross. Copyright © 2009 by Pearson Education, Inc. Published by Prentice Hall. All rights reserved.

57

Logic Logic is the discipline that deals with the methods of reasoning. On an elementary level, logic provides rules and techniques for determining whether a given argument is valid. Logical reasoning is used in mathematics to prove theorems, in computer science to verify the correctness of programs and to prove theorems, in the natural and physical sciences to draw conclusions from experiments, in the social sciences, and in our everyday lives to solve a multitude of problems. Indeed, we are constantly using logical reasoning. In this chapter we discuss a few of the basic ideas.

Looking Back In the 1840s Augustus De Morgan, a British mathematician, set out to extend the logic developed by the early Greeks and others and to correct some of the weaknesses in these ideas. De Morgan (1806–1871) was born in India but was educated in England. He taught at London University for many years and was the first to use the word “induction” for a method of proof that had been used in a rather informal manner and put it on a firm rigorous foundation. In 1847, a few years after De Morgan’s work on an extended system of logic had appeared, his countryman George Boole published the book entitled The Mathematical Analysis of Logic and followed it up a few years later by the book An Investigation of the Laws of Thought. Boole’s objective in these books was

should be studied in logic. George Boole (1815–1864) taught at Queen’s College in Ireland for many years. Thus, De Morgan started and Boole completed the task of folding a large part of the study of logic into mathematics. We shall briefly study the work of De Morgan and Boole in logic in this chapter.

to investigate the fundamental laws of those operations of the mind by which reasoning is performed; to give expression to them in the symbolical language of a Calculus; and upon this foundation to establish the science of Logic and construct its method.∗ Boole’s work in this area firmly established the point of view that logic should use symbols and that algebraic properties ∗ Quoted

58

Augustus De Morgan CORBIS BETTMANN

George Boole Mary Evans Picture Library Ltd.

in Victor J. Katz, A History of Mathematics, An Introduction, New York: HarperCollins, 1993, p. 619.

Logic

1

Propositions and Logical Operations A statement or proposition is a declarative sentence that is either true or false, but not both.

Example 1

Which of the following are statements? (a) (b) (c) (d) (e) (f) (g)

The earth is round. 2+3=5 Do you speak English? 3−x=5 Take two aspirins. The temperature on the surface of the planet Venus is 800 ◦ F. The sun will come out tomorrow.

Solution

(a) and (b) are statements that happen to be true. (c) is a question, so it is not a statement. (d) is a declarative sentence, but not a statement, since it is true or false depending on the value of x. (e) is not a statement; it is a command. (f) is a declarative sentence whose truth or falsity we do not know at this time; however, we can in principle determine if it is true or false, so it is a statement. (g) is a statement since it is either true or false, but not both, although we would have to wait until tomorrow to find out if it is true or false. ◆

Logical Connectives and Compound Statements

TABLE 1 p

∼p

T F

F T

Example 2

In mathematics, the letters x, y, z, . . . often denote variables that can be replaced by real numbers, and these variables can be combined with the familiar operations +, ×, −, and ÷. In logic, the letters p, q, r, . . . denote propositional variables; that is, variables that can be replaced by statements. Thus we can write p: The sun is shining today. q: It is cold. Statements or propositional variables can be combined by logical connectives to obtain compound statements. For example, we may combine the preceding statements by the connective and to form the compound statement p and q: The sun is shining today and it is cold. The truth value of a compound statement depends only on the truth values of the statements being combined and on the types of connectives being used. We shall look at the most important connectives. If p is a statement, the negation of p is the statement not p, denoted by ∼p. Thus ∼p is the statement “it is not the case that p.” From this definition, it follows that if p is true, then ∼p is false, and if p is false, then ∼p is true. The truth value of ∼p relative to p is given in Table 1. Such a table, giving the truth values of a compound statement in terms of its component parts, is called a truth table. Strictly speaking, not is not a connective, since it does not join two statements, and ∼p is not really a compound statement. However, not is a unary operation for the collection of statements and ∼p is a statement if p is. Give the negation of the following statements: (a) p: 2 + 3 > 1

(b) q: It is cold.

59

Logic Solution

(a) ∼p: 2 + 3 is not greater than 1. That is, ∼p: 2 + 3 ≤ 1. Since p is true in this case, ∼p is false. (b) ∼q: It is not the case that it is cold. More simply, ∼q: It is not cold. ◆

TABLE 2 p

q

p∧q

T T F F

T F T F

T F F F

Example 3

If p and q are statements, the conjunction of p and q is the compound statement “p and q,” denoted by p ∧ q. The connective and is denoted by the symbol ∧. The compound statement p ∧ q is true when both p and q are true; otherwise, it is false. The truth values of p ∧ q in terms of the truth values of p and q are given in the truth table shown in Table 2. Observe that in giving the truth table of p ∧ q we need to look at four possible cases. This follows from the fact that each of p and q can be true or false. Form the conjunction of p and q for each of the following. (a) p: It is snowing. (b) p: 2 < 3 (c) p: It is snowing.

q: I am cold. q: −5 > −8 q: 3 < 5

Solution

(a) p ∧ q: It is snowing and I am cold. (b) p ∧ q: 2 < 3 and −5 > −8 (c) p ∧ q: It is snowing and 3 < 5.

◆

Example 3(c) shows that in logic, unlike in everyday English, we may join two totally unrelated statements by the connective and. If p and q are statements, the disjunction of p and q is the compound statement “p or q,” denoted by p ∨ q. The connective or is denoted by the symbol ∨. The compound statement p ∨ q is true if at least one of p or q is true; it is false when both p and q are false. The truth values of p ∨ q are given in the truth table shown in Table 3. Example 4

TABLE 3

60

p

q

p∨q

T T F F

T F T F

T T T F

Form the disjunction of p and q for each of the following. √ q: 2 is a rational number. (a) p: 2 is a positive integer q: London is the capital of France. (b) p: 2 + 3 = 5 Solution

√ (a) p ∨ q: 2 is a positive integer or 2 is a rational number. Since p is true, the disjunction p ∨ q is true, even though q is false. (b) p ∨ q: 2 + 3 = 5 or London is the capital of France. Since both p and q ◆ are false, p ∨ q is false.

Example 4(b) shows that in logic, unlike in ordinary English, we may join two totally unrelated statements by the connective or. The connective or is more complicated than the connective and because it is used in two different ways in English. Suppose that we say “I left for Spain on Monday or I left for Spain on Friday.” In this compound statement we have the disjunction of the statements p: I left for Spain on Monday and q: I left for Spain on Friday. Of course, exactly one of the two possibilities could have occurred. Both could not have occurred, so the connective or is being used in an exclusive sense. On the other hand, consider the disjunction “I passed mathematics or I failed French.” In this case, at least one of the two possibilities occurred. However, both

Logic could have occurred, so the connective or is being used in an inclusive sense. In mathematics and computer science we agree to use the connective or always in the inclusive manner. In general, a compound statement may have many component parts, each of which is itself a statement, represented by some propositional variable. The statement s: p ∨ (q ∧ (p ∨ r)) involves three propositions, p, q, and r, each of which may independently be true or false. There are altogether 23 or 8 possible combinations of truth values for p, q, and r, and a truth table for s must give the truth or falsity of s in all these cases. If a compound statement s contains n component statements, there will need to be 2n rows in the truth table for s. Such a truth table may be systematically constructed in the following way. Step 1 The first n columns of the table are labeled by the component propositional variables. Further columns are included for all intermediate combinations of the variables, culminating in a column for the full statement. Step 2 Under each of the first n headings, we list the 2n possible n-tuples of truth values for the n component statements. Step 3 For each of the remaining columns, we compute, in sequence, the remaining truth values. Example 5 p

q

p∧q

∨

∼p

T T F F

T F T F

T F F F

T F T T

F F T T

(1)

(3)

(2)

Make a truth table for the statement (p ∧ q) ∨ (∼p). Solution

Because two propositions are involved, the truth table will have 22 or 4 rows. In the first two columns we list all possible pairs of truth values for p and q. The numbers below the remaining columns show the order in which the columns were ◆ filled.

Quantifiers Sets are defined by specifying a property P(x) that elements of the set have in common. Thus, an element of {x | P(x)} is an object t for which the statement P(t) is true. Such a sentence P(x) is called a predicate, because in English the property is grammatically a predicate. P(x) is also called a propositional function, because each choice of x produces a proposition P(x) that is either true or false. Another use of predicates is in programming. Two common constructions are “if P(x), then execute certain steps” and “while Q(x), do specified actions.” The predicates P(x) and Q(x) are called the guards for the block of programming code. Often the guard for a block is a conjunction or disjunction. Example 6

Let A = {x | x is an integer less than 8}. Here P(x) is the sentence “x is an integer less than 8.” The common property is “is an integer less than 8.” P(1) is ◆ the statement “1 is an integer < 8.” Since P(1) is true, 1 ∈ A.

Example 7

(a) Consider the following program fragment, 1. IF N < 10 THEN a. Replace N with N + 1 b. RETURN

Here the statement N < 10 is the guard.

61

Logic (b) Consider the following program fragment, 1. WHILE t ∈ T and s a. PRINT t + s b. RETURN

∈

S

Here the compound statement t ∈ T and s ∈ S is the guard.

◆

The universal quantification of a predicate P(x) is the statement “For all values of x, P(x) is true.” We assume here that only values of x that make sense in P(x) are considered. The universal quantification of P(x) is denoted ∀x P(x). The symbol ∀ is called the universal quantifier. Example 8

(a) The sentence P(x): −(−x) = x is a predicate that makes sense for real numbers x. The universal quantification of P(x), ∀x P(x), is a true statement, because for all real numbers, −(−x) = x. (b) Let Q(x): x + 1 < 4. Then ∀x Q(x) is a false statement, because Q(5) is not ◆ true. Universal quantification can also be stated in English as “for every x,” “every x,” or “for any x.” A predicate may contain several variables. Universal quantification may be applied to each of the variables. For example, a commutative property can be expressed as ∀x ∀y x 2 y = y 2 x. The order in which the universal quantifiers are considered does not change the truth value. Often mathematical statements contain implied universal quantifications. In some situations we only require that there be at least one value for which the predicate is true. The existential quantification of a predicate P(x) is the statement “There exists a value of x for which P(x) is true.” The existential quantification of P(x) is denoted ∃x P(x). The symbol ∃ is called the existential quantifier.

Example 9

(a) Let Q(x): x + 1 < 4. The existential quantification of Q(x), ∃x Q(x), is a true statement, because Q(2) is a true statement. (b) The statement ∃y y + 2 = y is false. There is no value of y for which the ◆ propositional function y + 2 = y produces a true statement. In English ∃x can also be read “there is an x,” “there is some x,” “there exists an x,” or “there is at least one x.” Occasionally a variation of a universal or existential quantifier is useful. The possible values of the variable may be restricted as in the following examples. The statement ∀x ∈ A P(x) represents “for all x in the set A, P(x) is true,” and the statement ∃k ≥ n Q(k) represents “there is a k greater than or equal to n such that Q(k) is true.”

Example 10

(a) Let D be the set of n × n diagonal matrices. Consider the statement ∃M ∈ D such that M −1 does not exist. To determine if this statement is true or false, we must decide if there is an n × n diagonal matrix that does not have an inverse. (b) For the statement ∀n ≥ 6, 2n > 2n, only integers greater than or equal to 6 ◆ need to be tested. Existential quantification may be applied to several variables in a predicate and the order in which the quantifications are considered does not affect the truth value. For a predicate with several variables we may apply both universal and existential quantification. In this case the order does matter.

62

Logic Example 11

Let A and B be n × n matrices. (a) The statement ∀A ∃B A + B = In is read“for every A there is a B such that A + B = In .” For a given A = aij , define B = bij as follows: bii = 1 − aii , 1 ≤ i ≤ n and bij = −aij , i = j, 1 ≤ i ≤ n, 1 ≤ j ≤ n. Then A + B = In and we have shown that ∀A ∃B A + B = In is a true statement. (b) ∃B ∀A A+B = In is the statement “there is a B such that for all A, A+B = In .” This statement is false; no single B has this property for all A’s. (c) ∃B ∀A A + B = A is true. What is the value for B that makes the statement ◆ true? Let p: ∀x P(x). The negation of p is false when p is true, and true when p is false. For p to be false there must be at least one value of x for which P(x) is false. Thus, p is false if ∃x ∼P(x) is true. On the other hand, if ∃x ∼P(x) is false, then for every x, ∼P(x) is false; that is, ∀x P(x) is true.

Example 12

(a) Let p: For all positive integers n, n2 + 41n + 41 is a prime number. Then ∼p is There is at least one positive integer n for which n2 + 41n + 41 is not prime. (b) Let q: There is some integer k for which 12 = 3k. Then ∼q: For all integers ◆ k, 12 = 3k.

Example 13

Let p: The empty set is a subset of any set A. For p to be false, there must be an ◆ element of ∅ that is not in A, but this is impossible. Thus, p is true.

1 Exercises 1. Which of the following are statements? (a) Is 2 a positive number? (b) x2 + x + 1 = 0 (c) Study logic. (d) There will be snow in January. (e) If stock prices fall, then I will lose money. 2. Give the negation of each of the following statements. (a) 2 + 7 ≤ 11 (b) 2 is an even integer and 8 is an odd integer. 3. Give the negation of each of the following statements. (a) It will rain tomorrow or it will snow tomorrow. (b) If you drive, then I will walk. 4. In each of the following, form the conjunction and the disjunction of p and q. q: 7 = 3 × 6 (a) p: 3 + 1 < 5 q: I am happy. (b) p: I am rich. 5. In each of the following, form the conjunction and the disjunction of p and q. q: I will be late. (a) p: I will drive my car. q: NUM ≤ 15 (b) p: NUM > 10 6. Determine the truth or falsity of each of the following statements. (a) 2 < 3 and 3 is a positive integer.

(b) 2 ≥ 3 and 3 is a positive integer. (c) 2 < 3 and 3 is not a positive integer. (d) 2 ≥ 3 and 3 is not a positive integer. 7. Determine the truth or falsity of each of the following statements. (a) 2 < 3 or 3 is a positive integer. (b) 2 ≥ 3 or 3 is a positive integer. (c) 2 < 3 or 3 is not a positive integer. (d) 2 ≥ 3 or 3 is not a positive integer. In Exercises 8 and 9, find the truth value of each proposition if p and r are true and q is false. 8. (a) ∼p ∧ ∼q

(b) (∼p ∨ q) ∧ r

(c) p ∨ q ∨ r

(d) ∼(p ∨ q) ∧ r

9. (a) ∼p ∧ (q ∨ r) (c) (r ∧ ∼q) ∨ (p ∨ r)

(b) p ∧ (∼(q ∨ ∼r)) (d) (q ∧ r) ∧ (p ∨ ∼r)

10. Which of the following statements is the negation of the statement “2 is even and −3 is negative”? (a) 2 is even and −3 is not negative. (b) 2 is odd and −3 is not negative. (c) 2 is even or −3 is not negative. (d) 2 is odd or −3 is not negative.

63

Logic 11. Which of the following statements is the negation of the statement “2 is even or −3 is negative”? (a) 2 is even or −3 is not negative. (b) 2 is odd or −3 is not negative. (c) 2 is even and −3 is not negative. (d) 2 is odd and −3 is not negative.

21. If Q(n): n + 3 = 6, then (a) Q(5) is the statement

(b) P(5) is the statement

.

In Exercises 12 and 13, use p: Today is Monday; q: The grass is wet; and r: The dish ran away with the spoon.

(c) P(k) is the statement 23. If Q(m): m ≤ 3m , then (a) Q(0) is the statement

.

(b) Q(2) is the statement

.

12. Write each of the following in terms of p, q, r, and logical connectives. (a) Today is Monday and the dish did not run away with the spoon. (b) Either the grass is wet or today is Monday. (c) Today is not Monday and the grass is dry. (d) The dish ran away with the spoon, but the grass is wet. 13. Write an English sentence that corresponds to each of the following. (b) ∼q ∨ r (a) ∼r ∧ q (d) p ∨ ∼r (c) ∼(p ∨ q) In Exercises 14 through 19, use P(x): x is even; Q(x): x is a prime number; R(x, y): x + y is even. The variables x and y represent integers. 14. Write an English sentence corresponding to each of the following. (b) ∃x Q(x) (a) ∀x P(x) 15. Write an English sentence corresponding to each of the following. (b) ∃x ∀y R(x, y) (a) ∀x ∃y R(x, y) 16. Write an English sentence corresponding to each of the following. (b) ∃y (∼P(y)) (a) ∀x (∼Q(x)) 17. Write an English sentence corresponding to each of the following. (b) ∼(∀x Q(x)) (a) ∼(∃x P(x)) 18. Write each of the following in terms of P(x), Q(x), R(x, y), logical connectives, and quantifiers. (a) Every integer is an odd integer. (b) The sum of any two integers is an even number. (c) There are no even prime numbers. (d) Every integer is even or a prime. 19. Determine the truth value of each statement given in Exercises 14 through 18. 20. If P(x): x2 < 12, then . (a) P(4) is the statement . (b) P(1.5) is the statement

64

.

. (b) Q(m) is the statement 22. If P(y): 1 + 2 + · · · + y = 0, then . (a) P(1) is the statement

.

. (c) Q(k) is the statement 24. Give a symbolic statement of the commutative property for addition of real numbers using appropriate quantifiers. 25. Give a symbolic statement of De Morgan’s laws for sets using appropriate quantifiers. 26. Give a symbolic statement of the multiplicative inverse property for real numbers using appropriate quantifiers. In Exercises 27 through 30, make a truth table for the statement. 27. (∼p ∧ q) ∨ p

28. (p ∨ q) ∨ ∼q

29. (p ∨ q) ∧ r

30. (∼p ∨ q) ∧ ∼r

For Exercises 31 through 33, define p ↓ q to be a true statement if neither p nor q is true. p

q

p↓q

T T F F

T F T F

F F F T

31. Make a truth table for (p ↓ q) ↓ r. 32. Make a truth table for (p ↓ q) ∧ (p ↓ r). 33. Make a truth table for (p ↓ q) ↓ (p ↓ r). For Exercises 34 through 36, define p q to be true if either p or q, but not both, is true. Make a truth table for the statement. 34. (a) p q

(b) p ∼p

35. (p ∧ q) p 36. (p q) (q r) In Exercises 37 through 40, revision of the given programming block is needed. Replace the guard P(x) with ∼P(x). 37. IF (x = max and y > 4) THEN take action 38. WHILE (key = “open” or t < limit) take action 39. WHILE (item = sought and index < 101) take action 40. IF (cell > 0 or found) THEN take action

Logic

2

Conditional Statements If p and q are statements, the compound statement “if p then q,” denoted p ⇒ q, is called a conditional statement, or implication. The statement p is called the antecedent or hypothesis, and the statement q is called the consequent or conclusion. The connective if . . . then is denoted by the symbol ⇒.

Example 1

Form the implication p ⇒ q for each of the following. (a) p: I am hungry. (b) p: It is snowing.

q: I will eat. q: 3 + 5 = 8.

Solution

(a) If I am hungry, then I will eat. (b) If it is snowing, then 3 + 5 = 8.

TABLE 4 p

q

p⇒q

T T F F

T F T F

T F T T

Example 2

◆

Example 1(b) shows that in logic we use conditional statements in a more general sense than is customary. Thus in English, when we say “if p then q,” we are tacitly assuming there is a cause-and-effect relationship between p and q. That is, we would never use the statement in Example 1(b) in ordinary English, since there is no way statement p can have any effect on statement q. In logic, implication is used in a much weaker sense. To say the compound statement p ⇒ q is true simply asserts that if p is true, then q will also be found to be true. In other words, p ⇒ q says only that we will not have p true and q false at the same time. It does not say that p “caused” q in the usual sense. Table 4 describes the truth values of p ⇒ q in terms of the truth of p and q. Notice that p ⇒ q is considered false only if p is true and q is false. In particular, if p is false, then p ⇒ q is true for any q. This fact is sometimes described by the statement “A false hypothesis implies any conclusion.” This statement is misleading, since it seems to say that if the hypothesis is false, the conclusion must be true, an obviously silly statement. Similarly, if q is true, then p ⇒ q will be true for any statement p. The implication “If 2 + 2 = 5, then I am the king of England” is true, simply because p: 2 + 2 = 5 is false, so it is not the case that p is true and q is false simultaneously. In the English language, and in mathematics, each of the following expressions is an equivalent form of the conditional statement p ⇒ q: p implies q; q, if p; p only if q; p is a sufficient condition for q; q is a necessary condition for p. If p ⇒ q is an implication, then the converse of p ⇒ q is the implication q ⇒ p, and the contrapositive of p ⇒ q is the implication ∼q ⇒ ∼p. Give the converse and the contrapositive of the implication “If it is raining, then I get wet.” Solution

TABLE 5 p

q

p⇔q

T T F F

T F T F

T F F T

We have p: It is raining; and q: I get wet. The converse is q ⇒ p: If I get wet, then it is raining. The contrapositive is ∼q ⇒ ∼p: If I do not get wet, then it is ◆ not raining. If p and q are statements, the compound statement p if and only if q, denoted by p ⇔ q, is called an equivalence or biconditional. The connective if and only if is denoted by the symbol ⇔. The truth values of p ⇔ q are given in Table 5. Observe that p ⇔ q is true only when both p and q are true or when both p and q are false. The equivalence p ⇔ q can also be stated as p is a necessary and sufficient condition for q.

65

Logic Example 3

Is the following equivalence a true statement? 3 > 2 if and only if 0 < 3 − 2. Solution

Let p be the statement 3 > 2 and let q be the statement 0 < 3 − 2. Since both p and q are true, we conclude that p ⇔ q is true. ◆ Example 4

Compute the truth table of the statement (p ⇒ q) ⇔ (∼q ⇒ ∼p). Solution

The following table is constructed using steps 1, 2, and 3 as given in Section 1. The numbers below the columns show the order in which they were constructed. p

q

p⇒q

∼q

∼p

∼q ⇒ ∼p

T T F F

T F T F

T F T T (1)

F T F T (2)

F F T T (3)

T F T T (4)

(p ⇒ q) ⇔ (∼q ⇒ ∼p) T T T T (5)

◆

A statement that is true for all possible values of its propositional variables is called a tautology. A statement that is always false is called a contradiction or an absurdity, and a statement that can be either true or false, depending on the truth values of its propositional variables, is called a contingency. Example 5

(a) The statement in Example 4 is a tautology. (b) The statement p ∧ ∼p is an absurdity. (Verify this.) (c) The statement (p ⇒ q) ∧ (p ∨ q) is a contingency.

◆

We have now defined a new mathematical structure with two binary operations and one unary operation, (propositions, ∧, ∨, ∼). It makes no sense to say two propositions are equal; instead we say p and q are logically equivalent, or simply equivalent, if p ⇔ q is a tautology. When an equivalence is shown to be a tautology, this means its two component parts are always either both true or both false, for any values of the propositional variables. Thus the two sides are simply different ways of making the same statement and can be regarded as “equal.” We denote that p is equivalent to q by p ≡ q. Now we can adapt our properties for operations to say this structure has a property if using equivalent in place of equal gives a true statement. Example 6

The binary operation ∨ has the commutative property; that is, p ∨ q ≡ q ∨ p. The truth table for (p ∨ q) ⇔ (q ∨ p) shows the statement is a tautology. p

q

p∨q

q∨p

(p ∨ q) ⇔ (q ∨ p)

T T F F

T F T F

T T T F

T T T F

T T T T

◆

Another way to use a truth table to determine if two statements are equivalent is to construct a column for each statement and compare these to see if they are identical. In Example 6 the third and fourth columns are identical, and this will guarantee that the statements they represent are equivalent.

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Logic Forming p ⇒ q from p and q is another binary operation for statements, but we can express it in terms of the operations in Section 1. Example 7

The conditional statement p ⇒ q is equivalent to (∼p) ∨ q. Columns 1 and 3 in the following table show that for any truth values of p and q, p ⇒ q and (∼p) ∨ q have the same truth values. p

q

p⇒q

∼p

(∼p) ∨ q

T T F F

T F T F

T F T T (1)

F F T T (2)

T F T T (3)

◆

The structure (propositions, ∧, ∨, ∼) has many of the same properties as the structure (sets, ∪, ∩, ). THEOREM 1

The operations for propositions have the following properties. Commutative Properties 1. p ∨ q ≡ q ∨ p 2. p ∧ q ≡ q ∧ p Associative Properties 3. p ∨ (q ∨ r) ≡ (p ∨ q) ∨ r 4. p ∧ (q ∧ r) ≡ (p ∧ q) ∧ r Distributive Properties 5. p ∨ (q ∧ r) ≡ (p ∨ q) ∧ (p ∨ r) 6. p ∧ (q ∨ r) ≡ (p ∧ q) ∨ (p ∧ r) Idempotent Properties 7. p ∨ p ≡ p 8. p ∧ p ≡ p Properties of Negation 9. ∼(∼p) ≡ p 10. ∼(p ∨ q) ≡ (∼p) ∧ (∼q) 11. ∼(p ∧ q) ≡ (∼p) ∨ (∼q)

Properties 10 and 11 are De Morgan’s laws.

Proof We have proved Property 1 in Example 6. The remaining properties may be proved ■ the same way and are left for the reader as exercises. Truth tables can be used to prove statements about propositions, because in a truth table all possible cases are examined. The implication operation also has a number of important properties. THEOREM 2

(a) (b) (c) (d) (e)

(p ⇒ q) ≡ ((∼p) ∨ q) (p ⇒ q) ≡ (∼q ⇒ ∼p) (p ⇔ q) ≡ ((p ⇒ q) ∧ (q ⇒ p)) ∼(p ⇒ q) ≡ (p ∧ ∼q) ∼(p ⇔ q) ≡ ((p ∧ ∼q) ∨ (q ∧ ∼p))

67

Logic Proof (a) was proved in Example 7 and (b) was proved in Example 4. Notice that (b) says a conditional statement is equivalent to its contrapositive. (d) gives an alternate version for the negation of a conditional statement. This could be proved using truth tables, but it can also be proved by using previously proven facts. Since (p ⇒ q) ≡ ((∼p) ∨ q), the negation of p ⇒ q must be equivalent to ∼((∼p) ∨ q). By De Morgan’s laws, ∼((∼p) ∨ q) ≡ (∼(∼p)) ∧ (∼q) or p ∧ (∼q). Thus, ∼(p ⇒ q) ≡ (p ∧ ∼q). ■ The remaining parts of Theorem 2 are left as exercises. Theorem 3 states two results from Section 1, and several other properties for the universal and existential quantifiers. THEOREM 3

(a) (b) (c) (d) (e) (f) (g)

∼(∀x P(x)) ≡ ∃x ∼P(x) ∼(∃x P(x)) ≡ ∀x (∼P(x)) ∃x (P(x) ⇒ Q(x)) ≡ ∀x P(x) ⇒ ∃x Q(x) ∃x (P(x) ∨ Q(x)) ≡ ∃x P(x) ∨ ∃x Q(x) ∀x (P(x) ∧ Q(x)) ≡ ∀x P(x) ∧ ∀x Q(x) ((∀x P(x)) ∨ (∀x Q(x))) ⇒ ∀x (P(x) ∨ Q(x)) is a tautology. ∃x (P(x) ∧ Q(x)) ⇒ ∃x P(x) ∧ ∃x Q(x) is a tautology.

■

The following theorem gives several important tautologies that are implications. These are used extensively in proving results in mathematics and computer science and we will illustrate them in Section 3. THEOREM 4

Each of the following is a tautology. (a) (p ∧ q) ⇒ p (c) p ⇒ (p ∨ q) (e) ∼p ⇒ (p ⇒ q) (g) (p ∧ (p ⇒ q)) ⇒ q (i) (∼q ∧ (p ⇒ q)) ⇒ ∼p

(b) (d) (f) (h) (j)

(p ∧ q) ⇒ q q ⇒ (p ∨ q) ∼(p ⇒ q) ⇒ p (∼p ∧ (p ∨ q)) ⇒ q ((p ⇒ q) ∧ (q ⇒ r)) ⇒ (p ⇒ r)

■

2 Exercises In Exercises 1 and 2, use the following: p: I am awake; q: I work hard; r: I dream of home. 1. Write each of the following statements in terms of p, q, r, and logical connectives. (a) I am awake implies that I work hard. (b) I dream of home only if I am awake. (c) Working hard is sufficient for me to be awake. (d) Being awake is necessary for me not to dream of home. 2. Write each of the following statements in terms of p, q, r, and logical connectives. (a) I am not awake if and only if I dream of home. (b) If I dream of home, then I am awake and I work hard.

68

(c) I do not work hard only if I am awake and I do not dream of home. (d) Not being awake and dreaming of home is sufficient for me to work hard. 3. State the converse of each of the following implications. (a) If 2 + 2 = 4, then I am not the Queen of England. (b) If I am not President of the United Sates, then I will walk to work. (c) If I am late, then I did not take the train to work. (d) If I have time and I am not too tired, then I will go to the store. (e) If I have enough money, then I will buy a car and I will buy a house. 4. State the contrapositive of each implication in Exercise 3.

Logic 5. Determine the truth value for each of the following statements. (a) If 2 is even, then New York has a large population.

17. Fill the grid so that each row, column, and marked 2 × 2 square contains the letters M, A, T, H, with no repeats. (b) T (a) A

(b) If 2 is even, then New York has a small population.

M

(c) If 2 is odd, then New York has a large population.

A

T

(d) If 2 is odd, then New York has a small population.

M H

In Exercises 6 and 7, let p, q, and r be the following statements: p: I will study discrete structure; q: I will go to a movie; r: I am in a good mood. 6. Write the following statements in terms of p, q, r, and logical connectives. (a) If I am not in a good mood, then I will go to a movie.

H

18. Fill the grid so that each row, column, and marked 2 × 3 block contains 1, 2, 3, 4, 5, 6, with no repeats. (a) 4

(b) I will not go to a movie and I will study discrete structures.

3

2

3

(c) I will go to a movie only if I will not study discrete structures. 6

(d) If I will not study discrete structures, then I am not in a good mood. 7. Write English sentences corresponding to the following statements. (b) r ⇒ (p ∨ q) (a) ((∼p) ∧ q) ⇒ r (c) (∼r) ⇒ ((∼q ∨ p)

(d) (q ∧ (∼p)) ⇔ r

In Exercises 8 and 9, let p, q, r, and s be the following statements: p: 4 > 1; q: 4 < 5; r: 3 ≤ 3; s: 2 > 2. 8. Write the following statements in terms of p, q, r, and logical connectives. (a) Either 4 > 1 or 4 < 5. (b) If 3 ≤ 3, then 2 > 2.

1

In Exercises 10 through 12, construct truth tables to determine whether the given statement is a tautology, a contingency, or an absurdity. 10. (a) p ∧ ∼p

(b) q ∨ (∼q ∧ p)

11. (a) p ⇒ (q ⇒ p)

(b) q ⇒ (q ⇒ p) (c) p ⇒ (q ∧ p)

13. If p ⇒ q is false, can you determine the truth value of (∼(p ∧ q)) ⇒ q? Explain your answer. 14. If p ⇒ q is false, can you determine the truth value of (∼p) ∨ (p ⇔ q)? Explain your answer.

(b)

1

6

5

5

1

6

2

2

3

1

5

1

6 3 2

5 6

4

19. Fill the grid so that each row, column, and marked 3 × 3 block contains 1, 2, 3, 4, 5, 6, 7, 8, 9, with no repeats.

6

9 12

1

6

3 8

1

7

3

4

6

5

2

8

1

4

6 8 9

3 5

5 7

9

4

1 4

3 7

9

2 1

6

15. If p ⇒ q is true, can you determine the truth value of (p ∧ q) ⇒ ∼q? Explain your answer. 16. If p ⇒ q is true, can you determine the truth value of ∼(p ⇒ q) ∧ ∼p? Explain your answer.

6

4

12. (a) (q ∧ p) ∨ (q ∧ ∼p) (b) (p ∧ q) ⇒ p

1 5

4

(c) It is not the case that 2 > 2 or 4 > 1. 9. Write English sentences corresponding to the following statements. (b) ∼(r ∧ q) (c) (∼r) ⇒ p (a) (p ∧ s) ⇒ q

M

3

2

5 2

1

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Logic 20. Fill the grid so that each row, column, and marked 3 × 3 block contains 1, 2, 3, 4, 5, 6, 7, 8, 9, with no repeats. 4

7

9 6

1 1

4

7

4

8

1

7

9

8

9

2

4 8

6

4

5

7 2

6

In Exercises 21 and 22, find the truth value of each statement if p and q are true and r, s, and t are false. 21. (a) ∼(p ⇒ q) (b) (∼p) ⇒ r (c) (p ⇒ s) ∧ (s ⇒ t) (d) t ⇒ ∼q 22. (a) (∼q) ⇒ (r ⇒ (r ⇒ (p ∨ s))) (b) p ⇒ (r ⇒ q) (c) (q ⇒ (r ⇒ s)) ∧ ((p ⇒ s) ⇒ (∼t)) (d) (r ∧ s ∧ t) ⇒ (p ∨ q) 23. Use the definition of p ↓ q given for Exercise 31 in Section 1 and show that ((p ↓ p) ↓ (q ↓ q)) is equivalent to p ∧ q. 24. Write the negation of each of the following in good English. (a) The weather is bad and I will not go to work. (b) If Carol is not sick, then if she goes to the picnic, she will have a good time. (c) I will not win the game or I will not enter the contest. 25. Write the negation of each of the following in good English.

3

(c) If Tom stole a pie and ran away, then the three pigs do not have any supper. 26. Consider the following conditional statement:

(a) Which of the following is the converse of p? (b) Which of the following is the contrapositive of p?

2 1

(b) Mary lost her lamb or the wolf ate the lamb.

p: If the flood destroys my house or the fire destroys my house, then my insurance company will pay me.

9

3 5

2 9

5

5

(a) Jack did not eat fat, but he did eat broccoli.

(i) If my insurance company pays me, then the flood destroys my house or the fire destroys my house. (ii) If my insurance company pays me, then the flood destroys my house and the fire destroys my house. (iii) If my insurance company does not pay me, then the flood does not destroy my house or the fire does not destroy my house. (iv) If my insurance company does not pay me, then the flood does not destroy my house and the fire does not destroy my house. 27. Prove Theorem 1 part 6. 28. Prove Theorem 1 part 11. 29. Prove Theorem 2 part (e). 30. Prove Theorem 3 part (d). 31. Prove Theorem 3 part (e). 32. Prove Theorem 4 part (a). 33. Prove Theorem 4 part (d). 34. Prove Theorem 4 part (g). 35. Prove Theorem 4 part ( j). 36. Explain why proving part (e) of Theorem 4 provides a one-line proof of part (f) of Theorem 4. 37. Explain why proving part (a) of Theorem 4 provides a one-line proof of part (b) of Theorem 4.

Methods of Proof Some methods of proof we have already used are direct proofs using generic elements, definitions, and previously proven facts, and proofs by cases, such as examining all possible truth value situations in a truth table. Here we look at proofs in more detail. If an implication p ⇒ q is a tautology, where p and q may be compound statements involving any number of propositional variables, we say that q logically follows from p. Suppose that an implication of the form (p1 ∧ p2 ∧ · · · ∧ pn ) ⇒ q is a tautology. Then this implication is true regardless of the truth values of any of its components. In this case, we say that q logically follows from p1 , p2 , . . . , pn .

70

Logic When q logically follows from p1 , p2 , . . . , pn , we write p1 p2 .. . ∴

pn q

where the symbol ∴ means therefore. This means if we know that p1 is true, p2 is true, . . . , and pn is true, then we know q is true. Virtually all mathematical theorems are composed of implications of the type (p1 ∧ p2 ∧ · · · ∧ pn ) ⇒ q. The pi ’s are called the hypotheses or premises, and q is called the conclusion. To “prove the theorem” means to show that the implication is a tautology. Note that we are not trying to show that q (the conclusion) is true, but only that q will be true if all the pi are true. For this reason, mathematical proofs often begin with the statement “suppose that p1 , p2 , . . . , and pn are true” and conclude with the statement “therefore, q is true.” The proof does not show that q is true, but simply shows if the pi are all true, then q has to be true. Arguments based on tautologies represent universally correct methods of reasoning. Their validity depends only on the form of the statements involved and not on the truth values of the variables they contain. Such arguments are called rules of inference. The various steps in a mathematical proof of a theorem must follow from the use of various rules of inference, and a mathematical proof of a theorem must begin with the hypotheses, proceed through various steps, each justified by some rule of inference, and arrive at the conclusion. Example 1

According to Theorem 4(j) of the last section, ((p ⇒ q) ∧ (q ⇒ r)) ⇒ (p ⇒ r) is a tautology. Thus the argument p⇒q q⇒r ∴ p⇒r ◆

is universally valid, and so is a rule of inference. Example 2

Is the following argument valid? If you invest in the stock market, then you will get rich. If you get rich, then you will be happy. ∴ If you invest in the stock market, then you will be happy. Solution

The argument is of the form given in Example 1, hence the argument is valid, although the conclusion may be false. ◆ Example 3

The tautology (p ⇔ q) ⇔ ((p ⇒ q) ∧ (q ⇒ p)) is Theorem 2(c), Section 2. Thus both of the following arguments are valid. p⇔q ∴ (p ⇒ q) ∧ (q ⇒ p)

p⇒q q⇒p ∴ p⇔q

◆

71

Logic Some mathematical theorems are equivalences; that is, they are of the form p ⇔ q. They are usually stated p if and only if q. By Example 3, the proof of such a theorem is logically equivalent with proving both p ⇒ q and q ⇒ p, and this is almost always the way in which equivalences are proved. We wish to show that both p ⇒ q and q ⇒ p are true. To do so, we first assume that p is true, and show that q must then be true; next we assume that q is true and show that p must then be true. A very important rule of inference is p p⇒q ∴ q. That is, p is true, and p ⇒ q is true, so q is true. This follows from Theorem 4(g), Section 2. Some rules of inference were given Latin names by classical scholars. Theorem 4(g) is referred to as modus ponens, or loosely, the method of asserting. Example 4

Is the following argument valid? Smoking is healthy. If smoking is healthy, then cigarettes are prescribed by physicians. ∴ Cigarettes are prescribed by physicians. Solution The argument is valid since it is of the form modus ponens. However, the conclusion is false. Observe that the first premise p: smoking is healthy is false. The second premise p ⇒ q is then true and (p∧(p ⇒ q)), the conjunction of the two premises, ◆ is false.

Example 5

Is the following argument valid? If taxes are lowered, then income rises. Income rises. ∴ Taxes are lowered. Solution

Let p: taxes are lowered and q: income rises. Then the argument is of the form p⇒q q ∴ p. Assume that p ⇒ q and q are both true. Now p ⇒ q may be true with p being false. Then the conclusion p is false. Hence the argument is not valid. Another approach to answering this question is to verify whether the statement ((p ⇒ q)∧q) logically implies the statement p. A truth table shows this is not the case. (Verify this.) ◆ An important proof technique, which is an example of an indirect method of proof, follows from the tautology (p ⇒ q) ⇔ ((∼q) ⇒ (∼p)). This states that an implication is equivalent to its contrapositive. Thus to prove p ⇒ q indirectly, we assume q is false (the statement ∼q) and show that p is then false (the statement ∼p). Example 6

72

Let n be an integer. Prove that if n2 is odd, then n is odd.

Logic Solution

Let p: n2 is odd and q: n is odd. We have to prove that p ⇒ q is true. Instead, we prove the contrapositive ∼q ⇒ ∼p. Thus suppose that n is not odd, so that n is even. Then n = 2k, where k is an integer. We have n2 = (2k)2 = 4k2 = 2(2k2 ), so n2 is even. We thus show that if n is even, then n2 is even, which is the contrapositive of the given statement. Hence the given statement has been proved. ◆ Another important indirect proof technique is proof by contradiction. This method is based on the tautology ((p ⇒ q) ∧ (∼q)) ⇒ (∼p). Thus the rule of inference p⇒q ∼q ∴ ∼p

Example 7

is valid. Informally, this states that if a statement p implies a false statement q, then p must be false. This is often applied to the case where q is an absurdity or contradiction, that is, a statement that is always false. An example is given by taking q as the contradiction r ∧ (∼r). Thus any statement that implies a contradiction must be false. In order to use proof by contradiction, suppose we wish to show that a statement q logically follows from statements p1 , p2 , . . . , pn . Assume that ∼q is true (that is, q is false) as an extra hypothesis, and that p1 , p2 , . . . , pn are also true. If this enlarged hypothesis p1 ∧ p2 ∧ · · · ∧ pn ∧ (∼q) implies a contradiction, then at least one of the statements p1 , p2 , . . . , pn , ∼q must be false. This means that if all the pi ’s are true, then ∼q must be false, so q must be true. Thus q follows from p1 , p2 , . . . , pn . This is proof by contradiction. √ Prove there is no rational number p/q whose square is 2. In other words, show 2 is irrational. Solution

This statement is a good candidate for proof by contradiction, because we could not check all possible rational numbers to demonstrate that none had a square equal to 2. Assume (p/q)2 = 2 for some integers p and q, which have no common factors. If the original choice of p/q is not in lowest terms, we can replace it with its equivalent lowest-term form. Then p2 = 2q2 , so p2 is even. This implies p is even, since the square of an odd number is odd. Thus, p = 2n for some integer n. We see that 2q2 = p2 = (2n)2 = 4n2 , so q2 = 2n2 . Thus q2 is even, and so q is even. We now have that both p and q are even, and therefore have a common factor 2. This is a contradiction to the assumption. Thus the assumption must be ◆ false. We have presented several rules of inference and logical equivalences that correspond to valid proof techniques. In order to prove a theorem of the (typical) form (p1 ∧ p2 ∧ · · · ∧ pn ) ⇒ q, we begin with the hypothesis p1 , p2 , . . . , pn and show that some result r1 logically follows. Then, using p1 , p2 , . . . , pn , r1 , we show that some other statement r2 logically follows. We continue this process, producing intermediate statements r1 , r2 , . . . , rk , called steps in the proof, until we can finally show that the conclusion q logically follows from p1 , p2 , . . . , pn , r1 , r2 , . . . , rk . Each logical step must be justified by some valid proof technique, based on the rules of inference we have developed, or on some other rules that come from tautological implications we have not discussed. At any stage, we can replace a statement that needs to be derived by its contrapositive statement, or any other equivalent form.

73

Logic In practice, the construction of proofs is an art and must be learned in part from observation and experience. The choice of intermediate steps and methods of deriving them is a creative activity that cannot be precisely described. But a few simple techniques are applicable to a wide variety of settings. We will focus on these techniques. The “Tips for Proofs” notes at the end of this chapter highlight the methods most useful for the chapter’s material. Example 8

Let m and n be integers. Prove that n2 = m2 if and only if n is m or n is −m. Solution

Let us analyze the proof as we present it. Suppose p is the statement n2 = m2 , q is the statement n is m, and r is the statement n is −m. Then we wish to prove the theorem p ⇔ (q ∨ r). We know from previous discussion that we may instead prove s: p ⇒ (q ∨ r) and t: (q ∨ r) ⇒ p are true. First, we assume that either q: n is m or r: n is −m is true. If q is true, then n2 = m2 , and if r is true, then n2 = (−m)2 = m2 , so in either case p is true. We have therefore shown that the implication t: (q ∨ r) ⇒ p is true. Now we must prove that s: p ⇒ (q ∨ r) is true; that is, we assume p and try to prove either q or r. If p is true, then n2 = m2 , so n2 − m2 = 0. But n2 − m2 = (n − m)(n + m). If r1 is the intermediate statement (n − m)(n + m) = 0, we have shown p ⇒ r1 is true. We now show that r1 ⇒ (q ∨ r) is true, by showing that the contrapositive ∼(q ∨ r) ⇒ (∼r1 ) is true. Now ∼(q ∨ r) is equivalent to (∼q) ∧ (∼r), so we show that (∼q) ∧ (∼r) ⇒ (∼r1 ). Thus, if (∼q): n is not m and (∼r): n is not −m are true, then (n − m) = 0 and (n + m) = 0, so (n − m)(n + m) = 0 and r1 is false. We have therefore shown that r1 ⇒ (q ∨ r) is true. Finally, from the truth of p ⇒ r1 and r1 ⇒ (q ∨ r), we can conclude that ◆ p ⇒ (q ∨ r) is true, and we are done. We do not usually analyze proofs in this detailed manner. We have done so only to illustrate that proofs are devised by piecing together equivalences and valid steps resulting from rules of inference. The amount of detail given in a proof depends on who the reader is likely to be. As a final remark, we remind the reader that many mathematical theorems actually mean that the statement is true for all objects of a certain type. Sometimes this is not evident. Thus the theorem in Example 8 really states that for all integers m and n, n2 = m2 if and only if n is m or n is −m. Similarly, the statement “If x and y are real numbers, and x = y, then x < y or y < x” is a statement about all real numbers x and y. To prove such a theorem, we must make sure that the steps in the proof are valid for every √ real number. We could not assume, for example, that x is 2, or that y is π or 3. This is why proofs often begin by selecting a generic element, denoted by a variable. On the other hand, we know from Section 2 that the negation of a statement of the form ∀x P(x) is ∃x ∼P(x), so we need only find a single example where the statement is false to disprove it. Example 9

Prove or disprove the statement that if x and y are real numbers, (x2 = y2 ) ⇔ (x = y). Solution The statement can be restated in the form ∀x ∀y R(x, y). Thus, to prove this result, we would need to provide steps, each of which would be true for all x and y. To disprove the result, we need only find one example for which the implication is false. Since (−3)2 = 32 , but −3 = 3, the result is false. Our example is called a ◆ counterexample, and any other counterexample would do just as well.

74

Logic In summary, if a statement claims that a property holds for all objects of a certain type, then to prove it, we must use steps that are valid for all objects of that type and that do not make references to any particular object. To disprove such a statement, we need only show one counterexample, that is, one particular object or set of objects for which the claim fails.

3 Exercises In Exercises 1 through 11, state whether the argument given is valid or not. If it is valid, identify the tautology or tautologies on which it is based. 1.

If I drive to work, then I will arrive tired. I am not tired when I arrive at work. ∴ I do not drive to work.

2.

If I drive to work, then I will arrive tired. I arrive at work tired. ∴ I drive to work.

3.

If I drive to work, then I will arrive tired. I do not drive to work. ∴ I will not arrive tired.

4.

5.

If I drive to work, then I will arrive tired. I drive to work. ∴ I will arrive tired. I will become famous or I will not become a writer. I will become a writer. ∴ I will become famous.

6.

I will become famous or I will be a writer. I will not be a writer. ∴ I will become famous.

7.

If I try hard and I have talent, then I will become a musician. If I become a musician, then I will be happy. ∴ If I will not be happy, then I did not try hard or I do not have talent.

8.

If I graduate this semester, then I will have passed the physics course. If I do not study physics for 10 hours a week, then I will not pass physics. If I study physics for 10 hours a week, then I cannot play volleyball. ∴ If I play volleyball, I will not graduate this semester.

9.

If my plumbing plans do not meet the construction code, then I cannot build my house. If I hire a licensed contractor, then my plumbing plans will meet the construction code. I hire a licensed contractor. ∴ I can build my house.

(b) p⇒q p∨q ∼p ∼q ∴ p ∴ ∼q 11. Write each argument in Exercise 10 as a single compound statement.

10. (a)

(p ⇒ q) ∧ (q ⇒ r) (b) ∼(p ⇒ q) (∼q) ∧ r p ∴ p ∴ ∼q 13. Write each argument in Exercise 12 as a single compound statement.

12. (a)

14. Prove that the sum of two even numbers is even. 15. Prove that the sum of two odd numbers is even. 16. Prove that the structure (even integers, +, ∗ ) is closed with respect to ∗ . 17. Prove that the structure (odd integers, +, ∗ ) is closed with respect to ∗ . 18. Prove that n2 is even if and only if n is even. 19. Prove that A = B if and only if A ⊆ B and B ⊆ A. 20. Let A and B be subsets of a universal set U. Prove that A ⊆ B if and only if B ⊆ A. 21. Show that (a) A ⊆ B is a necessary and sufficient condition for A ∪ B = B. (b) A ⊆ B is a necessary and sufficient condition for A ∩ B = A. 22. Show that k is odd is a necessary and sufficient condition for k3 to be odd. 23. Prove or disprove: n2 + 41n + 41 is a prime number for every integer n. 24. Prove or disprove: The sum of any five consecutive integers is divisible by 5. 25. Prove or disprove that 3 | (n3 − n) for every positive integer n. 26. Prove or disprove: 1 + 2n > 3n , for all n ∈ Z+ . 27. Determine if the following is a valid argument. Explain your conclusion. Prove: ∀x x3 > x2 . Proof: ∀x x2 > 0 so ∀x x2 (x − 1) > 0(x − 1) and ∀x x3 − x2 > 0. Hence ∀x x3 > x2 . 28. Determine if the following is a valid argument. Explain your conclusion.

75

Logic Prove: If A and B are matrices such that AB = 0, then either A = 0 or B = 0. Proof: There are two cases to consider: A = 0 or A = 0. If A = 0, then we are done. If A = 0, then A−1 (AB) = A−1 0 and (A−1 A)B = 0 and B = 0. 29. Determine if the following is a valid argument. Explain your conclusion. Let m and n be two relatively prime integers. Prove that if mn is a cube, then m and n are each cubes. Proof: We first note that in the factorization of any cube into prime factors, each prime must have an exponent that is a multiple of 3. Write m and n each as a product of b a primes; m = p1a1 pa22 · · · pkk and n = q1b1 q2b2 · · · qj j . Suppose m is not a cube. Then at least one ai is not a multiple of 3. Since each prime factor of mn must have an exponent that is a multiple of 3, n must have a factor pbi i such that bi = 0 and ai + bi is a multiple of 3. But this means that m and n share a factor, pi . This contradicts the fact that m and n are relatively prime.

4

30. Determine if the following is a valid argument. Explain your conclusion. Prove: If x is an irrational number, then 1 − x is also an irrational number. Proof: Suppose 1 − x is rational. Then we can write a a 1 − x as , with a, b ∈ Z. Now we have 1 − = x and b b b−a x= , a rational number. This is a contradiction. b Hence, if x is irrational, so is 1 − x. 31. Prove that the sum of two prime numbers, each larger than 2, is not a prime number. 32. Prove that if two lines are each perpendicular to a third line in the plane, then the two lines are parallel. 33. Prove that if x is a rational number and y is an irrational number, then x + y is an irrational number. 34. Prove that if 2y is an irrational number, then y is an irrational number.

Mathematical Induction Here we discuss another proof technique. Suppose the statement to be proved can be put in the form ∀n ≥ n0 P(n), where n0 is some fixed integer. That is, suppose we wish to show that P(n) is true for all integers n ≥ n0 . The following result shows how this can be done. Suppose that (a) P(n0 ) is true and (b) If P(k) is true for some k ≥ n0 , then P(k + 1) must also be true. Then P(n) is true for all n ≥ n0 . This result is called the principle of mathematical induction. Thus to prove the truth of a statement ∀n ≥ n0 P(n), using the principle of mathematical induction, we must begin by proving directly that the first proposition P(n0 ) is true. This is called the basis step of the induction and is generally very easy. Then we must prove that P(k) ⇒ P(k + 1) is a tautology for any choice of k ≥ n0 . Since the only case where an implication is false is if the antecedent is true and the consequent is false, this step is usually done by showing that if P(k) were true, then P(k + 1) would also have to be true. Note that this is not the same as assuming that P(k) is true for some value of k. This step is called the induction step, and some work will usually be required to show that the implication is always true.

Example 1

Show, by mathematical induction, that for all n ≥ 1, 1 + 2 + 3 + ··· + n =

n(n + 1) . 2

Solution

Let P(n) be the predicate 1 + 2 + 3 + · · · + n =

n(n + 1) . In this example, n0 = 1. 2

Basis Step We must first show that P(1) is true. P(1) is the statement 1= which is clearly true.

76

1(1 + 1) , 2

Logic

Induction Step We must now show that for k ≥ 1, if P(k) is true, then P(k + 1) must also be true. We assume that for some fixed k ≥ 1, 1 + 2 + 3 + ··· + k =

k(k + 1) . 2

(1)

We now wish to show the truth of P(k + 1): 1 + 2 + 3 + · · · + (k + 1) =

(k + 1)((k + 1) + 1) . 2

The left-hand side of P(k + 1) can be written as 1 + 2 + 3 + · · · + k + (k + 1) and we have (1 + 2 + 3 + · · · + k) + (k + 1) k(k + 1) + (k + 1) 2 k +1 = (k + 1) 2

=

using (1) to replace 1 + 2 + · · · + k factoring

=

(k + 1)(k + 2) 2

=

(k + 1)((k + 1) + 1) 2

the right-hand side of P(k + 1)

Thus, we have shown the left-hand side of P(k + 1) equals the right-hand side of P(k + 1). By the principle of mathematical induction, it follows that P(n) is true ◆ for all n ≥ 1. Example 2

Let A1 , A2 , A3 , . . . , An be any n sets. We show by mathematical induction that n i=1

Ai

=

n

Ai .

i=1

(This is an extended version of one of De Morgan’s laws.) Let P(n) be the predicate that the equality holds for any n sets. We prove by mathematical induction that for all n ≥ 1, P(n) is true.

Basis Step P(1) is the statement A1 = A1 , which is obviously true.

77

Logic

Induction Step We use P(k) to show P(k + 1). The left-hand side of P(k + 1) is k+1 Ai = A1 ∪ A2 ∪ · · · ∪ Ak ∪ Ak+1 i=1

= (A1 ∪ A2 ∪ · · · ∪ Ak ) ∪ Ak+1

associative property of ∪

= (A1 ∪ A2 ∪ · · · ∪ Ak ) ∩ Ak+1 k Ai ∩ Ak+1 =

by De Morgan’s law for two sets using P(k)

i=1

=

k+1

Ai

right-hand side of P(k + 1)

i=1

Thus, the implication P(k) ⇒ P(k + 1) is a tautology, and by the principle of ◆ mathematical induction P(n) is true for all n ≥ 1. Example 3

We show by mathematical induction that any finite, nonempty set is countable; that is, it can be arranged in a list. Let P(n) be the predicate that if A is any set with |A| = n ≥ 1, then A is countable.

Basis Step Here n0 is 1, so we let A be any set with one element, say A = {x}. In this case x forms a sequence all by itself whose set is A, so P(1) is true. Induction Step We want to use the statement P(k) that if A is any set with k elements, then A is countable. Now choose any set B with k + 1 elements and pick any element x in B. Since B − {x} is a set with k elements, the induction hypothesis P(k) tells us there is a sequence x1 , x2 , . . . , xk with B − {x} as its corresponding set. The sequence x1 , x2 , . . . , xk , x then has B as the corresponding set so B is countable. Since B can be any set with k + 1 elements, P(k + 1) is true if P(k) is. Thus, by the principle ◆ of mathematical induction, P(n) is true for all n ≥ 1. In proving results by induction, you should not start by assuming that P(k + 1) is true and attempting to manipulate this result until you arrive at a true statement. This common mistake is always an incorrect use of the principle of mathematical induction. A natural connection exists between recursion and induction, because objects that are recursively defined often use a natural sequence in their definition. Induction is frequently the best, maybe the only, way to prove results about recursively defined objects. Example 4

Consider the following recursive definition of the factorial function: 1! = 1, n! = n(n − 1)!, n > 1. Suppose we wish to prove for all n ≥ 1, n! ≥ 2n−1 . We proceed by mathematical induction. Let P(n): n! ≥ 2n−1 . Here n0 is 1.

Basis Step P(1) is the statement 1! ≥ 20 . Since 1! is 1, this statement is true.

78

Logic

Induction Step We want to show P(k) ⇒ P(k + 1) is a tautology. It will be a tautology if P(k) true guarantees P(k + 1) is true. Suppose k! ≥ 2k−1 for some k ≥ 1. Then by the recursive definition, the left side of P(k + 1) is (k + 1)! = (k + 1)k! ≥ (k + 1)2k−1

using P(k)

≥2×2

k + 1 ≥ 2, since k ≥ 1

=2

right-hand side of P(k + 1)

k−1

k

Thus, P(k + 1) is true. By the principle of mathematical induction, it follows that ◆ P(n) is true for all n ≥ 1. The following example shows one way in which induction can be useful in computer programming. Example 5

Consider the following function given in pseudocode. FUNCTION SQ(A) 1. C ← 0 2. D ← 0 3. WHILE (D = A) a. C ← C + A b. D ← D + 1 4. RETURN (C) END OF FUNCTION SQ

The name of the function, SQ, suggests that it computes the square of A. Step 3b shows A must be a positive integer if the looping is to end. A few trials with particular values of A will provide evidence that the function does carry out this task. However, suppose we now want to prove that SQ always computes the square of the positive integer A, no matter how large A might be. We shall give a proof by mathematical induction. For each integer n ≥ 0, let Cn and Dn be the values of the variables C and D, respectively, after passing through the WHILE loop n times. In particular, C0 and D0 represent the values of the variables before looping starts. Let P(n) be the predicate Cn = A × Dn . We shall prove by induction that ∀n ≥ 0 P(n) is true. Here n0 is 0.

Basis Step P(0) is the statement C0 = A × D0 , which is true since the value of both C and D is zero “after” zero passes through the WHILE loop.

Induction Step We must now use P(k) : Ck = A × Dk

(2)

to show that P(k + 1): Ck+1 = A × Dk+1 . After a pass through the loop, C is

79

Logic increased by A, and D is increased by 1, so Ck+1 = Ck + A and Dk+1 = Dk + 1. left-hand side of P(k + 1): Ck+1 = Ck + A = A × Dk + A = A × (Dk + 1) = A × Dk+1

using (2) to replace Ck factoring right-hand side of P(k + 1)

By the principle of mathematical induction, it follows that as long as looping occurs, Cn = A×Dn . The loop must terminate. (Why?) When the loop terminates, D = A, so C = A × A, or A2 , and this is the value returned by the function SQ. ◆ Example 5 illustrates the use of a loop invariant, a relationship between variables that persists through all iterations of the loop. This technique for proving that loops and programs do what is claimed they do is an important part of the theory of algorithm verification. In Example 5 it is clear that the looping stops if A is a positive integer, but for more complex cases, this may also be proved by induction. Example 6

Use the technique of Example 5 to prove that the pseudocode program does compute the greatest common divisor of two positive integers. Solution

Here is the pseudocode: FUNCTION GCD(X,Y) 1. WHILE (X = Y) a. IF (X > Y) THEN 1. X ← X − Y b. ELSE 1. Y ← Y − X 2. RETURN (X) END OF FUNCTION GCD

We claim that if X and Y are positive integers, then GCD returns GCD(X, Y). To prove this, let Xn and Yn be the values of X and Y after n ≥ 0 passes through the WHILE loop. We claim that P(n): GCD(Xn , Yn ) = GCD(X, Y) is true for all n ≥ 0, and we prove this by mathematical induction. Here n0 is 0.

Basis Step X0 = X, Y0 = Y , since these are the values of the variables before looping begins; thus P(0) is the statement GCD(X0 , Y0 ) = GCD(X, Y), which is true. Induction Step Consider the left-hand side of P(k + 1), that is, GCD(Xk+1 , Yk+1 ). After the k + 1 pass through the loop, either Xk+1 = Xk and Yk+1 = Yk − Xk or Xk+1 = Xk − Yk and Yk+1 = Yk . Then if P(k): GCD(Xk , Yk ) = GCD(X, Y) is true, we have that GCD(Xk+1 , Yk+1 ) = GCD(Xk , Yk ) = GCD(X, Y). Thus, by the principle of mathematical induction, P(n) is true for all n ≥ 0. The exit condition for the loop is Xn = Yn and we have GCD(Xn , Yn ) = Xn . Hence the function always returns ◆ the value GCD(X, Y).

80

Logic A discussion of how to construct a loop invariant for a given loop would take us too far from the main ideas of this section. In writing programs, loop invariant construction usually precedes decisions about the steps to be executed in the loop. In the exercises, loop invariants will be provided where needed.

Strong Induction A slightly different form of mathematical induction is easier to use in some proofs. In the strong form of mathematical induction, or strong induction, the induction step is to show that P(n0 ) ∧ P(n0 + 1) ∧ P(n0 + 2) ∧ · · · ∧ P(k) ⇒ P(k + 1) is a tautology. As before, the only case we need to check is that if each P(j), j = n0 , . . . , k, is true, then P(k + 1) is true. The strong form of induction is equivalent to the form we first presented, so it is a matter of convenience which one we use in a proof. Example 7

a

a

Prove that every positive integer n > 1 can be written uniquely as p11 p22 · · · pas s , where the pi are primes and p1 < p2 < · · · < ps . Proof (by strong induction)

Basis Step Here n0 is 2. P(2) is clearly true, since 2 is prime. Induction Step We use P(2), P(3), . . . , P(k) to show P(k + 1): k + 1 can be written uniquely a a as p11 p22 · · · pas s , where the pi are primes and p1 < p2 < · · · < ps . There are two cases to consider. If k + 1 is a prime, then P(k + 1) is true. If k + 1 is not prime, then k + 1 = lm, 2 ≤ l ≤ k, 2 ≤ m ≤ k. Using P(l) and P(m), we have b b c c a a k = lm = q11 q22 · · · qtbt r11 r22 · · · rucu = p11 p22 · · · pas s , where each pi = qj or rk , p1 < p2 < · · · < ps , and if qj = rk = pi , then ai = bj + ck , otherwise pi = qj and ai = bj or pi = rk and ai = ck . Since the factorization of l and m are unique, ◆ so is the factorization of k + 1.

4 Exercises In Exercises 1 through 7, prove the statement is true by using mathematical induction.

8. Let P(n): 13 + 23 + 33 + · · · + n3 =

1. 2 + 4 + 6 + · · · + 2n = n(n + 1) 2. 12 + 32 + 52 + · · · + (2n − 1)2 =

n(2n + 1)(2n − 1) 3

9.

3. 1 + 21 + 22 + · · · + 2n = 2n+1 − 1 5n(n + 1) 2 n(n + 1)(2n + 1) 5. 12 + 22 + 32 + · · · + n2 = 6 n a −1 6. 1 + a + a2 + · · · + an−1 = a−1 4. 5 + 10 + 15 + · · · + 5n =

7. a + ar + ar 2 + · · · + ar n−1 =

a(1 − r n ) for r = 1 1−r

10. 11. 12. 13. 14.

n2 (n + 1)2 + 4 . 4

(a) Use P(k) to show P(k + 1). (b) Is P(n) true for all n ≥ 1? Let P(n): 1 + 5 + 9 + · · · + (4n − 3) = (2n + 1)(n − 1). (a) Use P(k) to show P(k + 1). (b) Is P(n) true for all n ≥ 1? Prove 1 + 2n < 3n for n ≥ 2. Prove n < 2n for n > 1. (2n + 1)2 Prove 1 + 2 + 3 + · · · + n < 8 Find the least n for which the statement is true and then prove that (1 + n2 ) < 2n . Find the least n for which the statement is true and then prove that 10n < 3n .

81

Logic 15. Prove by mathematical induction that if a set A has n elements, then P(A) has 2n elements. 16. Prove by mathematical induction that 3 | (n3 −n) for every positive integer n. 17. Prove by mathematical induction that if A1 , A2 , . . . , An are any n sets, then n n Ai = Ai . i=1

i=1

18. Prove by mathematical induction that if A1 , A2 , . . . , An and B are any n + 1 sets, then n n Ai (Ai ∩ B). B= i=1

i=1

19. Prove by mathematical induction that if A1 , A2 , . . . , An and B are any n + 1 sets, then n n Ai (Ai ∪ B). B= i=1

i=1

20. Let P(n) be the statement 2 | (2n − 1). (a) Prove that P(k) ⇒ P(k + 1) is a tautology. (b) Show that P(n) is not true for any integer n. (c) Do the results in (a) and (b) contradict the principle of mathematical induction? Explain. 21. Let P(n) be the statement n2 + n is an odd number for n ∈ Z+ . (a) Prove that P(k) ⇒ P(k + 1) is a tautology. (b) Is P(n) true for all n? Explain. 22. Explain the flaw in the following “argument.” For z = 0, zn = 1, n ≥ 0. Proof: Basis Step: For n = 0, P(0): zn = 1 is true by definition. zk 1 Induction Step: zk+1 = k−1 · zk = · 1 or 1. z 1 23. Explain the flaw in the following “argument.” All trucks are the same color. Proof: Let P(n): Any set of n trucks consists of trucks of the same color. Basis Step: Certainly P(1) is true, since there is only one truck in this case. Induction Step: We use P(k): Any set of k trucks consists of trucks of the same color to show P(k + 1): Any set of k + 1 trucks consists of trucks of the same color. Choose one truck from the set of k + 1 trucks and consider the remaining set of k trucks. By P(k) these are all the same color. Now return the chosen truck and set aside another truck. The remaining trucks are all the same color by P(k). But trucks do not change color in this procedure, so all k + 1 trucks must be the same color. In Exercises 24 through 26, prove the given statement about matrices. Assume A is n × n.

82

24. (A1 + A2 + · · · + An )T = AT1 + AT2 + · · · + ATn 25. A2 An = A2+n 26. Let A and B be square matrices. If AB = BA, then (AB)n = An Bn , for n ≥ 1. 27. Prove that any restaurant bill of $n, n ≥ 5, can be paid exactly using only $2 and $5 bills. 28. Prove that every integer greater than 27 can be written as 5a + 8b, where a, b ∈ Z+ . 29. Use induction to show that if p is a prime and p | an for n > 1, then p | a. 30. Prove that if GCD(a, b) = 1, then GCD(an , bn ) = 1 for all n ≥ 1. (Hint: Use Exercise 29.) 31. (a) Find the smallest positive integer n0 such that 2n0 > n20 . (b) Prove 2n > n2 for all n ≥ n0 . 32. Prove or disprove: 2 + 8 + 18 + · · · + 2n2 = n2 + n. 33. Prove or disprove: x − y divides xn − yn for n ≥ 1. In Exercises 34 through 39, show that the given algorithm, correctly used, produces the output stated, by using mathematical induction to prove the relationship indicated is a loop invariant and checking values when the looping stops. All variables represent nonnegative integers. 34. SUBROUTINE COMP(X,Y;Z) 1. Z ← X 2. W ← Y 3. WHILE (W > 0) a. Z ← Z + Y b. W ← W − 1 4. RETURN END OF SUBROUTINE COMP COMPUTES: Z = X + Y 2 LOOP INVARIANT: (Y × W) + Z = X + Y 2

35. SUBROUTINE DIFF (X,Y;Z) 1. Z ← X 2. W ← Y 3. WHILE (W > 0) a. Z ← Z − 1 b. W ← W − 1 4. RETURN END OF SUBROUTINE DIFF COMPUTES: Z = X − Y LOOP INVARIANT: X − Z + W = Y

36. SUBROUTINE EXP2 (N,M;R) 1. R ← 1 2. K ← 2M 3. WHILE (K > 0) a. R ← R × N b. K ← K − 1 4. RETURN END OF SUBROUTINE EXP2 COMPUTES: R = N 2M LOOP INVARIANT: R × N K = N 2M

Logic 37. SUBROUTINE POWER (X,Y;Z) 1. Z ← 0 2. W ← Y 3. WHILE (W > 0) a. Z ← Z + X b. W ← W − 1 4. W ← Y − 1 5. U ← Z 6. WHILE (W > 0) a. Z ← Z + U b. W ← W − 1 7. RETURN END OF SUBROUTINE POWER COMPUTES: Z = X × Y 2 LOOP INVARIANT (first loop): X + (X × W) = X × Y LOOP INVARIANT (second loop): X + (X × Y × W) = X × Y 2 (Hint: Use the value of Z at the end of the first loop in

loop 2.) 38. SUBROUTINE DIV(X,Y) 1. IF (Y = 0) THEN a. PRINT (’error Y = 0’) 2. ELSE a. R ← X b. K ← 0 c. WHILE (K ≥ Y)

5

1. R ← R − Y 2. K ← K + 1 d. IF (R = 0) THEN 1. PRINT (’true’) e. ELSE 1. PRINT (’false’) 3. RETURN END OF SUBROUTINE DIV COMPUTES: TRUTH VALUE OF Y | X. LOOP INVARIANT: R + K × Y = X

39. SUBROUTINE SQS(X,Y;Z) 1. Z ← Y 2. W ← X 3. WHILE (W > 0) a. Z ← Z + X b. W ← W − 1 4. W ← Y − 1 5. WHILE (W > 0) a. Z ← Z + X b. W ← W − 1 6. RETURN END OF SUBROUTINE SQS COMPUTES: Z = X 2 × Y 2 LOOP INVARIANT (first loop): Z + (X × W) = Y + X 2 LOOP INVARIANT (second loop): Z + (Y × W) = X 2 + Y 2

Mathematical Statements The previous sections in this chapter lay the groundwork for learning to write proofs. The examples and exercises give many mathematical statements for which you are asked to write proofs. How are statements like this produced? Some of the most common ways to create a statement that may be mathematically true are from experience, by applying old ideas in new settings, by extending known facts to more general cases, and by experimentation. A mathematical statement that has not been proved or disproved yet is called a conjecture.

Example 1

(a) Young students soon notice that adding two odd numbers yields an even number. A conjecture that captures this experience is “The sum of two odd numbers is an even number.” (b) Suppose S is a set of objects and a binary operation has been defined for these objects. Then the truth of s t = t s, s, t ∈ S can be proved or disproved to determine whether is commutative. (c) If A, B, and C are n × n matrices, the fact that (A + B)T = AT + BT suggests ◆ the conjecture that (A + B + C)T = AT + BT + CT .

Example 2

Produce a reasonable conjecture about the sum of the powers of 2 beginning with 20 .

83

Logic Solution

We begin by experimenting with small cases and recording the results. 20 2 +2 0

1

2 +2 +2 0

1

2

2 +2 +2 +2 0

1

2

3

2 +2 +2 +2 +2 0

1

2

3

4

1

1

1+2

3

1+2+4

7

1+2+4+8

15

1 + 2 + 4 + 8 + 16

31

Note that each sum is one less than the next term to be added. (Recognizing a pattern may take more experimentation than shown here.) Thus, a reasonable conjecture is “The sum of the powers of 2 beginning with 20 is one less than the next power of 2 to be added.” This would be clearer in symbolic form. The sum of an arbitrary number of powers of 2 beginning with 20 can be represented by 20 + 21 + · · · + 2n . The result is then 2n+1 − 1, one less than the next power of 2 to be added. 20 + 21 + · · · + 2n = 2n+1 − 1 is given to be proved in Exercise 3, ◆ Section 4. Even when a mathematical statement is written in ordinary language, the proof often needs a way to represent a general case. For example, a proof of the conjecture “The sum of two odd numbers is an even number” requires a way to represent a generic odd number. One way to do this is with 2n + 1. (What is another one?) We need to identify what n can represent. If n is a whole number, then the original conjecture becomes “The sum of two odd whole numbers is an even whole number.” If n is an integer, the statement becomes “The sum of two odd integers is an even integer.” Let us examine a proof of this last version. The conjecture refers to two odd integers, so we must represent another odd integer. This requires another variable, say k. Now the two odd integers are represented as 2n + 1 and 2k + 1, where both n and k are integers. Note that this description includes the case that an odd integer is added to itself. Any even integer can be written as 2t, t ∈ Z. Example 3

The sum of two odd integers is an even integer. Proof

Let 2n + 1 and 2k + 1 be two odd integers. Their sum is (2n + 1) + (2k + 1) or 2n + 2k + 2. 2n + 2k + 2 = 2(n + k + 1). Since n + k + 1 is an integer, ◆ (2n + 1) + (2k + 1) is an even integer. This proof assumes that the reader knows the distributive property and that the sum of two integers is an integer. Depending on the audience for the proof, these facts might have to be proved first or spelled out. Remember that here a proof is a written communication, and the intended audience must be considered when writing a proof. Example 4

Let A, B, C, and D be n × n Boolean matrices. The fact that A ∧ (B ∨ C) = (A ∧ B) ∨ (A ∧ C) suggests that the extended case A ∧ (B ∨ C ∨ D) = (A ∧ B) ∨ (A ∧ C) ∨ (A ∧ D) may be true.

84

Logic Often the key to proving an extension is to apply the original statement. Here A ∧ (B ∨ C ∨ D) = A ∧ ((B ∨ C) ∨ D), because ∨ is associative. The original statement for three matrices implies that A∧((B∨C)∨D) = (A∧(B∨C))∨(A∧D) when we view (B ∧ C) as a single matrix. Now using the original statement again, we have A ∧ ((B ∨ C) ∨ D) = (A ∧ (B ∨ C)) ∨ (A ∧ D) = ((A ∧ B) ∨ (A ∧ C)) ∨ (A ∧ D) = (A ∧ B) ∨ (A ∧ C) ∨ (A ∧ D). The last equality follows again from the fact that ∨ is associative.

◆

Notice that in Example 4 we used more than the original statement to complete the proof. Thus, we cannot expect that extension statements will always be true. A variation of applying an old idea in a new setting is to identify a special situation where a statement that is false in general will now be true. For example, in general, matrix multiplication is not commutative, but for some special cases it may be true that AB = BA. Example 5

Let S be the set of matrices of the form ⎡

⎤ a 0 0 ⎣ 0 0 0 ⎦, 0 0 b where a, b ∈ R. Prove that for the mathematical structure (S, matrix multiplication) the commutative property holds. Solution

We need to represent two generic elements of S, and because we need to examine the results of matrix multiplication, the entries of these generic elements need to be shown. Let ⎤ ⎡ ⎤ ⎡ d1 0 0 c1 0 0 C = ⎣ 0 0 0 ⎦ and D = ⎣ 0 0 0 ⎦ . 0 0 c2 0 0 d2 (How do we know C and D are in S?) Then ⎤ 0 c1 d1 0 0 0 ⎦ CD = ⎣ 0 0 0 c2 d2

⎡

⎡

and

⎤ d1 c1 0 0 0 0 ⎦. DC = ⎣ 0 0 0 d2 c2

Hence, CD = DC, because multiplication in R is commutative; c1 d1 = d1 c1 and c2 d2 = d2 c2 . As noted after Example 3, the intended audience for the proof ◆ determines how much detail about R is included. There are many other ways to create conjectures.

85

Logic

5 Exercises 1. Prove or disprove that the cube of an even number is even. 2. Prove or disprove that the cube of an odd number is odd. 3. (a) Prove or disprove that the sum of three consecutive integers is divisible by 3. (b) Prove or disprove that the sum of four consecutive integers is divisible by 4. 4. Prove that the product of a nonzero rational number and number. an irrational number is a(n) 5. Prove that the quotient of a nonzero rational number and an irrational number is a(n) number. 6. State and prove the extension of A ∪ B = A ∩ B for three sets. 7. State and prove the extension of A ∩ B = A ∪ B for three sets. 8. Let A, B, C, and D be n × n Boolean matrices. State and prove the extension of A ∨ (B ∧ C) = (A ∨ B) ∧ (A ∨ C) to four matrices. 9. Modify the conjecture and proof in Example 4 to give the general case for m + 1 Boolean matrices. 10. Let p and q be propositions. State and prove an extension of ∼(p ∨ q) ≡ ∼p ∧ ∼q to the case of (b) n propositions. (a) 3 propositions 11. Let p and q be propositions. State and prove an extension of ∼(p ∧ q) ≡ ∼p ∨ ∼q to the case of (b) n propositions. (a) 3 propositions 12. Prove that the commutative property holds for the mathematical structure (3 × 3 diagonal matrices, matrix multiplication). 13. Prove that the commutative property holds for the mathematical structure (r × r diagonal matrices, matrix multiplication). 14. State and prove a conjecture about the sum of the first n positive odd integers.

6

For Exercises 15 through 17, use the sequence 3, 9, 15, 21, 27, 33, . . . . 15. Give both a recursive and an explicit formula for this sequence. 16. Experiment with the sums of the terms of the sequence to produce a reasonable conjecture. 17. Prove the conjecture from Exercise 16. 18. State and prove a reasonable conjecture about the sum of the first n terms of the sequence 6, 10, 14, 18, 22, . . . . For Exercises 19 through 22, use the recursively defined sequence g1 = 1, g2 = 3, gn = gn−1 + gn−2 . 19. Write the first ten terms of the sequence. 20. Experiment with the sums of terms in the sequence that occupy even-numbered positions to produce a reasonable conjecture. 21. Experiment with the sums of terms of the sequence to produce a reasonable conjecture about the sum of the first n terms of the sequence. 22. Experiment with the sums of terms in the sequence that occupy odd-numbered positions to produce a reasonable conjecture. 23. Prove the conjecture from Exercise 20. 24. Prove the conjecture from Exercise 21. 25. Prove the conjecture from Exercise 22. 26. Produce and prove a conjecture about the sum of the powers of 3 beginning with 30 . 27. Build on the work done in Exercise 3 to produce a conjecture about when the sum of k consecutive integers is divisible by k. 28. Prove the conjecture of Exercise 27.

Logic and Problem Solving In previous sections, we investigated the use of logic to prove mathematical theorems and to verify the correctness of computational algorithms. However, logic is also valuable in less formal settings. Logic is used every day to decide between alternatives and investigate consequences. It is, in many situations, essentially the same as precise and careful thinking, particularly in cases where the details of the situation are complex. One of the most important uses of logic is to develop correct and efficient algorithms to solve problems. Part of this process may be to express the solution in terms of a computation, even if the problem does not at first seem computational. In Section 5, we noted that one common way to create mathematical conjectures is to apply old ideas in new settings. This approach is also a powerful problem-solving technique. To demonstrate these two ideas, we present a variety of seemingly unrelated problems that can be solved by a single method.

86

Logic We begin with an easily stated mathematical question. Given a set A and a finite number of subsets of A, say A1 , A2 , . . . , Ak , is it possible to find a collection S of the Ai ’s with the following two properties: 1. Any two sets in S are disjoint. 2. The union of all sets in S is A? The collection S is called an exact cover of A with respect to the Ai ’s, because together the Ai ’s “cover” A with no overlaps. Example 1

Let A = {a, b, c, d, e, f, g, h, i, j} and A1 = {a, c, d}, A2 = {a, b, e, f }, A3 = {b, f, g}, A4 = {d, h, i}, A5 = {a, h, j}, A6 = {e, h}, A7 = {c, i, j}, A8 = {i, j}. Is there an exact cover of A with respect to A1 , A2 , A3 , A4 , A5 , A6 , A7 , A8 ? Solution

We start with A1 and consider each of the other sets in turn. We reject A2 , since A1 ∩ A2 = ∅, but A3 is suitable. A1 and A3 together contain the elements a, b, c, d, f , and g of A. The sets A4 and A5 contain d and a respectively, and so they are rejected. At this point we can see that A6 and A8 will complete the search, and A1 , ◆ A3 , A6 , A8 form an exact cover of A. If we did not find an exact cover that contained A1 , we would have started the search over with A2 as the first choice. This method would be continued until either we found an exact cover or failed for each choice of a starting set. A systematic way to try all possibilities, such as we just described, is a key step in writing algorithms and computer programs to solve many types of problems involving only a finite number of possible solutions. The technique is called backtracking, and is an essential part of many algorithms that search all possible outcomes of a process. This method is a way of implementing a brute force search of possibilities and will always produce a solution if one exists. Example 2

Let U = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} and let A1 = {2, 4, 6, 7}, A2 = {1, 2, 5}, A3 = {9, 10}, A4 = {5, 8, 1}, A5 = {1, 3, 5}. Find an exact cover of U with respect to the given subsets. Solution

A1 is not disjoint from A2 , so A2 cannot be used. A1 , A3 , and A4 are a collection of disjoint sets, but A4 has elements in common with A5 , so A5 is eliminated. However, A1 ∪ A3 ∪ A4 = A and we begin the search again with A2 . A2 is only disjoint from A3 . Again, this is not enough for a cover; A2 ∪ A3 = A. Continuing to check, starting with A3 , A4 , and A5 in turn, we find there is no exact cover of A with respect to these subsets. ◆ The exact cover problem can be restated in a way that leads to an algorithm that can be implemented by a computer program. If U is a set with n elements, then each subset of U can be represented by a unique sequence of 0’s and 1’s of length n. We can use this representation to restate the exact cover problem. If U is a set with n elements, and if S1 , S2 , . . . , Sk are a finite number of subsets of U, we form a matrix M whose rows correspond, in order, to the representations of S1 , S2 , . . . , Sk as sequences of 0’s and 1’s. Example 3

Represent the subsets A1 , A2 , . . . , A8 of Example 1 by a matrix M.

87

Logic Solution

The set A of Example 1 has 10 elements. The eight subsets can be represented in order by the rows of the following matrix ⎡ ⎤ 1 0 1 1 0 0 0 0 0 0 ⎢1 1 0 0 1 1 0 0 0 0⎥ ⎢ ⎥ ⎢0 1 0 0 0 1 1 0 0 0⎥ ⎢ ⎥ ⎢0 0 0 1 0 0 0 1 1 0⎥ M=⎢ ⎥. ⎢1 0 0 0 0 0 0 1 0 1⎥ ⎢0 0 0 0 1 0 0 1 0 0⎥ ⎢ ⎥ ⎣0 0 1 0 0 0 0 0 1 1⎦ 0 0 0 0 0 0 0 0 1 1 ◆ We can now use M to find an exact cover of A with respect to A1 , A2 , A3 , A4 , A5 , A6 , A7 , and A8 . Let 0= 0 0 0 0 0 0 0 0 0 0 and 1= 1 1 1 1 1 1 1 1 1 1 . In matrix terms the problem becomes: Find a collection r1 , r2 , . . . , rk of rows of M having the following properties: (a) ri ∧ rj = 0 for 1 ≤ i, j ≤ k and (b) r1 ∨ r2 ∨ · · · ∨ rk = 1 It is easy to see that one solution is given by rows 1, 3, 6, and 8. This is the solution found before. Once it is reduced to searching a Boolean matrix, the exact cover problem can be more easily tackled with computer-implemented algorithms. The best-known algorithm for solving the exact cover problem is Donald Knuth’s Algorithm X. Algorithm ALGORITHM X

Label the rows of the original matrix A and retain these labels throughout the application of the algorithm. Begin with M = A and L = { }. Step 1 If M has a column of 0’s, then there is no solution, since by definition the rows representing an exact cover must together have 1’s in every position. Step 2 Otherwise: (a) Choose the column c of M with the fewest 1’s. (b) Choose a row r with a 1 in column c, and place the number r in L. (c) Eliminate any row r1 having the property that r ∧ r1 = 0 (where 0 represents a row with all 0 entries). (d) Eliminate all columns in which r has a 1. (e) Eliminate row r. ● (f) Let M be the resulting matrix and repeat steps 1 and 2. The logic behind (c) is that any such row r1 represents a subset that is not disjoint from the subset of row r. The logic of (d) is that the deleted columns record elements that are already in the subset of row r, and so cannot play any further role in additional rows chosen.

88

Logic We repeat Steps 1 and 2 producing ever smaller matrices, choosing a row each time and accumulating corresponding row labels from A, until the process terminates. Let us call such a succession of choices and computations a path. In general, there may be many paths to follow, since we may have several rows available at each successive stage in the algorithm. A path is successful if the final matrix produced is empty (has no rows or columns), in which case L will contain the row labels of an exact cover. A path fails if the final matrix produced has a column of 0’s. Note that if all paths that start with row r fail, there may still be a solution. In this case, we remove r from L (and from any further consideration) and choose (if possible) another row with a 1 in column c. Example 4

Apply Algorithm X to find an exact cover for A = {a, b, c, d, e, f, g} with respect to A1 = {a, b, e}, A2 = {a, b}, A3 = {a, e, f }, A4 = {d, f, g}, A5 = {c, d, e, g}, A6 = {c, e}. Solution

The matrix A represents the original situation. ⎡ 1 1 1 0 0 2 ⎢1 1 0 0 ⎢ 3 ⎢1 0 0 0 A= 4 ⎢ ⎢0 0 0 1 5 ⎣0 0 1 1 6 0 0 1 0

1 0 1 0 1 1

0 0 1 1 0 0

⎤ 0 0⎥ ⎥ 0⎥ 1⎥ ⎥ 1⎦ 0

The rows of A have been numbered. Let M = A and L = { }. Step 1 M does not have a column of zeros. Step 2 The least number of 1’s in any column is 2, so we may choose the first such column, column 2. We need to choose a row with a 1 in column 2, say r = 1. Now L = {1}. We perform actions (c) through (e). Rows 2, 3, 5, and 6 are eliminated, since they each have a 1 in a common column with the first row. We then eliminate columns 1, 2, and 5, and row 1. The new matrix is what remains of row 4 of A. 4 0 1 1 1 Note: The original row label is still attached. However, this matrix has a column of 0’s, and so this path fails. That means that we backtrack, removing row 1 from further consideration. So L = ∅ again, and we return to the original matrix A. Since row 1 failed to start a successful path, we try row 2, the only other row to have a 1 in column 2. If this also fails, then no solution is possible. Perform Steps 1 and 2 on M, based on row 2. This first gives L = {2}, and the following matrix: ⎡ ⎤ 4 0 1 0 1 1 5 ⎣ 1 1 1 0 1 ⎦. 6 1 0 1 0 0 Column 4 of this matrix has the least number of 1’s, and the top row has a 1 in column 4, so we choose to work with it. Since that row corresponds to row 4 of A, we have L = {2, 4}. Actions (c) through (e) on the new matrix, using its top row, produce 6 1 1 .

89

Logic Now we can only choose the last remaining row and add 6 to L, so that L = {2, 4, 6}. Actions (c) through (e), applied to this matrix, eliminate all rows and columns, so the path is successful, and an exact cover consists of rows 2, 4, ◆ and 6 of A, as is easily verified. Many other problems and recreational puzzles are equivalent to the exact cover problem. Example 5

Figure 1(a) shows a 3 × 3 matrix with two positions filled. ⎤ 1 ⎣ 2 ⎦ ⎡

(a)

⎡

⎤ 1 2 3 ⎣ 2⎦ 1 (b)

⎡

⎤ 1 2 3 ⎣3 1 2⎦ 2 1 (c)

Figure 1

Fill the remaining positions with 1, 2, and 3, so that each row and each column has exactly one 1, one 2, and one 3. Solution

We reason as follows: Position (1, 3) must contain a 3, since there is already a 1 in the first row and a 2 in the third column, and duplicates are not permitted. The rest of row 1 and column 3 must then be as shown in Figure 1(b). Similar reasoning shows that position (2, 1) can only contain a 3, and so row 2 and column 1 must be as shown in Figure 1(c). Finally, we see that position (3, 2) ◆ must contain a 3, and we have found the unique solution. Example 6

Fill the remaining positions of Figure 2 with 1, 2, and 3, so that each row and each column has exactly one 1, one 2, and one 3. ⎤ ⎡ 1 ⎣ 2 ⎦ 2 Figure 2

Solution

As before, we reason that position (1, 3) must contain the number 3. However, this forces position (1, 2) to contain a 2. On the other hand, this is impossible, since there would then be two 2’s in the second column. Thus we see that no solution is ◆ possible. Consider again the puzzle given in Figure 1. Recall that we are trying to put the digits 1, 2, and 3 in a 3 × 3 matrix so that each row and column contains these digits exactly once. Suppose that we are free to choose nine digits, each of which is either 1, 2, or 3 (without restrictions on the number of each), and place them in the matrix at any location (where more than one number can occupy the same location, and some locations can be empty). In order for such a placement to constitute an allowable solution to the puzzle, three sets of constraints must be true. 1. Each cell in the matrix must contain a digit. (9 constraints) 2. Each digit must occur in each column. (9 constraints)

90

Logic 3. Each digit must occur in each row. (9 constraints) Thus, there are 27 constraints in all. With the complete freedom of placement described above, we could easily fill the array so that any two of the three conditions were true, but the third was not, even if we always had three of each number. Some cases are shown in Figure 3. ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ 1 2 3 1 1 1 1, 2, 3 ⎣1 2 3⎦ ⎣2 2 2⎦ ⎣ 1, 2, 3 ⎦ 1, 2, 3 1 2 3 3 3 3 Figure 3

Thus, we need all 27 constraints satisfied in order to have a solution. We now construct a Boolean matrix A with 27 columns, one for each constraint. The first 9 columns of A represent respectively the constraints that the positions (1, 1), (1, 2), (1, 3), (2, 1), . . . , (3, 3) contain one digit. The columns from 10 to 18 represent respectively the constraints that (a 1 is in row 1), (a 1 is in row 2), . . . , (a 3 is in row 3). Finally, the columns from 19 to 27 represent respectively the constraints that (a 1 is in column 1), (a 1 is in column 2), . . . , (a 3 is in column 3). Let us say that a “move” is the placement of one of the digits 1, 2, 3 in a particular location of A. Since there are three digits and nine cells, there are 27 moves. Hence, let matrix A have 27 rows, one for each possible move. The rows of A contain 0’s and 1’s and are constructed in the following way. If a row corresponds to putting a 2 in position (3, 1), then three constraints are satisfied—there is a digit in position (3, 1), there is a 2 in row 3, and there is a 2 in column 1. We place 1’s in the columns of that row to correspond to these three constraints, namely columns 7, 15, and 22, and place 0’s elsewhere. Then the row of A that corresponds to the placement of a 2 in position (3, 1) is 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0 . Each row of A corresponds to a move, thus satisfying three constraints: one in the range 1–9, one in the range 10–18, and one in the range 19–27. It therefore has exactly three 1’s in the positions corresponding to satisfied constraints. Thus, the row 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 corresponds to placing a 3 in position (2, 3). Example 7

(a) Give the row of A for the move that places a 1 in position (2, 2). (b) Describe the move represented by this row of A:

0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 . Solution

(a) 0 0 0 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 ◆ (b) The move consists of placing a 2 in the position (1, 2).

If we now solve the exact cover problem for a set represented by A, then the rows in the solution correspond to placements that are a legitimate solution to the

91

Logic

⎡

1 ⎢0 ⎢ ⎢0 ⎢ ⎢0 ⎢ ⎢0 ⎢0 ⎢ ⎢0 ⎢ ⎣0 0

0 1 0 0 0 0 0 0 0

0 0 1 0 0 0 0 0 0

0 0 0 1 0 0 0 0 0

puzzle, since each constraint is satisfied in exactly one way. If there is an initial placement of digits, then the corresponding rows must belong to the solution. The solution to the puzzle shown in Figure 1(a) is then represented by the following matrix. ⎤ 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0⎥ ⎥ 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1⎥ ⎥ 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0⎥ ⎥ 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0⎥ 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0⎥ ⎥ 0 0 1 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 0 0 0 0⎥ ⎥ 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 0⎦ 0 0 0 0 1 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 Rows 1 and 6 correspond to the initial placement of digits.

Sudoku A generalization of the simple puzzle shown in Example 5 is one of the most popular puzzles of recent times, Sudoku. We begin with a simplified version of Sudoku shown in Figure 4(a). It has 16 small squares, organized into 4 larger 2 × 2 squares, outlined more thickly.

2

3 4

(a)

2

1 4

1 4

2

3

1 3

1 2 3 1 3 4

1

1

1 4

4

2

1 3 4 1 2 4 1 3

1 3

(b)

1

4

2

3

2

3

1

4

3

1

4

2

4

2

3

1

(c)

Figure 4

The goal of the puzzle is to fill the grid with 1, 2, 3, and 4 in such a way that each row, each column, and each of the outlined 2 × 2 squares will contain these numbers exactly once. Often there is an initial assignment, as shown in Figure 4(a). The main hand techniques for solving a Sudoku puzzle are variations of what may be described as “guess and check the consequences.” The ideas are similar in nature to those for Example 1, but may involve more complex logical deductions. We could, for example, temporarily place in each cell all the numbers that might go there; that is, numbers not obviously ruled out by the initial placements. Thus in cell (1, 1), the top left cell, we would place 1 and 4, since both 2 and 3 are elsewhere in the top left large block. The resulting matrix is shown in Figure 4(b). Then we test these possible choices by investigating the logical consequences, as we did in the simpler 3 × 3 version. In Figure 4(b) we see that positions (2, 3) and (3, 2) can only contain 1. Having used 1’s in columns 2 and 3, we see that positions (1, 2) and (4, 3) must contain 4 and 3 respectively. In turn, we deduce that positions (1, 1) and (4, 4) must contain 1’s to fill in the top left and bottom right 2 × 2 squares. We continue this

92

Logic logical progression and find the solution shown in Figure 4(c). Sometimes we have to make an arbitrary choice in some positions, and follow its consequences. Occasionally this results in a contradiction, and we must then make an alternative choice and start over. The most common Sudoku puzzle consists of a 9 × 9 grid divided into 9 larger 3 × 3 grids, as shown in Figure 5. You must make each row, each column, and each larger grid contain the numbers 1, 2, 3, 4, 5, 6, 7, 8, and 9 exactly once. Solutions are found by methods similar to, but often more complex than, those used above. Sudoku-type puzzles can be reduced to the exact cover problem in ways similar to how the 3 × 3 case was handled. In general, the following constraints must be satisfied: (a) (b) (c) (d) Figure 5

Each cell in the matrix must contain a digit. Each digit must occur in each column. Each digit must occur in each row. Each digit must occur in certain submatrices.

For the 4 × 4 case, this gives a total of 64 constraints (16 for each category of constraint). There are also 64 possible moves (putting any of 4 digits into any of 16 cells), and so the Boolean matrix corresponding to this problem is 64 × 64. In the more common 9 × 9 Sudoku case, the corresponding matrix has 729 rows and 324 columns. (See Exercises 13 and 14.) In spite of these large sizes of these matrices, computer implementations of solution algorithms easily handle them. Dozens of other types of puzzles reduce by similar methods to the exact cover problem, and so can be solved by computers implementing Knuth’s Algorithm X in various ways. For example, the eight-queens puzzle is the problem of putting eight chess queens on an 8 × 8 chessboard such that none of them is able to capture any other using the standard chess queen’s moves. One solution is shown in Figure 6, and there are 91 others. Pentominoes are figures formed from 5 squares connected on one or more edges. There are 12 pentominoes, which are shown in Figure 7 with the standard letters used to identify them. A popular problem is to arrange sets of pentominoes into rectangular or other shaped figures. Two cases are shown in Figures 8 and 9.

Figure 6

L

I

F

T

N

U

V

V U

P

L

N

F

Y L

X W

Z

Y

Z

P

F

W

P U

Figure 7

I

X

X

Figure 8

T

Figure 9

Both the eight-queens puzzle and the pentominoes problems reduce to the exact cover problem, and so can be solved by any algorithm that solves this problem.

93

Logic A computer implementation of Algorithm X involves data structures such as trees and linked lists to facilitate searching and backtracking. One of the most efficient and interesting ways to do this implementation is shown in Donald Knuth’s “Dancing Links” paper. The interested reader should consult this paper for details and other applications of the exact cover problem. Because of the many applications of this algorithm, the underlying problem is often referred to as the open cover problem. Although this section has focused on one family of related problems, the technique of reducing a problem to one that has already been solved is widely used. In later sections some of these cases are pointed out.

6 Exercises 1. Find an exact cover for A = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10} with respect to A1 = {1, 2, 5}, A2 = {3, 5, 8, 9}, A3 = {1, 8, 10}, A4 = {4, 7, 10}, A5 = {3, 7}, and A6 = {1, 2, 6}. 2. Find an exact cover for B = {a, b, c, d, e, f, g, h, i} with respect to B1 = {a, b, d, h}, B2 = {c, f, i}, B3 = {a, e, i}, B4 = {e, g}, and B5 = {a, g, h}. 3. Represent the subsets of A in Exercise 1 in a matrix. 4. Represent the subsets of B in Exercise 2 in a matrix. 5. Apply Algorithm X to the sets in Exercise 1 to produce an exact cover of A. 6. Apply Algorithm X to the sets in Exercise 2 to produce an exact cover of B. In Exercises 7 through 10, consider the problem of filling a 3×3 matrix with the digits 1, 2, and 3, so that each digit appears exactly once in each row and each column. 7. Give the row for the move that places a 3 in position (1, 1).

15. (a) Name the columns where 1’s appear for the move of placing a 4 in position (5, 7). (b) Name the columns where 1’s appear for the move of placing a 6 in the upper left corner of the block in the center of the grid. 16. (a) Name the columns where 1’s appear for the move of placing a 2 in position (3, 8). (b) Name the columns where 1’s appear for the move of placing an 8 in the center position of the block in the center of the bottom row of the grid. 17. Suppose the problem is to cover a 5 × 6 rectangle using some of the twelve pentominoes of Figure 7 exactly once each. (a) What constitutes a move in this situation? (b) Name the columns needed for the Boolean matrix for this problem. 18. Use the results of Exercise 17 to give a matrix that describes the solution shown in Figure 10. I

8. Give the row for the move that places a 2 in position (2, 3).

F X

9. Give the row for the move that places a 1 in position (3, 2).

P

10. Give the row for the move that places a 3 in position (1, 3). 11. For the 4 × 4 Sudoku case, give an example that shows that the 4th category of constraints is necessary. 12. Describe a method for labeling the columns corresponding to the 4th category of constraints in the 4 × 4 Sudoku case.

U

L

Figure 10 The four tiles shown in Figure 11 are used in Exercises 19 through 28. In this situation, tiles may not be flipped or rotated when used to cover a given figure. For example, tile U will always have its dot in the up position.

Exercises 13 through 16 refer to the 9 × 9 Sudoku case. 13. (a) Describe the required categories of constraints. (b) How many columns are needed in each category of constraints? (c) Give a description of column names that correspond to the categories given in part (a). 14. Explain why there need to be 729 rows in the initial Boolean matrix.

94

U

D

R

L

Figure 11 19. If each tile is used exactly once, then only rectangles that are by can be covered. 20. Draw two figures that are not rectangles but are composed of squares and can be covered by using the four tiles exactly once.

Logic 21. (a) How many ways can tile D be placed in rectangle A? (b) How many ways can tile L be placed in rectangle A?

25. Give a matrix that records the initial placement shown in Figure 12. Include the column names.

Figure 12 26. Give a matrix that records the initial placement shown in Figure 13. Include the column names.

A

22. (a) How many ways can tile U be placed in rectangle B? (b) How many ways can tile R be placed in rectangle B?

Figure 13

B

23. Using tiles U, D, R, L exactly once each and rectangle A, what information is required to define a move?

27. Figure 12 cannot be covered with U, D, R, L. Use Algorithm X to prove this.

24. Using tiles U, D, R, L exactly once each and rectangle B, what information is required to define a move?

28. Figure 13 can be covered in two different ways using U, D, R, L exactly once each. Use Algorithm X to prove this.

Tips for Proofs

This chapter provides the formal basis for our proofs, although most proofs are not so formal as the patterns given in Section 3. Two new types of proofs are presented: indirect proofs and induction proofs. Indirect proofs are based either on the pattern (p ⇒ q) ∧ ∼q (proof by contradiction) or on the fact that (p ⇒ q) ≡ (∼q ⇒ ∼p) (prove the contrapositive). There are no hard and fast rules about when to use a direct or indirect proof. One strategy is to proceed optimistically with a direct proof. If that does not lead to anything useful, you may be able to identify a counterexample if the statement is in fact false or start a new proof based on one of the indirect models. Where the difficulty occurs in the attempted direct proof can often point the way to go next. Remember that a certain amount of creativity is required for any proof. Conjectures that are good candidates for proof by induction are ones that involve the counting of whole numbers in some way, either to count something or to describe a pattern. Examples of these are in Section 4, Exercises 11 and 15. Notice that for most of the induction proofs in Section 4, P(k) is used early and then properties of operations and arithmetic are used, but in proving loop invariants, the “arithmetic” comes first, then the use of P(k). In proving conjectures about propositions, try to use the properties of logical operations (see Section 2, Theorem 1 for some of these). Building truth tables should be your second-choice strategy. The proof of a conjecture that is the extension of a known fact usually requires the application of the original fact.

Key Ideas for Review • Statement: declarative sentence that is either true or false,

but not both • Propositional variable: letter denoting a statement • Compound statement: statement obtained by combining

two or more statements by a logical connective • Logical connectives: not (∼), and (∧), or (∨), if then (⇒),

if and only if (⇔) • Conjunction: p ∧ q (p and q)

95

Logic • Disjunction: p ∨ q (p or q) • Predicate (propositional function): a sentence of the form • • •

• • • • • • •

P(x) Universal quantification: ∀x P(x) [For all values of x, P(x) is true.] Existential quantification: ∃x P(x) [There exists an x such that P(x) is true.] Conditional statement or implication: p ⇒ q (if p then q); p is the antecedent or hypothesis and q is the consequent or conclusion Converse of p ⇒ q: q ⇒ p Contrapositive of p ⇒ q: ∼q ⇒ ∼p Equivalence: p ⇔ q Tautology: a statement that is true for all possible values of its propositional variables Absurdity: a statement that is false for all possible values of its propositional variables Contingency: a statement that may be true or false, depending on the truth values of its propositional variables p ≡ q (Logically equivalent statements p and q): p ⇔ q is a tautology

• Methods of proof:

• •

• • • • •

q logically follows from p Rules of inference Modus ponens Indirect method Proof by contradiction Counterexample: single instance that disproves a theorem or proposition Principle of mathematical induction: Let n0 be a fixed integer. Suppose that for each integer n ≥ n0 we have a proposition P(n). Suppose that (a) P(n0 ) is true and (b) If P(k), then P(k + 1) is a tautology for every k ≥ n0 . Then the principle of mathematical induction states that P(n) is true for all n ≥ n0 . Loop invariant: a statement that is true before and after every pass through a programming loop Strong form of mathematical induction Conjecture: a mathematical statement that has not been proved or disproved Exact cover Algorithm X

Chapter Self-Test 1. Why is it important to recognize the converse and the contrapositive of a conditional statement? 2. How does the strong form of induction differ from basic mathematical induction? 3. What mathematical structure previously studied has the same properties as (logical statements, ∨, ∧, ∼)? 4. How does an indirect proof technique differ from a direct proof? 5. What is the structure of a proof by contradiction? 6. Determine the truth value of the given statements if p is true and q is false. (b) ∼p ∨ ∼q (a) ∼p ∧ q 7. Determine the truth value for each of the following statements. Assume x, y ∈ Z. (a) ∀x, y x + y is even. (b) ∃x ∀y x + y is even. 8. Make a truth table for (p ∧ ∼p) ∨ (∼(q ∧ r)).

96

11. Give the truth value of each proposition in Problem 9. 12. The English word “or” is sometimes used in the exclusive sense meaning that either p or q, but not both, is true. Make a truth table for this exclusive or, xor. 13. Let p: An Internet business is cheaper to start, q: I will start an Internet business, and r: An Internet business makes less money. For each of the following write the argument in English sentences and also determine the validity of the argument. r ⇒ (q ⇒ p) (b) p⇒q (a) ∼p q⇒r ∴ (∼r) ∨ (∼q) p ∴ r 14. Suppose that m and n are integers such that n | m and m | n. Are these hypotheses sufficient to prove that m = n? If so, give a proof. If not, supply a simple additional hypothesis that will guarantee m = n and provide a proof.

For Problems 9 through 11, let p: 1 < −1, q: |2| = |−2|, r: −3 < −1, and s: 1 < 3.

15. Prove or disprove by giving a counterexample that the sum of any three consecutive odd integers is divisible by 6.

9. Write the symbolic version of the converse and of the contrapositive for each of the following propositions. (b) (∼r) ∨ (∼s) ⇒ q (a) p ⇒ q (c) q ⇒ p ∨ s 10. Write the converse and the contrapositive of the propositions in Problem 4 as English sentences.

16. Use mathematical induction to prove that 4n −1 is divisible by 3. 17. Use mathematical induction to prove that 1 + 2 + 3 + ··· + n

10 or NUM ≤ 15.

7. (a) True.

(b) True.

(c) True.

(d) False.

9. (a) False.

(b) True.

(c) True.

(d) False.

27. p

q

T T F F

T F T F

11. (d) is the negation. 13. (a) The dish did not run away with the spoon and the grass is wet. (b) The grass is dry or the dish ran away with the spoon. (c) It is not true that today is Monday or the grass is wet.

(∼p ∧ q) F F T T (1)

F F T F (2)

∨p T T T F ↑

(d) Today is Monday or the dish did not run away with the spoon. 15. (a) For all x there exists a y such that x + y is even. (b) There exists an x such that, for all y, x + y is even. 17. (a) It is not true that there is an x such that x is even. (b) It is not true that, for all x, x is a prime number. 19. 14: 15: 16: 17: 18:

(a) False. (a) True. (a) False. (a) False. (a) False.

21. (a) 5 + 3 = 6.

98

(b) True. (b) False. (b) True. (b) True. (b) False.

(c) False.

(b) m + 3 = 6.

(d) False.

29.

p

q

r

(p ∨ q)

∧r

T T T T F F F F

T T F F T T F F

T F T F T F T F

T T T T T T F F (1)

T F T F T F F F ↑

Logic 31.

33.

9. (a) (b) (c) 11. (a)

p

q

r

(p ↓ q)

↓r

T T T T F F F F

T T F F T T F F

T F T F T F T F

F F F F F F T T (1)

F T F T F T F F ↑

p

q

r

(p ↓ q)

↓

(p ↓ r)

T T T T F F F F

T T F F T T F F

T F T F T F T F

F F F F F F T T (1)

T T T T T F F F ↑

F F F F F T F T (2)

35. p

q

T T F F

T F T F

(p ∧ q) p T F T F (1)

F T T F ↑

37. (x = max) or (y ≤ 4). 39. WHILE (item = sought or index ≥ 101) take action Exercise Set 2 1. (a) p ⇒ q. (c) q ⇒ p.

(b) r ⇒ p. (d) ∼r ⇒ p.

3. (a) If I am not the Queen of England, then 2 + 2 = 4. (b) If I walk to work, then I am not the President of the United States. (c) If I did not take the train to work, then I am late. (d) If I go to the store, then I have time and I am not too tired. (e) If I buy a car and I buy a house, then I have enough money. 5. (a) True.

(b) False.

(c) True.

(d) True.

7. (a) If I do not study discrete structures and I go to a movie, then I am in a good mood. (b) If I am in a good mood, then I will study discrete structures or I will go to a movie. (c) If I am not in a good mood, then I will not go to a movie or I will study discrete structures. (d) I will go to a movie and I will not study discrete structures if and only if I am in a good mood.

If 4 > 1 and 2 > 2, then 4 < 5. It is not true that 3 ≤ 3 and 4 < 5. If 3 > 3, then 4 > 1. p

q

T T F F

T F T F

p ⇒ T T T T ↑

tautology (b) p q q ⇒ T T F F

13. 15.

T F T F

T T F T ↑

(q ⇒ p) T T F T

(q ⇒ p) T T F T

contingency Yes. If p ⇒ q is false, then p is true and q is false. Hence p ∧ q is false, ∼(p ∧ q) is true, and (∼(p ∧ q)) ⇒ q is false. No, because if p ⇒ q is true, it may be that both p and q are true, so (p ∧ q) ⇒ ∼q is false. But it could also be that p and q are both false, and then (p ∧ q) ⇒ ∼q is true. The solultions are self-checking. The solution is self-checking. (a) False. (b) True. (c) False. (d) True.

17. 19. 21. 23. p

q

T T F F

T F T F

p

q

r

T T T T F F F F

T T F F T T F F

T F T F T F T F

(p ∧ q)

(p ↓ p)

↓

(q ↓ q)

T F T F F F F T F T F F F T F T (A) (B) Since columns (A) and (B) are the same, the statements are equivalent. 25. (a) Jack did eat fat or he did not eat broccoli. (b) Mary did not lose her lamb and the wolf did not eat the lamb. (c) Tom stole a pie and ran away and the three pigs have some supper. 27.

p

∧

(q ∨ r)

(p ∧ q)

∨

(p ∧ r)

T T T T T T T T T F T T F T T F F F F F F T F F F F T F F F F T F F F F F F F F (A) (B) Because (A) and (B) are the same, the statements are equivalent.

99

Logic 29. p

q

T T F F

T F T F

∼ (p ⇔ q)

(p ∧ ∼ q)

∨

(q ∧ ∼ p)

F T F F F T F T T F T F F T T F T F F F (A) (B) Because columns (A) and (B) are the same, the statements are equivalent. 31. The statement ∀x(P(x) ∧ Q(x)) is true if and only if ∀x both P(x) and Q(x) are true, but this means ∀xP(x) is true and ∀xQ(x) is true. This holds if and only if ∀xP(x) ∧ ∀xQ(x) is true. 33. p

q

T T F F

T F T F

35. p

q

r

T T T T F F F F

T T F F T T F F

T F T F T F T F

T T T F

((p ⇒ q)

23. For n = 41, we have a counterexample. 412 + 41 · 41 + 41 is 41(41 + 41 + 1) or 41 · 83. 25. n3 − n = n(n − 1)(n + 1), the product of three consecutive integers. One of these must be a multiple of 3, so 3 | n3 − n.

31. Let x and y be prime numbers, each larger than 2. Then x and y are odd and their sum is even (Exercise 15). The only even prime is 2, so x + y is not a prime. ∧

(q ⇒ r))

⇒

(p ⇒ r)

Exercise Set 3 Valid: ((d ⇒ t) ∧ ∼t) ⇒ ∼d. Invalid. Valid: ((f ∨ ∼w) ∧ w) ⇒ f . Valid: [(ht ⇒ m) ∧ (m ⇒ hp)] ⇒ [∼hp ⇒ ∼ht]. Invalid. (a) ((p ∨ q) ∧ ∼q) ⇒ p. (b) ((p ⇒ q) ∧ ∼p) ⇒ ∼q. 13. (a) (((p ⇒ q) ∧ (q ⇒ r)) ∧ ((∼q) ∧ r)) ⇒ p. (b) (∼(p ⇒ q) ∧ p) ⇒ ∼q. 15. Suppose m and n are odd. Then there exist integers j and k such that m = 2j + 1 and n = 2k + 1. m + n = (2j + 1) + (2k + 1) = 2j + 2k + 2 = 2(j + k + 1). Since j + k + 1 is an integer, m + n is even. 17. Suppose that m and n are odd. Then there exist integers j and k such that m = 2j + 1 and n = 2k + 1. m · n = 2j · 2k + 2j + 2k + 1 = 2(2jk + j + k) + 1. Since 2jk + j + k is an integer, m · n is odd and the system is closed with respect to multiplication.

100

(b) If A ⊆ B, then A ⊆ A ∩ B. But A ∩ B ⊆ A. Hence A ∩ B = A. If A ∩ B = A, then since A ∩ B ⊆ B, we have A ⊆ B.

29. Valid.

T T T T T T F F T F F F T T T F F T T F T T T T T T F F T T T T T T T T T T T T (1) (3) (2) ↑ (4) 37. Because (p ∧ q) ≡ (q ∧ p), parts (a) and (b) of Theorem 4 say the same thing; that is, either p or q can be considered the first statement. 1. 3. 5. 7. 9. 11.

21. (a) If A ⊆ B, then A ∪ B ⊆ B. But B ⊆ A ∪ B. Hence A ∪ B = B. If A ∪ B = B, then since A ⊆ A ∪ B, we have A ⊆ B.

27. Invalid. Multiplying by x − 1 may or may not preserve the order of the inequality.

q ⇒ (p ∨ q) T T T T ↑

19. If A = B, then, clearly, A ⊆ B and B ⊆ A. If A ⊆ B and B ⊆ A, then A ⊆ B ⊆ A and B must be the same as A.

33. Suppose x + y is rational. Then there are integers a and b such that x + y = ba . Since x is rational, we can write x = dc with integers c and d. But now y = x + y − x = ba − dc . y = ad−bc . Both ad − bc and bd are integers. This is a bd contradiction since y is an irrational number and cannot be expressed as the quotient of two integers. Exercise Set 4 Note: Only the outlines of the induction proofs are given. These are not complete proofs. 1. Basis step: n = 1 P(1): 2(1) = 1(1 + 1) is true. Induction step: P(k): 2 + 4 + · · · + 2k = k(k + 1). P(k + 1): 2 + 4 + · · · + 2(k + 1) = (k + 1)(k + 2). LHS of P(k + 1): 2 + 4 + · · · + 2k + 2(k + 1) = k(k + 1) + 2(k + 1) = (k + 1)(k + 2) RHS of P(k + 1). 3. Basis step: n = 0 P(0): 20 = 20+1 − 1 is true. Induction step: LHS of P(k + 1): 1 + 21 + 22 + · · · + 2k + 2k+1 = (2k+1 − 1) + 2k+1 = 2 · 2k+1 − 1 = 2k+2 − 1. RHS of P(k + 1). 1(1 + 1)(2 + 1) is true. 5. Basis step: n = 1 P(1): 12 = 6 Induction step: LHS of P(k + 1): 12 + 22 + · · · + k2 + (k + 1)2 k(k + 1)(2k + 1) + (k + 1)2 = 6 k(2k + 1) + (k + 1) = (k + 1) 6 k+1 2 (2k + k + 6(k + 1)) = 6 k+1 2 (2k + 7k + 6) = 6

Logic (k + 1)(k + 2)(2k + 3) 6 (k + 1)((k + 1) + 1)(2(k + 1) + 1) . = 6 RHS of P(k + 1).

=

=

a(1 − r ) is true. 1−r Induction step: LHS of P(k+1): a+ar+· · ·+ar k−1 +ar k = a(1 − r k+1 ) a − ar k + ar k − ar k+1 a(1 − r k ) + ar k = = . 1−r 1−r 1−r RHS of P(k + 1).

7. Basis step: n = 1

1

P(1): a =

9. (a) LHS of P(k + 1): 1 + 5 + 9 + · · · + (4(k + 1) − 3) = (2k+1)(k−1)+4(k+1)−3 = 2k2 +3k = (2k+3)(k) = (2(k + 1) + 1)((k + 1) − 1). RHS of P(k + 1). (b) No; P(1): 1 = (2 · 1 + 1)(1 − 1) is false. 11. Basis step: n = 2 P(2): 2 < 2 is true. Induction step: LHS of P(k+1): k+1 < 2k +1 < 2k +2k = 2 · 2k = 2k+1 . RHS of P(k + 1). 2

13. Basis step: n = 5 P(5): 1 + 52 < 25 is true. Induction step: LHS of P(k + 1): 1 + (k + 1)2 = k2 + 1 + 2k+1 < 2k +2k+1 < 2k +k2 +1 < 2k +2k = 2·2k = 2k+1 . RHS of P(k + 1).

k ∩ (Ai ∪ B) ∩ (Ak+1 ∪ B)

i=1 k+1

= ∩ (Ai ∪ B). i=1

RHS of P(k + 1).

21. (a) (k+1)2 +(k+1) = k2 +2k+1+k+1 = k2 +k+2(k+1). Using P(k), k2 +k is odd; 2(k+1) is clearly even. Hence their sum is odd. (b) No, P(1) is false. 23. The flaw is that to carry out the procedure in the induction step, you must have at least three trucks, but the basis step was done for one truck. 25. Basis step: n = 1 P(1): A2 · A = A2+1 is true. Induction step: LHS of P(k + 1): A2 · Ak+1 = A2 (Ak · A) = (A2 · Ak ) · A = A2+k · A = A2+k+1 . RHS of P(k + 1). 27. Basis step: n = 5 P(5): A restaurant bill of $5 can be paid exactly with a $5 bill. Induction step: We use P(j): A restaurant bill of $j can be paid exactly using $2 and $5 bills for j = 5, 6, . . . , k to show P(k + 1). Write k + 1 as 2m + r, 0 ≤ r < 2. If r = 0, then the bill can be paid with m $2 bills. If r = 1, then 2m + r = 2m + 1 = 2(m − 2) + 5. Since k + 1 ≥ 7, m ≥ 2, and m − 2 ≥ 0. Thus a bill of k + 1 dollars can be paid exactly with (m − 2) $2 bills and a $5 bill.

15. Basis step: n = 0 A = { } and P(A) = {{ }}, so |P(A)| = 20 and P(0) is true. Induction step: Use P(k): If |A| = k, then |P(A)| = 2k to show P(k + 1): If |A| = k + 1, then |P(A)| = 2k+1 . Suppose that |A| = k + 1. Set aside one element x of A. Then |A − {x}| = k and A − {x} has 2k subsets. These subsets are also subsets of A. We can form another 2k subsets of A by forming the union of {x} with each subset of A − {x}. None of these subsets are duplicates. Now A has 2k + 2k , or 2k+1 , subsets.

29. Basis step: n = 1 P(1): If p is prime and p | a1 , then p | a is true. Induction step: If p | ak+1 , then p | ak · a and either p | ak or p | a. If p | ak , then (using P(k)), p | a.

17. Basis step: n = 1 P(1): A1 = A1 is true. Induction step: LHS of P(k + 1):

33. Basis step: n = 1 P(1): x − y divides x − y is true. Induction step: xk+1 − yk+1 = x · xk − y · yk = x · xk − x · yk + x · yk − y · yk = x(xk − yk ) + yk (x − y). This rewriting gives an expression where each term is divisible by x − y and so the sum is as well.

k+1

∩ Ai =

i=1

k ∩ Ai ∩ Ak+1

i=1 k

= ∩ Ai ∪ Ak+1 (De Morgan’s laws) i=1 k = ∪ Ai ∪ Ak+1 i=1

k+1

= ∪ Ai . RHS of P(k + 1). i=1

19. Basis step: n = 1 P(1): A1 ∪ B = A1 ∪ B is true. Induction step: LHS of P(k + 1): k+1 k ∩ Ai ∪ B = ∩ Ai ∩ Ak+1 ∪ B i=1 i=1 k ∩ Ai ∪ B ∩ (Ak+1 ∪ B) = i=1

(distributive property)

31. (a) 5. (b) Basis step: n = 5 P(5): 25 > 52 is true. Induction step: LHS of P(k + 1): 2k+1 = 2 · 2k = 2k + 2k > k2 + k2 > k2 + 3k > k2 + 2k + 1 (since k ≥ 5) = (k + 1)2 . RHS of P(k + 1).

35. Loop invariant check: Basis step: n = 0 P(0): X−Z0 +W0 = Y is true, because Z0 = X and W0 = Y . Induction step: LHS of P(k + 1): X − Zk+1 + Wk+1 = X − (Zk − 1) + (Wk − 1) = X − Zk + Wk = Y . RHS of P(k + 1). Exit condition check: W = 0 X − Z + W = Y yields X − Z = Y or Z = X − Y . 37. Loop invariant check: Basis step: n = 0 P(0): Z0 + (X × W0 ) = X × Y is true, because Z0 = 0 and W0 = Y . Induction step: LHS of P(k + 1): Zk+1 + (X × Wk+1 ) = Zk + X + (X × (Wk − 1)) = Zk + (X × Wk ) = X × Y . RHS of P(k + 1). Exit condition check: W = 0 Z + (X × W) = X × Y yields Z = X × Y .

101

Logic Loop invariant check: Basis step: n = 0 P(0): Z0 + (X × Y × W0 ) = X + Y 2 is true, because Z0 = X × Y and W0 = Y − 1. Induction step: LHS of P(k + 1): Zk+1 + (X × Y × Wk+1 ) = Zk +X×Y +(X×Y ×(Wk −1)) = Zk +X×Y ×Wk = X+Y 2 . RHS of P(k + 1). Exit condition check: W = 0 Z + X × Y × W = X + Y 2 yields Z = X + Y 2 . 39. Loop invariant check: Basis step: n = 0 P(0): Z0 + (X × W0 ) = Y + X2 is true, because Z0 = Y and W0 = X. Induction step: LHS of P(k + 1): Zk+1 + (X × Wk+1 ) = (Zk + X) + (X × (Wk − 1)) = Zk + X × Wk = Y + X2 RHS of P(k + 1). Exit condition check: W = 0 Z + (X × W) = Y + X2 yields Z = Y + X2 . Loop invariant check: Basis step: n = 0 P(0): Z0 + (Y × W0 ) = X2 + Y 2 is true, because Z0 = Y + X2 and W0 = Y − 1. Induction step: LHS of P(k + 1): Zk+1 + (Y × Wk+1 ) = Zk + Y + Y × (Wk − 1) = Zk + Y × Wk = X2 + Y 2 . RHS of P(k + 1). Exit condition check: W = 0 Z + Y × W = X2 + Y 2 yields Z = X2 + Y 2 . Exercise Set 5 1. Let 2k be any even number. (2k)3 = 8k3 = 2(4k3 ) is clearly an even number. 3. (a) Let n, n+1, and n+2 be any three consecutive integers. n + (n + 1) + (n + 2) = 3n + 3 = 3(n + 1), which is clearly divisible by 3. (b) Let n, n + 1, n + 2, and n + 3 be any four consecutive integers. n + (n + 1) + (n + 2) + (n + 3) = 4n + 6, which is not divisible by 4. 5. The quotient of a nonzero rational number and an irrational number is irrational. Proof: Let r = ba , a = 0, represent a rational number and j, an irrational number. Suppose that jr is rational, then = j; this is ∃c, d ∈ Z such that jr = dc = ba /j. But then ad bc a contradiction. 7. A ∩ B ∩ C = A ∪ B ∪ C Proof: A ∩ B ∩ C = A ∩ (B ∩ C) by the associative property of ∪. = A ∪ (B ∩ C) by De Morgan’s law = A ∪ (B ∪ C) by De Morgan’s law =A∪B∪C by the associative property of ∪ 9. Let A1 , A2 , . . . , Am be m Boolean matrices of the same size. Then A1 ∧ (A2 ∨ · · · ∨ Am ) = (A1 ∧ A2 ) ∨ (A1 ∧ A3 ) ∨ · · · ∨ (A1 ∧ Am ). Proof (by induction): Basis Step: n = 3

102

Induction Step: LHS of P(k + 1): A1 ∧ (A2 ∨ · · · ∨ Ak ∨ Ak+1 ) = A1 ∧ ((A2 ∨ · · · ∨ Ak ) ∨ Ak+1 ) = ((A1 ∧ A2 ) ∨ (A1 ∧ A3 ) ∨ · · · ∨(A1 ∧ Ak )) ∨ (A1 ∧ Ak+1 ) = (A1 ∧ A2 ) ∨ (A1 ∧ A3 ) ∨ · · · ∨ (A1 ∧ Ak+1 ). RHS of P(k + 1). 11. (a) ∼(p ∧ q ∧ r) ≡ ∼p ∨ ∼q ∨ ∼r. Proof: ∼(p ∧ q ∧ r) ≡ ∼(p ∧ (q ∧ r)) since ∧ is associative ≡ ∼p ∨ ∼(q ∧ r) ≡ ∼p ∨ (∼q ∨ ∼r) ≡ ∼p ∨ ∼q ∨ ∼r (b) ∼(p1 ∧ p2 ∧ · · · ∧ pn ) ≡ ∼p1 ∨ ∼p2 ∨ · · · ∨ ∼pn Basis step: n = 2 ∼(p ∧ q) ≡ ∼p ∨ ∼q by Theorem 1, part 11, Section 2. Induction step: LHS of P(k + 1): ∼(p1 ∧ p2 ∧ · · · ∧ pk ∧ pk+1 ) ≡ ∼((p1 ∧ p2 ∧ · · · ∧ pk ) ∧ pk+1 ) ≡ ∼(p1 ∧ p2 ∧ · · · ∧ pk ) ∨ ∼pk+1 ≡ (∼p1 ∨ ∼p2 ∨ · · · ∨ ∼pk ) ∨ ∼pk+1 ≡ ∼p1 ∨ ∼p2 ∨ · · · ∨ ∼pk+1 . RHS of P(k + 1). 13. Let A = aij and B = bij be two r × r diagonal matrices; thus, aij = 0 if i = j and bij = 0 if i = j. The ij-entry in AB is ai1 b1j + ai2 b2j + · · · + air brj , but this sum is aii bii . Similarly, the ij-entry in BA is bii aii . Hence, AB = BA. 15. recursive dn = dn−1 + 6, d1 = 3; explicit dn = 3(2n − 1) 17. Basis step: n = 1 3 = 3(2(1) − 1) = 3 · 12 is true. Induction step: LHS of P(k + 1): 3 + 9 + 15 + · · · + 3(2(k + 1) − 1) = 3 + 9 + 15 + · · · + 3(2k − 1) + 3(2k + 1) = 3k2 + 6k + 3 = 3(k + 1)2 RHS of P(k + 1). 19. 1, 3, 4, 7, 11, 18, 29, 47, 76, 123. 21. The sums are 1, 4, 8, 15, 26, 44, 73, 120. A reasonable conjecture would be g1 + g2 + · · · + gn = gn+2 − 3. 23. Proof: Basis step: n = 1 g2 = 3 = g3 − 1 = g2(1)+1 − 1. Induction step: LHS of P(k + 1): g2 + g4 + · · · + g2k + g2(k+1) = g2k+1 − 1 + g2(k+1) = g2k+1 + g2k+2 − 1 = g2k+3 − 1 = g2(k+1)+1 − 1. RHS of P(k + 1). 25. Proof: Basis step: n = 1 g1 = 1 = g2 − 2 = g2(1) − 2 Induction step: LHS of P(k + 1): g1 + g3 + · · · + g2k−1 + g2(k+1)−1 = g2k − 2 + g2(k+1)−1 = g2k + g2k+1 − 2 = g2k+2 − 2 = g2(k+1) − 2. RHS of P(k + 1). 27. Further experimentation suggests the conjecture that the sum of k consecutive integers is divisible by k if and only if k is odd.

Logic Exercise Set 6

1. A2 , A6 , A4 . ⎡

⎡

1 ⎢0 ⎢ ⎢1 3. ⎢ ⎢0 ⎣0 1

1 0 0 0 0 1

0 1 0 0 1 0

0 0 0 1 0 0

1 1 0 1 0 0 5. Pick A4 . ⎣ 0 0 1 1 0 1 1 1 0 0 1 0 Pick A2 . 1 1 1 A6 An exact cover is A4 , A2 , A6 . 7. 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 9. 0 0 0 0 0 0 0 1 0 0 0 0 0 1 0 0

1 1 0 0 0 0

0 0 0 0 0 1

⎤ 0 A1 1 ⎦ A2 0 A6

0 0 0 1 1 0

0 1 1 0 0 0

0 1 0 0 0 0

0 0

0 0

0 1

0 0

0 0 0

0 0

0 0

0 0

1 0

0 0 0

U 1 ⎢1 ⎢ ⎢1 ⎢ ⎢1 ⎢ ⎢0 ⎢ ⎢0 ⎢0 ⎢ ⎢0 ⎢ ⎣0 0 ⎡

D 0 0 0 0 1 1 1 1 0 0

R 0 0 0 0 0 0 0 0 1 0

(b) 81; 9 × 9; 9 × 9; 9 × 9 for a total of 324 columns. (c) 1. (1, 1), (1, 2), (1, 3), . . . , (1, 9), (2, 1), . . . , (9, 9) 2. a 1 in column 1, a 1 in column 2, . . . , a 1 in column 9, a 2 in column 1, . . . , a 9 in column 9 3. a 1 in row 1, a 1 in row 2, . . . , a 1 in row 9, a 2 in row 1, . . . , a 9 in row 9 4. a 1 in block 1, a 1 in block 2, . . . , a 1 in block 9, a 2 in block 1, . . . , a 9 in block 9 15. (a) (5, 7); 4 in column 7; 4 in row 5; 4 in block 6 (using the numbering of Exercise 13(c) 4) (b) (4, 4); 6 in column 4; 6 in row 4; 6 in block 5

1 2 3 4 2 3 4 1 11. One possible solution is 3 4 1 2 4 1 2 3 13. (a) 1. Each of the 81 cells must contain a digit. 2. Each digit 1 through 9 must appear in each of the nine columns. 3. Each digit 1 through 9 must appear in each of the nine rows.

1 2 3 4 5 25. 6 7 8 9 10

4. Each digit 1 through 9 must appear in each of the nine 3 × 3 blocks.

⎤ 0 0⎥ ⎥ 1⎥ ⎥. 1⎥ 0⎦ 0

L 0 0 0 0 0 0 0 0 0 1

27. Use the matrix from Exercise 25 and choose row 9 for Algorithm X. ⎡ ⎤ 1 1 0 0 1 0 0 0 1 0 2 ⎢1 0 0 0 1 0 0 0 1⎥ ⎢ ⎥ 5 ⎢0 1 0 1 0 0 0 1 0⎥ ⎣ 6 0 1 0 0 1 0 0 0 1⎦ 10 0 0 1 1 1 0 0 0 0

17. (a) A move consists of placing F, I, L, N, P, T, U, V, W, X, Y, or Z in five specified squares. Squares may be labelled by their positions (1, 1), (1, 2), . . . , (1, 6), (2, 1), . . . , (5, 6). (b) One solution is F, I, L, N, P, T, U, V, W, X, Y, Z, (1, 1), (1, 2), . . . , (1, 6), (2, 1), . . . , (5, 6) for a total of 42 columns. 19. 2 by 4

21. (a) 6

(b) 4

23. A move is placing U, D, R, or L in two specified squares. Squares may be labelled by position: (1, 1), (1, 2), (2, 1), . . . , (4, 2). (1, 1) 1 0 0 0 1 0 0 0 0 1

(1, 2) 0 1 0 0 0 1 0 0 0 1

(1, 3) 0 0 1 0 0 0 1 0 0 0

(1, 4) 0 0 0 1 0 0 0 1 0 0

(2, 1) 1 0 0 0 1 0 0 0 0 0

(2, 2) 0 1 0 0 0 1 0 0 0 0

(2, 3) (2, 4) ⎤ 0 0 0 0 ⎥ ⎥ 1 0 ⎥ ⎥ 0 1 ⎥ ⎥ 0 0 ⎥ ⎥. 0 0 ⎥ 1 0 ⎥ ⎥ 0 1 ⎥ ⎥ 1 1 ⎦ 0 0

The two columns of zeros signal that there is no solution using row 9, but row 9 represents an initial placement. Hence, there is no solultion with this initial placement.

Answers to Chapter Self-Tests 1. The converse is not equivalent to the original statement, but the contrapositive is. In some cases, the contrapositive may be easier to prove than the original statement.

2. In the strong form of induction all statements P(n0 ), P(n0 + 1), . . . , P(k) may be used to show P(k + 1), not just P(k).

103

Logic 3. The mathematical structure of sets, union, intersection, and complement has the same properties as (logical statements, ∨, ∧, ∼). 4. An indirect proof proves the contrapositive of the statement or assumes the statement is false and derives a contradiction. 5. A proof by contradiction proceeds by assuming the negation of the conclusion of the statement to be proved. Then definitions, previous theorems, and commonly known facts are used to derive a contradiction. 6. (a) False.

(b) True.

7. (a) False.

(b) True.

8. p

q

r

(p ∧

T T T T F F F F

T T F F T T F F

T F T F T F T F

F F F F F F F F (1)

9. (a) q ⇒ p.

∼p)

∨

(∼ (q ∧ r))

F T T T F T T T ↑

F T T T F T T T (3)

T F F F T F F F (2)

∼q ⇒ ∼p.

(b) q ⇒ (∼r) ∨ (∼s). ∼q ⇒ (r ∧ s). (c) (p ∨ s) ⇒ q. (∼p ∧ ∼s) ⇒ ∼q. 10. (a) If |2| = |−2|, then 1 < −1. If |2| = |−2|, then 1 ≥ −1. (b) If |2| = |−2|, then either −3 ≥ −1 or 1 ≥ 3. If |2| = |−2|, then −3 < −1 and 1 < 3. (c) If 1 < −1 or 1 < 3, then |2| = |−2|. If 1 ≥ −1 and 1 ≥ 3, then |2| = |−2|. 11. (a) False; True.

13. (a) If an Internet business makes less money, then if I start an Internet business, then an Internet business is cheaper to start. An Internet business is not cheaper to start. Therefore, either an Internet business does not make less money or I do not start an Internet business. Valid. (b) If an Internet business is cheaper to start, then I will start an Internet business. If I start an Internet business, then an Internet business will make less money. An Internet business is cheaper to start. Therefore, an Internet business will make less money. Valid. 14. No, consider 6 | −6 and −6 | 6. If both m and n are positive (or negative), then n | m and m | n guarantees that n = m. 15. Consider 7, 9, and 11. These are three consecutive odd integers whose sum is not divisible by 6. 16. Basis step: n = 0. P(0) : 40 − 1 is divisible by 3 is true since 3 | 0. Induction step: We use P(k): 3 divides 4k − 1 to show P(k + 1): 3 divides 4k+1 − 1. Consider 4k+1 − 1 = 4(4k − 1) + 3. By P(k), 3 | (4k − 1) and we have 4k+1 − 1 = 3(a + 1) where a = 4k − 1. So 3 | (4k+1 − 1). (1 + 1)2 17. Basis step: n = 1. P(1) : 1 < is true. 2 Induction step: We use P(k): 1 + 2 + 3 + ··· + k < to show P(k + 1): 1 + 2 + · · · + (k + 1)

b. Similarly, the complement of S is 5. (a) Draw the digraph of R. (b) Give MR . 10. Let B = {a, b, c, d} and R = {(a, a), (a, b), (b, c), (c, d), (d, b)}. (a) Draw the digraphs of R and R2 . (b) Give MR and MR2 . (c) Give MR∞ . 11. Determine whether the relation R on the set A is reflexive, irreflexive, symmetric, asymmetric, antisymmetric, or transitive, if A = Z+ ; x R y if and only if x ≤ 3y.

12. Let D = {1, 2, 3, 4, 5, 6} and R be the relation on D whose matrix is ⎡ ⎤ 1 0 0 0 0 1 ⎢0 1 1 0 1 0⎥ ⎢ ⎥ ⎢0 0 0 1 0 1⎥ MR = ⎢ ⎥. ⎢1 0 0 1 0 1⎥ ⎣0 0 1 0 1 0⎦ 0 1 0 0 1 1 Determine whether R is reflexive, irreflexive, symmetric, asymmetric, antisymmetric, or transitive. 13. Suppose R is a relation on a set A and that R is asymmetric. Can R also be antisymmetric? Must R be antisymmetric? Explain your answers. 14. Let B = {1, 2, 3, 4, 5}, A = B × B, and define R on A as follows: (u, v) R (x, y) if and only if u − v = x − y. (a) Prove that R is an equivalence relation. (b) Find [(2, 3)]. (c) Compute A/R. 15. The following arrays describe a relation R on the set A = {1, 2, 3, 4}. Give the matrix of R. VERT = [5, 3, 1, 8] TAIL = [3, 3, 2, 2, 1, 1, 4, 4] HEAD = [1, 4, 1, 3, 2, 3, 4, 2] NEXT = [2, 0, 4, 0, 6, 0, 0, 7] 16. Let R and S be relations on {a, b, c, d, e} where R = {(a, b), (a, c), (b, c), (c, e), (e, a), (a, a), (d, c)} and S = {(a, a), (a, b), (b, a), (c, c), (c, d), (d, e), (b, e), (e, d)}.

195

Relations and Digraphs (b) Compute R ◦ S. (a) Give R−1 . 17. Let R = {(1, 4), (2, 1), (2, 5), (2, 4), (4, 3), (5, 3), (3, 2)}.

Use Warshall’s algorithm to find the matrix of the connectivity relation based on R.

Experiment 4 Equivalence relations and partial orders are defined as relations with certain properties. In this experiment, you will investigate compatibility relations that are also defined by the relation properties they have. A compatibility relation is a relation that is reflexive and symmetric. Every equivalence relation is a compatibility relation, but here you will focus on compatibility relations that are not equivalence relations. Part I. 1. Verify that the relation R on A is a compatibility relation. (a) A is the set of students at your college; x R y if and only if x and y have taken the same course. (b) A is the set of all triangles; x R y if and only if x and y have an angle with the same measure. (c) A = {1, 2, 3, 4, 5}; R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (2, 3), (3, 2), (4, 1), (1, 4), (2, 4), (4, 2), (1, 2), (2, 1), (4, 5), (5, 4), (1, 3), (3, 1)}. 2. In Part I.1(c), the relation is given as a set of ordered pairs. A relation can also be represented by a matrix or a digraph. Describe how to determine if R is a compatibility relation using its (a) Matrix. (b) Digraph. 3. Give another example of a compatibility relation that is not an equivalence relation. Part II. Every relation has several associated relations that may or may not share its properties. 1. If R is a compatibility relation, is R−1 , the inverse of R, also a compatibility relation? If so, prove this. If not, give a counterexample. 2. If R is a compatibility relation, is R, the complement of R, also a compatibility relation? If so, prove this. If not, give a counterexample. 3. If R and S are compatibility relations, is R ◦ S also a compatibility relation? If so, prove this. If not, give a counterexample. Part III. In Section 5, we showed that each equivalence relation R on a set A gives a partition of A. A compatibility relation R on a set A gives instead a covering of A. A covering of A is a set of subsets of A, {A1 , A2 , . . . , Ak }, k such that Ai = A. We define a maximal compatibility block to be i=1

a subset B of A with each element of B related by R to every other element of B, and no element of A − B is R-related to every element of B. For example, in Part I.1(c), the sets {1, 2, 3} and {1, 2, 4} are maximal compatibility blocks. The set of all maximal compatibility blocks relative to a compatibility relation R forms a covering of A. 1. Give all maximal compatibility blocks for the relation in Part I.1(c). Verify that they form a covering of A.

196

Relations and Digraphs 2. Describe the maximal compatibility blocks for the relation in Part I.1(b). The set of all maximal compatibility blocks form a covering of A. Is this covering also a partition for this example? Explain. 3. The digraph of a compatibility relation R can be simplified by omitting the loop at each vertex and using a single edge with no arrow between related vertices. (a) Draw the simplified graph for the relation in Part I.1(c). (b) Describe how to find the maximal compatibility blocks of a compatibility relation, given its simplified graph. 4. Find the covering of A associated with the relation whose simplified graph is given in (a) Figure 1. (b) Figure 2. (c) Figure 3. 1

1

2

2 3

3

4 6

3

Figure 1

5

4

1

5

4

5 7

2

6

Figure 2

6

Figure 3

5. Given the following covering of A, produce an associated compatibility relation R; that is, one whose maximal compatibility blocks are the elements of the covering. {{1, 2}, {1, 3, 6, 7}, {4, 5, 11}, {5, 10}, {8, 5}, {2, 8, 9}, {3, 9}, {9, 10}} Is there another compatibility relation that would produce the same covering of A?

Coding Exercises 1. Write a program CROSS with input positive integers m and n and output the set A × B where A = {1, 2, 3, . . . , m} and B = {1, 2, 3, . . . , n}. 2. (a) Write a subroutine that has as input the matrix of a relation and determines whether the relation is reflexive. (b) Write a subroutine that has as input the matrix of a relation and determines whether the relation is symmetric. 3. Write a program that has as input the matrix of a relation and determines whether the relation is an equivalence relation. 4.

Let R and S be relations represented by matrices MR and MS , respectively. Write a subroutine to produce the matrix of (b) R ∩ S (c) R ◦ S (a) R ∪ S

5.

Let R be a relation represented by the matrix MR . Write a subroutine to produce the matrix of (b) R (a) R−1

197

Relations and Digraphs

Answers to Odd-Numbered Exercises Exercise Set 1 1. (a) x is 4. 3. (a) 5. (a) (b) 7. (a)

(b) y is 3.

x is 4; y is 6.

(b) x is 4; y is 2.

{(a, 4), (a, 5), (a, 6), (b, 4), (b, 5), (b, 6)}. {(4, a), (5, a), (6, a), (4, b), (5, b), (6, b)}.

{(Fine, president), (Fine, vice-president), (Fine, secretary), (Fine, treasurer), (Yang, president), (Yang, vice-president), (Yang, secretary), (Yang, treasurer)}. (b) {(president, Fine), (vice-president, Fine), (secretary, Fine), (treasurer, Fine), (president, Yang), (vice-president, Yang), (secretary, Yang), (treasurer, Yang)}. (c) {(Fine, Fine), (Fine, Yang), (Yang, Fine), (Yang, Yang)}.

9. gs, ds, gc, dc, gv, dv. 11. (Outline) Basis step: n = 1. P(1): If |A| = 3 and |B| = 1, then |A × B| = 3. A × B = {(a1 , b1 ), (a2 , b1 ), (a3 , b1 )}. Clearly, |A × B| = 3. Induction step: Suppose |B| = k > 1. Let x ∈ B and C = B − {x}. Then |C| = k − 1 ≥ 1 and using P(k), we have |A × C| = 3(k − 1). |A × {x}| = 3. Since (A×C)∩(A×{x}) = { } and (A×C)∪(A×{x}) = A×B, |A × B| = 3(k − 1) + 3 or 3k. 13.

y (3, 5) (1, 2)

(2, 1)

(1, 0)

x

(2, 0)

Exercise Set 2

(3, –4)

15.

y (1, 3)

(5, 3)

x (1, –2)

(5, –2)

17. Let (x, y) ∈ A × B, then x ∈ A and y ∈ B. Since A ⊆ C and B ⊆ D, x ∈ C and y ∈ D. Hence (x, y) ∈ C × D. 19. One answer is project(select Employees[Department = Human Resources]) [Last Name]. 21. One answer is project(select Employees[Department = Research]) [Years with Company]. 23. (a) Yes.

198

(b) No.

25. An exact cover of T must consist of subsets chosen from a specified collection of subsets of T . A partition of T may use any subsets of T . In both cases, the union of the subsets must form T . 27. Answers will vary. 29. No. |A| = 26. 31. {{1}, {2}, {3}}, {{1}, {2, 3}}, {{2}, {1, 3}}, {{3}, {1, 2}}, {{1, 2, 3}}. 33. 3. There are three 2-element partitions listed in the solution to Exercise 31. 35. 6. 37. Let (x, y) ∈ A × (B ∪ C). Then x ∈ A, y ∈ B ∪ C. Hence (x, y) ∈ A × B or (x, y) ∈ A × C. Thus A × (B ∪ C) ⊆ (A × B) ∪ (A × C). Let (x, y) ∈ (A × B) ∪ (A × C). Then x ∈ A, y ∈ B or y ∈ C. Hence y ∈ B ∪ C and (x, y) ∈ A×(B ∪C). So, (A×B)∪(A×C) ⊆ A×(B ∪C). 39. Let {B1 , B2 , . . . , Bm } be a partition of B. Form the set {B1 ∩ A, B2 ∩ A, . . . , Bm ∩ A}. Delete any empty intersections from this set. The resulting set is a partition of A. If a ∈ A, then a ∈ B. We know that as an element of B, a is in exactly one of the Bi and hence in exactly one of the Bi ∩ A. 41. (a) Number of subsets in partition # of elements 1 2 3 4 5 6 1 1 2 1 1 3 1 3 1 4 1 7 6 1 5 1 15 25 10 1 6 1 31 90 65 10 1 (b) It appears that each entry in the S(n, 2)-column can be calculated as 2n−1 − 1. 1. (a) No. (b) No. (c) Yes. (e) Yes. (f) Yes. (d) Yes. 3. (a) No. (b) No. (c) Yes. (e) No. (f) Only if n = 1. (d) Yes. 5. Domain: {daisy, violet, rose, daffodil}, Range: {red, pink, purple, white}; ⎡ ⎤ 1 0 0 0 0 0 0 ⎢0 0 1 0 0 0 0⎥ ⎢ ⎥ ⎢ 0 0 0 0 0 1 0 ⎥. ⎣0 1 0 0 0 0 0⎦ 0 0 0 0 0 0 0 7. Domain: {1, 2, 3, 4, 8}, Range: {1, 2, 3, 4, 8}; ⎡ ⎤ 1 0 0 0 0 1 2 ⎢0 1 0 0 0⎥ ⎢ ⎥ ⎢ 0 0 1 0 0 ⎥. 4 ⎣0 0 0 1 0⎦ 0 0 0 0 1 3 5

Relations and Digraphs 9. Domain: {1, 2, 3, 4, 6}, Range: {1, 2, 3, 4, 6}; ⎡

1 ⎢1 ⎢ ⎢1 ⎣1 1

0 1 0 1 1

0 0 1 0 1

0 0 0 1 0

1

⎤

0 0⎥ ⎥ 0 ⎥. 0⎦ 1

Exercise Set 3 1. 1, 2

2

4

13. (a) No.

(b) No.

(c) Yes.

(d) Yes.

(e) No.

(f) No.

15. Dom(R) = [−5, 5], Ran(R) = [−5, 5]. 19. (a) {3}.

(b) {1, 2, 3, 6}.

(c) {1, 2, 4, 3, 6}.

(b) {2, 4}. (c) {3, 2, 4}. (d) {2, 4}.

21. a R b if and only if 0 ≤ a ≤ 3 and 0 ≤ b ≤ 2. 23. R = {(1, 1), (1, 2), (1, 4), (2, 2), (2, 3), (3, 3), (3, 4), (4, 1)}.

5. One is 6, 4, 1, 6. ⎡ 0 0 1 1 ⎢0 0 1 1 ⎢ ⎢1 0 1 1 7. ⎢ ⎢0 1 1 1 ⎣0 0 0 0 1 0 1 0

11. (a) a, c, d, c a, b, b, f

3

25. R = {(1, 2), (2, 2), (2, 3), (3, 4), (4, 4), (5, 1), (5, 4)}. ⎡ ⎤ 0 1 0 0 0 ⎢0 1 1 0 0⎥ ⎢ ⎥ ⎢ 0 0 0 1 0 ⎥. ⎣0 0 0 1 0⎦ 1 0 0 1 0 1 1 1

2 2 2

3 1 1

4 3 1

5 0 2

29. The in-degree of a vertex is the number of ones in the column labeled by that vertex. The out-degree of a vertex is the number of ones in the row labeled by that vertex. 31. {(2, 3), (3, 6)}. 33. Delete any vertex labeled by an element of A − B. Then delete any edges that do not point to a vertex. 35. (a) The elements of R(ak ) are those elements of A that can be reached from ak in one step. (b) The elements of R({a1 , aj , an }) are those elements of A that can be reached from a1 , aj , or an in one step. 37. 2mn .

3, 4

4, 3

3, 3, 4, 5 3, 4, 3, 4

4, 5

4, 1

6, 4.

3, 4, 1, 6 3, 3, 4, 1

⎤ 0 0⎥ ⎥ 0⎥ ⎥. 1⎥ 0⎦ 0

b, f

c, d

a, c, d, b a, b, f, d.

c, e

a, c, e, f

d, c

d, b

a, b, b, b

(b) In addition to those in part (a), b, b, b, b b, b, b, f b, b, f, d b, f, d, b b, f, d, c c, d, c, d c, d, c, e c, d, b, b c, d, b, f c, e, f, d d, c, d, c d, c, e, f d, b, b, b d, b, b, f d, b, f, d d, c, e, f e, f, d, b e, f, d, c f, d, c, d f, d, c, e f, d, b, b f, d, b, f .

2

4

0 0 1 0 0 1

9. a, c a, b b, b e, f f, d.

1

27. Vertex In-degree Out-degree

3, 3

(b) In addition to those in part (a), 1, 2, 3, 3 1, 2, 3, 4 1, 6, 4, 1 1, 6, 4, 5 2, 3, 3, 3 2, 3, 3, 4 2, 3, 4, 3 2, 3, 4, 5 4, 1, 2, 3 4, 1, 6, 4 6, 4, 3, 3 6, 4, 3, 4 6, 4, 1, 2 6, 4, 1, 6 1, 6, 4, 3 2, 3, 4, 1 4, 3, 3, 3 4, 3, 4, 3 4, 3, 4, 1 4, 3, 4, 5 4, 3, 3, 4.

3

{3, 5, 7, 9}, Range: {2, 4, 6, 8}; ⎤ 0 0 0 0⎥ ⎥ 0 0 ⎥. 1 0⎦ 1 1

17. (a) {1, 3}.

2, 3

3. (a) 3, 3, 3, 3 3, 3, 4, 3 3, 4, 1, 2 3, 4, 3, 3 3, 3, 3, 4. 6

11. Domain: ⎡ 0 0 ⎢1 0 ⎢ ⎢1 1 ⎣1 1 1 1

1, 6

13. One is d, b, f , d. 15.

a c b

d

e f

⎡

0 ⎢0 ⎢ ⎢0 17. (a) ⎢ ⎢0 ⎣0 0

1 1 1 1 1 1

1 1 1 1 1 1

1 1 1 1 1 1

1 1 1 1 1 1

⎤ 1 1⎥ ⎥ 1⎥ ⎥. 1⎥ 1⎦ 1

(b) {(a, c), (a, d), (a, b), (a, e), (a, f), (b, b), (b, c), (b, d), (b, e), (b, f), (c, b), (c, c), (c, d), (c, e), (c, f), (d, b), (d, c), (d, d), (d, e), (d, f), (e, b), (e, c), (e, d), (e, e), (e, f), (f, b), (f, c), (f, d), (f, e), (f, f)}.

199

Relations and Digraphs 19. xi R∗ xj if and only if xi = xj or xi Rn xj for some n. The i, jth entry of MR∗ is 1 if and only if i = j or the ∞

i, jth entry of MRn is 1 for some n. Since R∞ = ∪ Rk , the k=1

i, jth entry of MR∗ is 1 if and only if i = j or the i, jth entry of MR∞ is 1. Hence MR∗ = In ∨ MR∞ . 21. 1, 7, 5, 6, 7, 4, 3.

35. (a) One answer is {(a, a), (b, b), (c, c), (d, d)}.

23. 2, 3, 5, 6, 7, 5, 6, 4.

(b) One answer is {(a, a), (b, b), (c, c), (d, d), (a, b)}.

25. 7, 4, 3, 5, 6, 7 is an answer.

37. (a) One answer is {(a, a), (b, b), (c, c), (d, d)}.

27. The ij-entry of MR · MR is the number of paths from i to j of length two, because it is also the number of k’s such that aik = bkj = 1. 29. Direct; Boolean multiplication. 31. Suppose each vertex has out-degree at least one. Choose a vertex, say vi . Construct a path R vi , vi+1 , vi+2 , . . . . This is possible since each vertex has an edge leaving it. But there are only a finite number of vertices so for some k and j, vj = vk and a cycle is created.

(b) One answer is {(a, b), (b, c), (a, c)}. Exercise Set 5 1. Yes.

3. Yes.

5. No.

7. No.

9. Yes.

11. Yes.

13. {(a, a), (a, c), (a, e), (c, a), (c, c), (c, e), (e, a), (e, c), (e, e), (b, b), (b, f), (b, d), (d, b), (d, d), (d, f), (f, b), (f, d), (f, f)}.

1. Reflexive, symmetric, transitive.

15. {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9), (10, 10), (1, 3), (3, 1), (1, 5), (5, 1), (1, 7), (7, 1), (1, 9), (9, 1), (3, 5), (5, 3), (3, 7), (7, 3), (3, 9), (9, 3), (5, 7), (7, 5), (5, 9), (9, 5), (7, 9), (9, 7), (2, 4), (4, 2), (2, 6), (6, 2), (2, 8), (8, 2), (2, 10), (10, 2), (4, 6), (6, 4), (4, 8), (8, 4), (4, 10), (10, 4), (6, 8), (8, 6), (6, 10), (10, 6), (8, 10), (10, 8)}.

3. None.

17. {{. . . , −3, −1, 1, 3, 5, . . . }, {. . . , −4, −2, 0, 2, 4, . . . }}.

5. Irreflexive, symmetric, asymmetric, antisymmetric, transitive.

19. (a) R is reflexive because a2 + b2 = a2 + b2 . R is clearly symmetric. R is transitive because if a2 + b2 = c2 + d 2 and c2 + d 2 = e2 + f 2 , certainly a2 + b2 = e2 + f 2 .

33. The essentials of the digraph are the connections made by the arrows. Compare the arrows leaving each vertex in turn to pairs in R with that vertex as first element. Exercise Set 4

7. Transitive.

(b) The equivalence classes of A/R are circles with center at (0, 0), including the circle with radius 0.

9. Antisymmetric, transitive. 11. Irreflexive, symmetric. 17. Symmetric. 19. Reflexive, symmetric, transitive. 21. Reflexive, symmetric, transitive. 23. 5

21. (a) (a, b) R (a, b) because ab = ba. Hence R is reflexive. If (a, b) R (a , b ), then ab = ba . Then a b = b a and (a , b ) R (a, b). Hence R is symmetric. Now suppose that (a, b) R (a , b ) and (a , b ) R (a , b ). Then ab = ba and a b = b a . ab = a

1 3 2 4

25. {(1, 5), (5, 1), (1, 6), (6, 1), (5, 6), (6, 5), (1, 2), (2, 1), (2, 7), (7, 2), (2, 3), (3, 2)}. 27. Let a1 , a2 , . . . , an be the elements of the base set. The graph of R is connected if for each ai , there is a 1 in the ith column of (MR )k for some k. 29. Let R be transitive and irreflexive. Suppose a R b and b R a. Then a R a since R is transitive. But this contradicts the fact that R is irreflexive. Hence R is asymmetric. 31. Let R = { } be symmetric and transitive. There exists (x, y) ∈ R and (y, x) ∈ R. Since R is transitive, we have (x, x) ∈ R, and R is not irreflexive.

200

33. (Outline) Basis step: n = 1 P(1): If R is symmetric, then R1 is symmetric is true. Induction step: Use P(k): If R is symmetric, then Rk is symmetric to show P(k + 1). Suppose that a Rk+1 b. Then there is a c ∈ A such that a Rk c and c R b. We have b R c and c Rk a. Hence b Rk+1 a.

b a a a = ab = ba = ba . a a a

Hence (a, b) R (a , b ) and R is transitive. (b) {{(1, 1), (2, 2), (3, 3), (4, 4), (5, 5)}, {(1, 2), (2, 4)}, {(1, 3)}, {(1, 4)}, {(1, 5)}, {(2, 1), (4, 2)}, {(2, 3)}, {(2, 5)}, {(3, 1)}, {(3, 2)}, {(3, 4)}, {(3, 5)}, {(4, 1)}, {(4, 3)}, {(4, 5)}, {(5, 1)}, {(5, 2)}, {(5, 3)}, {(5, 4)}}. 23. Let R be reflexive and circular. If a R b, then a R b and b R b, so b R a. Hence R is symmetric. If a R b and b R c, then c R a. But R is symmetric, so a R c, and R is transitive. Let R be an equivalence relation. Then R is reflexive. If a R b and b R c, then a R c (transitivity) and c R a (symmetry), so R is also circular. 25. a R b if and only if ab > 0.

Relations and Digraphs 27. If z is even (or odd), then R(z) is the set of even (or odd) integers. Thus, if a and b are both even (or odd), then R(a) + R(b) = {x | x = s + t, s ∈ R(a), t ∈ R(b)} = {x | x is even} = R(a + b). If a and b have opposite parity, then R(a) + R(b) = {x | x = s + t, s ∈ R(a), t ∈ R(b)} = {x | x is odd} = R(a + b).

11. VERT

TAIL

HEAD

NEXT

a a a b b c c c d d d d

a b d b c b c d a b c d

2 3 0 5 0 7 8 0 10 11 12 0

1 4 6 9

29. (1, 2) R (2, 4) and (1, 3) R (1, 3), but ((1, 2) + (1, 3)) R/ ((2, 4) + (1, 3)) so the set R((a, b)) + R((a , b )) is not an equivalence class. Exercise Set 6 1. VERT[1] = 9 NEXT[10] = 1 VERT[2] = 3 NEXT[2] = 0 VERT[3] = 6 NEXT[4] = 7 VERT[4] = 0 VERT[5] = 5 VERT[6] = 8

(1, 6) (1, 2) (2, 1)

NEXT[9] = 10 NEXT[1] = 0 NEXT[3] = 2

(1, 3)

(3, 4) (3, 6)

NEXT[6] = 4 NEXT[7] = 0

(3, 5)

(5, 4) (6, 1)

NEXT[5] = 0 NEXT[8] = 0

(2, 3)

⎡

1 ⎢0 ⎣1 0

⎡

0 ⎢0 7. ⎣ 0 1

9. VERT 1 3 5 6 9

1 0 0 1

1 1 0 0

1 1 0 0

1 1 0 1

0 1⎥ . 1⎦ 0

⎤ 0 0⎥ . 1⎦ 1

1

2

4

3

TAIL

HEAD

NEXT

1 1 2 2 3 4 4 4 5 5

1 4 2 3 4 1 3 5 2 5

2 0 4 0 0 7 8 0 10 0

(b) {(3, 1)}. (d) {(1, 2), (1, 3), (2, 3), (3, 3)}.

VERT TAIL HEAD NEXT 1 1 1 2 4 1 2 3 6 1 3 0 8 2 3 5 2 4 0 3 1 7 3 4 0 4 2 0

⎤

1. (a) {(1, 3), (2, 1), (2, 2), (3, 2), (3, 3)}. (c) {(1, 1), (1, 2), (2, 1), (2, 3), (3, 1), (3, 2), (3, 3)}.

3. On average, EDGE must look at the average number of edges from any vertex.If R has P edges and N vertices, P Pij = edges on average. then EDGE examines N N 5.

Exercise Set 7

3. {(a, b) | a, b are sisters or a, b are brothers}. 5. a (R ∪ S) b if and only if a is a parent of b. 7. (a) {(2, 1), (3, 1), (3, 2), (3, 3), (4, 2), (4, 3), (4, 4), (1, 4)}. (b) {(1, 1), (1, 2), (2, 2), (2, 3), (2, 4), (4, 1), (3, 4)}. (c) {(1, 1), (1, 2), (1, 3), (1, 4), (2, 2), (2, 3), (2, 4), (3, 1), (3, 2), (3, 4), (4, 1), (4, 4)}. (d) {(1, 1), (2, 1), (2, 2), (1, 4), (4, 1), (2, 3), (3, 2), (1, 3), (4, 2), (3, 4), (4, 4)}. 9. (a) {(1, 1), (1, 4), (2, 2), (2, 3), (3, 3), (3, 4)}. (b) {(1, 2), (2, 4), (3, 1), (3, 2)}. (c) {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 4), (3, 1), (3, 2), (3, 3)}. (d) {(1, 1), (2, 1), (4, 1), (4, 2), (1, 3), (2, 3), (3, 3)}. ⎡ ⎤ ⎡ ⎤ 0 0 0 1 1 1 ⎢0 0 1⎥ ⎢1 1 1⎥ 11. (a) ⎣ . (b) ⎣ . 0 0 0⎦ 1 1 1⎦ 1 0 1 1 1 1 ⎡ ⎤ ⎡ ⎤ 1 0 1 1 0 0 1 ⎢0 1 0⎥ (d) ⎣ . (c) ⎣ 0 1 1 0 ⎦. 0 1 0⎦ 1 1 0 1 0 0 0 ⎡ ⎤ ⎡ ⎤ 0 0 0 0 1 1 1 0 ⎢0 0 1 0⎥ ⎢0 0 1 0⎥ 13. (a) ⎣ . (b) ⎣ . 1 0 0 0⎦ 1 1 0 0⎦ 0 0 0 0 0 1 1 1 ⎡ ⎤ ⎡ ⎤ 1 0 1 0 1 0 1 1 ⎢0 0 0 1⎥ ⎢1 1 0 1⎥ . (d) ⎣ . (c) ⎣ 1 1 0 0⎦ 0 0 1 1⎦ 0 0 0 1 1 1 0 1 15. R ∩ S = {(a, a), (b, b), (b, c), (c, b), (c, c), (d, d), (e, e)}, {{a}, {b, c}, {d}, {e}}.

201

Relations and Digraphs 17. (a) {(a, a), (a, d), (a, e), (b, b), (b, c), (b, e), (c, a), (c, b), (c, c), (d, b), (d, c), (d, d), (e, c), (e, e)}. (b) {(a, a), (a, d), (d, a), (a, e), (e, a), (b, c), (c, b), (b, e), (e, b), (c, a), (a, c), (c, c), (d, b), (b, d), (d, c), (c, d), (e, c), (c, e), (e, e)}. 19. The definitions of irreflexive, asymmetric, and antisymmetric each require that a certain pair does not belong to R. We cannot “fix” this by including more pairs in R. 21. (a) Yes.

(b) Yes.

(c) x (S ◦ R) y if and only if x ≤ 6y. 23. (a) Reflexive. a R a ∧ a S a ⇒ a S ◦ R a. Irreflexive. No. 1 R 2 ∧ 2 S 1 ⇒ 1 S ◦ R 1. Symmetric. No. 1 R 3, 3 R 1, 3 S 2, 2 S 3 ⇒ 1. 1 S ◦ R 2, but 2 S ◦R Asymmetric. No. R = {(1, 2), (3, 4)} and S = {(2, 3), (4, 1)} provide a counterexample. Antisymmetric. No. R = {(a, b), (c, d)} and S = {(b, c), (d, a)} provide a counterexample. Transitive. No. R = {(a, d), (b, e)} and S = {(d, b), (e, c)} provide a counterexample. (b) No, symmetric and transitive properties are not preserved. ⎤ ⎡ 1 1 0 1 1 ⎢0 1 0 1 1⎥ ⎢ ⎥ 25. (a) ⎢ 1 1 0 1 1 ⎥. ⎣1 1 0 1 1⎦ 1 1 0 0 1 ⎡ ⎤ 1 0 1 1 1 ⎢1 1 0 1 1⎥ ⎢ ⎥ (b) ⎢ 1 0 1 1 1 ⎥. ⎣1 1 1 1 1⎦ 0 0 0 1 1 ⎡ ⎤ 1 1 0 1 1 ⎢1 1 0 0 1⎥ ⎢ ⎥ (c) ⎢ 0 1 0 1 1 ⎥. ⎣1 1 0 1 1⎦ 1 1 0 0 1 ⎡ ⎤ 1 1 1 1 1 ⎢1 0 1 1 1⎥ ⎢ ⎥ (d) ⎢ 1 1 0 1 1 ⎥. ⎣1 1 1 1 1⎦ 0 1 0 1 1 27. R ∩ S is antisymmetric. If a(R ∩ S)b and b(R ∩ S)a, then a R b and b R a. Hence a = b because R is antisymmetric. R ∪ S may not be antisymmetric. Let R = {(1, 2)}, S = {(2, 1)}. 29. Let R = {(x, z), (x, m)}, S = {(z, y)}, and T = {(m, y)}. Then (x, y) ∈ (S ◦ R) ∩ (T ◦ R), but (S ∩ T) ◦ R = { }. 31. (a) Let MR∩S = mij , MR = rij , MS = sij . mij = 1 if and only if (i, j) ∈ R ∩ S. (i, j) ∈ R if and only if rij = 1 and (i, j) ∈ S if and only if sij = 1. But this happens if and only if the i, jth entry of MR ∧ MS is 1.

202

(b) Let MR∪S = mij , MR = rij , MS = sij . mij = 1 if and only if (i, j) ∈ R ∪ S. (i, j) ∈ R if and only if rij = 1 or (i, j) ∈ S if and only if sij = 1. But this happens if and only if the i, jth entry of MR ∨ MS is 1. (c) The i, jth entry of MR−1 is 1 if and only if (i, j) ∈ R−1 if and only if (j, i) ∈ R if and only if the j, ith entry of MR is 1 if and only if the i, jth entry of MRT is 1. (d) The i, jth entry of MR is 1 if and only if (i, j) ∈ R if and only if (i, j) ∈ / R if and only if the i, jth entry of MR is 0 if and only if the i, jth entry of MR is 1. 33. To form the digraph of R−1 , reverse the arrows in the digraph of R. 35. The edges of the digraph of R ∪ S are the edges that appear on either the digraph of R or the digraph of S. 37. (a) R is symmetric if and only if x R y ⇒ y R x if and only if R ⊆ R−1 ⊆ R. (b) Suppose R is antisymmetric. Let (x, y) ∈ R∩R−1 , then x = y and (x, y) ∈ . Suppose R ∩ R−1 ⊆ . If x R y and y R x, then x R y and x R−1 y. Thus (x, y) ∈ , so x = y. (c) Suppose R is asymmetric. Let (x, y) ∈ R ∩ R−1 . This contradicts the fact that R is asymmetric. Hence R ∩ R−1 = ∅. Let x R y and y R x. Then (x, y) ∈ R ∩ R−1 = ∅. Hence R is asymmetric. Exercise Set 8 ⎡ 1 1 1. (a) ⎣ 1 1 1 1

⎤ 1 1 ⎦. 1

(b) {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)}. ⎡ ⎤ 1 0 0 1 0 ⎢0 1 0 0 0⎥ ⎢ ⎥ 3. W1 = ⎢ 0 0 0 1 1 ⎥, W2 = W1 = W3 . ⎣1 0 0 1 0⎦ 0 1 0 0 1 5. a R∞ b if and only if b > a. 7. Let R be reflexive and transitive. Suppose that x Rn y. Then x, a1 , a2 , . . . , an−1 , y is a path of length n from x to y. x R a1 ∧ a1 R a2 ⇒ x R a2 . Similarly, we have x R ak ∧ ak R ak+1 ⇒ x R ak+1 and finally x R an−1 ∧ an−1 R y ⇒ x R y. Hence Rn ⊆ R. If x R y, then since R is reflexive we can build a path of length n, x, x, x, . . . , x, y from x to y and x Rn y. ⎡ ⎤ 1 0 0 1 ⎢1 1 0 1⎥ 9. ⎣ . 0 0 1 0⎦ 0 0 0 1 ⎡ ⎤ 1 0 0 1 ⎢0 1 1 0⎥ 11. ⎣ . 0 1 1 0⎦ 1 0 0 1

Relations and Digraphs 13. The sets are the same. This can be shown by computing each set. 15. A × A. 17. A/R = {{1, 2, 3}, {4, 5}}, A/S = {{1}, {2, 3, 4}, {5}}, A/(R ∪ S)∞ = {A}. 19. The collection of elements in A/R and A/S can be separated into subcollections of nondisjoint sets. Each element

of A/(R ∪ S)∞ is the union of the sets in one of these subcollections. 21. {(1, 1), (1, 2), (1, 4), (2, 2), (3, 2), (3, 3), (4, 2), (4, 3), (4, 4), (1, 3)}. 23. We first show R∞ is transitive. Then we show it is the smallest relation that contains R. It is a direct proof. 25. R × R.

Answers to Chapter Self-Tests 1. A × B is a set of ordered pairs. 2. A partition of A is a set of nonempty subsets of A. 3. A relation may be given by a verbal description, a set of ordered pairs, a digraph (in some cases) and as a matrix (in some cases). 4. Unless |R| is very small, using the matrix representation will be the most efficient way to determine if R is transitive. 5. The transitive closure of R is the connectivity relation R∞ ; it is also the smallest transitive relation containing R. 6. (a) 12. (b) {(2, 1), (2, 2), (2, 3), (2, 4), (5, 1), (5, 2), (5, 3), (5, 4), (7, 1), (7, 2), (7, 3), (7, 4)}. 7. Let U = {1, 2, 3, 4, 5}, A = {1, 2, 3}, B = {2, 3}. Then (2, 5) ∈ A × B, but (2, 5) ∈ / A × B. 8. {{a, b, c}, {d, e}}, {{a, b, d}, {c, e}}, {{a, b, e}, {c, d}}, {{b, c, d}, {a, e}}, {{b, c, e}, {a, d}}, {{b, d, e}, {a, c}}, {{c, d, e}, {a, b}}, {{a, c, d}, {b, e}}, {{a, c, e}, {b, d}}, {{a, d, e}, {b, c}}, {{a, b, c, d}, {e}}, {{a, b, d, e}, {c}}, {{b, c, d, e}, {a}}, {{a, b, c, e}, {d}}, {{a, c, d, e}, {b}}. ⎡ ⎤ 0 0 0 0 ⎢1 0 0 0⎥ 9. (a) (b) ⎣ . 2 1 1 0 0⎦ 1 1 0 0 18 14

10. (a) a b

d c

⎡

1 ⎢0 (b) MR = ⎣ 0 0

c

R

1 0 0 1

b

d

0 1 0 0

⎤

0 0⎥ . 1⎦ 0

1 ⎢0 MR2 = ⎣ 0 0 ⎡ 1 ⎢0 (c) MR∞ = ⎣ 0 0

1 0 1 0

1 0 0 1

1 1 1 1

1 1 1 1

⎤ 0 1⎥ . 0⎦ 0 ⎤ 1 1⎥ . 1⎦ 1

11. Reflexive, not irreflexive, not symmetric, not asymmetric, not antisymmetric, not transitive. 12. Not reflexive, not irreflexive, not symmetric, not asymmetric, antisymmetric, not transitive. 13. Since a R b implies b R/ a, a R b ∧ b R a is always false. Hence a R b ∧ b R a ⇒ a = b is always true. R must be antisymmetric. 14. (a) (u, v) R (u, v) since u − v = u − v is true. Thus R is reflexive. If (u, v) R (x, y), then u − v = x − y and (x, y) R (u, v). Thus R is symmetric. If ((u, v) R (x, y)) ∧ ((x, y) R (w, z)), then u − v = x − y = w − z and (u, v) R (w, z). Hence, R is transitive. (b) [(2, 3)] = {(2, 3), (1, 2), (3, 4), (4, 5)}. (c) A/R = {[(2, 3)], [(2, 4)], [(2, 5)], [(2, 2)], [(2, 1)], [(3, 1)], [(4, 1)], [(5, 1)], [(1, 5)]}. ⎡ ⎤ 0 1 1 0 ⎢1 0 1 0⎥ 15. MR = ⎣ . 1 0 0 1⎦ 0 1 0 1

8

a

⎡

R2

16. (a) R−1 = {(a, a), (a, e), (b, a), (c, a), (c, b), (c, d), (e, c)}. (b) R◦S = {(a, a), (a, b), (a, c), (b, b), (b, a), (b, c), (c, e), (c, c), (d, a), (e, c)}. ⎡ ⎤ 1 1 1 1 1 ⎢1 1 1 1 1⎥ ⎢ ⎥ 17. MR∞ = ⎢ 1 1 1 1 1 ⎥. ⎣1 1 1 1 1⎦ 1 1 1 1 1

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Functions

From Discrete Mathematical Structures, Sixth Edition, Bernard Kolman, Robert C. Busby, Sharon Cutler Ross. Copyright © 2009 by Pearson Education, Inc. Published by Prentice Hall. All rights reserved.

205

Functions In this chapter we focus our attention on a special type of relation, a function, that plays an important role in mathematics, computer science, and many applications. We also define some functions used in computer science and examine the growth of functions.

Looking Back The origins of the notion of a function can be traced back to the great Italian philosopher, astronomer, and mathematician Galileo Galilei (1564–1642), who in the 1630s observed the relationship between two variables. The early work on functions in the second half of the seventeenth century concentrated on the study of special functions as curves. These included the power, exponential, logarithmic, and trigonometric functions. Gottfried Wilhelm Leibniz (1646–1716) was the first person to use the word function for a quantity whose value varies as a point moves on a curve. Leibniz was an extraordinary person who made brilliant contributions in a number of diverse areas, including logic, philosophy, law, metaphysics, religion, mathematics, diplomacy, and literature. He has often been called a “universal genius.” Leibniz was born in Leipzig, and died in Hanover, both in Germany. Early in his career, he developed the foundations for what would later be called symbolic logic. Leibniz began his study of advanced contemporary mathematics in 1672 at the age of 26. Three years later he discovered the Fundamental Theorem of Calculus independently of Newton, who had also discovered the same result. Indeed, a heated battle raged over a number of years between the supporters of Leibniz and Newton as to who had discovered calculus first. Today, both Newton and Leibniz are considered the fathers of calculus. It is quite surprising to learn what a visionary Leibniz was. In the 1670s he invented a mechanical calculator, known as the Leibniz wheel, capable of adding, subtracting, multiplying, and dividing. He almost envisioned the modern age of computing!

206

Gottfried Wilhelm Leibniz Library of Congress

Leonhard Euler Bavarian State Library, Munich

The commonly used notation for a function value, f(x), is due to Leonhard Euler (1707–1783), who was born in Basel, Switzerland, and died in St. Petersburg, Russia. Euler is one of the greatest and most prolific mathematicians in history. After his death, it took nearly 50 years to publish all his papers and his collected works comprise more than 75 volumes. He was also able to carry out complex calculations in his head. During the last 17 years of his life, Euler was totally blind, but his mathematical output remained undiminished. Euler made significant contributions to many areas of mathematics and used mathematics to solve a wide variety of problems in the sciences.

Functions

1

Functions In this section we define the notion of a function, a special type of relation. We study its basic properties and then discuss several special types of functions. A number of important applications of functions will occur in later sections so it is essential to get a good grasp of the material in this section. Let A and B be nonempty sets. A function f from A to B, which is denoted f : A → B, is a relation from A to B such that for all a ∈ Dom(f ), f(a), the f -relative set of a, contains just one element of B. Naturally, if a is not in Dom(f ), then f(a) = ∅. If f(a) = {b}, it is traditional to identify the set {b} with the element b and write f(a) = b. We will follow this custom, since no confusion results. The relation f can then be described as the set of pairs {(a, f(a)) | a ∈ Dom(f )}. Functions are also called mappings or transformations, since they can be geometrically viewed as rules that assign to each element a ∈ A the unique element f(a) ∈ B (see Figure 1). The element a is called an argument of the function f , and f(a) is called the value of the function for the argument a and is also referred to as the image of a under f . Figure 1 is a schematic or pictorial display of our definition of a function, and we will use several other similar diagrams. They should not be confused with the digraph of the relation f , which we will not generally display. f b = f (a)

a A

B

Figure 1

Example 1

Let A = {1, 2, 3, 4} and B = {a, b, c, d}, and let f = {(1, a), (2, a), (3, d), (4, c)}. Here we have f(1) = a f(2) = a f(3) = d f(4) = c. Since each set f(n) is a single value, f is a function. Note that the element a ∈ B appears as the second element of two different ordered pairs in f . This does not conflict with the definition of a function. Thus a ◆ function may take the same value at two different elements of A.

Example 2

Let A = {1, 2, 3} and B = {x, y, z}. Consider the relations R = {(1, x), (2, x)} and

S = {(1, x), (1, y), (2, z), (3, y)}.

The relation S is not a function since S(1) = {x, y}. The relation R is a function with Dom(R) = {1, 2} and Ran(R) = {x}. ◆ Example 3

Let P be a computer program that accepts an integer as input and produces an integer as output. Let A = B = Z. Then P determines a relation fP defined as

207

Functions follows: (m, n) ∈ fP means that n is the output produced by program P when the input is m. It is clear that fP is a function, since any particular input corresponds to a unique output. (We assume that computer results are reproducible; that is, they are ◆ the same each time the program is run.) Example 3 can be generalized to a program with any set A of possible inputs and set B of corresponding outputs. In general, therefore, we may think of functions as input-output relations. Example 4

Let A = R be the set of real numbers, and let p(x) = a0 + a1 x + · · · + an xn be a real polynomial. Then p may be viewed as a relation on R. For each r in R we determine the relative set p(r) by substituting r into the polynomial. Then, since all relative sets p(r) are known, the relation p is determined. Since a unique value ◆ is produced by this substitution, the relation p is actually a function. If the formula defining the function does not make sense for all elements of A, then the domain of the function is taken to be the set of elements for A for which the formula does make sense. In elementary mathematics, the formula (in the case of Example 4, the polynomial) is sometimes confused with the function it produces. This is not harmful, unless the student comes to expect a formula for every type of function. Suppose that, in the preceding construction, we used √ a formula that produced more than one element in p(x), for example, p(x) = ± x. Then the resulting relation would not be a function. For this reason, in older texts, relations were sometimes called multiple-valued functions.

Example 5

Worcester 64 Hartford 39

44

Boston 49

51

Providence 106

New Haven

Figure 2

Example 6

A labeled digraph is a digraph in which the vertices or the edges (or both) are labeled with information from a set. If V is the set of vertices and L is the set of labels of a labeled digraph, then the labeling of V can be specified to be a function f : V → L, where, for each v ∈ V , f(v) is the label we wish to attach to v. Similarly, we can define a labeling of the edges E as a function g : E → L, where, for each e ∈ E, g(e) is the label we wish to attach to e. An example of a labeled digraph is a map on which the vertices are labeled with the names of cities and the edges are labeled with the distances or travel times between the cities. Figure 2 shows an example of a labeled digraph. Another example is a flow chart of a program in which the vertices are labeled with the steps that are to be performed at that point in the program; the edges indicate the flow from one part of the program ◆ to another part. Let A = B = Z and let f : A → B be defined by f(a) = a + 1,

for a ∈ A.

Here, as in Example 4, f is defined by giving a formula for the values f(a). Example 7

◆

Let A = Z and let B = {0, 1}. Let f : A → B be found by 0 if a is even f(a) = 1 if a is odd. Then f is a function, since each set f(a) consists of a single element. Unlike the situation in Examples 4 and 6, the elements f(a) are not specified through an ◆ algebraic formula. Instead, a verbal description is given.

208

Functions Example 8

Let A be an arbitrary nonempty set. The identity function on A, denoted by 1A , is ◆ defined by 1A (a) = a. The reader may notice that 1A is the relation we previously called which stands for the diagonal subset of A × A. In the context of functions, the notation 1A is preferred, since it emphasizes the input-output or functional nature of the relation. Clearly, if A1 ⊆ A, then 1A (A1 ) = A1 . Suppose that f : A → B and g : B → C are functions. Then the composition of f and g, g◦f is a relation. Let a ∈ Dom(g◦f ). Then (g◦f )(a) = g(f(a)). Since f and g are functions, f(a) consists of a single element b ∈ B, so g(f(a)) = g(b). Since g is also a function, g(b) contains just one element of C. Thus each set (g ◦ f )(a), for a in Dom(g ◦ f ), contains just one element of C, so g ◦ f is a function. This is illustrated in Figure 3. g◦f f a

A

C

g b = f (a) B

c = g(b) = (g ◦ f )(a)

Figure 3

Example 9

Let A = B = Z, and C be the set of even integers. Let f : A → B and g : B → C be defined by f(a) = a + 1 g(b) = 2b. Find g ◦ f . Solution

We have

(g ◦ f )(a) = g(f(a)) = g(a + 1) = 2(a + 1).

Thus, if f and g are functions specified by giving formulas, then so is g ◦ f and the formula for g ◦ f is produced by substituting the formula for f into the formula ◆ for g.

Special Types of Functions Let f be a function from A to B. Then we say that f is everywhere defined if Dom(f ) = A. We say that f is onto if Ran(f ) = B. Finally, we say that f is one to one if we cannot have f(a) = f(a ) for two distinct elements a and a of A. The definition of one to one may be restated in the following equivalent form: If f(a) = f(a ),

then a = a .

The latter form is often easier to verify in particular examples. Example 10

Consider the function f defined in Example 1. Since Dom(f ) = A, f is everywhere defined. On the other hand, Ran(f ) = {a, c, d} = B; therefore, f is not onto. Since f(1) = f(2) = a,

209

Functions we can conclude that f is not one to one. Example 11

◆

Consider the function f defined in Example 6. Which of the special properties, if any, does f possess? Solution

Since the formula defining f makes sense for all integers, Dom(f ) = Z = A, and so f is everywhere defined. Suppose that f(a) = f(a ) for a and a in A. Then

a + 1 = a + 1

so

a = a .

Hence f is one to one. To see if f is onto, let b be an arbitrary element of B. Can we find an element a ∈ A such that f(a) = b? Since f(a) = a + 1, we need an element a in A such that a + 1 = b. Of course,

a=b−1

will satisfy the desired equation since b − 1 is in A. Hence Ran(f ) = B; therefore, f is onto. ◆ Example 12

Let A = {a1 , a2 , a3 }, B = {b1 , b2 , b3 }, C = {c1 , c2 }, and D = {d1 , d2 , d3 , d4 }. Consider the following four functions, from A to B, A to D, B to C, and D to B, respectively. (a) (b) (c) (d)

f1 f2 f3 f4

= {(a1 , b2 ), (a2 , b3 ), (a3 , b1 )} = {(a1 , d2 ), (a2 , d1 ), (a3 , d4 )} = {(b1 , c2 ), (b2 , c2 ), (b3 , c1 )} = {(d1 , b1 ), (d2 , b2 ), (d3 , b1 )}

Determine whether each function is one to one, whether each function is onto, and whether each function is everywhere defined. Solution

(a) (b) (c) (d)

f1 is everywhere defined, one to one, and onto. f2 is everywhere defined and one to one, but not onto. f3 is everywhere defined and onto, but is not one to one. f4 is not everywhere defined, not one to one, and not onto.

◆

If f : A → B is a one-to-one function, then f assigns to each element a of Dom(f ) an element b = f(a) of Ran(f ). Every b in Ran(f ) is matched, in this way, with one and only one element of Dom(f ). For this reason, such an f is often called a bijection between Dom(f ) and Ran(f ). If f is also everywhere defined and onto, then f is called a one-to-one correspondence between A and B.

210

Functions Example 13

Let R be the set of all equivalence relations on a given set A, and let be the set of all partitions on A. Then we can define a function f : R → as follows. For each equivalence relation R on A, let f(R) = A/R, the partition of A that corresponds ◆ to R; f is a one-to-one correspondence between R and .

Invertible Functions A function f : A → B is said to be invertible if its inverse relation, f −1 , is also a function. The next example shows that a function is not necessarily invertible. Example 14

Let f be the function of Example 1. Then f −1 = {(a, 1), (a, 2), (d, 3), (c, 4)}. We see that f −1 is not a function, since f −1 (a) = {1, 2}.

◆

The following theorem is frequently used. THEOREM 1

Let f : A → B be a function. (a) Then f −1 is a function from B to A if and only if f is one to one. If f −1 is a function, then (b) the function f −1 is also one to one. (c) f −1 is everywhere defined if and only if f is onto. (d) f −1 is onto if and only if f is everywhere defined. Proof (a) We prove the following equivalent statement. f −1 is not a function if and only if f is not one to one. Suppose first that f −1 is not a function. Then, for some b in B, f −1 (b) must contain at least two distinct elements, a1 and a2 . Then f(a1 ) = b = f(a2 ), so f is not one to one. Conversely, suppose that f is not one to one. Then f(a1 ) = f(a2 ) = b for two distinct elements a1 and a2 of A. Thus f −1 (b) contains both a1 and a2 , so f −1 cannot be a function. (b) Since (f −1 )−1 is the function f , part (a) shows that f −1 is one to one. (c) Recall that Dom(f −1 ) = Ran(f ). Thus B = Dom(f −1 ) if and only if B = Ran(f ). In other words, f −1 is everywhere defined if and only if f is onto. (d) Since Ran(f −1 ) = Dom(f ), A = Dom(f ) if and only if A = Ran(f −1 ). ■ That is, f is everywhere defined if and only if f −1 is onto. As an immediate consequence of Theorem 1, we see that if f is a one-toone correspondence between A and B, then f −1 is a one-to-one correspondence between B and A. Note also that if f : A → B is a one-to-one function, then the equation b = f(a) is equivalent to a = f −1 (b).

Example 15

Consider the function f defined in Example 6. Since it is everywhere defined, one to one, and onto, f is a one-to-one correspondence between A and B. Thus f is invertible, and f −1 is a one-to-one correspondence between B and A. ◆

211

Functions Example 16

Let R be the set of real numbers, and let f : R → R be defined by f(x) = x2 . Is f invertible? Solution

We must determine whether f is one to one. Since f(2) = f(−2) = 4, we conclude that f is not one to one. Hence f is not invertible.

◆

There are some useful results concerning the composition of functions. We summarize these in the following theorem. THEOREM 2

Let f : A → B be any function. Then (a) 1B ◦ f = f . (b) f ◦ 1A = f . If f is a one-to-one correspondence between A and B, then (c) f −1 ◦ f = 1A . (d) f ◦ f −1 = 1B . Proof (a) (1B ◦ f )(a) = 1B (f(a)) = f(a), for all a in Dom(f ). Thus, by Theorem 2 of Section 4.2, 1B ◦ f = f . (b) (f ◦ 1A )(a) = f(1A (a)) = f(a), for all a in Dom(f ), so f ◦ 1A = f . Suppose now that f is a one-to-one correspondence between A and B. As we pointed out, the equation b = f(a) is equivalent to the equation a = f −1 (b). Since f and f −1 are both everywhere defined and onto, this means that, for all a in A and b in B, f(f −1 (b)) = b and f −1 (f(a)) = a. (c) For all a in A, 1A (a) = a = f −1 (f(a)) = (f −1 ◦ f )(a). Thus 1A = f −1 ◦ f . (d) For all b in B, 1B (b) = b = f(f −1 (b)) = (f ◦ f −1 )(b). Thus 1B = ■ f ◦ f −1 .

THEOREM 3

(a) Let f : A → B and g : B → A be functions such that g ◦ f = 1A and f ◦ g = 1B . Then f is a one-to-one correspondence between A and B, g is a one-to-one correspondence between B and A, and each is the inverse of the other. (b) Let f : A → B and g : B → C be invertible. Then g ◦ f is invertible, and (g ◦ f )−1 = f −1 ◦ g−1 . Proof (a) The assumptions mean that g(f(a)) = a

and

f(g(b)) = b,

for all a in A and b in B.

This shows in particular that Ran(f ) = B and Ran(g) = A, so each function is onto. If f(a1 ) = f(a2 ), then a1 = g(f(a1 )) = g(f(a2 )) = a2 . Thus f is one to one. In a similar way, we see that g is one to one, so both f and g are invertible. Note that f −1 is everywhere defined since Dom(f −1 ) = Ran(f ) = B. Now, if b is any element in B, f −1 (b) = f −1 (f(g(b)) = (f −1 ◦ f )g(b)) = 1A (g(b)) = g(b).

212

Functions Thus g = f −1 , so also f = (f −1 )−1 = g−1 . Then, since g and f are onto, f −1 and g−1 are onto, so f and g must be everywhere defined. This proves all parts of part (a). (b) We know that (g◦f )−1 = f −1 ◦g−1 , since this is true for any two relations. Since g−1 and f −1 are functions by assumption, so is their composition, and then (g ◦ f )−1 is a function. Thus g ◦ f is invertible. ■ Example 17

Let A = B = R, the set of real numbers. Let f : A → B be given by the formula f(x) = 2x3 − 1 and let g : B → A be given by g(y) =

3 1

2

y + 12 .

Show that f is a bijection between A and B and g is a bijection between B and A. Solution

Let x ∈ A and y = f(x) = 2x3 − 1. Then 12 (y + 1) = x3 ; therefore, x=

3 1

2

y+

1 2

= g(y) = g(f(x)) = (g ◦ f )(x).

Thus g ◦ f = 1A . Similarly, f ◦ g = 1B , so by Theorem 3(a) both f and g are bijections. ◆ As Example 17 shows, it is often easier to show that a function, such as f , is one to one and onto by constructing an inverse instead of proceeding directly. Finally, we discuss briefly some special results that hold when A and B are finite sets. Let A = {a1 , . . . , an } and B = {b1 , . . . , bn }, and let f be a function from A to B that is everywhere defined. If f is one to one, then f(a1 ), f(a2 ), . . . , f(an ) are n distinct elements of B. Thus we must have all of B, so f is also onto. On the other hand, if f is onto, then f(a1 ), . . . , f(an ) form the entire set B, so they must all be different. Hence f is also one to one. We have therefore shown the following: THEOREM 4

Let A and B be two finite sets with the same number of elements, and let f : A → B be an everywhere defined function. (a) If f is one to one, then f is onto. ■ (b) If f is onto, then f is one to one. Thus for finite sets A and B with the same number of elements, and particularly if A = B, we need only prove that a function is one to one or onto to show that it is a bijection. This is an application of the pigeonhole principle. One-to-one functions are a fundamental tool in cryptology, because of the need to both encode and decode. Many secret codes are simple substitution codes created as follows. Let A = {a, b, . . . , z} be the English alphabet, and let f : A → A be a function agreed on in advance by each party to a correspondence. A message is encoded by replacing each letter with its f image. In order for the message to be decoded, the function f must have an inverse. The recipient decodes the message by applying f −1 to each letter. We know by Theorem 3(a) that f must therefore be one to one.

213

Functions Example 18

Suppose that f is defined by the following table: A D N J

B E O K

C S P L

D T Q M

E I R O

F N S P

G Y T Q

H A U R

I B V U

J C W V

K F X W

L G Y X

M H Z Z

Thus f(D) = T, f(R) = O, and so on. The rearrangement of the alphabet that defines this function is an example of using a keyword to begin, then listing all remaining letters in order. The phrase THE TRUCK ARRIVES TONIGHT is encoded as QAIQORSFDOOBUIPQKJBYAQ. In this case, the inverse function is easily found by using the table from bottom to top. Thus, we can decode the phrase CKAJADPDGKJYEIDOT as JOHNHASALONGBEARD; ◆

that is, JOHN HAS A LONG BEARD.

1 Exercises 1. Let A = {a, b, c, d} and B = {1, 2, 3}. Determine whether the relation R from A to B is a function. If it is a function, give its range. (a) R = {(a, 1), (b, 2), (c, 1), (d, 2)} (b) R = {(a, 1), (b, 2), (a, 2), (c, 1), (d, 2)} 2. Let A = {a, b, c, d} and B = {1, 2, 3}. Determine whether the relation R from A to B is a function. If it is a function, give its range. (a) R = {(a, 3), (b, 2), (c, 1)} (b) R = {(a, 1), (b, 1), (c, 1), (d, 1)} 3. Determine whether the relation R from A to B is a function. A = the set of all recipients of Medicare in the United States, B = {x | x is a nine-digit number}, a R b if b is a’s Social Security number. 4. Determine whether the relation R from A to B is a function. A = a set of people in the United States, B = {x | x is a nine-digit number}, a R b if b is a’s passport number. In Exercises 5 through 8, verify that the formula yields a function from A to B. 5. A = B = Z; f(a) = a2 6. A = B = R; f(a) = ea 7. A = R, B = {0, 1}; let Z be the set of integers and note

214

that Z ⊆ R. Then for any real number a, let 0 if a ∈ /Z f(a) = 1 if a ∈ Z. 8. A = R, B = Z; f(a) = the greatest integer less than or equal to a. 9. Let A = B = C = R, and let f : A → B, g : B → C be defined by f(a) = a − 1 and g(b) = b2 . Find (b) (g ◦ f )(2) (a) (f ◦ g)(2) (c) (g ◦ f )(x)

(d) (f ◦ g)(x)

(f) (g ◦ g)(y) (e) (f ◦ f )(y) 10. Let A = B = C = R, and let f : A → B, g : B → C be defined by f(a) = a + 1 and g(b) = b2 + 2. Find (b) (f ◦ g)(−2) (a) (g ◦ f )(−2) (c) (g ◦ f )(x)

(d) (f ◦ g)(x)

(f) (g ◦ g)(y) (e) (f ◦ f )(y) 11. In each part, sets A and B and a function from A to B are given. Determine whether the function is one to one or onto (or both or neither). (a) A = {1, 2, 3, 4} = B; f = {(1, 1), (2, 3), (3, 4), (4, 2)} (b) A = {1, 2, 3}; B = {a, b, c, d}; f = {(1, a), (2, a), (3, c)} 12. In each part, sets A and B and a function from A to B are given. Determine whether the function is one to one or onto (or both or neither). (a) A = 12 , 13 , 14 ; B = {x, y, z, w}; 1 1 1 f = 2, x , 4, y , 3, w

Functions (b) A = {1.1, 7, 0.06}; B = {p, q}; f = {(1.1, p), (7, q), (0.06, p)} 13. In each part, sets A and B and a function from A to B are given. Determine whether the function is one to one or onto (or both or neither). (a) A = B = Z; f(a) = a − 1 (b) A = R, B = {x | x is real and x ≥ 0}; f(a) = |a| 14. In each part, sets A and B and a function from A to B are given. Determine whether the function is one to one or onto (or both or neither). (a) A = R × R, B = R; f((a, b)) = a (b) Let S = {1, 2, 3}, T = {a, b}. Let A = B = S × T and let f be defined by f(n, a) = (n, b), n = 1, 2, 3, and f(n, b) = (1, a), n = 1, 2, 3. 15. In each part, sets A and B and a function from A to B are given. Determine whether the function is one to one or onto (or both or neither). (a) A = B = R × R; f((a, b)) = (a + b, a − b) (b) A = R, B = {x | x is real and x ≥ 0}; f(a) = a2 16. Let f(n) be the number of divisors of n, n ∈ Z+ . Determine whether f is one to one or onto (or both or neither). 17. Let f(n) be the maximum of n and 50, n ∈ Z+ . Determine whether f is one to one or onto (or both or neither). 18. Explain why Theorem 1(a) is equivalent to “f −1 is not a function if and only if f is not one to one.” 19. Let f : A → B and g : B → A. Verify that g = f −1 . , g(b) = 2b − 1 (a) A = B = R; f(a) = a+1 2 (b) A = {x | x is real and x ≥ 0}; B = √ {y | y is real and y ≥ −1}; f(a) = a2 − 1, g(b) = b + 1 20. Let f : A → B and g : B → A. Verify that g = f −1 . (a) A = B = P(S), where S is a set. If X ∈ P(S), let f(X) = X = g(X). (b) A = B = {1, 2, 3, 4}; f = {(1, 4), (2, 1), (3, 2), (4, 3)}; g = {(1, 2), (2, 3), (3, 4), (4, 1)} 21. Let f be a function from A to B. Find f −1 . (a) A = {x | x is real and x√≥ −1}; B = {x | x is real and x ≥ 0}; f(a) = a + 1 (b) A = B = R; f(a) = a3 + 1 22. Let f be a function from A to B. Find f −1 . (a) A = B = R; f(a) = 2a−1 3 (b) A = B = {1, 2, 3, 4, 5}; f = {(1, 3), (2, 2), (3, 4), (4, 5), (5, 1)} 23. Let f(x, y) = (2x − y, x − 2y), (x, y) ∈ R × R. (a) Show that f is one to one. (b) Find f −1 . In Exercises 24 and 25, let f be a function from A = {1, 2, 3, 4} to B = {a, b, c, d}. Determine whether f −1 is a function.

24. f = {(1, a), (2, a), (3, c), (4, d)} 25. f = {(1, a), (2, c), (3, b), (4, d)} 26. Let A = B = C = R and consider the functions f : A → B and g : B → C defined by f(a) = 2a + 1, g(b) = b/3. Verify Theorem 3(b): (g ◦ f )−1 = f −1 ◦ g−1 . 27. If a set A has n elements, how many functions are there from A to A? 28. If a set A has n elements, how many bijections are there from A to A? 29. If A has m elements and B has n elements, how many functions are there from A to B? 30. Complete the following proof. If f : A → B and g : B → C are one-to-one functions, then g ◦ f is one to one. Proof: Let a1 , a2 ∈ A. Suppose (g ◦ f )(a1 ) = (g ◦ f )(a2 ). Then g(f(a1 )) = g(f(a2 )) and f(a1 ) = f(a2 ), because . Thus a1 = a2 , because . Hence g ◦ f is one to one. 31. Complete the following proof. If f : A → B and g : B → C are onto functions, then g ◦ f is onto. Proof: Choose x ∈ . Then there exists y ∈ such that g(y) = x. (Why?) Then there exists z ∈ such that f(z) = y (why?) and (g ◦ f )(z) = x. Hence, g ◦ f is onto. 32. Let f : A → B and g : B → C be functions. Show that if g ◦ f is one to one, then f is one to one. 33. Let f : A → B and g : B → C be functions. Show that if g ◦ f is onto, then g is onto. 34. Let A be a set, and let f : A → A be a bijection. For any integer k ≥ 1, let f k = f ◦ f ◦ · · · ◦ f (k factors), and let f −k = f −1 ◦ f −1 ◦ · · · ◦ f −1 (k factors). Define f 0 to be 1A . Then f n is defined for all n ∈ Z. For any a ∈ A, let O(a, f ) = {f n (a) | n ∈ Z}. Prove that if a1 , a2 ∈ A, and O(a1 , f ) ∩ O(a2 , f ) = ∅, then O(a1 , f ) = O(a2 , f ). 35. Let f : A → B be a function with finite domain and range. Suppose that | Dom(f )| = n and | Ran(f )| = m. Prove that (a) if f is one to one, then m = n. (b) if f is not one to one, then m < n. 36. Let |A| = |B| = n and let f : A → B be an everywhere defined function. Prove that the following three statements are equivalent. (b) f is onto. (a) f is one to one. (c) f is a one-to-one correspondence (that is, f is one to one and onto). 37. Give a one-to-one correspondence between Z+ , the set of positive integers, and A = {x | x is a positive even integer}. 38. Give a one-to-one correspondence between Z+ , the set of positive integers, and A = {x | x is a positive odd integer}.

215

Functions 39. Based on Exercises 37 and 38, does |Z+ | = |A| = |B|? Justify your conclusion. 40. (a) Let A = R and f : A → R be defined by f(a) = a2 . Prove or disprove that f(a1 + a2 ) = f(a1 ) + f(a2 ). (b) Let A = {a, b} and f : A∗ → Z be defined by f(s) is the length of the string s. Prove or disprove that f(s1 · s2 ) = f(s1 ) + f(s2 ). 41. Let A = {0, 1} and define a b = (a + b) mod 2. Let B = {true, false}. Define f : A → B by f(0) = true and f(1) = false. (a) Prove or disprove that f(a b) = f(a) ∨ f(b). (b) Prove or disprove that f(a b) = f(a) ∧ f(b).

2

42. (a) Use the function in Example 18 to encode the message COME BACK AT ONCE. (b) Decode the following message that was encoded using the function of Example 18. QODLLITSDJJKQOIQROJ 43. Use the method of Example 18 and the keyword JOURNALISM to encode the message ALL PROJECTS ARE ON TRACK. 44. Substitution codes like the one in Example 18 are not very secure. Describe a commonsense method to break such a code.

Functions for Computer Science In this section we review some functions commonly used in computer science applications and define some others.

Example 1

Let A be a subset of the universal set U = {u1 , u2 , u3 , . . . , un }. The characteristic function of A is defined as a function from U to {0, 1} by the following: 1 if ui ∈ A fA (ui ) = 0 if ui ∈ / A. If A = {4, 7, 9} and U = {1, 2, 3, . . . , 10}, then fA (2) = 0, fA (4) = 1, fA (7) = 1, and fA (12) is undefined. It is easy to check that fA is everywhere defined and onto, ◆ but is not one to one.

Example 2

We define a family of mod-n functions. As one for each positive integer n. We call these functions fn ; that is, fn (m) = m (mod n). Each fn is a function from the nonnegative integers to the set {0, 1, 2, 3, . . . , n−1}. For a fixed n, any nonnegative integer z can be written as z = kn + r with 0 ≤ r < n. Then fn (z) = r. We can also express this relation as z ≡ r (mod n). Each member of the mod function ◆ family is everywhere defined and onto, but not one to one.

Example 3

Let A be the set of nonnegative integers, B = Z+ , and let f : A → B be defined ◆ by f(n) = n!.

Example 4

The general version of the pigeonhole principle requires the floor function, which is defined for rational numbers as f(q) is the largest integer less than or equal to q. Here again is an example of a function that is not defined by a formula. The notation q is often used for f(q). Thus f(1.5) = 1.5 = 1,

Example 5

f(−3) = −3 = −3,

f(−2.7) = −2.7 = −3. ◆

A function similar to that in Example 4 is the ceiling function, which is defined for rational numbers as c(q) is the smallest integer greater than or equal to q. The notation q is often used for c(q). Thus c(1.5) = 1.5 = 2,

c(−3) = −3 = −3,

c(−2.7) = −2.7 = −2.

◆

Many common algebraic functions are used in computer science, often with domains restricted to subsets of the integers.

216

Functions Example 6

(a) Any polynomial with integer coefficients, p, can be used to define a function on Z as follows: If p(x) = a0 + a1 x + a2 x2 + · · · + an xn and z ∈ Z, then f(z) is the value of p evaluated at z. (b) Let A = B = Z+ and let f : A → B be defined by f(z) = 2z . We call f the base 2 exponential function. Other bases may be used to define similar functions. (c) Let A = B = R and let fn : A → B be defined for each positive integer n > 1 as fn (x) = logn (x), the logarithm to the base n of x. In computer science applications, the bases 2 and 10 are particularly useful. ◆ In general, the unary operations discussed in previous sections can be used to create functions similar to the function in Example 3. The sets A and B in the definition of a function need not be sets of numbers, as seen in the following examples.

Example 7

(a) Let A be a finite set and define l : A∗ → Z as l(w) is the length of the string w. (b) Let B be a finite subset of the universal set U and define pow(B) to be the power set of B. Then pow is a function from V , the power set of U, to the power set of V . (c) Let A = B = the set of all 2 × 2 matrices with real number entries and let t(M) = MT , the transpose of M. Then t is everywhere defined, onto, and one ◆ to one.

Example 8

(a) For elements of Z+ × Z+ , define g(z1 , z2 ) to be GCD(z1 , z2 ). Then g is a function from Z+ × Z+ to Z+ . ◆ (b) In a similar fashion we can define m(z1 , z2 ) to be LCM(z1 , z2 ). Another type of function, a Boolean function, plays a key role in nearly all computer programs. Let B = {true, false}. Then a function from a set A to B is called a Boolean function.

Example 9

Let P(x) : x is even and Q(y) : y is odd. Then P and Q are functions from Z to B. We see that P(4) is true and Q(4) is false. The predicate R(x, y) : x is even or y is odd is a Boolean function of two variables from Z × Z to B. Here R(3, 4) is false ◆ and R(6, 4) is true.

Hashing Functions Here we consider a more general problem of storing data. Suppose that we must store and later examine a large number of data records, customer accounts for example. In general we do not know how many records we may have to store at any given time. This suggests that linked-list storage is appropriate, because storage space is only used when we assign a record to it and we are not holding idle storage space. In order to examine a record we will have to be able to find it, so storing the data in a single linked list may not be practical because looking for an item may take a very long time (relatively speaking). One technique for handling such storage problems is to create a number of linked lists and to provide a method for deciding onto which list a new item should be linked. This method will also determine which list to search for a desired item. A key

217

Functions point is to attempt to assign an item to one of the lists at random. This will have the effect of making the lists roughly the same size and thus keep the searching time about the same for any item. Suppose we must maintain the customer records for a large company and will store the information as computer records. We begin by assigning each customer a unique seven-digit account number. A unique identifier for a record is called its key. For now we will not consider exactly how and what information will be stored for each customer account, but will describe only the storage of a location in the computer’s memory where this information will be found. In order to determine to which list a particular record should be assigned, we create a hashing function from the set of keys to the set of list numbers. Hashing functions frequently use a mod-n function, as shown in the next example. Example 10

Suppose that (approximately) 10,000 customer account records must be stored and processed. The company’s computer is capable of searching a list of 100 items in an acceptable amount of time. We decide to create 101 linked lists for storage, because if the hashing function works well in “randomly” assigning records to lists, we would expect to see roughly 100 records per list. We define a hashing function from the set of seven-digit account numbers to the set {0, 1, 2, 3, . . . , 100} as follows: h(n) = n (mod 101). That is, h is the mod-101 function. Thus, h(2473871) = 2473871 (mod 101) = 78. This means that the record with account number 2473871 will be assigned to list ◆ 78. Note that the range of h is the set {0, 1, 2, . . . , 100}. Because the function h in Example 10 is not one to one, different account numbers may be assigned to the same list by the hashing function. If the first position on list 78 is already occupied when the record with key 2473871 is to be stored, we say a collision has occurred. There are many methods for resolving collisions. One very simple method that will be sufficient for our work is to insert the new record at the end of the existing list. Using this method, when we wish to find a record, its key will be hashed and the list h(key) will be searched sequentially. Many other hashing functions are suitable for this situation. For example, we may break the seven-digit account number into a three-digit number and a fourdigit number, add these, and then apply the mod-101 function. Chopping the key into pieces to create the function is the origin of the name hashing function. Many factors are considered in addition to the number of records to be stored; the speed with which an average-length list can be searched and the time needed to compute the list number for an account are two possible factors to be taken into account. For reasons that will not be discussed here, the modulus used in the mod function should be a prime. Thus, in Example 10, we chose 101 as the modulus, rather than 100 or 102. Determining a “good” hashing function for a particular application is a challenging task. Hashing functions are also employed in other applications such as cryptology where they are used to produce digital fingerprints and other electronic means to verify the authenticity of messages.

218

Functions

Fuzzy Sets Sets are precisely defined objects. If A is a subset of a universal set U, then given any element x of U, it must be possible, in principle at least, to say that x is a member of A or x is not a member of A. However, the descriptions of many collections used in daily life and in practical technology are not so precise, but “fuzzy” in nature. An example is the “collection of tall humans” in the universal set U of human beings. The adjective “tall” is imprecise; its meaning is likely to depend on the people using it and the situations in which it is used. Given a fixed range of height, the difficulty centers around the boundary of the collection. We may regard some people as definitely tall, and some as definitely not tall, but in between there will be people that we feel are “somewhat tall” or “a little tall.” Membership in the collection of tall people is not a yes-or-no question, but has degrees. One can be a member, somewhat of a member, or definitely not a member. This is the idea of a fuzzy set. Other examples are • • • •

the collection of rich people, the collection of automobile speeds that are too fast, the collection of dangerously close distances between moving objects, and the collection of useful laws.

What is needed is a precise way to deal with these useful “fuzzy” collections that are not sets. A set is completely described by its characteristic function. By definition, the set A determines the characteristic function fA . On the other hand, A = {x | fA (x) = 1}, so the function, in turn, determines the set. Which representation of a set we choose to use depends on the task at hand. Any mathematical operation that we can perform on sets has a corresponding operation on the characteristic functions. One way to define a fuzzy set precisely is to use a function representation that is a generalization of a characteristic function. A fuzzy set in a universal set U is a function f defined on U and having values in the interval [0, 1]. This function is a generalization of the characteristic function of a set. If f(x) = 0, then we say that x is not in the fuzzy set. If f(x) = 1, then we say that x is in the fuzzy set. But if 0 < f(x) < 1, then f(x) can be thought of as the degree to which x is in the fuzzy set. If the range of f is only the numbers 0 and 1, then f is the characteristic function of an ordinary set. Thus, ordinary sets are special cases of fuzzy sets. Here, a function such as f is the fuzzy set, but some other authors call a function such as f the membership function for the fuzzy set. Example 11

We define a universal set U to consist of all yearly incomes between $0 and $1,000,000, in thousands. Thus, U is the interval [0, 1000]. Suppose we wish to define the (fuzzy) set of rich people, based on yearly income. We may consider that a person whose yearly income is greater than $250,000 is definitely rich, a person with an income less than $30,000 is definitely not, and a person with an income in between has some degree of “richness.” With these initial conditions, a possible fuzzy set of rich people is the function f in Figure 4. (The horizontal line continues to 1000.) An equation of the slanted line segment is f(x) = (x−30)/220, 30 ≤ x ≤ 250. (Why?) Based on this, a person with a yearly salary of $100,000 (or x = 100)

219

Functions

Figure 4

has degree of membership (100 − 30)/220 = 7/22 or about 0.32 in the fuzzy set f , a person making less than $30,000 a year is not in this fuzzy set at all, and a person making $300,000 a year is definitely rich. Many other fuzzy sets with the same initial conditions could define the collection of rich people. The choice of a defining function is highly subjective and may depend on the situation under discussion. Figure 5 shows another appropriate fuzzy set, g, where 0 ≤ x ≤ 1000, g(x) = 0 ◆ for 0 ≤ x ≤ 30, and g(x) = 1 for x ≥ 250.

g(x )

Degree of richness

1 0.8 0.6 0.4 0.2

x 30

250 Income

500

Figure 5

Example 12

Determine in which fuzzy set f , or g, shown in Figures 4 and 5, the given income has a higher degree of richness. (a) $30,000

(b) $75,000

(c) $200,000

(d) $260,000

Solution

Although no explicit definition of g is provided, a comparison of the graphs shows that f(30) = g(30) = 0 and f(260) = g(260) = 1, so the answer for parts (a) and (d) is neither. It is geometrically clear that g(75) < f(75) and g(200) > f(200). ◆ Thus, (b) and (c) have greater degrees of richness in f and g, respectively.

220

Functions

Fuzzy Set Operations Just as we can form unions, intersections, complements, and so on for sets, we can define similar operations for fuzzy sets. Here is a generalization of this theorem that also applies to fuzzy sets. THEOREM 1

Let A and B be subsets of the same universal set U. Then (a) fA∩B (x) = min{fa (x), fb (x)}. (b) fA∪B (x) = max{fA (x), fB (x)}. (c) fA (x) = 1 − fA (x) (characteristic function of the complement of A). ■ The proof of this theorem is left to the reader. Parts (a) and (c) can easily be combined to express the characteristic function of the complement of one set with respect to another or the symmetric difference of two sets. We note that the computations on the right-hand side of these equations are also meaningful for fuzzy sets. We now use these computations to define intersection, union, and complement for fuzzy sets. If f and g are fuzzy sets, then the fuzzy sets f ∩ g, f ∪ g, and f are defined, respectively, to be the functions defined by the extension of parts (a), (b), and (c) of Theorem 1.

Example 13

Two fuzzy sets, f and g, in the universe of real numbers between 0 and 10 are shown in Figure 6 and their explicit definitions are also given. Compute the degree of membership of (a) 3.5 in f ∩ g

f (x)

⎧1 2 x ⎪ 16 ⎪ ⎪ ⎨1 f (x) = 1 ⎪ ⎪– 2 x + 4 ⎪ ⎩ 0

(b) 7 in f ∩ g

(c) 3.5 in f ∪ g

for 0 ≤ x < 4 for 4 ≤ x < 6

g(x) =

for 6 ≤ x < 8 for 8 ≤ x ≤ 10

⎧1 x ⎪ 3 ⎪ ⎪ ⎨ 1 –

1 ⎪ ⎪ ⎪ ⎩2

–

g(x)

4

(d) 7 in f ∪ g.

for 0 ≤ x < 3 x+

for 3 ≤ x < 5

7 4

for 5 ≤ x < 7 1 x 6

+

for 7 ≤ x ≤ 10

10 6

1

1

0.5

x

x 0

1

2

3

4

5 (a)

6

7

8

9

10

0

1

2

3

4

5

6

7

8

9

10

(b)

Figure 6

Solution

By using the algebraic definitions, we have f(3.5) = (3.52 )/16 = 49/64 and g(3.5) = (−3.5/4) + (7/4) = 56/64. So (f ∩ g)(3.5) = min{49/64, 56/64} or 49/64 and (f ∪ g)(3.5) = 56/64. We also see that f(7) = 4 − 7/2 = 1/2 and g(7) = 1/2. Thus, the degree of membership of 7 in both the union and intersection ◆ of f and g is 1/2.

221

Functions Because f and g are in the same universal set, we can quickly sketch the graph of f ∩ g (or f ∪ g) by graphing f and g on the same axes, as shown in Figure 7, and tracing the lower (or upper) graph segments.

1

f

g

x 1

0

2

3

4

5

6

7

8

9

10

Figure 7

Using this method, we construct the intersection and union of f and g and show the results in Figure 8. ( f ∩ g)(x)

( f ∪ g)(x)

x 0

1

2

3

4

5

6

7

(a)

8

9

10

x 0

1

2

3

4

5

6

7

8

9

10

(b)

Figure 8

Fuzzy Logic There are many similarities between sets and propositions. These similarities are not surprising since many sets can be defined by specifying a property P(x) that elements of the set have in common. A simple case is A = {x | x is an even number}; here P(x) is “x is an even number.” Such a P(x) is called a prepositional function or a predicate. The range of P is {false, true}. If false is assigned the value 0 and true, the value 1, then the predicate is the same as the characteristic function of the set. By allowing a predicate to have values in the interval [0, 1], we create the notion of a fuzzy predicate. This allows us to work with statements that are not simply true or false, but may be true to varying degrees. Fuzzy logic is the name given to the analysis and manipulation of fuzzy predicates, and to the process of forming arguments, drawing conclusions, and taking actions based on them. Fuzzy logic plays an important role in control theory and expert systems. For example, some computers use fuzzy logic to control a machine in a way similar to that used by a human, reacting to conditions such as “too close” or “too hot,”

222

Functions and interpreting “fuzzy” implications such as “If you are too close, slow down and move a little to the right.” Today many automobiles contain devices that are controlled by fuzzy logic. Consider the following simplified example. An automobile has sensors that measure the distance between the front bumper and the rear of the car ahead, and sensors that can determine if the tires are skidding on the road surface. Automobile engineers want to design a device that accepts these inputs and applies the brakes automatically to various degrees. First the engineers state in general terms what they want to happen. For example: If the car is too close to the car in front, then apply the brakes. If the car gets even closer to the car in front, apply the brakes more. If the car is near the point where the tires will skid, then apply the brakes less. One simple way to approach this problem is to construct fuzzy sets for the predicates f(x): “x is too close” and g(x): “x is near to skidding.” Then the pressure on the brakes can be computed from the degree of memberships in the fuzzy sets f and g, possibly by looking at the fuzzy set f − g. If (f − g)(x) is 0, no additional brakes are applied, and higher degrees of membership will translate into greater pressure on the brakes. This example is too simplified for practical situations, but more complex versions are in use in many ways today.

2 Exercises 1. Let f be the mod-10 function. Compute (b) f(38) (c) f(253) (a) f(417) 2. Let f be the mod-10 function. Compute (b) f(316) (c) f(1057) (a) f(81) In Exercises 3 and 4, use the universal set U = {a, b, c, . . . , y, z} and the characteristic function for the specified subset to compute the requested function values. 3. A = {a, e, i, o, u} (b) fA (y) (c) fA (o) (a) fA (i) 4. B = {m, n, o, p, q, r, z} (b) fB (m) (c) fB (s) (a) fB (a) 5. Compute each of the following. (b) −2.78 (c) 14 (a) 2.78 (e) 21.5 (d) −17.3 6. Compute each of the following. (b) −2.78 (c) 14 (a) 2.78 (e) 21.5 (d) −17.3 7. Let k, n be positive integers with k ≤ n. Prove n that the . number of multiples of k between 1 and n is k

2 n n2 + 3 8. Prove that if n is odd, then = . 4 4 In Exercises 9 and 10, compute the values indicated. Note that if the domain of these functions is Z+ , then each function is the explicit formula for an infinite sequence. Thus sequences can be viewed as a special type of function.

9. f(n) = 3n2 − 1 (b) f(17) (c) f(5) (d) f(12) (a) f(3) 10. g(n) = 5 − 2n (b) g(14) (c) g(129) (d) g(23) (a) g(4) 11. Let f2 (n) = 2n . Compute each of the following. (b) f2 (3) (c) f2 (5) (d) f2 (10) (a) f2 (1) 12. Let f3 (n) = 3n . Compute each of the following. (b) f3 (3) (c) f3 (6) (d) f3 (8) (a) f3 (2) In Exercises 13 through 16, let lg(x) = log2 (x). 13. Compute each of the following. (b) lg(128) (c) lg(512) (d) lg(1024) (a) lg(16) 14. For each of the following find the largest integer less than or equal to the function value and the smallest integer greater than or equal to the function value. (b) lg(25) (a) lg(10) 15. For each of the following find the largest integer less than or equal to the function value and the smallest integer greater than or equal to the function value. (b) lg(100) (a) lg(50) 16. For each of the following find the largest integer less than or equal to the function value and the smallest integer greater than or equal to the function value. (b) lg(500) (a) lg(256) 17. Prove that the function in Example 7(c), t : {2 × 2 matrices with real entries} → { 2 × 2 matrices with real entries} is

223

Functions everywhere defined, onto, and one to one. 18. Let A = {a, b, c, d}. Let l be the function in Example 7(a). (a) Prove that l is everywhere defined. (b) Prove that l is not one to one. (c) Prove or disprove that l is onto. 19. Let A be a set with n elements, S be the set of relations on A, and M the set of n × n Boolean matrices. Define f : S → M by f(R) = MR . Prove that f is a bijection between S and M.

In Exercises 30 through 36, let f and g be the fuzzy sets whose graphs and definitions are shown in Figure 9. ⎧1 2 ⎪ ⎪ 9 x + 3 2 for 0 ≤ x < 3 ⎪ ⎪ (x 3) – ⎪ for 3 ≤ x < 5 ⎪ ⎨1 – 4 f (x ) = 0 for 5 ≤ x < 7 ⎪ ⎪ ⎪ (x – 7)2 ⎪ for 7 ≤ x < 9 ⎪ ⎪ ⎩ 24 – 3x + 7 for 9 ≤ x ≤ 10 f (x) 1

20. Let p be a Boolean variable. How many different Boolean functions of p are there? How many different Boolean functions of two Boolean variables are there? 21. Build a table to represent the Boolean function f(x, y, z) = (∼x ∧ y) ∨ z for all possible values of x, y, and z.

x 0

22. Let P be the propositional function defined by P(x, y) = (x ∨ y) ∧ ∼y. Evaluate each of the following. (b) P(false, true) (a) P(true, true)

2

3

4

5

6

7

8

9

10

(a)

(c) P(true, false) 23. Let Q be the propositional function defined by Q(x) : ∃(y ∈ Z+ )(xy = 60). Evaluate each of the following. (b) Q(7) (c) Q(−6) (d) Q(15) (a) Q(3) In Exercises 24 through 26, use the hashing function h, which takes the first three digits of the account number as one number and the last four digits as another number, adds them, and then applies the mod-59 function.

1

g(x)

⎧ ⎪ 0 ⎪ ⎪ ⎪1 ⎪ x– 1 ⎪ ⎪ ⎪2 ⎨ – x +5 g(x ) = ⎪ 0 ⎪ ⎪ ⎪ ⎪ x– 7 ⎪ ⎪ ⎪ ⎩– 1 x + 5 2

for 0 ≤ for 2 ≤ for 4 ≤ for 5 ≤ for 7 ≤ for 8 ≤

x x x x x x

k and N 2 > c. Then N 7 ≤ cN 5 < N 2 · N 5 , but this is a contradiction. Hence f is O(g), but g is not O(f ), and f is lower order than g. ◆ This agrees with our experience that n5 grows more slowly than n7 . We define a relation , big-theta, on functions whose domains are subsets of

Z+ as f g if and only if f and g have the same order.

THEOREM 1

The relation , big-theta, is an equivalence relation. Proof Clearly, is reflexive since every function has the same order as itself. Because the definition of same order treats f and g in the same way, this definition is symmetric and the relation is symmetric. To see that is transitive, suppose f and g have the same order. Then there exist c1 and k1 with |f(n)| ≤ c1 · |g(n)| for all n ≥ k1 , and there exist c2 and k2 with |g(n)| ≤ c2 · |f(n)| for all n ≥ k2 . Suppose that g and h have the same order; then there exist c3 , k3 with |g(n)| ≤ c3 · |h(n)| for all n ≥ k3 , and there exist c4 , k4 with |h(n)| ≤ c4 · |g(n)| for all n ≥ k4 . Then |f(n)| ≤ c1 · |g(n)| ≤ c1 (c3 · |h(n)|) if n ≥ k1 and n ≥ k3 . Thus |f(n)| ≤ c1 c3 · |h(n)| for all n ≥ maximum of k1 and k3 . Similarly, |h(n)| ≤ c2 c4 · |f(n)| for all n ≥ maximum of k2 and k4 . Thus f ■ and h have the same order and is transitive. The equivalence classes of consist of functions that have the same order. We use any simple function in the equivalence class to represent the order of all functions in that class. One -class is said to be lower than another -class if a representative function from the first is of lower order than one from the second class. This means functions in the first class grow more slowly than those in the

227

Functions second. It is the -class of a function that gives the information we need for algorithm analysis. Example 5

All functions that have the same order as g(n) = n3 are said to have order (n3 ). The most common orders in computer science applications are (1), (n), (n2 ), (n3 ), (lg(n)), (nlg(n)), and (2n ). Here (1) represents the class of constant functions and lg is the base 2 log function. The continuous versions of some of these functions are shown in Figure 10. ◆

y y = 2x y = x2 y = lg(x)

x

y=x

y = x3

Figure 10

Example 6

Every logarithmic function f(n) = logb (n) has the same order as g(n) = lg(n). There is a logarithmic change-of-base identity logb (x) =

loga (x) loga (b)

in which loga (b) is a constant. Thus log (n) ≤ b

1 |lg(n)| lg(b)

and, conversely, |lg(n)| ≤ lg(b) · | logb (n)|. Hence g is O(f ) and f is O(g).

◆

It is sometimes necessary to combine functions that give the number of steps required for pieces of an algorithm as is done in the analysis of TRANS where functions are added, and in the analysis of NEWTRANS, where functions are multiplied. There are some general rules regarding the ordering of the -equivalence classes that can be used to determine the class of many functions and the class of the sum or product of previously classified functions.

228

Functions

Rules for Determining the -Class of a Function 1. (1) functions are constant and have zero growth, the slowest growth possible. 2. (lg(n)) is lower than (nk ) if k > 0. This means that any logarithmic function grows more slowly than any power function with positive exponent. 3. (na ) is lower than (nb ) if and only if 0 < a < b. 4. (an ) is lower than (bn ) if and only if 0 < a < b. 5. (nk ) is lower than (an ) for any power nk and any a > 1. This means that any exponential function with base greater than 1 grows more rapidly than any power function. 6. If r is not zero, then (rf ) = (f ) for any function f . 7. If h is a nonzero function and (f ) is lower than (or the same as) (g), then (fh) is lower than (or the same as) (gh). 8. If (f ) is lower than (g), then (f + g) = (g). Example 7

Determine the -class of each of the following. (a) f(n) = 4n4 − 6n7 + 25n3 (b) g(n) = lg(n) − 3n (c) h(n) = 1.1n + n15 Solution

(a) By Rules 3, 6, and 8, the degree of the polynomial determines the -class of a polynomial function. (f ) = (n7 ). (b) Using Rules 2, 6, and 8, we have that (g) = (n). ◆ (c) By Rules 5 and 8, (h) = (1.1n ). Example 8

Using the rules for ordering -classes, arrange the following in order from lowest to highest. (nlg(n))

(1000n2 − n)

(n0.2 )

(1,000,000)

(1.3n )

(n + 107 )

Solution

(1,000,000) is the class of constant functions, so it is the first on the list. By Rules 5 and 8, (n + 107 ) is lower than (1000n2 − n), but higher than (n0.2 ). To determine the position of (nlg(n)) on the list, we apply Rules 2 and 7. These give that (nlg(n)) is lower than (n2 ) and higher than (n). Rule 5 says that (1.3n ) is the highest class on this list. In order, the classes are (1,000,000) (nlg(n))

(n0.2 )

(n + 107 )

(1000n2 − n)

(1.3n ).

◆

The -class of a function that describes the number of steps performed by an algorithm is frequently referred to as the running time of the algorithm. For example, the algorithm TRANS has an average running time of n3 . In general, algorithms with exponential running times are impractical for all but very small values of n. In many cases the running time of an algorithm is estimated by examining best, worst, or average cases.

229

Functions

3 Exercises In Exercises 1 and 2, let f be a function that describes the number of steps required to carry out a certain algorithm. The number of items to be processed is represented by n. For each function, describe what happens to the number of steps if the number of items is doubled. 1. (a) f(n) = 1001 (b) f(n) = 3n (d) f(n) = 2.5n3 (c) f(n) = 5n2 2. (a) f(n) = 1.4lg(n) (b) f(n) = 2n (d) f(n) = 100n4 (c) f(n) = nlg(n) 3. Show that g(n) = n! is O(nn ). 4. Show that h(n) = 1 + 2 + 3 + · · · + n is O(n2 ). 5. Show that f(n) = 8n + lg(n) is O(n). 6. Show that g(n) = n2 (7n − 2) is O(n3 ). 7. Show that f(n) = nlg(n) is O(g) for g(n) = n2 , but that g is not O(f ). 8. Show that f(n) = n100 is O(g) for g(n) = 2n , but that g is not O(f ). 9. Show that f and g have the same order for f(n) = 5n2 + 4n + 3 and g(n) = n2 + 100n. 10. Show that f and g have the same order for f(n) = lg(n3 ) and g(n) = log5 (6n). 11. Determine which of the following are in the same -class. A function may be in a class by itself. f1 (n) = 5nlg(n),

f2 (n) = 6n2 − 3n + 7,

f3 (n) = 1.5n , f4 (n) = lg(n4 ), f5 (n) = 13,463, f6 (n) = −15n, f7 (n) = lg(lg(n)), f8 (n) = 9n0.7 , f9 (n) = n!, f10 (n) = n + lg(n), √ f11 (n) = n + 12n, f12 (n) = lg(n!) 12. Order the -classes in Exercise 11 from lowest to highest. 13. Consider the functions f1 , f2 , f4 , f5 , f6 , f10 , f11 in Exercise 11. Match each of the functions with its -class from the following list: (1), (n), (nlg(n)), (lg(n)), √ (n2 ), ( n), (2n ). In Exercises 14 through 21, analyze the operation performed by the given piece of pseudocode and write a function that describes the number of steps required. Give the -class of the function. 14. 1. A ← 1 2. B ← 1 3. UNTIL (B > 100) a. B ← 2A − 2 b. A ← A + 3

15. 1. X ← 1 2. Y ← 100 3. WHILE (X < Y) a. X ← X + 2

230

b. Y ← 1 2Y

16. 1. I ← 1 2. X ← 0

3. WHILE (I ≤ N) a. X ← X + 1 b. I ← I + 1

17. 1. SUM ← 0

2. FOR I = 0 THRU 2(N − 1) BY 2 a. SUM ← SUM + I

18. Assume that N is a power of 2. 1. X ← 1 2. K ← N 3. WHILE (K ≥ 1) a. X ← 3X b. K ← K/2 19. 1. I ← 1 2. SUM ← 0

3. WHILE (I ≤ N) a. FOR K = 1 THRU I 1. SUM ← SUM + K b. I ← I + 1

20. 1. K ← 0 2. FOR I = 0 THRU N a. WHILE K ≤ I 1. K ← K + 1

21. SUBROUTINE MATMUL(A,B,N,M,P,Q;C) 1. IF (M = P) THEN a. FOR I = 1 THRU N 1. FOR J = 1 THRU Q a. C[I,J] ← 0 b. FOR K = 1 THRU M 1. C[I,J] ← C[I,J] + (A[I,K] × B[K,J]) 2. ELSE a. CALL PRINT (’INCOMPATIBLE’) 3. RETURN END OF SUBROUTINE MATMUL

22. (a) Write a recurrence relation to count the number of ways a 3 × 3 square can be placed on an n × n square with the edges of the squares parallel. (b) What is the running time of an algorithm that uses the recurrence relation in (a) to count the number of placements? 23. Prove Rule 3. 24. Prove Rule 4. 25. Prove Rule 6. 26. Prove Rule 7.

Functions 27. Prove that if (f ) = (g) = (h), then f + g is O(h).

4

28. Prove that if (f ) = (g) and c = 0, then (cf ) = (g).

Permutation Functions In this section we discuss bijections from a set A to itself. Of special importance is the case when A is finite. Bijections on a finite set occur in a wide variety of applications in mathematics, computer science, and physics. A bijection from a set A to itself is called a permutation of A.

Example 1

Let A = R and let f : A → A be defined by f(a) = 2a + 1. Since f is one to one ◆ and onto (verify), it follows that f is a permutation of A. If A = {a1 , a2 , . . . , an } is a finite set and p is a bijection on A, we list the elements of A and the corresponding function values p(a1 ), p(a2 ), . . . , p(an ) in the following form:

a2 ··· an a1 . (1) p(a1 ) p(a2 ) · · · p(an ) Observe that (1) completely describes p since it gives the value of p for every element of A. We often write

a1 a2 ··· an . p= p(a1 ) p(a2 ) · · · p(an ) Thus, if p is a permutation of a finite set A = {a1 , a2 , . . . , an }, then the sequence p(a1 ), p(a2 ), . . . , p(an ) is just a rearrangement of the elements of A and so corresponds exactly to a permutation of A.

Example 2

Example 3

Let A = {1, 2, 3}. 1 1A = 1 1 p3 = 2

Then all the permutations of A are 2 3 1 2 3 1 2 , p1 = , p2 = 2 3 1 3 2 2 1 2 3 1 2 3 1 2 , p4 = , p5 = 3 1 3 1 2 3 2

3 3 3 1

, .

◆

Using the permutations of Example 2, compute (a) p−1 4 ; (b) p3 ◦ p2 . Solution

(a) Viewing p4 as a function, we have p4 = {(1, 3), (2, 1), (3, 2)}. Then

p−1 4 = {(3, 1), (1, 2), (2, 3)}

or, when written in increasing order of the first component of each ordered pair, we have p−1 4 = {(1, 2), (2, 3), (3, 1)}.

Thus p−1 4

=

1 2 3 2 3 1

= p3 .

231

Functions (b) The function p2 takes 1 to 2 and p3 takes 2 to 3, so p3 ◦ p2 takes 1 to 3. Also, p2 takes 2 to 1 and p3 takes 1 to 2, so p3 ◦ p2 takes 2 to 2. Finally, p2 takes 3 to 3 and p3 takes 3 to 1, so p3 ◦ p2 takes 3 to 1. Thus 1 2 3 . p3 ◦ p2 = 3 2 1 We may view the process of forming p3 ◦ p2 as shown in Figure 11. Observe that p3 ◦ p2 = p5 . ◆ ⎛ p3 ◦ p 2 = ⎝

1 2

? ⎞ ⎛ 2 3 1 2 ⎠ ◦ ⎝ ? ? 3 1 2 1

3 3

⎞

⎛

⎠ = ⎝

1

2

3

3

2

1

⎞ ⎠

Figure 11

The composition of two permutations is another permutation, usually referred to as the product of these permutations. In the remainder of this chapter, we will follow this convention. THEOREM 1

If A = {a1 , a2 , . . . , an } is a set containing n elements, then there are n! = n · (n − 1) · · · 2 · 1

permutations of A.

Proof This result follows by letting r = n.

(2) ■

Let b1 , b2 , . . . , br be r distinct elements of the set A = {a1 , a2 , . . . , an }. The permutation p : A → A defined by p(b1 ) = b2 p(b2 ) = b3 .. . p(br−1 ) = br p(br ) = b1 p(x) = x,

if x ∈ A, x ∈ / {b1 , b2 , . . . , br },

is called a cyclic permutation of length r, or simply a cycle of length r, and will be denoted by (b1 , b2 , . . . , br ). Do not confuse this terminology with that used for cycles in a digraph. The two concepts are different and we use slightly different notations. If the elements b1 , b2 , . . . , br are arranged uniformly on a circle, as shown in Figure 12, then a cycle p of length r moves these elements in a clockwise direction so that b1 is sent to b2 , b2 to b3 , . . . , br−1 to br , and br to b1 . All the other elements of A are left fixed by p. Example 4

232

Let A = {1, 2, 3, 4, 5}. The cycle (1, 3, 5) denotes the permutation 1 2 3 4 5 . 3 2 5 4 1

◆

Functions Observe that if p = (b1 , b2 , . . . , br ) is a cycle of length r, then we can also write p by starting with any bi , 1 ≤ i ≤ r, and moving in a clockwise direction, as shown in Figure 12. Thus, as cycles,

b1

br

b2 br–1

(3, 5, 8, 2) = (5, 8, 2, 3) = (8, 2, 3, 5) = (2, 3, 5, 8).

b3

Figure 12

Example 5

Note also that the notation for a cycle does not include the number of elements in the set A. Thus the cycle (3, 2, 1, 4) could be a permutation of the set {1, 2, 3, 4} or of {1, 2, 3, 4, 5, 6, 7, 8}. We need to be told explicitly the set on which a cycle is defined. It follows from the definition that a cycle on a set A is of length 1 if and only if it is the identity permutation, 1A . Since cycles are permutations, we can form their product. However, as we show in the following example, the product of two cycles need not be a cycle. Let A = {1, 2, 3, 4, 5, 6}. Compute (4, 1, 3, 5)◦(5, 6, 3) and (5, 6, 3)◦(4, 1, 3, 5). Solution

We have (4, 1, 3, 5) =

and (5, 6, 3) = Then

(4, 1, 3, 5) ◦ (5, 6, 3) = = and

1 2 3 4 5 6 1 2 5 4 6 3

1 2 3 4 5 6 3 2 5 1 4 6 1 2 3 4 5 6 3 2 4 1 6 5

.

1 2 3 4 5 6 ◦ 1 2 5 4 6 3

1 2 3 4 5 6 (5, 6, 3) ◦ (4, 1, 3, 5) = ◦ 3 2 5 1 4 6 1 2 3 4 5 6 = . 5 2 6 1 4 3

Observe that

1 2 3 4 5 6 3 2 5 1 4 6

1 2 3 4 5 6 1 2 5 4 6 3

(4, 1, 3, 5) ◦ (5, 6, 3) = (5, 6, 3) ◦ (4, 1, 3, 5)

and that neither product is a cycle.

◆

Two cycles of a set A are said to be disjoint if no element of A appears in both cycles. Example 6

Let A = {1, 2, 3, 4, 5, 6}. Then the cycles (1, 2, 5) and (3, 4, 6) are disjoint, ◆ whereas the cycles (1, 2, 5) and (2, 4, 6) are not. It is not difficult to show that if p1 = (a1 , a2 , . . . , ar ) and p2 = (b1 , b2 , . . . , bs ) are disjoint cycles of A, then p1 ◦ p2 = p2 ◦ p1 . This can be seen by observing that p1 affects only the a’s, while p2 affects only the b’s. We shall now present a fundamental theorem and, instead of giving its proof, we shall give an example that imitates the proof.

233

Functions THEOREM 2 Example 7

A permutation of a finite set that is not the identity or a cycle can be written as a ■ product of disjoint cycles of length ≥ 2. Write the permutation

p=

1 2 3 4 5 6 7 8 3 4 6 5 2 1 8 7

of the set A = {1, 2, 3, 4, 5, 6, 7, 8} as a product of disjoint cycles. Solution

We start with 1 and find that p(1) = 3, p(3) = 6, and p(6) = 1, so we have the cycle (1, 3, 6). Next we choose the first element of A that has not appeared in a previous cycle. We choose 2, and we have p(2) = 4, p(4) = 5, and p(5) = 2, so we obtain the cycle (2, 4, 5). We now choose 7, the first element of A that has not appeared in a previous cycle. Since p(7) = 8 and p(8) = 7, we obtain the cycle (7, 8). We can then write p as a product of disjoint cycles as p = (7, 8) ◦ (2, 4, 5) ◦ (1, 3, 6).

◆

It is not difficult to show that in Theorem 2, when a permutation is written as a product of disjoint cycles, the product is unique except for the order of the cycles. We saw in Section 1 how a permutation of the alphabet produces a substitution code. Permutations are also used to produce transposition codes. Unlike a substitution code in which each letter is replaced by a substitute, the letters in transposition coded messages are not changed, but are rearranged. Thus if a message TEST THE WATERS is subjected to the permutation (1, 2, 3) ◦ (4, 7) ◦ (5, 10, 11) ◦ (6, 8, 12, 13, 9), where the numbers refer to the positions of the letters in the message, the message becomes STEEEATHSTTWR. If the permutation is known to both parties, then the receiver of the message has only to apply the inverse permutation to decode. Example 8

One commonly used transposition code is the keyword columnar transposition. For this it is only necessary to remember a keyword, say JONES. The message to be encoded is written under the keyword in successive rows, padding at the end if necessary. For example, the message THE FIFTH GOBLET CONTAINS THE GOLD would be arranged as shown: J T F B O N G

O H T L N S O

N E H E T T L

E F G T A H D

S I O C I E X

Note that the message has been padded with an X to fill out the row. Then the coded message is constructed by writing the columns in succession, beginning with column 4 (since E is the first keyword letter to appear in the alphabet) and following with the letters in columns 1, 3, 2, 5. The encoded message is thus FGTAHDTFBOGCEHETTLHTLNSOIOCIEX. The recipient of the message divides the number of letters by 5 to find that there are 6 rows. She writes the coded message, six letters at a time, in columns 4, ◆ 1, 3, 2, 5, then reads the original message from the rows.

234

Functions Notice that although the encoded message is a permutation of the original message string, this permutation depends on the length of the message. In Example 8, the permutation of the 30 positions begins 1 2 3 ··· . 7 19 13 · · · But using the keyword JONES to encode MAKE ME AN OFFER produces a permutation of 15 positions that begin 1 2 3 ··· . 4 10 7 · · · A common variation of this idea, used in the U.S. Civil War, is to transpose words rather than letters and add some superfluous known words for confusion. These extra words were called arbitraries during the Civil War.

Even and Odd Permutations A cycle of length 2 is called a transposition. That is, a transposition is a cycle p = (ai , aj ), where p(ai ) = aj and p(aj ) = ai . Observe that if p = (ai , aj ) is a transposition of A, then p ◦ p = 1A , the identity permutation of A. Every cycle can be written as a product of transpositions. In fact, (b1 , b2 , . . . , br ) = (b1 , br ) ◦ (b1 , br−1 ) ◦ · · · ◦ (b1 , b3 ) ◦ (b1 , b2 ). This case can be verified by induction on r, as follows:

Basis Step If r = 2, then the cycle is just (b1 , b2 ), which already has the proper form. Induction Step We use P(k) to show P(k + 1). Let (b1 , b2 , . . . , bk , bk+1 ) be a cycle of length k + 1. Then (b1 , b2 , . . . , bk , bk+1 ) = (b1 , bk+1 ) ◦ (b1 , b2 , . . . , bk ), as may be verified by computing the composition. Using P(k), (b1 , b2 , . . . , bk ) = (b1 , bk ) ◦ (b1 , bk−1 ) ◦ · · · ◦ (b1 , b2 ). Thus, by substitution, (b1 , b2 , . . . , bk+1 ) = (b1 , bk+1 ) ◦ (b1 , bk ) ◦ · · · ◦ (b1 , b3 )(b1 , b2 ). This completes the induction step. Thus, by the principle of mathematical induction, the result holds for every cycle. For example, (1, 2, 3, 4, 5) = (1, 5) ◦ (1, 4) ◦ (1, 3) ◦ (1, 2). We now obtain the following corollary of Theorem 2. Corollary 1

Every permutation of a finite set with at least two elements can be written as a ■ product of transpositions. Observe that the transpositions in Corollary 1 need not be disjoint.

Example 9

Write the permutation p of Example 7 as a product of transpositions.

235

Functions Solution

We have p = (7, 8) ◦ (2, 4, 5) ◦ (1, 3, 6). Since we can write (1, 3, 6) = (1, 6) ◦ (1, 3) (2, 4, 5) = (2, 5) ◦ (2, 4), we have p = (7, 8) ◦ (2, 5) ◦ (2, 4) ◦ (1, 6) ◦ (1, 3).

◆

We have observed that every cycle can be written as a product of transpositions. However, this can be done in many different ways. For example, (1, 2, 3) = (1, 3) ◦ (1, 2) = (2, 1) ◦ (2, 3) = (1, 3) ◦ (3, 1) ◦ (1, 3) ◦ (1, 2) ◦ (3, 2) ◦ (2, 3). It then follows that every permutation on a set of two or more elements can be written as a product of transpositions in many ways. However, the following theorem, whose proof we omit, brings some order to the situation. THEOREM 3

If a permutation of a finite set can be written as a product of an even number of transpositions, then it can never be written as a product of an odd number of ■ transpositions, and conversely. A permutation of a finite set is called even if it can be written as a product of an even number of transpositions, and it is called odd if it can be written as a product of an odd number of transpositions.

Example 10

Is the permutation

p=

1 2 3 4 5 6 7 2 4 5 7 6 3 1

even or odd? Solution

We first write p as a product of disjoint cycles, obtaining p = (3, 5, 6) ◦ (1, 2, 4, 7).

(Verify this.)

Next we write each of the cycles as a product of transpositions: (1, 2, 4, 7) = (1, 7) ◦ (1, 4) ◦ (1, 2) (3, 5, 6) = (3, 6) ◦ (3, 5). Then p = (3, 6) ◦ (3, 5) ◦ (1, 7) ◦ (1, 4) ◦ (1, 2). Since p is a product of an odd ◆ number of transpositions, it is an odd permutation. From the definition of even and odd permutations, it follows (see Exercises 22 through 24) that (a) the product of two even permutations is even. (b) the product of two odd permutations is even. (c) the product of an even and an odd permutation is odd. THEOREM 4

236

Let A = {a1 , a2 , . . . , an } be a finite set with n elements, n ≥ 2. There are n!/2 even permutations and n!/2 odd permutations.

Functions Proof Let An be the set of all even permutations of A, and let Bn be the set of all odd permutations. We shall define a function f : An → Bn , which we show is one to one and onto, and this will show that An and Bn have the same number of elements. Since n ≥ 2, we can choose a particular transposition q0 of A. Say that q0 = (an−1 , an ). We now define the function f : An → Bn by f(p) = q0 ◦ p,

p ∈ An .

Observe that if p ∈ An , then p is an even permutation, so q0 ◦p is an odd permutation and thus f(p) ∈ Bn . Suppose now that p1 and p2 are in An and f(p1 ) = f(p2 ). Then q0 ◦ p1 = q0 ◦ p2 .

(3)

We now compose each side of equation (3) with q0 : q0 ◦ (q0 ◦ p1 ) = q0 ◦ (q0 ◦ p2 ); so, by the associative property, (q0 ◦ q0 ) ◦ p1 = (q0 ◦ q0 ) ◦ p2 or, since q0 ◦ q0 = 1A , 1A ◦ p1 = 1A ◦ p2 p1 = p2 . Thus f is one to one. Now let q ∈ Bn . Then q0 ◦ q ∈ An , and f(q0 ◦ q) = q0 ◦ (q0 ◦ q) = (q0 ◦ q0 ) ◦ q = 1A ◦ q = q, which means that f is an onto function. Since f : An → Bn is one to one and onto, we conclude that An and Bn have the same number of elements. Note that An ∩ Bn = ∅ since no permutation can be both even and odd. Also, by Theorem 1, |An ∪ Bn | = n!. Thus, n! = |An ∪ Bn | = |An | + |Bn | − |An ∩ Bn | = 2|An |. We then have |An | = |Bn | =

n! . 2

■

4 Exercises 1. Which of the following functions f : R → R are permutations of R? (a) f is defined by f(a) = a − 1. (b) f is defined by f(a) = a2 . 2. Which of the following functions f : R → R are permutations of R? (a) f is defined by f(a) = a3 . (b) f is defined by f(a) = ea . 3. Which of the following functions f : Z → Z are permutations of Z? (a) f is defined by f(a) = a + 1. (b) f is defined by f(a) = (a − 1)2 .

4. Which of the following functions f : Z → Z are permutations of Z? (a) f is defined by f(a) = a2 + 1. (b) f is defined by f(a) = a3 − 3. In Exercises 5 through 8, let A = {1, 2, 3, 4, 5, 6} and p1 = p2 = p3 =

1 3

2 4

3 1

4 2

5 6

6 5

1 2

2 3

3 1

4 5

5 4

6 6

1 6

2 3

3 2

4 5

5 4

6 1

, , .

237

Functions 5. Compute (b) p3 ◦ p1 (a) p−1 1 6. Compute (b) p1 ◦ (p3 ◦ p−1 (a) (p2 ◦ p1 ) ◦ p2 2 ) 7. Compute −1 (b) p−1 (a) p−1 3 1 ◦ p2 8. Compute (b) p3 ◦ (p2 ◦ p1 )−1 (a) (p3 ◦ p2 ) ◦ p1

In Exercises 20 and 21, let A = {1, 2, 3, 4, 5, 6, 7, 8}. Determine whether the permutation is even or odd. 1 2 3 4 5 6 7 8 20. (a) 4 2 1 6 5 8 7 3 1 2 3 4 5 6 7 8 (b) 7 3 4 2 1 8 6 5

In Exercises 9 and 10, let A = {1, 2, 3, 4, 5, 6, 7, 8}. Compute the products.

21. (a) (6, 4, 2, 1, 5)

9. (a) (3, 5, 7, 8) ◦ (1, 3, 2) (b) (2, 6) ◦ (3, 5, 7, 8) ◦ (2, 5, 3, 4) 10. (a) (1, 4) ◦ (2, 4, 5, 6) ◦ (1, 4, 6, 7) (b) (5, 8) ◦ (1, 2, 3, 4) ◦ (3, 5, 6, 7) 11. Let A = {a, b, c, d, e, f, g}. Compute the products. (a) (a, f, g) ◦ (b, c, d, e) (b) (f, g) ◦ (b, c, f ) ◦ (a, b, c)

22. Prove that the product of two even permutations is even.

In Exercises 12 and 13, let A = {1, 2, 3, 4, 5, 6, 7, 8}. Write each permutation as the product of disjoint cycles. 1 2 3 4 5 6 7 8 12. (a) 4 3 2 5 1 8 7 6 1 2 3 4 5 6 7 8 (b) 2 3 4 1 7 5 8 6 1 2 3 4 5 6 7 8 13. (a) 6 5 7 8 4 3 2 1 1 2 3 4 5 6 7 8 (b) 2 3 1 4 6 7 8 5 14. Let A = {a, b, c, d, e, f, g}. Write each permutation as the product of disjoint cycles. a b c d e f g (a) g d b a c f e a b c d e f g (b) d e a b g f c 15. Let A = {1, 2, 3, 4, 5, 6, 7, 8}. Write each permutation as a product of transpositions. (b) (3, 1, 6) ◦ (4, 8, 2, 5) (a) (2, 1, 4, 5, 8, 6) 16. Code the message WHERE ARE YOU by applying the permutation (1, 7, 3, 5, 11) ◦ (2, 6, 9) ◦ (4, 8, 10). 17. Decode the message ATEHAOMOMNTI, which was encoded using the permutation (3, 7, 1, 12) ◦ (2, 5, 8) ◦ (4, 10, 6, 11, 9). 18. (a) Give the complete permutation of the positions for the message in Example 8. (b) Write the permutation found in part (a) as the product of disjoint cycles. 19. (a) Encode the message MAKE ME AN OFFER using the keyword JONES and the method of Example 8.

238

(b) Write the permutation of the positions for the message in part (a).

(b) (4, 8) ◦ (3, 5, 2, 1) ◦ (2, 4, 7, 1) 23. Prove that the product of two odd permutations is even. 24. Prove that the product of an even and an odd permutation is odd. 25. Let A = {1, 2, 3, 4, 5}. Let f = (5, 2, 3) and g = (3, 4, 1) be permutations of A. Compute each of the following and write the result as the product of disjoint cycles. (b) f −1 ◦ g−1 (a) f ◦ g 26. Show that if p is a permutation of a finite set A, then p2 = p ◦ p is a permutation of A. 27. Let A = {1, 2, 3, 4, 5, 6} and 1 2 3 p= 2 4 3

4 1

5 5

6 6

be a permutation of A. (a) Write p as a product of disjoint cycles. (b) Compute p−1 .

(c) Compute p2 .

(d) Find the period of p, that is, the smallest positive integer k such that pk = 1A . 28. Let A = {1, 2, 3, 4, 5, 6} and 1 2 3 4 5 6 p= 4 3 5 1 2 6 be a permutation of A. (a) Write p as a product of disjoint cycles. (b) Compute p−1 .

(c) Compute p2 .

(d) Find the period of p, that is, the smallest positive integer k such that pk = 1A . 29. (a) Use mathematical induction to show that if p is a permutation of a finite set A, then pn = p ◦ p ◦ · · · ◦ p (n factors) is a permutation of A for n ∈ Z+ . (b) If A is a finite set and p is a permutation of A, show that pm = 1A for some m ∈ Z+ . 30. Let p be a permutation of a set A. Define the following relation R on A: a R b if and only if pn (a) = b for some n ∈ Z. [p0 is defined as the identity permutation and p−n is defined as (p−1 )n .] Show that R is an equivalence relation and describe the equivalence classes.

Functions 31. Build a composition table for the permutations of A = {1, 2, 3} given in Example 2.

length 3 can be formed using the elements of A and such that a1 < a2 < a3 ?

32. Describe how to use the composition table in Exercise 31 to identify p−1 for any permutation p of A.

37. We call a permutation p = (a1 , a2 , . . . , an ) up-down if the elements in the odd positions form an increasing sequence and the elements in the even positions form a decreasing sequence. (a) Let A = {1, 2, 3}. How many up-down permutations of A are there?

33. Find all subsets of {1A , p1 , p2 , p3 , p4 , p5 }, the permutations in Example 2, that satisfy the closure property for composition. 34. For each permutation, p, of A in Example 2, determine its period. How does this relate to the subset in Exercise 33 to which p belongs? 35. Let A = {1, 2, 3, . . . , n}. How many permutations of A, p = (a1 , a2 , . . . , an ), are there for which ai < ai+1 , 1 ≤ i ≤ n − 1? How many permutations of A, p = (a1 , a2 , . . . , an ), are there for which ai > ai+1 , 1 ≤ i ≤ n − 1? 36. Let A = {1, 2, 3, 4, 5}. How many different sequences of

(b) Let A = {1, 2, 3, 4}. How many up-down permutations of A are there? 38. Let A = {1, 2, 3, 4, 5}. How many up-down permutations of A are there? 39. Prove that the number of up-down permutations for A = {1, 2, 3, . . . , n} is the as the number of increasing same sequences of length n2 that can be formed from elements of A.

Tips for Proofs

Before beginning a proof, you might find it helpful to consider what the statement does not say. This can help clarify your thinking about what facts and tools are available for the proof. Consider Theorem 4, Section 1. It does not say that if f is one to one, then f is onto. The additional facts that |A| = |B| and that f is everywhere defined will need to be used in the proof. To show that a function is one to one or onto, we need to use generic elements. See Example 11, Section 1. Either the definition of one-to-oneness or its contrapositive may be used to prove this property. We also have the fact that if f : A → B is everywhere defined and |A| = |B| = n, then f is one to one if and only if f is onto. In addition, if we wish to show f is one to one and onto, we may do this by constructing the inverse function f −1 . Establishing a one-to-one correspondence is a powerful counting strategy, because it allows us to count a different set than the original one. For example, see Theorem 4, Section 4, and Exercises 32 through 34, Section 2. To prove that f and g have the same order or one is of lower order than the other, the principal tools are the rules for -classes or manipulation of inequalities (Section 3, Examples 2 and 3).

Key Ideas for Review • Function • Identity function, 1A : 1A (a) = a • One-to-one function f from A to B: a = a implies • • • • • •

f(a) = f(a ) Onto function f from A to B: Ran(f ) = B Bijection: one-to-one and onto function One-to-one correspondence: onto, one-to-one, everywhere defined function If f is a function from A to B, 1B ◦ f = f ; f ◦ 1A = f If f is an invertible function from A to B, f −1 ◦ f = 1A ; f ◦ f −1 = 1B (g ◦ f )−1 = f −1 ◦ g−1

• Substitution code • Boolean function f : Ran(f ) ⊆ {true, false} • Hashing function • Fuzzy set • Fuzzy predicate • O(g) (big Oh of g) • f and g of the same order: f is O(g) and g is O(f ) • Theorem: The relation , f g if and only if f and g have

the same order, is an equivalence relation. • Lower -class • Rules for determining -class of a function

239

Functions • Running time of an algorithm: -class of a function that • • • • • • • •

describes the number of steps performed by the algorithm Permutation function: a bijection from a set A to itself Theorem: If A is a set that has n elements, then there are n! permutations of A. Cycle of length r: (b1 , b2 , . . . , br ) Theorem: A permutation of a finite set that is not the identity or a cycle can be written as a product of disjoint cycles. Transposition: a cycle of length 2 Corollary: Every permutation of a finite set with at least two elements can be written as a product of transpositions. Transposition code Keyword columnar transposition code

• Even (odd) permutation: one that can be written as a product

of an even (odd) number of transpositions • Theorem: If a permutation of a finite set can be written as

a product of an even number of transpositions, then it can never be written as a product of an odd number of transpositions, and conversely. • The product of

(a) Two even permutations is even. (b) Two odd permutations is even. (c) An even and an odd permutation is odd. • Theorem: If A is a set that has n elements, then there are n!/2 even permutations and n!/2 odd permutations of A.

Chapter Self-Test 1. How does a function differ from a general relation? 2. What is a common strategy to prove that a function is one to one? 3. What is a common strategy to prove that a function is onto? 4. Why are mod functions often used in constructing hashing functions? 5. What does the -class of a function represent? 6. Let A = {a, b, c, d}, B = {1, 2, 3}, and R = {(a, 2), (b, 1), (c, 2), (d, 1)}. Is R a function? Is R−1 a function? Explain your answers. 7. Let A = B = R. Let f : A → B be the function defined by f(x) = −5x + 8. Show that f is one to one and onto. 8. Compute (b) −1.6 (a) 16.29 9. Compute (b) −1.6 (a) 16.29 10. Compute (b) lg(64) (a) lg(1) 11. Let Q be the propositional function defined by 1 0 Q(x) : ∃y xy = . 0 1 2 1 2 3 Evaluate Q and Q . 0 3 4 6 12. Assume that 9500 account records need to be stored using the hashing function h, which takes the first two digits of the account number as one number and the last four digits as another number, adds them, and then applies the mod-89 function.

240

(a) How many linked lists will be needed? (b) If an approximately even distribution of records is achieved, roughly how many records will be stored in each linked list? (c) Compute h(473810), h(125332), and h(308691). 13. Show that f(n) = 2n2 + 9n + 5 is O(n2 ). 14. Determine the -class of f(n) = lg(n) + n2 + 2n . 15. Consider the following pseudocode. 1. X ← 10 2. I ← 0 3. UNTIL (I > N) a. X ← X + I b. I ← I + 2

Write a function of N that describes the number of steps required and give the -class of the function. 16. Let A = {1, 2, 3, 4, 5, 6} and let p1 = (3, 6, 2) and p2 = (5, 1, 4) be permutations of A. (a) Compute p1 ◦ p2 and write the result as a product of cycles and as the product of transpositions. −1 (b) Compute p−1 1 ◦ p2 . 1 2 3 4 17. Let p1 = 7 3 2 1 1 2 3 4 5 p2 = 6 3 2 1 5 (a) Compute p1 ◦ p2 .

5 4 6 4

6 7 5 6 7 . 7

and

(b) Compute p−1 1 . (c) Is p1 an even or odd permutation? Explain.

Functions

Experiment 5 The -class of a function that describes the number of steps performed by an algorithm is referred to as the running time of the algorithm. In this experiment you will analyze several algorithms, presented in pseudocode, to determine their running times. Part I. The first algorithm is one method for computing the product of two n × n matrices. Assume that the matrices are each stored in an array of dimension 2 and that A[i, j] holds the element of A in row i and column j. Algorithm MATMUL(A, B; C) 1. FOR I = a. FOR 1. 2.

1 THRU N J = 1 THRU N C[I,J] ← 0 FOR K = 1 THRU N a. C[I,J] ← C[I,J]

+

A[I,K]

×

B[K,J]

●

Assume that each assignment of a value, each addition, and each element multiplication take the same fixed amount of time. 1. How many assignments will be done in the second FOR loop? 2. How many element multiplications are done in the third FOR loop? 3. What is the running time of MATMUL? Justify your answer. Part II. The following recursive algorithm will compute n! for any positive integer n. Algorithm FAC(N) 1. IF (N = 1) THEN a. A ← 1 2. ELSE a. A ← N × FAC(N 3. RETURN (A)

−

1)

●

1. Let Sn be the number of steps needed to calculate n! using FAC. Write a recurrence relation for Sn in terms of Sn−1 . 2. Solve the recurrence relation in question 1 and use the result to determine the running time of FAC. Part III. The function SEEK will give the cell in which a specified value is stored in cells i through i + n − 1 (inclusive) of an array A. Assume that i ≥ 1. FUNCTION SEEK(ITEM, I, I + N 1. CELL ← 0 2. FOR J = I THRU I + N − 1 a. IF (A[J] = ITEM) THEN b. CELL ← J 3. RETURN (CELL) END OF FUNCTION SEEK

−

1)

●

1. How many cells are there from A[i] to A[i + n − 1] (inclusive)? 2. Give a verbal description of how SEEK operates.

241

Functions 3. What is the running time of SEEK? Justify your answer. Part IV. The algorithm HUNT will give the cell in which a specified value is stored in cells i through i + n − 1 (inclusive) of an array A. Assume that i ≥ 1. To simplify the analysis of this algorithm, assume that n, the number of cells to be inspected, is a power of 2. Algorithm HUNT (ITEM, I, I + N − 1) 1. CELL ← 0 2. IF (N = 1 AND A[I] = ITEM) THEN a. CELL ← I 3. ELSE a. CELL1 ← HUNT(ITEM, I, I + N/2 − 1) b. CELL2 ← HUNT(ITEM, I + N/2, I + N − 1) 4. IF (CELL1 = 0) THEN a. CELL ← CELL1 5. ELSE a. CELL ← CELL2 6. RETURN (CELL)

●

1. Give a verbal description of how HUNT operates. 2. What is the running time of HUNT? Justify your answer. 3. Under what circumstances would it be better to use SEEK (Part III) rather than HUNT? When would it be better to use HUNT rather than SEEK?

Coding Exercises 1.

Let U = {u1 , u2 , . . . , un } be the universal set for possible input sets. Write a function CHARFCN that given a set as input returns the characteristic function of the set as a sequence.

2. Write a function TRANSPOSE that, given an n×n matrix, returns its transpose. 3. Write a program that writes a given permutation as a product of disjoint cycles. 4. Write a program that writes a given permutation as a product of transpositions. 5.

Use the program in Exercise 4 as a subroutine in a program that determines whether a given permutation is even or odd.

Answers to Odd-Numbered Exercises Exercise Set 1

11. (a) Both.

(b) Neither.

13. (a) Both.

(b) Onto.

3. Yes.

15. (a) Both.

(b) Onto.

5. Each integer has a unique square that is also an integer.

17. Neither.

1. (a) Yes. Ran(R) = {1, 2}.

(b) No.

7. Each r ∈ R is either an integer or it is not. 9. (a) 3. (d) x2 − 1.

242

(b) 1.

(c) (x − 1)2 .

(e) y − 2.

(f) y4 .

a+1 =2 − 1 = a. 2 √ (b) (g ◦ f )(a) = g(a2 − 1) = a2 + 1 − 1 = |a| = a, since a ≥ 0.

a+1 19. (a) (g ◦ f )(a) = g 2

Functions 21. (a) f −1 (b) = b2 − 1.

(b) f −1 (b) =

√ 3

b − 1.

23. (a) Suppose f(x, y) = f(w, z). Then 2x − y = 2w − z and x − 2y = w − 2z. Solving this system of equations gives x = w, y = z. Therefore, f is one to one. 2x − y x − 2y , . (b) f −1 (x, y) = 3 3 29. nm . 25. Yes. 27. nn . 31. C; B; because g is onto; A; because f is onto. 33. Suppose g ◦ f is onto. Let c ∈ C. Then ∃a ∈ A such that (g ◦ f )(a) = c. But (g ◦ f )(a) = g(f(a)), f(a) ∈ B, so g is onto. 35. (a) Let Dom(f ) = {a1 , a2 , . . . , an }. Then Ran(f ) = {f(a1 ), f(a2 ), . . . , f(an )}. Since f is one to one, the f(ai ) are all distinct, so Ran(f ) has n elements. Thus m = n. (b) If f is not one to one, some f(ai ) = f(aj ) and m < n. 37. One one-to-one correspondence is f : Z+ → B, where f(z) = 2z. 39. Yes, the cardinality of a set is determined by creating a oneto-one correspondence between the set and a subset of Z+ . Here we use all of Z+ . 41. (a) and (b). Consider the table for . 0 1

0 0 1

21. x T T T T F F F F

y T T F F T T F F

f(x, y, z) T F T F T T T F

z T F T F T F T F

23. (a) True.

(b) False.

(c) False.

(d) True.

25. (a) 31. (b) 0. (c) 36. m +2 . 27. k 2 −1 29. f (1) is the set of elements of A. 31. (a)

3 . 4

(b)

15 . 16

(c)

1 . 16

33. (a) 3 and 9. (b) 4 and 8. ⎧ 35. ⎪ 0 for 0 ≤ x < 2 ⎪ √ ⎪ ⎪1 – ⎪ x 1 for 2≤ x k with N > c · lg(N). Then N 2 ≤ c · N · lg(N) < N 2 , a contradiction. 9. |5n2 +4n+3| ≤ |5n2 +500n|, n ≥ 1; |5n2 +500n| ≤ 5|n2 + 100n|. We have |n2 + 100n| = |n2 + 4 · 25n| ≤ |n2 + 4n2 |, n ≥ 25. But |5n2 | ≤ |5n2 + 4n + 3|. 11. {f5 }, {f6 , f10 , f11 }, {f7 }, {f4 }, {f8 }, {f1 }, {f2 }, {f3 }, {f9 }, {f12 }. 13. f1 , (n lg n); f2 , (n2 ); f4 , (lg n); f5 , (1); f6 , (n); f10 , (n); f11 , (n). 15. f(n) = 2 + 4 · 5 + 1 or f(n) = 23. (1). 17. f(n) = 1 + n · 2. (n). . (n2 ). 19. f(n) = 2 + 2n + n(n+1) 2 21. f(n, m, q) = 1 + nq + 3nmq + 1. Let N = max(n, m, q), then f is (N 3 ). 22. |rf(n)| ≤ |r| · |f(n)|, n ≥ 1, so rf is O(f ). Choose c such that c · |r| ≥ 1, then |f(n)| ≤ c · |r| · |f(n)| = c · |rf(n)|, n ≥ 1 and f is O(rf ). 23. There exist c1 , k1 such that |f(n)| ≤ c1 |g(n)|, ∀n ≥ k1 , so |cf(n)| = |c| · |f(n)| ≤ c1 |c| · |g(n)|, ∀n ≥ k1 . Also, there exist c2 , k2 such that |g(n)| ≤ c2 |f(n)|, ∀n ≥ k2 and so c |g(n)| ≤ c2 |f(n)| = |c|2 · |cf(n)|, ∀n ≥ k2 . 24. Suppose h(n) > 0, ∀n, (f ) lower than (or the same as) (g). |f(n)| ≤ c · |g(n)|, n ≥ k (and |g(n)| ≤ d · |f(n)|, n ≥ l). h(n)|f(n)| ≤ c · h(n) · |g(n)|, n ≥ k (and h(n)|g(n)| ≤ d · h(n) · |f(n)|, n ≥ l). Hence |f(n) · h(n)| ≤ c · |g(n) · h(n)|, n ≥ k (and |g(n) · h(n)| ≤ d · |f(n) · h(n)|, n ≥ l). Hence (fh) is lower than (or the same as) (gh). Note that if (f ) is strictly lower than (g), then (fh) must be strictly lower than (gh). 25. (a) Pn = Pn−1 + (n − 2) + (n − 3), P3 = 1, P4 = 4. (b) θ(n2 ). Exercise Set 4

244

(b)

1 6

2 3

3 2

4 5

5 4

1 1

2 3

3 4

4 6

5 2

6 . 1 6 . 5

9. (a) (1, 5, 7, 8, 3, 2).

Exercise Set 3

1. (a) Yes. 1 5. (a) 3 1 (b) 2

7. (a)

(b) (2, 7, 8, 3, 4, 6).

11. (a) (a, f, g) ◦ (b, c, d, e).

(b) (a, c) ◦ (b, g, f ).

13. (a) (1, 6, 3, 7, 2, 5, 4, 8). (b) (5, 6, 7, 8) ◦ (1, 2, 3). 15. (a) (2, 6) ◦ (2, 8) ◦ (2, 5) ◦ (2, 4) ◦ (2, 1). (b) (3, 6) ◦ (3, 1) ◦ (4, 5) ◦ (4, 2) ◦ (4, 8). 17. I AM NOT AT HOME. 19. (a) EOXMEFKNRAAEMFX. 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 . (b) 4 10 7 1 13 5 11 8 2 14 6 12 9 3 15 21. (a) Even.

(b) Odd.

23. Suppose p1 is the product of 2k1 + 1 transpositions and p2 is the product of 2k2 + 1 transpositions. Then p2 ◦ p1 can be written as the product of 2(k1 + k2 ) + 2 transpositions. By Theorem 3, p2 ◦ p1 is even. 25. (a) (1, 5, 2, 3, 4). 27. (a) (1, 2, 4). 1 2 (b) 4 1 1 2 (c) 4 1

(b) (1, 4, 2, 5, 3).

3 3

4 2

5 5

3 3

4 2

5 5

6 . 6 6 . 6

(d) 3. 29. (a) Basis step: n = 1. If p is a permutation of a finite set A, then p1 is a permutation of A is true. Induction step: The argument in Exercise 26 also shows that if pn−1 is a permutation of A, then pn−1 ◦ p is a permutation of A. Hence pn is a permutation of A. (b) If |A| = n, then there are n! permutations of A. Hence, the sequence 1A , p, p2 , p3 , . . . is finite and pi = pj for some i = j. Suppose i < j. Then p−i ◦ pi = 1A = p−i ◦ pj . So pj−i = 1A , j − i ∈ Z. 31. 1A p1 p2 p3 p4 p5

1A 1A p1 p2 p3 p4 p5

p1 p1 1A p3 p2 p5 p4

p2 p2 p4 1A p5 p1 p3

p3 p3 p5 p1 p4 1A p2

p4 p4 p2 p5 1A p3 p1

p5 p5 p3 p4 p1 p2 1A

33. {1A }, {1A , p1 }, {1A , p2 }, {1A , p1 , p2 , p3 , p4 , p5 }.

{1A , p5 },

{1A , p3 , p4 },

35. There is exactly one of each kind. (b) No. 3. (a) Yes. 2 3 4 5 6 . 4 1 2 6 5 2 3 4 5 6 . 5 6 3 1 4

(b) No.

37. (a) 3.

(b) 6.

39. For each increasing sequence of length n2 , there is exactly one associated up-down permutation of A, because there is just one way to arrange the remaining elements of A in decreasing order and insert them to fill the even positions.

Functions

Answers to Chapter Self-Tests 1. Let f : A → B be a function, then |f(a)| ≤ 1, a ∈ A. 2. Assume that f(a1 ) = f(a2 ) and show that a1 = a2 . 3. Let f : A → B. Choose b ∈ B and find a ∈ A such that f(a) = b. 4. A hashing function is designed to assign items to a limited number of storage places, and any mod-n function has only n outputs. 5. The -class of a function f represents an approximation of how values f(n) grow as n grows. 6. (a) Yes, |R(x)| = 1, x ∈ A. (b) No, (1, b), (1, d) ∈ R−1 . 7. Suppose f(a) = f(b), then −5a + 8 = −5b + 8. But then a = b so f is one to one. Let r ∈ R. Then r−8 r−8 = −5 + 8 = r − 8 + 8 = r. f −5 −5 So f is onto. 8. (a) 16. (b) −2.

9. (a) 17. (b) −1. 10. (a) 0.

(b) 6.

11. True; false. 12. (a) 89.

(b) 107.

(c) 30; 4; 88.

13. 2n + 9n + 5 ≤ 2n + n + n , n ≥ 9. Choose 4 for c and 9 for k. Then |2n2 + 9n + 5| ≤ 4|n2 |, n ≥ 9. 2

14. (2n ). 15. f(N) = 2 + 5

2

N 2

2

2

+ 1; (n).

16. (a) (1, 4, 5) ◦ (2, 3, 6); (b) (2, 6, 3) ◦ (1, 5, 4). 1 2 3 4 5 17. (a) 5 2 3 7 4 1 2 3 4 5 (b) 4 3 2 5 6

(1, 5) ◦ (1, 4) ◦ (2, 6) ◦ (2, 3). 6 1 6 7

7 . 6 7 . 1

(c) p1 = (1, 7, 6, 5, 4) ◦ (2, 3) = (1, 4) ◦ (1, 5) ◦ (1, 6) ◦ (1, 7) ◦ (2, 3);

odd.

245

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Order Relations and Structures In this chapter we study partially ordered sets, including lattices and Boolean algebras. These structures are useful in set theory, algebra, sorting and searching, and, especially in the case of Boolean algebras, in the construction of logical representations for computer circuits.

Looking Back George Boole is known for his development of symbolic logic. In 1938, Claude E. Shannon, an American mathematician wrote a Master’s thesis at MIT entitled A Symbolic Analysis of Relay and Switching Circuits, in which he showed how to use the work of Boole in the design of circuits. The heart of a computer consists of circuits that perform the many different operations and tasks of the computer. Every electronic circuit can be written as a combination of the connectors and, or, and not, and by using a device that can be turned on or off, we can construct each type of connector. Since this

combination is not unique, it is desirable to write it in as simple a manner as possible (a problem that will be examined briefly in this chapter). The benefits of doing so include greater speed, smaller circuits, and lower power consumption. The field in computer science that deals with this problem is logic design, and Boolean algebra plays a major role in this field. As electronic circuits have grown enormously in size and complexity, the field of computer-aided design (CAD) has been developed, whereby very sophisticated computer programs carry out the logic design.

Printed Circuit Board Printed Circuit Board Repair Specialists

From Discrete Mathematical Structures, Sixth Edition, Bernard Kolman, Robert C. Busby, Sharon Cutler Ross. Copyright © 2009 by Pearson Education, Inc. Published by Prentice Hall. All rights reserved.

247

Order Relations and Structures

1

Partially Ordered Sets A relation R on a set A is called a partial order if R is reflexive, antisymmetric, and transitive. The set A together with the partial order R is called a partially ordered set, or simply a poset, and we will denote this poset by (A, R). If there is no possibility of confusion about the partial order, we may refer to the poset simply as A, rather than (A, R).

Example 1

Let A be a collection of subsets of a set S. The relation ⊆ of set inclusion is a ◆ partial order on A, so (A, ⊆) is a poset.

Example 2

Let Z+ be the set of positive integers. The usual relation ≤ (less than or equal to) ◆ is a partial order on Z+ , as is ≥ (greater than or equal to).

Example 3

The relation of divisibility (a R b if and only if a | b) is a partial order on Z+ . ◆

Example 4

Let R be the set of all equivalence relations on a set A. Since R consists of subsets of A × A, R is a partially ordered set under the partial order of set containment. If R and S are equivalence relations on A, the same property may be expressed in relational notation as follows. R ⊆ S if and only if x R y

implies

x S y for all x, y in A.

Then (R, ⊆) is a poset.

◆

Example 5

The relation < on Z+ is not a partial order, since it is not reflexive.

◆

Example 6

Let R be a partial order on a set A, and let R−1 be the inverse relation of R. Then R−1 is also a partial order. If R has these three properties, then ⊆ R, R ∩ R−1 ⊆ , and R2 ⊆ R. By taking inverses, we have = −1 ⊆ R−1 ,

R−1 ∩ (R−1 )−1 = R−1 ∩ R ⊆ ,

and (R−1 )2 ⊆ R−1 ,

so, R−1 is reflexive, antisymmetric, and transitive. Thus R−1 is also a partial order. The poset (A, R−1 ) is called the dual of the poset (A, R), and the partial order R−1 is called the dual of the partial order R. ◆ The most familiar partial orders are the relations ≤ and ≥ on Z and R. For this reason, when speaking in general of a partial order R on a set A, we shall often use the symbols ≤ or ≥ for R. This makes the properties of R more familiar and easier to remember. Thus the reader may see the symbol ≤ used for many different partial orders on different sets. Do not mistake this to mean that these relations are all the same or that they are the familiar relation ≤ on Z or R. If it becomes absolutely necessary to distinguish partial orders from one another, we may also use symbols such as ≤1 , ≤ , ≥1 , ≥ , and so on, to denote partial orders. We will observe the following convention. Whenever (A, ≤) is a poset, we will always use the symbol ≥ for the partial order ≤−1 , and thus (A, ≥) will be the dual poset. Similarly, the dual of poset (A, ≤1 ) will be denoted by (A, ≥1 ), and the dual of the poset (B, ≤ ) will be denoted by (B, ≥ ). Again, this convention is to remind us of the familiar dual posets (Z, ≤) and (Z, ≥), as well as the posets (R, ≤) and (R, ≥). If (A, ≤) is a poset, the elements a and b of A are said to be comparable if a≤b

248

or

b ≤ a.

Order Relations and Structures Observe that in a partially ordered set every pair of elements need not be comparable. For example, consider the poset in Example 3. The elements 2 and 7 are not comparable, since 2 7 and 7 2. Thus the word “partial” in partially ordered set means that some elements may not be comparable. If every pair of elements in a poset A is comparable, we say that A is a linearly ordered set, and the partial order is called a linear order. We also say that A is a chain. Example 7

◆

The poset of Example 2 is linearly ordered.

The following theorem is sometimes useful since it shows how to construct a new poset from given posets. THEOREM 1

If (A, ≤) and (B, ≤) are posets, then (A × B, ≤) is a poset, with partial order ≤ defined by (a, b) ≤ (a , b ) if a ≤ a in A and b ≤ b in B. Note that the symbol ≤ is being used to denote three distinct partial orders. The reader should find it easy to determine which of the three is meant at any time. Proof If (a, b) ∈ A × B, then (a, b) ≤ (a, b) since a ≤ a in A and b ≤ b in B, so ≤ satisfies the reflexive property in A × B. Now suppose that (a, b) ≤ (a , b ) and (a , b ) ≤ (a, b), where a and a ∈ A and b and b ∈ B. Then

and

a ≤ a

and

a ≤ a

in A

b ≤ b

and

b ≤ b

in B.

Since A and B are posets, the antisymmetry of the partial orders on A and B implies that a = a and b = b . Hence ≤ satisfies the antisymmetry property in A × B. Finally, suppose that (a, b) ≤ (a , b )

and

(a , b ) ≤ (a , b ),

where a, a , a ∈ A, and b, b , b ∈ B. Then a ≤ a

and

a ≤ a ,

so a ≤ a , by the transitive property of the partial order on A. Similarly, b ≤ b

and

b ≤ b ,

so b ≤ b , by the transitive property of the partial order on B. Hence (a, b) ≤ (a , b ). Consequently, the transitive property holds for the partial order on A × B, and we ■ conclude that A × B is a poset. The partial order ≤ defined on the Cartesian product A×B is called the product partial order. If (A, ≤) is a poset, we say that a < b if a ≤ b but a = b. Suppose now that (A, ≤) and (B, ≤) are posets. In Theorem 1 we have defined the product partial

249

Order Relations and Structures order on A × B. Another useful partial order on A × B, denoted by ≺, is defined as follows: (a, b) ≺ (a , b )

if a < a or if a = a and b ≤ b .

This ordering is called lexicographic, or “dictionary” order. The ordering of the elements in the first coordinate dominates, except in case of “ties,” when attention passes to the second coordinate. If (A, ≤) and (B, ≤) are linearly ordered sets, then the lexicographic order ≺ on A × B is also a linear order. Example 8

p3

Let A = R, with the usual ordering ≤. Then the plane R2 = R × R may be given lexicographic order. This is illustrated in Figure 1. We see that the plane is linearly ordered by lexicographic order. Each vertical line has the usual order, and points on one line are less than points on a line farther to the right. Thus, in Figure 1, p1 ≺ p2 , p1 ≺ p3 , and p2 ≺ p3 . ◆

p1 p2 x x = x1

x = x2

Figure 1

Lexicographic ordering is A1 × A2 × · · · × An as follows:

easily

extended

to

(a1 , a2 , . . . , an ) ≺ (a1 , a2 , . . . , an ) if and only if a1 < a1 or a1 = a1 and a2 < a2 or a1 = a1 , a2 = a2 , and a3 < a3 or . . . and a1 = a1 , a2 = a2 , . . . , an−1 = an−1

Cartesian

products

an ≤ an .

Thus the first coordinate dominates except for equality, in which case we consider the second coordinate. If equality holds again, we pass to the next coordinate, and so on.

Example 9

Let S = {a, b, . . . , z} be the ordinary alphabet, linearly ordered in the usual way (a ≤ b, b ≤ c, . . . , y ≤ z). Then S n = S × S × · · · × S (n factors) can be identified with the set of all words having length n. Lexicographic order on S n has the property that if w1 ≺ w2 (w1 , w2 ∈ S n ), then w1 would precede w2 in a dictionary listing. This fact accounts for the name of the ordering. Thus park ≺ part, help ≺ hind, jump ≺ mump. The third is true since j < m; the second, since h = h, e < i; and the first is true since p = p, a = a, r = r, ◆ k < t. If S is a poset, we can extend lexicographic order to S ∗ in the following way. If x = a1 a2 · · · an and y = b1 b2 · · · bk are in S ∗ with n ≤ k, we say that x ≺ y if (a1 , . . . , an ) ≺ (b1 , . . . , bn ) in S n under lexicographic ordering of S n . In the previous paragraph, we use the fact that the n-tuple (a1 , a2 , . . . , an ) ∈ S n , and the string a1 a2 · · · an ∈ S ∗ are really the same sequence of length n, written in two different notations. The notations differ for historical reasons, and we will use them interchangeably depending on context.

250

Order Relations and Structures Example 10

Let S be {a, b, . . . , z}, ordered as usual. Then S ∗ is the set of all possible “words” of any length, whether such words are meaningful or not. Thus we have help ≺ helping in S ∗ since

help ≺ help

in S 4 . Similarly, we have since

helper ≺ helping helper ≺ helpin

6

in S . As the example

help ≺ helping

shows, this order includes prefix order; that is, any word is greater than all of its prefixes (beginning parts). This is also the way that words occur in the dictionary. Thus we have dictionary ordering again, but this time for words of any finite ◆ length. Since a partial order is a relation, we can look at the digraph of any partial order on a finite set. We shall find that the digraphs of partial orders can be represented in a simpler manner than those of general relations. The following theorem provides the first result in this direction. THEOREM 2

The digraph of a partial order has no cycle of length greater than 1. Proof Suppose that the digraph of the partial order ≤ on the set A contains a cycle of length n ≥ 2. Then there exist distinct elements a1 , a2 , . . . , an ∈ A such that a1 ≤ a2 ,

a2 ≤ a3 ,

...,

an−1 ≤ an ,

an ≤ a1 .

By the transitivity of the partial order, used n − 1 times, a1 ≤ an . By antisymmetry, an ≤ a1 and a1 ≤ an imply that an = a1 , a contradiction to the assumption that ■ a1 , a2 , . . . , an are distinct.

Hasse Diagrams Let A be a finite set. Theorem 2 has shown that the digraph of a partial order on A has only cycles of length 1. Indeed, since a partial order is reflexive, every vertex in the digraph of the partial order is contained in a cycle of length 1. To simplify matters, we shall delete all such cycles from the digraph. Thus the digraph shown in Figure 2(a) would be drawn as shown in Figure 2(b). We shall also eliminate all edges that are implied by the transitive property. Thus, if a ≤ b and b ≤ c, it follows that a ≤ c. In this case, we omit the edge from a to c; however, we do draw the edges from a to b and from b to c. For example, the digraph shown in Figure 3(a) would be drawn as shown in Figure 3(b). We also agree to draw the digraph of a partial order with all edges pointing upward, so that arrows may be omitted from the edges. Finally, we replace the circles representing the vertices by dots. Thus the diagram shown in Figure 4 gives the final form of the digraph shown in Figure 2(a). The resulting diagram of a partial order, much simpler than its digraph, is called the Hasse diagram of the partial order of the poset. Since the Hasse diagram completely describes the associated partial order, we shall find it to be a very useful tool.

251

Order Relations and Structures c

c

b

b a

a

(a)

(b)

Figure 2

c

c

b

b

a

a a

a b

(a)

(b)

c

c (a)

Figure 3

Example 11

b

(b)

Figure 4

Let A = {1, 2, 3, 4, 12}. Consider the partial order of divisibility on A. That is, if a and b ∈ A, a ≤ b if and only if a | b. Draw the Hasse diagram of the poset (A, ≤). Solution

The Hasse diagram is shown in Figure 5. To emphasize the simplicity of the Hasse ◆ diagram, we show in Figure 6 the digraph of the poset in Figure 5.

12

12

4

4

3

3 2

2

1 1

Figure 5

Example 12

Figure 6

Let S = {a, b, c} and A = P(S). Draw the Hasse diagram of the poset A with the partial order ⊆ (set inclusion). Solution We first determine A, obtaining

A = {∅, {a}, {b}, {c}, {a, b}, {a, c}, {b, c}, {a, b, c}}. The Hasse diagram can then be drawn as shown in Figure 7.

252

◆

Order Relations and Structures {a, b, c} b {b, c}

{a, b}

{c}

{b}

{a, c} c d {a} e

∅

f

Figure 7

Figure 8

Observe that the Hasse diagram of a finite linearly ordered set is always of the form shown in Figure 8. It is easily seen that if (A, ≤) is a poset and (A, ≥) is the dual poset, then the Hasse diagram of (A, ≥) is just the Hasse diagram of (A, ≤) turned upside down. Example 13

Figure 9(a) shows the Hasse diagram of a poset (A, ≤), where A = {a, b, c, d, e, f }. Figure 9(b) shows the Hasse diagram of the dual poset (A, ≥). Notice that, as stated, each of these diagrams can be constructed by turning the other upside down. ◆ a

f

b

c

e

d

d

a

b (a)

c

e

f (b)

Figure 9

Topological Sorting If A is a poset with partial order ≤, we sometimes need to find a linear order ≺ for the set A that will merely be an extension of the given partial order in the sense that if a ≤ b, then a ≺ b. The process of constructing a linear order such as ≺ is called topological sorting. This problem might arise when we have to enter a finite poset A into a computer. The elements of A must be entered in some order, and we might want them entered so that the partial order is preserved. That is, if a ≤ b, then a is entered before b. A topological sorting ≺ will give an order of entry of the elements that meets this condition. Example 14

Give a topological sorting for the poset whose Hasse diagram is shown in Figure 10.

253

Order Relations and Structures Solution

The partial order ≺ whose Hasse diagram is shown in Figure 11(a) is clearly a linear order. It is easy to see that every pair in ≤ is also in the order ≺, so ≺ is a topological sorting of the partial order ≤. Figures 11(b) and (c) show two other ◆ solutions to this problem. f

f

g

g

f

f

e d

c

e e

g

b

d

b

c

g

b

c

e b d

d c a (a)

a

Figure 10

a (b)

a (c)

Figure 11

Isomorphism Let (A, ≤) and (A , ≤ ) be posets and let f : A → A be a one-to-one correspondence between A and A . The function f is called an isomorphism from (A, ≤) to (A , ≤ ) if, for any a and b in A, a≤b

if and only if

f(a) ≤ f(b).

If f : A → A is an isomorphism, we say that (A, ≤) and (A , ≤ ) are isomorphic posets. Example 15

Let A be the set Z+ of positive integers, and let ≤ be the usual partial order on A (see Example 2). Let A be the set of positive even integers, and let ≤ be the usual partial order on A . The function f : A → A given by f(a) = 2a is an isomorphism from (A, ≤) to (A , ≤ ). First, f is one to one since, if f(a) = f(b), then 2a = 2b, so a = b. Next, Dom(f) = A, so f is everywhere defined. Finally, if c ∈ A , then c = 2a for some a ∈ Z+ ; therefore, c = f(a). This shows that f is onto, so we see that f is a one-to-one correspondence. Finally, if a and b are elements of A, then it is clear ◆ that a ≤ b if and only if 2a ≤ 2b. Thus f is an isomorphism. Suppose that f : A → A is an isomorphism from a poset (A, ≤) to a poset (A , ≤ ). Suppose also that B is a subset of A, and B = f(B) is the corresponding subset of A . Then we see from the definition of isomorphism that the following principle must hold.

THEOREM 3 Principle of Correspondence

254

If the elements of B have any property relating to one another or to other elements of A, and if this property can be defined entirely in terms of the relation ≤, then the elements of B must possess exactly the same property, defined in terms of ≤ . ■

Order Relations and Structures Example 16

a

b c

d

Figure 12

Let (A, ≤) be the poset whose Hasse diagram is shown in Figure 12, and suppose that f is an isomorphism from (A, ≤) to some other poset (A , ≤ ). Note first that d ≤ x for any x in A (later we will call an element such as d a “least element” of A). Then the corresponding element f(d) in A must satisfy the property f(d) ≤ y for all y in A . As another example, note that a b and b a. Such a pair is called incomparable in A. It then follows from the principle of correspondence that f(a) and f(b) must be incomparable in A . ◆ For a finite poset, one of the objects that is defined entirely in terms of the partial order is its Hasse diagram. It follows from the principle of correspondence that two finite isomorphic posets must have the same Hasse diagrams. To be precise, let (A, ≤) and (A , ≤ ) be finite posets, let f : A → A be a one-to-one correspondence, and let H be any Hasse diagram of (A, ≤). Then 1. If f is an isomorphism and each label a of H is replaced by f(a), then H will become a Hasse diagram for (A , ≤ ). Conversely, 2. If H becomes a Hasse diagram for (A , ≤ ), whenever each label a is replaced by f(a), then f is an isomorphism. This justifies the name “isomorphism,” since isomorphic posets have the same (iso-) “shape” (morph) as described by their Hasse diagrams.

Example 17

Let A = {1, 2, 3, 6} and let ≤ be the relation | (divides). Figure 13(a) shows the Hasse diagram for (A, ≤). Let A = P({a, b}) = {∅, {a}, {b}, {a, b}}, and let ≤ be set containment, ⊆. If f : A → A is defined by f(1) = ∅,

f(2) = {a},

f(3) = {b},

f(6) = {a, b},

then it is easily seen that f is a one-to-one correspondence. If each label a ∈ A of the Hasse diagram is replaced by f(a), the result is as shown in Figure 13(b). Since this is clearly a Hasse diagram for (A , ≤ ), the function f is an ◆ isomorphism.

{a, b}

6

3

2

1 (a)

{a}

{b}

∅ (b)

Figure 13

255

Order Relations and Structures

1 Exercises 1. Determine whether the relation R is a partial order on the set A. (a) A = Z, and a R b if and only if a = 2b.

13. a

(b) A = Z, and a R b if and only if b2 | a. 2. Determine whether the relation R is a partial order on the set A. (a) A = Z, and a R b if and only if a = bk for some k ∈ Z+ . Note that k depends on a and b.

b

d

c

(b) A = R, and a R b if and only if a ≤ b. 3. Determine whether the relation R is a linear order on the set A. (a) A = R, and a R b if and only if a ≤ b. (b) A = R, and a R b if and only if a ≥ b. 4. Determine whether the relation R is a linear order on the set A. (a) A = P(S), where S is a set. The relation R is set inclusion.

e

Figure 16 14. 4

3

(b) A = R × R, and (a, b) R (a , b ) if and only if a ≤ a and b ≤ b , where ≤ is the usual partial order on R. 5. On the set A = {a, b, c}, find all partial orders ≤ in which a ≤ b. 6. What can you say about the relation R on a set A if R is a partial order and an equivalence relation?

5 2

1

7. Outline the structure of the proof given for Theorem 1. 8. Outline the structure of the proof given for Theorem 2. In Exercises 9 and 10, determine the Hasse diagram of the relation R. 9. A = {1, 2, 3, 4}, R = {(1, 1), (1, 2), (2, 2), (2, 4), (1, 3), (3, 3), (3, 4), (1, 4), (4, 4)}. 10. A = {a, b, c, d, e}, R = {(a, a), (b, b), (c, c), (a, c), (c, d), (c, e), (a, d), (d, d), (a, e), (b, c), (b, d), (b, e), (e, e)}. In Exercises 11 and 12, describe the ordered pairs in the relation determined by the Hasse diagram on the set A = {1, 2, 3, 4} (Figures 14 and 15). 11.

12.

4

Figure 17 15. Determine the Hasse diagram of the relation on A = {1, 2, 3, 4, 5} whose matrix is shown. ⎡

1 ⎢0 ⎢ ⎢0 ⎣0 0

1 1 0 0 0

1 1 1 0 0

1 1 1 1 0

⎤ 1 1⎥ ⎥ 1⎥ 1⎦ 1

4 3

3

16. Determine the Hasse diagram of the relation on A = {1, 2, 3, 4, 5} whose matrix is shown.

2 2

1

Figure 14

1

Figure 15

In Exercises 13 and 14, determine the Hasse diagram of the partial order having the given digraph (Figures 16 and 17).

256

⎡

1 ⎢0 ⎢ ⎢0 ⎣0 0

0 1 0 0 0

1 1 1 0 0

1 1 1 1 0

⎤ 1 1⎥ ⎥ 1⎥ 0⎦ 1

Order Relations and Structures In Exercises 17 and 18, determine the matrix of the partial order whose Hasse diagram is given (Figures 18 and 19).

In Exercises 29 and 30, draw the Hasse diagram of a topological sorting of the given poset (Figures 20 and 21).

17. 2

29.

3

4

5

18.

4

5

30.

8

2

7 2

3

1

Figure 18

6 4

1

19. Let A = { , A, B, C, E, O, M, P, S} have the usual alphabetical order, where represents a “blank” character and ≤ x for all x ∈ A. Arrange the following in lexicographic order (as elements of A × A × A × A). (b) MOPE (c) CAP (a) MOP (e) BASE (f) ACE (d) MAP (g) MACE (h) CAPE 20. Let A = Z+ × Z+ have lexicographic order. Mark each of the following as true or false. (b) (3, 6) ≺ (3, 24) (a) (2, 12) ≺ (5, 3) (d) (15, 92) ≺ (12, 3) (c) (4, 8) ≺ (4, 6) In Exercises 21 through 24, consider the partial order of divisibility on the set A. Draw the Hasse diagram of the poset and determine which posets are linearly ordered. A = {1, 2, 3, 5, 6, 10, 15, 30} A = {2, 4, 8, 16, 32} A = {3, 6, 12, 36, 72} A = {1, 2, 3, 4, 5, 6, 10, 12, 15, 30, 60} Describe how to use MR to determine if R is a partial order. A partial order may or may not be a linear order, but any poset can be partitioned into subsets that are each a linear order. Give a partition into linearly ordered subsets with as few subsets as possible for each of the following. The poset whose Hasse diagram is given in (b) Figure 9(a) (a) Figure 5 (d) Figure 20 (c) Figure 15 (e) Figure 21 27. For each of the posets whose Hasse diagram is indicated below, give as large a set of elements as possible that are incomparable to one another. (b) Figure 9(a) (a) Figure 5 (d) Figure 20 (c) Figure 15 (e) Figure 21 28. What is the relationship between the smallest number of subsets needed to partition a partial order into linearly ordered subsets and the cardinality of the largest set of incomparable elements in the partial order? Justify your response.

8

7 3

5

Figure 19

21. 22. 23. 24. 25. 26.

9

2

1 3

1

6 5 4

Figure 21

Figure 20 31. If (A, ≤) is a poset and A is a subset of A, show that (A , ≤ ) is also a poset, where ≤ is the restriction of ≤ to A . 32. Show that if R is a linear order on the set A, then R−1 is also a linear order on A. 33. A relation R on a set A is called a quasiorder if it is transitive and irreflexive. Let A = P(S) be the power set of a set S, and consider the following relation R on A: U R T if and only if U ⊂ T (proper containment). Show that R is a quasiorder. 34. Let A = {x | x is a real number and −5 ≤ x ≤ 20}. Show that the usual relation < is a quasiorder (see Exercise 33) on A. 35. If R is a quasiorder on A (see Exercise 33), prove that R−1 is also a quasiorder on A. 36. Modify the relation in Example 3 to produce a quasiorder on Z+ . 37. Let B = {2, 3, 6, 9, 12, 18, 24} and let A = B ×B. Define the following relation on A: (a, b) ≺ (a , b ) if and only if a | a and b ≤ b , where ≤ is the usual partial order. Show that ≺ is a partial order. 38. Let A be the set of 2 × 2 Boolean matrices with M R N if and only if mij ≤ nij 1 ≤ i ≤ 2, 1 ≤ j ≤ 2. Prove that R is a partial order on A. 39. Let A = {1, 2, 3, 5, 6, 10, 15, 30} and consider the partial order ≤ of divisibility on A. That is, define a ≤ b to mean that a | b. Let A = P(S), where S = {e, f, g}, be the poset with partial order ⊆. Show that (A, ≤) and (A , ⊆) are isomorphic. 40. Let A = {1, 2, 4, 8} and let ≤ be the partial order of divisibility on A. Let A = {0, 1, 2, 3} and let ≤ be the usual relation “less than or equal to” on integers. Show that (A, ≤) and (A , ≤ ) are isomorphic posets. 41. Show that the partial order (A, R) of Exercise 38 is isomorphic to (P({a, b, c, d}), ⊆).

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2

a3 a1

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b2 b3

Figure 22

Extremal Elements of Partially Ordered Sets Certain elements in a poset are of special importance for many of the properties and applications of posets. In this section we discuss these elements, and in later sections we shall see the important role played by them. In this section we consider a poset (A, ≤). An element a ∈ A is called a maximal element of A if there is no element c in A such that a < c (see Section 1). An element b ∈ A is called a minimal element of A if there is no element c in A such that c < b. It follows immediately that, if (A, ≤) is a poset and (A, ≥) is its dual poset, an element a ∈ A is a maximal element of (A, ≥) if and only if a is a minimal element of (A, ≤). Also, a is a minimal element of (A, ≥) if and only if it is a maximal element of (A, ≤).

Example 1

Consider the poset A whose Hasse diagram is shown in Figure 22. The elements a1 , a2 , and a3 are maximal elements of A, and the elements b1 , b2 , and b3 are minimal elements. Observe that, since there is no line between b2 and b3 , we can conclude ◆ neither that b3 ≤ b2 nor that b2 ≤ b3 .

Example 2

Let A be the poset of nonnegative real numbers with the usual partial order ≤. Then ◆ 0 is a minimal element of A. There are no maximal elements of A.

Example 3

The poset Z with the usual partial order ≤ has no maximal elements and has no ◆ minimal elements.

THEOREM 1

Let A be a finite nonempty poset with partial order ≤. Then A has at least one maximal element and at least one minimal element. Proof Let a be any element of A. If a is not maximal, we can find an element a1 ∈ A such that a < a1 . If a1 is not maximal, we can find an element a2 ∈ A such that a1 < a2 . This argument cannot be continued indefinitely, since A is a finite set. Thus we eventually obtain the finite chain a < a1 < a2 < · · · < ak−1 < ak , which cannot be extended. Hence we cannot have ak < b for any b ∈ A, so ak is a maximal element of (A, ≤). This same argument says that the dual poset (A, ≥) has a maximal element, so ■ (A, ≤) has a minimal element. By using the concept of a minimal element, we can give an algorithm for finding a topological sorting of a given finite poset (A, ≤). We remark first that if a ∈ A and B = A−{a}, then B is also a poset under the restriction of ≤ to B×B. We then have the following algorithm, which produces a linear array named SORT. We assume that SORT is ordered by increasing index, that is, SORT[1] ≺ SORT[2] ≺ · · · . The relation ≺ on A defined in this way is a topological sorting of (A, ≤).

258

Order Relations and Structures Algorithm

For finding a topological sorting of a finite poset (A, ≤). Step 1 Choose a minimal element a of A. Step 2 Make a the next entry of SORT and replace A with A − {a}. Step 3 Repeat steps 1 and 2 until A = { }. Example 4

●

Let A = {a, b, c, d, e}, and let the Hasse diagram of a partial order ≤ on A be as shown in Figure 23(a). A minimal element of this poset is the vertex labeled d (we could also have chosen e). We put d in SORT[1] and in Figure 23(b) we show the Hasse diagram of A − {d}. A minimal element of the new A is e, so e becomes SORT[2], and A − {e} is shown in Figure 23(c). This process continues until we have exhausted A and filled SORT. Figure 23(f) shows the completed array SORT and the Hasse diagram of the poset corresponding to SORT. This is a topological ◆ sorting of (A, ≤).

a

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Figure 23

An element a ∈ A is called a greatest element of A if x ≤ a for all x ∈ A. An element a ∈ A is called a least element of A if a ≤ x for all x ∈ A. As before, an element a of (A, ≤) is a greatest (or least) element if and only if it is a least (or greatest) element of (A, ≥). Example 5

Consider the poset defined in Example 2. Then 0 is a least element; there is no ◆ greatest element.

Example 6

Let S = {a, b, c} and consider the poset A = P(S) defined in Example 12 of Section 1. The empty set is a least element of A, and the set S is a greatest element ◆ of A.

Example 7

The poset Z with the usual partial order has neither a least nor a greatest element. ◆

259

Order Relations and Structures THEOREM 2

A poset has at most one greatest element and at most one least element. Proof Suppose that a and b are greatest elements of a poset A. Then, since b is a greatest element, we have a ≤ b. Similarly, since a is a greatest element, we have b ≤ a. Hence a = b by the antisymmetry property. Thus, if the poset has a greatest element, it only has one such element. Since this fact is true for all posets, the dual poset (A, ≥) has at most one greatest element, so (A, ≤) also has at most one least ■ element. The greatest element of a poset, if it exists, is denoted by I and is often called the unit element. Similarly, the least element of a poset, if it exists, is denoted by 0 and is often called the zero element. Consider a poset A and a subset B of A. An element a ∈ A is called an upper bound of B if b ≤ a for all b ∈ B. An element a ∈ A is called a lower bound of B if a ≤ b for all b ∈ B.

Example 8 h f

g

d

e c a

b

Figure 24

Example 9

Consider the poset A = {a, b, c, d, e, f, g, h}, whose Hasse diagram is shown in Figure 24. Find all upper and lower bounds of the following subsets of A: (a) B1 = {a, b}; (b) B2 = {c, d, e}. Solution

(a) B1 has no lower bounds; its upper bounds are c, d, e, f , g, and h. (b) The upper bounds of B2 are f , g, and h; its lower bounds are c, a, and b. ◆ As Example 8 shows, a subset B of a poset may or may not have upper or lower bounds (in A). Moreover, an upper or lower bound of B may or may not belong to B itself. Let A be a poset and B a subset of A. An element a ∈ A is called a least upper bound of B, (LUB(B)), if a is an upper bound of B and a ≤ a , whenever a is an upper bound of B. Thus a = LUB(B) if b ≤ a for all b ∈ B, and if whenever a ∈ A is also an upper bound of B, then a ≤ a . Similarly, an element a ∈ A is called a greatest lower bound of B, (GLB(B)), if a is a lower bound of B and a ≤ a, whenever a is a lower bound of B. Thus a = GLB(B) if a ≤ b for all b ∈ B, and if whenever a ∈ A is also a lower bound of B, then a ≤ a. As usual, upper bounds in (A, ≤) correspond to lower bounds in (A, ≥) (for the same set of elements), and lower bounds in (A, ≤) correspond to upper bounds in (A, ≥). Similar statements hold for greatest lower bounds and least upper bounds. Let A be the poset considered in Example 8 with subsets B1 and B2 as defined in that example. Find all least upper bounds and all greatest lower bounds of (a) B1 ; (b) B2 . Solution

(a) Since B1 has no lower bounds, it has no greatest lower bounds. However, LUB(B1 ) = c. (b) Since the lower bounds of B2 are c, a, and b, we find that GLB(B2 ) = c.

260

Order Relations and Structures The upper bounds of B2 are f , g, and h. Since f and g are not comparable, ◆ we conclude that B2 has no least upper bound. THEOREM 3 11

Proof The proof is similar to the proof of Theorem 2.

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5 2

Let (A, ≤) be a poset. Then a subset B of A has at most one LUB and at most one GLB.

We conclude this section with some remarks about LUB and GLB in a finite poset A, as viewed from the Hasse diagram of A. Let B = {b1 , b2 , . . . , br }. If a = LUB(B), then a is the first vertex that can be reached from b1 , b2 , . . . , br by upward paths. Similarly, if a = GLB(B), then a is the first vertex that can be reached from b1 , b2 , . . . , br by downward paths.

4

1

Figure 25

Example 10

Let A = {1, 2, 3, 4, 5, . . . , 11} be the poset whose Hasse diagram is shown in Figure 25. Find the LUB and GLB of B = {6, 7, 10}, if they exist. Solution

Exploring all upward paths from vertices 6, 7, and 10, we find that LUB(B) = 10. Similarly, by examining all downward paths from 6, 7, and 10, we find that ◆ GLB(B) = 4. The next result follows immediately from the principle of correspondence (see Section 1). THEOREM 4

Suppose that (A, ≤) and (A , ≤ ) are isomorphic posets under the isomorphism f : A → A . (a) If a is a maximal (minimal) element of (A, ≤), then f(a) is a maximal (minimal) element of (A , ≤ ). (b) If a is the greatest (least) element of (A, ≤), then f(a) is the greatest (least) element of (A , ≤ ). (c) If a is an upper bound (lower bound, least upper bound, greatest lower bound) of a subset B of A, then f(a) is an upper bound (lower bound, least upper bound, greatest lower bound) for the subset f(B) of A . (d) If every subset of (A, ≤) has a LUB (GLB), then every subset of (A , ≤ ) ■ has a LUB (GLB).

Example 11

Show that the posets (A, ≤) and (A , ≤ ), whose Hasse diagrams are shown in Figures 26(a) and (b), respectively, are not isomorphic.

a

a

b

Solution b

c

c (a)

Figure 26

(b)

The two posets are not isomorphic because (A, ≤) has a greatest element a, while (A , ≤ ) does not have a greatest element. We could also argue that they are not isomorphic because (A, ≤) does not have a least element, while (A , ≤ ) does have ◆ a least element.

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2 Exercises In Exercises 1 through 8, determine all maximal and minimal elements of the poset. 1.

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15. A = {2, 4, 6, 8, 12, 18, 24, 36, 72} with the partial order of divisibility. e

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Figure 27 3. e

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16. A = {2, 3, 4, 6, 12, 18, 24, 36} with the partial order of divisibility.

c

Figure 28

In Exercises 17 and 18, determine if the statements are equivalent. Justify your conclusion.

4

17. (a) If a ∈ A is a maximal element, then there is no c ∈ A such that a < c.

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Figure 29

(b) If a ∈ A is a minimal element, then for all b ∈ A, a ≤ b.

5. A = R with the usual partial order ≤. 6. A = {x | x is a real number and 0 ≤ x < 1} with the usual partial order ≤. 7. A = {x | x is a real number and 0 < x ≤ 1} with the usual partial order ≤. 8. A = {2, 3, 4, 6, 8, 24, 48} with the partial order of divisibility. In Exercises 9 through 16, determine the greatest and least elements, if they exist, of the poset. 9.

10.

f d

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b

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(b) If a ∈ A is a maximal element, then for all b ∈ A, b ≤ a. 18. (a) If a ∈ A is a minimal element, then there is no c ∈ A such that c < a.

Figure 30

a

14. A = {x | x is a real number and 0 ≤ x ≤ 1} with the usual partial order ≤.

e

19. Determine if the given statement is true or false. Explain your reasoning. (a) A nonempty finite poset has a maximal element. (b) A nonempty finite poset has a greatest element. (c) A nonempty finite poset has a minimal element. (d) A nonempty finite poset has a least element. 20. Prove that if (A, ≤) has a greatest element, then (A, ≤) has a unique greatest element. 21. Prove that if (A, ≤) has a least element, then (A, ≤) has a unique least element.

d c

22. Prove Theorem 3. a

a

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In Exercises 23 through 32 find, if they exist, (a) all upper bounds of B; (b) all lower bounds of B; (c) the least upper bound of B; (d) the greatest lower bound of B.

Figure 32

Figure 31 11. 4

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Figure 34

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Figure 35

Figure 36

Order Relations and Structures 25.

f

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Figure 38

Figure 37

27. (A, ≤) is the poset in Exercise 23; B = {b, g, h}. 28. (a) (A, ≤) is the poset in Exercise 26; B = {4, 6, 9}. (b) (A, ≤) is the poset in Exercise 26; B = {3, 4, 8}. 29. A = R and ≤ denotes the usual partial order; B = {x | x is a real number and 1 < x < 2}. 30. A = R and ≤ denotes the usual partial order; B = {x | x is a real number and 1 ≤ x < 2}. 31. A is the set of 2 × 2 Boolean matrices and ≤ denotes the relation R with M R N if and only if mij ≤ nij , 1 ≤ i ≤ 2, 1 ≤ j ≤ 2; B is the set of matrices in A with exactly two ones.

3

32. A is the set of 2 × 2 Boolean matrices and ≤ denotes the ≤ i ≤ 2, relation R with M R N if and

only if m

ij ≤ nij , 1 0 1 1 1 1 0 1 ≤ j ≤ 2; B = , , . 0 1 0 1 0 1 33. Construct the Hasse diagram of a topological sorting of the poset whose Hasse diagram is shown in Figure 35. Use the algorithm SORT. 34. Construct the Hasse diagram of a topological sorting of the poset whose Hasse diagram is shown in Figure 36. Use the algorithm SORT. 35. Let R be a partial order on a finite set A. Describe how to use MR to find the least and greatest elements of A if they exist. 36. Give an example of a partial order on A = {a, b, c, d, e} that has two maximal elements and no least element. 37. Let A = {2, 3, 4, . . . , 100} with the partial order of divisibility. (a) How many maximal elements does (A, ≤) have? (b) Give a subset of A that is a linear order under divisibility and is as large as possible. 38. Let (A, ≤) be as in Exercise 37. How many minimal elements does (A, ≤) have?

Lattices A lattice is a poset (L, ≤) in which every subset {a, b} consisting of two elements has a least upper bound and a greatest lower bound. We denote LUB({a, b}) by a ∨ b and call it the join of a and b. Similarly, we denote GLB({a, b}) by a ∧ b and call it the meet of a and b. Lattice structures often appear in computing and mathematical applications. Observe that a lattice is a mathematical structure with two binary operations, join and meet.

Example 1

Let S be a set and let L = P(S). As we have seen, ⊆, containment, is a partial order on L. Let A and B belong to the poset (L, ⊆). Then A ∨ B is the set A ∪ B. To see this, note that A ⊆ A ∪ B, B ⊆ A ∪ B, and, if A ⊆ C and B ⊆ C, then it follows that A ∪ B ⊆ C. Similarly, we can show that the element A ∧ B in (L, ⊆) ◆ is the set A ∩ B. Thus, L is a lattice.

Example 2

Consider the poset (Z+ , ≤), where for a and b in Z+ , a ≤ b if and only if a | b. Then L is a lattice in which the join and meet of a and b are their least common multiple and greatest common divisor, respectively. That is, a ∨ b = LCM(a, b)

Example 3

and a ∧ b = GCD(a, b).

◆

Let n be a positive integer and let Dn be the set of all positive divisors of n. Then Dn is a lattice under the relation of divisibility as considered in Example 2. Thus, if n = 20, we have D20 = {1, 2, 4, 5, 10, 20}. The Hasse diagram of D20 is shown in Figure 39(a). If n = 30, we have D30 = {1, 2, 3, 5, 6, 10, 15, 30}. The Hasse ◆ diagram of D30 is shown in Figure 39(b).

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Order Relations and Structures 20

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Figure 39

Example 4

Which of the Hasse diagrams in Figure 40 represent lattices? Solution

Hasse diagrams (a), (b), (d), and (e) represent lattices. Diagram (c) does not represent a lattice because f ∨ g does not exist. Diagram (f) does not represent a lattice because neither d ∧ e nor b ∨ c exist. Diagram (g) does not represent a ◆ lattice because c ∧ d does not exist. f

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Figure 40

Example 5

We have already observed in Example 4 of Section 1 that the set R of all equivalence relations on a set A is a poset under the partial order of set containment. We can now conclude that R is a lattice where the meet of the equivalence relations R and S is their intersection R ∩ S and their join is (R ∪ S)∞ , the transitive closure of ◆ their union. Let (L, ≤) be a poset and let (L, ≥) be the dual poset. If (L, ≤) is a lattice, we can show that (L, ≥) is also a lattice. In fact, for any a and b in L, the least

264

Order Relations and Structures upper bound of a and b in (L, ≤) is equal to the greatest lower bound of a and b in (L, ≥). Similarly, the greatest lower bound of a and b in (L, ≤) is equal to the least upper bound of a and b in (L, ≥). If L is a finite set, this property can easily be seen by examining the Hasse diagrams of the poset and its dual. Example 6

Let S be a set and L = P(S). Then (L, ⊆) is a lattice, and its dual lattice is (L, ⊇), where ⊆ is “contained in” and ⊇ is “contains.” The discussion preceding this example then shows that in the poset (L, ⊇) the join A ∨ B is the set A ∩ B, and ◆ the meet A ∧ B is the set A ∪ B.

THEOREM 1

If (L1 , ≤) and (L2 , ≤) are lattices, then (L, ≤) is a lattice, where L = L1 × L2 , and the partial order ≤ of L is the product partial order. Proof We denote the join and meet in L1 by ∨1 and ∧1 , respectively, and the join and meet in L2 by ∨2 and ∧2 , respectively. We already know from Theorem 1 of Section 1 that L is a poset. We now need to show that if (a1 , b1 ) and (a2 , b2 ) ∈ L, then (a1 , b1 ) ∨ (a2 , b2 ) and (a1 , b1 ) ∧ (a2 , b2 ) exist in L. We leave it as an exercise to verify that (a1 , b1 ) ∨ (a2 , b2 ) = (a1 ∨1 a2 , b1 ∨2 b2 ) (a1 , b1 ) ∧ (a2 , b2 ) = (a1 ∧1 a2 , b1 ∧2 b2 ). ■

Thus L is a lattice. Example 7

Let L1 and L2 be the lattices shown in Figures 41(a) and (b), respectively. Then ◆ L = L1 × L2 is the lattice shown in Figure 41(c). (I1, I2)

(I1, a) I1

(01, I2)

I2

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Figure 41

Let (L, ≤) be a lattice. A nonempty subset S of L is called a sublattice of L if a ∨ b ∈ S and a ∧ b ∈ S whenever a ∈ S and b ∈ S. Example 8

The lattice Dn of all positive divisors of n (see Example 3) is a sublattice of the lattice Z+ under the relation of divisibility (see Example 2). ◆

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Order Relations and Structures Example 9

Consider the lattice L shown in Figure 42(a). The partially ordered subset Sb shown in Figure 42(b) is not a sublattice of L since a∧b ∈ / Sb . The partially ordered subset / Sc . Observe, however, that Sc in Figure 42(c) is not a sublattice of L since a ∨ b ∈ Sc is a lattice when considered as a poset by itself. The partially ordered subset Sd in Figure 42(d) is a sublattice of L. ◆

I

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Figure 42

Isomorphic Lattices If f : L1 → L2 is an isomorphism from the poset (L1 , ≤1 ) to the poset (L2 , ≤2 ), then Theorem 4 of Section 2 tells us that L1 is a lattice if and only if L2 is a lattice. In fact, if a and b are elements of L1 , then f(a ∧ b) = f(a) ∧ f(b) and f(a ∨ b) = f(a) ∨ f(b). If two lattices are isomorphic, as posets, we say they are isomorphic lattices. Example 10

Let L be the lattice D6 , and let L be the lattice P(S) under the relation of containment, where S = {a, b}. These posets were discussed in Example 16 of Section 1, where they were shown to be isomorphic. Thus, since both are lattices, they are ◆ isomorphic lattices. If f : A → B is a one-to-one correspondence from a lattice (A, ≤) to a set B, then we can use the function f to define a partial order ≤ on B. If b1 and b2 are in B, then b1 = f(a1 ) and b2 = f(a2 ) for some unique elements a1 and a2 of A. Define b1 ≤ b2 (in B) if and only if a1 ≤ a2 (in A). If A and B are finite, then we can describe this process geometrically as follows. Construct the Hasse diagram for (A, ≤). Then replace each label a by the corresponding element f(a) of B. The result is the Hasse diagram of the partial order ≤ on B. When B is given the partial order ≤ , f will be an isomorphism from the poset (A, ≤) to the poset (B, ≤ ). To see this, note that f is already assumed to be a one-to-one correspondence. The definition of ≤ states that, for any a1 and a2 in A, a1 ≤ a2 if and only if f(a1 ) ≤ f(a2 ). Thus f is an isomorphism. Since (A, ≤) is a lattice, so is (B, ≤ ), and they are isomorphic lattices.

Example 11

266

If A is a set, let R be the set of all equivalence relations on A and let be the set of all partitions on A. In Example 4 of Section 1, we considered the partial order ⊆ on R. From this partial order we can construct, using f as explained before, a partial order ≤ on . By construction, if P1 and P2 are partitions of A, and R1 and R2 , respectively, are the equivalence relations corresponding to

Order Relations and Structures these partitions, then P1 ≤ P2 will mean that R1 ⊆ R2 . Since we showed in Example 5 that (R, ⊆) is a lattice, and we know that f is an isomorphism, it follows that (, ≤ ) is also a lattice. In Exercise 35 we describe the partial order ◆ ≤ directly in terms of the partitions themselves.

Properties of Lattices Before proving a number of the properties of lattices, we recall the meaning of a∨b and a ∧ b. 1. 2. 1 . 2 . THEOREM 2

a ≤ a ∨ b and b ≤ a ∨ b; a ∨ b is an upper bound of a and b. If a ≤ c and b ≤ c, then a ∨ b ≤ c; a ∨ b is the least upper bound of a and b. a ∧ b ≤ a and a ∧ b ≤ b; a ∧ b is a lower bound of a and b. If c ≤ a and c ≤ b, then c ≤ a ∧ b; a ∧ b is the greatest lower bound of a and b.

Let L be a lattice. Then for every a and b in L, (a) a ∨ b = b if and only if a ≤ b. (b) a ∧ b = a if and only if a ≤ b. (c) a ∧ b = a if and only if a ∨ b = b. Proof (a) Suppose that a ∨ b = b. Since a ≤ a ∨ b = b, we have a ≤ b. Conversely, if a ≤ b, then, since b ≤ b, b is an upper bound of a and b; so by definition of least upper bound we have a ∨ b ≤ b. Since a ∨ b is an upper bound, b ≤ a ∨ b, so a ∨ b = b. (b) The proof is analogous to the proof of part (a), and we leave it as an exercise for the reader. ■ (c) The proof follows from parts (a) and (b).

Example 12

Let L be a linearly ordered set. If a and b ∈ L, then either a ≤ b or b ≤ a. It follows from Theorem 2 that L is a lattice, since every pair of elements has a least ◆ upper bound and a greatest lower bound.

THEOREM 3

Let L be a lattice. Then 1. Idempotent Properties (a) a ∨ a = a (b) a ∧ a = a 2. Commutative Properties (a) a ∨ b = b ∨ a (b) a ∧ b = b ∧ a 3. Associative Properties (a) a ∨ (b ∨ c) = (a ∨ b) ∨ c (b) a ∧ (b ∧ c) = (a ∧ b) ∧ c 4. Absorption Properties (a) a ∨ (a ∧ b) = a (b) a ∧ (a ∨ b) = a

267

Order Relations and Structures Proof 1. The statements follow from the definition of LUB and GLB. 2. The definition of LUB and GLB treat a and b symmetrically, so the results follow. 3. (a) From the definition of LUB, we have a ≤ a ∨ (b ∨ c) and b ∨ c ≤ a ∨ (b ∨ c). Moreover, b ≤ b ∨ c and c ≤ b ∨ c, so, by transitivity, b ≤ a ∨ (b ∨ c) and c ≤ a ∨ (b ∨ c). Thus a ∨ (b ∨ c) is an upper bound of a and b, so by definition of least upper bound we have a ∨ b ≤ a ∨ (b ∨ c). Since a ∨ (b ∨ c) is an upper bound of a ∨ b and c, we obtain (a ∨ b) ∨ c ≤ a ∨ (b ∨ c). Similarly, a∨(b∨c) ≤ (a∨b)∨c. By the antisymmetry of ≤, property 3(a) follows. (b) The proof is analogous to the proof of part (a) and we omit it. 4. (a) Since a ∧ b ≤ a and a ≤ a, we see that a is an upper bound of a ∧ b and a; so a ∨ (a ∧ b) ≤ a. On the other hand, by the definition of LUB, we have a ≤ a ∨ (a ∧ b), so a ∨ (a ∧ b) = a. ■ (b) The proof is analogous to the proof of part (a) and we omit it. It follows from property 3 that we can write a ∨ (b ∨ c) and (a ∨ b) ∨ c merely as a ∨ b ∨ c, and similarly for a ∧ b ∧ c. Moreover, we can write LUB({a1 , a2 , . . . , an }) GLB({a1 , a2 , . . . , an })

as as

a1 ∨ a2 ∨ · · · ∨ an a1 ∧ a2 ∧ · · · ∧ an ,

since we can show by induction that these joins and meets are independent of the grouping of the terms. THEOREM 4

Let L be a lattice. Then, for every a, b, and c in L, 1. If a ≤ b, then (a) a ∨ c ≤ b ∨ c. (b) a ∧ c ≤ b ∧ c. 2. a ≤ c and b ≤ c if and only if a ∨ b ≤ c. 3. c ≤ a and c ≤ b if and only if c ≤ a ∧ b. 4. If a ≤ b and c ≤ d, then (a) a ∨ c ≤ b ∨ d. (b) a ∧ c ≤ b ∧ d. Proof The proof is left as an exercise.

■

Special Types of Lattices A lattice L is said to be bounded if it has a greatest element I and a least element 0 (see Section 2). Example 13

268

The lattice Z+ under the partial order of divisibility, as defined in Example 2, is not a bounded lattice since it has a least element, the number 1, but no greatest ◆ element.

Order Relations and Structures Example 14

The lattice Z under the partial order ≤ is not bounded since it has neither a greatest ◆ nor a least element.

Example 15

The lattice P(S) of all subsets of a set S, as defined in Example 1, is bounded. Its ◆ greatest element is S and its least element is ∅. If L is a bounded lattice, then for all a ∈ A, 0≤a≤I a ∨ 0 = a, a ∧ 0 = 0 a ∨ I = I, a ∧ I = a.

THEOREM 5

Let L = {a1 , a2 , . . . , an } be a finite lattice. Then L is bounded. Proof The greatest element of L is a1 ∨ a2 ∨ · · · ∨ an , and its least element is a1 ∧ a2 ∧ ■ · · · ∧ an . Note that the proof of Theorem 5 is a constructive proof. We show that L is bounded by constructing the greatest and the least elements. A lattice L is called distributive if for any elements a, b, and c in L we have the following distributive properties:

I

b

d

a

c

1. a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c) 2. a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) If L is not distributive, we say that L is nondistributive. We leave it as an exercise to show that the distributive property holds when any two of the elements a, b, or c are equal or when any one of the elements is 0 or I. This observation reduces the number of cases that must be checked in verifying that a distributive property holds. However, verification of a distributive property is generally a tedious task.

0

Figure 43

Example 16

For a set S, the lattice P(S) is distributive, since union and intersection (the join and meet, respectively) each satisfy a distributive property. ◆

Example 17

The lattice shown in Figure 43 is distributive, as can be seen by verifying the distributive properties for all ordered triples chosen from the elements a, b, c, ◆ and d.

Example 18

Show that the lattices pictured in Figure 44 are nondistributive. Solution

(a) We have

I I a

while

b

c 0 (a)

Figure 44

a

b

c

(b) Observe that 0 (b)

a ∧ (b ∨ c) = a ∧ I = a (a ∧ b) ∨ (a ∧ c) = b ∨ 0 = b. a ∧ (b ∨ c) = a ∧ I = a

while (a ∧ b) ∨ (a ∧ c) = 0 ∨ 0 = 0.

◆

269

Order Relations and Structures The nondistributive lattices discussed in Example 18 are useful for showing that a given lattice is nondistributive, as the following theorem, whose proof we omit, asserts. THEOREM 6

A lattice L is nondistributive if and only if it contains a sublattice that is isomorphic ■ to one of the two lattices of Example 18. Theorem 6 can be used quite efficiently by inspecting the Hasse diagram of L. Let L be a bounded lattice with greatest element I and least element 0, and let a ∈ L. An element a ∈ L is called a complement of a if

Observe that

a ∨ a = I

and

a ∧ a = 0.

0 = I

and

I = 0.

Example 19

The lattice L = P(S) is such that every element has a complement, since if A ∈ L, then its set complement A has the properties A ∨ A = S and A ∧ A = ∅. That is, ◆ the set complement is also the complement in the lattice L.

Example 20

The lattices in Figure 44 each have the property that every element has a comple◆ ment. The element c in both cases has two complements, a and b.

Example 21

Consider the lattices D20 and D30 discussed in Example 3 and shown in Figure 39. Observe that every element in D30 has a complement. For example, if a = 5, then ◆ a = 6. However, the elements 2 and 10 in D20 have no complements. Examples 20 and 21 show that an element a in a lattice need not have a complement, and it may have more than one complement. However, for a bounded distributive lattice, the situation is more restrictive, as shown by the following theorem.

THEOREM 7

Let L be a bounded distributive lattice. If a complement exists, it is unique. Proof Let a and a be complements of the element a ∈ L. Then a ∨ a = I, a ∧ a = 0,

a ∨ a = I a ∧ a = 0.

Using the distributive laws, we obtain a = a ∨ 0 = a ∨ (a ∧ a ) = (a ∨ a) ∧ (a ∨ a ) = I ∧ (a ∨ a ) = a ∨ a . Also, a = a ∨ 0 = a ∨ (a ∧ a ) = (a ∨ a) ∧ (a ∨ a ) = I ∧ (a ∨ a ) = a ∨ a . Hence a = a .

270

■

Order Relations and Structures The proof of Theorem 7 is a direct proof, but it is not obvious how the representations of a and a were chosen. There is some trial and error involved in this sort of proof, but we expect to use the hypothesis that L is bounded and that L is distributive. An alternative proof is outlined in Exercise 36. A lattice L is called complemented if it is bounded and if every element in L has a complement. Example 22

The lattice L = P(S) is complemented. Observe that in this case each element of L has a unique complement, which can be seen directly or is implied by Theorem 7. ◆

Example 23

The lattices discussed in Example 20 and shown in Figure 44 are complemented. ◆ In this case, the complements are not unique.

3 Exercises In Exercises 1 through 6 (Figures 45 through 50), determine whether the Hasse diagram represents a lattice. 1.

2. e

i g

h

d

e

9. Let A be the set of 2 × 2 matrices with M R N if and only if mij ≤ nij , 1 ≤ i ≤ 2, 1 ≤ j ≤ 2. Is (A, R) a lattice?

d

10. Amplify the explanations in the solution of Example 4 by explaining why the specified object does not exist.

c

c a

a

Figure 45 3.

8. Let A be the set of 2 × 2 Boolean matrices with M R N if and only if mij ≤ nij , 1 ≤ i ≤ 2, 1 ≤ j ≤ 2. Is (A, R) a lattice?

f

f

b

11. If L1 and L2 are the lattices shown in Figure 51, draw the Hasse diagram of L1 × L2 with the product partial order.

b

Figure 46 4.

g

h

f

d

e

Figure 47 6.

e

d c

c a

Figure 49

13. Let L = P(S) be the lattice of all subsets of a set S under the relation of containment. Let T be a subset of S. Show that P(T) is a sublattice of L.

f

d b

L2

12. Complete the proof of Theorem 1 by verifying that (a1 , b1 ) ∨ (a2 , b2 ) = (a1 ∨1 a2 , b1 ∨2 b2 ) and (a1 , b1 ) ∧ (a2 , b2 ) = (a1 ∧1 a2 , b1 ∧2 b2 ).

c

Figure 48

f

a2

Figure 51

b

e

a1

d b

g

b2

L1

f

a

5.

b1

g

e

c

a

7. Is the poset A = {2, 3, 6, 12, 24, 36, 72} under the relation of divisibility a lattice?

b a

Figure 50

14. Let L be a lattice and let a and b be elements of L such that a ≤ b. The interval [a, b] is defined as the set of all x ∈ L such that a ≤ x ≤ b. Prove that [a, b] is a sublattice of L. 15. Show that a subset of a linearly ordered poset is a sublattice. 16. Find all sublattices of D24 that contain at least five elements. 17. Give the Hasse diagrams of all nonisomorphic lattices that have one, two, three, four, or five elements.

271

Order Relations and Structures 18. Show that if a bounded lattice has two or more elements, then 0 = I. 19. Prove Theorem 2(b). 20. Show that the lattice Dn is distributive for any n. 21. (a) In Example 17 (Figure 43), how many ordered triples must be checked to see if the lattice is distributive? (b) In Example 18 (Figure 44), what is the maximum number of ordered triples that would need to be checked to show that the lattice is not distributive? 22. Show that a sublattice of a distributive lattice is distributive. 23. Show that if L1 and L2 are distributive lattices, then L = L1 × L2 is also distributive, where the order of L is the product of the orders in L1 and L2 . 24. Prove that if a and b are elements in a bounded, distributive lattice and if a has a complement a , then a ∨ (a ∧ b) = a ∨ b a ∧ (a ∨ b) = a ∧ b.

I b

Figure 52 27. Find the complement of each element in D42 . In Exercises 28 through 31 (Figures 53 through 56), determine whether each lattice is distributive, complemented, or both. 29.

e

g

d e

c

f

b

d

a b

Figure 55

c a

Figure 56 32. Prove Theorem 4, part (2). 33. Let L be a bounded lattice with at least two elements. Show that no element of L is its own complement.

e d b

c a

Figure 57 35. Let P1 = {A1 , A2 , . . . }, P2 = {B1 , B2 , . . . } be two partitions of a set S. Show that P1 ≤ P2 (see the definition in Example 11) if and only if each Ai is contained in some Bj .

c

0

28.

31.

34. Consider the complemented lattice shown in Figure 57. Give the complements of each element.

25. Let L be a distributive lattice. Show that if there exists an a with a ∧ x = a ∧ y and a ∨ x = a ∨ y, then x = y. 26. A lattice is said to be modular if, for all a, b, c, a ≤ c implies that a ∨ (b ∧ c) = (a ∨ b) ∧ c. (a) Show that a distributive lattice is modular. (b) Show that the lattice shown in Figure 52 is a nondistributive lattice that is modular.

a

30. e

f

36. Complete the following proof of Theorem 7. Proof: Let a and a be complements of a ∈ L. Then a = a ∧ I = . Also, a = a ∧ I = . Hence a =a . 37. Prove or disprove that a sublattice of a complemented lattice is also complemented.

d c

f

b

g

e

c

d b a a

Figure 53

272

38. Prove or disprove that a sublattice of a bounded lattice is also bounded.

Figure 54

39. Prove or disprove that a sublattice of a modular lattice is also modular. 40. Prove that any linear order is a distributive lattice.

Order Relations and Structures

4

Finite Boolean Algebras In this section we discuss a certain type of lattice that has a great many applications in computer science. We have seen in Example 6 of Section 3 that if S is a set, L = P(S), and ⊆ is the usual relation of containment, then the poset (L, ⊆) is a lattice. These lattices have many properties that are not shared by lattices in general. For this reason they are easier to work with, and they play a more important role in various applications. We will restrict our attention to the lattices (P(S), ⊆), where S is a finite set, and we begin by finding all essentially different examples.

THEOREM 1

If S1 = {x1 , x2 , . . . , xn } and S2 = {y1 , y2 , . . . , yn } are any two finite sets with n elements, then the lattices (P(S1 ), ⊆) and (P(S2 ), ⊆) are isomorphic. Consequently, the Hasse diagrams of these lattices may be drawn identically. Proof Arrange the sets as shown in Figure 58 so that each element of S1 is directly over the correspondingly numbered element in S2 . For each subset A of S1 , let f(A) be the subset of S2 consisting of all elements that correspond to the elements of A. Figure 59 shows a typical subset A of S1 and the corresponding subset f(A) of S2 . It is easily seen that the function f is a one-to-one correspondence from subsets of S1 to subsets of S2 . Equally clear is the fact that if A and B are any subsets of S1 , then A ⊆ B if and only if f(A) ⊆ f(B). We omit the details. Thus the lattices ■ (P(S1 ), ⊆) and (P(S2 ), ⊆) are isomorphic. A

S1 :

S2 :

x1

y1

x2

y2

···

···

S1 : x1

x2

x3

x4

· · · xn

S2 : y1

y2

y3

y4

· · · yn

xn

yn

Figure 58

f (A)

Figure 59

The essential point of this theorem is that the lattice (P(S), ⊆) is completely determined as a poset by the number |S| and does not depend in any way on the nature of the elements in S. Example 1

Figures 60(a) and (b) show Hasse diagrams for the lattices (P(S), ⊆) and (P(T), ⊆), respectively, where S = {a, b, c} and T = {2, 3, 5}. It is clear from this figure that the two lattices are isomorphic. In fact, we see that one possible isomorphism f : S → T is given by f({a}) = {2}, f({a, b}) = {2, 3},

f({b}) = {3}, f({b, c}) = {3, 5}, f({a, b, c}) = {2, 3, 5},

f({c}) = {5}, f({a, c}) = {2, 5}, f(∅) = ∅.

◆

Thus, for each n = 0, 1, 2, . . . , there is only one type of lattice having the form (P(S), ⊆). This lattice depends only on n, not on S, and it has 2n elements. If a set S has n elements, then all subsets of S can be represented by sequences of 0’s and

273

Order Relations and Structures {a, b, c}

{a, b}

{2, 3, 5}

{b, c}

{2, 3}

111

{3, 5}

110

011

{a, c}

{2, 5}

101

{b}

{3}

010

{c}

{a}

{2}

{5}

100

∅

∅ (a)

001

000 (c)

(b)

Figure 60

1’s of length n. We can therefore label the Hasse diagram of a lattice (P(S), ⊆) by such sequences. In doing so, we free the diagram from dependence on a particular set S and emphasize the fact that it depends only on n. Example 2

Figure 60(c) shows how the diagrams that appear in Figures 60(a) and (b) can be labeled by sequences of 0’s and 1’s. This labeling serves equally well to describe the lattice of Figure 60(a) or (b), or for that matter the lattice (P(S), ⊆) that arises ◆ from any set S having three elements. If the Hasse diagram of the lattice corresponding to a set with n elements is labeled by sequences of 0’s and 1’s of length n, as described previously, then the resulting lattice is named Bn . The properties of the partial order on Bn can be described directly as follows. If x = a1 a2 · · · an and y = b1 b2 · · · bn are two elements of Bn , then 1. 2. 3. 4.

x ≤ y if and only if ak ≤ bk (as numbers 0 or 1) for k = 1, 2, . . . , n. x ∧ y = c1 c2 · · · cn , where ck = min{ak , bk }. x ∨ y = d1 d2 · · · dn , where dk = max{ak , bk }. x has a complement x = z1 z2 · · · zn , where zk = 1 if xk = 0, and zk = 0 if xk = 1.

The truth of these statements can be seen by noting that (Bn , ≤) is isomorphic with (P(S), ⊆), so each x and y in Bn correspond to subsets A and B of S. Then x ≤ y, x ∧ y, x ∨ y, and x correspond to A ⊆ B, A ∩ B, A ∪ B, and A (set complement), respectively (verify). Figure 61 shows the Hasse diagrams of the lattices Bn for n = 0, 1, 2, 3. We have seen that each lattice (P(S), ⊆) is isomorphic with Bn , where n = |S|. Other lattices may also be isomorphic with one of the Bn and thus possess all the special properties that the Bn possess. Example 3

In Example 17 of Section 1, we considered the lattice D6 consisting of all positive integer divisors of 6 under the partial order of divisibility. The Hasse diagram of D6 is shown in that example, and we now see that D6 is isomorphic with B2 . In fact, f : D6 → B2 is an isomorphism, where f(1) = 00,

f(2) = 10,

f(3) = 01,

f(6) = 11.

◆

We are therefore led to make the following definition. A finite lattice is called a Boolean algebra if it is isomorphic with Bn for some nonnegative integer n.

274

Order Relations and Structures 111

110 1

011 101

11

010 10

n= 0

0 n= 1

01

100

001

000 n= 3

00 n= 2

Figure 61

Thus each Bn is a Boolean algebra and so is each lattice (P(S), ⊆), where S is a finite set. Example 3 shows that D6 is also a Boolean algebra. We will work only with finite posets in this section. For the curious, however, we note that there are infinite posets that share all the relevant properties of the lattices (P(S), ⊆) (for infinite sets S, of course), but that are not isomorphic with one these lattices. This necessitates the restriction of our definition of Boolean algebra to the finite case, which is sufficient for the applications that we present. Example 4

Consider the lattices D20 and D30 of all positive integer divisors of 20 and 30, respectively, under the partial order of divisibility. These posets were introduced in Example 3 of Section 3, and their Hasse diagrams were shown in Figure 39. Since D20 has six elements and 6 = 2n for any integer n ≥ 0, we conclude that D20 is not a Boolean algebra. The poset D30 has eight elements, and since 8 = 23 , it could be a Boolean algebra. By comparing Figure 39(b) and Figure 61, we see that D30 is isomorphic with B3 . In fact, we see that the one-to-one correspondence f : D30 → B3 defined by f(1) = 000, f(5) = 001, f(15) = 011,

f(2) = 100, f(6) = 110, f(30) = 111

f(3) = 010, f(10) = 101,

is an isomorphism. Thus D30 is a Boolean algebra.

◆

If a finite lattice L does not contain 2n elements for some nonnegative integer n, we know that L cannot be a Boolean algebra. If |L| = 2n , then L may or may not be a Boolean algebra. If L is relatively small, we may be able to compare its Hasse diagram with the Hasse diagram of Bn . In this way we saw in Example 4 that D30 is a Boolean algebra. However, this technique may not be practical if L is large. In that case, we may be able to show that L is a Boolean algebra by directly constructing an isomorphism with some Bn or, equivalently, with (P(S), ⊆) for some finite set S. Suppose, for example, that we want to know whether a lattice Dn is a Boolean algebra, and we want a method that works no matter how large n is. The following theorem gives a partial answer.

275

Order Relations and Structures THEOREM 2

Let

n = p1 p2 · · · pk ,

where the pi are distinct primes. Then Dn is a Boolean algebra. Proof Let S = {p1 , p2 , . . . , pk }. If T ⊆ S and aT is the product of the primes in T , then aT | n. Any divisor of n must be of the form aT for some subset T of S (where we let a∅ = 1). The reader may verify that if V and T are subsets of S, V ⊆ T if and only if aV | aT . Also, it follows that aV ∩T = aV ∧ aT = GCD(aV , aT ) and aV ∪T = aV ∨ aT = LCM(aV , aT ). Thus the function f : P(S) → Dn given by f(T) = aT is an isomorphism from P(S) to Dn . Since P(S) is a Boolean algebra, ■ so is Dn . Example 5

Since 210 = 2 · 3 · 5 · 7, 66 = 2 · 3 · 11, and 646 = 2 · 17 · 19, we see from Theorem ◆ 2 that D210 , D66 , and D646 are all Boolean algebras. In other cases of large lattices L, we may be able to show that L is not a Boolean algebra by showing that the partial order of L does not have the necessary properties. A Boolean algebra is isomorphic with some Bn and therefore with some lattice (P(S), ⊆). Thus a Boolean algebra L must be a bounded lattice and a complemented lattice (see Section 3). In other words, it will have a greatest element I corresponding to the set S and a least element 0 corresponding to the subset ∅. Also, every element x of L will have a complement x . According to Example 16, Section 3, L must also be distributive. The principle of correspondence (see Section 1) then tells us that the following rule holds.

THEOREM 3 Substitution Rule for Boolean Algebras

Any formula involving ∪ or ∩ that holds for arbitrary subsets of a set S will continue to hold for arbitrary elements of a Boolean algebra L if ∧ is substituted for ∩ and ■ ∨ for ∪.

Example 6

If L is any Boolean algebra and x, y, and z are in L, then the following three properties hold. Involution Property 1. (x ) = x 2. (x ∧ y) = x ∨ y De Morgan’s Laws 3. (x ∨ y) = x ∧ y This is true by the substitution rule for Boolean algebras, since we know that the corresponding formulas 1 . (A) = A 2 . (A ∩ B) = A ∪ B 3 . (A ∪ B) = A ∩ B hold for arbitrary subsets A and B of a set S.

◆

In a similar way, we can list other properties that must hold in any Boolean algebra by the substitution rule. Next we summarize all the basic properties of a Boolean algebra (L, ≤) and, next to each one, we list the corresponding property for subsets of a set S. We suppose that x, y, and z are arbitrary elements in L, and A, B, and C are arbitrary subsets of S. Also, we denote the greatest and least elements of L by I and 0, respectively.

276

Order Relations and Structures 1. x ≤ y if and only if x ∨ y = y. 2. x ≤ y if and only if x ∧ y = x. 3. (a) x ∨ x = x. (b) x ∧ x = x. 4. (a) x ∨ y = y ∨ x. (b) x ∧ y = y ∧ x. 5. (a) x ∨ (y ∨ z) = (x ∨ y) ∨ z. (b) x ∧ (y ∧ z) = (x ∧ y) ∧ z. 6. (a) x ∨ (x ∧ y) = x. (b) x ∧ (x ∨ y) = x. 7. 0 ≤ x ≤ I for all x in L. 8. (a) x ∨ 0 = x. (b) x ∧ 0 = 0. 9. (a) x ∨ I = I. (b) x ∧ I = x. 10. (a) x ∧ (y ∨ z) = (x ∧ y) ∨ (x ∧ z). (b) x ∨ (y ∧ z) = (x ∨ y) ∧ (x ∨ z). 11. Every element x has a unique complement x satisfying (a) x ∨ x = I. (b) x ∧ x = 0.

I a

f

e d b

c 0

Figure 62

Example 7

12. (a) 0 = I. (b) I = 0. 13. (x ) = x. 14. (a) (x ∧ y) = x ∨ y . (b) (x ∨ y) = x ∧ y .

1 . A ⊆ B if and only if A ∪ B = B. 2 . A ⊆ B if and only if A ∩ B = A. 3 . (a) A ∪ A = A. (b) A ∩ A = A. 4 . (a) A ∪ B = B ∪ A. (b) A ∩ B = B ∩ A. 5 . (a) A ∪ (B ∪ C) = (A ∪ B) ∪ C. (b) A ∩ (B ∩ C) = (A ∩ B) ∩ C. 6 . (a) A ∪ (A ∩ B) = A. (b) A ∩ (A ∪ B) = A. 7 . ∅ ⊆ A ⊆ S for all A in P(S). 8 . (a) A ∪ ∅ = A. (b) A ∩ ∅ = ∅. 9 . (a) A ∪ S = S. (b) A ∩ S = A. 10 . (a) A ∩ (B ∪ C) = (A ∩ B) ∪ (A ∩ C). (b) A ∪ (B ∩ C) = (A ∪ B) ∩ (A ∪ C). 11 . Every element A has a unique complement A satisfying (a) A ∪ A = S. (b) A ∩ A = ∅. 12 . (a) ∅ = S. (b) S = ∅. 13 . (A) = A. 14 . (a) (A ∩ B) = A ∪ B. (b) (A ∪ B) = A ∩ B.

Thus we may be able to show that a lattice L is not a Boolean algebra by showing that it does not possess one or more of these properties. Show that the lattice whose Hasse diagram is shown in Figure 62 is not a Boolean algebra. Solution

Elements a and e are both complements of c; that is, they both satisfy properties 11(a) and 11(b) with respect to the element c. But property 11 says that such an element is unique in any Boolean algebra. Thus the given lattice cannot be a ◆ Boolean algebra. Example 8

Show that if n is a positive integer and p2 | n, where p is a prime number, then Dn is not a Boolean algebra. Solution

Suppose that p2 | n so that n = p2 q for some positive integer q. Since p is also a divisor of n, p is an element of Dn . Thus, by the remarks given previously, if Dn is a Boolean algebra, then p must have a complement p . Then GCD(p, p ) = 1 277

Order Relations and Structures and LCM(p, p ) = n. Similarly, pp = n, so p = n/p = pq. This shows that GCD(p, pq) = 1, which is impossible, since p and pq have p as a common divisor. ◆ Hence Dn cannot be a Boolean algebra. If we combine Example 8 and Theorem 2, we see that Dn is a Boolean algebra if and only if n is the product of distinct primes, that is, if and only if no prime divides n more than once. If n = 40, then n = 23 · 5, so 2 divides n three times. If n = 75, then n = 3 · 52 , so 5 divides n twice. Thus neither D40 nor D75 are Boolean algebras. ◆

Example 9

Let us summarize what we have shown about Boolean algebras. We may attempt to show that a lattice L is a Boolean algebra by examining its Hasse diagram or constructing directly an isomorphism between L and Bn or (P(S), ⊆). We may attempt to show that L is not a Boolean algebra by checking the number of elements in L or the properties of its partial order. If L is a Boolean algebra, then we may use any of the properties 1 through 14 to manipulate or rewrite expressions involving elements of L. Simply proceed as if the elements were subsets and the manipulations were those that arise in set theory. We call such a lattice an algebra, because we use properties 1 through 14 just as the properties of real numbers are used in high school algebra. From now on we will denote the Boolean algebra B1 simply as B. Thus B contains only the two elements 0 and 1. It is a fact that any of the Boolean algebras Bn can be described in terms of B. The following theorem gives this description. For any n ≥ 1, Bn is the product B×B×· · ·×B of B, n factors, where B×B×· · ·×B is given the product partial order.

THEOREM 4

Proof By definition, Bn consists of all n-tuples of 0’s and 1’s, that is, all n-tuples of elements from B. Thus, as a set, Bn is equal to B × B × · · · × B (n factors). Moreover, if x = x1 x2 · · · xn and y = y1 y2 · · · yn are two elements of Bn , then we know that for all k. x ≤ y if and only if xk ≤ yk Thus Bn , identified with B × B × · · · × B (n factors), has the product partial order. ■

4 Exercises In Exercises 1 through 10, determine whether the poset is a Boolean algebra. Explain. 1.

2.

f d

e

b

c

3. c

g

e

f

e

b a

d b

Figure 63

a

Figure 64

278

f

c a

c

a

d

d

b

4. e

Figure 65

Figure 66

Order Relations and Structures 5.

6.

n

f

21. Show that in a Boolean algebra, for any a, b, and c, if a ≤ b, then

h f

g

a ∨ (b ∧ c) = b ∧ (a ∨ c).

g

e

22. Explain the connection between Examples 7 and 8.

e

For Exercises 23 through 26, let A = {a, b, c, d, e, f, g, h} and R be the relation defined by ⎤ ⎡ 1 1 1 1 0 0 0 0 ⎢0 1 0 1 0 0 0 0⎥ ⎥ ⎢ ⎢0 0 1 1 0 0 0 0⎥ ⎥ ⎢ ⎢0 0 0 1 0 0 0 0⎥ MR = ⎢ ⎥. ⎢1 1 1 1 1 1 1 1⎥ ⎢0 1 0 1 0 1 0 1⎥ ⎥ ⎢ ⎣0 0 1 1 0 0 1 1⎦ 0 0 0 1 0 0 0 1

d b

c

b

d

a

a

Figure 68

Figure 67 7. b

c

8.

d

b

c

a

a

Figure 69

Figure 70

9. D385 10. D60 11. Are there any Boolean algebras having three elements? Why or why not? 12. Show that in a Boolean algebra, for any a and b, a ≤ b if and only if b ≤ a . 13. Show that in a Boolean algebra, for any a and b, a = b if and only if (a ∧ b ) ∨ (a ∧ b) = 0. 14. Show that in a Boolean algebra, for any a, b, and c, if a ≤ b, then a ∨ c ≤ b ∨ c. 15. Show that in a Boolean algebra, for any a, b, and c, if a ≤ b, then a ∧ c ≤ b ∧ c. 16. Show that in a Boolean algebra the following statements are equivalent for any a and b. (b) a ∧ b = a (a) a ∨ b = b (d) a ∧ b = 0 (c) a ∨ b = I (e) a ≤ b 17. Show that in a Boolean algebra, for any a and b,

23. Show that (A, R) is a poset. 24. Does the poset (A, R) have a least element? a greatest element? If so, identify them. 25. Show that the poset (A, R) is complemented and give all pairs of complements. 26. Prove or disprove that (A, R) is a Boolean algebra. 27. Let A = {a, b, c, d, e, f, g, h} and R be the relation defined by ⎤ ⎡ 1 1 1 1 1 1 1 1 ⎢0 1 0 0 1 1 1 1⎥ ⎥ ⎢ ⎢0 0 1 0 1 1 1 1⎥ ⎥ ⎢ ⎢0 0 0 1 0 1 1 1⎥ MR = ⎢ ⎥. ⎢0 0 0 0 1 1 1 1⎥ ⎢0 0 0 0 0 1 0 1⎥ ⎥ ⎢ ⎣0 0 0 0 0 0 1 1⎦ 0 0 0 0 0 0 0 1

28.

29.

(a ∧ b) ∨ (a ∧ b ) = a. 18. Show that in a Boolean algebra, for any a and b, b ∧ (a ∨ (a ∧ (b ∨ b ))) = b. 19. Show that in a Boolean algebra, for any a, b, and c, (a ∧ b ∧ c) ∨ (b ∧ c) = b ∧ c. 20. Show that in a Boolean algebra, for any a, b, and c, ((a ∨ c) ∧ (b ∨ c)) = (a ∨ b) ∧ c .

30. 31. 32.

Prove or disprove that (A, R) is a Boolean algebra. Let A be the set of 2 × 2 Boolean matrices with M R N if and only if mij ≤ nij , 1 ≤ i ≤ 2, 1 ≤ j ≤ 2. Prove that (A, R) is a Boolean algebra. An atom of a Boolean algebra, B, is a nonzero element a such that there is no x in B with 0 < x < a. (a) Identify the atoms of the Boolean algebra in Figure 60(a). (b) Identify the atoms of D30 . How many atoms does Bn have? Give the atoms of (A, R) as defined in Exercise 28. Like physical atoms, the atoms of a Boolean algebra serve as building blocks for the elements of the Boolean algebra. Write each nonzero element of the given Boolean algebra as the join of distinct atoms. (b) D42 (a) B3 (c) (A, R) in Exercise 28

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Order Relations and Structures

5

Functions on Boolean Algebras Tables listing the values of a function f for all elements of Bn , such as shown in Figure 71(a), are often called truth tables for f . This is because they are analogous to tables that arise in logic. Suppose that the xk represent propositions, and f(x1 , x2 , . . . , xn ) represents a compound sentence constructed from the xk ’s. If we think of the value 0 for a sentence as meaning that the sentence is false, and 1 as meaning that the sentence is true, then tables such as Figure 71(a) show us how truth or falsity of f(x1 , x2 , . . . , xn ) depends on the truth or falsity of its component sentences xk . Thus such tables are called truth tables, even when they arise in areas other than logic, such as in Boolean algebras. x2

x3

f (x1, x2, x3)

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

0 1 1 0 1 0 1 0

(a)

x1 x2 f

f (x1, x2, x3)

···

x1

xn (b)

Figure 71

The reason that such functions are important is that, as shown schematically in Figure 71(b), they may be used to represent the output requirements of a circuit for all possible input values. Thus each xi represents an input circuit capable of carrying two indicator voltages (one voltage for 0 and a different voltage for 1). The function f represents the desired output response in all cases. Such requirements occur at the design stage of all combinational and sequential computer circuitry. Note carefully that the specification of a function f : Bn → B simply lists circuit requirements. It gives no indication of how these requirements can be met. One important way of producing functions from Bn to B is by using Boolean polynomials, which we now consider.

Boolean Polynomials Let x1 , x2 , . . . , xn be a set of n symbols or variables. A Boolean polynomial p(x1 , x2 , . . . , xn ) in the variables xk is defined recursively as follows: 1. x1 , x2 , . . . , xn are all Boolean polynomials. 2. The symbols 0 and 1 are Boolean polynomials. 3. If p(x1 , x2 , . . . , xn ) and q(x1 , x2 , . . . , xn ) are two Boolean polynomials, then so are p(x1 , x2 , . . . , xn ) ∨ q(x1 , x2 , . . . , xn ) and p(x1 , x2 , . . . , xn ) ∧ q(x1 , x2 , . . . , xn ). 4. If p(x1 , x2 , . . . , xn ) is a Boolean polynomial, then so is (p(x1 , x2 , . . . , xn )) . By tradition, (0) is denoted 0 , (1) is denoted 1 , and (xk ) is denoted xk .

280

Order Relations and Structures 5. There are no Boolean polynomials in the variables xk other than those that can be obtained by repeated use of rules 1, 2, 3, and 4. Boolean polynomials are also called Boolean expressions. Example 1

The following are Boolean polynomials in the variables x, y, and z. p1 (x, y, z) = (x ∨ y) ∧ z p2 (x, y, z) = (x ∨ y ) ∨ (y ∧ 1) p3 (x, y, z) = (x ∨ (y ∧ z)) ∨ (x ∧ (y ∧ 1)) p4 (x, y, z) = (x ∨ (y ∨ z )) ∧ ((x ∧ z) ∧ (y ∨ 0))

◆

Ordinary polynomials in several variables, such as x2 y + z4 , xy + yz + x2 y2 , x y +xz4 , and so on, are generally interpreted as expressions representing algebraic computations with unspecified numbers. As such, they are subject to the usual rules of arithmetic. Thus the polynomials x2 + 2x + 1 and (x + 1)(x + 1) are considered equivalent, and so are x(xy + yz)(x + z) and x3 y + 2x2 yz + xyz2 , since in each case we can turn one into the other with algebraic manipulation. Similarly, Boolean polynomials may be interpreted as representing Boolean computations with unspecified elements of B, that is, with 0’s and 1’s. As such, these polynomials are subject to the rules of Boolean arithmetic, that is, to the rules obeyed by ∧, ∨, and in Boolean algebras. As with ordinary polynomials, two Boolean polynomials are considered equivalent if we can turn one into the other with Boolean manipulations. Ordinary polynomials can produce functions by substitution. This process works whether the polynomials involve one or several variables. Thus the polynomial xy + yz3 produces a function f : R3 → R by letting f(x, y, z) = xy + yz3 . For example, f(3, 4, 2) = (3)(4) + (4)(23 ) or 44. In a similar way, Boolean polynomials involving n variables produce functions from Bn to B. 3 3

Example 2

Consider the Boolean polynomial p(x1 , x2 , x3 ) = (x1 ∧ x2 ) ∨ (x1 ∨ (x2 ∧ x3 )). Construct the truth table for the Boolean function f : B3 → B determined by this Boolean polynomial. Solution

The Boolean function f : B3 → B is described by substituting all the 23 ordered triples of values from B for x1 , x2 , and x3 . The truth table for the resulting function ◆ is shown in Figure 72. Boolean polynomials can also be written in a graphical or schematic way. If x and y are variables, then the basic polynomials x ∨ y, x ∧ y, and x are shown schematically in Figure 73. Each symbol has lines for the variables on the left and a line on the right representing the polynomial as a whole. The symbol for x ∨ y is called an or gate, that for x ∧ y is called an and gate, and the symbol for x is called an inverter. The logical names arise because the truth tables showing the functions represented by x ∨ y and x ∧ y are exact analogs of the truth table for the connectives “or” and “and,” respectively. Recall that functions from Bn to B can be used to describe the desired behavior of circuits with n 0-or-1 inputs and one 0-or-1 output. In the case of the

281

Order Relations and Structures x1

x2

x3

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

f (x1, x2, x3) = (x1 ∧ x2) ∨ (x1 ∨ (x 2 ∧ x3)) 0 1 0 0 1 1 1 1

Figure 72 x

x

x∨y

y

x∧y

y

(a)

x

x

(b)

(c)

Figure 73

functions corresponding to the Boolean polynomials x ∨ y, x ∧ y, and x , the desired circuits can be implemented, and the schematic forms of Figure 73 are also used to represent these circuits. By repeatedly using these schematic forms for ∨, ∧, and , we can make a schematic form to represent any Boolean polynomial. For the reasons given previously, such diagrams are called logic diagrams for the polynomial. Example 3

Let

p(x, y, z) = (x ∧ y) ∨ (y ∧ z ).

Figure 74(a) shows the truth table for the corresponding function f : B3 → B. ◆ Figure 74(b) shows the logic diagram for p.

x

y

z

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

f (x, y, z) = (x ∧ y) ∨ (y ∧ z) 0 0 1 0 0 0 1 1 (a)

x

x∧y

y

z

z

y ∧ z

(b)

Figure 74

282

(x ∧ y) ∨ (y ∧ z)

Order Relations and Structures Suppose that p is a Boolean polynomial in n variables, and f is the corresponding function from Bn to B. We know that f may be viewed as a description of the behavior of a circuit having n inputs and one output. In the same way, the logic diagram of p can be viewed as a description of the construction of such a circuit, at least in terms of and gates, or gates, and inverters. Thus if the function f , describing the desired behavior of a circuit, can be produced by a Boolean polynomial p, then the logic diagram for p will give one way to construct a circuit having that behavior. In general, many different polynomials will produce the same function. The logic diagrams of these polynomials will represent alternative methods for constructing the desired circuit. It is almost impossible to overestimate the importance of these facts for the study of computer circuitry. Nearly all circuit design is done by software packages, but it is important to understand the basic principles of circuit design outlined in this section and the next.

5 Exercises 12. (z ∨ (y ∧ (x ∨ x ))) ∧ (y ∧ z ) ; one variable

1. Consider the Boolean polynomial p(x, y, z) = x ∧ (y ∨ z ).

13. (y ∧ z) ∨ x ∨ (w ∧ w ) ∨ (y ∧ z ); two variables and two operations

If B = {0, 1}, compute the truth table of the function f : B3 → B defined by p.

14. (x ∧ y ∧ z ∧ w) ∨ (x ∧ z ∧ w ∧ y ) ∨ (w ∧ x ∧ y ∧ z ) ∨ (w ∧ x ∧ y ∧ z ); two variables and three operations

2. Consider the Boolean polynomial p(x, y, z) = (x ∨ y) ∧ (z ∨ x ). If B = {0, 1}, compute the truth table of the function f : B3 → B defined by p. 3. Consider the Boolean polynomial p(x, y, z) = (x ∧ y ) ∨ (y ∧ (x ∨ y)). If B = {0, 1}, compute the truth table of the function f : B3 → B defined by p.

15. Construct a logic diagram implementing the function f of (b) Exercise 2. (a) Exercise 1 16. Construct a logic diagram implementing the function f of (b) Exercise 4. (a) Exercise 3 17. Construct a logic diagram implementing the function f for (a) f(x, y, z) = (x ∨ (y ∧ z)) ∨ (x ∧ (y ∧ 1)). (b) f(x, y, z) = (x ∨ (y ∨ z )) ∧ ((x ∧ z) ∧ (y ∨ 0)). 18. Give the Boolean function described by the logic diagram in Figure 75. x

4. Consider the Boolean polynomial p(x, y, z) = (x ∧ y) ∨ (x ∧ (y ∧ z )).

y z

If B = {0, 1}, compute the truth table of the function f : B3 → B defined by p. In Exercises 5 through 10, apply the rules of Boolean arithmetic to show that the given Boolean polynomials are equivalent.

19. Give the Boolean function described by the logic diagram in Figure 76.

5. (x ∨ y) ∧ (x ∨ y); y 6. x ∧ (y ∨ (y ∧ (y ∨ y ))); x

x

7. (z ∨ x) ∧ ((x ∧ y) ∨ z) ∧ (z ∨ y); x ∧ y 8. [(x ∧ z) ∨ (y ∨ z) ] ∨ [(y ∧ z) ∨ (x ∧ z )]; x ∨ y

Figure 75

y

9. (x ∨ y) ∨ z ∨ (x ∧ ((y ∧ z) ∨ (y ∧ z ))); (x ∧ y ) ∨ z 10. (x ∧ z) ∨ (y ∨ (y ∧ z)) ∨ ((x ∧ y ) ∧ z ); (x ∧ z) ∨ y In Exercises 11 through 14, rewrite the given Boolean polynomial to obtain the requested format. 11. (x ∧ y ∧ z) ∨ (x ∧ y ∧ z); two variables and one operation

z

Figure 76 20. Give the Boolean function described by the logic diagram in Figure 77.

283

Order Relations and Structures x

y

w z

Figure 77 21. Use the properties of a Boolean algebra to refine the function in Exercise 18 to use the minimal number of variables and operations. Draw the logic diagram for the new function. 22. Use the properties of a Boolean algebra to refine the function in Exercise 19 to use the minimal number of variables

6

and operations. Draw the logic diagram for the new function. 23. Use the properties of a Boolean algebra to refine the function in Exercise 20 to use the minimal number of variables and operations. Draw the logic diagram for the new function.

Circuit Design In Section 5 we considered functions from Bn to B, where B is the Boolean algebra {0, 1}. We noted that such functions can represent input-output requirements for models of many practical computer circuits. We also pointed out that if the function is given by some Boolean expression, then we can construct a logic diagram for it and thus model the implementation of the function. In this section we show that all functions from Bn to B are given by Boolean expressions, and thus logic diagrams can be constructed for any such function. Our discussion illustrates a method for finding a Boolean expression that produces a given function. If f : Bn → B, we will let S(f) = {b ∈ Bn | f(b) = 1}. We then have the following result.

THEOREM 1

Let f , f1 , and f2 be three functions from Bn to B. (a) If S(f) = S(f1 ) ∪ S(f2 ), then f(b) = f1 (b) ∨ f2 (b) for all b in B. (b) If S(f) = S(f1 ) ∩ S(f2 ), then f(b) = f1 (b) ∧ f2 (b) for all b in B. (∨ and ∧ are LUB and GLB, respectively, in B.) Proof (a) Let b ∈ Bn . If b ∈ S(f), then, by the definition of S(f), f(b) = 1. Since S(f) = S(f1 ) ∪ S(f2 ), either b ∈ S(f1 ) or b ∈ S(f2 ), or both. In any case, / S(f), then f(b) = 0. This means that f1 (b) ∨ f2 (b) = 1. Now, if b ∈ f1 (b) ∨ f2 (b) = 0. Thus, for all b ∈ Bn , f(b) = f1 (b) ∨ f2 (b). (b) This part is proved in a manner completely analogous to that used in part ■ (a).

284

Order Relations and Structures Recall that a function f : Bn → B can be viewed as a function f(x1 , x2 , . . . , xn ) of n variables, each of which may assume the values 0 or 1. If E(x1 , x2 , . . . , xn ) is a Boolean expression, then the function that it produces is generated by substituting all combinations of 0’s and 1’s for the xi ’s in the expression. Example 1

Let f1 : B2 → B be produced by the expression E(x, y) = x , and let f2 : B2 → B be produced by the expression E(x, y) = y . Then the truth tables of f1 and f2 are shown in Figures 78(a) and (b), respectively. Let f : B2 → B be the function whose truth table is shown in Figure 78(c). Clearly, S(f) = S(f1 ) ∪ S(f2 ), since f1 is 1 at the elements (0, 0) and (0, 1) of B2 , f2 is 1 at the elements (0, 0) and (1, 0) of B2 , and f is 1 at the elements (0, 0), (0, 1), and (1, 0) of B2 . By Theorem 1, f = f1 ∨ f2 , so a Boolean expression that produces f is x ∨ y . This is easily verified. ◆ x

y

f1 (x, y)

x

y

f2 (x, y)

x

y

f (x, y)

0 0 1 1

0 1 0 1

1 1 0 0

0 0 1 1

0 1 0 1

1 0 1 0

0 0 1 1

0 1 0 1

1 1 1 0

(a)

(b)

(c)

Figure 78

It is not hard to show that any function f : Bn → B for which S(f) has exactly one element is produced by a Boolean expression. Table 1 shows the correspondence between functions of two variables that are 1 at just one element and the Boolean expression that produces these functions.

TABLE 1

Example 2

S(f )

Expression Producing f

{(0, 0)}

x ∧ y

{(0, 1)}

x ∧ y

{(1, 0)}

x ∧ y

{(1, 1)}

x∧y

Let f : B2 → B be the function whose truth table is shown in Figure 79(a). This function is equal to 1 only at the element (0, 1) of B2 ; that is, S(f) = {(0, 1)}. Thus f(x, y) = 1 only when x = 0 and y = 1. This is also true for the expression E(x, y) = x ∧ y, so f is produced by this expression. The function f : B3 → B whose truth table is shown in Figure 79(b) has S(f) = {(0, 1, 1)}; that is, f equals 1 only when x = 0, y = 1, and z = 1. This is also true for the Boolean expression x ∧ y ∧ z, which must therefore produce f . ◆ If b ∈ Bn , then b is a sequence (c1 , c2 , . . . , cn ) of length n, where each ck is 0 or 1. Let Eb be the Boolean expression x1 ∧ x2 ∧ · · · ∧ xn , where xk = xk when ck = 1 and xk = xk when ck = 0. Such an expression is called a minterm. Example 2 illustrates the fact that any function f : Bn → B for which S(f) is a single element of Bn is produced by a minterm expression. In fact, if S(f) = {b}, it is easily seen that the minterm expression Eb produces f . We then have the following result.

285

Order Relations and Structures

x

y

f (x, y)

0 0 1 1

0 1 0 1

0 1 0 0

x

y

z

f (x, y, z)

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

0 0 0 1 0 0 0 0

(b)

(a)

Figure 79

THEOREM 2

Any function f : Bn → B is produced by a Boolean expression. Proof Let S(f ) = {b1 , b2 , . . . , bk }, and for each i, let fi : Bn → B be the function defined by fi (bi ) = 1 fi (b) = 0,

if b = bi .

Then S(fi ) = {bi }, so S(f) = S(f1 ) ∪ · · · ∪ S(fn ) and by Theorem 1, f = f1 ∨ f2 ∨ · · · ∨ fn . By the preceding discussion, each fi is produced by the minterm Ebi . Thus f is produced by the Boolean expression Eb1 ∨ Eb2 ∨ · · · ∨ Ebn and this completes the proof. Example 3

x

y

z

f (x, y, z)

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

0 0 0 1 0 0 0 1

Figure 80

286

■

Consider the function f : B3 → B whose truth table is shown in Figure 80. Since S(f) = {(0, 1, 1), (1, 1, 1)}, Theorem 2 shows that f is produced by the Boolean expression E(x, y, z) = E(0,1,1) ∨E(1,1,1) = (x ∧y∧z)∨(x∧y∧z). This expression, however, is not the simplest Boolean expression that produces f . Using properties of Boolean algebras, we have (x ∧ y ∧ z) ∨ (x ∧ y ∧ z) = (x ∨ x) ∧ (y ∧ z) = 1 ∧ (y ∧ z) = y ∧ z. Thus f is also produced by the simple expression y ∧ z.

◆

The process of writing a function as an “or” combination of minterms and simplifying the resulting expression can be systematized in various ways. We will demonstrate a graphical procedure utilizing what is known as a Karnaugh map. This procedure is easy for human beings to use with functions f : Bn → B, if n is not too large. We will illustrate the method for n = 2, 3, and 4. If n is large or if a programmable algorithm is desired, other techniques may be preferable. The Exercises explore a nongraphical method. We consider first the case where n = 2 so that f is a function of two variables, say x and y. In Figure 81(a), we show a 2 × 2 matrix of squares with each square containing one possible input b from B2 . In Figure 81(b), we have replaced each input b with the corresponding minterm Eb . The labeling of the squares in Figure

Order Relations and Structures 81 is for reference only. In the future we will not exhibit these labels, but we will assume that the reader remembers their locations. In Figure 81(b), we note that x appears everywhere in the first row and x appears everywhere in the second row. We label these rows accordingly, and we perform a similar labeling of the columns. y

y

00

01

x

x∧ y

x∧ y

10

11

x

x ∧ y

x∧y

(a)

(b)

Figure 81

Example 4

Let f : B2 → B be the function whose truth table is shown in Figure 82(a). In Figure 82(b), we have arranged the values of f in the appropriate squares, and we have kept the row and column labels. The resulting 2 × 2 array of 0’s and 1’s is called the Karnaugh map of f . Since S(f) = {(0, 0), (0, 1)}, the corresponding ◆ expression for f is (x ∧ y ) ∨ (x ∧ y) = x ∧ (y ∨ y) = x . x

y

f (x, y)

0 0 1 1

0 1 0 1

1 1 0 0

Truth table of f (a)

y

y

x

1

1

x

0

0

Karnaugh map of f (b)

Figure 82

The outcome of Example 4 is typical. When the 1-values of a function f : B2 → B exactly fill one row or one column, the label of that row or column gives the Boolean expression for f . Of course, we already know that if the 1-values of f fill just one square, then f is produced by the corresponding minterm. It can be shown that the larger the rectangle of 1-values of f , the smaller the expression for f will be. Finally, if the 1-values of f do not lie in a rectangle, we can decompose these values into the union of (possibly overlapping) rectangles. Then, by Theorem 1, the Boolean expression for f can be found by computing the expressions corresponding to each rectangle and combining them with ∨. Example 5

Consider the function f : B2 → B whose truth table is shown in Figure 83(a). In Figure 83(b), we show the Karnaugh map of f and decompose the 1-values x

y

f (x, y)

0 0 1 1

0 1 0 1

1 1 1 0 (a)

y

y

x

1

1

x

1

0 (b)

y

y

x

1

1

x

1

0 (c)

Figure 83

287

Order Relations and Structures into the two indicated rectangles. The expression for the function having 1’s in the horizontal rectangle is x (verify). The function having all its 1’s in the vertical rectangle corresponds to the expression y (verify). Thus f corresponds to the expression x ∨ y . In Figure 83(c), we show a different decomposition of the 1-values of f into rectangles. This decomposition is also correct, but it leads to the more complex expression y ∨ (x ∧ y). We see that the decomposition into rectangles is not unique and that we should try to use the largest possible ◆ rectangles. We now turn to the case of a function f : B3 → B, which we consider to be a function of x, y, and z. We could proceed as in the case of two variables and construct a cube of side 2 to contain the values of f . This would work, but three-dimensional figures are awkward to draw and use, and the idea would not generalize. Instead, we use a rectangle of size 2 × 4. In Figures 84(a) and (b), respectively, we show the inputs (from B3 ) and corresponding minterms for each square of such a rectangle. 0

0

0

1

1

1

1

0

0 0

0

0

0

0

1

0

1

1

0

1

0

1 1

0

0

1

0

1

1

1

1

1

1

0

(a) y

y

x x ∧ y ∧ z

x ∧ y ∧ z

x ∧ y ∧ z

x ∧ y ∧ z

x ∧ y ∧ z

x ∧ y ∧ z

x∧y∧z

x ∧ y ∧ z

x

z z (b)

Figure 84

Consider the rectangular areas shown in Figure 85. If the 1-values for a function f : B3 → B exactly fill one of the rectangles shown, then the Boolean expression for this function is one of the six expressions x, y, z, x , y , or z , as indicated in Figure 85. Consider the situation shown in Figure 85(a). Theorem 1(a) shows that f can be computed by joining all the minterms corresponding to squares of the region with the symbol ∨. Thus f is produced by (x ∧ y ∧ z ) ∨ (x ∧ y ∧ z) ∨ (x ∧ y ∧ z ) ∨ (x ∧ y ∧ z) = ((x ∨ x) ∧ (y ∧ z )) ∨ ((x ∨ x) ∧ (y ∧ z)) = (1 ∧ (y ∧ z )) ∨ (1 ∧ (y ∧ z)) = (y ∧ z ) ∨ (y ∧ z) = y ∧ (z ∨ z) = y ∧ 1 = y . A similar computation shows that the other five regions are correctly labeled. If we think of the left and right edges of our basic rectangle as glued together to make a cylinder, as we show in Figure 86, we can say that the six large regions

288

Order Relations and Structures

Shaded region is y

Shaded region is y

(a)

(b)

y ∧ z y ∧ z

y ∧ z

x Shaded region is z

Shaded region is z

(c)

(d)

y

Shaded region is x

Shaded region is x

(e)

(f)

Figure 85

y∧z

Figure 86

shown in Figure 85 consist of any two adjacent columns of the cylinder, or of the top or bottom half-cylinder. The six basic regions shown in Figure 85 are the only ones whose corresponding Boolean expressions need be considered. That is why we used them to label Figure 84(b), and we keep them as labels for all Karnaugh maps of functions from B3 to B. Theorem 1(b) tells us that, if the 1-values of a function f : B3 → B form exactly the intersection of two or three of the basic six regions, then a Boolean expression for f can be computed by combining the expressions for these basic regions with ∧ symbols. Thus, if the 1-values of the function f are as shown in Figure 87(a), then we get them by intersecting the regions shown in Figures 85(a) and (d). The Boolean expression for f is therefore y ∧ z . Similar derivations can be given for the other three columns. If the 1-values of f are as shown in Figure 87(b), we get them by intersecting the regions of Figures 85(c) and (e), so a Boolean expression for f is z ∧ x . In a similar fashion, we can compute the expression for any function whose 1-values fill two horizontally adjacent squares. There are eight such functions if we again consider the rectangle to be formed into a cylinder. Thus we include the case where the 1-values of f are as shown in Figure 87(c). The resulting Boolean expression is z ∧ x . If we intersect three of the basic regions and the intersection is not empty, the intersection must be a single square, and the resulting Boolean expression is a minterm. In Figure 87(d), the 1-values of f form the intersection of the three regions shown in Figures 85(a), (c), and (f). The corresponding minterm is y ∧ z ∧ x. Thus we need not remember the placement of minterms in Figure 84(b), but instead may reconstruct it. We have seen how to compute a Boolean expression for any function f : B3 → B whose 1-values form a rectangle of adjacent squares (in the cylinder) of size 2n × 2m , n = 0, 1; m = 0, 1, 2. In general, if the set of 1-values of f do not form such a rectangle, we may write this set as the union of such rectangles. Then a Boolean expression for f is computed by combining the expression associated with each rectangle with ∨ symbols. This is true by Theorem 1(a).

289

Order Relations and Structures The preceding discussion shows that the larger the rectangles that are chosen, the simpler will be the resulting Boolean expression. y

y

y

y

x

1

0

0

0

x

0

1

1

0

x

1

0

0

0

x

0

0

0

0

z z

z z

(a)

(b)

y

y

y

y

x

1

0

0

1

x

0

0

0

0

x

0

0

0

0

x

0

1

0

0

z z (c)

z z (d)

Figure 87

Example 6

Consider the function f whose truth table and corresponding Karnaugh map are shown in Figure 88. The placement of the 1’s can be derived by locating the corresponding inputs in Figure 84(a). One decomposition of the 1-values of f is shown in Figure 88(b). From this we see that a Boolean expression for f is ◆ (y ∧ z ) ∨ (x ∧ y ) ∨ (y ∧ z).

x

y

z

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1 (a)

f (x, y, z) 1 1 0 1 1 0 0 1

y

y

x

1

1

1

0

x

1

0

1

0

z z (b)

Figure 88

Example 7

290

Figure 89 shows the truth table and corresponding Karnaugh map for a function f . The decomposition into rectangles shown in Figure 89(b) uses the idea that the first and last columns are considered adjacent (by wrapping around the cylinder). Thus the symbols are left open ended to signify that they join in one 2 × 2 rectangle corresponding to z . The resulting Boolean expression is z ∨ (x ∧ y) (verify). ◆

Order Relations and Structures x

y

z

f (x, y, z)

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

1 0 1 0 1 0 1 1

y

y

x

1

0

0

1

x

1

0

1

1

z z (b)

(a)

Figure 89

Finally, without additional comment, we present in Figure 90 the distribution of inputs and corresponding labeling of rectangles for the case of a function f : B4 → B, considered as a function of x, y, z, and w. Here again, we consider the first and last columns to be adjacent, and the first and last rows to be adjacent, both by wraparound, and we look for rectangles with sides of length some power of 2, so the length is 1, 2, or 4. The expression corresponding to such rectangles is given by intersecting the large labeled rectangles of Figure 91. z 00

01

11

10

00

0000

0001

0011

0010

01

0100

0101

0111

0110

11

1100

1101

1111

1110

10

1000

1001

1011

1010

(a)

z

x y y x

w w (b)

Figure 90

Example 8

Figure 92 shows the Karnaugh map of a function f : B4 → B. The 1-values are placed by considering the location of inputs in Figure 90(a). Thus f(0101) = 1, f(0001) = 0, and so on. The center 2 × 2 square represents the Boolean expression w ∧ y (verify). The four corners also form a 2 × 2 square, since the right and left edges and the top and bottom edges are considered adjacent. From a geometric point of view, we can see that if we wrap the rectangle around horizontally, getting a cylinder, then when we further wrap around vertically, we will get a torus or inner tube. On this inner tube, the four corners form a 2 × 2 square, which represents the Boolean expression w ∧ y (verify). It then follows that the decomposition leads to the Boolean expression (w ∧ y) ∨ (w ∧ y ) for f .

◆

291

Order Relations and Structures

Shaded region is z

Shaded region is z

Shaded region is x

Shaded region is x

(a)

(b)

(c)

(d)

Shaded region is y

Shaded region is y

Shaded region is w

Shaded region is w

(e)

(f)

(g)

(h)

Figure 91 z

1

z

z

0

0

1

z

1

1

1

1

0

0

0

0

x

x 0

1

1

0

y y

y y 0

1

1

0

x

0

0

1

0

1

1

0

0

x 1

0

0

1

w w

Figure 92

Example 9

w w

Figure 93

In Figure 93, we show the Karnaugh map of a function f : B4 → B. The decomposition of 1-values into rectangles of sides 2n , shown in this figure, again uses the wraparound property of top and bottom rows. The resulting expression for f is (verify) (z ∧ y ) ∨ (x ∧ y ∧ z) ∨ (x ∧ y ∧ z ∧ w). The first term comes from the 2 × 2 square formed by joining the 1 × 2 rectangle in the upper-left corner and the 1 × 2 rectangle in the lower-left corner. The second comes from the rectangle of size 1 × 2 in the upper-right corner, and the last is a ◆ minterm corresponding to the isolated square.

292

Order Relations and Structures

6 Exercises In Exercises 1 through 6, construct Karnaugh maps for the functions whose truth tables are given. 1. x 0 0 1 1

y 0 1 0 1

f(x, y) 1 0 0 1

2. x 0 0 1 1

y 0 1 0 1

f(x, y) 1 0 1 0

3. x 0 0 0 0 1 1 1 1

y 0 0 1 1 0 0 1 1

z 0 1 0 1 0 1 0 1

f(x, y, z) 1 1 0 0 1 0 1 0

4. x 0 0 0 0 1 1 1 1

y 0 0 1 1 0 0 1 1

z 0 1 0 1 0 1 0 1

f(x, y, z) 0 1 1 1 0 0 0 1

5. x 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

y 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

z 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

w 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

f(x, y, z, w) 1 0 1 0 0 1 1 0 0 0 0 0 1 0 1 0

6. x 0 0 0 0 0 0 0 0 1 1 1 1 1 1 1 1

y 0 0 0 0 1 1 1 1 0 0 0 0 1 1 1 1

z 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1

w 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1

f(x, y, z, w) 0 0 1 0 0 0 1 0 0 0 0 1 0 0 1 1

7. Construct a Karnaugh map for the function f for which S(f) = {(0, 0, 1), (0, 1, 1), (1, 0, 1), (1, 1, 1)}. 8. Construct a Karnaugh map for the function f for which S(f) = {(0, 0, 0, 1), (0, 0, 1, 1), (1, 0, 1, 0), (1, 1, 0, 1), (0, 1, 0, 0), (1, 0, 0, 0)}. In Exercises 9 through 16 (Figures 94 through 101), Karnaugh maps of functions are given, and a decomposition of 1-values into rectangles is shown. Write the Boolean expression for these functions, which arise from the maps and rectangular decompositions.

293

Order Relations and Structures 9.

y

y

x

1

0

x

0

1

10.

Figure 94

14.

y

y

y

x

1

1

x

1

1

1

1

x

1

0

x

0

0

1

0

y

Figure 95

z z

Figure 99 11.

y x

1

15.

y

1

1

z

1 x

x

1

0

0

z

0

0

1

1

0

0

1

1

1

y y z z

1

0

0

1

0

1

1

0

x

Figure 96

w w

12.

y

y

Figure 100 16.

x

1

1

0

1

x

0

1

0

1

z

z

1

1

0

1

1

1

0

1

x

z

y y

z

0

0

0

0

1

0

0

1

x

Figure 97

w

13.

y

w

y

Figure 101 x

0

1

0

1

In Exercises 17 through 24, use the Karnaugh map method to find a Boolean expression for the function f .

x

1

1

0

1

17. Let f be the function of Exercise 1. 18. Let f be the function of Exercise 2.

z z

Figure 98

294

19. Let f be the function of Exercise 3. 20. Let f be the function of Exercise 4. 21. Let f be the function of Exercise 5.

Order Relations and Structures 22. Let f be the function of Exercise 6. 23. Let f be the function of Exercise 7. 24. Let f be the function of Exercise 8. The following exercises develop another way to produce an expression for a Boolean function. For Exercises 25 and 26, let f : B2 → B with S(f ) = {(0, 0), (0, 1), (1, 0)}. 25. (a) Give the corresponding minterm for each element of S(f ). (b) If two elements s1 , s2 of S(f ) differ in exactly one position, the variable xi in that position is not needed in the join of the minterms corresponding to s1 and s2 . Explain why. 26. For each pair in S(f ) that differ in exactly one position, form a reduced minterm as follows. Suppose the corresponding minterms a1 ∧ a2 ∧ · · · ∧ an and b1 ∧ b2 ∧ · · · ∧ bn differ only at position i. Replace this pair with the reduced minterm a1 ∧ a2 ∧ · · · ∧ ai−1 ∧ ai+1 ∧ · · · ∧ an , where xi does not appear.

(a) Give the reduced minterms for S(f ) = {(0, 0), (0, 1), (1, 0)}. (b) Verify that f is produced by the join of the reduced minterms in part (a). The process in Exercise 26 can be repeated. For Exercises 27 through 29, let f : B3 → B with S(f ) = {(0, 0, 0), (0, 1, 0), (1, 0, 0), (1, 1, 0), (1, 1, 1)}. 27. Give the reduced minterms for each pair of elements in S(f ) that differ in exactly one position. 28. Using the reduced minterms produced for Exercise 27, again replace each pair of reduced minterms that differ in exactly one position. 29. (a) Form the join of the expressions represented by the reduced minterms of Exercise 28 and any (reduced) minterms that have not been replaced. (b) Verify that the expression formed in part (a) produces f .

Tips for Proofs

Statements of the form ∀x P(x) or ∼(∃x Q(x)) are candidates for proof by contradiction, since it is natural to explore what would happen if there were an object that did not have property P or if there were an object with property Q. This situation is seen in Theorem 2, Section 1. Statements about finite sets also often lend themselves to indirect proofs, as seen in Theorem 1, Section 2. Many examples and exercises in this chapter ask that you check whether a specified lattice is a Boolean algebra or not. Remember that this is the same as proving that the lattice is a Boolean algebra. The fact that the number of elements is a power of 2 is a necessary condition, but not a sufficient condition. This means that if the number of elements is not a power of 2, the lattice is not a Boolean algebra, but there are lattices with 2n elements that are not Boolean algebras. The check on a divisibility lattice Dn is an easy one and worth memorizing. If the Hasse diagram is “small,” then comparing it with those of the Bn is an efficient way to carry out the check. Attempting to construct an isomorphism between the given lattice and a known Boolean algebra is the next most efficient method. As a last resort, try to verify that the properties of a Boolean algebra are satisfied by the lattice. But be sensible about this. Try the “single” cases first—there is a unique least element 0; there is a unique greatest element I; 0 = I; and so on—not those that require working through lots of cases, such as the associative property for ∧.

Key Ideas for Review • Partial order on a set: relation that is reflexive, antisymmet-

ric, and transitive • Partially ordered set or poset: set together with a partial order • Linearly ordered set: partially ordered set in which every

pair of elements is comparable • Theorem: If A and B are posets, then A × B is a poset with

the product partial order. • Dual of a poset (A, ≤): the poset (A, ≥), where ≥ denotes

the inverse of ≤

• Hasse diagram • Topological sorting • Isomorphism of posets • Maximal (minimal) element of a poset • Theorem: A finite nonempty poset has at least one maximal

element and at least one minimal element. • Greatest (least) element of a poset A • Theorem: A poset has at most one greatest element and at

most one least element.

295

Order Relations and Structures • Upper (lower) bound of subset B of poset A: element a ∈ A •

• • •

•

•

such that b ≤ a (a ≤ b) for all b ∈ B Least upper bound (greatest lower bound) of subset B of poset A: element a ∈ A such that a is an upper (lower) bound of B and a ≤ a (a ≤ a), where a is any upper (lower) bound of B Lattice: a poset in which every subset consisting of two elements has a LUB and a GLB Theorem: If L1 and L2 are lattices, then L = L1 × L2 is a lattice. Theorem: Let L be a lattice, and a, b ∈ L. Then (a) a ∨ b = b if and only if a ≤ b. (b) a ∧ b = a if and only if a ≤ b. (c) a ∧ b = a if and only if a ∨ b = b. Theorem: Let L be a lattice. Then 1. (a) a ∨ a = a (b) a ∧ a = a 2. (a) a ∨ b = b ∨ a (b) a ∧ b = b ∧ a 3. (a) a ∨ (b ∨ c) = (a ∨ b) ∨ c (b) a ∧ (b ∧ c) = (a ∧ b) ∧ c 4. (a) a ∨ (a ∧ b) = a (b) a ∧ (a ∨ b) = a Theorem: Let L be a lattice, and a, b, c ∈ L. 1. If a ≤ b, then (a) a ∨ c ≤ b ∨ c (b) a ∧ c ≤ b ∧ c 2. a ≤ c and b ≤ c if and only if a ∨ b ≤ c 3. c ≤ a and c ≤ b if and only if c ≤ a ∧ b 4. If a ≤ b and c ≤ d, then

• • • •

(a) a ∨ c ≤ b ∨ d (b) a ∧ c ≤ b ∧ d Isomorphic lattices Bounded lattices: lattice that has a greatest element I and a least element 0 Theorem: A finite lattice is bounded. Distributive lattice: lattice that satisfies the distributive laws: a ∧ (b ∨ c) = (a ∧ b) ∨ (a ∧ c), a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c).

• Complement of a: element a ∈ L (bounded lattice) such

that

a ∨ a = I

and

a ∧ a = 0.

• Theorem: Let L be a bounded distributive lattice. If a com-

plement exists, it is unique. • Complemented lattice: bounded lattice in which every ele-

ment has a complement • Boolean algebra: a lattice isomorphic with (P(S), ⊆) for

some finite set S • Properties of a Boolean algebra • Truth tables • Boolean expression • Minterm: a Boolean expression of the form x1 ∧x2 ∧· · ·∧xn ,

where each xk is either xk or xk • Theorem: Any function f : Bn → B is produced by a Boolean expression. • Karnaugh map

Chapter Self-Test 1. Upon what common relation is the concept of partial order based?

x3

a

x4 x2

c

b

f

2. How does a partial order differ from a lattice? 3. What is a model for any Boolean algebra? 4. What kind of mathematical object is the Karnaugh map of a function and how is it used? 5. What is the relationship between Boolean functions and Boolean expressions? 6. Determine whether the given relation is a partial order. Explain your answer. (a) A is any set; a R b if and only if a = b. (b) A is the set of parallel lines in the plane; l1 R l2 if and only if l1 coincides with l2 or l1 is parallel to l2 .

x1

Figure 102

d

e

Figure 103

8. Let (A, ≤) be the poset whose Hasse diagram is given in Figure 103. (a) Find all minimal and maximal elements of A. (b) Find the least and greatest elements of A. 9. Let A = {2, 3, 4, 6, 8, 12, 24, 48} and ≤ denote the partial order of divisibility; B = {4, 6, 12}. Find, if they exist, (a) all upper bounds of B (b) all lower bounds of B (c) the least upper bound of B

7. Given the Hasse diagram in Figure 102, (a) Draw the digraph of the partial order R defined; (b) Draw the Hasse diagram of the partial order R−1 .

296

(d) the greatest lower bound of B. 10. Show that a linearly ordered poset is a distributive lattice. 11. Find the complement of each element in D105 .

Order Relations and Structures 12. Let A = {a, b, c, d} and R be a relation on A whose matrix is ⎡ ⎤ 1 0 1 1 ⎢0 1 1 1⎥ . MR = ⎣ 0 0 1 1⎦ 0 0 0 1 (a) Prove that R is a partial order. (b) Draw the Hasse diagram of R. 13. Let L be a lattice. Prove that for every a, b, and c in L, if a ≤ b and c ≤ d, then a ∨ c ≤ b ∨ d and a ∧ c ≤ b ∧ d. 14. Consider the Hasse diagrams given in Figure 104. (a) Which of these posets are not lattices? Explain. (b) Which of these posets are not Boolean algebras? Explain. f

e b

c

d b

a

c a

d

e

b

c a

d b

c a

Figure 104 15. Let (D63 , ≤) be the lattice of all positive divisors of 63 and x ≤ y means x | y. (a) Draw the Hasse diagram of the lattice. (b) Prove or disprove the statement: (D63 , ≤) is a Boolean algebra.

16. (a) Write the Boolean expression represented by the logic diagram in Figure 105. (b) Use the rules of Boolean arithmetic to find an expression using fewer operations that is equivalent to the expression found in part (a). (c) Draw a logic diagram for the expression found in part (b). x y z

Figure 105 17. Use the Karnaugh map method to find a Boolean expression for the function f whose truth table is as follows. x y z f(x, y, z) 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 1 1 0 1 1 1 1 0 1 1 1 1 0

Experiment 6 Many real-life and computer processes can be modeled by a partial order that shows the necessary sequence of actions to complete the process. Section 1 contains a brief description of how to convert such a partial order into a linear order that could be followed by a person or a computer program. In many cases, a strict linear order is not necessary, for example, if several actions can take place concurrently. In this experiment you will learn the basic ideas of another modeling technique for concurrent actions. Part I. To obtain a degree in computer science at Ole U, the following courses are required: Computer Science I, Computer Science II, Discrete Mathematics, Data Structures, Compiler Design, AlgorithmAnalysis, Assembly Architecture, Formal Languages, Operating Systems, and Complexity Theory. Computer Science II has Computer Science I and Discrete Mathematics as prerequisites. Computer Science II is a prerequisite for Data Structures, Assembly Architecture, and Formal Languages. Data Structures is a prerequisite for Algorithm Analysis and for Operating Systems. Algorithm Analysis is a prerequisite for Complexity Theory; Compiler Design requires both Assembly Architecture and Formal Languages. 1. Give a Hasse diagram that shows the prerequisite relationships among these courses.

297

Order Relations and Structures 2. Suppose you were able to take only one course each term. Give a topological sorting of the partial order in Question 1.

p2 p1

t

p3

Figure 1 p1 p2 p3

p4 t1

p6 p5 t3

t2

Figure 2

p2 p1

t

p3

Figure 3

p1 p2 p3

p4 t1

t2

Figure 4

p6 p5 t3

Part II. Progressing through the required courses in a linear fashion is not necessary and certainly not optimal in terms of how long it will take to obtain your degree. Here we present a different type of digraph to model conditions (prerequisites) and possibly concurrent actions. A Petri net is a digraph whose set of vertices can be partitioned into P, the set of places (or conditions), and T , the set of transitions (or actions), along with a function f that associates a nonnegative number with each element of P. Any edge in the digraph is between an element of P and an element of T . That is, the digraph represents a relation R, where R = I ∪O, I ⊆ P ×T , O ⊆ T ×P. Usually the elements of P are denoted by circles and those of T by vertical bars. As a way to indicate that a condition pi ∈ P is satisfied, one or more dots, called tokens, are marked inside the circle for pi . Allowing more than one token at a place permits Petri nets to model a broad range of situations. The function f is defined by f(pi ) is the number of tokens at pi ; hence, f is sometimes called a marking of the Petri net. 1. For the Petri net depicted by Figure 1, give P, T , R, and f . 2. For the Petri net depicted by Figure 2, give P, T , R, and f . Completing the action represented by t ∈ T is called firing the transition. A transition can be fired only when the conditions for that action are met. Formally, a transition t can be fired only when every pi such that (pi , t) ∈ I contains at least one token. When this is true, we say that t is enabled. Firing a transition t creates a new Petri net in which a token is removed from each pi such that (pi , t) ∈ I and a token is placed in each pj such that (t, pj ) ∈ O. In Figure 1, firing t produces the net in Figure 3. Only t2 is enabled in Figure 2. The result of firing t2 is shown in Figure 4. If several transitions are enabled, then any one of them may fire at any time. This models the sometimes random nature of choices for concurrent actions. 1. Create a Petri net to represent the course prerequisite situation of Part I. Identify the set of places and the set of transitions. Based on your academic record, mark the net appropriately. Which transitions are enabled in this digraph? 2. (a) Give the new marking function for the net represented by Figure 3. (b) Give the new marking function for the net represented by Figure 4. Part III. Our simple example of course prerequisites lacks a common feature of concurrent processes, repeatability. Consider the following Petri net, N, shown in Figure 5.

N

Figure 5

298

Order Relations and Structures We see here that each transition in T can be enabled and fired repeatedly. A marking of a Petri net is live if every transition can be fired by some sequence of transition firings. Not every marking is live. In the net shown in Figure 6, once t1 or t2 is fired, either t3 or t4 is enabled (but not both). After that transition is fired, no further action is possible. t3

t4

t1

t2

Figure 6

If no transitions can be fired, we say there is a deadlock. 1. Is it possible to mark Figure 5 using some tokens so that deadlock occurs? If so, draw this marked Petri net. If not, explain why. 2. Is it possible to mark Figure 6 using some tokens so a deadlock never occurs? If so, draw this marked Petri net. If not, explain why. Part IV. Figure 7 gives a model of a simple computer network where two workers share a high-speed Internet connection and a database. Only one user of a resource is allowed at a time, but both resources are needed for some tasks the workers perform. Internet available

B releases Internet Internet in use by B

Internet in use by A A does task A acquires Internet

B acquires Internet

Task done

B does task Database in use by B

B releases database

A releases Internet

Task done B acquires database

A acquires database

Database available

Database in use by A

A releases database

Figure 7

1. Give a sequence of transition firings that returns the net to its original marking. Explain in words what the workers have done. 2. Give a sequence of transition firings that produces deadlock in this net. Explain in words what the workers have done. 3. If they were in the same room, the workers might choose to communicate to avoid deadlock, but this is not always possible. Redesign the Petri net in Figure 7 so that deadlock cannot occur. Consider using conditions such as p: The system is available; q: A is in control of the system; and r: B is in control of the system. Clearly identify all conditions and actions.

299

Order Relations and Structures

Coding Exercises 1. Write a subroutine that determines if a relation R represented by its matrix is a partial order. For Exercises 2 through 4, let Bn = {(x1 , x2 , x3 , . . . , xn ) | xi ∈ {0, 1}} and x, y ∈ Bn . 2. Write a subroutine that determines if x ≤ y. 3.

(a) Write a function that computes x ∧ y. (b) Write a function that computes x ∨ y. (c) Write a function that computes x .

4. Write a subroutine that given x produces the corresponding minterm. 5.

Let B = {0, 1}. Write a program that prints a truth table for the function f : B3 → B defined by p(x, y, z) = (x ∧ y ) ∨ (y ∧ (x ∨ y)).

Answers to Odd-Numbered Exercises Exercise Set 1

21.

1. (a) No.

(b) No.

3. (a) Yes.

(b) Yes.

30 6

10

5. {(a, a), (b, b), (c, c), (a, b)}, {(a, a), (b, b), (c, c), (a, b), (a, c)}, {(a, a), (b, b), (c, c), (a, b), (c, b)}, {(a, a), (b, b), (c, c), (a, b), (b, c), (a, c)}, {(a, a), (b, b), (c, c), (a, b), (c, b), (c, a)}, {(a, a), (b, b), (c, c), (a, c), (c, b), (a, b)}.

2

15 3

5 1

23. Linear.

72 36

7. The structure of the proof is to check directly each of the three properties required for a partial order.

12

9.

6

4

3 2

25. If the main diagonal of MR is all 1’s, then R is reflexive. If MR MR = MR , then R is transitive. If aij = 1, i = j, in MR , then aji must be 0 in order for R to be antisymmetric.

3 1

11. {(1, 1), (2, 2), (3, 3), (4, 4), (1, 3), (1, 4), (2, 3), (2, 4), (3, 4)}.

27. (a) {2, 3}. (b) {b, c, d}. (e) {2, 3, 7, 8}.

13.

29.

15.

a b

e

5 4 3

c

d

2 1

⎡

1 ⎢0 ⎢ 17. ⎢ 0 ⎣0 0

1 1 0 0 0

1 0 1 0 0

1 0 0 1 0

⎤ 1 0⎥ ⎥ 0 ⎥. 0⎦ 1

19. ACE, BASE, CAP, CAPE, MACE, MAP, MOP, MOPE.

300

(c) {3}.

(d) {2, 3}.

8 7 5 6 4 2 3 1

31. a ≤ a gives a ≤ a for all a ∈ A . ≤ is reflexive. Suppose a ≤ b and b ≤ a. Then a ≤ b, b ≤ a, and a = b. Hence ≤ is antisymmetric.

Order Relations and Structures Suppose a ≤ b and b ≤ c. Then a ≤ b, b ≤ c, and a ≤ c. Hence a ≤ c and ≤ is transitive. 33. Suppose U ⊂ T and T ⊂ V . Then U ⊂ V and R is transitive. No set is a proper subset of itself so R is irreflexive. Hence R is a quasiorder. 35. Suppose a R−1 b and b R−1 c. Then c R b, b R a, and c R a. Hence a R−1 c and R−1 is transitive. Suppose that x R−1 x. Then x R x, but this is a contradiction. Hence R−1 is irreflexive and a quasiorder. 37. (a, b) ≺ (a, b) since a | a and b ≤ b. Thus ≺ is reflexive. Suppose (a, b) ≺ (c, d) and (c, d) ≺ (a, b). Then a | c and c | a. This means c = ka = k(mc) and for a and c in B, km = 1 implies k = m = 1. Hence a = c. Also, b ≤ d and d ≤ b so b = d. Thus ≺ is antisymmetric. 39. 30 {e, f, g} 10

6

15

2

5

3

{e, f } {e}

{e, g} {f}

23. (a) f , g, h.

(b) a, b, c.

(c) f .

(d) c.

25. (a) d, e, f .

(b) b, a

(c) d.

(d) b.

27. (a) None.

(b) b.

(c) None.

(d) b.

29. (a) x ∈ [2, ∞).

(b) x ∈ (−∞, 1].

(c) 2. 1 31. (a) 1 1 (c) 1

(d) 1. 0 (b) 0 0 (d) 0

33.

1 . 1

1 . 1

0 . 0

0 . 0

h g f e

{ f, g}

d

{g}

c b

1

{}

Define F as follows: F(1) = { }; F(2) = {e}; F(5) = {f }; F(3) = {g}; F(10) = {e, f }; F(6) = {e, g}; F(15) = {f, g}; F(30) = {e, f, g}. 41. Let U = {a, b, c, d}, S a subset of U, and fS be the characteristic function of S (relative to U). Define

fS (a) fS (b) g(S) = . fS (c) fS (d) Then g is a one-to-one correspondence between P(U) and A. If S ≤ T in (U, ⊆), then fS (x) ≤ fT (x) for all x in U. Hence g(S) ≤ g(T ). Exercise Set 2 1. 3. 5. 7. 9. 11. 13. 15. 17.

Maximal: 3, 5; minimal: 1, 6. Maximal: e, f ; minimal: a. Maximal: none; minimal: none. Maximal: 1; minimal: none. Greatest: f ; least: a. No greatest or least. Greatest: none; least: none. Greatest: 72; least: 2. No, a may be maximal and there exists an element of A, b, such that a and b are incomparable. 19. (a) True. There cannot be a1 < a2 < · · · since A is finite. (b) False. Not all elements have to be comparable. (c) True. There cannot be · · · < a2 < a1 since A is finite. (d) False. Not all elements have to be comparable. 21. Suppose a and b are least elements of (A, ≤). Then a ≤ b and b ≤ a. Since ≤ is antisymmetric, a = b. Note: This is a restatement of Theorem 2.

a

35. The least element of A is the label on the row that is all ones. The greatest element of A is the label on the column that is all ones. (b) {2, 4, 8, 16, 32, 64}.

37. (a) 49. Exercise Set 3

1. Yes, all the properties are satisfied. 3. No, GLB({e, b}) does not exist. 5. Yes, all the properties are satisfied. 7. No.

9. Yes, LUB(M, N) = aij = max{mij , nij } and

GLB(M, N) = bij = min{mij , nij } 11.

(b1, b2) (a1, b2)

(b1, a2) (a1, a2)

13. For each T1 , T2 ⊆ T , T1 ∩ T2 , and T1 ∪ T2 are subsets of T so P(T ) is a sublattice of P(S ). 15. For any elements x, y of a linearly ordered poset, x ≤ y or y ≤ x. Say x ≤ y. Then x = x ∧ y and y = x ∨ y. Hence any subset of a linearly ordered poset is a sublattice. 17.

301

Order Relations and Structures 19. Suppose a∧b = a. a ≤ a∨b = (a∧b)∨b = (a∨b)∧b ≤ b. Thus a ≤ b. Suppose a ≤ b. a ∧ b ≤ a and a ≤ a, a ≤ b gives a ≤ a ∧ b. Hence a ∧ b = a. 21. (a) 12. (b) Figure 44(a): 3;

25. Suppose a ∧ x = a ∧ y and a ∨ x = a ∨ y. Then y ≤ y ∨ (y ∧ a) = (y ∧ y) ∨ (y ∧ a) = y ∧ (y ∨ a) = y ∧ (a ∨ x) = (y ∧ a) ∨ (y ∧ x) = (a ∧ x) ∨ (y ∧ x) = x ∧ (a ∨ y) ≤ x. Hence y ≤ x. A similar argument shows x ≤ y. Thus x = y.

15. Suppose a ≤ b. Then a ∧ c ≤ a ≤ b and a ∧ c ≤ c so a ∧ c ≤ b ∧ c. 17. (a ∧ b) ∨ (a ∧ b ) = a ∧ (b ∨ b ) = a ∧ I = a.

Figure 44(b): 3.

23. (a1 , a2 ) ∧ ((b1 , b2 ) ∨ (c1 , c2 )) = (a1 , a2 ) ∧ (b1 ∨ c1 , b2 ∨ c2 ) = (a1 ∧ (b1 ∨ c1 ), a2 ∧ (b2 ∨ c2 )) = ((a1 ∧ b1 ) ∨ (a1 ∧ c1 ), (a2 ∧ b2 ) ∨ (a2 ∧ c2 )) = ((a1 , a2 ) ∧ (b1 , b2 )) ∨ ((a1 , a2 ) ∧ (c1 , c2 )). A similar argument establishes the other distributive property.

Suppose (a ∧ b ) ∨ (a ∧ b) = 0. Then a ∧ b = 0 and a ∧ b = 0. We have I = 0 = (a ∧ b ) = a ∨ b. So a is the complement of b; b = a .

27. 1 = 42, 42 = 1, 2 = 21, 21 = 2, 3 = 14, 14 = 3, 7 = 6, 6 = 7. 29. Neither. 31. Distributive, but not complemented. 33. If x = x , then x = x ∨ x = I and x = x ∧ x = 0. But by Exercise 18, 0 = I. Hence, x = x . 35. Suppose P1 ≤ P2 . Then R1 ⊆ R2 . Let x ∈ Ai . Then Ai = {y | a R1 y} and Ai ⊆ {y | x R2 y} = Bj , where x ∈ Bj . Suppose each Ai ⊆ Bj . Then x R1 y implies x R2 y and R1 ⊆ R2 . Thus P1 ≤ P2 . 37. The sublattice {a, b, d} of Figure 57 is not complemented. 39. For any a, b, c in the sublattice with a ≤ c, a ∨ (b ∧ c) = (a ∨ b) ∧ c, because this is true in the full lattice. Exercise Set 4

19. (a∧b ∧c)∨(b ∧c) = (a∨I)∧(b ∧c) = I ∧(b ∧c) = b ∧c. 21. Suppose a ≤ b. Then a ∨ (b ∧ c) = (a ∨ b) ∧ (a ∨ c) = b ∧ (a ∨ c). 23. R is reflexive because mii = 1, i = 1, 2, . . . , 8. R is antisymmetric since if mij = 1 and i = j, then mji = 0. R is transitive, because MR MR shows that R2 ⊆ R. 25. Complement pairs are a, h; b, g; c, f ; d, e. Since each element has a unique complement, (A, R) is complemented. 27. (A, R) is not a Boolean algebra; complements are not unique. 29. (a) {a}, {b}, {c}.

31. Matrices with exactly one 1. Exercise Set 5 1. x

y

z

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

3. x

y

z

(x ∨ y )

∨

(y ∧

0 0 0 0 1 1 1 1

0 0 1 1 0 0 1 1

0 1 0 1 0 1 0 1

0 0 0 0 1 1 0 0 (1)

0 0 1 1 1 1 1 1 (4)

0 0 1 1 0 0 1 1 (3)

1. No, it has 6 elements, not 2n elements.

x ∧

(y ∨ z )

0 0 0 0 1 0 1 1 ↑

1 0 1 1 1 0 1 1 (x ∨ y) 1 1 1 1 0 0 1 1 (2)

3. No, it has 6 elements, not 2n elements.

5. (x ∨ y) ∧ (x ∨ y) = (x ∧ x ) ∨ y = 0 ∨ y = y.

5. Yes, it is B3 .

7. (z ∨ x) ∧ ((x ∧ y) ∨ z) ∧ (z ∨ y)) = (z ∨ (x ∧ y)) ∧ ((x ∧ y) ∨ z) = (x ∧ y) ∨ (z ∧ z) = (x ∧ y) ∨ 0 = x ∧ y.

7. Yes, it is B1 . 9. Yes; 385 = 5 · 7 · 11. 11. No, each Boolean algebra must have 2n elements. 13. Suppose a = b. (a ∧ b ) ∨ (a ∧ b) = (b ∧ b ) ∨ (a ∧ a) = 0 ∨ 0 = 0.

302

(b) 2, 3, 5.

9. (x ∨y) ∨z∨x∧((y ∧z)∨(y ∧z )) = (x∧y )∨z∨(x∧y ∧ z) ∨ (x ∧ y ∧ z ) = ((x ∧ y ) ∧ (1 ∨ z )) ∨ z ∧ (1 ∨ (x ∧ y)) = (x ∧ y ) ∨ z. 11. x ∧ z.

13. y ∨ x .

Order Relations and Structures 15. (a) x

(b) y

x ∧ (y ∨ z )

y

x

y ∨ z

z

(x ∨ y) ∧ (x ∨ z)

x∨y x ∨ z

x

z z

17. (a)

x

x ∧ (y ∧ 1) (y ∨ y )

y∧1 (x ∨ (z ∧ y)) ∨(x ∧ (y ∧ 1))

y

x ∨ (z ∧ y)

y z

z ∧ y

(b) x

x ∨ (y ∨ z )

(y ∨ z )

(x ∨ (y ∨ z )( ∧ ( (z ∧ x) ∧ (y ∨ 0)(

y z (y ∨ 0)

y

(z ∧ x) ∧ (y ∨ 0)

z ∧ x (z ∧ x )

x z

(z ∧ x )

z ∧ x

19. ((x ∧ y) ∨ (y ∧ z)) .

3.

y

y

21. ((x ∨ y) ∧ z) . x y

x∨y

(x ∨ y) ∧ z

((x ∨ y) ∧ z)

x

1

1

0

0

x

1

0

0

1

z

z z

23. y ∨ z . y ∨ z'

y

5.

z

z'

z

x

z

1

0

0

1

0

1

0

1

1

0

0

1

0

0

0

0

Exercise Set 6 1.

y

y

x

1

0

x

0

1

y y

x

w w

303

Order Relations and Structures 7.

y

(x ∧ y ) ∨ (x ∧ y). (x ∧ y ) ∨ (x ∧ z ). z. (x ∧ y ∧ w ) ∨ (y ∧ z ∧ w ) ∨ (x ∧ z ∧ y ∧ w). (a) x ∧ y , x ∧ y, x ∧ y (b) Since ∧ is commutative and associative, we need only consider the case (w1 ∧ w2 ∧ · · · ∧ wn ∧ y) ∨ (w1 ∧ w2 ∧ · · ·∧wn ∧y ). But this is equivalent to w1 ∧w2 ∧· · ·∧wn . 27. x ∧ z , y ∧ z , y ∧ z , x ∧ z , x ∧ y. 29. (a) (x ∧ y) ∨ z . (b) A simple check of the values of f(x, y, z) will verify this.

17. 19. 21. 23. 25.

y

x

0

1

1

0

x

0

1

1

0

z z

9. (x ∧ y ) ∨ (x ∧ y). 11. z ∨ (x ∧ z). 13. (z ∧ y) ∨ (x ∧ y ) ∨ (y ∧ z). 15. (z ∧ x ) ∨ (w ∧ x ∧ y) ∨ (w ∧ x ∧ y ).

Answers to Chapter Self-Tests 1. Partial order is a generalization of less than or equal for the real numbers.

a if a ≤ b ∨ c or a ≤ b, c (1) b ∨ c if b ∨ c ≤ a or b, c ≤ a (2) a ∨ a or a (1) Thus, (a ∧ b) ∨ (a ∧ c) = b∨c (2)

10. (a) a ∧ (b ∨ c) =

2. In a partial order not every pair of elements must have a least upper bound and a greatest lower bound. 3. A model for any Boolean algebra is the power set of a set and the subset relation.

a if (b ∧ c) ≤ a or b, c ≤ a (3) b ∧ c if a ≤ (b ∧ c) or a ≤ b, c (4) a ∧ a or a (3) Thus, (a ∨ b) ∧ (a ∨ c) = b∧c (4)

(b) a ∨ (b ∧ c) =

4. The Karnaugh map of a function is an n × n array of 0’s and 1’s used to create a Boolean expression that produces the function. 5. Every Boolean function can be produced by a Boolean expression; every Boolean expression produces a Boolean function. 6. (a) R is reflexive, antisymmetric, and transitive. Hence, R is a partial order on A. (b) R is reflexive and transitive, but not antisymmetric. R is not a partial order. 7. (a)

(b)

11. 1 = 105; 3 = 35; 5 = 21; 7 = 15; 15 = 7; 21 = 5; 35 = 3; 105 = 1. 12. (a) R is reflexive, because the main diagonal of MR is all ones. R is antisymmetric, because if mij = 1, then mji = 0. MR2 = MR , so R is transitive. (b)

d

x1

x2

c x2

x3

x1

x3

x4

8. (a) Minimal: d, e; (b) Least: none;

maximal: a. greatest: a.

9. (a) Upper bounds: 12, 24, 48. (b) Lower bounds: 2. (c) LUB(B) = 12.

304

(d) GLB(B) = 2.

a x4

b

13. Since a ≤ b ≤ b ∨ d and c ≤ d ≤ b ∨ d, we have a ∨ c ≤ b ∨ d. (a ∨ c is the LUB of a and c.) Also, a ∧ c ≤ a ≤ b and a ∧ c ≤ c ≤ d since a ∧ c is the GLB of a and c. Thus, a ∧ c ≤ b ∧ d. 14. (a) (1) is not a lattice; b ∨ c does not exist. (2), (3), and (4) are lattices. (b) (1), (2), and (3) are not Boolean algebras; the number of vertices is not a power of 2. (4) is B3 .

Order Relations and Structures 15. (a)

16. (a) ((x ∧ y) ∨ (y ∧ z )) .

63 9

21

3

7 1

(b) D63 is not a Boolean algebra; there are 6 elements.

(b) (y ∧ (x ∨ z )) .

(c) x z y

17. (x ∧ y ) ∨ (y ∧ z ).

305

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Trees

From Discrete Mathematical Structures, Sixth Edition, Bernard Kolman, Robert C. Busby, Sharon Cutler Ross. Copyright © 2009 by Pearson Education, Inc. Published by Prentice Hall. All rights reserved.

307

Trees In this chapter we study a special type of relation that is exceptionally useful in a variety of biology and computer science applications and is usually represented by its digraph. These relations are essential for the construction of databases and language compilers, to name just two important computer science areas. They are called trees or sometimes rooted trees, because of the appearance of their digraphs.

Looking Back Trees were first used in 1847 by the German mathematician Karl Georg Christian von Staudt in his work on projective geometry, and in the same year by the German physicist Gustav Robert Kirchhoff in a paper on electrical networks. However, it is Arthur Cayley who first used the word tree in a mathematics paper on differential transformations. Trees are used in a wide variety of applications, such as family trees, organizational charts, sports competitions, biology, and computer file structures.

The great German composer Johann Sebastian Bach was very proud of his long family musical heritage, dating back to the late 1500s. In 1735, he prepared a genealogy of his family entitled, Ursprung der musicalisch-Bachischen Familie (“Origin of the Musical Bach Family”). A small part of this family is shown in the tree below. During their annual family reunions they performed together, and on some occasions there were more than 100 family members present.

Veit Johannes Johann

Cristoph

Heinrich

Johann Egidius Georg Christoph Johann Christoph Johann Michael

Johann Bernhard

Johann Ernst

Johann Christoph Johann Ambrosius Johann Nikolaus

Johann Sebastian

Johann Christoph

Wilhelm Friedemann Carl Philipp Emanuel Johann Christoph Friedrich

Family tree of Johann Sebastian Bach Bernard Kolman

308

Trees

1

Trees Let A be a set, and let T be a relation on A. We say that T is a tree if there is a vertex v0 in A with the property that there exists a unique path in T from v0 to every other vertex in A, but no path from v0 to v0 . We show below that the vertex v0 , described in the definition of a tree, is unique. It is often called the root of the tree T , and T is then referred to as a rooted tree. We write (T, v0 ) to denote a rooted tree T with root v0 . If (T, v0 ) is a rooted tree on the set A, an element v of A will often be referred to as a vertex in T. This terminology simplifies the discussion, since it often happens that the underlying set A of T is of no importance. To help us see the nature of trees, we will prove some simple properties satisfied by trees.

THEOREM 1

Let (T, v0 ) be a rooted tree. Then (a) There are no cycles in T . (b) v0 is the only root of T . (c) Each vertex in T , other than v0 , has in-degree one, and v0 has in-degree zero. Proof (a) Suppose that there is a cycle q in T , beginning and ending at vertex v. By definition of a tree, we know that v = v0 , and there must be a path p from v0 to v. Then q ◦ p is a path from v0 to v that is different from p, and this contradicts the definition of a tree. (b) If v0 is another root of T , there is a path p from v0 to v0 and a path q from v0 to v0 (since v0 is a root). Then q ◦ p is a cycle from v0 to v0 , and this is impossible by definition. Hence the vertex v0 is the unique root. (c) Let w1 be a vertex in T other than v0 . Then there is a unique path v0 , . . . , vk , w1 from v0 to w1 in T . This means that (vk , w1 ) ∈ T , so w1 has in-degree at least one. If the in-degree of w1 is more than one, there must be distinct vertices w2 and w3 such that (w2 , w1 ) and (w3 , w1 ) are both in T . If w2 = v0 and w3 = v0 , there are paths p2 from v0 to w2 and p3 from v0 to w3 , by definition. Then (w2 , w1 ) ◦ p2 and (w3 , w1 ) ◦ p3 are two different paths from v0 to w1 , and this contradicts the definition of a tree with root v0 . Hence, the in-degree of w1 is one. We leave it as an exercise to complete the proof if w2 = v0 or w3 = v0 and to show that v0 has in-degree zero. ■ Theorem 1 summarizes the geometric properties of a tree. With these properties in mind, we can see how the digraph of a typical tree must look. Let us first draw the root v0 . No edges enter v0 , but several may leave, and we draw these edges downward. The terminal vertices of the edges beginning at v0 will be called the level 1 vertices, while v0 will be said to be at level 0. Also, v0 is sometimes called the parent of these level 1 vertices, and the level 1 vertices are called the offspring of v0 . This is shown in Figure 1(a). Each vertex at level 1 has no other edges entering it, by part (c) of Theorem 1, but each of these vertices may have edges leaving the vertex. The edges leaving a vertex of level 1 are drawn downward and terminate at various vertices, which are said to be at level 2. Figure 1(b) shows the situation at this point. A parent-offspring relationship holds also for these levels (and at every consecutive pair of levels). For example, v3 would

309

Trees v0 Level 0

v1

v2

v0 Level 0

v1

v3 Level 1

v4 (a)

v2

v5

v3 Level 1

v6

v7

v8

v9 Level 2

(b)

Figure 1

be called the parent of the three offspring v7 , v8 , and v9 . The offspring of any one vertex are sometimes called siblings. The preceding process continues for as many levels as are required to complete the digraph. If we view the digraph upside down, we will see why these relations are called trees. The largest level number of a tree is called the height of the tree. We should note that a tree may have infinitely many levels and that any level other than level 0 may contain an infinite number of vertices. In fact, any vertex could have infinitely many offspring. However, in all our future discussions, trees will be assumed to have a finite number of vertices. Thus the trees will always have a bottom (highest-numbered) level consisting of vertices with no offspring. The vertices of the tree that have no offspring are called the leaves of the tree. The vertices of a tree that lie at any one level simply form a subset of A. Often, however, it is useful to suppose that the offspring of each vertex of the tree are linearly ordered. Thus, if a vertex v has four offspring, we may assume that they are ordered, so we may refer to them as the first, second, third, or fourth offspring of v. Whenever we draw the digraph of a tree, we automatically assume some ordering of the offspring of each vertex by arranging them from left to right. Such a tree will be called an ordered tree. Generally, ordering of offspring in a tree is not explicitly mentioned. If ordering is needed, it is usually introduced at the time when the need arises, and it often is specified by the way the digraph of the tree is drawn. The following relational properties of trees are easily verified. THEOREM 2

Let (T, v0 ) be a rooted tree on a set A. Then (a) T is irreflexive. (b) T is asymmetric. (c) If (a, b) ∈ T and (b, c) ∈ T , then (a, c) ∈ / T , for all a, b, and c in A. Proof The proof is left as an exercise.

310

■

Example 1

Let A be the set consisting of a given woman v0 and all of her female descendants. We now define the following relation T on A: If v1 and v2 are elements of A, then v1 T v2 if and only if v1 is the mother of v2 . The relation T on A is a rooted tree ◆ with root v0 .

Example 2

Let A = {v1 , v2 , v3 , v4 , v5 , v6 , v7 , v8 , v9 , v10 } and let T = {(v2 , v3 ), (v2 , v1 ), (v4 , v5 ), (v4 , v6 ), (v5 , v8 ), (v6 , v7 ), (v4 , v2 ), (v7 , v9 ), (v7 , v10 )}. Show that T is a rooted tree and identify the root.

Trees Solution v4

v6

v2

v7

v9

v1

v5

v3

v8

If n is a positive integer, we say that a tree T is an n-tree if every vertex has at most n offspring. If all vertices of T , other than the leaves, have exactly n offspring, we say that T is a complete n-tree. In particular, a 2-tree is often called a binary tree, and a complete 2-tree is often called a complete binary tree. Binary trees are extremely important, since there are efficient methods of implementing them and searching through them on computers. We will see some of these methods in Section 3, and we will also see that any tree can be represented as a binary tree. Let (T, v0 ) be a rooted tree on the set A, and let v be a vertex of T . Let B be the set consisting of v and all its descendants, that is, all vertices of T that can be reached by a path beginning at v. Observe that B ⊆ A. Let T(v) be the restriction of the relation T to B, that is, T ∩ (B × B). In other words, T(v) is the relation that results from T in the following way. Delete all vertices that are not descendants of v and all edges that do not begin and end at any such vertex. Then we have the following result.

v10

Figure 2

THEOREM 3

v5

v8

v2

v1

Since no paths begin at vertices v1 , v3 , v8 , v9 , and v10 , these vertices cannot be roots of a tree. There are no paths from vertices v6 , v7 , v2 , and v5 to vertex v4 , so we must eliminate these vertices as possible roots. Thus, if T is a rooted tree, its root must be vertex v4 . It is easy to show that there is a path from v4 to every other vertex. For example, the path v4 , v6 , v7 , v9 leads from v4 to v9 , since (v4 , v6 ), (v6 , v7 ), and (v7 , v9 ) are all in T . We draw the digraph of T , beginning with vertex v4 , and with edges shown downward. The result is shown in Figure 2. A quick inspection of this digraph shows that paths from vertex v4 to every other vertex are ◆ unique, and there are no paths from v4 to v4 . Thus T is a tree with root v4 .

Proof By the definition of T(v), we see that there is a path from v to every other vertex in T(v). If there is a vertex w in T(v) such that there are two distinct paths q and q from v to w, and if p is the path in T from v0 to v, then q ◦ p and q ◦ p would be two distinct paths in T from v0 to w. This is impossible, since T is a tree with root v0 . Thus each path from v to another vertex w in T(v) must be unique. Also, if q is a cycle at v in T(v), then q is also a cycle in T . This contradicts Theorem 1(a); ■ therefore, q cannot exist. It follows that T(v) is a tree with root v.

v6

v3

v7

v9

If (T, v0 ) is a rooted tree and v ∈ T , then T(v) is also a rooted tree with root v. We will say that T(v) is the subtree of T beginning at v.

v10

Figure 3

Example 3

Subtrees and sublattices are examples of substructures of a mathematical structure. In general, if a set A and a collection of operations and their properties form a mathematical structure, then a substructure of the same type is a subset of A with the same operations that satisfies all the properties that define this type of structure. Consider the tree T of Example 2. This tree has root v4 and is shown in Figure 2. ◆ In Figure 3 we have drawn the subtrees T(v5 ), T(v2 ), and T(v6 ) of T .

311

Trees

1 Exercises In Exercises 1 through 8, each relation R is defined on the set A. In each case determine if R is a tree and, if it is, find the root. 1. A = {a, b, c, d, e} R = {(a, d), (b, c), (c, a), (d, e)}

13. Is (T, v0 ) an n-tree? If so, for what integer n? Is (T, v0 ) a complete n-tree? If so, for what integer n? In Exercises 14 through 18, consider the rooted tree (T, v0 ) shown in Figure 5. v0

2. A = {a, b, c, d, e} R = {(a, b), (b, e), (c, d), (d, b), (c, a)} 3. A = {a, b, c, d, e, f } R = {(a, b), (c, e), (f, a), (f, c), (f, d)}

v3

v1

4. A = {1, 2, 3, 4, 5, 6} R = {(2, 1), (3, 4), (5, 2), (6, 5), (6, 3)}

v4

5. A = {1, 2, 3, 4, 5, 6} R = {(1, 1), (2, 1), (2, 3), (3, 4), (4, 5), (4, 6)}

v11

v5

v12

v2

v6

v7

v9

v10

v8

6. A = {1, 2, 3, 4, 5, 6} R = {(1, 2), (1, 3), (4, 5), (4, 6)} 7. A = {t, u, v, w, x, y, z} R = {(t, u), (u, w), (u, x), (u, v), (v, z), (v, y)}

v13

8. A = {u, v, w, x, y, z} R = {(u, x), (u, v), (w, v), (x, z), (x, y)}

v15

In Exercises 9 through 13, consider the rooted tree (T, v0 ) shown in Figure 4.

v4

v2

v5

v11

v6

v3

v7

v12

v10 (T, v0)

Figure 4 9. (a) List all level-3 vertices. (b) List all leaves. 10. (a) What are the siblings of v8 ? (b) What are the descendants of v8 ? (b) Compute the tree T(v3 ). 12. (a) What is the height of (T, v0 )? (b) What is the height of T(v3 )?

312

v8

v13

v15

11. (a) Compute the tree T(v2 ).

v16 (T, v0)

Figure 5

v0

v1

v14

v9

v14

14. (a) List all level-4 vertices. (b) List all leaves. 15. (a) What are the siblings of v2 ? (b) What are the descendants of v2 ? 16. (a) Compute the tree T(v4 ). (b) Compute the tree T(v2 ). 17. (a) What is the height of (T, v0 )? (b) What is the height of T(v4 )? 18. What is the minimal number of vertices that would need to be added to make (T, v0 ) a complete 3-tree? Draw the new tree. 19. Give all the subtrees of the rooted tree (T, v0 ) shown in Figure 2. 20. Show that the maximum number of vertices in a binary tree of height n is 2n+1 − 1. 21. Prove that the largest possible number of leaves in an ntree of height k is nk . 22. If T is a complete n-tree with exactly three levels, prove that the number of vertices of T must be 1 + kn, where 2 ≤ k ≤ n + 1. 23. Let T be a complete n-tree with m vertices of which k are nonleaves and l are leaves. (That is, m = k + l.) Prove that m = nk + 1 and l = (n − 1)k + 1. 24. Prove Theorem 2(a). 25. Prove Theorem 2(b). 26. Prove Theorem 2(c).

Trees 27. Let T be a tree. Suppose that T has r vertices and s edges. Find a formula relating r to s. 28. Draw all possible unordered trees on the set S = {a, b, c}. 29. What is the maximum height for a tree on S = {a, b, c, d, e}? Explain. 30. What is the maximum height for a complete binary tree on S = {a, b, c, d, e}? 31. Show that if (T, v0 ) is a rooted tree, then v0 has in-degree zero.

2

32. An n-tree of height k is balanced if all the leaves are at level k or k − 1. Which of the following are balanced trees? The tree shown in (b) Figure 1(b) (a) Figure 1(a) (d) Figure 5. (c) Figure 4 33. Let T be a balanced n-tree with 125 leaves. (a) What are the possible values of n? (b) What are the possible values for the height of T ? 34. Let T be a balanced n-tree of height 4 with 125 leaves. What are the possible values for n?

Labeled Trees It is sometimes useful to label the vertices or edges of a digraph to indicate that the digraph is being used for a particular purpose. This is especially true for many uses of trees in computer science and biology. We will now give a series of examples in which the sets of vertices of the trees are not important, but rather the utility of the tree is best emphasized by the labels on these vertices. Thus we will represent the vertices simply as dots and show the label of each vertex next to the dot representing that vertex. Consider the fully parenthesized, algebraic expression (3 − (2 × x)) + ((x − 2) − (3 + x)).

+ –

– +

–

3

2

Figure 6

x

x

2

3

x

We assume, in such an expression, that no operation such as −, +, ×, or ÷ can be performed until both of its arguments have been evaluated, that is, until all computations inside both the left and right arguments have been performed. Therefore, we cannot perform the central addition until we have evaluated (3 − (2 × x)) and ((x − 2) − (3 + x)). We cannot perform the central subtraction in ((x − 2) − (3 + x)) until we evaluate (x − 2) and (3 + x), and so on. It is easy to see that each such expression has a central operator, corresponding to the last computation that can be performed. Thus + is central to the main expression, − is central to (3 − (2 × x)), and so on. An important graphical representation of such an expression is as a labeled binary tree. In this tree the root is labeled with the central operator of the main expression. The two offspring of the root are labeled with the central operator of the expressions for the left and right arguments, respectively. If either argument is a constant or variable, instead of an expression, this constant or variable is used to label the corresponding offspring vertex. This process continues until the expression is exhausted. Figure 6 shows the tree for the original expression of this example. To illustrate the technique further, we have shown in Figure 7 the tree corresponding to the full parenthesized expression ((3 × (1 − x)) ÷ ((4 + (7 − (y + 2))) × (7 + (x ÷ y)))). Our next example of a labeled tree is important for the computer implementation of a tree data structure. We start with an n-tree (T, v0 ). Each vertex in T has at most n offspring. We imagine that each vertex potentially has exactly n offspring, which would be ordered from 1 to n, but that some of the offspring in the sequence may be missing. The remaining offspring are labeled with the position that they occupy in the hypothetical sequence. Thus the offspring of any vertex are labeled with distinct numbers from the set {1, 2, . . . , n}.

313

Trees

–

3

+

+

x

1

–

4

7 +

7

y

x 2

y

Figure 7

Such a labeled digraph is sometimes called positional, and we will also use this term. Note that positional trees are also ordered trees. When drawing the digraphs of a positional tree, we will imagine that the n offspring positions for each vertex are arranged symmetrically below the vertex, and we place in its appropriate position each offspring that actually occurs. Figure 8 shows the digraph of a positional 3-tree, with all actually occurring positions labeled. If offspring 1 of any vertex v actually exists, the edge from v to that offspring is drawn sloping to the left. Offspring 2 of any vertex v is drawn vertically downward from v, whenever it occurs. Similarly, offspring labeled 3 will be drawn to the right. Naturally, the root is not labeled, since it is not an offspring.

L

2

1

1

2

2

3

1

3

L

3

L

3

3

1

Figure 8

R

2

R

L

R

3

L

L

R

R

R

R

R

L

L

Figure 9

The positional binary tree is of special importance. In this case, for obvious reasons, the positions for potential offspring are often labeled left and right, instead of 1 and 2. Figure 9 shows the digraph of a positional binary tree, with offspring labeled L for left and R for right. Labeled trees may have several sets of labels, all in force simultaneously. We will usually omit the left-right labels on a positional binary tree in order to emphasize other useful labels. The positions of the offspring will then be indicated by the direction of the edges, as we have drawn them in Figure 9.

314

Trees

Binary Positional Trees as Data Structures A cell contains two items. One is data of some sort and the other is a pointer to the next cell, that is, an address where the next cell is located. A collection of such cells, linked together by their pointers, is called a linked list. We need here an extended version of this concept, called a doubly linked list, in which each cell contains two pointers and a data item. We use the pictorial to represent these new cells. The center space represents symbol data storage and the two pointers, called the left pointer and the right pointer, are for represented as before by dots and arrows. Once again we use the symbol a pointer signifying no additional data. Sometimes a doubly linked list is arranged so that each cell points to both the next cell and the previous cell. This is useful if we want to search through a set of data items in either direction. Our use of doubly linked lists here is very different. We will use them to represent binary positional labeled trees. Each cell will correspond to a vertex, and the data part can contain a label for the vertex or a pointer to such a label. The left and right pointers will direct us to the left and right offspring vertices, if they exist. If either offspring fails to exist, the corresponding pointer will be . We implement this representation by using three arrays: LEFT holds pointers to the left offspring, RIGHT holds the pointers to the right offspring, and DATA holds information or labels related to each vertex, or pointers to such information. The value 0, used as a pointer, will signify that the corresponding offspring does not exist. To the linked list and the arrays we add a starting entry that points to the root of the tree. Example 1

We consider again the positional binary tree shown in Figure 6. In Figure 10(a), we represent this tree as a doubly linked list, in symbolic form. In Figure 10(b), we show the implementation of this list as a sequence of three arrays. The first row of these arrays is just a starting point whose left pointer points to the root of the tree.

Start

+

–

+

x

3

x

–

3

–

x

2 (a)

2

INDEX LEFT 2 1 3 2 3 4 4 0 5 6 6 0 7 0 8 9 9 10 10 0 11 0 12 13 13 0 14 0

DATA RIGHT 0 8 + – 5 3 0 7 0 2 x 0 – 12 – 11 x 0 0 2 + 14 3 0 x 0 (b)

Figure 10

315

Trees As an example of how to interpret the three arrays, consider the fifth entry in the array DATA, which is ×. The fifth entry in LEFT is 6, which means that the left offspring of × is the sixth entry in DATA, or 2. Similarly, the fifth entry in RIGHT ◆ is 7, so the right offspring of × is the seventh entry in DATA, or x. Example 2

Now consider the tree of Figure 7. We represent this tree in Figure 11(a) as a doubly linked list. As before, Figure 11(b) shows the implementation of this linked list in three arrays. Again, the first entry is a starting point whose left pointer points to the root of the tree. We have listed the vertices in a somewhat unnatural order to show that, if the pointers are correctly determined, any ordering of vertices can ◆ be used. INDEX LEFT 1 7 2 0 3 2 4 0 5 4 6 0 7 3 8 0 9 8 10 0 11 10

Start

←

+

y

7

x

+

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+

2

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x

4

7

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1

(a)

y

DATA RIGHT 0 3 0 5 1 0 6 – x 0 15 4 0 + 11 7 0 13 –

12 13 14

0 12 0

y + 2

0 14 0

15 16 17

9 0 16

7 +

17 0 19

18 19 20

0 18 0

x y

0 20 0

(b)

Figure 11

Huffman Code Trees

0

1 1

0

E

0

A

1 0

S R

1 C

Figure 12

Example 3

316

The simple ASCII code for the letters of the English alphabet represents each letter by a string of 0’s and 1’s of length 7. A Huffman code also uses strings of 0’s and 1’s, but the strings are of variable length. The more frequently used letters are assigned shorter strings. In this way messages may be significantly shorter than with ASCII code. A labeled positional binary tree is an efficient way to represent a Huffman code. The string assigned to a letter gives a path from the root of the tree to the leaf labeled with that letter; a 0 indicates that we should go to the left, a 1 that we should go to the right. To decode a message, we follow the path indicated. When a leaf is reached, the label is recorded, we return to the root and continue with the remainder of the string. We omit the discussion of how to create a Huffman code. Use the Huffman code tree in Figure 12 to decode the string 0101100.

Trees Solution

We begin at the root and move to the left (using the first 0). This leaf has label E. We return to the root and use the string 101100. We go to the right, then to the left and record A. Repeating the process with 1100 produces S and finally another E. ◆ The string 0101100 represents EASE. Example 4

One disadvantage of a Huffman code is the potential for errors in transmission. If an error was made in transmitting string 0101100 and string 0101110 is received, ◆ then the word read is EAR, not EASE.

2 Exercises In Exercises 1 through 10, construct the tree of the algebraic expression. 1. (7 + (6 − 2)) − (x − (y − 4)) 2. (x + (y − (x + y))) × ((3 ÷ (2 × 7)) × 4) 3. 3 − (x + (6 × (4 ÷ (2 − 3)))) 4. (((2 × 7) + x) ÷ y) ÷ (3 − 11) 5. ((2 + x) − (2 × x)) − (x − 2) 6. (11 − (11 × (11 + 11))) + (11 + (11 × 11)) 7. (3 − (2 − (11 − (9 − 4)))) ÷ (2 + (3 + (4 + 7))) 8. (x ÷ y) ÷ ((x × 3) − (z ÷ 4)) 9. ((2 × x) + (3 − (4 × x))) + (x − (3 × 11)) 10. ((1 + 1) + (1 − 2)) ÷ ((2 − x) + 1) 11. The annual NCAA Basketball Tournament begins with 64 teams. One loss eliminates a team. Suppose the results of the tournament are represented in a binary tree whose leaves are labeled with the original teams and the interior vertices are labeled with the winner of the game between the children of the vertex. What is the height of the tree created? 12. Make a “family” tree that shows the descendants of one of your great grandmothers. 13. Construct the digraphs of all distinct binary positional trees having three or fewer edges and height 2. 14. How many distinct binary positional trees are there with height 2? 15. How many distinct positional 3-trees are there with height 2? 16. Construct the digraphs of all distinct positional 3-trees having two or fewer edges. 17. The following is the doubly linked list representation of a binary positional labeled tree. Construct the digraph of this tree with each vertex labeled as indicated.

INDEX

LEFT

D AT A

RIGHT

1 2 3 4 5 6 7 8 9

8 5 9 2 0 0 0 6 0

D E C F B G A H

0 7 0 3 0 4 0 0 0

18. The following is the doubly linked list representation of a binary positional labeled tree. Construct the digraph of this tree with each vertex labeled as indicated. INDEX

LEFT

D AT A

RIGHT

1 2 3 4 5 6 7 8 9 10

9 10 0 8 3 0 0 0 6 0

M Q T V X K D G C

0 7 0 0 4 2 0 0 5 0

19. The following is the doubly linked list representation of a binary positional labeled tree. Construct the digraph of this tree with each vertex labeled as indicated. INDEX

LEFT

DATA

RIGHT

1 2 3 4 5 6 7 8 9 10 11 12 13

7 0 2 0 0 10 3 4 0 0 0 0 12

c a t s a n d f r o g s

0 9 6 0 0 11 8 13 0 0 0 0 5

317

Trees 20. Give arrays LEFT, DATA, and RIGHT describing the tree given in Figure 13 as a doubly linked list. 1

2

3 5

4

6 10 9

7

8 12 11

Figure 13

3

In Exercises 21 through 24, give arrays LEFT, DATA, and RIGHT describing the tree created in the indicated exercise. 21. Exercise 1 22. Exercise 4 23. Exercise 5 24. Exercise 8 25. Use the Huffman code tree in Figure 12 to decode each of the following messages. (b) 1100101110 (a) 1111101110 (d) 1101111101110 (c) 11101011110 26. Use the Huffman code tree in Figure 12 to find the string that represents the given word. (b) SEA (a) CARE (d) CASE (c) ACE 27. Construct the labeled tree that represents this Huffman code. A 000 B 01 C 001 D 1100 E 1101

Tree Searching There are many occasions when it is useful to consider each vertex of a tree T exactly once in some specific order. As each successive vertex is encountered, we may wish to take some action or perform some computation appropriate to the application being represented by the tree. For example, if the tree T is labeled, the label on each vertex may be displayed. If T is the tree of an algebraic expression, then at each vertex we may want to perform the computation indicated by the operator that labels that vertex. Performing appropriate tasks at a vertex will be called visiting the vertex. This is a convenient, nonspecific term that allows us to write algorithms without giving the details of what constitutes a “visit” in each particular case. The process of visiting each vertex of a tree in some specific order will be called searching the tree or performing a tree search. In some texts, this process is called walking or traversing the tree. Let us consider tree searches on binary positional trees. Recall that in a binary positional tree each vertex has two potential offspring. We denote these potential offspring by vL (the left offspring) and vR (the right offspring), and either or both may be missing. If a binary tree T is not positional, it may always be labeled so that it becomes positional. Let T be a binary positional tree with root v. Then, if vL exists, the subtree T(vL ) (see Section 1) will be called the left subtree of T , and if vR exists, the subtree T(vR ) will be called the right subtree of T . Note that T(vL ), if it exists, is a positional binary tree with root vL , and similarly T(vR ) is a positional binary tree with root vR . This notation allows us to specify searching algorithms in a natural and powerful recursive form. Recall that recursive algorithms are those that refer to themselves. We first describe a method of searching called a preorder search. For the moment, we leave the details of visiting a vertex of a tree unspecified. Consider the following algorithm for searching a positional binary tree T with root v.

318

Trees Algorithm PREORDER

Step 1 Visit v. Step 2 If vL exists, then apply this algorithm to (T(vL ), vL ). Step 3 If vR exists, then apply this algorithm to (T(vR ), vR ).

●

Informally, we see that a preorder search of a tree consists of the following three steps: 1. Visit the root. 2. Search the left subtree if it exists. 3. Search the right subtree if it exists. Example 1

Let T be the labeled, positional binary tree whose digraph is shown in Figure 14(a). The root of this tree is the vertex labeled A. Suppose that, for any vertex v of T , visiting v prints out the label of v. Let us now apply the preorder search algorithm to this tree. Note first that if a tree consists only of one vertex, its root, then a search of this tree simply prints out the label of the root. In Figure 14(b), we have placed boxes around the subtrees of T and numbered these subtrees (in the corner of the boxes) for convenient reference. A B

H E

C D

F

I

K

G

J

L

(a) A

H

B E

C

D 3 2

K

G

F 5 4

I

L

J

6

9 8

1

11 10

7 (b)

Figure 14

According to PREORDER, applied to T , we will visit the root and print A, then search subtree 1, and then subtree 7. Applying PREORDER to subtree 1 results in visiting the root of subtree 1 and printing B, then searching subtree 2, and finally searching subtree 4. The search of subtree 2 first prints the symbol C

319

Trees and then searches subtree 3. Subtree 3 has just one vertex, and so, as previously mentioned, a search of this tree yields just the symbol D. Up to this point, the search has yielded the string ABCD. Note that we have had to interrupt the search of each tree (except subtree 3, which is a leaf of T ) in order to apply the search procedure to a subtree. Thus we cannot finish the search of T by searching subtree 7 until we apply the search procedure to subtrees 2 and 4. We could not complete the search of subtree 2 until we search subtree 3, and so on. The bookkeeping brought about by these interruptions produces the labels in the desired order, and recursion is a simple way to specify this bookkeeping. Returning to the search, we have completed searching subtree 2, and we now must search subtree 4, since this is the right subtree of tree 1. Thus we print E and search 5 and 6 in order. These searches produce F and G. The search of subtree 1 is now complete, and we go to subtree 7. Applying the same procedure, we can see that the search of subtree 7 will ultimately produce the string HIJKL. The result, ◆ then, of the complete search of T is to print the string ABCDEFGHIJKL. Example 2

Consider the completely parenthesized expression (a−b)×(c+(d÷e)). Figure 15(a) shows the digraph of the labeled, positional binary tree representation of this expression. We apply the search procedure PREORDER to this tree, as we did to the tree in Example 1. Figure 15(b) shows the various subtrees encountered in the search. Proceeding as in Example 1 and supposing again that visiting v simply prints out the label of v, we see that the string × − a b + c ÷ d e is the result of the search. This is the prefix or Polish form of the given algebraic expression. Once again, the numbering of the boxes in Figure 15(b) shows the order in which the algorithm ◆ PREORDER is applied to subtrees.

+

– +

–

a 2 a

b 3

c 5

1

b c

e

d 7 d

(a)

8 6

e

4 (b)

Figure 15

The Polish form of an algebraic expression is interesting because it represents the expression unambiguously, without the need for parentheses. To evaluate an expression in Polish form, proceed as follows. Move from left to right until we find a string of the form Fxy, where F is the symbol for a binary operation (+, −, ×, and so on) and x and y are numbers. Evaluate x F y and substitute the answer for the string Fxy. Continue this procedure until only one number remains. For example, in the preceding expression, suppose that a = 6, b = 4, c = 5, d = 2, and e = 2. Then we are to evaluate × − 6 4 + 5 ÷ 2 2. This is done in the following sequence of steps.

320

Trees 1. × − 6 4 + 5 ÷ 2 2 2. × 2 + 5 ÷ 2 2 since the first string of the correct type is − 6 4 and 6−4=2 3. × 2 + 5 1 replacing ÷ 2 2 by 2 ÷ 2 or 1 4. × 2 6 replacing + 5 1 by 5 + 1 or 6 5. 12 replacing × 2 6 by 2 × 6 This example is one of the primary reasons for calling this type of search the preorder search, because here the operation symbol precedes the arguments. Consider now the following informal descriptions of two other procedures for searching a positional binary tree T with root v. Algorithm INORDER

Step 1 Search the left subtree (T(vL ), vL ), if it exists. Step 2 Visit the root, v. Step 3 Search the right subtree (T(vR ), vR ), if it exists.

●

Algorithm POSTORDER

Step 1 Search the left subtree (T(vL ), vL ), if it exists. Step 2 Search the right subtree (T(vR ), vR ), if it exists. Step 3 Visit the root, v.

●

As indicated by the naming of the algorithms, these searches are called, respectively, the inorder and postorder searches. The names indicate when the root of the (sub)tree is visited relative to when the left and right subtrees are searched. Informally, in a preorder search, the order is root, left, right; for an inorder search, it is left, root, right; and for a postorder search, it is left, right, root. Example 3

Consider the tree of Figure 14(b) and apply the algorithm INORDER to search it. First we must search subtree 1. This requires us to first search subtree 2, and this in turn requires us to search subtree 3. As before, a search of a tree with only one vertex simply prints the label of the vertex. Thus D is the first symbol printed. The search of subtree 2 continues by printing C and then stops, since there is no right subtree at C. We then visit the root of subtree 1 and print B, and then proceed to the search of subtree 4, which yields F , E, and G, in that order. We then visit the root of T and print A and proceed to search subtree 7. The reader may complete the analysis of the search of subtree 7 to show that the subtree yields the string IJHKL. Thus the complete search yields the string DCBFEGAIJHKL. Suppose now that we apply algorithm POSTORDER to search the same tree. Again, the search of a tree with just one vertex will yield the label of that vertex. In general, we must search both the left and the right subtrees of a tree with root v before we print out the label at v. Referring again to Figure 14(b), we see that both subtree 1 and subtree 7 must be searched before A is printed. Subtrees 2 and 4 must be searched before B is printed, and so on. The search of subtree 2 requires us to search subtree 3, and D is the first symbol printed. The search of subtree 2 continues by printing C. We now search subtree 4, yielding F , G, and E. We next visit the root of subtree 1 and print B. Then we proceed with the search of subtree 7 and print the symbols J, I, L, K, and H. Finally, we visit the root of T and print A. Thus we print out the string ◆ DCFGEBJILKHA.

321

Trees Example 4

Let us now apply the inorder and postorder searches to the algebraic expression tree of Example 2 [see Figure 15(a)]. The use of INORDER produces the string a − b × c + d ÷ e. Notice that this is exactly the expression that we began with in Example 2, with all parentheses removed. Since the algebraic symbols lie between their arguments, this is often called the infix notation, and this explains the name INORDER. The preceding expression is ambiguous without parentheses. It could have come from the expression a − (b × ((c + d) ÷ e)), which would have produced a different tree. Thus the tree cannot be recovered from the output of search procedure INORDER, while it can be shown that the tree is recoverable from the Polish form produced by PREORDER. For this reason, Polish notation is often better for computer applications, although infix form is more familiar to human beings. The use of search procedure POSTORDER on this tree produces the string ab − cde ÷ + ×. This is the postfix or reverse Polish form of the expression. It is evaluated in a manner similar to that used for Polish form, except that the operator symbol is after its arguments rather than before them. If a = 2, b = 1, c = 3, d = 4, and e = 2, the preceding expression is evaluated in the following sequence of steps. 1. 2. 3. 4. 5.

21−342÷+× 1342÷+× replacing 2 1 − by 2 − 1 or 1 132+× replacing 4 2 ÷ by 4 ÷ 2 or 2 15× replacing 3 2 + by 3 + 2 or 5 5 replacing 1 5 × by 1 × 5 or 5

◆

Reverse Polish form is also parentheses free, and from it we can recover the tree of the expression. It is used even more frequently than the Polish form and is the method of evaluating expressions in some calculators.

Searching General Trees Until now, we have only shown how to search binary positional trees. We now show that any ordered tree T (see Section 1) may be represented as a binary positional tree that, although different from T , captures all the structure of T and can be used to re-create T . With the binary positional description of the tree, we may apply the computer implementation and search methods previously developed. Since any tree may be ordered, we can use this technique on any (finite) tree. Let T be any ordered tree and let A be the set of vertices of T . Define a binary positional tree B(T ) on the set of vertices A, as follows. If v ∈ A, then the left offspring vL of v in B(T ) is the first offspring of v in T (in the given order of siblings in T ), if it exists. The right offspring vR of v in B(T ) is the next sibling of v in T (in the given order of siblings in T ), if it exists. Example 5

322

Figure 16(a) shows the digraph of a labeled tree T . We assume that each set of siblings is ordered from left to right, as they are drawn. Thus the offspring of vertex 1, that is, vertices 2, 3, and 4, are ordered with vertex 2 first, 3 second, and 4 third. Similarly, the first offspring of vertex 5 is vertex 11, the second is vertex 12, and the third is vertex 13. In Figure 16(b), we show the digraph of the corresponding binary positional tree, B(T ). To obtain Figure 16(b), we simply draw a left edge from each vertex v to its first offspring (if v has offspring). Then we draw a right edge from each vertex v to its next sibling (in the order given), if v has a next sibling. Thus the

Trees 1

1 2

2

4

3

3

5 8 11

5 6 7 8 9

9

10 12

11

12

4

6

7 10

13

13 T (a)

B(T) (b)

Figure 16

left edge from vertex 2, in Figure 16(b), goes to vertex 5, because vertex 5 is the first offspring of vertex 2 in the tree T . Also, the right edge from vertex 2, in Figure 16(b), goes to vertex 3, since vertex 3 is the next sibling in line (among all offspring of vertex 1). A doubly-linked-list representation of B(T ) is sometimes ◆ simply referred to as a linked-list representation of T. Example 6

Figure 17(a) shows the digraph of another labeled tree, with siblings ordered from left to right, as indicated. Figure 17(b) shows the digraph of the corresponding tree B(T ), and Figure 17(c) gives an array representation of B(T ). As mentioned, the ◆ data in Figure 17(c) would be called a linked-list representation of T . a

a b

b

c

d

e

c f

e

f

g

d g

T

B(T)

(a)

(b)

INDEX

LEFT

DATA

RIGHT

1 2 3 4 5 6 7 8

2 3 6 0 0 0 0 0

a b c d e f g

0 0 4 5 0 7 8 0

(c)

Figure 17

Pseudocode Versions The three search algorithms in this section have straightforward pseudocode versions, which we present here. In each, we assume that the subroutine VISIT has been previously defined. SUBROUTINE

PREORDER(T,v) 1. CALL VISIT (v) 2. IF (vL exists) THEN a. CALL PREORDER(T(vL ),vL ) 3. IF (vR exists) THEN a. CALL PREORDER(T(vR ),vR ) 4. RETURN END OF SUBROUTINE PREORDER

323

Trees SUBROUTINE

INORDER(T,v) 1. IF (vL exists) THEN a. CALL INORDER(T(vL ),vL ) 2. CALL VISIT(v) 3. IF (vR exists) THEN a. CALL INORDER(T(vR ),vR ) 4. RETURN END OF SUBROUTINE INORDER

SUBROUTINE

POSTORDER(T,v) 1. IF (vL exists) THEN a. CALL POSTORDER(T(vL ),vL ) 2. IF (vR exists) THEN a. CALL POSTORDER(T(vR ),vR ) 3. CALL VISIT(v) 4. RETURN END OF SUBROUTINE POSTORDER

3 Exercises In Exercises 1 through 5 (Figures 18 through 22), the digraphs of labeled, positional binary trees are shown. In each case we suppose that visiting a node results in printing out the label of that node. For each exercise, show the result of performing a preorder search of the tree whose digraph is shown. 1.

2.

x y

←

z

–

s u

2

S

M

A

R E

L G

M D Q Z

K

C

J

W

F

1 3

Figure 19

v

T

2

+

t

5.

N

T

Figure 22

Figure 18 3.

In Exercises 6 through 15, visiting a node means printing out the label of the node.

a

6. Show the result of performing an inorder search of the tree shown in Figure 18.

d

b c

e

i k

g

h

7. Show the result of performing an inorder search of the tree shown in Figure 19. f

j

8. Show the result of performing an inorder search of the tree shown in Figure 20.

Figure 20 4. 2 4 6

Figure 21

324

9. Show the result of performing an inorder search of the tree shown in Figure 21.

1

10. Show the result of performing a postorder search of the tree shown in Figure 22.

3

11. Show the result of performing a postorder search of the tree shown in Figure 18.

5 7

12. Show the result of performing a postorder search of the tree shown in Figure 19.

Trees 13. Show the result of performing a postorder search of the tree shown in Figure 20. 14. Show the result of performing a postorder search of the tree shown in Figure 21.

29. (a) Every binary tree whose preorder search produces the string JBACDIHEGF must have 10 vertices. What else do the trees have in common? (b) Every binary tree whose postorder search produces the string SEARCHING must have 9 vertices. What else do the trees have in common?

15. Show the result of performing an inorder search of the tree shown in Figure 22. For Exercises 16 through 18, consider the tree whose digraph is shown in Figure 23 and the accompanying list of words. Suppose that visiting a node means printing out the word corresponding to the number that labels the node. 1 2

3

4

5 7 8

6 10

9 11

12

1. 2. 3. 4. 5. 6.

ONE COW SEE NEVER PURPLE NEVER

7. 8. 9. 10. 11. 12.

I A I I SAW HOPE

30. Show that any element of the string ABCDEF may be the root of a binary tree whose inorder search produces this string. In Exercises 31 and 32 (Figures 24 and 25), draw the digraph of the binary positional tree B(T ) that corresponds to the tree shown. Label the vertices of B(T ) to show their correspondence to the vertices of T . 31.

s

t v

Figure 23

u

w

x

y

z

G

O

Figure 24 16. Give the sentence that results from doing a preorder search of the tree. 17. Give the sentence that results from doing an inorder search of the tree.

32.

18. Give the sentence that results from doing a postorder search of the tree.

A

L

In Exercises 19 and 20, evaluate the expression, which is given in Polish, or prefix, notation.

R

19. × − + 3 4 − 7 2 ÷ 12 × 3 − 6 4

Figure 25

20. ÷ − × 3 x × 4 y + 15 × 2 − 6 y, where x is 2 and y is 3.

I T H M

S

21. 4 3 2 ÷ − 5 × 4 2 × 5 × 3 ÷ ÷

In Exercises 33 and 34, we give, in array form, the doublylinked-list representation of a labeled tree T (not binary). Draw the digraph of both the labeled binary tree B(T ) actually stored in the arrays and the labeled tree T of which B(T ) is the binary representation.

22. 3 7 × 4 − 9 × 6 5 × 2 + ÷

33.

In Exercises 21 through 24, evaluate the expression, which is given in reverse Polish, or postfix, notation.

23. x 2 − 3 + 2 3 y + − w 3 − × ÷, where x is 7, y is 2, and w is 1. 24. 7 x × y − 8 x × w + × where x is 7, y is 2, and w is 1. 25. Draw a binary tree whose preorder search produces the string JBACDIHEGF. 26. Draw a binary tree whose preorder search produces the string CATSANDDOGS. 27. Draw a binary tree whose postorder search produces the string SEARCHING. 28. Draw a binary tree whose postorder search produces the string TREEHOUSE.

INDEX 1 2 3 4 5 6 7 8 9 10 11 12

LEFT 2 3 4 6 8 0 0 0 0 0 0 0

DATA

H a b c d e f g h i j k

RIGHT 0 0 5 7 0 10 0 11 0 9 12 0

325

Trees 34.

INDEX 1 2 3 4 5 6 7 8 9 10 11 12

LEFT 12 0 0 2 0 0 0 6 5 4 0 11

DATA

H T W O B R A N C H E S

RIGHT 0 0 0 3 0 0 0 7 8 9 10 0

In Exercises 35 and 36 (Figures 26 and 27), consider the digraph of the labeled binary positional tree shown. If this tree is the binary form B(T ) of some tree T , draw the digraph of the labeled tree T . 35.

a b d

c e

j g

36.

D E1 N1 S2

S1 D

A

C

N2

E2 T

Figure 27 37. Finding information stored in a binary tree is a common task. We can improve the efficiency of such a search by having a “balanced” tree. An AVL tree, (T, v0 ), is a binary tree where for each v ∈ T , the height of T(vL ) and the height of T(vR ) differ by at most one. For the given height, draw an AVL tree using the smallest possible number of vertices. (b) ht = 1 (a) ht = 0 (d) ht = 3 (c) ht = 2 38. Is an AVL-tree a balanced 2-tree as defined in Section 1, Exercise 32? 39. Write a recurrence relation for AVL n , the minimum number of vertices needed for an AVL tree of height n. Justify your recurrence relation by giving a procedure for forming an AVL tree of height k with a minimum number of vertices from smaller AVL trees with minimal numbers of vertices.

f h

i

Figure 26

4

Undirected Trees An undirected tree is simply the symmetric closure of a tree; that is, it is the relation that results from a tree when all the edges are made bidirectional. As is the custom with symmetric relations, we represent an undirected tree by its graph, rather than by its digraph. The graph of an undirected tree T will have a single line without arrows connecting vertices a and b whenever (a, b) and (b, a) belong to T . The set {a, b}, where (a, b) and (b, a) are in T , is called an undirected edge of T . In this case, the vertices a and b are called adjacent vertices. Thus each undirected edge {a, b} corresponds to two ordinary edges, (a, b) and (b, a). The lines in the graph of an undirected tree T correspond to the undirected edges in T .

Example 1

Figure 28(a) shows the graph of an undirected tree T . In Figures 28(b) and (c), we show digraphs of ordinary trees T1 and T2 , respectively, which have T as symmetric closure. This shows that an undirected tree will, in general, correspond to many directed trees. Labels are included to show the correspondence of underlying vertices in the three relations. Note that the graph of T in Figure 28(a) has six lines ◆ (undirected edges), although the relation T contains 12 pairs. We want to present some useful alternative definitions of an undirected tree, and to do so we must make a few remarks about symmetric relations.

326

Trees b b

d

a c

c d

e f

e

f

d

g

a a

g (a)

c b e

f

g (b)

(c)

Figure 28

Let R be a symmetric relation, and let p : v1 , v2 , . . . , vn be a path in R. We will say that p is simple if no two edges of p correspond to the same undirected edge. If, in addition, v1 equals vn (so that p is a cycle), we will call p a simple cycle. Example 2 a

f

e

b

c

d

Figure 29

THEOREM 1

v4

v3

v2

v5

v1 v6

v7

vi

vn

(a) v0 p

q

p–1

a

a b (c)

Figure 30

Figure 29 shows the graph of a symmetric relation R. The path a, b, c, e, d is simple, but the path f , e, d, c, d, a is not simple, since d, c and c, d correspond to the same undirected edge. Also, f , e, a, d, b, a, f and d, a, b, d are simple cycles, but f , e, d, c, e, f is not a simple cycle, since f , e and e, f correspond to the same ◆ undirected edge. We will say that a symmetric relation R is acyclic if it contains no simple cycles. A symmetric relation R is connected if there is a path in R from any vertex to any other vertex. The following theorem provides a useful statement equivalent to the previous definition of an undirected tree. Let R be a symmetric relation on a set A. Then the following statements are equivalent. (a) R is an undirected tree. (b) R is connected and acyclic.

Proof We will prove that part (a) implies part (b), and we will omit the proof that part (b) implies part (a). We suppose that R is an undirected tree, which means that R is the symmetric closure of some tree T on A. Note first that if (a, b) ∈ R, we must vk have either (a, b) ∈ T or (b, a) ∈ T . In geometric terms, this means that every vj undirected edge in the graph of R appears in the digraph of T , directed one way or the other. We will show by contradiction that R has no simple cycles. Suppose that R has (b) a simple cycle p : v1 , v2 , . . . , vn , v1 . For each edge (vi , vj ) in p, choose whichever v0 pair (vi , vj ) or (vj , vi ) is in T . The result is a closed figure with edges in T , where each edge may be pointing in either direction. Now there are three possibilities. Either all arrows point clockwise, as in Figure 30(a), all point counterclockwise, q or some pair must be as in Figure 30(b). Figure 30(b) is impossible, since in a tree T every vertex (except the root) has in-degree 1 (see Theorem 1 of Section 1). But either of the other two cases would mean that T contains a cycle, which is also impossible. Thus the existence of the cycle p in R leads to a contradiction and so b is impossible. We must also show that R is connected. Let v0 be the root of the tree T . Then, (d) if a and b are any vertices in A, there must be paths p from v0 to a and q from v0 to b, as shown in Figure 30(c). Now all paths in T are reversible in R, so the path

327

Trees q ◦ p−1 , shown in Figure 30(d), connects a with b in R, where p−1 is the reverse path of p. Since a and b are arbitrary, R is connected, and part (b) is proved. ■ There are other useful characterizations of undirected trees. We state two of these without proof in the following theorem. THEOREM 2

Let R be a symmetric relation on a set A. Then R is an undirected tree if and only if either of the following statements is true. (a) R is acyclic, and if any undirected edge is added to R, the new relation will not be acyclic. (b) R is connected, and if any undirected edge is removed from R, the new ■ relation will not be connected. Note that Theorems 1 and 2 tell us that an undirected tree must have exactly the “right” number of edges; one too many and a cycle will be created; one too few and the tree will become disconnected. The following theorem will be useful in finding certain types of trees.

THEOREM 3

A tree with n vertices has n − 1 edges. Proof Because a tree is connected, there must be at least n − 1 edges to connect the n vertices. Suppose that there are more than n − 1 edges. Then either the root has in-degree 1 or some other vertex has in-degree at least 2. But by Theorem 1, ■ Section 1, this is impossible. Thus there are exactly n − 1 edges.

Spanning Trees of Connected Relations If R is a symmetric, connected relation on a set A, we say that a tree T on A is a spanning tree for R if T is a tree with exactly the same vertices as R and which can be obtained from R by deleting some edges of R. Example 3

The symmetric relation R whose graph is shown in Figure 31(a) has the tree T , whose digraph is shown in Figure 31(b), as a spanning tree. Also, the tree T , whose digraph is shown in Figure 31(c), is a spanning tree for R. Since R, T , and T are all relations on the same set A, we have labeled the vertices to show the correspondence of elements. As this example illustrates, spanning trees ◆ are not unique. a b

c e

d f R (a)

a

a b

c e

d

b

c e

d f

f

T (b)

a b

c e

d f

T (c)

(d)

Figure 31

An undirected spanning tree for a symmetric connected relation R is useful in some applications. This is just the symmetric closure of a spanning tree. Figure 31(d) shows an undirected spanning tree for R that is derived from the

328

Trees spanning tree of Figure 31(c). If R is a complicated relation that is symmetric and connected, it might be difficult to devise a scheme for searching R, that is, for visiting each of its vertices once in some systematic manner. If R is reduced to a spanning tree, the searching algorithms discussed in Section 3 can be used. Theorem 2(b) suggests an algorithm for finding an undirected spanning tree for a relation R. Simply remove undirected edges from R until we reach a point where removal of one more undirected edge will result in a relation that is not connected. The result will be an undirected spanning tree. Example 4

In Figure 32(a), we repeat the graph of Figure 31(a). We then show the result of successive removal of undirected edges, culminating in Figure 32(f), the undirected ◆ spanning tree, which agrees with Figure 31(d). a

a b

c d

b

c d

e

a b

e

c e

d

f

f

f

(a)

(b)

(c)

a

a

a

b

c d f (d)

e

b

c d f (e)

e

b

c d

e

f (f)

Figure 32

This algorithm is fine for small relations whose graphs are easily drawn. For large relations, perhaps stored in a computer, it is inefficient because at each stage we must check for connectedness, and this in itself requires a complicated algorithm. We now introduce a more efficient method, which also yields a spanning tree, rather than an undirected spanning tree. Let R be a relation on a set A, and let a, b ∈ A. Let A0 = A − {a, b}, and A = A0 ∪ {a }, where a is some new element not in A. Define a relation R on A as follows. Suppose u, v ∈ A , u = a , v = a . Let (a , u) ∈ R if and only if (a, u) ∈ R or (b, u) ∈ R. Let (u, a ) ∈ R if and only if (u, a) ∈ R or (u, b) ∈ R. Finally, let (u, v) ∈ R if and only if (u, v) ∈ R. We say that R is a result of merging the vertices a and b. Imagine, in the digraph of R, that the vertices are pins, and the edges are elastic bands that can be shrunk to zero length. Now physically move pins a and b together, shrinking the edge between them, if there is one, to zero length. The resulting digraph is the digraph of R . If R is symmetric, we may perform this operation on the graph of R. The result is the graph of the symmetric relation R . Example 5

Figure 33(a) shows the graph of a symmetric relation R. In Figure 33(b), we show the result of merging vertices v0 and v1 into a new vertex v0 . In Figure 33(c), we show the result of merging vertices v0 and v2 of the relation whose graph is shown in Figure 33(b) into a new vertex v0 . Notice in Figure 33(c) that the undirected edges that were previously present between v0 and v5 and between v2 and v5 have ◆ been combined into one undirected edge.

329

Trees v0

v0 v1

v 0

v2

v2

v5 v4 v3

v5

v5 v4 v3

v6

v4 v6

v3

(b)

(a)

v6

(c)

Figure 33

The algebraic form of this merging process is also very important. Let us restrict our attention to symmetric relations and their graphs. If R is a relation on A, we will temporarily refer to elements of A as vertices of R. This will facilitate the discussion. Suppose now that vertices a and b of a relation R are merged into a new vertex a that replaces a and b to obtain the relation R . To determine the matrix of R , we proceed as follows. Step 1 Let row i represent vertex a and row j represent vertex b. Replace row i by the join of rows i and j. The join of two n-tuples of 0’s and 1’s has a 1 in some position exactly when either of those two n-tuples has a 1 in that position. Step 2 Replace column i by the join of columns i and j. Step 3 Restore the main diagonal to its original values in R. Step 4 Delete row j and column j. We make the following observation regarding Step 3. If e = (a, b) ∈ R and we merge a and b, then e would become a cycle of length 1 at a . We do not want to create this situation, since it does not correspond to “shrinking (a, b) to zero.” Step 3 corrects for this occurrence. Example 6

v0 v1 v2 v3 v4 v5 v6

v0 0 ⎢1 ⎢ ⎢1 ⎢ ⎢0 ⎢0 ⎢ ⎣1 0 ⎡

v1 1 0 0 1 1 0 0

v2 1 0 0 0 0 1 1

Figure 34 gives the matrices for the corresponding symmetric relations whose graphs are given in Figure 33. In Figure 34(b), we have merged vertices v0 and v1 into v0 . Note that this is done by taking the join of the first two rows and entering the result in row 1, doing the same for the columns, then restoring the diagonal, and removing row 2 and column 2. If vertices v0 and v2 in the graph whose matrix is given by Figure 34(b) are merged, the resulting graph has the matrix given by ◆ Figure 34(c).

v3 0 1 0 0 0 0 0 (a)

Figure 34

330

v4 0 1 0 0 0 0 0

v5 1 0 1 0 0 0 0

v6 ⎤ 0 0⎥ ⎥ 1⎥ ⎥ 0⎥ 0⎥ ⎥ 0⎦ 0

v0 v2 v3 v4 v5 v6

v0 0 ⎢1 ⎢ ⎢1 ⎢ ⎢1 ⎣1 0 ⎡

v2 1 0 0 0 1 1

v3 1 0 0 0 0 0

(b)

v4 1 0 0 0 0 0

v5 1 1 0 0 0 0

v6 ⎤ 0 1⎥ ⎥ 0⎥ ⎥ 0⎥ 0⎦ 0

v0 v3 v4 v5 v6

v0 0 ⎢1 ⎢ ⎢1 ⎣1 1 ⎡

v3 1 0 0 0 0

v4 1 0 0 0 0

(c)

v5 1 0 0 0 0

v6 ⎤ 1 0⎥ ⎥ 0⎥ 0⎦ 0

Trees We can now give an algorithm for finding a spanning tree for a symmetric, connected relation R on the set A = {v1 , v2 , . . . , vn }. The method is equivalent to a special case of an algorithm called Prim’s algorithm. The steps are as follows: Step 1 Choose a vertex v1 of R, and arrange the matrix of R so that the first row corresponds to v1 . Step 2 Choose a vertex v2 of R such that (v1 , v2 ) ∈ R, merge v1 and v2 into a new vertex v1 , representing {v1 , v2 }, and replace v1 by v1 . Compute the matrix of the resulting relation R . Call the vertex v1 a merged vertex. Step 3 Repeat Steps 1 and 2 on R and on all subsequent relations until a relation with a single vertex is obtained. At each stage, keep a record of the set of original vertices that is represented by each merged vertex. Step 4 Construct the spanning tree as follows. At each stage, when merging vertices a and b, select an edge in R from one of the original vertices represented by a to one of the original vertices represented by b. Example 7

We apply Prim’s algorithm to the symmetric relation whose graph is shown in Figure 35. In Table 1, we show the matrices that are obtained when the original set of vertices is reduced by merging until a single vertex is obtained, and at each stage we keep track of the set of original vertices represented by each merged vertex, as well as of the new vertex that is about to be merged.

TABLE 1

Matrix

a

b

a 0 ⎢0 ⎣1 1 ⎡

a b c d

a 0 a b ⎣1 d 1

c

d

Figure 35

a d

a

a

b

c

d

Figure 36

⎡

b 0 0 1 1

c d ⎤ 1 1 1 1⎥ 0 0⎦ 0 0

b 1 0 1

a d 0 1 1 0 a

0

d ⎤ 1 1⎦ 0

Original Vertices Represented by Merged Vertices

New Vertex to Be Merged (with First Row)

—

c

a ↔ {a, c}

b

a ↔ {a, c, b}

d

a ↔ {a, c, d, b}

—

The first vertex chosen is a, and we choose c as the vertex to be merged with a, since there is a 1 at vertex c in row 1. We also select the edge (a, c) from the original graph. At the second stage, there is a 1 at vertex b in row 1, so we merge b with vertex a . We select an edge in the original relation R from a vertex of {a, c} to b, say (c, b). At the third stage, we have to merge d with vertex a . Again, we need an edge in R from a vertex of {a, c, b} to d, say (a, d). The selected edges (a, c), (c, b), and (a, d) form the spanning tree for R, which is shown in Figure 36. Note that the first vertex selected becomes the root of the spanning tree that is ◆ constructed.

331

Trees

4 Exercises In Exercises 1 through 6, construct an undirected spanning tree for the connected graph G by removing edges in succession. Show the graph of the resulting undirected tree.

6. Let G be the graph shown in Figure 42. b

1. Let G be the graph shown in Figure 37.

a

e c

a

d

f b

f

g

Figure 42 c

e

In Exercises 7 through 12 (Figures 37 through 42), use Prim’s algorithm to construct a spanning tree for the connected graph shown. Use the indicated vertex as the root of the tree and draw the digraph of the spanning tree produced.

d

Figure 37

7. Figure 37; use e as the root.

2. Let G be the graph shown in Figure 38. 1

8. Figure 38; use 5 as the root.

2

9. Figure 39; use c as the root. 10. Figure 40; use 4 as the root. 11. Figure 41; use e as the root.

5 3

12. Figure 42; use d as the root.

4

13. Consider the connected graph shown in Figure 43. Show the graphs of three different undirected spanning trees.

Figure 38

b

3. Let G be the graph shown in Figure 39. a

e

b f

a

d g

c c

d

Figure 43 14. For the connected graph shown in Figure 44, show the graphs of all undirected spanning trees.

Figure 39 4. Let G be the graph shown in Figure 40. 1

2

5 3

1

4

6

1

3

3

7

2

4

Figure 40 5. Let G be the graph shown in Figure 41. a

b

d e

Figure 41

332

2

c

4

5

Figure 44 Figure 45 15. For the undirected tree shown in Figure 45, show the digraphs of all spanning trees. How many are there? Two trees, T1 and T2 , are isomorphic if they are isomorphic posets; that is, there is a one-to-one correspondence, f , between the vertices of T1 and T2 and (a, b) is an edge in T1 if and only if (f(a), f(b)) is an edge in T2 . 16. For the graph shown in Figure 44, show the graphs of all nonisomorphic undirected spanning trees.

Trees A1

17. For the graph shown in Figure 45, show the graphs of all nonisomorphic spanning trees. 18. For the graph in Figure 46, give all spanning trees. A1

A1

A5

19. 20. 21. 22.

A3

A3

A3

A3

A5

Figure 48

A4

A4

Figure 49

23. For the graph in Figure 49, how many different spanning trees are there?

Figure 46 Figure 47 For the graph in Figure 47, give all spanning trees. For the graph in Figure 47, show the graphs of all nonisomorphic spanning trees. For the graph in Figure 48, how many different spanning trees are there? For the graph shown in Figure 48, show the graphs of all nonisomorphic spanning trees.

5

A2

An

A2 A4

A2

A1

A2

24. State your conclusion for Figure 49 as a theorem and prove it. 25. For the graph shown in Figure 49, how many nonisomorphic spanning trees are there? Justify your answer. 26. Prove that a symmetric connected relation has an undirected spanning tree.

Minimal Spanning Trees In many applications of symmetric connected relations, the (undirected) graph of the relation models a situation in which the edges as well as the vertices carry information. A weighted graph is a graph for which each edge is labeled with a numerical value called its weight.

Example 1

The small town of Social Circle maintains a system of walking trails between the recreational areas in town. The system is modeled by the weighted graph in Figure 50, where the weights represent the distances in kilometers between sites.◆

Example 2

A communications company is investigating the costs of upgrading links between the relay stations it owns. The weighted graph in Figure 51 shows the stations and ◆ the cost in millions of dollars for upgrading each link. The weight of an edge (vi , vj ) is sometimes referred to as the distance between vertices vi and vj . A vertex u is a nearest neighbor of vertex v if u and v are B 2.6 2.7

3

3 A

2

2 5

C

F

6 3 4

H

1.8

I

D

2.8 J

1.7

3.2

2.5 5.3

E 2.2

3.3

H

E

4

5

Figure 50

2

4.2 2.9

2.4

3.4 D

6

C

3.6

A

B

2.1

4.4

2.1 F

G

G

Figure 51

333

Trees adjacent and no other vertex is joined to v by an edge of lesser weight than (u, v). Notice that, ungrammatically, v may have more than one nearest neighbor. Example 3

In the graph shown in Figure 50, vertex C is a nearest neighbor of vertex A. Vertices ◆ E and G are both nearest neighbors of vertex F . A vertex v is a nearest neighbor of a set of vertices V = {v1 , v2 , . . . , vk } in a graph if v is adjacent to some member vi of V and no other vertex adjacent to a member of V is joined by an edge of lesser weight than (v, vi ). This vertex v may belong to V .

Example 4

Referring to the graph given in Figure 51, let V = {C, E, J}. Then vertex D is a nearest neighbor of V , because (D, E) has weight 2.2 and no other vertex adjacent to C, E, or J is linked by an edge of lesser weight to one of these vertices. ◆ With applications of weighted graphs, it is often necessary to find an undirected spanning tree for which the total weight of the edges in the tree is as small as possible. Such a spanning tree is called a minimal spanning tree. Prim’s algorithm (Section 4) can easily be adapted to produce a minimal spanning tree for a weighted graph. We restate Prim’s algorithm as it would be applied to a symmetric, connected relation given by its undirected weighted graph. Algorithm PRIM’S ALGORITHM

Let R be a symmetric, connected relation with n vertices. Step 1 Choose a vertex v1 of R. Let V = {v1 } and E = { }. Step 2 Choose a nearest neighbor vi of V that is adjacent to vj , vj ∈ V , and for which the edge (vi , vj ) does not form a cycle with members of E. Add vi to V and add (vi , vj ) to E. Step 3 Repeat Step 2 until |E| = n − 1. Then V contains all n vertices of R, and ● E contains the edges of a minimal spanning tree for R. In this version of Prim’s algorithm, we begin at any vertex of R and construct a minimal spanning tree by adding an edge to a nearest neighbor of the set of vertices already linked, as long as adding this edge does not complete a cycle. This is an example of a greedy algorithm. At each stage we chose what is “best” based on local conditions, rather than looking at the global situation. Greedy algorithms do not always produce optimal solutions, but we can show that for Prim’s algorithm the solution is optimal. THEOREM 1

Prim’s algorithm, as given, produces a minimal spanning tree for the relation. Proof Let R have n vertices. Let T be the spanning tree for R produced by Prim’s algorithm. Suppose that the edges of T , in the order in which they were selected, are t1 , t2 , . . . , tn−1 . For each i from 1 to n − 1, we define Ti to be the tree with edges t1 , t2 , . . . , ti and T0 = { }. Then T0 ⊂ T1 ⊂ · · · ⊂ Tn−1 = T . We now prove, by mathematical induction, that each Ti is contained in a minimal spanning tree for R.

Basis Step Clearly P(0): T0 = { } is contained in every minimal spanning tree for R.

334

Trees

s1

Induction Step Let P(k): Tk is contained in a minimal spanning tree T for R. We use P(k) to show P(k + 1): Tk+1 is contained in a minimal spanning tree for R. By definition we have {t1 , t2 , . . . , tk } ⊆ T . If tk+1 also belongs to T , then Tk+1 ⊆ T and we have P(k + 1) is true. If tk+1 does not belong to T , then T ∪ {tk+1 } must contain a cycle. (Why?) This cycle would be as shown in Figure 52 for some edges s1 , s2 , . . . , sr in T . Now the edges of this cycle cannot all be from Tk , or Tk+1 would contain this cycle. Let sl be the edge with smallest index l that is not in Tk . Then sl has one vertex in the tree Tk and one not in Tk . This means that when tk+1 was chosen by Prim’s algorithm, sl was also available. Thus the weight of sl is at least as large as that of tk+1 . The spanning tree (T − {sl }) ∪ {tk+1 } contains Tk+1 . The weight of this tree is less than or equal to the weight of T , so it is a minimal spanning tree for R. Thus, P(k + 1) is true. So Tn−1 = T is contained in a minimal spanning tree ■ and must in fact be that minimal spanning tree. (Why?)

s2

ti+1 sr

Figure 52

Example 5

Social Circle, the town in Example 1, plans to pave some of the walking trails to make them bicycle paths as well. As a first stage, the town wants to link all the recreational areas with bicycle paths as cheaply as possible. Assuming that construction costs are the same on all parts of the system, use Prim’s algorithm to find a plan for the town’s paving. Solution

Referring to Figure 50, if we choose A as the first vertex, the nearest neighbor is C, 2 kilometers away. So (A, C) is the first edge selected. Considering the set of vertices {A, C}, B is the nearest neighbor, and we may choose either (A, B) or (B, C) as the next edge. Arbitrarily, we choose (B, C). B is a nearest neighbor for {A, B, C}, but the only edge available (A, B) would make a cycle, so we must move to a next nearest neighbor and choose (C, F ) [or (C, E)]. Figures 53(a) through (c) show the beginning steps and Figure 53(d) shows a possible final result. Figure 53(e) shows a minimal spanning tree using Prim’s algorithm beginning with vertex E. In either case, the bicycle paths would cover 21 kilometers. ◆ Example 6

Aminimal spanning tree for the communication network in Example 2 may be found by using Prim’s algorithm beginning at any vertex. Figure 54 shows a minimal spanning tree produced by beginning at I. The total cost of upgrading these links ◆ would be $20,200,000. B

A

A 2

3

2

(a)

2

D

(c) H

4

5 F

A 3 G

(d)

D

2 H

A

I

5 2 C F

4 (e)

H

3

E

2.6

G

C

2.1

2.4

2

3

E

B

B

2

3 C

Figure 53

2

5

C

(b)

B

F

3

2

C

C

A

B A

D 2.8 J

1.8

E

2.1 2.5

1.7

2.2

F

G

Figure 54

335

Trees How is Prim’s algorithm in Section 4 related to Prim’s algorithm given in this section? The earlier version is a special case where each edge has weight 1 and so any spanning tree is minimal. (Remember that each spanning tree will have n − 1 edges to reach n vertices.) This means that we can modify the matrix version of Prim’s algorithm in Section 4 to handle the general case of a weighted graph. A matrix version might be more easily programmed. The key to the modification is to consider an entry in the matrix as representing not only the existence of an edge, but also the weight of the edge. Thus, we construct a matrix containing the weights of edges, with a 0 to show there is no edge between a pair of vertices as before. Figure 55 shows a connected weighted graph and the associated matrix of weights. 4

A 3

B

3 5

D 2

C

A ⎡ A 0 B ⎢4 C ⎣3 D 0

B 4 0 5 3

C 3 5 0 2

D ⎤ 0 3⎥ 2⎦ 0

Figure 55

Algorithm PRIM’S ALGORITHM (Matrix Version)

Let R be a symmetric, connected relation with n vertices and M be the associated matrix of weights. Step 1 Choose the smallest entry in M, say mij . Let a be the vertex that is represented by row i and b the vertex represented by column j. Step 2 Merge a with b as follows: Replace row i with ⎧ ⎪ mik if mjk = 0 ⎪ ⎪ ⎨m if mik = 0 jk 1≤k≤n mik = ⎪ min(m , m ) if mik = 0, mjk = 0 ik jk ⎪ ⎪ ⎩ 0 if mik = mjk = 0. Replace column i with ⎧ ⎪ mki ⎪ ⎪ ⎨m kj mki = ⎪ min(m ki , mkj ) ⎪ ⎪ ⎩ 0

if mkj if mki if mki if mki

=0 =0 = 0, mkj = 0 = mkj = 0.

1≤k≤n

Replace the main diagonal with the original entries of M. Delete row j and column j. Call the resulting matrix M . Step 3 Repeat Steps 1 and 2 on M and subsequent matrices until a single vertex is obtained. At each stage, keep a record of the set of original vertices that is represented by each merged vertex. Step 4 Construct the minimal spanning tree as follows: At each stage, when merging vertices a and b, select the edge represented by the minimal weight from one of the original vertices represented by a to one of the original vertices ● represented by b.

336

Trees Example 7

Apply the matrix version of Prim’s algorithm to the graph and associated matrix in Figure 55. Solution

Either 2 may be selected as mij . We choose m3,4 and merge C and D. This produces ⎡A A 0 M = B ⎣ 4 C 3 A

B 3

A B 2

C

Figure 56

C ⎤ 3 3⎦ 0

with C ↔ {C, D} and the first edge (C, D). We repeat Steps 1 and 2 on M using m1,3 = 3. This gives

3

D

B 4 0 3

A 0 3

B 3 0

with A ↔ {A, C, D} and the selection of edge (A, C). A final merge yields the ◆ edge (B, D). The minimal spanning tree is shown in Figure 56. If a symmetric connected relation R has n vertices, then Prim’s algorithm has running time (n2 ). (This can be improved somewhat.) If R has relatively few edges, a different algorithm may be more efficient. This is similar to the case for determining whether a relation is transitive. Kruskal’s algorithm is another example of a greedy algorithm that produces an optimal solution. Algorithm KRUSKAL’S ALGORITHM

Let R be a symmetric, connected relation with n vertices and let S = {e1 , e2 , . . . , ek } be the set of weighted edges of R. Step 1 Choose an edge e1 in S of least weight. Let E = {e1 }. Replace S with S − {e1 }. Step 2 Select an edge ei in S of least weight that will not make a cycle with members of E. Replace E with E ∪ {ei } and S with S − {ei }. ● Step 3 Repeat Step 2 until |E| = n − 1. Since R has n vertices, the n − 1 edges in E will form a spanning tree. The selection process in Step 2 guarantees that this is a minimal spanning tree. (We omit the proof.) Roughly speaking, the running time of Kruskal’s algorithm is (k lg(k)), where k is the number of edges in R. Example 8

A minimal spanning tree from Kruskal’s algorithm for the walking trails in Example 1 is given in Figure 57. One sequence of edge selections is (D, E), (D, H ), (A, C), (A, B), (E, G), (E, F), and (C, E) for a total weight of 21 kilometers. Naturally either of the algorithms for minimal spanning trees should produce trees of the ◆ same weight.

Example 9

Use Kruskal’s algorithm to find a minimal spanning tree for the relation given by the graph in Figure 58.

337

Trees B

5 3

4 E

2 C

Figure 57

D 12

20

12 G

F

15

15

20

A

2

3 A

2 H

C

16

12 D

B

10

14

G

F

10

E

Figure 58

Solution

Initially, there are two edges of least weight, (B, C) and (E, F). Both of these are selected. Next there are three edges, (A, G), (B, G), and (D, E), of weight 12. All of these may be added without creating any cycles. Edge (F, G) of weight 14 is the remaining edge of least weight. Adding (F, G) gives us six edges for a 7-vertex ◆ graph, so a minimal spanning tree has been found.

5 Exercises In Exercises 1 through 6, use Prim’s algorithm as given in this section to find a minimal spanning tree for the connected graph indicated. Use the specified vertex as the initial vertex.

In Exercises 7 through 9, use the matrix version of Prim’s algorithm to find a minimal spanning tree for the indicated graph. 7. Let G be the graph shown in Figure 61.

1. Let G be the graph shown in Figure 50. Begin at F . 2. Let G be the graph shown in Figure 51. Begin at A. 3. Let G be the graph shown in Figure 58. Begin at G.

A

4

2 5

3

4. Let G be the graph shown in Figure 59. Begin at E. B

2

3

4

A

2

F

D 5

G

5

2

1

5

E 4 H

C

6. Let G be the graph shown in Figure 60. Begin at M. A

6.8

H 2.2

3.5 G

3.7 3.5

4.9

K

2.8

2.3

2.1

F

L

1.9

E

2.6

7.2

1.7 4.3

Figure 60

338

2.3

D 2.7

N

M 1.7

In Exercises 10 through 12, use Kruskal’s algorithm to find a minimal spanning tree for the indicated graph. 10. Let G be the graph shown in Figure 51.

J 2.3

2.7 I

3.2

C

Figure 61

9. Let G be the graph shown in Figure 59.

5. Let G be the graph shown in Figure 60. Begin at K.

3.7

D

8. Let G be the graph shown in Figure 50.

Figure 59

B

2 C

4

3

B

11. Let G be the graph shown in Figure 59. 12. Let G be the graph shown in Figure 60. 13. The distances between eight cities are given in the following table. Use Kruskal’s algorithm to find a minimal spanning tree whose vertices are these cities. What is the total distance for the tree? 14. Suppose that in constructing a minimal spanning tree a certain edge must be included. Give a modified version of Kruskal’s algorithm for this case. 15. Redo Exercise 13 with the requirement that the route from Atlanta to Augusta must be included. How much longer does this make the tree?

Trees Abbeville Abbeville Aiken Allendale Anderson Asheville Athens Atlanta Augusta

69 121 30 113 70 135 63

Aiken

Allendale

Anderson

Asheville

Athens

Atlanta

Augusta

69

121 52

30 97 149

113 170 222 92

70 117 160 63 155

135 163 206 122 204 66

63 16 59 93 174 101 147

52 97 170 117 163 16

149 222 160 206 59

16. Modify Prim’s algorithm to handle the case of finding a maximal spanning tree if a certain edge must be included in the tree. 17. Use the modification of Prim’s algorithm developed in Exercise 16 on the graph shown in Figure 60 if the edge from D to L must be included in the tree. 18. Modify Kruskal’s algorithm so that it will produce a maximal spanning tree, that is, one with the largest possible sum of the weights. 19. Suppose that the graph in Figure 60 represents possible flows through a system of pipes. Find a spanning tree that gives the maximum possible flow in this system. 20. Modify Prim’s algorithm as given in this section to find a maximal spanning tree. 21. Use the modified Prim’s algorithm from Exercise 20 to find a maximal spanning tree for the graph in Figure 60. 22. In Example 5, two different minimal spanning trees for the same graph were displayed. When will a weighted graph have a unique minimal spanning tree? Give reasons for your answer. 23. Give a simple condition on the weights of a graph that will guarantee that there is a unique maximal spanning tree for the graph.

92 63 122 93

155 204 174

66 101

147

24. Prove or disprove that an edge of unique least weight in a connected graph must be included in any minimal spanning tree. 25. A greedy algorithm does not always produce an optimal solution. Consider the problem of selecting a specified amount of money from an unlimited supply of coins using as few coins as possible. A greedy algorithm would be to select as many of the largest denomination coins first, then as many of the next largest as possible, and so on until the specified amount has been selected. If the coins in Acia are the 1-xebec, the 7-xebec, and the 11-xebec, construct an example to show that the greedy algorithm does not always give a minimal solution. 26. Expand the proof of Theorem 1 by completing the following. . (a) T must have n − 1 edges, because (b) T ∪ {tk+1 } must contain a cycle, because

.

(c) (T − {sl }) ∪ {tk+1 } is a spanning tree for R, because . (d) If T is contained in a minimal spanning tree for R, then T must be that tree. Why?

Tips for Proofs

The uniqueness of the root or of paths from the root to other vertices forms the basis of many indirect proofs for statements about trees. [See Theorem 1(c), Section 1.] Counting arguments are also common in proofs about trees; for example, Theorem 3, Section 4. Because a tree, like a lattice, is a relation with certain properties, the various facts and representations for relations are available to create proofs for tree theorems. We see this in Theorem 2, Section 1. The development of Prim’s algorithm in Section 4 uses the matrix representation of a relation; this representation could be used in a proof that the running time of the algorithm is (n2 ). 339

Trees

Key Ideas for Review • Tree: relation on a finite set A such that there exists a vertex

• • •

• • • •

• • • • • • • •

v0 ∈ A with the property that there is a unique path from v0 to any other vertex in A and no path from v0 to v0 . Root of tree: vertex v0 in the preceding definition Rooted tree (T, v0 ): tree T with root v0 Theorem: Let (T, v0 ) be a rooted tree. Then (a) There are no cycles in T . (b) v0 is the only root of T . (c) Each vertex in T , other than v0 , has in-degree one, and v0 has in-degree zero. Level Height of a tree: the largest level number of a tree Leaves: vertices having no offspring Theorem: Let T be a rooted tree on a set A. Then (a) T is irreflexive. (b) T is asymmetric. (c) If (a, b) ∈ T and (b, c) ∈ T , then (a, c) ∈ / T , for all a, b, and c in A. n-tree: tree in which every vertex has at most n offspring Complete n-tree Binary tree: 2-tree Theorem. If (T, v0 ) is a rooted tree and v ∈ T , then T(v) is also a rooted tree with root v. T(v): subtree of T beginning at v Positional binary tree Computer representation of trees Huffman code tree

• Preorder search • Inorder search • Postorder search • Reverse Polish notation • Searching general trees • Linked-list representation of a tree • Undirected tree: symmetric closure of a tree • Simple path: No two edges correspond to the same undi-

rected edge. • Connected symmetric relation R: There is a path in R from

any vertex to any other vertex. • Theorem: A tree with n vertices has n − 1 edges. • Spanning tree for symmetric connected relation R: tree

• • • • • • • • •

reaching all the vertices of R and whose edges are edges of R Undirected spanning tree: symmetric closure of a spanning tree Prim’s algorithm Weighted graph: a graph whose edges are each labeled with a numerical value Distance between vertices vi and vj : weight of (vi , vj ) Nearest neighbor of v Minimal spanning tree: undirected spanning tree for which the total weight of the edges is as small as possible Prim’s algorithm (second version) Greedy algorithm Kruskal’s algorithm

Chapter Self-Test v0

1. In this chapter, what kind of mathematical object is a tree? 2. What changes in performing a preorder, inorder, or postorder search?

v3

3. What are the other three possible sequences for searching a binary tree?

v1

v7

4. What advantages might a complete n-tree have over a general n-tree as a data storage structure? 5. What information would help you decide between using Prim’s and Kruskal’s algorithms to find a minimal spanning tree? 6. Determine if the relation R = {(1, 7), (2, 3), (4, 1), (2, 6), (4, 5), (5, 3), (4, 2)} is a tree on the set A = {1, 2, 3, 4, 5, 6, 7}. If it is a tree, what is the root? If it is not a tree, make the least number of changes necessary to make it a tree and give the root. 7. Consider (T, v0 ), the tree whose digraph is given in Figure 62. (a) What is the height of T ? (b) List the leaves of T .

340

v8

v6 v2

v4

v9

v5

v10

Figure 62 (c) How many subtrees of T contain v4 ? (d) List the siblings of v7 . 8. Let (T, v0 ) be a rooted tree. Prove that if any edge is removed from T , then the resulting relation cannot be a tree.

Trees q

9. Construct the labeled tree representing the algebraic expression s

(((x + 3)(x + 3) − (x − 2)(x − 2)) ÷ (6x − 5)) + (13 − πx). v

t

10. The arrays LEFT, DATA, RIGHT give a doubly-linked-list representation of a labeled binary, positional tree. Construct the digraph of this tree. INDEX

LEFT

1 2 3 4 5 6 7 8

8 0 2 0 4 0 0 6

9 10

0 0

DATA S T R U C T U R E

w u z

RIGHT 0 5 0 0 7 9 0 3 10 0

11. Here, to visit a vertex means to print the contents of the vertex. (a) Show the result of performing a preorder search on the tree constructed in Problem 10. (b) Show the results of performing a postorder search on the tree constructed in Problem 10. 12. Create a binary tree for which the results of performing a preorder search are S1 TRE1 S2 S3 E2 D and for which the results of performing a postorder search are DE2 S3 S2 E1 RT S1 . Assume that to visit a vertex means to print the contents of the vertex. 13. Draw a complete 3-tree with height 3 using the smallest possible number of vertices. 14. The digraph of a labeled, binary positional tree is shown in Figure 63. Construct the digraph of the labeled ordered tree T such that T = B(T ).

y

x r T

Figure 63 15. Give an undirected spanning tree for the relation whose graph is given in Figure 64. G

B

A

E

H

F

C D

I

J

Figure 64 16. Use Prim’s greedy algorithm to find a minimal spanning tree for the graph in Figure 65. Use vertex E as the initial vertex and list the edges in the order in which they are chosen. A 4

3 2

E 5

D

C

6

4

2 G 7

3

5 4

F 4

B

Figure 65 17. Use Kruskal’s algorithm to find a minimal spanning tree for the graph in Figure 65. List the edges in the order in which they are chosen.

Experiment 7 Ways to store and retrieve information in binary trees are presented in Sections 2 and 3. In this experiment you will investigate another type of tree that is frequently used for data storage. A B-tree of degree k is a tree with the following properties: 1. All leaves are on the same level. 2. If it is not a leaf, the root has at least two children and at most k children. 3. Any vertex that is not a leaf or the root has at least k/2 children and at most k children. Figure 1

The tree in Figure 1 is a B-tree of degree 3.

341

Trees Part I. Recall that the height of a tree is the length of the longest path from the root to a leaf. 1. Draw three different B-trees of degree 3 with height 2. Your examples should not also be of degree 2 or 1. 2. Draw three different B-trees of degree 4 (but not less) with height 3. 3. Give an example of a B-tree of degree 5 (but not less) with height 4. 4. Discuss the features of your examples in Questions 1 through 3 that suggest that a B-tree would be a good storage structure. Part II. The properties that define a B-tree of degree k not only restrict how the tree can look, but also limit the number of leaves for a given height and the height for a given number of leaves. 1. If T is a B-tree of degree k and T has height h, what is the maximum number of leaves that T can have? Explain your reasoning. 2. If T is a B-tree of degree k and T has height h, what is the minimum number of leaves that T can have? Explain your reasoning. 3. If T is a B-tree of degree k and T has n leaves, what is the maximum height that T can have? Explain your reasoning. 4. If T is a B-tree of degree k and T has n leaves, what is the minimum height that T can have? Explain your reasoning. 5. Explain how your results in Part II, Questions 3 and 4, support your conclusions in Part I, Question 4.

Coding Exercises 1.

Use the arrays LEFT, DATA, RIGHT (Section 2) in a program to store letters so that a postorder traversal of the tree created will print the letters out in alphabetical order.

2. Write a program that with input an ordered tree has as output the corresponding binary positional tree (as described in Section 3). 3. Write a subroutine to carry out the merging of vertices as described in Prim’s algorithm. 4. Write code for the second version of Prim’s algorithm (Section 5). 5. Write code for Kruskal’s algorithm.

Answers to Odd-Numbered Exercises Exercise Set 1

11. (a)

v2

1. Yes, the root is b. 3. Yes, the root is f . 5. No. 7. Yes, the root is t. 9. (a) v12 , v10 , v11 , v13 , v14 . (b) v10 , v11 , v5 , v12 , v7 , v15 , v14 , v9 .

342

v5

v6

v12

Trees (b)

Exercise Set 2

v3

1. v7

–

v9

v8

+ v14

v13

7

–

v15

3.

15. (a) v1 , v3 .

+ x

(b) 2.

v8

v9

÷

6 v10

v5

–

4 2

v8

5. v7

v6 v9

v7

v1 v10

v2

–

– +

v3

v10

x x 2

23. The total number of vertices is 1 + kn, where k is the number of nonleaves and 1 counts the root, because every vertex except the root is an offspring. Since l = m − k, l = 1 + kn − k or 1 + k(n − 1).

÷ +

+

3

–

3

4

–

2

7

–

11 9

4

9.

+ +

–

27. Each vertex except the root has in-degree 1. Thus s = r − 1.

– x

29. 4. The tree of maximum height has one vertex on each level.

(b) 1 ≤ k ≤ 7.

+

2

–

25. If both v T u and u T v, then v, u, v is a cycle in T . Thus v T u implies u T/ v. T is asymmetric.

31. Assume that the in-degree of v0 = 0. Then there is a cycle that begins and ends at v0 . This is impossible. Hence the in-degree of v0 must be 0.

2

x

7.

21. Basis step: n = 1. An n-tree of height 1 can have at most n leaves by definition. Induction step: Use P(i): An n-tree of height i has at most ni leaves to show P(i + 1): An n-tree of height i + 1 has at most ni+1 leaves. The leaves of a tree T of height i + 1 belong to the subtrees of T whose roots are at level 1. Each of these subtrees has height at most i, and there are at most n of them. Hence the maximum number of leaves of T is n · ni or ni+1 .

33. (a) 2 ≤ n.

3

–

2 v9

4

–

(b) v6 , v7 , v8 , v13 , v14 , v16 , v10 .

v3

y

2

3

19. v 1

–

x

6

13. (T, v0 ) may be an n-tree for n ≥ 3. It is not a complete 3-tree.

17. (a) 4.

–

2

x 3

3 4

11

x

11. 6.

343

Trees 13.

27.

1

0

1

1

0

B 0

1

0

C

A

1

0

E

D

15. 721. 17.

Exercise Set 3

A

1. x y s z t u v.

B

3. a b c g h i d k e j f.

C D

5. TSAMZWEDQMLCKFNTRGJ.

E

7. 2 + 3 − 1 × 2. F

G H

9. 6 4 2 1 3 5 7.

19.

n

11. s y v u t z x. 13. g h c i b k j f e d a.

a

d

c

a

15. ZWMADQESCNTFKLJGRMT. s

17. NEVER I COW A SAW PURPLE ONE SEE I NEVER HOPE I.

t f r

21. LEFT

o DATA

RIGHT

– + – 7 – 6 2 x – y 4

0 4 6 10 0 8 0 0 0 12 0 0

DATA

RIGHT

2 3 5 9 0 7 0 0 0 11 0 0

23. LEFT 2 3 5 9 7 11 0 0 0 0 0 0

25. (a) CAR.

344

g

– – – + 2 x x 2 2 x

s

0 4 6 10 8 12 0 0 0 0 0 0

(b) SEAR. (c) RACE. (d) SCAR.

19. 4. 21.

15 . 16

23.

8 . 6

25.

J

H

B

G

A I C

E F

D

27.

G

C

N

S

R H E

I

A

29. (a) The root must be labeled J; if J has a left offspring, it must be labeled B, otherwise the right offspring is labeled B. (b) The root must be labeled G; if G has a left offspring, it must be labeled N, otherwise the right offspring is labeled N.

Trees 31.

5. a

s

b

7.

c

d

t

y x

33.

a c

u w

f

d

e

y

e

9. c

d

a

b

11.

b e

c

b

z

B(T)

d

T a

a

a

b b c

d

b

e

g

j

d

e

j

i

i

35.

f

k g

c

g

h b

e d

a

k

h

e d

a

f

c

g

b

e

a

c f

f

13.

d

c

a

f g

15. There are 5 spanning trees. b d c

e

1

f

4

h

1 j

g

i

37. (a) v0

(b)

3

v0

(c)

v0

3

4

2

1

5

4

2

1

5

4

17. 1

2

3

39. AVL n = AVL n−1 + AVL n−2 + 1. Let v be a new root. Let T(vL ) be an AVL tree of height n − 1 using a minimal number of vertices. Let T(vR ) be an AVL tree of height n − 2 using a minimal number of vertices. Then T(v), where the left offspring of v is vL and its right offspring is vR , is an AVL tree of height n using a minimal number of vertices. Exercise Set 4 1.

a b

f

c

e d

5

4

3

2 5

5 1

2 3

4

v0

1

2

3

3

(d)

2

5

4

5

19. A 1

A2

A1

A2

A4

A3

A4

A3

A1

A2

A1

A2

A4

A3

A4

A3

21. Five.

23. n.

25. There is only one. Any spanning tree is formed by omitting one edge from the graph. A clockwise shift of the labels gives an isomorphism between any spanning tree and the tree formed by omitting the edge (vn , v1 ). Exercise Set 5

3. a

b

c

d

1.

B

H

D C

A

E F

G

345

Trees 3.

B

A

15.

C

Asheville

D G

F

92

5.

H

J

63

K

I

B

Atlanta

L

C M

D A

17.

N

A

C

F 3.7

D D 4

2

11. B

2

C

H

D

19.

G

1.7

A

4

2.3 L 2.3 M

7.2 D

5

E

I 2.1

1.9

E

C

F

3 A

2.7

3.5

B

J 2.3 K

2.2

G

2

2

16 Augusta

H

3.5

B

2

9. B

147

N

1.7

H

J

G B

F

K I L

F E

G

E A

13.

C

H

C

Athens 66 Atlanta

63

Allendale

Anderson 30

52 Abbeville 63

M D

N

21. The maximal spanning tree is the same as that in Exercise 19.

Asheville 92

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23. If each edge has a distinct weight, there will be a unique maximal spanning tree since only one choice can be made at each step. 25. One example is the sum 50. The greedy algorithm would select four 11-xebec coins and then six 1-xebec coins for a total of 10 coins. But the amount can be made from seven 7-xebec coins and one 1-xebec coin, a total of eight coins.

Answers to Chapter Self-Tests 1. A tree is a relation with certain conditions. 2. The only change in performing a preorder, inorder, or postorder search is when the root is visited. 3. The other three sequences are (i) right, root, left, (ii) right, left, root, and (iii) root, right, left. 4. A complete n-tree usually has smaller height than a general n-tree with the same number of vertices. This would shorten a search of the tree that begins at the root. 5. An estimate of how many vertices and edges the graph has would help in the decision process.

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6. It is not a tree; deleting either (2, 3) or (5, 3) will give a tree with root 4.

7. (a) 4. (c) 4.

(b) v4 , v10 , v6 , v9 , v8 . (d) v6 , v8 .

8. Every edge (vi , vj ) in (T, v0 ) belongs to a unique path from v0 to vj . Hence removing (vi , vj ) would mean there is no path from v0 to vj .

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Topics in Graph Theory Graph theory begins with very simple geometric ideas and has many powerful applications. Here we give an alternate definition of graph that includes the more general multigraphs and is more appropriate for the applications developed in this chapter.

Looking Back Graph theory began in 1736 when Leonhard Euler solved a problem that had been puzzling the good citizens of the town of Königsberg in Prussia (now Kaliningrad in Russia). The river Pregel divides the town into four sections, and in Euler’s days seven bridges connected these sections. The people wanted to know if it were possible to start at any location in town, cross every bridge exactly once, and return to the starting location. Euler showed that it is impossible to take such a walk. A problem in graph theory that attracted considerably more attention is the four-color map problem. Frank Guthrie, a former student of Augustus De Morgan, observed that he was able to color the map of England so that no two adjacent counties have the same color by using four different colors. He asked his brother, who in 1852 was a student of De Morgan, to ask De Morgan whether his conjecture that four colors will suffice to color every map so that no two adjacent counties have the same color was true. De Morgan publicized the problem, and for years many people worked on the problem. In 1879, Alfred Bray Kempe, a lawyer who had studied mathematics at Cambridge University, published a proof of the four-color conjecture that was highly acclaimed. Unfortunately, his proof had a fatal error. The theorem was finally proved by the American mathematicians Kenneth Appel and Wolfgang

Haken at the University of Illinois. Their proof used as a basis the work of Kempe and over 1000 hours of computer time to check 1936 different map configurations. Their proof, the first of a major mathematical result by using a computer, was initially met with considerable skepticism. Some years later, the number of configurations was reduced, but to date there is no proof that does not involve significant use of a computer.

Wolfgang Haken Kenneth Appel

From Discrete Mathematical Structures, Sixth Edition, Bernard Kolman, Robert C. Busby, Sharon Cutler Ross. Copyright © 2009 by Pearson Education, Inc. Published by Prentice Hall. All rights reserved.

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Topics in Graph Theory

1

Graphs A graph G consists of a finite set V of objects called vertices, a finite set E of objects called edges, and a function γ that assigns to each edge a subset {v, w}, where v and w are vertices (and may be the same). We will write G = (V, E, γ) when we need to name the parts of G. If e is an edge, and γ(e) = {v, w}, we say that e is an edge between v and w and that e is determined by v and w. The vertices v and w are called the end points of e. If there is only one edge between v and w, we often identify e with the set {v, w}. This should cause no confusion. The restriction that there is only a finite number of vertices may be dropped, but for the discussion here all graphs have a finite number of vertices.

Example 1

Let V = {1, 2, 3, 4} and E = {e1 , e2 , e3 , e4 , e5 }. Let γ be defined by γ(e1 ) = γ(e5 ) = {1, 2},

γ(e2 ) = {4, 3},

γ(e3 ) = {1, 3},

Then G = (V, E, γ) is a graph. 1

2

4

3

Figure 1

Example 2

γ(e4 ) = {2, 4}. ◆

Graphs are usually represented by pictures, using a point for each vertex and a line for each edge. G in Example 1 is represented in Figure 1. We usually omit the names of the edges, since they have no intrinsic meaning. Also, we may want to put other more useful labels on the edges. We sometimes omit the labels on vertices as well if the graphical information is adequate for the discussion. Graphs are often used to record information about relationships or connections. An edge between vi and vj indicates a connection between the objects vi and vj . In a pictorial representation of a graph, the connections are the most important information, and generally a number of different pictures may represent the same graph. Figures 2 and 3 also represent the graph G given in Example 1. 1

2

3

4

Figure 2

1

2

3

◆

4

Figure 3

The degree of a vertex is the number of edges having that vertex as an end point. A graph may contain an edge from a vertex to itself; such an edge is referred to as a loop. A loop contributes 2 to the degree of a vertex, since that vertex serves as both end points of the loop. Example 3

(a) In the graph in Figure 4, the vertex A has degree 2, vertex B has degree 4, and vertex D has degree 3. (b) In Figure 5, vertex a has degree 4, vertex e has degree 0, and vertex b has degree 2. ◆ (c) Each vertex of the graph in Figure 6 has degree 2. A vertex with degree 0 is called an isolated vertex. A pair of vertices that determine an edge are adjacent vertices.

350

Topics in Graph Theory p A

a

B

q

b

s

r

5

2

C

E

c

t

3 4 D

Figure 4

Example 4

u d

Figure 5

e

1

6

Figure 6

In Figure 5, vertex e is an isolated vertex. In Figure 5, a and b are adjacent vertices; ◆ vertices a and d are not adjacent. A path π in a graph G consists of a pair (Vπ , Eπ ) of sequences: a vertex sequence Vπ : v1 , v2 , . . . , vk and an edge sequence Eπ : e1 , e2 , . . . , ek−1 for which 1. Each successive pair vi , vi+1 of vertices is adjacent in G, and edge ei has vi and vi+1 as end points, for i = 1, . . . , k − 1. 2. No edge occurs more than once in the edge sequence. Thus we may begin at v1 and travel through the edges e1 , e2 , . . . , ek−1 to vk without using any edge twice. A circuit is a path that begins and ends at the same vertex. Such paths may be called cycles; the word “circuit” is more common in general graph theory. A path is called simple if no vertex appears more than once in the vertex sequence, except possibly if v1 = vk . In this case the path is called a simple circuit. This expanded definition of path is needed to handle paths in graphs that may have multiple edges between vertices. For the graph in Figure 4 we define a path π1 by sequences Vπ1 : A, B, E, D, D and Eπ1 : p, r, t, u, and the path π2 by the sequences Vπ2 : A, B, A and Eπ2 : p, q. The vertices alone would not be sufficient to define these paths. For π1 we would not know which edge to travel from A to B, p or q, and for π2 we would not know which edge to use first and which to use second. The edges alone would not always be enough either, for if we only knew the edge sequence p, q for π2 , we would not know if the vertex sequence was A, B, A or B, A, B. If the vertices of a path have only one edge between each adjacent pair, then the edge sequence is completely determined by the vertex sequence. In this case we specify the path by the vertex sequence alone and write π : v1 , v2 , . . . , vk . Thus for the graph in Figure 2 we can write π3 : 1, 3, 4, 2, since this path contains no adjacent vertices with two edges between them. In cases such as this, it is not necessary to label the edges.

Example 5

(a) Paths π1 and π2 in the graph of Figure 4 were defined previously. Path π1 is not simple, since vertex D appears twice, but π2 is a simple circuit, since the only vertex appearing twice occurs at the beginning and at the end. (b) The path π4 : D, E, B, C in the graph of Figure 4 is simple. Here no mention of edges is needed. (c) Examples of paths in the graph of Figure 5 are π5 : a, b, c, a and π6 : d, c, a, a. Here π5 is a simple circuit, but π6 is not simple. (d) In Figure 6 the vertex sequence 1, 2, 3, 2 does not specify a path, since the single edge between 2 and 3 would be traveled twice. ◆ (e) The path π7 : c, a, b, c, d in Figure 5 is not simple.

351

Topics in Graph Theory A graph is called connected if there is a path from any vertex to any other vertex in the graph. Otherwise, the graph is disconnected. If the graph is disconnected, the various connected pieces are called the components of the graph. Example 6

U2

The graphs in Figures 1 and 4 are connected. Those in Figures 5 and 6 are discon◆ nected. The graph of Figure 6 has two components. Some important special families of graphs will be useful in our discussions. We present them here.

U5

1. For each integer n ≥ 1, we let Un denote the graph with n vertices and no edges. Figure 7 shows U2 and U5 . We call Un the discrete graph on n vertices. 2. For each integer n ≥ 1, let Kn denote the graph with vertices {v1 ,v2 , . . . ,vn} and with an edge {vi , vj } for every i and j. In other words, every vertex in Kn is connected to every other vertex. In Figure 8 we show K3 , K4 , and K5 . The graph Kn is called the complete graph on n vertices. More generally, if each vertex of a graph has the same degree as every other vertex, the graph is called regular. The graphs Un are also regular. 3. For each integer n ≥ 1, we let Ln denote the graph with n vertices {v1 , v2 , . . . , vn } and with edges {vi , vi+1 } for 1 ≤ i < n. We show L2 and L4 in Figure 9. We call Ln the linear graph on n vertices.

Figure 7

K3

K4

K5

Figure 8

Example 7

L2

L4

Figure 9

All the Kn and Ln are connected, while the Un are disconnected for n > 1. In fact, ◆ the graph Un has exactly n components.

Subgraphs and Quotient Graphs Suppose that G = (V, E, γ) is a graph. Choose a subset E1 of the edges in E and a subset V1 of the vertices in V , so that V1 contains (at least) all the end points of edges in E1 . Then H = (V1 , E1 , γ1 ) is also a graph where γ1 is γ restricted to edges in E1 . Such a graph H is called a subgraph of G. Subgraphs play an important role in analyzing graph properties. Example 8

The graphs shown in Figures 11, 12, and 13 are each a subgraph of the graph shown ◆ in Figure 10. One of the most important subgraphs is the one that arises by deleting one edge and no vertices. If G = (V, E, γ) is a graph and e ∈ E, then we denote by Ge the subgraph obtained by omitting the edge e from E and keeping all vertices. If G is the graph of Figure 10, and e = {a, b}, then Ge is the graph shown in Figure 13.

352

Topics in Graph Theory a

b i

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Figure 10

h c

Figure 11

a

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Figure 12

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Figure 13

Our second important construction is defined for graphs without multiple edges between the same vertices. Suppose that G = (V, E, γ) is such a graph and that R is an equivalence relation on the set V . Then we construct the quotient graph GR in the following way. The vertices of GR are the equivalence classes of V produced by R. If [v] and [w] are the equivalence classes of vertices v and w of G, then there is an edge in GR from [v] to [w] if and only if some vertex in [v] is connected to some vertex in [w] in the graph G. Informally, this just says that we get GR by merging all the vertices in each equivalence class into a single vertex and combining any edges that are superimposed by such a process. Example 9

Let G be the graph of Figure 14 (which has no multiple edges), and let R be the equivalence relation on V defined by the partition {{a, m, i}, {b, f, j}, {c, g, k}, {d, h, l}}. Then GR is shown in Figure 15. [a] a

[b]

b m

f i

j

l

k

h d

[a]

[i] [h]

g c

Figure 14

[b]

[d]

[c]

Figure 15

[g]

[d]

Figure 16

If S is also an equivalence relation on V defined by the partition b

a m

f

{{i, j, k, l}, {a, m}, {f, b, c}, {d}, {g}, {h}}, then the quotient graph GS is shown in Figure 16.

◆

[i] l h d

Figure 17

k g c

Again, one of the most important cases arises from using just one edge. If e is an edge between vertex v and vertex w in a graph G = {V, E, γ}, then we consider the equivalence relation whose partition consists of {v, w} and {vi }, for each vi = v, vi = w. That is, we merge v and w and leave everything else alone. The resulting quotient graph is denoted Ge . If G is the graph of Figure 14, and e = {i, j}, then Ge is the graph shown in Figure 17.

353

Topics in Graph Theory

1 Exercises In Exercises 1 through 4 (Figures 18 through 21), give V , the set of vertices, and E, the set of edges, for the graph G = (V, E, γ). 1. a

b

b a

2.

d

For Exercises 19 through 22, use the graph G in Figure 22.

a

b f c

c d

Figure 18 3.

b a

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e

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d

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Figure 22

Figure 19 4. 1

2

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4

19. If R is the equivalence relation defined by the partition {{a, f }, {e, b, d}, {c}}, find the quotient graph, GR . 5

20. If R is the equivalence relation defined by the partition {{a, b}, {e}, {d}, {f, c}}, find the quotient graph, GR . 21. (a) Give the largest subgraph of G that does not contain f .

d

Figure 20

6

7

Figure 21

5. Give two subgraphs with three vertices for the graph shown in Figure 20.

(b) Give the largest subgraph of G that does not contain a. 22. Let e1 be the edge between c and f . Draw the graph of (b) Ge1 (a) Ge1 For Exercises 23 and 24, use the graph G in Figure 23.

7. Draw a picture of the graph G = (V, E, γ), where V = {a, b, c, d, e}, E = {e1 , e2 , e3 , e4 , e5 , e6 }, and γ(e1 ) = γ(e5 ) = {a, c}, γ(e2 ) = {a, d}, γ(e3 ) = {e, c}, γ(e4 ) = {b, c}, and γ(e6 ) = {e, d}. 8. Draw a picture of the graph G = {V, E, γ), where V = {a, b, c, d, e, f, g, h}, E = {e1 , e2 , . . . , e9 }, and γ(e1 ) = {a, c}, γ(e2 ) = {a, b}, γ(e3 ) = {d, c}, γ(e4 ) = {b, d}, γ(e5 ) = {e, a}, γ(e6 ) = {e, d}, γ(e7 ) = {f, e}, γ(e8 ) = {e, g}, and γ(e9 ) = {f, g}. 9. Give the degree of each vertex in Figure 18. 10. Give the degree of each vertex in Figure 20. 11. List all paths that begin at a in Figure 19. 12. List three circuits that begin at 5 in Figure 21. 13. Draw the complete graph on seven vertices. 14. Consider Kn , the complete graph on n vertices. What is the degree of each vertex? 15. Which of the graphs in Exercises 1 through 4 are regular?

354

11

3

6. Give three subgraphs with four vertices and four edges for the graph shown in Figure 21.

10

4

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12

1 5 6

8 7

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9 16

14 15

Figure 23 23. Let R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9), (10, 10), (11, 11), (12, 12), (13, 13), (14, 14), (15, 15), (16, 16), (1, 10), (10, 1), (3, 12), (12, 3), (5, 14), (14, 5), (2, 11), (11, 2), (4, 13), (13, 4), (6, 15), (15, 6), (7, 16), (16, 7), (8, 9), (9, 8)}. Draw the quotient graph GR . 24. Let R = {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (7, 7), (8, 8), (9, 9), (10, 10), (11, 11), (12, 12), (13, 13), (14, 14), (15, 15), (16, 16), (1, 2), (2, 1), (3, 4), (4, 3), (5, 6), (6, 5), (7, 8), (8, 7), (9, 16), (16, 9), (10, 11), (11, 10), (12, 13), (13, 12), (14, 15), (15, 14)}. Draw the quotient graph GR .

16. Give an example of a regular, connected graph on six vertices that is not complete.

25. Complete the following statement. Every linear graph on n vertices must have edges. Explain your answer.

17. Give an example of a graph on five vertices with exactly two components.

26. What is the total number of edges in Kn , the complete graph on n vertices? Justify your answer.

18. Give an example of a graph that is regular, but not complete, with each vertex having degree three.

27. Restate the definition of isomorphic trees to define isomorphic graphs.

Topics in Graph Theory m c

2

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b

d

n

q

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e

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r

(a)

p

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t

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s

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(d)

Figure 24 28. Prove or disprove that the graphs shown in Figures 24(a) and (b) are isomorphic. 29. Prove or disprove that the graphs shown in Figures 24(a) and (c) are isomorphic. 30. Prove or disprove that the graphs shown in Figures 24(a) and (d) are isomorphic.

2

31. Prove that if a graph G has no loops or multiple edges, then twice the number of edges is equal to the sum of the degrees of all vertices. 32. Use the result of Exercise 31 to prove that if a graph G has no loops or multiple edges, then the number of vertices of odd degree is an even number.

Euler Paths and Circuits In this section and the next, we consider broad categories of problems for which graph theory is used. In the first type of problem, the task is to travel a path using each edge of the graph exactly once. It may or may not be necessary to begin and end at the same vertex. A simple example of this is the common puzzle problem that asks the solver to trace a geometric figure without lifting pencil from paper or tracing an edge more than once. A path in a graph G is called an Euler path if it includes every edge exactly once. An Euler circuit is an Euler path that is a circuit. Leonhard Euler (1707– 1783) worked in many areas of mathematics. The names “Euler path” and “Euler circuit” recognize his work with the Königsberg Bridge problem.

Example 1

Figure 25 shows the street map of a small neighborhood. A recycling ordinance has been passed, and those responsible for picking up the recyclables must start and end each trip by having the truck in the recycling terminal. They would like

cling Recy nal i Term

Figure 25

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Topics in Graph Theory to plan the truck route so that the entire neighborhood can be covered and each street need be traveled only once. A graph can be constructed having one vertex for each intersection and an edge for each street between any two intersections. The ◆ problem then is to find an Euler circuit for this graph. Example 2

(a) An Euler path in Figure 26 is π : E, D, B, A, C, D. (b) One Euler circuit in the graph of Figure 27 is π : 5, 3, 2, 1, 3, 4, 5.

◆

E 2

4

D B

C

3 5

1

A

Figure 26

Figure 27

A little experimentation will show that no Euler circuit is possible for the graph in Figure 26. We also see that an Euler path is not possible for the graph in Figure 6 in Section 1. (Why?) a

b d A

c

d

Room A

Room B

a c

h

D (Outside)

Figure 28

Example 3

f

i

D

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g Room C

B

b

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Figure 29

Consider the floor plan of a three-room structure that is shown in Figure 28. Each room is connected to every room that it shares a wall with and to the outside along each wall. The problem is this: Is it possible to begin in a room or outside and take a walk that goes through each door exactly once? This diagram can also be formulated as a graph where each room and the outside constitute a vertex and an edge corresponds to each door. A possible graph for this structure is shown in Figure 29. The translation of the problem is whether or not there exists an Euler ◆ path for this graph. We will solve this problem later. Two questions arise naturally at this point. Is it possible to determine whether an Euler path or Euler circuit exists without actually finding it? If there must be an Euler path or circuit, is there an efficient way to find one? Consider again the graphs in Example 2. In Figure 26 the edge {D, E} must be either the first or the last traveled, because there is no other way to travel to or from vertex E. This means that if G has a vertex of degree 1, there cannot be an

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Topics in Graph Theory Euler circuit, and if there is an Euler path, it must begin or end at this vertex. A similar argument applies to any vertex v of odd degree, say 2n + 1. We may travel in on one of these edges and out on another one n times, leaving one edge from v untraveled. This last edge may be used for leaving v or arriving at v, but not both, so a circuit cannot be completed. We have just shown the first of the following results. THEOREM 1

(a) If a graph G has a vertex of odd degree, there can be no Euler circuit in G. (b) If G is a connected graph and every vertex has even degree, then there is an Euler circuit in G. Proof (b) Suppose that there are connected graphs where every vertex has even degree, but there is no Euler circuit. Choose such a G with the smallest number of edges. G must have more than one vertex since, if there were only one vertex of even degree, there is clearly an Euler circuit. We show first that G must have at least one circuit. If v is a fixed vertex of G, then since G is connected and has more than one vertex, there must be an edge between v and some other vertex v1 . This is a simple path (of length 1) and so simple paths exist. Let π0 be a simple path in G having the longest possible length, and let its vertex sequence be v1 , v2 , . . . , vs . Since vs has even degree and π0 uses only one edge that has vs as a vertex, there must be an edge e not in π0 that also has vs as a vertex. If the other vertex of e is not one of the vi , then we could construct a simple path longer than π0 , which is a contradiction. Thus e has some vi as its other vertex, and therefore we can construct a circuit vi , vi+1 , . . . , vs , vi in G. Since we now know that G has circuits, we may choose a circuit π in G that has the longest possible length. Since we assumed that G has no Euler circuits, π cannot contain all the edges of G. Let G1 be the graph formed from G by deleting all edges in π (but no vertices). Since π is a circuit, deleting its edges will reduce the degree of every vertex by 0 or 2, so G1 is also a graph with all vertices of even degree. The graph G1 may not be connected, but we can choose a largest connected component (piece) and call this graph G2 (G2 may be G1 ). Now G2 has fewer edges than G, and so (because of the way G was chosen), G2 must have an Euler path π . If π passes through all the vertices on G, then π and π clearly have vertices in common. If not, then there must be an edge e in G between some vertex v in π , and some vertex v not in π . Otherwise we could not get from vertices in π to the other vertices in G, and G would not be connected. Since e is not in π , it must have been deleted when G1 was created from G, and so must be an edge in π. Then v is also in the vertex sequence of π, and so in any case π and π have at least one vertex v in common. We can then construct a circuit in G that is longer than π by combining π and π at v . This is a contradiction, since π was chosen to be the longest possible circuit in G. Hence the existence of the graph G ■ always produces a contradiction, and so no such graph is possible. The strategy of this proof is one we have used before: Suppose there is a largest (smallest) object and construct a larger (smaller) object of the same type thereby creating a contradiction. Here we have π, the longest possible circuit that

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Topics in Graph Theory begins and ends at v, in G, and we construct a longer circuit that begins and ends at v. We have proved that if G has vertices of odd degree, it is not possible to construct an Euler circuit for G, but an Euler path may be possible. Our earlier discussion noted that a vertex of odd degree must be either the beginning or the end of any possible Euler path. We have the following theorem. THEOREM 2

(a) If a graph G has more than two vertices of odd degree, then there can be no Euler path in G. (b) If G is connected and has exactly two vertices of odd degree, there is an Euler path in G. Any Euler path in G must begin at one vertex of odd degree and end at the other. Proof (a) Let v1 , v2 , v3 be vertices of odd degree. Any possible Euler path must leave (or arrive at) each of v1 , v2 , v3 with no way to return (or leave) since each of these vertices has odd degree. One vertex of these three vertices may be the beginning of the Euler path and another the end, but this leaves the third vertex at one end of an untraveled edge. Thus there is no Euler path. (b) Let u and v be the two vertices of odd degree. Adding the edge {u, v} to G produces a connected graph G all of whose vertices have even degree. By Theorem 1(b), there is an Euler circuit π in G . Omitting {u, v} from π produces an Euler path that begins at u (or v) and ends at v (or u). ■

Example 4

Which of the graphs in Figures 30, 31, and 32 have an Euler circuit, an Euler path but not an Euler circuit, or neither?

Figure 30

Figure 31

Figure 32

Solution

(a) In Figure 30, each of the four vertices has degree 3; thus, by Theorems 1 and 2, there is neither an Euler circuit nor an Euler path. (b) The graph in Figure 31 has exactly two vertices of odd degree. There is no Euler circuit, but there must be an Euler path. (c) In Figure 32, every vertex has even degree; thus the graph must have an ◆ Euler circuit. Example 5

Let us return to Example 3. We see that the four vertices have degrees 4, 4, 5, and 7, respectively. Thus the problem can be solved by Theorem 2; that is, there is an Euler path. One is shown in Figure 33. Using the labels of Figure 29, this path π is specified by Vπ : C, D, C, A, D, A, B, D, B, C and Eπ : i, h, f, c, a, d, b, e, ◆ g, j. Theorems 1 and 2 are examples of existence theorems. They guarantee the existence of an object of a certain type, but they give no information on how to

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Topics in Graph Theory

Room A

Room B

Room C

D (Outside)

Figure 33

b A

a

c D

B e

f d

C

Figure 34

produce the object. There is one hint in Theorem 2(b) about how to proceed. In Figure 34, an Euler path must begin (or end) at B and end (or begin) at C. One possible Euler path for this graph is specified by Vπ : B, A, D, C, A, B, C and Eπ : a, c, d, f, b, e. We next give an algorithm that produces an Euler circuit for a connected graph with no vertices of odd degree. We require an additional definition before stating the algorithm. An edge is a bridge in a connected graph G if deleting it would create a disconnected graph. For example, in the graph of Figure 4, Section 1, {B, E} is a bridge. Algorithm FLEURY’S ALGORITHM

Let G = (V, E, γ) be a connected graph with each vertex of even degree. Step 1 Select an edge e1 that is not a bridge in G. Let its vertices be v1 , v2 . Let π be specified by Vπ : v1 , v2 and Eπ : e1 . Remove e1 from E and let G1 be the resulting subgraph of G. Step 2 Suppose that Vπ : v1 , v2 , . . . , vk and Eπ : e1 , e2 , . . . , ek−1 have been constructed so far, and that all of these edges and any resulting isolated vertices have been removed from V and E to form Gk−1 . Since vk has even degree, and ek−1 ends there, there must be an edge ek in Gk−1 that also has vk as a vertex. If there is more than one such edge, select one that is not a bridge for Gk−1 . Denote the vertex of ek other than vk by vk+1 , and extend Vπ and Eπ to Vπ : v1 , v2 , . . . , vk , vk+1 and Eπ : e1 , e2 , . . . , ek−1 , ek . Then delete ek and any isolated vertices from Gk−1 to form Gk . ● Step 3 Repeat Step 2 until no edges remain in E. Example 6

Use Fleury’s algorithm to construct an Euler circuit for the graph in Figure 35.

B

F C

A

D

E

G

H

Figure 35

359

Topics in Graph Theory Solution

According to Step 1, we may begin anywhere. Arbitrarily choose vertex A. We summarize the results of applying Step 2 repeatedly in Table 1.

TABLE 1 Current Path π: π: π: π: π: π: π: π: π: π: π: π: π:

Next Edge

Reasoning

{A, B} {B, C} {C, A} {A, D} {D, C} {C, E} {E, G} {G, F } {F, E} {E, H} {H, G} {G, A}

No edge from A is a bridge. Choose any one. Only one edge from B remains. No edge from C is a bridge. Choose any one. No edge from A is a bridge. Choose any one. Only one edge from D remains. Only one edge from C remains. No edge from E is a bridge. Choose any one. {A, G} is a bridge. Choose {G, F } or {G, H}. Only one edge from F remains. Only one edge from E remains. Only one edge from H remains. Only one edge from G remains.

A A, B A, B, C A, B, C, A A, B, C, A, D A, B, C, A, D, C A, B, C, A, D, C, E A, B, C, A, D, C, E, G A, B, C, A, D, C, E, G, F A, B, C, A, D, C, E, G, F, E A, B, C, A, D, C, E, G, F, E, H A, B, C, A, D, C, E, G, F, E, H, G A, B, C, A, D, C, E, G, F, E, H, G, A

The edges in Figure 36 have been numbered in the order of their choice in applying Step 2. In several places, other choices could have been made. In general, if a graph has an Euler circuit, it is likely to have several different Euler circuits.◆ 12 B 1 A 4

F 2

3

C

E

9

8

6 5

7

10

D

G 11

H

Figure 36

2 Exercises In Exercises 1 through 8, (Figures 37 through 44), tell whether the graph has an Euler circuit, an Euler path but no Euler circuit, or neither. Give reasons for your choice.

3.

b a

1. a

b

d

2.

a

c

Figure 37

360

Figure 39

c d

2

3

4

5

b d

c

4. 1

e

Figure 38

6

Figure 40

7

Topics in Graph Theory 5.

12. Use Fleury’s algorithm to produce an Euler circuit for the graph in Figure 45.

Figure 41

13. An art museum arranged its current exhibit in the five rooms shown in Figure 48. Is there a way to tour the exhibit so that you pass through each door exactly once? If so, give a sketch of your tour.

6.

Figure 42 7.

Figure 48 14. At the door of an historical mansion, you receive a copy of the floor plan for the house (Figure 49). Is it possible to visit every room in the house by passing through each door exactly once? Explain your reasoning.

Figure 43 8.

Figure 44 In Exercises 9 and 10, (Figures 45 and 46), tell if it is possible to trace the figure without lifting the pencil. Explain your reasoning. 9.

Figure 45

Figure 49 In Exercises 15 through 18 (Figures 50 through 53), no Euler circuit is possible for the graph given. For each graph, show the minimum number of edges that would need to be traveled twice in order to travel every edge and return to the starting vertex. 15.

16.

10.

Figure 50 Figure 46 11. Use Fleury’s algorithm to produce an Euler circuit for the graph in Figure 47.

Figure 47

Figure 51 17.

Figure 52

361

Topics in Graph Theory circuit for the modified version of the graph. Begin at A.

18.

21. For which values of n does the complete graph on n vertices, Kn , have an Euler circuit? A

22. Let G and H be a pair of isomorphic graphs. Prove or disprove that if there is an Euler circuit in G, then there is one in H also. 23. Consider the Hasse diagram of the Boolean algebra Bn as an undirected graph with the vertices labeled as usual by strings of 0’s and 1’s of length n. Prove that there is an edge between two vertices in this graph if and only if the strings labeling the end points differ in exactly one position.

Figure 53 19. Modify the graph in Figure 52 by adding the minimum number of duplicate edges needed to make all vertices have even degree. Use Fleury’s algorithm to find an Euler circuit for the modified version of the graph. Begin at the upper-left corner. 20. Modify the graph in Figure 53 by adding the minimum number of duplicate edges needed to make all vertices have even degree. Use Fleury’s algorithm to find an Euler

3

24. With Bn as described in Exercise 23, which of the following graphs have an Euler circuit? (b) B3 (c) B4 (d) B5 (a) B2 25. In general, the Hasse diagrams of which Bn have an Euler circuit? Justify your answer.

Hamiltonian Paths and Circuits We turn now to the second category of graph problems in which the task is to visit each vertex exactly once, with the exception of the beginning vertex if it must also be the last vertex. For example, such a path would be useful to someone who must service a set of vending machines on a regular basis. Each vending machine could be represented by a vertex. A Hamiltonian path is a path that contains each vertex exactly once. A Hamiltonian circuit is a circuit that contains each vertex exactly once except for the first vertex, which is also the last. This sort of path is named for the mathematician Sir William Hamilton, who developed and marketed a game consisting of a wooden graph in the shape of a regular dodecahedron and instructions to find what we have called a Hamiltonian circuit. A planar version of this solid is shown in Figure 54(a), with a Hamiltonian circuit (one of many) shown in Figure 54(b) by the consecutively numbered vertices. It is clear that loops and multiple edges are of no use in finding Hamiltonian circuits, since loops could not be used, and only one edge can be used between any two vertices. Thus we will suppose that any graph we mention has no loops or multiple edges. 1 20

13 2

12 11

14

19 18 17

15

16

10 3 (a)

Figure 54

362

6

9 (b)

7

8 4

5

Topics in Graph Theory Example 1

Consider the graph in Figure 55. The path a, b, c, d, e is a Hamiltonian path because it contains each vertex exactly once. It is not hard to see, however, that there is no Hamiltonian circuit for this graph. For the graph shown in Figure 56, the path A, D, C, B, A is a Hamiltonian circuit. In Figures 57 and 58, no Hamiltonian path is ◆ possible. (Verify this.)

d 5

2 e

c

A

3

Figure 55

Example 2

D

Figure 56

B

E

C

B

b a

A

C

1

Figure 57

4 6

D

Figure 58

Any complete graph Kn has Hamiltonian circuits. In fact, starting at any vertex, ◆ you can visit the other vertices sequentially in any desired order. Questions analogous to those about Euler paths and circuits can be asked about Hamiltonian paths and circuits. Is it possible to determine whether a Hamiltonian path or circuit exists? If there must be a Hamiltonian path or circuit, is there an efficient way to find it? Surprisingly, considering Theorems 1 and 2 of Section 2, the first question about Hamiltonian paths and circuits has not been completely answered and the second is still unanswered as well. However, we can make several observations based on the examples. If a graph G on n vertices has a Hamiltonian circuit, then G must have at least n edges. We now state some partial answers that say if a graph G has “enough” edges, a Hamiltonian circuit can be found. These are again existence statements; no method for constructing a Hamiltonian circuit is given.

THEOREM 1

Let G be a connected graph with n vertices, n > 2, and no loops or multiple edges. G has a Hamiltonian circuit if for any two vertices u and v of G that are not adjacent, ■ the degree of u plus the degree of v is greater than or equal to n. We omit the proof of this result, but from it we can prove the following:

Corollary 1

G has a Hamiltonian circuit if each vertex has degree greater than or equal to n/2. Proof The sum of the degrees of any two vertices is at least n2 + n2 = n, so the hypotheses ■ of Theorem 1 hold.

THEOREM 2

Let the number of edges of G be m. Then G has a Hamiltonian circuit if m ≥ 1 2 (n − 3n + 6) (recall that n is the number of vertices). 2 Proof Suppose that u and v are any two vertices of G that are not adjacent. We write deg(u) for the degree of u. Let H be the graph produced by eliminating u and v from G along with any edges that have u or v as end points. Then H has n − 2

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Topics in Graph Theory vertices and m − deg(u) − deg(v) edges (one fewer edge would have been removed if u and v had been adjacent). The maximum number of edges that H could possibly have is n−2 C2 . This happens when there is an edge connecting every distinct pair of vertices. Thus the number of edges of H is at most n−2 C2

=

(n − 2)(n − 3) 2

or

1 2 (n − 5n + 6). 2

We then have m − deg(u) − deg(v) ≤ 12 (n2 − 5n + 6). Therefore, deg(u) + deg(v) ≥ m − 12 (n2 − 5n + 6). By the hypothesis of the theorem, deg(u) + deg(v) ≥

1 1 2 (n − 3n + 6) − (n2 − 5n + 6) = n. 2 2 ■

Thus the result follows from Theorem 1. Example 3 A H

B

G

C

F

D

The converses of Theorems 1 and 2 given previously are not true; that is, the conditions given are sufficient, but not necessary, for the conclusion. Consider the graph represented by Figure 59. Here n, the number of vertices, is 8, each vertex has degree 2, and deg(u) + deg(v) = 4 for every pair of nonadjacent vertices u and v. The total number of edges is also 8. Thus the premises of Theorems 1 and 2 fail to be satisfied, but there are certainly Hamiltonian circuits for this graph. ◆ The problem we have been considering has a number of important variations. In one case, the edges may have weights representing distance, cost, and the like. The problem is then to find a Hamiltonian circuit (or path) for which the total sum of weights in the path is a minimum. For example, the vertices might represent cities; the edges, lines of transportation; and the weight of an edge, the cost of traveling along that edge. This version of the problem is often called the traveling salesperson problem.

E

Figure 59

3 Exercises In Exercises 1 through 6 (Figures 60 through 65), determine whether the graph shown has a Hamiltonian circuit, a Hamiltonian path but no Hamiltonian circuit, or neither. If the graph has a Hamiltonian circuit, give the circuit. 1. A

2.

B

6.

5.

Figure 64

b

Figure 65 a

C

D

c

d

E

Figure 61

Figure 60 3. 1 3

In Exercises 8 through 11 (Figures 66 through 69), find a Hamiltonian circuit for the graph given. 8.

4.

2

7. Give two Hamiltonian circuits in K5 that have no edges in common.

B 3

5

2

D 5

4

3

4 6

Figure 62

364

E

2 7

Figure 63

A

2

C

Figure 66

F

6

5

4

G 5

6

4 H

Topics in Graph Theory 9.

B

In Exercises 12 through 15, find a Hamiltonian circuit of minimal weight for the graph represented by the given figure.

H

2

3

3 A

D

6

2

2

5

C

E

5

6

4

3 4

F

G

A

2.6 2.1

2.4

C 4.2 3.3 2.9

2.1 2.5

5.3

1.7

(b) how many different Hamiltonian circuits, beginning at a fixed vertex, are there? 19. Prove that Kn , the complete graph on n vertices with n ≥ 3, has (n − 1)! Hamiltonian circuits.

E

2.2 D

J

3.2

H

15. Figure 69

18. For the complete graph on n vertices, Kn , n ≥ 3, (a) how many edges must a Hamiltonian circuit have?

3.6

2.8

I 1.8

3.4

14. Figure 68

17. Find a Hamiltonian circuit of minimal weight for the graph represented by Figure 67 if you must begin and end at F .

B 2.7

13. Figure 67

16. Find a Hamiltonian circuit of minimal weight for the graph represented by Figure 66 if you must begin and end at D.

Figure 67 10.

12. Figure 66

4.4

20. Give an example of a graph with at least four vertices with a circuit that is both an Euler and a Hamiltonian circuit.

F

G

21. Give an example of a graph that has an Euler circuit and a Hamiltonian circuit that are not the same.

Figure 68

22. Let G = (V, E, γ) be a graph with |V | = n that has no multiple edges. The relation R on V defined by G can be represented by a matrix, MR . Explain how to use MR∞ to determine if G is connected.

11.

15

B 12

10 C 16

20

A

10

14

G

F

Figure 69

4 4

4 2

1 5

2

Figure 70

3

5

4

2

3

12

20

12

6 3

3

D

15

E

23. Using the usual vertex labels of strings of 0’s and 1’s, give a Hamiltonian circuit for the Hasse diagram of B3 . 24. Using the usual vertex labels of strings of 0’s and 1’s, give a Hamiltonian circuit for the Hasse diagram of B4 . 25. The problem of finding a Hamiltonian circuit for the Hasse diagram of Bn is equivalent to a problem about strings of 0’s and 1’s of length n. State this problem.

Transport Networks We have previously examined several uses of labeled graphs. In this section we return to the idea of a directed graph (digraph). An important use of labeled digraphs is to model what are commonly called transport networks. Consider the labeled digraph shown in Figure 70. This might represent a pipeline that carries water from vertex 1 to vertex 6 as part of a municipal water system. The label on an edge represents the maximum flow that can be passed through that edge and is called the capacity of the edge. Many situations can be modeled in this way. For instance, Figure 70 might as easily represent an oil pipeline, a highway system, a communications network, or an electric power grid. The vertices of a network are usually called nodes and may denote pumping stations, shipping depots, relay stations, or highway interchanges. More formally, a transport network, or a network, is a connected digraph N with the following properties: (a) There is a unique node, the source, that has in-degree 0. We generally label the source node 1.

365

Topics in Graph Theory (b) There is a unique node, the sink, that has out-degree 0. If N has n nodes, we generally label the sink as node n. (c) The graph N is labeled. The label, Cij , on edge (i, j) is a nonnegative number called the capacity of the edge. For simplicity we also assume that all edges carry material in one direction only; that is, if (i, j) is in N, then (j, i) is not.

Flows The purpose of a network is to implement a flow of water, oil, electricity, traffic, or whatever the network is designed to carry. Mathematically, a flow in a network N is a function that assigns to each edge (i, j) of N a nonnegative number Fij that does not exceed Cij . Intuitively, Fij represents the amount of material passing through the edge (i, j) when the flow is F . Informally, we refer to Fij as the flow through edge (i, j). We also require that for each node other than the source and sink, the sum of the Fik on edges entering node k must be equal to the sum of the Fkj on edges leaving node k. This means that material cannot accumulate, be created, dissipate, or be lost at any node other than the source or the sink. This is called conservation of flow. A consequence of this requirement is that the sum of the flows leaving the source must equal the sum of the flows entering the sink. This sum is called the value of the flow, written value(F). We can represent a flow F by labeling each edge (i, j) with the pair (Cij , Fij ). A flow F in the network represented by Figure 70 is shown in Figure 71. Example 1

(4, 2) 1

4

(2, 1)

(5, 3)

2

Figure 71

(3, 3)

5

(4, 3)

(2, 0)

6

3

(3, 2)

(3, 2)

Example 2

In Figure 71, flow is conserved at node 4 since there are input flows of size 2 and 1, and an output flow of size 3. (Verify that flow is conserved properly at the other ◆ nodes.) Here value(F) = 5.

Maximum Flows For any network an important problem is to determine the maximum value of a flow through the network and to describe a flow that has the maximum value. For obvious reasons this is commonly referred to as the maximum flow problem. Figure 72(a) shows a flow that has value 8. Three of the five edges are carrying their maximum capacity. This seems to be a good flow function, but Figure 72(b) ◆ shows a flow with value 10 for the same network. 2

(6, 4) 1

(6, 6)

(2, 2)

(4, 4)

3 (a)

(6, 6) 4

(4, 2)

1

2

(6, 6)

(2, 0) (4, 4)

3

4 (4, 4)

(b)

Figure 72

Example 2 shows that even for a small network, we need a systematic procedure for solving the maximum flow problem. Examining the flow in Figure 72(a) shows that using the edge from node 3 to node 2 as we did was a mistake. We should reduce flow in edge (3, 2) so that we can increase it in other edges. Suppose that in some network N we have an edge (i, j) that is carrying a flow of 5 units. If we want to reduce this flow to 3 units, we can imagine that it is

366

Topics in Graph Theory (6, 4) 1

2

(2, 2) 3

Figure 73

4 (4, 2)

combined with a flow of two units in the opposite direction. Although edge (j, i) is not in N, there is no harm in considering such a virtual flow as long as it only has the effect of reducing the existing flow in the actual edge (i, j). Figure 73 displays a portion of the flow shown in Figure 72(a). The path π : 1, 2, 3, 4 is not an actual path in this network, since (2, 3) is not an actual edge. However, π is a path in the symmetric closure of the network. Moreover, if we consider a virtual flow of two units through π, the effect on the network is to increase the flows through edges (1, 2) and (3, 4) by two units and decrease the flow through edge (3, 2) by two units. Thus, the flow of Figure 72(a) becomes the flow of Figure 72(b). We now describe this improvement in general terms. Let N be a network and let G be the symmetric closure of N. Choose a path in G and an edge (i, j) in this path. If (i, j) belongs to N, then we say this edge has positive excess capacity if eij = Cij − Fij > 0. If (i, j) is not an edge of N, then we are traveling this edge in the wrong direction. In this case we say (i, j) has excess capacity eij = Fji if Fji > 0. Then increasing flow through edge (i, j) will have the effect of reducing Fji . We now give a procedure for solving a maximum flow problem.

A Maximum Flow Algorithm The algorithm we present is due to Ford and Fulkerson and is often called the labeling algorithm. The labeling referred to is an additional labeling of nodes. We have used integer capacities for simplicity, but Ford and Fulkerson show that this algorithm will stop in a finite number of steps if the capacities are rational numbers. Let N be a network with n nodes and G be the symmetric closure of N. All edges and paths used are in G. Begin with all flows set to 0. As we proceed, it will be convenient to track the excess capacities in the edges and how they change rather than tracking the increasing flows. When the algorithm terminates, it is easy to find the maximum flow from the final excess capacities. Algorithm THE LABELING ALGORITHM

Step 1 Let N1 be the set of all nodes connected to the source by an edge with positive excess capacity. Label each j in N1 with [Ej , 1], where Ej is the excess capacity e1j of edge (1, j). The 1 in the label indicates that j is connected to the source, node 1. Step 2 Let node j in N1 be the node with smallest node number and let N2 (j) be the set of all unlabeled nodes, other than the source, that are joined to node j and have positive excess capacity. Suppose that node k is in N2 (j) and (j, k) is the edge with positive excess capacity. Label node k with [Ek , j], where Ek is the minimum of Ej and the excess capacity ejk of edge (j, k). in this way, repeat this process for When all the nodes in N2 (j) are labeled the other nodes in N1 . Let N2 = j∈N1 N2 (j). Note that after Step 1, we have labeled each node j in N1 with Ej , the amount of material that can flow from the source to j through one edge and with the information that this flow came from node 1. In Step 2, previously unlabeled nodes k that can be reached from the source by a path π : 1, j, k are labeled with [Ek , j]. Here Ek is the maximum flow that can pass through π since it is the smaller of the amount that can reach j and the amount that can then pass on to k. Thus when Step 2 is finished, we have constructed two-step paths to all nodes in N2 . The label for each of these nodes records the total flow that can reach the

367

Topics in Graph Theory node through the path and its immediate predecessor in the path. We attempt to continue this construction increasing the lengths of the paths until we reach the sink (if possible). Then the total flow can be increased and we can retrace the path used for this increase. Step 3 Repeat Step 2, labeling all previously unlabeled nodes N3 that can be reached from a node in N2 by an edge having positive excess capacity. Continue this process forming sets N4 , N5 , . . . until after a finite number of steps either (i) the sink has not been labeled and no other nodes can be labeled. It can happen that no nodes have been labeled; remember that the source is not labeled. or (ii) the sink has been labeled. Step 4 In case (i), the algorithm terminates and the total flow then is a maximum flow. (We show this later.) Step 5 In case (ii) the sink, node n, has been labeled with [En , m], where En is the amount of extra flow that can be made to reach the sink through a path π. We examine π in reverse order. If edge (i, j) ∈ N, then we increase the flow in (i, j) by En and decrease the excess capacity eij by the same amount. Simultaneously, we increase the excess capacity of the (virtual) edge (j, i) by En since there is that much more flow in (i, j) to reverse. If, on the other hand, (i, j) ∈ / N, we decrease the flow in (j, i) by En and increase its excess capacity by En . We simultaneously decrease the excess capacity in (i, j) by the same amount, since there is less flow in (i, j) to reverse. We now have a new flow that is En units greater than before and ● we return to Step 1. Example 3

Use the labeling algorithm to find a maximum flow for the network in Figure 70. Solution

Figure 74 shows the network with initial capacities of all edges in G. The initial flow in all edges is zero. 4

e14 = 4 e41 = 0 1

e45 = 3

e24 = 2 e42 = 0

e12 = 5 e21 = 0

e54 = 0

5

e25 = 2 e52 = 0

2

e23 = 3

e32 = 0

e56 = 4 e65 = 0 6

3

e36 = 3 e63 = 0

Figure 74

Step 1 Starting at the source, we can reach nodes 2 and 4 by edges having excess capacity, so N1 = {2, 4}. We label nodes 2 and 4 with the labels [5, 1] and [4, 1], respectively, as shown in Figure 75. Step 2 From node 2 we can reach nodes 5 and 3 using edges with positive excess capacity. Node 5 is labeled with [2, 2] since only two additional units of flow can pass through edge (2, 5). Node 3 is labeled with [3, 2] since

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Topics in Graph Theory [4, 1] 4

e14 = 4 e41 = 0 1

e45 = 3

e24 = 2

5

e25 = 2 e52 = 0

e42 = 0

e12 = 5 e21 = 0

e54 = 0

2 [5, 1]

e23 = 3

e32 = 0

e56 = 4 e65 = 0 6

3

e36 = 3 e63 = 0

Figure 75

only 3 additional units of flow can pass through edge (2, 3). The result of this step is shown in Figure 76. We cannot travel from node 4 to any unlabeled node by one edge. Thus, N2 = {3, 5} and Step 2 is complete. [4, 1] 4

e14 = 4 e41 = 0 1

e45 = 3

e24 = 2 e42 = 0

e12 = 5 e21 = 0

e54 = 0

[2, 2] 5

e25 = 2 e52 = 0

2 [5, 1]

e23 = 3

e32 = 0

e56 = 4 e65 = 0 6

3

e36 = 3 e63 = 0

[3, 2]

Figure 76

Step 3 We repeat Step 2 using N2 . We can reach the sink from node 3 and 3 units through edge (3, 6). Thus the sink is labeled with [3, 3]. Step 5 We work backward through the path 1, 2, 3, 6 and subtract 3 from the excess capacity of each edge, indicating an increased flow through that edge, and adding an equal amount to the excess capacities of the (virtual) edges. We now return to Step 1 with the situation shown in Figure 77. 4

e14 = 4 e41 = 0 1

e45 = 3

e24 = 2 e42 = 0

e12 = 2 e21 = 3

e54 = 0

5

e25 = 2 e52 = 0

2

e23 = 0

e32 = 3

e56 = 4 e65 = 0 6

3

e36 = 0 e63 = 3

Figure 77

Proceeding as before, nodes 2 and 4 are labeled [2, 1] and [4, 1], respectively. Note that E2 is now only 2 units, the new excess capacity of edge (1, 2). Node 2 can no longer be used to label node 3, since there is no excess capacity in the edge (2, 3). But node 5 now will be labeled [2, 2]. Once again no unlabeled node can be reached from node 4, so we move

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Topics in Graph Theory to Step 3. Here we can reach node 6 from node 5 so node 6 is labeled with [2, 5]. The final result of Step 3 is shown in Figure 78, and we have increased the flow by 2 units to a total of 5 units. [4, 1] 4

e14 = 4 e41 = 0 1

e45 = 3

e24 = 2 e42 = 0

e12 = 2 e21 = 3

e54 = 0

[2, 2] 5

e25 = 2 e52 = 0

2 [2, 1]

e23 = 0

e32 = 3

e56 = 4 e65 = 0 6 [2, 5]

3

e36 = 0 e63 = 3

Figure 78

We move to Step 5 again and work back along the path 1, 2, 5, 6, subtracting 2 from the excess capacities of these edges and adding 2 to the capacities of the corresponding (virtual) edges. We return to Step 1 with Figure 79. This time Steps 1 and 2 produce the following results. Only node 4 is labeled from node 1, with [4, 1]. Node 5 is the only node labeled from node 4, with [3, 4]. Step 3 begins with Figure 80. At this point, node 5 could label node 2 using the excess capacity of edge (5, 2). (Verify that this would label node 2 with [2, 5].) However, node 5 can also be used to label the sink. The sink is labeled [2, 5] and the total flow is increased to 7 units. In Step 5, we work back along the path 1, 4, 5, 6, adjusting excess capacities. We return to Step 1 with the configuration shown in Figure 81. 4

e14 = 4 e41 = 0 1

e45 = 3

e24 = 2 e42 = 0

e12 = 0 e21 = 5

e54 = 0

5

e25 = 0 e52 = 2

2

e23 = 0

e32 = 3

e45 = 3

e54 = 0

e56 = 2 e65 = 2 6

3

e36 = 0 e63 = 3

Figure 79

[4, 1] 4

e14 = 4 e41 = 0 1

e24 = 2 e42 = 0

e12 = 0 e21 = 5

Figure 80

370

[3, 4] 5

e25 = 0 e52 = 2

2

e23 = 0

e32 = 3

e56 = 2 e65 = 2 6

3

e36 = 0 e63 = 3

Topics in Graph Theory 4

e14 = 2 e41 = 2 1

e45 = 1

e54 = 2

e24 = 2 e42 = 0

e12 = 0 e21 = 5

5

e56 = 0 e65 = 4

e25 = 0 e52 = 2

2

e23 = 0

e32 = 3

6

3

e36 = 0 e63 = 3

Figure 81

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There remains the problem of showing that the labeling algorithm produces a maximum flow. First, we define a cut in a network N as a set K of edges having the property that every path from the source to the sink contains at least one edge from K. In effect, a cut does “cut” a digraph into two pieces, one containing the source and one containing the sink. If the edges of a cut were removed, nothing could flow from the source to the sink. The capacity of a cut K, c(K), is the sum of the capacities of all edges in K. Figure 84 shows two cuts for the network given by Figure 70. Each cut is marked by a jagged line and consists of all edges touched by the jagged line. Verify that ◆ c(K1 ) = 10 and c(K2 ) = 7.

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Topics in Graph Theory If F is any flow and K is any cut, then value(F) ≤ c(K). This is true because all parts of F must pass through the edges of K, and c(K) is the maximum amount that can pass through the edges of K. Now suppose for some flow F and some cut K, value(F) = c(K); in other words, the flow F uses the full capacity of all edges in K. Then F would be a flow with maximum value, since no flow can have value bigger than c(K). Similarly, K must be a minimum capacity cut, because every cut must have capacity at least equal to value(F). From this discussion we conclude the following. THEOREM 1 The Max Flow Min Cut Theorem

A maximum flow F in a network has value equal to the capacity of a minimum cut ■ of the network.

Example 5

The minimum cut corresponding to the maximum flow found in Example 3 is ◆ K = {(5, 6), (3, 6)} with c(K) = 7 = value(F).

We now show that the labeling algorithm results in a maximum flow by finding a minimum cut whose capacity is equal to the value of the flow. Suppose that the algorithm has been run and has stopped at Step 4. Then the sink has not been labeled. Divide the nodes into two sets, M1 and M2 , where M1 contains the source and all nodes that have been labeled, and M2 contains all unlabeled nodes, other than the source. Let K consist of all edges of the network N that connect a node in M1 with a node in M2 . Any path π in N from the source to the sink begins with a node in M1 and ends with a node in M2 . If i is the last node in π that belongs to M1 and j is the node that follows i in the path, then j belongs to M2 and so by definition (i, j) is in K. Therefore, K is a cut. Now suppose that (i, j) is an edge in K, so that i ∈ M1 and j ∈ M2 . The final flow F produced by the algorithm must result in (i, j) carrying its full capacity; otherwise, we could use node i and the excess capacity to label j, which by definition is not labeled. Thus the value of the final flow of the algorithm is equal to the capacity c(K), and so F is a maximum flow.

4 Exercises In Exercises 1 through 4 (Figures 85 through 88), label the network in the given figure with a flow that conserves flow at each node, except the source and the sink. Each edge is labeled with its maximum capacity. 1.

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Matching Problems The definition of a transport network can be extended, and the concept of a maximal flow in a network can be used to model situations that, at first glance, do not seem to be network problems. We consider two examples in this section. The first example is to allow a network to have multiple sources or multiple sinks as many real networks do. In the network N shown in Figure 92, nodes 1, 2, and 3 are all sources, and nodes 6 and 7 are sinks. For example, the sources could carry water from different pumping stations on a lake to two tanks (nodes 6 and 7) that supply two towns’ drinking water. In a case like this, we want to maximize the flow from all sources taken together to all the sinks taken together. As before, a flow F consists of the quantities Fij assigned to each edge (i, j). We require that flow be conserved at each node that is not a source or a sink and that the flow in each edge not exceed the capacity of that edge. The value of the flow F , value(F), is the sum of the flows in all edges that come from any source. It is not hard to show that this value must equal the sum of flows in all edges that lead to any sink. To find the maximal flow in a general network N, we change the problem into a single-source, single-sink network problem by enlarging N to N as follows. We add two nodes, which we call a and b. Node a is the source for N and is connected to all nodes that are sources in N. Similarly, node b is the sink for N , and all nodes that were sinks in

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Topics in Graph Theory N are connected to it. Nodes a and b are called, respectively, a supersource and a supersink. To complete the new network we set the capacities of all new edges to some number C0 that is so large that no real flow through N can pass more than C0 units through any of the new edges. Some authors choose C0 to be infinity. Example 1

An enlarged network N for the network of Figure 92 is shown in Figure 93. We have set C0 to be 11, the sum of all capacities leaving sources of N. ◆ 1

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Find the maximal flow for the network N given in Figure 92. Solution

Applying the labeling algorithm to the enlarged network N given in Figure 93 produces the result shown in Figure 94. (Verify.) The value of this flow is 9. This ◆ is the maximal flow from nodes 1, 2, and 3 to nodes 6 and 7 in N. 1 (11, 2)

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The Matching Problem We consider now an important class of problems, matching problems, that can also be modeled by network flows. We begin with two finite sets A and B and a relation R from A to B. A matching function M is a one-to-one function from a subset of A to a subset of B. We say a is matched with b if M(a) = b. A matching function M is compatible with R if M ⊆ R; that is, if M(a) = b, then a R b. Example 3

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Let A be a set of girls and B a set of boys attending a school dance. Define R by a R b if and only if a knows b. A matching function M is defined from A to B by M(a) = b if a and b dance the third dance together. M is compatible with R if ◆ each girl knows her partner for the third dance.

Topics in Graph Theory Example 4

Let A = {s1 , s2 , s3 , s4 , s5 } be a set of students working on a research project and B = {b1 , b2 , b3 , b4 , b5 } be a set of reference books on reserve in the library for the project. Define R by si R bk if and only if student si wants to sign out book bk . A matching of students to books would be compatible with R if each student is ◆ matched with a book that he or she wants to sign out. Given any relation R from A to B, a matching M that is compatible with R is called maximal if its domain is as large as possible and is complete if its domain is A. In general, a matching problem is, given A, B, and R, find a maximal matching from A to B that is compatible with R. Somewhat surprisingly, matching problems can be solved using networks. We create a network to model the situation by using the elements of A as sources and the elements of B as sinks. There is a directed edge (a, b) if and only if a R b. To complete the network, each edge is assigned capacity 1.

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Every flow in N provides a matching of students to books that is compatible with R, and every compatible matching arises from a flow in N. To see this, suppose that F is a flow in N. If the flow into node sm is 1, then the flow out must be 1, so the flow from sm (if any) can go to only one node, say bn . Similarly, flow can enter a node bn from at most one sm since the flow out of bn is 1. We match sm to bn if and only if there is flow between these two nodes. The matching function M that we construct is clearly compatible with R. Conversely, if we have an R-compatible matching M, we can define a flow F by letting the flow be 1 between any two matched nodes, 1 from x to each student matched with a book, and from each matched book to y, and 0 on all other edges. This flow yields the matching M again. Hence, there is a one-to-one correspondence between flows in N and matching functions compatible with R. This means that we can use the labeling algorithm to solve the matching problem by constructing N and N as in this example.

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Topics in Graph Theory We saw that no complete matching is possible in the student-book example, but whether or not there is a complete matching for a particular problem is generally not obvious. A condition that will guarantee the existence of a complete matching was first found for the matching problem for a set of men and a set of women with the relation “suitable for marriage.” Because the question posed was “Is it possible for each man to marry a suitable woman?” this is often referred to as the marriage problem. The notion of an R-relative set defined in Section 2 simplifies the statement of the next theorem. THEOREM 1 Hall’s Marriage Theorem

Let R be a relation from A to B. Then there exists a complete matching M if and only if for each X ⊆ A, |X| ≤ |R(X)|. Proof If a complete matching M exists, then M(X) ⊆ R(X) for every subset X of A. But M is one to one, so |X| = |M(X)| ≤ |R(X)|. Conversely, suppose that for any X ⊆ A, |X| ≤ |R(X)|. Construct the network N that corresponds to R. Suppose |A| = n. We want to show that there is a flow in N with value n, which will correspond to a complete matching. We know by the Max Flow Min Cut Theorem that it is sufficient to show that the minimal cut in N has value n. A typical situation is shown in Figure 97, with the wavy line representing a cut in N.

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Remember that all edges are directed to the right and have capacity 1. We say that two vertices of N are connected if there is a path from one to the other; if not, the two vertices are disconnected. With this language, a cut in N is a set of edges whose removal will make the supersource x and the supersink y disconnected. Let n = |A| and p = |B|. Suppose K is a minimal cut of N. Divide the edges of K into three sets: S1 contains the edges that begin at x, S2 contains the edges that correspond to pairs in R, S3 contains the edges that end with y. Consider removing the edges of K one set at a time. Suppose that |S1 | = k so x is connected to k elements of A. Let A1 be this subset of A. When the edges in S1 are removed, no path from x to y can pass through an element of A1 . Since K is minimal, we can suppose that no edges in K begin with elements in A1 . Let A2 = A − A1 so |A2 | = n−k. Let B2 = R(A2 ). Thus each element of B2 labels the terminal node of an edge that begins in A2 . Since the supersource x is connected to each element in A2 , x is also connected to each element in B2 . We know that |A2 | ≤ |R(A2 )| = |B2 | so there are at least n − k elements in B2 . Let |S2 | = r. Each of these edges connects some element a ∈ A2 to some element b ∈ B2 . Removing an edge in S2 may or may not disconnect x from any element of B2 , but it certainly cannot disconnect x from more than one element

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Topics in Graph Theory of B2 . Thus, when the edges in S2 are removed, x is still connected to at least (n − k) − r elements of B2 . These elements must then be disconnected from y to ensure there is no path from x to y, and so S3 must contain at least (n − k) − r edges. Hence the capacity of K = the number of edges in K = |S1 | + |S2 | + |S3 | ≥ k + r + ((n − k) − r) = n. Since the cut that removes all edges beginning at x has capacity n, we see that ■ the minimal cut, and therefore the maximum flow, is exactly n. Note that this is an example of an existence theorem; no method is given for finding a complete matching, if one exists. Example 6

Let MR be the matrix of a marriage suitability relation between five men and five women. Can each man marry a suitable woman? ⎤ ⎡ 1 1 0 0 0 ⎢0 0 0 1 1⎥ ⎥ ⎢ MR = ⎢ 1 0 1 0 0 ⎥ ⎣0 0 1 0 1⎦ 0 1 0 1 0 Solution

We could construct an enlarged network N to model this problem and apply the labeling algorithm in order to answer the question. Instead, we use Hall’s marriage theorem. Note that each man considers exactly two women suitable and each woman is considered suitable by exactly two men. Consider the network that represents this problem. Let S be any subset of the men and E be the set of edges that begin in S. Clearly |E| = 2|S|. Each edge in E must terminate in a node of R(S). But we know the number of edges terminating at elements of R(S) is exactly 2|R(S)|. Thus, 2|S| ≤ 2|R(S)|, and so |S| ≤ |R(S)|. By Hall’s marriage theorem, ◆ a complete match is possible.

5 Exercises 1. For Example 5, find a different maximal matching. Is it possible to find a maximal matching where s5 is matched with a book?

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Topics in Graph Theory In Exercises 6 through 10, the matrix MR for a relation from A to B is given. Find a maximal matching for A, B, and R. ⎡ ⎤ ⎡ ⎤ 1 0 1 0 1 0 1 0 ⎢0 0 1 0⎥ ⎢0 1 0 1⎥ 6. ⎣ 7. ⎣ 1 0 0 0⎦ 1 0 0 1⎦ 1 1 0 1 0 1 1 0 ⎡ ⎤ ⎡ ⎤ 1 0 1 0 1 1 0 1 0 1 ⎢0 1 1 0 0⎥ ⎢0 1 1 0 1⎥ ⎢ ⎥ ⎢ ⎥ 8. ⎢ 0 0 1 0 0 ⎥ 9. ⎢ 1 0 1 1 0 ⎥ ⎣1 1 0 1 1⎦ ⎣1 1 0 1 0⎦ 0 0 1 0 0 0 1 0 1 1 ⎡ ⎤ 1 0 1 0 1 1 ⎢0 1 1 0 1 0⎥ ⎢ ⎥ ⎢1 0 1 1 0 0⎥ 10. ⎢ ⎥ ⎢1 1 0 1 0 1⎥ ⎣0 1 0 1 1 1⎦ 0 0 1 0 0 1 11. Which of the matchings found in Exercises 6 through 10 are complete matchings? 12. Ann, Bing, Carmen, D’Andrea, Emilio, and Henri have inherited six paintings (of roughly equal value). Each heir has written on a slip of paper the letters that identify the paintings he or she would like to have. Ann: E; Bing: B, D; Carmen: A, E; D’Andrea: B, D; Emilio: C; Henri: A, D, F. As the executor of the estate, can you give each heir a painting he or she wants using the information on the slips of paper? If so, give the matching of heir to painting. 13. Eight friends are taking a dog-sledding trip with four sleds. Of course, not everyone is a good partner for everyone else. Here is a list of the friends each with the people with whom they can be paired. Sam: Jane, Gail Jane: Sam, Kip, Rufus Kip: Kirk, Jane, Gail Gail: Sam, Kip, Rufus Homer: Kirk, Rufus, Stacey Stacey: Homer, Kirk Kirk: Kip, Homer, Stacey Rufus: Sam, Homer, Jane, Gail (a) Give a set of four good partnerships for the trip. (b) A mistake was made and you’ve been given the list of people with whom people cannot be paired. Is a matching of good partners still possible? If so, give one.

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14. Let R be a relation from A to B with |A| = |B| = n. Prove that if the number of ones in each column of MR is k and the number of ones in each row of MR is k, then there is a complete matching for A, B, and R. 15. Let R be a relation from A to B with |A| = |B| = n. Let j be the maximum number of ones in any column of MR and k be the minimum number of ones in any row of MR . Prove that if j ≤ k, then there is a complete matching for A, B, and R. 16. Let R be a relation from A to B with |A| = |B| = n. Prove that a complete matching exists for A, B, and R if each node in the network corresponding to A, B, and R has degree at least n2 . (Hint: Use a result from Section 3.) Figure 95 is an example of a bipartite graph; there is a two-set partition of the set of vertices such that all edges in the graph connect members from different sets in the partition. 17. Which of the following graphs are bipartite? If the graph is bipartite, give a partition to show this. If not, explain why this is not possible. (a) 2 3 1

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Coloring Graphs Suppose that G = (V, E, γ) is a graph with no multiple edges, and C = {c1 , c2 , . . . , cn } is any set of n “colors.” Any function f : V → C is called a coloring of the graph G using n colors (or using the colors of C). For each vertex v, f(v) is the color of v. As we usually present a graph pictorially, so we also think of a coloring in the intuitive way of simply painting each vertex with a color from C.

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Topics in Graph Theory However, graph-coloring problems have a wide variety of practical applications in which “color” may have almost any meaning. For example, if the graph represents a connected grid of cities, each city can be marked with the name of the airline having the most flights to and from that city. In this case, the vertices are cities and the colors are airline names. Other examples are given later. A coloring is proper if any two adjacent vertices v and w have different colors. Example 1 w r

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Let C = {r, w, b, y} so that n = 4. Figure 103 shows a graph G properly colored with the colors from C in two different ways, one using three colors from C and one using all four. We show the colors as labels, which helps to explain why we avoid giving names to vertices. There are many ways to color this graph properly with three or four colors, but it is not hard to see that this cannot be done with two or fewer colors. (Experiment to convince yourself that this is true.) ◆ The smallest number of colors needed to produce a proper coloring of a graph G is called the chromatic number of G, denoted by χ(G). For the graph G of Figure 103, our discussion leads us to believe that χ(G) = 3. Of the many problems that can be viewed as graph-coloring problems, one of the oldest is the map-coloring problem. Consider the map shown in Figure 104.

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A coloring of a map is a way to color each region (country, state, county, province, etc.) so that no two distinct regions sharing a common border have the same color. The map-coloring problem is to find the smallest number of colors that can be used. We can view this problem as a proper graph-coloring problem as follows. Given a map M, construct a graph GM with one vertex for each region and an edge connecting any two vertices whose corresponding regions share a common boundary. Then the proper colorings of GM correspond exactly to the colorings of M. Example 2

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The map-coloring problem dates back to the mid-nineteenth century and has been an active subject of research at various times since then. A conjecture was that four colors are always enough to color any map drawn on a plane. This conjecture was proved to be true in 1976 with the aid of computer computations performed on almost 2000 configurations of graphs. There is still no proof known that does not depend on computer checking. The graph corresponding to a map is an example of a planar graph, meaning that it can be drawn in a plane so that no edges cross except at vertices. Figure 105 illustrates the planarity of the graph corresponding to the map of Figure 104. The complete graph K5 is not planar, so graph-coloring problems are more

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Topics in Graph Theory general than map-coloring problems. Later, we will see that five colors are required to color K5 . Graph-coloring problems also arise from counting problems. Example 3

Fifteen different foods are to be held in refrigerated compartments within the same refrigerator. Some of them can be kept together, but other foods must be kept apart. For example, spicy meats and cheeses should be kept separate from bland meats and vegetables. Apples, eggs, and onions should be isolated or they will contaminate many other foods. Butter, margarine, and cream cheese can be kept together, but must be separated from foods with strong odors. We can construct a graph G as follows. Construct one vertex for each food and connect two with an edge if they must be kept in separate compartments in the refrigerator. Then χ(G) is the smallest number of separate containers needed to store the 15 foods properly. ◆ A similar method could be used to calculate the minimum number of laboratory drawers needed to store chemicals if we need to separate chemicals that will react with one another if stored close to each other.

Chromatic Polynomials Closely related to the problem of computing χ(G) is the problem of computing the total number of different proper colorings of a graph G using a set C = {c1 , c2 , . . . , cn } of colors. If G is a graph and n ≥ 0 is an integer, let PG (n) be the number of ways to color G properly using n or fewer colors. Since PG (n) is a definite number for each n, we see that PG is a function. What may not be obvious is that PG is a polynomial in n. This can be shown in general and is clearly seen in the examples of this section. We call PG the chromatic polynomial of G. Example 4

Consider the linear graph L4 defined in Section 1 and shown in Figure 9. Suppose that we have x colors. The first vertex can be colored with any color. No matter how this is done, the second can be colored with any color that was not chosen for vertex 1. Thus there are x − 1 choices for vertex 2. Vertex 3 can then be colored with any of the x − 1 colors not used for vertex 2. A similar result holds for vertex 4. By the multiplication principle of counting, the total number of proper colorings ◆ is x(x − 1)3 . Thus PL4 = x(x − 1)3 . We can see from Example 4 that PL4 (0) = 0, PL4 (1) = 0, and PL4 (2) = 2. Thus there are no proper colorings of L4 using zero colors (obviously) or one color, and there are two using two colors. From this we see that χ(L4 ) = 2. This connection holds in general, and we have the following principle: If G is a graph with no multiple edges, and PG is the chromatic polynomial of G, then χ(G) is the smallest positive integer x for which PG (x) = 0. An argument similar to the one given in Example 4 shows that for Ln , n ≥ 1, PLn (x) = x(x − 1)n−1 . Thus, χ(Ln ) = 2 for every n.

Example 5

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For any n ≥ 1, consider the complete graph Kn defined in Section 1. Suppose that we again have x colors to use in coloring Kn . If x < n, no proper coloring is possible. So let x ≥ n. Vertex v1 can be colored with any of the x colors. For vertex v2 , only x − 1 remain since v2 is connected to v1 . We can only color v3 with x − 2 colors, since v3 is connected to v1 and v2 and so the colors of v1 and v2 cannot be used again. Similarly, only x − 3 colors remain for v4 and so on. Again using the multiplication principle of counting, we find that PKn (x) =

Topics in Graph Theory x(x − 1)(x − 2) · · · (x − n + 1). This shows that χ(Kn ) = n. Note that if there are at least n colors, then PKn (x) is the number of permutations of x objects taken n at ◆ a time (see Section 3.1). Suppose that a graph G is not connected and that G1 and G2 are two components of G. This means that no vertex in G1 is connected to any vertex in G2 . Thus any coloring of G1 can be paired with any coloring of G2 . This can be extended to any number of components, so the multiplication principle of counting gives the following result. THEOREM 1

If G is a disconnected graph with components G1 , G2 , . . . , Gm , then PG (x) = PG1 (x)PG2 (x) · · · PGm (x), the product of the chromatic polynomials for each com■ ponent.

Example 6

Let G be the graph shown in Figure 6. Then G has two components, each of which is K3 . The chromatic polynomial of K3 is x(x−1)(x−2), x ≥ 3. Thus, by Theorem 1, PG (x) = x2 (x − 1)2 (x − 2)2 . We see that χ(G) = 3 and that the number of distinct ways to color G using three colors is PG (3) = 36. If x is 4, then the total ◆ number of proper colorings of G is 42 · 32 · 22 or 576.

Example 7

Consider the discrete graph Un of Section 1, having n vertices and no edges. All n components are single points. The chromatic polynomial of a single point is x, so, ◆ by Theorem 1, PUn (x) = xn . Thus χ(Dn ) = 1 as can also be seen directly. There is a useful theorem for computing chromatic polynomials using the subgraph and quotient graph constructions of Section 1. Let G = {V, E, γ} be a graph with no multiple edges, and let e ∈ E, say e = {a, b}. As in Section 1, let Ge be the subgraph of G obtained by deleting e, and let Ge be the quotient graph of G obtained by merging the end points of e. Then we have the following result.

THEOREM 2

With the preceding notation and using x colors, PG (x) = PGe (x) − PGe (x).

(1)

Proof Consider all the proper colorings of Ge . They are of two types, those for which a and b have different colors and those for which a and b have the same color. Now a coloring of the first type is also a proper coloring for G, since a and b are connected in G, and this coloring gives them different colors. On the other hand, a coloring of Ge of the second type corresponds to a proper coloring of Ge . In fact, since a and b are combined in Ge , they must have the same color there. All other vertices of Ge have the same connections as in G. Thus we have proved that ■ PGe (x) = PG (x) + PGe (x) or PG (x) = PGe (x) − PGe (x). Example 8

Let us compute PG (x) for the graph G shown in Figure 106, using the edge e. Then Ge is K3 and Ge has two components, one being a single point and the other being K3 . By Theorem 1,

e

PGe (x) = x(x(x − 1)(x − 2)) = x2 (x − 1)(x − 2), if x ≥ 2. Also,

Figure 106

PGe (x) = x(x − 1)(x − 2).

Thus, by Theorem 2, we see that PG (x) = x2 (x − 1)(x − 2) − x(x − 1)(x − 2)

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Topics in Graph Theory or

x(x − 1)2 (x − 2).

Clearly, PG (1) = PG (2) = 0, and PG (3) = 12. This shows that χ(G) = 3.

◆

Using this model allows us to use chromatic polynomials as tools for counting the number of solutions and to answer the question of whether there is a unique solution.

6 Exercises In Exercises 1 through 4 (Figures 107 through 110), construct a graph for the map given as done in Example 2. 1.

A

B

7.

8.

V1 Vn

V2

2.

C

C

B

E

A

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D E G

Figure 107

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Figure 108

Figure 113

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Figure 114

In Exercises 9 through 12, find the chromatic polynomial for the graph represented by the given figure. Confirm each of these by using the results of Exercises 5 through 8. 3.

4. NE ME VT NH

UT

CO

AZ

NM

KS

Figure 110

9. Figure 111 11. Figure 113

10. Figure 112 12. Figure 114

In Exercises 13 through 16 (Figures 115 through 118), find the chromatic polynomial PG for the given graph and use PG to find χ(G). 13.

14.

MA CT

RI

Figure 115

Figure 109

Figure 116 In Exercises 5 through 8 (Figures 111 through 114), determine the chromatic number of the graph by inspection. 5.

15.

16.

6.

Figure 117 Figure 111 Figure 112

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Figure 118

17. Find PG and χ(G) for the graph G of the map in Exercise 1. 18. Find PG and χ(G) for the graph G of the map in Exercise 3.

Topics in Graph Theory 19. Find PG and χ(G) for the graph G of the map in Exercise 4. Consider using Theorem 2 to do this. 20. Let G be the graph represented by Figure 64, in Section 3, Exercise 5. Find PG and compute χ(G). 21. Let G be a bipartite graph. What is χ(G)? Explain your answer. 22. Prove that if G is a graph where the longest cycle has odd length n and n is greater than 2, then χ(G) ≤ n + 1. 23. Prove by mathematical induction that PLn (x) = x(x − 1)n−1 ,

n ≥ 1.

24. (a) Give an example of a connected graph with five vertices that is planar. (b) Give an example of a connected graph with five vertices that is not planar. 25. Show that the graph given is planar. (b) (a)

Figure 119 Figure 120 26. The problem of constructing a final exam schedule with a minimum number of time slots so that students do not have time conflicts can be modeled as a graph-coloring problem. If the vertices represent the courses,

(a) What do the edges represent? (b) What should the colors represent? (c) What question does χ(G) answer for the graph constructed? 27. The Pet Fishe Store has 20 species of fish in stock. Since some species eat others, the store owners would like to know the minimum number of tanks needed to prevent any fish tragedies. Model this problem for them as a graphcoloring problem. 28. To model an n2 ×n2 Sudoku grid as a graph-coloring problem, choose the vertices to represent the cells of the grid. To color a vertex means to fill the corresponding cell with a digit. (a) How many vertices does the graph have? (b) In order to represent the requirements that each digit appears once in each row, column, and subgrid, what should the edges represent? (c) How many colors are needed for a solution to the puzzle? 29. (a) For the model described in Exercise 28, explain why the graph produced is regular. (b) For n = 2 what is the degree of each vertex in the model described in Exercise 28? (c) For n = 3 what is the degree of each vertex in the model described in Exercise 28?

Tips for Proofs

For statements about graphs, a typical structure for a contradiction proof is to examine the largest (or smallest) object with certain properties and show that if the statement were not true, then a larger (or smaller) object with the same properties could be constructed. This is the structure of the proof of Theorem 1(b), Section 2. Often indirect proofs are also existence proofs since they tell us only that an object with certain properties must exist, but not how to construct one. The proofs of Theorems 1 and 2 in Section 2 are of this type. Induction proofs are common in graph theory, because many theorems are stated in terms of the number of vertices or the number of edges. Analysis of all possible cases is also common; the proof of the four-color theorem that was mentioned earlier is an extreme case of this technique.

Key Ideas for Review • Graph: G = (V, E, γ), where V is a finite set of objects,

called vertices, E is a set of objects, called edges, and γ is a function that assigns to each edge a two-element subset of V . • Degree of a vertex: number of edges at the vertex • Path

• Circuit: path that begins and ends at the same vertex • Simple path or circuit • Connected graph: There is a path from any vertex to any

other vertex. • Subgraph

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Topics in Graph Theory • Euler path (circuit): path (circuit) that contains every edge •

•

• • • • •

•

• •

of the graph exactly once Theorem: (a) If a graph G has a vertex of odd degree, there can be no Euler circuit in G. (b) If G is a connected graph and every vertex has even degree, then there is an Euler circuit in G. Theorem: (a) If a graph G has more than two vertices of odd degree, then there can be no Euler path in G. (b) If G is connected and has exactly two vertices of odd degree, there is an Euler path in G. Bridge: edge whose deletion would cause the graph to become disconnected Fleury’s algorithm Hamiltonian path: path that includes each vertex of the graph exactly once Hamiltonian circuit: circuit that includes each vertex exactly once except for the first vertex, which is also the last Theorem: Let G be a graph on n vertices with no loops or multiple edges, n > 2. If for any two vertices u and v of G, the degree of u plus the degree of v is at least n, then G has a Hamiltonian circuit. Theorem: Let G be a graph on n vertices that has no loops or multiple edges, n > 2. If the number of edges in G is at least 12 (n2 − 3n + 6), then G has a Hamiltonian circuit. Transport network Capacity: maximum flow that can be passed through an edge

• Flow in a network: function that assigns a flow to each edge

of a network • Value of a flow: the sum of flows entering the sink • Labeling algorithm • Cut in a network: set of edges in a network such that every

• •

• •

• • • • •

path from the source to the sink contains at least one edge in the set. Capacity of a cut Max Flow Min Cut Theorem: A maximum flow in a network has value equal to the capacity of a minimum cut of the network Matching (function) M compatible with a relation R: oneto-one function such that M ⊆ R. Hall’s Marriage Theorem: Let R be a relation from A to B. There exists a complete matching M if and only if for each X ⊆ A, |X| ≤ |R(X)|. Coloring of a graph using n colors Proper coloring of a graph: adjacent edges have different colors. Chromatic number of a graph G, χ(G): smallest number of colors needed for a proper coloring of G Planar graph: graph that can be drawn in a plane with no crossing edges Chromatic polynomial of a graph G, PG : number of proper colorings of G in terms of the number of colors available

Chapter Self-Test v5

1. What is the relationship between a quotient graph and a quotient set? 2. What is the difference between an Euler circuit and a Hamiltonian circuit?

v1

v2

v9

v3

v4

v6

3. What is needed for a subset of vertices and a subset of edges to form a subgraph?

v7

4. How are maximum flow problems related to matching problems?

Figure 121

5. How does the chromatic polynomial of a graph aid us in finding the chromatic number of the graph?

v8

Use Figures 122 and 123 for Problems 9 and 10.

6. Draw a picture of the graph G = (V, E, γ), where V = {j, k, l, m, n}, E = {e1 , e2 , e3 , e4 , e5 }, and γ(e1 ) = {m, k}, γ(e2 ) = {j, l}, γ(e3 ) = {n, k}, γ(e4 ) = {m, j}, and γ(e5 ) = {n, l}. 7. Give an example of a graph with seven vertices and exactly three components. 8. If R is the equivalence relation defined by the partition {{v2 , v4 , v5 }, {v1 , v3 }, {v7 , v8 }, {v6 , v9 }}, find the quotient graph GR of the graph G represented by Figure 121.

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Figure 122

Figure 123

9. Tell whether the graph in the specified figure has an Euler circuit, an Euler path but no Euler circuit, or neither. (b) Figure 123 (a) Figure 122

Topics in Graph Theory 10. Tell whether the graph in the specified figure has a Hamiltonian circuit, a Hamiltonian path but no Hamiltonian circuit, or neither. (b) Figure 123 (a) Figure 122 11. (a) The graph represented by Figure 124 does not have an Euler circuit. Mark the minimal number of edges that would have to be traveled twice in order to travel every edge and return to the starting vertex. (b) For each edge you marked in part (a), add an edge between the same vertices. Use Fleury’s algorithm to find an Euler circuit for the modified version of Figure 124.

2 2

5 2

3 2 3

1

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4

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2

Figure 126 15. Find a maximal flow in the network shown in Figure 127. 2 1

3 1

4

3

3 1

2

5 2

4

Figure 124 12. Give an Euler circuit, by numbering the edges, for the graph represented by Figure 125. Begin at A.

Figure 127 16. (a) Construct a graph for the map given in Figure 128. (b) Determine the number of colors required for a proper coloring of the map. A

B

A C E

Figure 125

D

13. Give a Hamiltonian circuit, by numbering the edges, for the graph represented by Figure 125. 14. Find a Hamiltonian circuit of minimal weight for the graph represented by Figure 126.

Figure 128 17. Compute the chromatic polynomial for the graph constructed in Problem 16(a) and use it to prove the result of Problem 16(b).

Experiment 8 Suppose that there are n individuals P1 , P2 , . . . , Pn some of whom can influence each other in making decisions. If P3 influences P5 , it may or may not be true that P5 influences P3 . In Figure 1 we have drawn a digraph to describe the influence relations among the six members of a design team. Notice that the digraph has no loops; an individual does not influence herself or himself.

P3 P2

P1 P5

P6 P4

Figure 1

1. (a) Give the matrix for this relation. (b) Is there a leader in this design group? Justify your answer. The relation described by the digraph in Figure 1 is not transitive, but we can speak of two-stage influence. We say Pi has two-stage influence on Pj if there is a path of length 2 from Pi to Pj . Similarly, Pi has r-stage influence on Pj if there is a path of length r from Pi to Pj .

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Topics in Graph Theory 2. Use the matrix for the relation described by Figure 1 to determine whether Pi has two-stage influence on Pj for each ordered pair of distinct members of the design team. 3. Consider a communication network among five sites with matrix ⎡ ⎤ 0 1 0 0 0 ⎢0 0 1 0 1⎥ ⎢ ⎥ ⎢ 0 0 0 1 0 ⎥. ⎣0 0 0 0 0⎦ 0 0 1 1 0 (a) Can P3 get a message to P5 in at most two stages? (b) What is the minimum number of stages that will guarantee that every site can get a message to any other different site? (c) What is the minimum number of stages that will guarantee that every site can get a message to any site including itself? A dictionary defines a clique as a small exclusive group of people. In studying organizational structures, we often find subsets of people in which any pair of individuals is related, and we borrow the word clique for such a subset. A clique in an influence digraph is a subset S of vertices such that (1) |S| ≥ 3.

P1 P3

P2

P5 P4

Figure 2

(2) If Pi and Pj are in S, then Pi influences Pj and Pj influences Pi . (3) S is the largest subset that satisfies (2). 4. Identify all cliques in the digraph in Figure 2. If the digraph is small, cliques can be identified by inspection of the digraph. In general, it can be difficult to determine cliques using only the digraph. The algorithm CLIQUE identifies which vertices belon