SOLUTIONS MANUAL for Discrete Mathematics, 8th Edition by Richard Johnsonbaugh

SOLUTIONS MANUAL Discrete Mathematics, 8th Edition by Richard Johnsonbaugh Solutions to Selected Exercises Section 1.1 2. {2, 4} 8. A

3. {7, 10} 9. ∅

15. {2, 3, 4, 5, 6, 7, 8, 9, 10}

5. {2, 3, 5, 6, 8, 9} 11. B

12. {1, 4} 18. {n ∈ Z+ | n ≥ 6}

21. {n ∈ Z+ | n ≤ 5 or n = 2m, m ≥ 3} 25. {n ∈ Z+ | n ≤ 5 or n = 2m + 1, m ≥ 3} 29. 1

6. {1, 3, 5, 7, 9, 10} 14. {1} 19. {2n − 1 | n ∈ Z+ }

22. {2n | n ≥ 3}

24. {1, 3, 5}

27. {n ∈ Z+ | n ≥ 6 or n = 2 or n = 4}

30. 3

33. We find that B = {2, 3}. Since A and B have the same elements, they are equal.

34. Let x ∈ A. Then x = 1, 2, 3. If x = 1, since 1 ∈ Z+ and 12 < 10, then x ∈ B. If x = 2, since 2 ∈ Z+ and 22 < 10, then x ∈ B. If x = 3, since 3 ∈ Z+ and 32 < 10, then x ∈ B. Thus if x ∈ A, then x ∈ B.

Now suppose that x ∈ B. Then x ∈ Z+ and x2 < 10. If x ≥ 4, then x2 > 10 and, for these values of x, x∈ / B. Therefore x = 1, 2, 3. For each of these values, x2 < 10 and x is indeed in B. Also, for each of the values x = 1, 2, 3, x ∈ A. Thus if x ∈ B, then x ∈ A. Therefore A = B.

37. Since (−1)3 − 2(−1)2 − (−1) + 2 = 0, −1 ∈ B. Since −1 ∈ / A, A 6= B. 38. Since 32 − 1 > 3, 3 ∈ / B. Since 3 ∈ A, A 6= B.

41. Equal

42. Not equal

45. Let x ∈ A. Then x = 1, 2. If x = 1, x3 − 6x2 + 11x = 13 − 6 · 12 + 11 · 1 = 6.

Thus x ∈ B. If x = 2,

x3 − 6x2 + 11x = 23 − 6 · 22 + 11 · 2 = 6.

Again x ∈ B. Therefore A ⊆ B. 46. Let x ∈ A. Then x = (1, 1) or x = (1, 2). In either case, x ∈ B. Therefore A ⊆ B. 49. Since (−1)3 − 2(−1)2 − (−1) + 2 = 0, −1 ∈ A. However, −1 ∈ / B. Therefore A is not a subset of B. 50. Consider 4, which is in A. If 4 ∈ B, then 4 ∈ A and 4 + m = 8 for some m ∈ C. However, the only value of m for which 4 + m = 8 is m = 4 and 4 ∈ / C. Therefore 4 ∈ / B. Since 4 ∈ A and 4 ∈ / B, A is not a subset of B. Copyright c 2018 Pearson Education, Inc.

2

SOLUTIONS 53.

U A

B

54.

U A

B

56.

A

B

U

C

57.

U A B C

59.

A

B

U

C

62. 32

63. 105

65. 51

67. Suppose that n students are taking both a mathematics course and a computer science course. Then 4n students are taking a mathematics course, but not a computer science course, and 7n students are taking a computer science course, but not a mathematics course. The following Venn diagram depicts the situation: Copyright c 2018 Pearson Education, Inc.

3

SOLUTIONS '$ '$ CompSci

Math

4n

n

7n

&% &%

Thus, the total number of students is 4n + n + 7n = 12n.

The proportion taking a mathematics course is 5 5n = , 12n 12 which is greater than one-third.

69. {(a, 1), (a, 2), (b, 1), (b, 2), (c, 1), (c, 2)} 70. {(1, 1), (1, 2), (2, 1), (2, 2)}

73. {(1, a, a), (2, a, a)}

74. {(1, 1, 1), (1, 2, 1), (2, 1, 1), (2, 2, 1), (1, 1, 2), (1, 2, 2), (2, 1, 2), (2, 2, 2)} 77. Vertical lines (parallel) spaced one unit apart extending infinitely to the left and right.

79. Consider all points on a horizontal line one unit apart. Now copy these points by moving the horizontal line n units straight up and straight down for all integers n > 0. The set of all points obtained in this way is the set Z × Z. 80. Ordinary 3-space 82. Take the lines described in the instructions for this set of exercises and copy them by moving n units out and back for all n > 0. The set of all points obtained in this way is the set R × Z × Z. 84. {1, 2} {1}, {2} 85. {a, b, c} {a, b}, {c} {a, c}, {b} {b, c}, {a} {a}, {b}, {c} 88. False

89. True

91. False

92. True

94. ∅, {a}, {b}, {c}, {d}, {a, b}, {a, c}, {a, d}, {b, c}, {b, d}, {c, d}, {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d}, {a, b, c, d}. All except {a, b, c, d} are proper subsets.

95. 210 = 1024; 210 − 1 = 1023

98. B ⊆ A

99. A = U

102. The symmetric difference of two sets consists of the elements in one or the other but not both.

103. A 4 A = ∅, A 4 A = U , U 4 A = A, ∅ 4 A = A 105. The set of primes Copyright c 2018 Pearson Education, Inc.

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SOLUTIONS

Section 1.2 2. Is a proposition. Negation: 6 + 9 6= 15.

3. Not a proposition 4. Is a proposition. Negation: π 6= 3.14.

6. Is a proposition. Negation: For every positive integer n, 19340 6= n · 17.

7. Is a proposition. Negation: Audrey Meadows was not the original “Alice” in the “Honeymooners.”

9. Is a proposition. Negation: The line “Play it again, Sam” does not occur in the movie Casablanca.

10. Is a proposition. Some even integer greater than 4 is not the sum of two primes.

12. Not a proposition. The statement is neither true nor false.

13. No heads were obtained.

19. True 24.

25.

27.

21. False

p T T F F

q T F T F

(¬p ∨ ¬q) ∨ p T T T T

p T T F F

q T F T F

(p ∨ q) ∧ ¬p F F T F

p T T F F

q T F T F

(p ∧ q) ∨ (¬p ∨ q) T F T T

p T T T T F F F F

q T T F F T T F F

r T F T F T F T F

15. No heads or no tails were obtained.

22. False

28. ¬(p ∧ q) ∨ (r ∧ ¬p) F F T T T T T T Copyright c 2018 Pearson Education, Inc.

18. True

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SOLUTIONS 30. p T T T T F F F F

q T T F F T T F F

r T F T F T F T F

¬(p ∧ q) ∨ (¬q ∨ r) T F T T T T T T

32. ¬(p ∧ q). True.

33. p ∨ ¬(q ∧ r). True.

35. Lee takes computer science and mathematics.

36. Lee takes computer science or mathematics.

38. Lee takes computer science but not mathematics.

39. Lee takes neither computer science nor mathematics.

41. You do not miss the midterm exam and you pass the course.

42. You play football or you miss the midterm exam or you pass the course.

44. Either you play football and you miss the midterm exam or you do not miss the midterm exam and you pass the course.

46. It is not Monday and either it is raining or it is hot.

47. It is not the case that today is Monday or it is raining, and it is hot.

50. Today is Monday and either it is raining or it is hot, and it is hot or either it is raining or today is Monday.

51. p ∧ q

52. p ∧ ¬q

54. p ∨ q

58. (p ∨ r) ∧ q

60. (q ∨ ¬p) ∧ ¬r

65. ¬p ∧ ¬q ∧ r

66. ¬(p ∨ q ∨ ¬r)

67.

p T T F F

q T F T F

55. (p ∨ q) ∧ ¬p 62. p ∧ ¬r

57. p ∧ r ∧ q

63. p ∧ q ∧ r

p exor q F T T F

69. Inclusive-or

70. Inclusive-or

72. Exclusive-or

77. ”lung disease” -cancer 78. ”minor league” baseball team illinois -”midwest league” Copyright c 2018 Pearson Education, Inc.

73. Exclusive-or

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SOLUTIONS

Section 1.3 2. If Rosa has 160 quarter-hours of credits, then she may graduate.

3. If Fernando buys a computer, then he obtains $2000.

5. If a person gets that job, then that person knows someone who knows the boss.

6. If you go to the Super Bowl, then you can afford the ticket.

8. If a better car is built, then Buick will build it.

9. If the chairperson gives the lecture, then the audience will go to sleep.

11. If the switch is not turned properly, then the light will not be on.

13. Contrapositive of Exercise 2: If Rosa does not graduate, then she does not have 160 quarter-hours of credits.

15. False

16. False

18. False

24. Unknown

25. Unknown

31. Unknown

34. True

40. True

41. False

48. (¬p ∨ ¬r) → ¬q

19. True

27. True 35. True 44. (p ∧ r) → q

49. r → q

21. True

28. Unknown 37. True

22. True 30. Unknown

38. False

45. ¬((r ∧ ¬q) → r)

51. q → (p ∨ r)

52. (q ∧ p) → ¬r

54. If it is not raining, then it is hot and today is Monday.

55. If today is not Monday, then either it is raining or it is hot.

57. If today is Monday and either it is raining or it is hot, then either it is hot, it is raining, or today is Monday.

58. If today is Monday or (it is not Monday and it is not the case that (it is raining or it is hot)), then either today is Monday or it is not the case that (it is hot or it is raining).

60. Let p: 4 > 6 and q: 9 > 12. Given statement: p → q; true. Converse: q → p; if 9 > 12, then 4 > 6; true.

Contrapositive: ¬q → ¬p; if 9 ≤ 12, then 4 ≤ 6; true.

61. Let p: |1| < 3 and q: −3 < 1 < 3. Given statement: q → p; true. Converse: p → q; if |1| < 3, then −3 < 1 < 3; true. Contrapositive: ¬p → ¬q; if |1| ≥ 3, then either −3 ≥ 1 or 1 ≥ 3; true.

64. P 6≡ Q

65. P ≡ Q

67. P 6≡ Q

68. P ≡ Q

71. P 6≡ Q

74. Either Dale is not smart or not funny.

70. P 6≡ Q

75. Shirley will not take the bus and not catch a ride to school.

78. (a) If p and q are both false, (p imp2 q) ∧ (q imp2 p) is false, but p ↔ q is true. (b) Making the suggested change does not alter the last line of the imp2 table.

79.

p T T F F

q T F T F

¬(p ∧ q) F T T T

¬p ∨ ¬q F T T T Copyright c 2018 Pearson Education, Inc.

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SOLUTIONS

Section 1.4 2. Invalid p→q ¬r → ¬q ... r 3. Valid p↔r r ... p 5. Valid p → (q ∨ r) ¬q ∧ ¬r ... ¬p 7. Valid (p ∨ q) → (r ∨ s) p ∧ ¬r ... s 8. Invalid p→r q→s ¬(q ∧ p) ¬p ... s 11. If 4 megabytes of memory is better than no memory at all, then either we will buy a new computer or we will buy more memory. If we will buy a new computer, then we will not buy more memory. Therefore if 4 megabytes of memory is better than no memory at all, then we will buy a new computer. Invalid.

12. If 4 megabytes of memory is better than no memory at all, then we will buy a new computer. If we will buy a new computer, then we will buy more memory. Therefore, we will buy more memory. Invalid.

14. If 4 megabytes of memory is better than no memory at all, then we will buy a new computer. If we will buy a new computer, then we will buy more memory. 4 megabytes of memory is better than no memory at all. Therefore we will buy more memory. Valid.

16. If the hardware is unreliable or the output is correct, then the while loop is not faulty. If the output is correct, then the while loop is faulty. Either the for loop is faulty or the output is correct. Therefore the hardware is unreliable. Invalid.

17. If, if the for loop is faulty, then the hardware is unreliable, then the while loop is faulty. If, if the while loop is faulty, then the output is correct, then the for loop is faulty. The hardware is unreliable and the output is correct. Either the for loop is faulty or the while loop is faulty. Therefore the for loop is faulty and the while loop is faulty. Invalid.

19. If the for loop is faulty, then the while loop is faulty or the hardware is unreliable. If the while loop is faulty, then the for loop is faulty or the output is correct. Either the for loop is faulty or the while loop is not faulty. The output is not correct. Therefore the for loop is faulty or the hardware is unreliable.

Invalid.

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SOLUTIONS 21. Valid

22. Valid

24. Valid

25. Suppose that p1 , p2 , . . . , pn are all true. Since the argument p1 , p2 / ... p is valid, p is true. Since p, p3, . . . , pn are all true and the argument p, p3 , . . . , pn / ... c is valid, c is true. Therefore the argument p1 , p2 , . . . , pn / ... c is valid.

28. Modus ponens

29. Disjunctive syllogism

31. Let p denote the proposition “there is gas in the car,” let q denote the proposition “I go to the store,” let r denote the proposition “I get a soda,” and let s denote the proposition “the car transmission is defective.” Then the hypotheses are: p → q, q → r, ¬r.

From p → q and q → r, we may use the hypothetical syllogism to conclude p → r. From p → r and ¬r, we may use modus tollens to conclude ¬p. From ¬p, we may use addition to conclude ¬p ∨ s. Since ¬p ∨ s represents the proposition “there is not gas in the car or the car transmission is defective,” we conclude that the conclusion does follow from the hypotheses.

32. Let p denote the proposition “Jill can sing,” let q denote the proposition “Dweezle can play,” let r denote the proposition “I’ll buy the compact disk,” and let s denote the proposition “I’ll buy the compact disk player.” Then the hypotheses are: (p ∨ q) → r, p, s.

From p, we may use addition to conclude p ∨ q. From p ∨ q and (p ∨ q) → r, we may use modus ponens to conclude r. From r and s, we may use conjunction to conclude r ∧ s. Since r ∧ s represents the proposition “I’ll buy the compact disk and the compact disk player,” we conclude that the conclusion does follow from the hypotheses.

34. The truth table p T T F F

q T F T F

p∨q T T T F

shows that whenever p is true, p ∨ q is also true. Therefore addition is a valid argument.

35. The truth table p T T F F

q T F T F

p∧q T F F F

shows that whenever p ∧ q is true, p is also true. Therefore simplification is a valid argument.

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SOLUTIONS 37. The truth table p T T T T F F F F

q T T F F T T F F

r T F T F T F T F

p→q T T F F T T T T

q→r T F T T T F T T

p→r T F T F T T T T

shows that whenever p → q and q → r are true, p → r is also true. Therefore hypothetical syllogism is a valid argument.

38. The truth table p T T F F

q T F T F

p∨q T T T F

¬p F F T T

shows that whenever p ∨ q and ¬p are true, q is also true. Therefore disjunctive syllogism is a valid argument.

Section 1.5 2. The statement is a command, not a propositional function.

3. The statement is a command, not a propositional function.

5. The statement is not a propositional function since it has no variables.

6. The statement is a propositional function. The domain of discourse is the set of real numbers.

8. 1 divides 77. True.

9. 3 divides 77. False.

11. For some n, n divides 77. True.

12. For every n, n does not divide 77. False.

14. It is false that for every n, n divides 77. True.

15. It is false that for some n, n divides 77. False.

17. False

18. True

20. True

21. True

23. False

26. ¬P (1) ∧ ¬P (2) ∧ ¬P (3) ∧ ¬P (4)

27. ¬(P (1) ∧ P (2) ∧ P (3) ∧ P (4))

29. ¬P (1) ∨ ¬P (2) ∨ ¬P (3) ∨ ¬P (4)

30. ¬(P (1) ∨ P (2) ∨ P (3) ∨ P (4))

33. Some student is taking a math course.

34. Every student is not taking a math course.

36. It is not the case that every student is taking a math course.

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24. True

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SOLUTIONS

37. It is not the case that some student is taking a math course.

40. There is some person such that if the person is a professional athlete, then the person plays soccer. True.

41. Every soccer player is a professional athlete. False.

43. Every person is either a professional athlete or a soccer player. False.

44. Someone is either a professional athlete or a soccer player. True.

46. Someone is a professional athlete and a soccer player. True.

49. ∃x(P (x) ∧ Q(x)) 50. ∀x(Q(x) → P (x)) 54. True

55. True

57. False

58. True

60. No. The suggested replacement returns false if ¬P (d1) is true, and true if ¬P (d1) is false.

62. Literal meaning: Every old thing does not covet a twenty-something. Intended meaning: Some old thing does not covet a twenty-something. Let P (x) denote the statement “x is an old thing” and Q(x) denote the statement “x covets a twenty-something.” The intended statement is ∃x(P (x) ∧ ¬Q(x)).

63. Literal meaning: Every hospital did not report every month. (Domain of discourse: the 74 hospitals.)

Intended meaning (most likely): Some hospital did not report every month. Let P (x) denote the statement “x is a hospital” and Q(x) denote the statement “x reports every month.” The intended statement is ∃x(P (x) ∧ ¬Q(x)).

65. Literal meaning: Everyone does not have a degree. (Domain of discourse: People in Door County.)

Intended meaning: Someone does not have a degree. Let P (x) denote the statement “x has a degree.”

The intended statement is ∃x¬P (x).

66. Literal meaning: No lampshade can be cleaned. Intended meaning: Some lampshade cannot be cleaned.

Let P (x) denote the statement “x is a lampshade” and Q(x) denote the statement “x can be cleaned.”

The intended statement is ∃x(P (x) ∧ ¬Q(x)).

68. Literal meaning: No person can afford a home. Intended meaning: Some person cannot afford a home.

Let P (x) denote the statement “x is a person” and Q(x) denote the statement “x can afford a home.”

The intended statement is ∃x(P (x) ∧ ¬Q(x)).

69. The literal meaning is as Mr. Bush spoke. He probably meant: Someone in this country doesn’t agree with the decisions I’ve made. Let P (x) denote the statement “x agrees with the decisions I’ve made.”

Symbolically, the clarified statement is ∃x ¬P (x).

71. Literal meaning: Every move does not work out. Intended meaning: Some move does not work out. Let P (x) denote the statement “x is a move” and Q(x) denote the statement “x works out .” The intended statement is ∃x(P (x) ∧ ¬Q(x)).

74. Let p(x) : x is good. q(x) : x is too long. r(x) : x is short enough.

The domain of discourse is the set of movies. The assertions are Copyright c 2018 Pearson Education, Inc.

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SOLUTIONS ∀x(p(x) → ¬q(x)) ∀x(¬p(x) → ¬r(x)) p(Love Actually) q(Love Actually).

By universal instantiation, p(Love Actually) → ¬q(Love Actually).

Since p(Love Actually) is true, then ¬q(Love Actually) is also true. But this contradicts, q(Love Actually).

77. Let P (x) denote the propositional function “x is a member of the Titans,” let Q(x) denote the propositional function “x can hit the ball a long way,” and let R(x) denote the propositional function “x can make a lot of money.” The hypotheses are P (Ken), Q(Ken), ∀x Q(x) → R(x).

By universal instantiation, we have Q(Ken) → R(Ken). From Q(Ken) and Q(Ken) → R(Ken), we may use modus ponens to conclude R(Ken). From P (Ken) and R(Ken), we may use conjunction to conclude P (Ken) ∧ R(Ken). By existential generalization, we have ∃x P (x) ∧ R(x) or, in words, someone is a member of the Titans and can make a lot of money. We conclude that the conclusion does follow from the hypotheses.

78. Let P (x) denote the propositional function “x is in the discrete mathematics class,” let Q(x) denote the propositional function “x loves proofs,” and let R(x) denote the propositional function “x has taken calculus.” The hypotheses are ∀x P (x) → Q(x), ∃x P (x) ∧ ¬R(x).

By existential instantiation, we have P (d) ∧ ¬R(d) for some d in the domain of discourse. From P (d) ∧ ¬R(d), we may use simplification to conclude P (d) and ¬R(d). By universal instantiation, we have P (d) → Q(d). From P (d) → Q(d) and P (d), we may use modus ponens to conclude Q(d). From Q(d) and ¬R(d), we may use conjunction to conclude Q(d) ∧ ¬R(d). By existential generalization, we have ∃ Q(x) ∧ ¬R(x) or, in words, someone who loves proofs has never taken calculus. We conclude that the conclusion does follow from the hypotheses.

80. By definition, the proposition ∃x ∈ D P (x) is true when P (x) is true for some x in the domain of discourse. Taking x equal to a d ∈ D for which P (d) is true, we find that P (d) is true for some d ∈ D. 81. By definition, the proposition ∃x ∈ D P (x) is true when P (x) is true for some x in the domain of discourse. Since P (d) is true for some d ∈ D, ∃x ∈ D P (x) is true.

Section 1.6 2. Everyone is taller than someone.

3. Someone is taller than everyone.

The solutions for 7–20 are for Exercise 2.

7. False

8. False

10. False

11. False

16. False

17. True

19. True

20. False

13. False

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14. False

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SOLUTIONS

23. Everyone is taller than or the same height as someone.

24. Someone is taller than or the same height as everyone.

29. For every person, there is a person such that if the persons are distinct, the first is taller than the second.

30. There is a person such that, for every person, if the persons are distinct, the first is taller than the second.

35. ∀x∀yL(x, y). False.

36. ∃x∃yL(x, y). True.

41. ∀x∃yE(x) → A(x, y)

44. True

40. ∀x ¬A(x, Profesor Sandwich)

45. False

52. False

53. False

55. False

56. False

61. True

62. False

64. True

65. True

49. True

50. False

58. True

59. True

67. for i = 1 to n if (forall dj (i)) return true return false forall dj (i) { for j = 1 to n if (¬P (di , dj )) return false return true } 68. for i = 1 to n for j = 1 to n if (P (di , dj )) return true return false 70. Since the first two quantifiers are universal and the last quantifier is existential, Farley chooses x and y, after which, you choose z. If Farley chooses values that make x ≥ y, say x = y = 0, whatever value you choose for z, (z > x) ∧ (z < y) is false. Since Farley can always win the game, the quantified propositional function is false.

71. Since the first two quantifiers are universal and the last quantifier is existential, Farley chooses x and y, after which, you choose z. Whatever values Farley chooses, you can choose z to be one less than the minimum of x and y; thus making (z < x) ∧ (z < y) true. Since you can always win the game, the quantified propositional function is true.

73. Since the first two quantifiers are universal and the last quantifier is existential, Farley chooses x and y, after which, you choose z. If Farley chooses values such that x ≥ y, the proposition (x < y) → ((z > x) ∧ (z < y)) is true by default (i.e., it is true regardless of what value you choose for z). If Farley chooses values such that x < y, you can choose z = (x + y)/2 and again the proposition (x < y) → ((z > x) ∧ (z < y)) is true. Since you can always win the game, the quantified propositional function is true.

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SOLUTIONS

75. The proposition must be true. P (x, y) is true for all x and y; therefore, no matter which value for x we choose, the proposition ∀yP (x, y) is true.

76. The proposition must be true. Since P (x, y) is true for all x and y, we may choose any values for x and y to make P (x, y) true.

78. The proposition can be false. Let N denote the set of persons James James, Terry James, and Lee James; let the domain of discourse be N × N ; and let P (x, y) be the statement “x’s first name is the same as y’s last name.” Then ∃x∀yP (x, y) is true, but ∀x∃yP (x, y) is false.

79. The proposition must be true. Since ∃x∀yP (x, y) is true, there is some value for x for which ∀yP (x, y) is true. Choosing any value for y whatsoever makes P (x, y) true. Therefore ∃x∃yP (x, y) is true.

81. The proposition can be false. Let P (x, y) be the statement x > y and let the domain of discourse be Z+ × Z+ . Then ∃x∃yP (x, y) is true, but ∀x∃yP (x, y) is false.

82. The proposition can be false. Let P (x, y) be the statement x > y and let the domain of discourse be Z+ × Z+ . Then ∃x∃yP (x, y) is true, but ∃x∀yP (x, y) is false.

84. The proposition can be true. Let P (x, y) be the statement x ≤ y and let the domain of discourse be Z+ × Z+ . Then ∀x∀yP (x, y) is false, but ∃x∀yP (x, y) is true.

85. The proposition can be true. Let P (x, y) be the statement x ≤ y and let the domain of discourse be Z+ × Z+ . Then ∀x∀yP (x, y) is false, but ∃x∃yP (x, y) is true.

87. The proposition can be true. Let N denote the set of persons James James, Terry James, and Lee James; let the domain of discourse be N × N ; and let P (x, y) be the statement “x’s first name is different from y’s last name.” Then ∀x∃yP (x, y) is false, but ∃x∀yP (x, y) is true.

88. The proposition can be true. Let P (x, y) be the statement x > y and let the domain of discourse be Z+ × Z+ . Then ∀x∃yP (x, y) is false, but ∃x∃yP (x, y) is true.

90. The proposition can be true. Let P (x, y) be the statement x < y and let the domain of discourse be Z+ × Z+ . Then ∃x∀yP (x, y) is false, but ∀x∃yP (x, y) is true.

91. The proposition can be true. Let P (x, y) be the statement x ≤ y and let the domain of discourse be Z × Z. Then ∃x∀yP (x, y) is false, but ∃x∃yP (x, y) is true.

93. ∀x ∃y P (x, y) must be false. Since ∃x ∃y P (x, y) is false, for every x and for every y, P (x, y) is false.

Choose x = x0 in the domain of discourse. For this choice of x, P (x, y) is false for every y. Therefore ∀x ∃y P (x, y) is false.

94. ∃x ∀y P (x, y) must be false. Since ∃x ∃y P (x, y) is false, for every x and for every y, P (x, y) is false.

Choose y = y0 in the domain of discourse. Now, for any choice of x, P (x, y) is false for y = y0 . Therefore ∃x ∀y P (x, y) is false.

96. Not equivalent. Let P (x, y) be the statement x > y and let the domain of discourse be Z+ × Z+ . Then ¬(∀x∃yP (x, y)) is true, but ∀x¬(∃yP (x, y)) is false.

97. Equivalent by De Morgan’s law 100. ∃ε > 0 ∀δ > 0 ∃x ((0 < |x − a| < δ) ∧ (|f(x) − L| ≥ ε)) 101. ∀L ∃ε > 0 ∀δ > 0 ∃x ((0 < |x − a| < δ) ∧ (|f(x) − L| ≥ ε)) 102. Literal meaning: No school may be right for every child. Intended meaning: Some school may not be right for some child. Let P (x, y) denote the statement “school x is right for child y.” The intended statement is ∃x∃y¬P (x, y).

Copyright c 2018 Pearson Education, Inc.

14

SOLUTIONS

Problem-Solving Corner: Quantifiers 1. The statement of Example 1.6.6 is ∀x∃y(x + y = 0).

As was pointed out in Example 1.6.6, this statement is true. Now ∀x∀y(x + y = 0) is false; a counterexample is x = y = 1. Also ∃x∀y(x + y = 0) is false since, given any x, if y = 1 − x, then x + y 6= 0.

2. Yes; the statement ∀m∃n(m < n) with domain of discourse Z × Z of Example 1.6.1 also solves problems (a) and (b).

Section 2.1 2. For all x, for all y, x + y = y + x.

3. An isosceles trapezoid is a trapezoid with equal legs.

5. The medians of any triangle intersect at a single point.

6. If 0 < x < 1 and ε > 0, there exists a positive integer n satisfying xn < ε.

8. Let m and n be odd integers. Then there exist k1 and k2 such that m = 2k1 + 1 and n = 2k2 + 1. Now m + n = (2k1 + 1) + (2k2 + 1) = 2(k1 + k2 + 1).

Therefore, m + n is even.

9. Let m and n be even integers. Then there exist k1 and k2 such that m = 2k1 and n = 2k2 . Now mn = (2k1 )(2k2 ) = 2(2k1 k2 ).

Therefore, mn is even.

11. Let m be an odd integer and n be an even integer. Then there exist k1 and k2 such that m = 2k1 + 1 and n = 2k2. Now mn = (2k1 + 1)(2k2 ) = 2(2k1 k2 + k2 ).

Therefore, mn is even.

12. Let m and n be integers such that m and m + n are even. Then there exist k1 and k2 such that m = 2k1 and m + n = 2k2 . Now n = (m + n) − m = 2k2 − 2k1 = 2(k2 − k1 ).

Therefore, n is even.

14. Let x and y be rational numbers. Then there exist integers m1 , n1 , m2 , n2 such that x = m1 /n1 and y = m2 /n2 . Now xy = (m1 m2 )/(n1 n2 ). Therefore xy is rational.

15. Let x be a nonzero rational number. Then there exist integers m 6= 0 and n 6= 0 such that x = m/n.

Now 1/x = n/m. Therefore 1/x is rational.

Copyright c 2018 Pearson Education, Inc.

15

SOLUTIONS 17. Let m = 3k1 + 2 and n = 3k2 + 2 be integers of the prescribed from. Then mn = 9k1 k2 + 6k1 + 6k2 + 4 = 3(3k1 k2 + 2k1 + 2k2 + 1) + 1 is of the form 3k3 + 1, where k3 = 3k1 k2 + 2k1 + 2k2 + 1.

19. x · 0 + 0 = x · 0 because b + 0 = b for all real numbers b = x · (0 + 0) because b + 0 = b for all real numbers b = x · 0 + x · 0 because a(b + c) = ab + ac for all real numbers a, b, c

Taking a = c = x · 0 and b = 0, the preceding equation becomes a + b = a + c; therefore, 0 = b = c = x · 0.

20. We must have X = Y . To prove this, suppose that x ∈ X. Since Y is nonempty, choose y ∈ Y . Then (x, y) ∈ X × Y . Since X × Y = Y × X, (x, y) ∈ Y × X. Therefore x ∈ Y . Similarly, if x ∈ Y , then x ∈ X. Thus X = Y .

22. Let x ∈ X. Then x ∈ X ∪ Y . Therefore X ⊆ X ∪ Y .

23. Let x ∈ X ∪ Z. Then x ∈ X or x ∈ Z. If x ∈ X, since X ⊆ Y , x ∈ Y . Therefore x ∈ Y ∪ Z. If x ∈ Z, then x ∈ Y ∪ Z. In either case, x ∈ Y ∪ Z. Therefore X ∪ Z ⊆ Y ∪ Z. 25. Let x ∈ Z − Y . Then x ∈ Z and x ∈ / Y . Now x cannot be in X, for if x ∈ X, since X ⊆ Y , then x ∈ Y , which is not the case. Since x ∈ Z and x ∈ / X, x ∈ Z − X. Therefore Z − Y ⊆ Z − X. 26. Let x ∈ Y − (Y − X). Then x ∈ Y and x ∈ / Y − X. Since x ∈ Y , we must have x ∈ X (if x ∈ / X, we would have x ∈ Y − X). Therefore Y − (Y − X) ⊆ X. Now let x ∈ X. Then x ∈ / Y − X. Since X ⊆ Y , x ∈ Y . Thus x ∈ Y − (Y − X). Therefore X ⊆ Y − (Y − X). We have shown that Y − (Y − X) = X. 28. Let Z ∈ P(X) ∪ P(Y ). Then Z ∈ P(X) or Z ∈ P(Y ). If Z ∈ P(X), then Z is a subset of X and, thus, Z is also a subset of X ∪ Y . Therefore Z ∈ P(X ∪ Y ). Similarly, if Z ∈ P(Y ), Z ∈ P(X ∪ Y ). In either case, Z ∈ P(X ∪ Y ). Therefore P(X) ∪ P(Y ) ⊆ P(X ∪ Y ).

29. Let Z ∈ P(X ∩ Y ). Then Z is a subset of X ∩ Y . Therefore Z is a subset of X and a subset of Y . Thus Z ∈ P(X) ∩ P(Y ). We have proved that P(X ∩ Y ) ⊆ P(X) ∩ P(Y ).

Let Z ∈ P(X) ∩ P(Y ). Then Z ∈ P(X) and Z ∈ P(Y ). Since Z ∈ P(X), Z is a subset of X. Since Z ∈ P(Y ), Z is a subset of Y . Since Z is a subset of X and Y , Z is a subset of X ∩Y . Thus Z ∈ P(X ∩Y ).

Therefore P(X) ∩ P(Y ) ⊆ P(X ∩ Y ). It follows that P(X ∩ Y ) = P(X) ∩ P(Y ).

31. Let X = {a} and Y = {b}. Then P(X) = {∅, {a}},

P(Y ) = {∅, {b}},

so P(X) ∪ P(Y ) = {∅, {a}, {b}}.

Since X ∪ Y = {a, b},

P(X ∪ Y ) = {∅, {a}, {b}, {a, b}}.

Now {a, b} ∈ P(X ∪ Y ), but {a, b} ∈ / P(X) ∪ P(Y ). Therefore P(X ∪ Y ) ⊆ P(X) ∪ P(Y ) is false in general.

Copyright c 2018 Pearson Education, Inc.

16 32.

SOLUTIONS (X ∩ Y ) − (X ∩ Z)

= =

(X ∩ Y ) ∩ (X ∩ Z) (X ∩ Y ) ∩ (X ∪ Z)

=

((X ∩ Y ) ∩ X) ∪ ((X ∩ Y ) ∩ Z)

=

((Y ∩ X) ∩ X) ∪ ((X ∩ Y ) ∩ Z)

=

(Y ∩ (X ∩ X)) ∪ (X ∩ (Y ∩ Z))

=

(Y ∩ ∅) ∪ (X ∩ (Y ∩ Z))

=

∅ ∪ (X ∩ (Y ∩ Z))

=

(X ∩ (Y ∩ Z)) ∪ ∅

=

X ∩ (Y ∩ Z)

=

X ∩ (Y − Z)

[A − B = A ∩ B] [De Morgan’s law; Theorem 1.1.21, part (k)] [Distributive law; Theorem 1.1.21, part (c)] [Commutative law; Theorem 1.1.21, part (b)] [Associative law; Theorem 1.1.21, part (a)] [Complement law; Theorem 1.1.21, part (e)] [Bound law; Theorem 1.1.21, part (g)] [Commutative law; Theorem 1.1.21, part (b)] [Identity law; Theorem 1.1.21, part (d)] [A − B = A ∩ B]

34. False. Let X = {a} and Y = Z = {b}. Then X ∪ (Y − Z) = {a},

(X ∪ Y ) − (X ∪ Z) = ∅.

35. True. Y − X = Y ∩ X = Y ∪ X = Y ∪ X = X ∪ Y .

36. False. Let X = {a} and Y = Z = {b}. Then X − (Y ∪ Z) = {a},

(X − Y ) ∪ Z = {a, b}.

38. False. Let X = {a}, Y = {b}, and U = {a, b}. Then X − Y = {b},

Y − X = {a}.

40. True. Let x ∈ (X ∩ Y ) ∪ (Y − X). Now either x ∈ X ∩ Y or x ∈ Y − X. In either case, x ∈ Y . Therefore (X ∩ Y ) ∪ (Y − X) ⊆ Y .

Now suppose that x ∈ Y . Either x ∈ X or x ∈ / X. If x ∈ X, then x ∈ X ∩Y . Thus x ∈ (X ∩Y )∪(Y −X).

If x ∈ / X, then x ∈ Y − X. Again x ∈ (X ∩ Y ) ∪ (Y − X). Thus Y ⊆ (X ∩ Y ) ∪ (Y − X). Therefore (X ∩ Y ) ∪ (Y − X) = Y .

41. True. Let a ∈ X × (Y ∪ Z). Then a = (x, y) where x ∈ X and y ∈ Y ∪ Z. Now y ∈ Y or y ∈ Z. If y ∈ Y , then a = (x, y) ∈ X × Y . Thus a ∈ (X × Y ) ∪ (X × Z). If y ∈ Z, then a = (x, y) ∈ X × Z. Again a ∈ (X × Y ) ∪ (X × Z). Therefore X × (Y ∪ Z) ⊆ (X × Y ) ∪ (X × Z).

Now suppose that a ∈ (X × Y ) ∪ (X × Z). Then either a ∈ X × Y or a ∈ X × Z. If a ∈ X × Y , then a = (x, y) where x ∈ X and y ∈ Y . In particular, y ∈ Y ∪ Z. Thus a = (x, y) ∈ X ×(Y ∪ Z). If a ∈ X ×Z, then a = (x, z) where x ∈ X and z ∈ Z. In particular, z ∈ Y ∪ Z. Thus a = (x, z) ∈ X × (Y ∪ Z).

Therefore (X × Y ) ∪ (X × Z) ⊆ X × (Y ∪ Z). We have proved that X × (Y ∪ Z) = (X × Y ) ∪ (X × Z).

43. True. Let a ∈ X × (Y − Z). Then a = (x, y), where x ∈ X and y ∈ Y − Z. Thus y ∈ Y and y ∈ / Z and, so, (x, y) ∈ X × Y and (x, y) ∈ / X × Z. Therefore a = (x, y) ∈ (X × Y ) − (X × Z). We have shown that X × (Y − Z) ⊆ (X × Y ) − (X × Z).

Now suppose that a ∈ (X × Y ) − (X × Z). Then a ∈ X × Y and a ∈ / X × Z. Thus a = (x, y), where x ∈ X, y ∈ Y , and y ∈ / Z. Therefore a = (x, y) ∈ X × (Y − Z). We have shown that (X × Y ) − (X × Z) ⊆ X × (Y − Z). It follows that X × (Y − Z) = (X × Y ) − (X × Z).

Copyright c 2018 Pearson Education, Inc.

17

SOLUTIONS 44. False. Take X = {1, 2}, Y = {1}, Z = {2}. Then Y × Z = {(1, 2)},

X − Y = {2},

X − Z = {1}.

Thus X − (Y × Z) = {1, 2} and (X − Y ) × (X − Z) = {(2, 1)}.

47–56. Argue as in the proof given in the book of the first associative law [Theorem 1.1.21, part (a)].

58. By definition (A 4 B) 4 A = [(A 4 B) ∪ A] − [(A 4 B) ∩ A].

Show that (A 4 B) ∪ A = A ∪ B

and

(A 4 B) ∩ A = A ∩ B.

The statement then follows easily.

59. The statement is true. We first prove that A ⊆ B. Let x ∈ A.

We divide the proof into two cases. First, we consider the case that x ∈ C. Then x 6∈ A 4 C. Therefore x 6∈ B 4 C. Thus x ∈ B (since if x 6∈ B, then x ∈ B 4 C).

Next, we consider the case that x 6∈ C. Then x ∈ A 4 C. Therefore x ∈ B 4 C. Thus x ∈ B. In either case, x ∈ B, and so A ⊆ B. Similarly, B ⊆ A, and so A = B.

61. The statement is false. Let A = {1, 2}, Since B ∩ C = {3},

B = {2, 3},

C = {1, 3}.

A 4 (B ∩ C) = {1, 2, 3}.

Now A 4 B = {1, 3} and A 4 C = {2, 3}, thus (A 4 B) ∩ (A 4 C) = {3}.

62. The statement is false. Let A = {1, 2}, Since B 4 C = {1, 2},

B = {2, 3},

C = {1, 3}.

A ∪ (B 4 C) = {1, 2}.

Since A ∪ B = A ∪ C = {1, 2, 3},

(A ∪ B) 4 (A ∪ C) = ∅.

64. Yes, 4 is commutative: A 4 B = (A ∪ B) − (A ∩ B) = (B ∪ A) − (B ∩ A) = B 4 A. 65. Yes, 4 is associative. We first prove that (A 4 B) 4 C = (A ∩ B ∩ C) ∪ (A ∩ B ∩ C) ∪ (A ∩ B ∩ C) ∪ (A ∩ B ∩ C).

[For the motivation of this formula, draw the Venn diagram of (A 4 B) 4 C.] By Exercise 57, (A 4 B) 4 C = [(A 4 B) − C] ∪ [C − (A 4 B)].

Again using Exercise 57 and the fact that X − Y = X ∩ Y , we have (A 4 B) − C = [(A − B) ∪ (B − A)] − C = [(A ∩ B) ∪ (B ∩ A)] ∩ C. Copyright c 2018 Pearson Education, Inc.

(1)

18

SOLUTIONS Using the definition of 4, the fact that X − Y = X ∩ Y , and De Morgan’s laws, we have A 4 B = (A ∪ B) − (A ∩ B) = (A ∪ B) ∩ (A ∩ B) = (A ∪ B) ∪ (A ∩ B) = (A ∩ B) ∪ (A ∩ B).

Thus C − (A 4 B) = C ∩ (A 4 B) = C ∩ [(A ∩ B) ∪ (A ∩ B)].

Combining the preceding equations and using Theorem 1.1.21, we obtain equation (1) (A 4 B) 4 C

= = =

[(A 4 B) − C] ∪ [C − (A 4 B)] {[(A ∩ B) ∪ (B ∩ A)] ∩ C} ∪ {C ∩ [(A ∩ B) ∪ (A ∩ B)]}

(A ∩ B ∩ C) ∪ (A ∩ B ∩ C) ∪ (A ∩ B ∩ C) ∪ (A ∩ B ∩ C).

By Exercise 64, 4 is commutative. Thus A 4 (B 4 C) = (B 4 C) 4 A. We can obtain a formula for (B 4 C) 4 A using equation (1) with A replaced by B, B replaced by C, and C replaced by A. However, noting that the right-hand side of equation (1) is symmetric in A, B, and C, we see that the two expressions (A 4 B) 4 C

and A 4 (B 4 C)

are equal. Therefore, 4 is associative.

Section 2.2 2. False; x =

√ 2 is a counterexample.

3. We prove the contrapositive: If x is rational, then x3 is rational.

Suppose that x is rational. Then there exist integers p and q such that x = p/q. Now x3 = p3 /q 3 . Thus x3 is rational.

5. Suppose, by way of contradiction, that x < 1 and y < 1 and z < 1. Adding these inequalities gives x + y + z < 3, which is a contradiction. √ √ 6. Suppose, by way of contradiction, that x > 2 and y > 2. Multiplying these inequalities gives xy > 2, which is a contradiction.

8. Suppose, by way of contradiction, that x + y is rational. Since x and x + y are rational, there exist integers p1 , p2 , q1, q2 such that x = p1 /q1 and x + y = p2 /q2. Now y = (x + y) − x =

p2 p1 p2 q 1 − p1 q 2 − = . q2 q1 q1 q2

Therefore y is rational, which is a contradiction. √ 9. False; a counterexample is x = 0, y = 2. √ 11. √ Since the integers increase without bound, there exists n ∈ Z such that 2/(b − a) < n. Therefore √ 2/n < b −√a. Choose m ∈ Z as large as possible satisfying m 2/n ≤ a. Then, by the choice of m, a < (m + 1) 2/n. Also √ √ √ (m + 1) 2 m 2 2 = + < a + (b − a) = b. n n n √ √ Therefore x √ = (m + 1) 2/n is an irrational number satisfying √ a < x < b. (If (m + 1) 2/n is rational, say (m + 1) 2/n = p/q where p and q are integers, then 2 = np/[(m + 1)q] is rational, which is not the case.)

Copyright c 2018 Pearson Education, Inc.

19

SOLUTIONS

√ √2 √ 2 is rational, then we have found irrational numbers a and b (namely a = b = √ 2) such that ab is √ √ √ 2 √ 2 √ √ 2 √ √ rational. Suppose that 2 is irrational. Let a = 2 and b = 2. Now ab = ( 2 ) 2 = ( 2)2 = 2 is rational. We have found irrational numbers a and b such that ab is rational. √ This proof is nonconstructive since it does not show whether the desired pair is a = b = 2 or a = √ √ 2 √ 2 , b = 2. √ 14. Let a = 2 and b = 1/2. Then a and b are rational. Now ab = 21/2 = 2 is irrational. This proof is a constructive existence proof.

12. If

15. Suppose, by way of contradiction, that x > y. Let ε = (x − y)/2. Then y +ε = y+

x−y x+y x+x = < = x, 2 2 2

which is a contradiction.

17. First prove that if b is a rational number, then bn is rational for every positive integer n. To this end, let b = p/q, where p and q are integers. Then bn = pn /q n . Since b is the quotient of integers, b is rational.

Now suppose, but way of contradiction, that ar is rational for some positive rational number r = p/q, where p and q are positive integers. Because ar is rational, (ar )q is rational by the result of the first paragraph. Since (ar )q = ap , ap is rational. This contradicts the hypothesis that an is irrational for every positive integer n.

18. We show that Abby, Cary, Dale, and Edie went to the concert, but not Bosco. Suppose that Bosco went.

Then Cary and Dale also went. Since Cary went, Edie went; and since Dale went, Abby went. But this contradicts the hypothesis that exactly four went to the concert. Therefore, Bosco did not go. This means Abby, Cary, Dale, and Edie went to the concert.

20. Suppose, by way of contradiction, that X × ∅ is not empty. Then there exists (x, y) ∈ X × ∅. Now y ∈ ∅, which is a contradiction.

21. Suppose that every box contains less than 12 balls. Then each box contains at most 11 balls and the maximum number of balls contained by the nine boxes is 9 · 11 = 99. Contradiction.

23. Suppose, by way of contradiction, that each of the other three suits contains at most six cards. Then these three suits together contain at most 3 · 6 = 18 cards. Together with the other suit, which contains exactly seven cards, we can account for at most 25 cards. Since S contains 26 cards, we have a contradiction.

Therefore there is another suit in which S has at least seven cards.

24. Since there is a suit in which S1 has at least nine cards, S2 has at most four cards in this suit. Now suppose, by way of contradiction, that S2 has at most seven cards in the other three suits. These three suits contain at most 3 · 7 = 21 cards. Together with at most four cards in the other suit, we can account for at most 25 cards in S2 . Since S2 contains 26 cards, we have a contradiction. Therefore there is a suit in which S2 has at least eight cards.

26. For n = 3, we have n2 > 2n .

28. The statement is false. Let s1 = s2 = 3. Then A = 3. For no i do we have si > A. The proof is by counterexample.

29. The statement is true and we prove it using proof by contradiction. Suppose that for every j, sj ≤ A. Since sj ≤ A for all j and si < A, s1 + · · · + si + · · · + sn < A + · · · + A + · · · + A = nA. Copyright c 2018 Pearson Education, Inc.

20

SOLUTIONS Dividing by n, we obtain

s1 + · · · + sn < A, n

which is a contradiction.

31. Since si 6= sj , either si 6= A or sj 6= A. By changing the notation, if necessary, we may assume that si 6= A. Either si < A or si > A. If si > A, the proof is complete; so assume that si < A. We show that there exists k such that sk > A. Suppose, by way of contradiction, that sm ≤ A for all m, that is, s1 s2 sn

≤ ≤ .. . ≤

A A

A.

Adding these inequalities yields s1 + s2 + · · · + si + · · · + sn < nA since si < A. Dividing by n gives

s1 + s2 + · · · + sn < A, n which is a contradiction. Therefore there exists k such that sk > A.

33. If m and n are positive integers and m > 3, then m3 + 2n2 > 36. If m and n are positive integers and n > 4, then m3 + 2n2 > 36. Thus it suffices to consider the cases 1 ≤ m ≤ 3 and 1 ≤ n ≤ 4. The following table, which shows the values of m3 + 2n2 , shows that there is no solution to m3 + 2n2 = 36:

n

1 2 3 4

1 3 9 19 33

m 2 10 16 26 40

3 29 35 45 59

34. Notice that 2m2 + 4n2 − 1 is odd and 2(m + n) is even. Therefore 2m2 + 4n2 − 1 6= 2(m + n) for all positive integers m and n.

36. We consider two cases: n is even, n is odd. First suppose that n is even. By Exercise 9, Section 2.1, the product of even integers is even. Therefore n2 = n · n is even. Again by Exercise 9, Section 2.1, n3 = n2 · n is even. By Exercise 7, Section 2.1, the sum of even integers is even. Therefore n3 + n is even.

Now suppose that n is odd. By Exercise 10, Section 2.1, the product of odd integers is odd. Therefore n2 = n · n is odd. Again by Exercise 10, Section 2.1, n3 = n2 · n is odd. By Exercise 8, Section 2.1, the sum of odd integers is even. Therefore n3 + n is even. In either case, n3 + n is even.

38. First, note that from Exercise 37, for all x, |−x| = |(−1)x| = |−1||x| = |x|.

Example 2.2.7 states that for all x, x ≤ |x|. Using these results, we consider two cases: x + y ≥ 0 and x + y < 0. If x + y ≥ 0, we have |x + y| = x + y ≤ |x| + |y|.

If x + y < 0, we have

|x + y| = −(x + y) = −x + −y ≤ |−x| + |−y| = |x| + |y|.

Copyright c 2018 Pearson Education, Inc.

21

SOLUTIONS 40. Suppose that xy > 0. Then either x > 0 and y > 0 or x < 0 and y < 0. If x > 0 and y > 0, sgn(xy) = 1 = 1 · 1 = sgn(x)sgn(y).

If x < 0 and y < 0, sgn(xy) = 1 = −1 · −1 = sgn(x)sgn(y).

Next, suppose that xy = 0. Then either x = 0 or y = 0. Thus either sgn(x) = 0 or sgn(y) = 0. In either case, sgn(x)sgn(y) = 0. Therefore sgn(xy) = 0 = sgn(x)sgn(y).

Finally, suppose that xy < 0. Then either x > 0 and y < 0 or x < 0 and y > 0. If x > 0 and y < 0, sgn(xy) = −1 = 1 · −1 = sgn(x)sgn(y).

If x < 0 and y > 0, sgn(xy) = −1 = −1 · 1 = sgn(x)sgn(y).

41. |xy| = sgn(xy)xy = sgn(x)sgn(y)xy = [sgn(x)x][sgn(y)y] = |x||y| 43. Suppose that x ≥ y. Then max{x, y} = x Thus max{x, y} = x =

and

|x − y| = x − y.

2x x+y+x−y x + y + |x − y| = = .

2 2 2

The other case is x < y. Then max{x, y} = y Thus max{x, y} = y =

and

|x − y| = y − x.

x+y+y−x x + y + |x − y| 2y = = .

2 2 2

44. Suppose that x ≥ y. Then min{x, y} = y Thus min{x, y} = y =

and

|x − y| = x − y.

2y x + y − (x − y) x + y − |x − y| = = .

2 2 2

The other case is x < y. Then min{x, y} = x Thus min{x, y} = x =

and

|x − y| = y − x.

2x x + y − (y − x) x + y − |x − y| = = .

2 2 2

x + y + |x − y| x + y − |x − y| + 2 2 x + y + |x − y| + x + y − |x − y| = 2 2x + 2y = = x + y.

2

45. max{x, y} + min{x, y} =

Copyright c 2018 Pearson Education, Inc.

22

SOLUTIONS

47. Suppose that n is odd. Then n = 2k + 1. Now n + 2 = (2k + 1) + 2 = 2(k + 1) + 1 is odd.

Now suppose that n + 2 is odd. Then n + 2 = 2k + 1. Now n = (2k + 1) − 2 = 2(k − 1) + 1 is odd.

Therefore n is odd if and only if n + 2 is odd.

49. Suppose that A ⊆ C and B ⊆ C. Let x ∈ A ∪ B. Then either x ∈ A or x ∈ B. If x ∈ A, since A ⊆ C, x ∈ C. If x ∈ B, since B ⊆ C, x ∈ C. In either case, x ∈ C. Therefore A ∪ B ⊆ C. Now suppose that A ∪ B ⊆ C. Let x ∈ A. Then x ∈ A ∪ B. Since A ∪ B ⊆ C, x ∈ C. Therefore A ⊆ C. Let x ∈ B. Then x ∈ A ∪ B. Since A ∪ B ⊆ C, x ∈ C. Therefore B ⊆ C. We conclude that A ⊆ C and B ⊆ C. It follows that A ⊆ C and B ⊆ C if and only if A ∪ B ⊆ C.

50. Suppose that C ⊆ A and C ⊆ B. Let x ∈ C. Since C ⊆ A, x ∈ A. Since C ⊆ B, x ∈ B. Since x ∈ A and x ∈ B, x ∈ A ∩ B. Therefore C ⊆ A ∩ B.

Now suppose that C ⊆ A ∩ B. Let x ∈ C. Then x ∈ A ∩ B. In particular, x ∈ A. Therefore C ⊆ A. Again let x ∈ C. Then x ∈ A ∩ B. In particular, x ∈ B. Therefore C ⊆ B. Thus C ⊆ A and C ⊆ B. It follows that C ⊆ A and C ⊆ B if and only if C ⊆ A ∩ B.

53. [(a) → (b)] We assume that A ∩ B = ∅ and prove that B ⊆ A. Let x ∈ B. If x ∈ A, we obtain the contradiction A ∩ B 6= ∅. Thus x ∈ / A. Hence x ∈ A. Therefore B ⊆ A. [(b) → (c)] We assume that B ⊆ A and prove that A 4 B = A ∪ B.

Let x ∈ A 4 B. By definition, A 4 B = (A ∪ B) − (A ∩ B), thus x ∈ A ∪ B. Therefore A 4 B ⊆ A ∪ B.

Let x ∈ A ∪ B. We first prove that x ∈ / A ∩ B. Suppose, by way of contradiction, that x ∈ A ∩ B. Then x ∈ A and x ∈ B. Since B ⊆ A, x ∈ A, which implies that x ∈ / A. We have the desired contradiction.

Therefore x ∈ / A ∩ B. Now x ∈ (A ∪ B) − (A ∩ B) = A 4 B. Therefore A ∪ B ⊆ A 4 B. It follows that A 4 B = A ∪ B.

[(c) → (a)] We assume that A 4 B = A ∪ B and prove that A ∩ B = ∅. Suppose, by way of contradiction, that A ∩ B is not empty. Then there exists x ∈ A ∩ B. Then x ∈ A ∪ B. This implies that x ∈ / A 4 B. Since x ∈ A ∪ B, A 4 B 6= A ∪ B, which is a contradiction. Therefore A ∩ B = ∅.

54. [(a) → (b)] We assume that A ∪ B = U and prove that A ∩ B = ∅. Taking the complement of both sides of the equation A ∪ B = U and using De Morgan’s law and the 0/1 law (Theorem 1.1.22), we obtain A ∩ B = A ∪ B = U = ∅. [(b) → (c)] We assume that A ∩ B = ∅ and prove that A ⊆ B. Replace A by B and B by A in Exercise 53(a) to obtain B ∩ A = ∅. Since Exercise 53(a) is equivalent to Exercise 53(b), we obtain A ⊆ B or A ⊆ B.

[(c) → (a)] We assume that A ⊆ B and prove that A∪B = U . Since U is a universal set, we automatically have A ∪ B ⊆ U .

Let x ∈ U . If x ∈ A, then x ∈ A ∪ B. If x ∈ / A, then x ∈ A. Since A ⊆ B, x ∈ B. Again x ∈ A ∪ B. Therefore U ⊆ A ∪ B. It follows that A ∪ B = U .

Problem-Solving Corner: Proofs 1. The least upper bound of a nonempty finite set of real numbers is the maximum number in the set.

2. Call the given set X. We prove that the least upper bound of X is 1. Since 1−

1 <1 n

Copyright c 2018 Pearson Education, Inc.

23

SOLUTIONS

for all positive integers n, 1 is an upper bound of X. Let a be an upper bound for X. Suppose, by way of contradiction, that a < 1. Since the integers are unbounded, there exists a positive integer k such that 1 < k.

1−a Multiplying by (1 − a)/k gives

1 < 1 − a, k

which, in turn, is equivalent to

1 . k This contradicts the fact that a is an upper bound of X. Thus 1 ≤ a and 1 is the least upper bound of X. a<1−

3. Let b be the least upper bound of Y . If x ∈ X, then x ∈ Y and x ≤ b. Thus b is an upper bound of X. If a is the least upper bound of X, a ≤ b.

4. 0 5. Let Z = {x + y | x ∈ X and y ∈ Y } and let z ∈ Z. Then z = x + y for some x ∈ X, y ∈ Y . Now z = x + y ≤ a + b. Therefore Z is bounded above by a + b.

Let c be an upper bound of Z. Suppose, by way of contradiction, that c < a + b. Let ε = a + b − c. Now a − ε/2 is not an upper bound of X so there exists x ∈ X such that

Similarly, there exists y ∈ Y such that

a−

ε < x.

2

b−

ε < y.

2

Adding the previous inequalities gives c = a + b − ε < x + y, which contradicts the fact that c is an upper bound of Z. Therefore c ≥ a + b and a + b is the least upper bound of Z. 6. Since a is a greatest lower bound for X and b is a lower bound for X, b ≤ a. Since b is a greatest lower bound for X and a is a lower bound for X, a ≤ b. Therefore a = b.

7. Let X be a nonempty set of real numbers bounded below. Let Y be the set of lower bounds of X. The set Y is nonempty since X is bounded below. Let x be an element of X. For every y ∈ Y , we have y ≤ x since y is a lower bound of X. Therefore Y is bounded above by x. Thus Y is has a least upper bound, say a.

Next we show that a is a lower bound of X. Suppose, by way of contradiction, that a is not a lower bound of X. Then there exists x ∈ X such that x < a. Then x is not an upper bound of Y . Therefore there exists y ∈ Y such that x < y. But this contradicts the fact that y is a lower bound of X. Therefore a is a lower bound of X. Finally, we show that a is the greatest lower bound of X. Let b be a lower bound of X. Then b ∈ Y .

Since a is an upper bound of Y , b ≤ a. Therefore a is the greatest lower bound of X. 8. Since a + ε > a, a + ε is not a lower bound of X. Therefore there exists x ∈ X such that a + ε > x. Since a is a lower bound of X, x ≥ a.

Copyright c 2018 Pearson Education, Inc.

24

SOLUTIONS 9. Let tX denote the set {tx | x ∈ X}.

We must prove that (a) z ≥ ta for every z ∈ tX (i.e., ta is an lower bound for tX),

(b) if b is an lower bound for tX, then b ≤ ta (i.e., ta is the greatest lower bound for tX).

We first prove part (a). Let z ∈ tX. Then z = tx for some x ∈ X. Since a is an upper bound for X, x ≤ a. Multiplying by t and noting that t < 0, we have z = tx ≥ ta. Therefore, z ≥ ta for every z ∈ tX and the proof of part (a) is complete.

Next we prove part (b). Let b be a lower bound for tX. Then tx ≥ b for every x ∈ X. Dividing by t and noting that t < 0, we have x ≤ b/t for every x ∈ X. Therefore b/t is an upper bound for X. Since a is the least upper bound for X, b/t ≥ a. Multiplying by t and noting again that t < 0, we have b ≤ ta.

Therefore ta is the greatest lower bound for tX. The proof is complete.

Section 2.3 3.

4.

1.

2.

3.

4.

5.

¬p ∨ r ¬r ∨ q p ¬p ∨ q q

1.

2.

3.

4.

5.

6.

7.

8.

¬p ∨ t ¬q ∨ s ¬r ∨ s ¬r ∨ t p∨q∨r∨u t∨q∨r∨u s∨t∨r∨u s∨t∨u

from 1,2 from 3,4

from 1,5 from 2,6 from 3,7

6. (p ↔ r) ≡ (p → r)(r → p) ≡ (¬p ∨ r)(¬r ∨ p)

8.

1.

2.

3.

4.

¬p ∨ r ¬r ∨ p r p

1.

2.

3.

4.

5.

6.

a ∨ ¬b a∨c ¬a ¬d b ¬b

from 2,3

negated conclusion from 1,3

Now 5 and 6 combine to give a contradiction.

Section 2.4 In some of these solutions, the Basis Steps are omitted.

Copyright c 2018 Pearson Education, Inc.

25

SOLUTIONS 2. 1 · 2 + 2 · 3 + · · · + n(n + 1) + (n + 1)(n + 2) (n + 1)(n + 2)(n + 3) n(n + 1)(n + 2) + (n + 1)(n + 2) = = 3 3 3. 1(1!) + 2(2!) + · · · + n(n!) + (n + 1)(n + 1)! = (n + 1)! − 1 + (n + 1)(n + 1)! = (n + 2)! − 1 5. 12 − 22 + · · · + (−1)n+1 n2 + (−1)n+2 (n + 1)2 (−1)n+1 n(n + 1) (−1)n+2 (n + 1)(n + 2) = + (−1)n+2 (n + 1)2 = 2 2 6. 13 + 23 + · · · + n3 + (n + 1)3

2

2 n(n + 1) (n + 1)(n + 2) 3 = + (n + 1) = 2 2 8.

9.

1·3 1 · 3 · · · (2n − 1) 1 · 3 · · · (2n − 1)(2n + 1) 1 + + ···+ + 2·4 2·4·6 2 · 4 · · · (2n + 2) 2 · 4 · · · (2n + 2)(2n + 4) 1 1 · 3 · · · (2n + 1) 1 · 3 · · · (2n − 1)(2n + 1) = − + 2 2 · 4 · · · (2n + 2) 2 · 4 · · · (2n + 2)(2n + 4) 1 1 · 3 · · · (2n + 3) = − 2 2 · 4 · · · (2n + 4) 1 22 − 1

+

1

1 1 + 2 (n + 1) − 1 (n + 2)2 − 1 3 1 1 1 = − − + 4 2(n + 1) 2(n + 2) (n + 2)2 − 1 1 1 3 − = − 4 2(n + 2) 2(n + 3) 32 − 1

+ ···+

n(n + 1)(n + 2)(3n + 5) + (n + 1)(n + 2)2 12 n(n + 1)(n + 2)(3n + 5) + 12(n + 1)(n + 2)2 = 12 (n + 1)(n + 2)[3n2 + 5n + 12(n + 2)] = 12 (n + 1)(n + 2)(3n2 + 17n + 24) = 12 (n + 1)(n + 2)(n + 3)(3n + 8) = 12

10. 1 · 22 + 2 · 32 + · · · + (n + 1)(n + 2)2 =

12. The solution is similar to that for Exercise 11, which is given in the book.

14. First note that 1 · 3 · · · (2n − 1)(2n + 1) 1 2n + 1 ≤ √ .

2 · 4 · · · (2n)(2n + 2) n + 1 2n + 2 The proof will be complete if we can show that 2n + 1 1 √ ≤ √ .

(2n + 2) n + 1 n+2 Copyright c 2018 Pearson Education, Inc.

26

SOLUTIONS This last inequality is successively equivalent to

1/2 n+2 n+1 n+2 n+1

≤

(n + 2)(2n + 1)2

≤

3

2

2n + 2 2n + 1 4(n + 1)2 (2n + 1)2

≤

4(n + 1)3 4n3 + 12n2 + 12n + 4 3n.

4n + 12n + 9n + 2 ≤ −2 ≤ This last inequality is true for all n ≥ 1.

15. 2(n + 1) + 1 = (2n + 1) + 2 ≤ 2n + 2 ≤ 2n + 2n = 2n+1 17. By the inductive assumption, (a1 · · · a2n )1/2

n

(a2n +1 · · · a2n+1 )1/2

n

a1 + · · · + a2 n 2n a2n +1 + · · · + a2n+1 .

2n

≤ ≤

(2) (3)

Let

a1 + · · · + a2 n a2n +1 + · · · + a2n+1 and B = . n 2 2n Multiplying inequalities (2) and (3), we have A=

(a1 · · · a2n+1 )1/2

n

≤ AB.

(4)

Applying the Basis Step to the numbers A and B, we have (AB)1/2 ≤

A+B 2

or, equivalently,

a1 + · · · + a2n+1 AB ≤ 2n+1

2

.

(5)

Combining inequalities (4) and (5), we have n

(a1 · · · a2n+1 )1/2 ≤

a1 + · · · + a2n+1 2n+1

2

.

Taking the square root of both sides of the last inequality gives the desired result.

18.

(1 + x)n+1

= ≥ = ≥

(1 + x)n (1 + x) (1 + nx)(1 + x) 1 + nx + x + nx2 1 + (n + 1)x

20. If we sum the terms in the diagonal direction, we obtain one r, two r 2 ’s, three r 3 ’s, and so on; that is, we obtain the sum 1 · r 1 + 2 · r 2 + · · · + nr n .

Multiplying the inequality of Exercise 19 by r yields r 1 + r 2 + · · · + r n+1 <

r 1−r

for all n ≥ 0.

Copyright c 2018 Pearson Education, Inc.

(6)

27

SOLUTIONS

Thus, the sum of the entries in the first column is less than r/(1 − r). Similarly, the sum of the entries in the second column is less than r 2 /(1 − r), and so on. It follows from the preceding discussion that 1 · r 1 + 2 · r 2 + · · · + nr n <

1 (r 1 + r 2 + · · · + r n ).

1−r

Using inequality (6), we obtain the desired result 1 (r 1 + r 2 + · · · + r n ) < 1 · r + 2 · r + · · · + nr < 1−r 1

2

n

1 1−r

r 1−r

=

r .

(1 − r)2

21. Take r = 1/2 in Exercise 20.

23. Assume that 11n − 6 is divisible by 5.

11n+1 − 6 = 11n · 11 − 6 = 11n (10 + 1) − 6 = 10 · 11n + 11n − 6, which is divisible by 5.

24. Suppose that 4 divides 6 · 7n − 2 · 3n . Now 6 · 7n+1 − 2 · 3n+1

= = =

7 · 6 · 7n − 3 · 2 · 3n 6 · 7n − 2 · 3n + 6 · 6 · 7n − 2 · 2 · 3n

6 · 7n − 2 · 3n + 36 · 7n − 4 · 3n .

Since 4 divides 6 · 7n − 2 · 3n , 36 · 7n , and −4 · 3n , it divides their sum, which is 6 · 7n+1 − 2 · 3n+1 .

26. We prove part (a) only. The Basis Step is immediate.

Assume that X ∩ (X1 ∪ X2 ∪ · · · ∪ Xn ) = (X ∩ X1 ) ∪ (X ∩ X2 ) ∪ · · · ∪ (X ∩ Xn ).

We must prove that X ∩ (X1 ∪ X2 ∪ · · · ∪ Xn ∪ Xn+1 ) = (X ∩ X1 ) ∪ (X ∩ X2 ) ∪ · · · ∪ (X ∩ Xn ) ∪ (X ∩ Xn+1 ).

Let Y = Xn ∪ Xn+1 . By the inductive assumption, X ∩ (X1 ∪ X2 ∪ · · · ∪ Xn−1 ∪ Y ) = (X ∩ X1 ) ∪ (X ∩ X2 ) ∪ · · · ∪ (X ∩ Xn−1 ) ∪ (X ∩ Y ).

By the associative law, X ∩ (X1 ∪ X2 ∪ · · · ∪ Xn−1 ∪ Y ) = X ∩ (X1 ∪ X2 ∪ · · · ∪ Xn ∪ Xn+1 ).

By the distributive law, X ∩ Y = X ∩ (Xn ∪ Xn+1 ) = (X ∩ Xn ) ∪ (X ∩ Xn+1 ).

Therefore (X ∩ X1 ) ∪ (X ∩ X2 ) ∪ · · · ∪ (X ∩ Xn−1) ∪ (X ∩ Y ) = (X ∩ X1 ) ∪ (X ∩ X2 ) ∪ · · · ∪ (X ∩ Xn ) ∪ (X ∩ Xn+1 ), and the Inductive Step is complete.

29.

n n+1 Copyright c 2018 Pearson Education, Inc.

28

SOLUTIONS

31. We use induction on n, the number of lines, to prove the result. If there is one line, the result is certainly true. Suppose that there are n > 1 lines. Remove one line. By the inductive hypothesis, the regions that result may be colored red and green so that no two regions that share an edge are the same color. Now restore the removed line. The regions above (or, if the line is vertical, to the left of) the restored line are colored red and green so that no two regions that share an edge are the same color, and the regions below (or, if the line is vertical, to the right of) the restored line are also colored red and green so that no two regions that share an edge are the same color. Now reverse the color of every region below (or, if the line is vertical, to the right of) the restored line. The regions below (or, if the line is vertical, to the right of) the restored line are still colored red and green so that no two regions that share an edge are the same color. Since the colors below the restored line have been reversed, regions that share an edge that is part of the restored line do not have the same color. Therefore the regions may be colored red and green so that no two regions that share an edge are the same color, and the inductive proof is complete.

32. The proof is by induction on the number n of zeros with the Basis Step, as usual, omitted.

Suppose that the result is true for n zeros, and we are given n + 1 zeros and n + 1 ones distributed around a circle. Find a zero followed, in clockwise order, by a one. Temporarily remove these two numbers. By the inductive assumption, it is possible to start at some number and proceed around the circle to the original starting position in such a way that, at any point during the cycle, one has seen at least as many zeros as ones. Notice that this last statement remains true if we restore the removed zero and one.

34. A tromino can cover the square to the left of the missing square as shown

or in a symmetric fashion by reversing “up” and “down.” In the first case, it is impossible to cover the two squares in the top row at the extreme left. In the second case, it is impossible to cover the two squares in the bottom row at the extreme left. Therefore, it is impossible to tile the board with trominoes.

35. The only 5 × 5 deficient boards that can be tiled with trominoes are those with the corner square or center square missing.

36. Such a board can be tiled with ij 2 × 3 rectangles of the form

38 By symmetry, we may assume that the missing square is located in the 7 × 7 subboard shown in the following figure. Exercise 37 shows how to tile this subboard. Exercise 34 shows that the two 6 × 4 subboards can be tiled. Exercise 32 shows that the 5 × 5 subboard with a corner square can be tiled.

Thus the deficient 11 × 11 board can be tiled with trominoes.

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29

SOLUTIONS 11

6×4

7×7 11

5×5

6×4

39. Basis Step (n = 0). In this case, the 2n × 2n L-shape is a tromino and, so, it is tiled.

Inductive Step. Assume that we can tile a 2n−1 × 2n−1 L-shape with trominoes. Given a 2n × 2n L-shape, divide it into four 2n−1 × 2n−1 L-shapes:

By the inductive assumption, we can tile each of the four 2n−1 × 2n−1 L-shapes with trominoes. The inductive step is complete.

42. Arguing as in the solution to Exercise 41, the numberings 1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

2

3

1

show that the only possibility for the missing square is the center square. This board can be tiled:

43. An argument like those in the solutions to Exercises 41 and 42 shows that the only board that can be tiled with straight trominoes is the one with the missing square in row 3, column 3 (and the three boards symmetric to it).

44. See D. W. Walkup, “Covering a rectangle with T-tetrominoes,” Amer. Math. Mo., (November 1965) 986–988.

Copyright c 2018 Pearson Education, Inc.

30

SOLUTIONS

46. Basis Step (k = i + 1) Case 1 a[k] < val . Now h becomes i + 1 and the swap changes nothing. Thus h ≤ k. The only p satisfying i < p ≤ h is p = i + 1. For this value of p, we do have a[p] < val . No p satisfies h < p ≤ k, so the third condition is vacuously satisfied.

Case 2 a[k] ≥ val . At the bottom of the loop, we have k = i + 1 and h = i. Thus h ≤ k. Since no p satisfies i < p ≤ h, the second condition is vacuously satisfied. The only p satisfying h < p ≤ k is p = i + 1. For this value of p, we do have a[p] ≥ val .

Inductive Step

By the inductive assumption, the invariant holds for the previous iteration of the loop. We denote these values of h and k as hold and kold . Thus we have hold ≤ kold ; for all p, i < p ≤ hold , a[p] < val ; and for p, hold < p ≤ kold , a[p] ≥ val . We denote the new values of h and k as hnew and knew . At the top of the loop, k is incremented, so we have knew = kold + 1.

Case 1 a[knew] ≥ val . At the bottom of the loop, we have hnew = hold . Thus hnew = hold ≤ kold < knew , so the first condition holds. Suppose that i < p ≤ hnew . Since hnew = hold , p satisfies i < p ≤ hold .

Therefore a[p] < val and the second condition holds. Suppose that hnew < p ≤ knew . If hnew < p < knew , then hold < p ≤ kold . In this case, a[p] < val . If p = knew , also a[p] ≥ val . Therefore the third condition holds.

Case 2 a[knew ] < val . First suppose that hold = kold . Then hnew = hold + 1 = kold + 1 = knew . Thus the first condition holds. The swap changes no values. Suppose that i < p ≤ hnew . If p < hnew , then i < p ≤ hold , so a[p] = val . If p = hnew , a[p] = a[hnew] = a[knew] < val , so the second condition holds.

Since no p satisfies hnew < p ≤ knew , the third condition is vacuously satisfied.

Now suppose that hold 6= kold . Then hold < kold . After h and k are incremented, hnew < knew , so the first condition is satisfied. Now hnew = hold + 1. Since hold < hnew ≤ kold , a[hnew] ≥ val . It follows that after the swap, a[hnew ] < val and a[knew] ≥ val . Suppose that i < p ≤ hnew . If i < p < hnew , then i < p ≤ hold , so a[p] < val . If p = hnew , we also have a[p] < val . Thus the second condition holds.

Suppose that hnew < p ≤ knew . If hnew < p < knew , then hold < p ≤ kold and p 6= hnew , so a[p] ≥ val . If p = knew , a[p] ≥ val. Thus the third condition holds. The Inductive Step is complete.

Before the final swap, for all p, i < p ≤ h, a[p] < val and for all p, h < p ≤ j, a[p] ≥ val . Note that a[i] = val and a[h] < val . Thus, after swapping a[i] and a[h], a[h] = val ; for all p, i ≤ p < h, a[p] < val ; and for all p, h < p ≤ j, a[p] ≥ val .

47. The argument is essentially identical to that of Example 2.4.7 that shows that any 2n ×2n deficient board can be tiled with trominoes.

48. Notice that k 3 − 1 = (k − 1)[(k − 2)(k − 4) + 7(k − 1)].

Since 7 divides k 3 − 1, 7 divides k − 1 or (k − 2)(k − 4) + 7(k − 1). If 7 divides the latter expression, 7 also divides (k − 2)(k − 4). If 7 divides (k − 2)(k − 4), 7 divides either k − 2 or k − 4.

50. The Inductive Step fails if either a or b is 1. In this case, the inductive hypothesis is erroneously applied to the pair a − 1, b − 1, which includes a nonpositive integer.

51. To argue by contradiction, one must assume that the proposition fails for some n ≥ 2. The alleged proof assumes that the proposition fails for all n ≥ 2.

52. For n = 2, the inequality becomes 21 + 23 < 43 , which is true. Thus the Basis Step is true.

Assume that the given statement holds for n. Now 1 2 n n+1 n2 n+1 + +···+ + < + .

2 3 n+1 n+2 n+1 n+2 Copyright c 2018 Pearson Education, Inc.

31

SOLUTIONS The Inductive Step will be proved provided n2 n+1 (n + 1)2 + < . n+1 n+2 n+2 If we multiply the last inequality by (n + 1)(n + 2), we obtain n2 (n + 2) + (n + 1)2 < (n + 1)3 , which is readily verified as true.

54. In the following figure r r r r

r r a b a and b are both survivors.

55. Suppose that there are three persons. The two persons closest together throw at each other, and the third person throws at one of the two closest. Therefore the third person survives. This complete the Basis Step.

Suppose that the assertion is true for n, and consider n + 2 persons. Again, the closest pair throws at each other. There are now two cases to consider. If the remaining n persons all throw at one another, by the inductive assumption, there is a survivor. If at least one of the remaining n persons throws at one of the closest pair, among the remaining n persons, at most n − 1 pies are thrown at one another. In this case, someone survives because there are not enough pies to go around. The Inductive Step is complete.

57. The statement is false. In the following figure r a

r

r r r

a throws a pie the greatest distance, but is not a survivor.

59. Let x1 be a common point of X2 , X3 , X4 ; let x2 be a common point of X1 , X3 , X4 ; let x3 be a common point of X1 , X2 , X4 ; and let x4 be a common point of X1 , X2 , X3 . Since x1 , x2 , x3 ∈ X4 , the triangle x1 x2 x3 (perimeter and interior) is in X4 . Similarly, the triangle x1 x2 x4 is in X3 ; the triangle x1 x3 x4 is in X2 ; and the triangle x2 x3 x4 is in X1 . We consider two cases: Case 1: One of the points x1 , x2 , x3 , x4 is in the triangle whose vertices are the other three points. For example, suppose that x1 is in triangle x2 x3 x4 . Since triangle x2 x3 x4 is in X1 , x1 ∈ X1 . By definition, x1 ∈ X2 ∩ X3 ∩ X4 . Therefore, x1 ∈ X1 ∩ X2 ∩ X3 ∩ X4 .

Case 2: None of the points x1 , x2 , x3 , x4 is in the triangle whose vertices are the other three points. In this case, x1 , x2 , x3 , x4 are the vertices of a convex quadrilateral: x1 rP

r2 P A PP J PPr A J P x PP A Jr x3 A

A Ar x

x4

Now the intersection, x, of the diagonals of this quadrilateral belongs to each of the triangles and, thus, to each of X1 , X2 , X3 , X4 .

Copyright c 2018 Pearson Education, Inc.

32

SOLUTIONS

60. The proof is by induction on n. The Basis Step is n = 4, which is Exercise 59.

We turn to the Inductive Step. Assume that if X1 , . . . , Xn are convex sets, each three of which have a common point, then all n sets have a common point.

Let X1 , . . . , Xn , Xn+1 be convex sets, each three of which have a common point. We must show that all n + 1 sets have a common point. By Exercise 55, X1 , . . . , Xn−1, Xn ∩ Xn+1

(7)

are convex sets. We claim that any three of the sets in (7) have a common point. The claim is true by hypothesis if the three sets are any of X1 , . . . , Xn−1 . Consider Xi , Xj , Xn ∩ Xn+1 , i < j ≤ n − 1. By hypothesis, any three of Xi , Xj , Xn , Xn+1 have a common point. By Exercise 56, Xi , Xj , Xn , Xn+1 have a common point. Therefore, Xi , Xj , Xn ∩ Xn+1 have a common point. Thus, any three of the sets in (7) have a common point. By the inductive assumption, the sets in (7) have a common point. The Inductive Step is complete.

62. We first prove the result for n = 3. Let A1 , A2 , A3 be open intervals such that each pair has a nonempty intersection. Choose x1 ∈ A1 ∩ A2 , x2 ∈ A1 ∩ A3 , x3 ∈ A2 ∩ A3 . Note that if any pair (x1 , x2 or x1 , x3 or x3 , x3 ) is equal, it is in A1 ∩ A2 ∩ A3 . We may assume x1 < x2 . We consider three cases. First suppose that x3 < x1 . Since x2 , x3 ∈ A3 , [x3 , x2] ⊆ A3 . ([a, b] is the set of all x satisfying a ≤ x ≤ b.) Thus x1 ∈ A3 . Therefore x1 ∈ A1 ∩ A2 ∩ A3 .

Next suppose that x1 < x3 < x2 . Since x1 , x2 ∈ A1 , [x1, x2 ] ⊆ A1 . Thus x3 ∈ A1 . Therefore x3 ∈ A1 ∩ A2 ∩ A3 .

Finally suppose that x1 < x2 < x3 . Since x1 , x3 ∈ A2 , [x1, x3 ] ⊆ A2 . Thus x2 ∈ A2 . Therefore x2 ∈ A1 ∩ A2 ∩ A3 . We have shown that if A1 , A2 , A3 are open intervals such that each pair has a nonempty intersection, then A1 ∩ A2 ∩ A3 is nonempty.

We now prove that given statement using induction on n. The Basis Step (n = 2) is trivial.

Assume that if I1 , . . . , In is a set of open intervals such that each pair has a nonempty intersection, then I1 ∩ · · · ∩ In is nonempty. Let I1 , . . . , In+1 be a set of open intervals such that each pair has a nonempty intersection. Since In ∩ In+1 is nonempty, it is an open interval. We claim that I1 , . . . , In−1, In ∩ In+1 is a set of open intervals such that each pair has a nonempty intersection. This is certainly true for pairs of the form Ii , Ij , 1 ≤ i < j ≤ n − 1. Consider a pair of the form Ii , i ≤ n − 1, and In ∩ In+1 . Since each pair among Ii , In , In+1 has nonempty intersection, by the case n = 3 proved previously, Ii ∩ In ∩ In+1 is nonempty. Therefore, I1 , . . . , In−1, In ∩ In+1 is a set of open intervals such that each pair has a nonempty intersection. By the inductive assumption Ii ∩ · · · ∩ In−1 ∩ (In ∩ In+1 ) is nonempty. The inductive step is complete.

64. 5

65. 5

67. After j rounds, 2, 4, . . . , 2j have been eliminated. At this point, there are 2i persons. This is exactly the Josephus problem when the number of persons is a power of 2, except that the round begins with person 2j + 1, rather than with person 1. By Exercise 66, person 2j + 1 is the survivor.

65. 977 Copyright c 2018 Pearson Education, Inc.

33

SOLUTIONS 71. ∆an = an+1 − an = (n + 1)2 − n2 = 2n + 1. Let bn = ∆an . Then b1 + b2 + · · · + bn

= = =

(2 · 1 + 1) + (2 · 2 + 1) + · · · + (2n + 1) 2(1 + 2 + · · · + n) + (1 + 1 + · · · + 1) 2(1 + 2 + · · · + n) + n.

By Exercise 70, b1 + b2 + · · · + bn = an+1 − a1 = (n + 1)2 − 12 = n2 + 2n.

Combining the previous equations, we obtain n2 + 2n = 2(1 + 2 + · · · + n) + n.

Solving for 1 + 2 + · · · + n, we obtain 1+2+···+n =

n2 + 2n − n n2 + n n(n + 1) = = .

2 2 2

72. Let an = n!. Then ∆an = an+1 − an = (n + 1)! − n! = n![(n + 1) − 1] = n(n!).

Let bn = ∆an . Then b1 + b2 + · · · + bn = 1(1!) + 2(2!) + · · · + n(n!).

By Exercise 70, b1 + b2 + · · · + bn = an+1 − a1 = (n + 1)! − 1!.

Combining the previous equations, we obtain 1(1!) + 2(2!) + · · · + n(n!) = (n + 1)! − 1.

74. Since p is divisible by k, there exists t1 such that p = t1 k. Since q is divisible by k, there exists t2 such that p = t2 k. Now p + q = t1 k + t2 k = (t1 + t2 )k.

Therefore, p + q is divisible by k.

Problem-Solving Corner: Mathematical Induction 1. The Basis Step (n = 0) is H1 ≤ 1 + 0. Since H1 = 1, the Basis Step is true.

Now assume that H2n ≤ 1 + n. Then H2n+1

= ≤ =

1 1 + · · · + n+1 2n + 1 2 1 1 1+n+ n +···+ n 2 +1 2 +1 2n 1+n+ n ≤ 1 + (n + 1).

2 +1 H2n +

The Inductive Step is complete.

Copyright c 2018 Pearson Education, Inc.

34

SOLUTIONS 2. The Basis Step (n = 1) is H1 = 2H1 − 1. Since H1 = 1, the Basis Step is true.

Now assume that

H1 + H2 + · · · + Hn = (n + 1)Hn − n.

Then H1 + H2 + · · · + Hn + Hn+1

= =

=

(n + 1)Hn − n + Hn+1

1 (n + 1) Hn+1 − n+1 − n + Hn+1 (n + 2)Hn+1 − (n + 1).

by Exercise 3

The Inductive Step is complete.

1 1 1 1 1 1 1 1 1 = + + ···+ + − = + + · · · + = Hn 3. Hn+1 − n+1 1 2 n n+1 n+1 1 2 n 4. We prove the assertion by induction. The Basis Step is n = 1: 1 · H1 = 1 =

3 1 1·2 1·2 − = H2 − .

2 2 2 4

For the Inductive Step, assume that the assertion is true for n. Now 1 · H1 + · · · + nHn + (n + 1)Hn+1

= = = =

= =

n(n + 1) n(n + 1) Hn+1 − + (n + 1)Hn+1 2 4 hn i n(n + 1) +1 − (n + 1)Hn+1 2 4

(n + 1)(n + 2) n(n + 1) Hn+1 − 2 4

1 (n + 1)(n + 2) Hn+2 −

by Exercise 3 n+2 2 n(n + 1) − 4

(n + 1)(n + 2) n + 1 n(n + 1) − − Hn+2 2 2 4

(n + 1)(n + 2) (n + 1)(n + 2) Hn+2 − .

2 4

5. We prove the assertion by induction. The Basis Step is n = 1: H1 1 1 H2 1 1 =1= + = 1 + .

1 2 2 2 2 12 We proceed to the Inductive Step. First, notice that 2 Hn+1 =

Hn +

1 n+1

2

= Hn2 +

2Hn 1 + . n + 1 (n + 1)2

Assume that the assertion is true for n. Then

H1 Hn Hn+1 Hn2 1 1 1 Hn+1 +···+ + = + +···+ 2 + 1 n n+1 2 2 12 n n+1 Copyright c 2018 Pearson Education, Inc.

(8)

by the Inductive Assumption

35

SOLUTIONS

= = = =

2 Hn+1 Hn 1 1 1 1 Hn+1 − − + + · · · + + 2 n + 1 2(n + 1)2 2 12 n2 n+1

2 Hn+1 1 1 1 1 1 + − + + ···+ 2 2 (n + 1)2 2(n + 1)2 2 12 n

2 Hn+1 1 1 1 1 + + + ···+ 2 2 2(n + 1)2 2 12 n

2 Hn+1 1 1 1 + +···+ .

2 2 12 (n + 1)2

by (8)

Section 2.5 2. Verify directly the cases n = 24, . . . , 28. Assume that the statement is true for postage i satisfying 24 ≤ i < n. We must show that we can make n cents postage using only 5-cent and 7-cent stamps. We may assume that n > 28. Then n > n − 5 ≥ 24. By the inductive assumption, we can make n − 5 cents postage using 5-cent and 7-cent stamps. Add a 5-cent stamp to obtain n cents postage.

3. Verify directly the cases n = 12, 13, 14. Assume that the statement is true for postage i satisfying 12 ≤ i < n. We must show that we can make n cents postage using only 3-cent and 7-cent stamps. We may assume that n > 14. Then n > n − 3 ≥ 12. By the inductive assumption, we can make n − 3 cents postage using 3-cent and 7-cent stamps. Add a 3-cent stamp to obtain n cents postage.

5. The Basis Step (n = 6) is proved by using three 2-cent stamps. Now assume that we can make postage for n cents. If there is at least one 7-cent stamp, replace it by four 2-cent stamps to make n + 1 cents postage. If there are no 7-cent stamps, there are at least three 2-cent stamps (because n ≥ 6). Replace three 2-cent stamps by one 7-cent stamp to make n + 1 cents postage. The Inductive Step is complete.

6. The Basis Step (n = 24) is proved by using two 5-cent stamps and two 7-cent stamps. Now assume that we can make postage for n cents. If there are at least two 7-cent stamps, replace two 7-cent stamps with three 5-cent stamps to make n + 1 cents postage. If there is exactly one 7-cent stamp, then there are at least four 5-cent stamps (because n ≥ 24). Replace one 7-cent stamp and four 5-cent stamps with four 7-cent stamps to make n + 1 cents postage. If there are no 7-cent stamps, then there are at least five 5-cent stamps (again because n ≥ 24). Replace five 5-cent stamps with three 7-cent stamps and one 5-cent to make n + 1 cents postage. The Inductive Step is complete.

8. We must have 4 ≤ bn/2c. Since this inequality fails for n = 5, 6, 7, the Basis Steps are n = 4, 5, 6, 7.

9. We must have 2 ≤ bn/3c. Since this inequality fails for n = 3, 4, 5, the Basis Steps are n = 2, 3, 4, 5.

11. We omit the Basis Step. For the Inductive Step, we have cn = cbn/2c + n2 < 4

j n k2 2

+ n2 ≤ 4

n 2 2

+ n2 = 2n2 < 4n2 .

13. We omit the Basis Step. For the Inductive Step, we have cn = 4cbn/2c + n

≤

≤ = ≤

4[4(bn/2c − 1)2 ] + n

4[4(n/2 − 1)2 ] + n 4n2 − 15n + 16 4(n − 1)2 .

The last inequality reduces to 12 ≤ 7n, which is true since n > 1.

Copyright c 2018 Pearson Education, Inc.

36

SOLUTIONS

14. We omit the Basis Steps (n = 2, 3). We turn to the Inductive Step. Assume that n ≥ 4. Then n/2 ≥ 2, so bn/2c ≥ 2. Then cn = 4cbn/2c + n > ≥ = >

4(bn/2c + 1)2 /8 + n 4[(n − 1)/2 + 1]2 /8 + n (n + 1)2 /8 + n (n + 1)2 /8.

We used the fact that bn/2c ≥ (n − 1)/2 for all n.

16. Basis Step (n = 1). The first player removes one card (from either pile). The second player then removes the last card and wins the game.

Inductive Step. Suppose that n > 1 and whenever there are two piles of k < n cards, the second player can always win the game.

Suppose that there are two piles of n cards. The first player removes i cards from one of the piles. If i = n (i.e., the first player removes all of the cards from one pile), the second player can win by removing all of the cards from the remaining pile. If i < n, the second player removes i cards from the other pile leaving two piles each with n − i cards. The game then resumes with the first player facing two piles each with k = n − i < n cards. By the inductive assumption, the second player can win the game. The inductive proof is complete.

18. q = −6, r = 7 24. If

19. q = 0, r = 7

21. q = 0, r = 0

22. q = 1, r = 0

p 1 1 1 = + +···+ q n1 n2 nk

where n1 < n2 < . . . < nk , another representation is p 1 1 1 1 1 = + +···+ + + q n1 n2 nk−1 nk + 1 nk (nk + 1) 25. (b) Since p/q < 1, n > 1. Since n is the smallest positive integer satisfying 1/n ≤ p/q and n − 1 is a positive integer less than n, p/q < 1/(n − 1). (d) We have

p1 np − q p 1 = = − . q1 nq q n Since 1/n < p/q, equation (9) shows that 0< Since

p1 . q1

p 1 < , q n−1

we have

np − p < q or p1 = np − q < p.

The third inequality is established.

Copyright c 2018 Pearson Education, Inc.

(9)

37

SOLUTIONS Now p p 1p 1 1 p1 < = = < ·1= . q1 q1 nq nq n n In particular,

(10)

p1 < 1. q1

We have established the second inequality.

By the inductive assumption, p1 /q1 can be expressed in Egyptian form. The last equation follows. (e) See (10). (f) The equation is true because of (d). For any i = 1, . . . , k, 1 1 p1 1 1 ≤ + ···+ = < . ni n1 nk q1 n It follows that n, n1 , . . . , nk are distinct.

1 5 1 13 1 26. 38 = 13 + 24 , 7 = 12 + 15 + 70 , 19 = 21 + 16 + 57

29. Enclose the missing square in a corner (n − 3) × (n − 3) subboard as shown in the following figure. Since 3 divides n2 − 1, 3 also divides (n − 3)2 − 1. Now n − 3 is odd, n − 3 > 5, and 3 divides (n − 3)2 − 1, so by Exercise 28, we may tile this subboard. Tile the two 3 × (n − 4) subboards using the result of Exercise 34, Section 2.4. Tile the deficient 4 × 4 subboard using Example 2.4.7. The n × n board is tiled. n

(n − 3) × (n − 3)

3×(n−4)

n

3 × (n − 4)

4×4

30. See I. P. Chu and R. Johnsonbaugh, “Tiling boards with trominoes,” J. Recr. Math., 18 (1985–86) 188–193.

32.

33. See C. Jepsen, “On tiling deficient boards with trominoes,” J. Recr. Math., 27 (1995) 125–130.

34. If n = 0, d · 1 = d > 0, and 1 is in X. If n > 0, d(2n) = n(2d) > n; thus 2n is in X. In either case X is nonempty. Since d > 0 and n ≥ 0, X contains only positive integers. By the Well-Ordering Property, X contains a least element q 0 > 0. Then dq 0 > n. Let q = q 0 − 1. We cannot have dq > n (for then q 0 would not be the least element in X); therefore, dq ≤ n. Let r = n − dq. Then r ≥ 0. Also r = n − dq = n − d(q 0 − 1) < dq 0 − d(q 0 − 1) = d.

Therefore, we have found q and r satisfying n = dq + r

0 ≤ r < d.

Copyright c 2018 Pearson Education, Inc.

38

SOLUTIONS

35. We first prove Theorem 2.5.6 for n > 0. The Basis Step is n = 1. If d = 1, we have n = dq + r, where q = n and r = 0, 0 ≤ r < d. If d > 1, we have n = dq + r, where q = 0 and r = 1, 0 ≤ r < d. Thus Theorem 2.5.6 is true for n = 1.

Assume that Theorem 2.5.6 holds for n. Then there exists q 0 and r 0 such that n = dq 0 + r 0

0 ≤ r 0 < d.

Now n + 1 = dq 0 + (r 0 + 1).

If r 0 < d − 1, then r 0 + 1 < d. In this case, if we take q = q 0 and r = r 0 + 1, we have n + 1 = dq + r If r 0 = d − 1, we have

0 ≤ r < d.

n + 1 = d(q 0 + 1).

In this case, if we take q = q 0 + 1 and r = 0, we have n + 1 = dq + r

0 ≤ r < d.

The Inductive Step is complete. Therefore, Theorem 2.5.6 is true for all n > 0.

If n = 0, we may write n = dq + r, where q = r = 0. Therefore, Theorem 2.5.6 is true for n = 0.

Finally, suppose that n < 0. Then −n > 0, so by the first part of the proof, there exist q 0 and r 0 such that −n = dq 0 + r 0 0 ≤ r 0 < d.

If r 0 = 0, we may take q = −q 0 and r = 0 to obtain

n = dq + 0.

If r 0 > 0, we take q = −q 0 − 1 and r = d − r 0 . Then 0 < r < d and n = d(−q 0 ) − r 0 = d(q + 1) + (r − d) = dq + r.

Therefore, Theorem 2.5.6 is true for n < 0.

37. Suppose that we have a propositional function S(n) whose domain of discourse is the set of integers greater than or equal to n0 . Suppose that S(n0 ) is true and, for all n > n0 , if S(k) is true for all k, n0 ≤ k < n, then S(n) is true. We must prove that S(n) is true for every integer n ≥ n0 . We first assume that n0 ≥ 0.

We argue by contradiction. So assume that S(n) is false for some integer n1 ≥ n0 . Let X be the set of nonnegative integers for which S(n) is false. Then X is nonempty. By the Well-Ordering Property, X has a least element n2 . Since S(n0 ) is true, n2 > n0 . Furthermore, for any k, n0 ≤ k < n2 , S(k) is true [otherwise n2 would not be the least integer n for which S(n) is false]. Since S(k) is true for all k, n0 ≤ k < n2 , by hypothesis, S(n2 ) is true. Contradiction.

If n0 < 0, apply the previous argument to the propositional function S 0 (n) : S(n + n0 ) with domain of discourse the set of nonnegative integers.

38. The strong form of induction clearly implies the form of induction where the Inductive Step is: “If S(n) is true, then S(n + 1) is true.” For the converse, use Exercises 36 and 37.

Copyright c 2018 Pearson Education, Inc.

39

SOLUTIONS

Section 3.1 2. Check digit is incorrect.

3. Check digit is incorrect.

6. Check digit is correct.

5. Check digit is correct.

9. Not a function

10. It is a function from X to Y ; domain = X, range = Y ; it is both one-to-one and onto. Its arrow diagram is 1 r ra S 7

2 r S r b S S7

3 r S w Sr c S 4 r w Sr d X Y

The inverse function is {(c, 1), (d, 2), (a, 3), (b, 4)}. For the inverse function, domain = Y , range = X. Its arrow diagram is a r r1 S 7

b r S r 2 S S7

c r S w Sr 3 S d r w Sr 4 Y X

12. It is a function from X to Y ; domain = X, range = {b}. Its arrow diagram is ra 1 r Q Q s -rb 2 r Q 3

7 3 r r c

4 r rd X Y

It is neither one-to-one nor onto.

14. y

6

2 -

r

−4

r

−3

r

−2

r

−1

1

r

−1 -

r

1

r

2

r

3

r - x

4

−2 -

Copyright c 2018 Pearson Education, Inc.

40

SOLUTIONS

15.

4

···

(

3

]

(

(

r r r

√ √ −2 − 3 − 2

]

( −1

(

r

(

r

(

1 (

]

2

]

r6

]

r

1

]

···

]

]

r √r r -

√

2

3 2

18. Since f(1) = f(−1), f is not one-to-one. Since f(n) = n2 − 1 ≥ −1, for no n do we have f(n) = −2. Therefore f is not onto.

19. Since f(1) = f(2), f is not one-to-one. f is onto. To prove this, let m ∈ Z. Taking n = 2m, we have f(n) = f(2m) = d(2m)/2e = dme = m.

Therefore f is onto.

21. f is one-to-one. To prove this, suppose that f(m) = f(n). Then 2m = 2n. Canceling 2, we have m = n.

Therefore f is one-to-one.

Since f(n) = 2n is an even integer for every n, f(n) 6= 3 for every n. Therefore f is not onto.

22. f is one-to-one. To prove this, suppose that f(m) = f(n). Then m3 = n3 . Therefore m = n and f is one-to-one.

Since f(n) = n3 is a cube for every n and 5 is not a cube, f(n) 6= 5 for every n. Therefore f is not onto.

24. Since f(1, 0) = 1 = f(1, 1), f is not one-to-one. f is onto. To prove this, let m ∈ Z. Then f(m, 0) = m.

Therefore f is onto.

25. Since f(1, 2) = 2 = f(2, 1), f is not one-to-one. f is onto. To prove this, let m ∈ Z. Then f(m, 1) = m.

Therefore f is onto.

27. Since f(0, 1) = f(0, −1), f is not one-to-one. Since f(m, n) = n2 + 1 ≥ 1 for all m and n, f(m, n) 6= 0 for all m and n. Therefore f is not onto.

28. Since f(1, 0) = f(0, 1), f is not one-to-one. f is onto. To prove this, let k ∈ Z. Then f(k − 2, 0) = (k − 2) + 0 + 2 = k.

Therefore f is onto.

31. Not one-to-one. f(4/3) = f(−2/3). Not onto. f(x) 6= 0 for any real x.

Copyright c 2018 Pearson Education, Inc.

41

SOLUTIONS 32. Not one-to-one. sin 0 = sin 2π. Not onto. sin x 6= 2 for any real x.

34. One-to-one. Not onto. f(x) 6= −2 for any real x.

35. Not one-to-one. Notice that f(x) = f(1/x). Thus any value of x, x 6= 0, x 6= 1, shows that f is not one-to-one. Not onto. f(x) 6= 1 for any real x. (In fact, −1/2 ≤ f(x) ≤ 1/2 for all real x.)

37. Let f be the function from X = {a, b} to Y = {y} given by f = {(a, y), (b, y)}.

38. The function {(1, 1), (2, 1)} from {1, 2} to {1, 2}.

40. ¬(∀x1 ∀x2 (f(x1 ) = f(x2 )) → (x1 = x2 )) ≡ ∃x1 ∃x2 f(x1 ) = f(x2 ) ∧ x1 6= x2 41. ∀y ∈ Y ∃x ∈ Xf(x) = y 44. f

−1

(y) = log3 y

45. f

−1

(y) = 2

y/3

47. f

−1

(y) =

y+5 4

1/3

48. f −1 (y) = [log2 (y − 6) + 1]/7 50. (f ◦ f)(x) = 2(2n + 1) + 1, (g ◦ g)(x) = 3(3n − 1) − 1, (f ◦ g)(x) = 2(3n − 1) + 1, (g ◦ f)(x) = 3(2n + 1) − 1 n

51. (f ◦ f)(x) = n4 , (g ◦ g)(x) = 22 , (f ◦ g)(x) = 22n , (g ◦ f)(x) = 2n

2

53. Let p be the original price of the item; let A be the amount that amount off that the coupon specifies; let D, as a decimal, be the discount; let C(p) be the cost if only the coupon is used; and let S(p) be the cost if only the discount is used. Then C(p) = p−A and S(p) = (1 −D)p. Therefore S(C(p)) = (1 −D)(p−A) is the cost if first the coupon is used then the discount is used, and C(S(p)) = (1 − D)p − A is the cost if first the discount is used and then the coupon is used. Since (1 − D)(p − A) < (1 − D)p − A, it is always cheapest to use the discount first.

55. g(x) = 1/x, h(x) = 2x, w(x) = x2 , (g ◦ h ◦ w)(x) = f(x) 56. g(x) = 2x, h(x) = sin x, f(x) = (h ◦ g)(x) 58. g(x) = x4 , h(x) = 3 + x, w(x) = sin x, (g ◦ h ◦ w)(x) = f(x) 59. g(x) = 1/x3 , h(x) = 6x, t(x) = cos x, f(x) = (g ◦ t ◦ h)(x) 61. 4; one-to-one functions: {(1, a), (2, b)} and {(1, b), (2, a)}. In this case, the onto and one-to-one functions are the same.

62. (a) f ◦ f = {(a, a), (b, b), (c, a)}, f ◦ f ◦ f = {(a, b), (b, a), (c, b)} (b) f 9 = f, f 623 = f

64. f = {(0, 0), (1, 4), (2, 2), (3, 0), (4, 4), (5, 2)}. f is neither one-to-one nor onto. The arrow diagram of f is 0 r 1 r

-r 0

r1

X

X

2 r A - r 2

A

3 r

A A r 3 A UAr 4 4 r r f r5 5 Copyright c 2018 Pearson Education, Inc.

42

SOLUTIONS

67. Incorrect check digit

68. Correct check digit

70. Correct check digit

71. Incorrect check digit 74. 714 : 0, 631 : 2, 26 : 9, 373 : 16, 775 : 10, 906 : 5, 509 : 1, 2032 : 11, 42 : 8, 4 : 4, 136 : 3, 1028 : 12 75. 53 : 4, 13 : 5, 281 : 3, 743 : 6, 377 : 9, 20 : 7, 10 : 1, 796 : 8 77. During a search if we stop the search at an empty cell, we may not find the item even if it is present.

The cell may be empty because an item was deleted. One solution is to mark deleted cells and consider them nonempty during a search.

78. No. If the data item is present, it will be found before an empty cell is encountered.

81. False. Let X = {1}, Y = {a, b}, Z = {α, β}. A counterexample is f = {(a, α), (b, β)}, g = {(1, a)}.

82. False. Let X = {1, 2}, Y = {a, b}, Z = {α, β}. A counterexample is f = {(a, α), (b, α)}, g = {(1, a), (2, b)}.

84. True 85. False. Let X = {1}, Y = {a, b}, Z = {α}. A counterexample is f = {(a, α), (b, α)}, g = {(1, a)}.

87. True. Let z ∈ Z. Since f ◦ g is onto, there exists x ∈ X such that f(g(x)) = z. Let y = g(x). Then f(y) = z. Therefore f is onto Z. 89. Suppose that f is not one-to-one. Then, for some x and y, f(x) = f(y), but x 6= y. Let A = {x}, B = {y}.

Suppose that f is one-to-one. Let y ∈ f(A∩B). Then y = f(x) for some x ∈ A∩B. Thus y ∈ f(A)∩f(B).

Let y ∈ f(A) ∩ f(B). Then y = f(a) = f(b), for some a ∈ A, b ∈ B. Since f is one-to-one, a = b.

Therefore, y ∈ f(A ∩ B).

90. [The case: If g is one-to-one, then f ◦ g is one-to-one implies that f is one-to-one.]

Suppose that f is not one-to-one. Then there exist distinct x1 , x2 ∈ X with f(x1 ) = f(x2 ). Let A = {1, 2}, and let g = {(1, x1 ), (2, x2)}. Now g is one-to-one, but f ◦ g is not one-to-one, which is a contradiction.

91. Suppose that f is onto Y . Let g be a function from Y onto Z. We must show that g ◦ f is onto Z. Let z ∈ Z. Since g is onto Z, there exists y ∈ Y , with g(y) = z. Since f is onto Y , there exists x ∈ X, with f(x) = y. Now g ◦ f(x) = z. Therefore, g ◦ f is onto.

Suppose that whenever g is a function from Y onto Z, g ◦ f is onto Z. Suppose that f is not onto Y .

Then there exists y0 ∈ Y such that for no x ∈ X do we have f(x) = y0 . Let Z = {0, 1}. Define g from Y to Z by g(y0 ) = 1, and g(y) = 0 if y 6= y0 . Then g is onto Z, but g ◦ f is not onto Z.

92. If x ∈ X, then x ∈ f −1 (f({x})). Thus ∪{S | S ∈ S} = X. Suppose that

a ∈ f −1 ({y}) ∩ f −1 ({z}) for some y, z ∈ Y . Then f(a) = y and f(a) = z. Thus y = z. Therefore, S is a partition of X. 94. E1 is onto. To prove this, let y ∈ R. Define f : R → R by the rule f(x) = x − 1 + y. Now E1 (f) = f(1) = 1 − 1 + y = y.

Therefore f is onto.

Copyright c 2018 Pearson Education, Inc.

43

SOLUTIONS

95. The proof is by induction on n. We omit the Basis Step. For the Inductive Step, assume that the assertion is true for n. Replacing x by r n x in f(rx) = rf(x), we obtain f(r(r n x)) = rf(r n x). By the inductive assumption, f(r n x) = r n f(x). Thus f(r n+1 x) = f(r(r n x)) = rf(r n x) = r(r n f(x)) = r n+1 f(x).

97. S(∅) = 000

98. S(X) = 111

100. Let s1 s2 s3 be a bit string of length 3. Define Y as follows: Put a in Y if and only if s1 = 1; put b in Y if and only if s2 = 1; and put c in Y if and only if s3 = 1. Then Y ⊆ X and S(Y ) = s1 s2 s3 . Therefore S is onto.

102. If x ∈ X − Y , then CX∪Y (x) = 1 = 1 + 0 − 1 · 0 = CX (x) + CY (x) − CX (x)CY (x).

Similarly, if x ∈ Y − X, the equation holds. If x ∈ X ∩ Y , then CX∪Y (x) = 1 = 1 + 1 − 1 · 1 = CX (x) + CY (x) − CX (x)CY (x).

If x 6∈ X ∪ Y , then CX∪Y (x) = 0 = 0 + 0 − 0 · 0 = CX (x) + CY (x) − CX (x)CY (x).

Thus the equation holds for all x ∈ U .

103. If x ∈ X, then x 6∈ X; thus, If x 6∈ X, then x ∈ X; thus,

CX (x) = 0 = 1 − 1 = 1 − CX (x).

CX (x) = 1 = 1 − 0 = 1 − CX (x).

105. If CX (x) = 0, the inequality obviously holds. If CX (x) = 1, then x ∈ X. Since X ⊆ Y , x ∈ Y . Thus CY (x) = 1 also. Again, the inequality holds.

106. CX∆Y (x) = CX (x) + CY (x) − 2CX (x)CY (x) 108. Suppose that there is a one-to-one function f from X to Y . Let R be the range of f and choose a ∈ X. If y ∈ R, let g(y) = f −1 (y). If y ∈ Y − R, let g(y) = a. Then g is a function from Y onto X.

Suppose that there is a function g from Y onto X. For each x ∈ X, choose one y ∈ Y with g(y) = x.

Define f(x) = y. Then f is a one-to-one function from X to Y .

110. f is not a binary operator since the range of f is not contained in X. 111. f is a commutative, binary operator.

113. f is a commutative, binary operator. To see this, note that if x, y ∈ X, then 1 ≤ xy. Now 0 ≤ (x − y)2 = x2 − 2xy + y2 ; hence, 1 ≤ xy ≤ x2 − xy + y2 = f(x, y).

115. f(X) = X ∪ {1} Copyright c 2018 Pearson Education, Inc.

44

SOLUTIONS

116. Let R denote the set {(y, x) | (x, y) ∈ f}.

The set of y such that (y, x) ∈ R is Y since f is onto. If (y, x), (y, x0 ) ∈ R, then x = x0 since f is one-to-one. Thus R is a function from Y to X. For each x ∈ X, there is exactly one y ∈ Y with R(y) = x since f is a function. Therefore R is one-to-one and onto.

118. False. A counterexample is x = y = 1.5.

119. False. A counterexample is x = 2, y = 2.6.

121. Since n is an odd integer, n = 2k + 1 for some integer k. Now 2 2

n 4k + 4k + 1 1 = = k2 + k + = k2 + k + 1 4 4 4 and

n2 + 3 (4k 2 + 4k + 1) + 3 = = k 2 + k + 1.

4 4

122. x = 1.5 124. Suppose that dxe = n. By definition, n is the least integer greater than or equal to x. Thus x ≤ n and n − 1 is less than x, that is, n − 1 < x ≤ n. Thus, if we let ε = n − x, then 0 ≤ ε < 1 and x + ε = n.

Now suppose that there exists ε satisfying 0 ≤ ε < 1 and x + ε = n. Then n − 1 < x ≤ n and so n is the least integer greater than or equal to x. Therefore dxe = n.

125. For all real numbers x and integers n, bxc = n if and only if there exists ε, 0 ≤ ε < 1, such that x −ε = n.

The proof is similar to the proof in the solution to Exercise 124.

128. January, April, July

Section 3.2 4. We must show that si < sj for j = i + 1, i + 2, . . . . The Basis Step is j = i + 1, which is given. For the Inductive Step, suppose that si < sj . Then si < sj < sj+1 . Therefore si < sj for all j, i < j and j ∈ D. 6. (c) nonincreasing

7. (b) decreasing and (c) nonincreasing

9. (a) increasing and (d) nondecreasing 14. 9

15. 45

16. 15

10. 5 17. 3465

11. 13

12. 199

13. 4153

18. sn = 2(n + 1) − 1

19. Yes

24. 26

26. 8

20. No

21. No

22. Yes

23. 8

27. Yes

28. No

29. No

30. Yes

37. No

40. Yes

41. Yes

42. Yes

43. Yes

44. Yes

25. 41 38. No

39. Yes

57. 2

58. 5

59. cn = n/2, if n is even; cn = (n − 1)/2 − n, if n is odd.

60. dn = (−1)n/2 n!, if n is even; dn = (−1)(n+1)/2 n!, if n is odd.

61. No

62. No

63. No

64. No

65. 9

68. 3n

69. No

70. No

71. Yes

72. Yes

Copyright c 2018 Pearson Education, Inc.

66. 30 80. 3/4

67. 3n

45

SOLUTIONS 81. 10/11 86. Yes

83. 1/[(n + 1)(n!)2 ]

82. 1 − 1/(n + 1) 87. No

89. 3/64

84. No

85. Yes

90. 2/9

92. We omit a proof of the formula for an . Now

1 1 1 1 1 z100 = − + − 4 982 1002 972 992

1 1 1 1 + − + − +··· 962 982 952 972 .. .

1 1 1 1 − 2 + − 2 + 22 4 12 3

1 1 1 1 1 = − − + + = 0.312449492.

4 1002 992 22 12 93. No: a3 = 2/9, a4 = 3/64.

95. Yes, see the solution to Exercise 94.

96. No, see the solution to Exercise 93.

98. No, see the solution to Exercise 97.

99. No, see the solution to Exercise 97.

103. s3 = 8. For 8, we have 3 < 6 < 8. Consider replacing 8 with a smaller value. 4 is a perfect square, so it can be eliminated. 5 would require another multiple of 5 greater than 8. 6 would require a multiple of 2 greater than 6. 7 would require a multiple of 7 greater than 8. Therefore s3 = 8.

104. s5 = 10. For 10, we have 5 < 8 < 10. Any choices of ai must include a multiple of 5. Since 10 is the smallest multiple of 5, s5 = 10.

106. First note that for all p, sp ≥ 2p since a multiple of p must occur among the ai and 2p is the least multiple of p. Also note that if we can find an integer k satisfying p < 2k 2 < 2p, then sp = 2p since p · 2k 2 · 2p is a perfect square.

For p = 5, we have 5 < 2 · 22 < 2 · 5. For p = 7, we have 7 < 2 · 22 < 2 · 7. For p = 11, we have 11 < 2 · 32 < 2 · 11. Suppose that p ≥ 13. Then

√ 1 1 p 1− √ ≥ 13 1 − √ > 1.

2 2 √ √ √ √ √ √ Therefore the interval ( p/ 2, p) contains an integer, say k; so p/ 2 < k < p. Squaring and 2 multiplying by 2 gives p < 2k < 2p. The proof is complete.

107. Since s11 = 22 > 20 = s12 , s is not increasing.

108. 3n n!

113. 21 , 22, 23 , 24 , 25 , 26 , 27

k(k − 1) + 2 2 123. −88 124. −476

115. nk =

126. 3 · 2n−1 − 4 · 5n−1

114. 21 , 22 , 24 , 27, 211 , 216, 222

116. tnk = 2nk = 2[k(k−1)+2]/2

121. −1

125. 3 · 2p − 4 · 5p 127. 3 · 2n−2 − 4 · 5n−2

128. 7rn−1 − 10rn−2 = 7(3 · 2n−1 − 4 · 5n−1 ) − 10(3 · 2n−2 − 4 · 5n−2) = 3(7 · 2n−1 − 10 · 2n−2 ) − 4(7 · 5n−1 − 10 · 5n−2 )

n n = 3 72 2n − 10 − 4 75 5n − 10 4 2 25 5 = 3 · 2n − 4 · 5n = rn Copyright c 2018 Pearson Education, Inc.

122. −14

46

SOLUTIONS

129. 2

130. 9

131. 36

133. (2 + i)3i

132. 135

134. (1 + n)3n−1

135. n3n−2 136.

138.

6zn−1 − 9zn−2

= =

6(1 + n)3n−1 − 9(n3n−2 ) = 2(1 + n)3n − n3n 3n [2(1 + n) − n] = (2 + n)3n = zn

n−1 X

(k + 1)2 r n−k−1

139.

n−1 X

Ci Cn−i−1

i=0

k=0

141. The first sum is the sum by rows of the matrix  a11 a12  a22   

a13 a23

... ... .. .

 a1n a2n   ,  ann

and the second sum is the sum by columns of the same array. Thus the two sums are equal.

142. (a) baabcaaba (e) 9 (f) 9 (k) baabcaababbab

(b) caababaab (c) baabbaab (d) caabacaaba (g) 8 (h) 10 (i) baab (j) caaba (l) caabacaababbabbaab

144. λ, 0, 1, 00, 01, 10, 11

145. 000, 001, 010, 011, 100, 101, 110, 111

147. λ, b, a, c, ba, ab, bc, bab, abc, babc 148. λ, a, b, aa, ab, ba, bb, aab, aba, baa, abb, aaba, abaa, baab, aabb, aabaa, abaab, baabb, aabaab, abaabb, aabaabb 150. The Basis Steps (n = 1, 2) are omitted.

For the Inductive Step, assume that n ≥ 3 and the formula is true for 1 ≤ i < n. Then an n!

= = =

n−1 (an−1 + an−2 ) n! " # n−1 n−2 X X (−1)k n−1 (−1)k (n − 1)! + (n − 2)! n! k! k! k=0 k=0 " n # " n # n − 1 X (−1)k (−1)n 1 X (−1)k (−1)n−1 (−1)n − − − + n k! n! n k! (n − 1)! n! k=0

=

n X (−1)k

k=0

k!

k=0

.

152. We prove only the case x < x1 . The case x > xn is treated similarly.

Since x < x1 , |x − x1 | = x1 − x and |x − xn | = xn − x. Thus n X i=1

|x − xi | =

(x1 − x) +

>

(x − x1 ) +

=

n−1 X i=2

n−1 X

i=2 n−1 X i=2

|x − xi | + (xn − x) |x − xi | + (xn − x)

|x − xi | + (xn − x1 ).

Copyright c 2018 Pearson Education, Inc.

47

SOLUTIONS 153. We use strong induction.

Basis Steps (n = 1, 2).

P1 If n = 1, i=1 |x − xi | = |x − x1 |, which is a minimum when x = x1 , the median.

Now suppose that n = 2. If x1 ≤ x ≤ x2 , then 2 X i=1

|x − xi | = (x − x1 ) + (x2 − x) = x2 − x1 .

However, if x < x1 , 2 X

|x − xi | = (x1 − x) + (x2 − x) > (x − x1 ) + (x2 − x) = x2 − x1 ,

2 X

|x − xi | = (x − x1 ) + (x − x2 ) > (x − x1 ) + (x2 − x) = x2 − x1 .

i=1

and if x > x2 ,

i=1

Thus the minimum error, x2 − x1 , is achieved when x is a median, a value between x1 and x2 . The proof of the Basis Step is complete.

Inductive Step. Suppose that the minimum error for k < n values, n ≥ 3, is achieved when x is a median. We prove that the minimum error for n values is achieved when x is a median.

By the inductive assumption, the minimum error, which we denote M E, for x2 , x3, . . . , xn−1, is achieved by taking x to be a median of x2 , x3 , . . . , xn−1. If x1 ≤ x ≤ xn , by Exercise 151, we have n X i=1

|x − xi | =

n−1 X i=2

|x − xi | + (xn − x1 ) ≥ M E + (xn − x1 ).

The minimum value, M E + (xn − x1 ), can be achieved by taking x to be a median of x2 , x3 , . . . , xn−1 . Since a median Pnof x2 , x3, . . . , xn−1 is equal to a median of x1 , x2 , . . . , xn , for x satisfying x1 ≤ x ≤ xn the minimum of i=1 |x − xi | is achieved by taking x to be a median of x1 , x2 , . . . , xn .

If x < x1 or x > xn , by Exercise 152 n X i=1

|x − xi | >

n−1 X i=2

|x − xi | + (xn − x1 ) ≥ M E + (xn − x1 ).

Pn Therefore, if x < x1 or x > xn , the sum i=1 |x − xi | is not a minimum.

Pn Therefore, the minimum of i=1 |x − xi | is obtained by taking x to be a median of x1 , x2 , . . . , xn . The proof of the Inductive Step is complete.

156. f is one-to-one. Suppose that f(α) = f(β). We then have αα = ββ. Now the first half of the first string, namely α, must equal the first half of the second string, namely β. Therefore f is one-to-one.

Notice that |f(α)| is an even integer for all α ∈ X ∗ . Therefore f(α) 6= aaa for all α ∈ X ∗ . Thus f is not onto.

157. f is one-to-one. Suppose that f(α) = f(β). We then have ααR = ββ R . Now the first half of the first string, namely α, must equal the first half of the second string, namely β. Therefore f is one-to-one.

Notice that |f(α)| is an even integer for all α ∈ X ∗ . Therefore f(α) 6= aaa for all α ∈ X ∗ . Thus f is not onto.

Copyright c 2018 Pearson Education, Inc.

48

SOLUTIONS

159. Taking α = λ, the first rule tells us that aαb = ab ∈ L and bαa = ba ∈ L. Taking α = ba and β = ab, the second rule tells us that αβ = baab ∈ L. Finally, taking α = baab and β = ab, the second rule tells us that αβ = baabab ∈ L. 160. Exercise 161 shows that if α ∈ L, α has equal numbers of a’s and b’s. Since aab has more a’s than b’s, aab is not in L. 162. We use strong induction on the length n of α to show that if α has equal numbers of a’s and b’s, then α ∈ L. The Basis Step is n = 0. In this case, α is the null string, and the null string is in L by the definition of L. Suppose that α has length n > 0, and α has equal numbers of a’s and b’s. Notice that, because of the first rule, the length of α is at least two. First suppose that α starts with a and ends with b, that is, α = aβb. Then β has length less than n, and β has equal numbers of a’s and b’s. By the inductive assumption, β ∈ L. By the first rule, α = aβb ∈ L. Similarly, if α starts with a and ends with b, then α ∈ L.

Now suppose that α has equal numbers of a’s and b’s and α starts with a and ends with a, that is, α = aβa. We claim that some proper substring of α starting at the beginning contains equal numbers of a’s and b’s; that is, we claim that α = γδ, where γ and δ have equal numbers of a’s and b’s, and neither γ nor δ is the null string. Assuming that this claim is true, by the inductive assumption γ and δ are in L, and it follows from the second rule that α ∈ L. The Inductive Step is complete.

To prove the claim, for each substring ε of α starting at the beginning, consider val(ε) = number of a’s − number of b’s.

Consider val(ε) for substrings ε of increasing length. For the first substring, a, val(a) = 1. For the next-to-last substring (α with the trailing a omitted), we have val(ε) = −1. When the length of the substring increases by one, val(ε) increases or decreases by one. Since val’s first value is 1 and its last value is −1, for some substring β, val(β) = 0. Thus β has equal numbers of a’s and b’s, and the claim is proved.

163. Let ε > 0. Since L − ε is not an upper bound of {an | n = 1, 2, . . .}, there exists N such that L − ε < aN .

Since {an }∞ n=1 is a nondecreasing sequence, for all n ≥ N we have L − ε < aN ≤ an . Since L is an upper bound of {an | n = 1, 2, . . .}, an ≤ L for all n.

Section 3.3 2. {(Sally,Math), (Ruth,Physics), (Sam,Econ)} 6.

1 2 3 4 2 3 4 2 3 4 2 3 4

3. {(a, 3), (b, 1), (b, 4), (c, 1)}

1 1 1 1 2 2 2 3 3 3 4 4 4 Copyright c 2018 Pearson Education, Inc.

49

SOLUTIONS 7.

Roger Pat Ben Pat

1 r 6

10.

4

r

Music History Math PolySci

-r 2

11. i r 1

r?

3

2 ri 3r

i i

14. {(1, 1), (2, 2), (3, 3), (3, 5), (4, 3), (4, 4), (5, 5), (5, 4)}

15. ∅

19. {(1, 1), (4, 1), (2, 2), (5, 2), (3, 3), (1, 4), (4, 4), (2, 5), (5, 5)} 21. R = {(1, 2), (2, 3), (3, 4), (4, 5)}

R−1 = {(2, 1), (3, 2), (4, 3), (5, 4)}

22. Symmetric

25. Symmetric

28. Antisymmetric, transitive

26. Reflexive, symmetric, transitive 29. Reflexive, antisymmetric, transitive, partial order

31. Reflexive, symmetric, transitive

32. Reflexive, symmetric, transitive

34. Symmetric

36. Let R be a relation on a set X. Suppose that R is antisymmetric. Let (x, y) ∈ R with x 6= y. Suppose, by way of contradiction, (y, x) ∈ R. By the definition of antisymmetric, since (x, y), (y, x) ∈ R, x = y, which is a contradiction. Therefore (y, x) ∈ / R.

Now suppose that for all x, y ∈ X, if (x, y) ∈ R and x 6= y, then (y, x) ∈ / R. We show that R is antisymmetric. Let (x, y), (y, x) ∈ R. Suppose, by way of contradiction, that x 6= y. Since (x, y) ∈ R and x 6= y, (y, x) ∈ / R, which is a contradiction. Therefore x = y and so R is antisymmetric.

37. Reflexive, symmetric 39. R1 ◦ R2 = {(1, 1), (1, 2), (2, 1), (2, 2), (3, 1), (3, 2), (4, 2)}, R2 ◦ R1 = {(1, 1), (1, 2), (3, 4), (4, 1), (4, 2)} 41. {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 3)} 42. {(1, 1), (2, 2), (3, 3), (4, 4), (1, 2), (2, 3)} 44. {(1, 2), (2, 3), (1, 3)} 46. True

47. True

49. True

52. False. Let R = {(2, 3), (3, 2)}, S = {(1, 2), (2, 1)}.

50. True 53. True

55. True

56. False. Let R = {(2, 3), (1, 1)}, S = {(1, 2), (3, 1)}.

58. Let X be an n-element set. A relation on X is a subset of X × X. Since |X × X| = n2 and |P(X × X)| = 2 2 2n , there are 2n relations on an n-element set.

60. R is reflexive, not symmetric, not antisymmetric, transitive, and not a partial order. To see that R is not symmetric, consider A = {1} and B = {1, 2}. To see that R is not antisymmetric, consider A = all real numbers and B = all rational numbers.

Copyright c 2018 Pearson Education, Inc.

50

SOLUTIONS

61. R is reflexive, symmetric, not antisymmetric, transitive, and not a partial order. To see that R is not antisymmetric, consider A = all real numbers and B = all rational numbers.

62. If may be the case that for x ∈ X, there is no y ∈ X such that (x, y) ∈ R. Consider, for example, X = {1, 2, 3}, R = {(1, 1), (2, 2), (1, 2), (2, 1)}, and x = 3.

Section 3.4 2. Not an equivalence relation (not transitive) 3. Not an equivalence relation (not reflexive) 5. Equivalence relation. [1] = [2] = [3] = [4] = [5] = {1, 2, 3, 4, 5}.

6. Equivalence relation. [1] = [5] = {1, 5}, [2] = {2}, [3] = {3}, [4] = {4}.

8. Not an equivalence relation (not reflexive, not symmetric, not transitive) 9. Not an equivalence relation (not reflexive) 13. Equivalence relation

12. Not an equivalence relation (not transitive)

15. Equivalence relation

16. Equivalence relation

18. {(1, 1), (2, 2), (3, 3), (4, 4), (3, 4), (4, 3)}, [1] = {1}, [2] = {2}, [3] = [4] = {3, 4} 19. {(1, 1), (2, 2), (3, 3), (4, 4)}, [i] = {i} for i = 1, . . . , 4 21. {(i, j) | i, j ∈ {1, 2, 3, 4}}, [1] = [2] = [3] = [4] = {1, 2, 3, 4} 22. {(1, 1), (2, 2), (2, 4), (4, 2), (4, 4), (3, 3)}, [1] = {1}, [2] = [4] = {2, 4}, [3] = {3} 23. Reflexive: ARA since A ∪ Y = A ∪ Y .

Symmetric: If ARB, then A ∪ Y = B ∪ Y . Now B ∪ Y = A ∪ Y , so BRA.

Transitive: Suppose that ARB and BRC. Then A ∪ Y = B ∪ Y and B ∪ Y = C ∪ Y . Therefore A ∪ Y = C ∪ Y . Thus ARC. 25. Eight. An equivalence class is determined by the presence or absence of 1, 2, and 5.

27. If R is a relation on X, then R = X × X. 29. {(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6), (1, 2), (2, 1), (3, 4), (4, 3)} 30. Five, corresponding to the partitions {{1}, {2}, {3}}, {{1}, {2, 3}}, {{1, 2}, {3}}, {{1, 3}, {2}}, {{1, 2, 3}} 32. Reflexive: For all f ∈ RR , f(0) = f(0). Therefore fRf and R is reflexive.

Symmetric: Suppose that fRg. Then f(0) = g(0). Since g(0) = f(0), gRf and R is symmetric.

Transitive: Suppose that fRg and gRh. Then f(0) = g(0) and g(0) = h(0). Therefore f(0) = g(0) and fRh. Thus R is transitive.

If f(x) = x for all x ∈ R, [f] consists of all functions g from R to R satisfying g(0) = 0.

33. Reflexive: Taking a = b = 0, we have f(x) = f(x + a) + b for all x ∈ R. Therefore fRg for all f ∈ RR.

Symmetric: Suppose that fRg. Then there exist a, b ∈ R such that f(x) = g(x + a) + b for all x ∈ R. Now g(x) = g((x − a) + a) = f(x − a) − b = f(x + (−a)) + (−b) for all x ∈ R. Therefore gRf. Copyright c 2018 Pearson Education, Inc.

51

SOLUTIONS Transitive: Suppose that fRg and gRh. Then there exist a, b, c, d ∈ R such that f(x) = g(x + a) + b for all x ∈ R. Now

and

g(x) = h(x + c) + d

f(x) = g(x + a) + b = h(x + a + c) + b + d

for all x ∈ R. Therefore fRh.

All functions in an equivalence class are translates of one another.

35. (a) Reflexive: (a, b)R(a, b) for all a, b ∈ X since ab = ba for all a, b ∈ X. Symmetric: Suppose that (a, b)R(c, d). Then ad = bc. Since cb = da, (c, d)R(a, b).

Transitive: Suppose that (a, b)R(c, d) and (c, d)R(e, f). Then ad = bc and cf = de. Now af = adf/d = bcf/d = bde/d = be. Therefore (a, b)R(e, f). (b) (1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (1, 7), (1, 8), (1, 9), (1, 10), (2, 1), (2, 3), (2, 5), (2, 7), (2, 9), (3, 1), (3, 2), (3, 4), (3, 5), (3, 7), (3, 8), (3, 10), (4, 1), (4, 3), (4, 5), (4, 7), (4, 9), (5, 1), (5, 2), (5, 3), (5, 4), (5, 6), (5, 7), (5, 8), (5, 9), (6, 1), (6, 5), (6, 7), (7, 1), (7, 2), (7, 3), (7, 4), (7, 5), (7, 6), (7, 8), (7, 9), (7, 10), (8, 1), (8, 3), (8, 5), (8, 7), (8, 9), (9, 1), (9, 2), (9, 4), (9, 5), (9, 7), (9, 8), (9, 10), (10, 1), (10, 3), (10, 7), (10, 9) a c (c) (a, b)R(c, d) if and only if = . b d 36. We show symmetry only. Let (a, b) ∈ R ∩ R−1 . Then (a, b) ∈ R, so (b, a) ∈ R−1 . Since (a, b) ∈ R−1 , (b, a) ∈ R. Thus (b, a) ∈ R ∩ R−1 and R ∩ R−1 is symmetric.

38. R is reflexive since for every x, x ∈ S for some S ∈ S. R is also symmetric, for suppose that xRy. Then x, y ∈ S for some S ∈ S. Thus y, x ∈ S and yRx. R need not be transitive. Let X = {1, 2, 3}, S1 = {1, 2}, S2 = {2, 3}. Then X = S1 ∪ S2 . Now 1R2 and 2R3, but 16 R3.

39. (b) Cylinder 41. Reflexive: For every x ∈ X, by the definition of a function f(x) is defined. Since f(x) = f(x), xRx for every x ∈ X. Symmetric: Suppose that xRy. Then f(x) = f(y). Since f(y) = f(x), yRx.

Transitive: Suppose that xRy and yRz. Then f(x) = f(y) and f(y) = f(z). Therefore f(x) = f(z) and xRz. 42. Suppose that f = CY . The equivalence classes are Y and Y .

44. When x and y are in the same equivalence class 45. Suppose that [x] = [y]. Then xRy. Therefore, g(x) = g(y).

47. Assume that the sets [x] partition X. Since R is not symmetric, there exists (b, c) ∈ R and (c, b) ∈ / R. Therefore c ∈ / [b]. There exists d such that c ∈ [d]. Therefore (c, d) ∈ R. Because R is transitive, (b, d) ∈ R. Therefore b ∈ [d]. Since b ∈ [c], [c] = [d]. There exists f ∈ [b]. Therefore (f, b) ∈ R. Because R is transitive, (f, c) ∈ R. Therefore f ∈ [c]. It follows that [b] = [c] = [d]. Therefore c ∈ / [b] and c ∈ [d], which is a contradiction. Thus the collection of pseudo equivalence classes does not partition X. 48. Since R is not symmetric, there exist (a, b) ∈ R and (b, a) ∈ / R. Then a ∈ [b], but b ∈ / [a]. Since R is reflexive, a ∈ [a] and b ∈ [b]. Therefore [a] ∩ [b] 6= ∅ but [a] 6= [b]. Thus the collection of pseudo equivalence classes does not partition X. 50. Let X = {a, b, c} and R = {(a, b), (b, c), (c, a)}. Then [a] = {c}, [b] = {a}, [c] = {b} partitions X. Copyright c 2018 Pearson Education, Inc.

52

SOLUTIONS

51. Let X = {a, b, c, d} and R = {(b, a), (b, c), (d, a), (d, c), (a, b), (a, d), (c, b), (c, d)}.

Then [a] = [c] = {b, d} and [b] = [d] = {a, c} partitions X. 52. (b) Sequences are equivalent if their domains have the same cardinality and their first range values agree, their second range values agree, and so on. (c) For two sequences to be equal, their domains must be equal and their first range values agree, their second range values agree, and so on. Equivalent sequences can have different domains.

54. Since (y, y) ∈ {(x, x) | x ∈ X} for all y ∈ X, (y, y) ∈ ρ(R) for all y ∈ X. Thus ρ(R) is reflexive.

55. Let (x, y) be in R ∪ R−1 . If (x, y) is in R, (y, x) is in R−1 , so (y, x) is in R ∪ R−1 . If (x, y) is in R−1 , then (y, x) is in R, so (y, x) is in R ∪ R−1 . In any case, if (x, y) is in R ∪ R−1 , (y, x) is in R ∪ R−1 , so R ∪ R−1 is symmetric.

57. If R is any relation on a set X, R ⊆ ρ(R), R ⊆ σ(R), R ⊆ τ (R).

(11)

Thus, for any relation R, R ⊆ τ (σ(ρ(R))).

By (11), ρ(R) ⊆ τ (σ(ρ(R))) and by Exercise 54, ρ(R) is reflexive. Therefore, τ (σ(ρ(R))) is reflexive.

By Exercise 55, σ(ρ(R)) is symmetric. We show that if R0 is any symmetric relation, τ (R0 ) is symmetric.

We can then conclude that τ (σ(ρ(R))) is symmetric.

Let R0 be a symmetric relation. Let (x, y) ∈ τ (R0 ). Then there exist x = x0 , . . . , xn = y ∈ X such that (xi−1 , xi ) ∈ R0 for i = 1, . . . , n. Since R0 is symmetric, (xi , xi−1) ∈ R0 for i = 1, . . . , n. Thus (y, x) ∈ τ (R0 ) and τ (R0 ) is symmetric.

By Exercise 56, τ (σ(ρ(R))) is transitive; hence τ (σ(ρ(R))) is an equivalence relation containing R. 58. By Exercise 57, τ (σ(ρ(R))) is an equivalence relation containing R. We first observe that if R1 ⊆ R2 ⊆ X × X, then ρ(R1 ) ⊆ ρ(R2 ), σ(R1 ) ⊆ σ(R2 ), τ (R1 ) ⊆ τ (R2 ).

It follows that if R0 is a relation on X and R0 ⊇ R, τ (σ(ρ(R0 ))) ⊇ τ (σ(ρ(R))).

The conclusion will follow if we show that if R0 is an equivalence relation on X, R0 = τ (σ(ρ(R0 ))).

Suppose that R0 is an equivalence relation. Since R0 is reflexive, ρ(R0 ) = R0 . Since R0 is symmetric, σ(R0 ) = R0 . Thus σ(ρ(R0 )) = R0 . We show that τ (R0 ) = R0 . Clearly, R0 ⊆ τ (R0 ). Let (x, y) ∈ τ (R0 ).

Then (x, y) ∈ R0n for some positive integer n. Thus there exist x0 , . . . , xn ∈ X with x = x0 , y = xn , and (xi−1 , xi ) ∈ R0 for i = 1, . . . , n. Since R is transitive, (x, y) = (x0 , xn ) ∈ R0 . Thus R0 ⊇ τ (R0 ).

Therefore, R0 = τ (R0 ). Now R0 = τ (R0 ) = τ (σ(ρ(R0 ))).

61. False. Let R1 = {(1, 1), (1, 2)}, R2 = {(2, 2), (2, 1)}.

62. False. Let X = {1, 2, 3}, R1 = {(1, 2)}, R2 = {(2, 3)}.

64. False. Let R1 = {(1, 2), (2, 3)}.

65. True

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53

SOLUTIONS 68. The function f(n) = 2n is a one-to-one, onto function from {1, 2, . . .} to {2, 4, . . .}.

69. Assume that X is equivalent to P(X). Then there is a one-to-one, onto function f from X to P(X). Let Y = {x ∈ X | x ∈ / f(x)}.

Then f(y) = Y for some y ∈ X. Consider the possibilities y ∈ Y and y ∈ / Y.

Section 3.5 



1 1 0 0 2.  1 0 1 0  0 0 0 1 

0  0 6.   0 0

1 0 0 0

1 1 0 0

 1 1   1  0





0  0  5.   0  0 1



1 0 1 3.  1 1 0 0 0 0

0 0  1 

0  1 7. [For Exercise 13]   0 0

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

12. Symmetric and transitive

1 0 0 0

1 0 1 0

0 0 0 1 0

 0 1   1  0

0 0 0 0 0

0 0 1 0 0

 0 1   0   0  0

10. {(w, w), (w, y), (y, w), (y, y), (z, z)}

13. Take the transpose of the given matrix.

15. [For Exercise 4] The matrix of R is

and its square is



0  0   0   0 0

1 0 0 0 0

0 1 0 0 0

0 0 1 0 0

 0 0   0  , 1  0



0 0 0 0 0

1 0 0 0 0

0 1 0 0 0

 0 0   1  .

0  0

0  0   0   0 0

The entry in row 1, column 3 of the square is nonzero, but the entry in row 1, column 3 of the original matrix is zero. Therefore the relation is not transitive.       1 0 1 0 1 0 0 0 2 1 1 0  0 1 0 0   1 1 0 0   1 1 0 0       17. (a)  (b) (c)  0 0 1 0   1 1 1 0   1 1 1 0  0 0 0 1 1 1 1 1 1 1 1 1   1 1 1 0  1 1 0 0    (d)  1 1 1 0  1 1 1 1 (e) {(2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1), (4, 2), (4, 3), (5, 1), (5, 2), (5, 3), (5, 4)} Copyright c 2018 Pearson Education, Inc.

54

SOLUTIONS 

1  1  18. (a)   1  1 1

1 1 1 1 0

1 1 1 0 0

1 1 0 0 0

 1 0   0   0  0



0  1  (b)   0  0 0

0 0 1 0 0

0 0 0 1 0

0 0 0 0 1

 0 0   0   0  0

(d) The matrix of part (c) is the matrix of the relation R2 ◦ R1 .



1  1  (c)   1  1 0

1 1 1 0 0

1 1 0 0 0

1 0 0 0 0

 0 0   0   0  0

(e) {(1, 1), (1, 2), (1, 3), (1, 4), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (4, 1)}

20. Let aij denote the ijth entry of A1 and bij denote the ijth entry of A2 . Let n be the number of elements in Y . The ikth entry of A1 A2 is found by taking the product of the ith row of A1 and the kth column of A2 . Thus if cik is the ikth entry of the product, we have cik =

n X

aij bjk .

j=1

Now a term in this sum is nonzero only when both factors in the term are nonzero. This will be the case for all j such that aij = bjk = 1. This happens only when (i, j) ∈ R1 and (j, k) ∈ R2 . cik is then precisely the number of these j’s.

22. Suppose that the ij th entry of A is 1. Then the ij th entries of both A1 and A2 are 1. Thus (i, j) ∈ R1 and (i, j) ∈ R2 . Therefore (i, j) ∈ R1 ∩ R2 . Now suppose that (i, j) ∈ R1 ∩ R2 . Then the ij th entries of both A1 and A2 are 1. Therefore the ij th entry of A is 1. It follows that A is the matrix of R1 ∩ R2 .

23. The matrix A of Exercise 21 is



 1 1 0  0 1 1 , 1 0 1

whose relation is

R1 ∪ R2 = {(1, 1), (1, 2), (2, 2), (2, 3), (3, 1), (3, 3)}.

24. The matrix A of Exercise 22 is



 0 0 0  0 1 0 , 1 0 1

whose relation is

R1 ∩ R2 = {(2, 2), (3, 1), (3, 3)}.

26. Every column must contain at least one 1.

27. Every column must contain at most one 1.

Section 3.6 2. {(23,Jones), (04,Yu), (96, Zamora), (66,Washington)} 3. {(04,335B2,220), (23,2A,14), (04,8C200,302), (66,42C,3), (04,900,7720), (96,20A8,200), (96,1199C,296), (23,772,39)} 4. {(United Supplies,2A), (ABC Unlimited,8C200), (United Supplies,1199C), (JCN Electronics,2A), (United Supplies,335B2), (ABC Unlimited,772), (Danny’s,900), (United Supplies,772), (Underhanded Sales,20A8), (Danny’s,20A8), (DePaul University,42C), (ABC Unlimited,20A8)} Copyright c 2018 Pearson Education, Inc.

55

SOLUTIONS 6. DEPARTMENT[Manager] Jones, Yu, Zamora, Washington 7. SUPPLIER[Part No] 335B2, 2A, 8C200, 42C, 900, 20A8, 1199C, 772 9. TEMP1 := EMPLOYEE[Manager = Jones] TEMP2 := TEMP1[Name] Kaminski, Schmidt, Manacotti 10. TEMP := SUPPLIER[Dept = 96] TEMP[Part No] 20A8, 1199C 12. TEMP1 := DEPARTMENT[Dept = 04] TEMP2 := TEMP1[TEMP1.Manager = EMPLOYEE.Manager] EMPLOYEE TEMP3 := TEMP2[Name] Jones, Beaulieu 13. TEMP := SUPPLIER[Amount ≥ 100] TEMP[Part No] 335B2, 8C200, 900, 20A8, 1199C 15. TEMP1 := BUYER[Name = United Supplies] TEMP2 := TEMP1[TEMP1.Part No = SUPPLIER.Part No] SUPPLIER TEMP3 := TEMP2[Part No, Amount] Part No 335B2 2A 1199C 772

Amount 220 14 296 39

16. TEMP1 := BUYER[Name = ABC Unlimited] TEMP2 := TEMP1 [Part No = Part No] SUPPLIER TEMP3 := TEMP2 [Dept = Dept] DEPARTMENT TEMP3[Manager] Yu, Jones, Zamora 18. TEMP1 := DEPARTMENT[Manager = Jones] TEMP2 := TEMP1[TEMP1.Dept = SUPPLIER.Dept] SUPPLIER TEMP3 := TEMP2[TEMP2.Part No = BUYER.Part No] BUYER TEMP4 := TEMP3[Name] United Supplies, JCN Electronics ABC Unlimited 19. TEMP1 := EMPLOYEE[Name = Suzuki] TEMP2 := TEMP1 [Manager = Manager] DEPARTMENT TEMP3 := TEMP2 [Dept = Dept] SUPPLIER TEMP4 := TEMP3 [Part No = Part No] BUYER TEMP4[Name] Underhanded Sales, Danny’s, ABC Unlimited, United Supplies Copyright c 2018 Pearson Education, Inc.

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SOLUTIONS

23. The intersection operator will operate on two relations with the same set of attributes (arranged in the same order). The relation resulting from the intersection will have the same set of attributes. A tuple in the new relation will be a tuple in both of the two relations operated on. We will express the intersection operation using the set intersection symbol.

TEMP1 := BUYER[Part No = 2A] TEMP2 := TEMP1[Name] TEMP3 := BUYER[Part No = 1199C] TEMP4 := TEMP3[Name] TEMP5 := TEMP2 ∩ TEMP4 24. Let R1 and R2 be two n-ary relations. The difference of R1 and R2 is the n-ary relation R1 − R2 .

TEMP1 := EMPLOYEE [Manager = Manager] DEPARTMENT TEMP2 := TEMP1[Dept = 04] TEMP3 := TEMP1 − TEMP2 TEMP3[Name] Suzuki, Kaminski, Ryan, Schmidt, Manacotti

Section 4.1 4. min(a, b, c) { small = a if (b < small ) small = b if (c < small ) small = c return small } 5. second smallest (a, b, c) { x=a y=b z=c if (x > y){ temp = x x=y y = temp } if (y > z) { temp = y y=z z = temp } if (x > y) { temp = x x=y y = temp } return y } Copyright c 2018 Pearson Education, Inc.

57

SOLUTIONS 7. find large 2nd large(s, n, large, second largest) { if (s1 < s2 ) { large = s2 second largest = s1 } else { large = s1 second largest = s2 } for i = 3 to n if (si > second largest) if (si > large) { second largest = large large = si } else second largest = si }

8. find small 2nd small (s, n, small, second smallest) { if (s1 < s2 ) { small = s1 second smallest = s2 } else { small = s2 second smallest = s1 } for i = 3 to n if (si < second smallest) if (si < small) { second smallest = small small = si } else second smallest = si }

10. find largest element (s, n) { large = s1 index large = 1 for i = 2 to n if (si > large) { large = si index large = i } return index large } Copyright c 2018 Pearson Education, Inc.

58

SOLUTIONS

11. find last largest element (s, n) { large = s1 index large = 1 for i = 2 to n if (si ≥ large) { large = si index large = i } return index large } 13. find out of order1 (s, n) { for i = 2 to n if (si < si−1 ) return i return 0 } 14. find out of order2 (s, n) { for i = 2 to n if (si > si−1 ) return i return 0 } 16. Assume that sn , sn−1 , . . . , s1 and tn , tn−1 , . . . , t1 are the decimal representations of the two numbers to be added. The output is un+1 , un , . . . , u1. add(s, t, u, n) { c=0 for i = 1 to n { Let xy be the decimal representation of the sum c + si + ti . ui = y c=x } un+1 = c } 17. transpose(A, n) { for i = 1 to n − 1 for j = i + 1 to n swap(Aij , Aji ) } 19. The input is the n × n matrix A of the relation, and n. is symmetric(A, n) { for i = 1 to n − 1 for j = i + 1 to n if (Aij ¬ = Aji ) return false return true } Copyright c 2018 Pearson Education, Inc.

59

SOLUTIONS 20. The input is the n × n matrix A of the relation, and n. is transitive(A, n) { // first compute B = A2 for i = 1 to n for j = 1 to n { Bij = 0 for k = 1 to n Bij = Bij + Aik ∗ Akj } // if an entry in A2 is nonzero, but the corresponding entry // in A is zero, the relation is not transitive for i = 1 to n for j = 1 to n { if (Bij ¬ = 0 ∧ Aij == 0) return false return true } 22. The input is A, the m × n matrix of the relation, and m and n. is function(A, m, n) { for i = 1 to m { sum = 0 for j = 1 to n sum = sum + Aij if (sum ¬ = 1) return false } return true } 23. The input is the n × n matrix A of the relation, and n. inverse(A, n) { for i = 1 to n − 1 for j = i + 1 to n swap(Aij , Aji ) } 25. pair sum(s, n, x) { for i = 1 to n − 1 for j = i + 1 to n if (x == si + sj ) return true return false }

Section 4.2 2. First i and j are set to 1. The while loop then compares t1 t2 t3 = “bal” with p = “lai”. Since “b” and “l” are not equal, i increments to 2 and j remains 1.

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60

SOLUTIONS The while loop then compares t2 t3 t4 = “ala” with p = “lai”. Since “a” and “l” are not equal, i increments to 3 and j remains 1.

The while loop then compares t3 t4 t5 = “lal” with p = “lai”. Since “l” and “l” are equal, j increments.

Since “a” and “a” are equal, j increments again. Since “l” and “i” are not equal, i increments to 4 and j is reset to 1.

The while loop then compares t4 t5 t6 = “ala” with p = “lai”. Since “a” and “l” are not equal, i increments to 5 and j remains 1.

The while loop then compares t5 t6 t7 = “lai” with p = “lai”. Since the comparison succeeds, the algorithm returns i = 5 to indicate that p was found in t starting at index 5 in t.

3. First i and j are set to 1. The while loop then compares t1 t2 t3 = “000” with p = “001”. Since “0” and “0” are equal, j increments. Since “0” and “0” are equal, j increments again. Since “0” and “1” are not equal, i increments to 2 and j is reset to 1.

The while loop then compares t2 t3 t4 = “000” with p = “001”. Since “0” and “0” are equal, j increments.

Since “0” and “0” are equal, j increments again. Since “0” and “1” are not equal, i increments to 3 and j is reset to 1.

This pattern repeats until the first for loop terminates. The algorithm then returns 0 to indicate the p was not found in t.

5. First 20 is inserted in 34 Since 20 < 34, 34 must move one position to the right 34 Now 20 is inserted 20 34 Since 19 < 34, 34 must move one position to the right 20

34

Since 19 < 20, 20 must move one position to the right 20 34 Now 19 is inserted 19 20 34 Since 5 < 34, 34 must move one position to the right 19 20

34

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61

SOLUTIONS Since 5 < 20, 20 must move one position to the right 19

20 34

Since 5 < 19, 19 must move one position to the right 19 20 34 Now 5 is inserted 5

19 20 34

The sequence is now sorted.

6. Since 55 > 34, it is immediately inserted to 34’s right 34 55 Since 144 > 55, it is immediately inserted to 55’s right 34 55 144 Since 259 > 144, it is immediately inserted to 144’s right 34 55 144 259 The sequence is now sorted.

9. We first swap ai and aj , where i = 1 and j = rand (1, 5) = 2. After the swap we have 57 34 72 101 135 ↑ i

↑ j

We next swap ai and aj , where i = 2 and j = rand (2, 5) = 5. After the swap we have 57 135 72 101 34 ↑ i

↑ j

We next swap ai and aj , where i = 3 and j = rand (3, 5) = 3. The sequence is unchanged.

We next swap ai and aj , where i = 4 and j = rand (4, 5) = 4. The sequence is again unchanged.

10. We first swap ai and aj , where i = 1 and j = rand (1, 5) = 5. After the swap we have 135 57 72 101 34 ↑ i

↑ j

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62

SOLUTIONS We next swap ai and aj , where i = 2 and j = rand (2, 5) = 5. After the swap we have 135 34 72 101 57 ↑ i

↑ j

We next swap ai and aj , where i = 3 and j = rand (3, 5) = 4. After the swap we have 135 34 101 72 57 ↑ i

↑ j

We next swap ai and aj , where i = 4 and j = rand (4, 5) = 4. The sequence is unchanged.

12. Yes: rand(1, 5) = 4, rand (2, 5) = 4, rand (3, 5) = 5, rand (4, 5) = 4.

14. Use the invariant: s1 , . . . , si is sorted.

15. find first key(s, n, key) { for i = 1 to n if (key == si ) return i return 0 } 17. insert(s, n, x) { i=1 while (i ≤ n ∧ x > si ) i = i+1 // move si , . . . , sn down to make room for x j=n while (j ≥ i) { sj+1 = sj j = j −1 } // insert x si = x } 18. Replace the line while (ti+j−1 == pj ) { by while (j ≤ m ∧ ti+j−1 == pj ) { Replace the line return i by println(i) Copyright c 2018 Pearson Education, Inc.

63

SOLUTIONS and remove the line return 0

20 The worst case occurs when the for loop and the while loop run as long as possible. This situation is achieved when t consists of n 0’s and p consists of m − 1 0’s followed by one 1.

21. insertion sort nonincreasing(s, n) { for i = 2 to n { val = si // save si so it can be inserted into the correct place j = i−1 // if val > sj , move sj right to make room for si while (j ≥ 1 ∧ val > sj ) { sj+1 = sj j = j −1 } sj+1 = val // insert val } } 23. [For Exercise 4] Selection sort first finds the smallest item, s2 , and places it first by swapping s1 and s2 .

The result is 20 34 144 55 ↑ 1

↑ 2

Selection sort next finds the smallest item, s2 , in s2 , s3 , s4 , and places it second by swapping s2 and s2 .

The sequence is unchanged.

Selection sort next finds the smallest item, s4 , in s3 , s4 , and places it third by swapping s3 and s4 . The result is 20 34 55 144 ↑ 3

↑ 4

The sequence is now sorted.

24. In the pseudocode selection sort (s, n) { for i = 1 to n − 1 { // find smallest in si , . . . , sn small index = i for j = i + 1 to n if (sj < ssmall index ) small index = j swap(si , ssmall index ) } } the for loops always run to completion regardless of the input.

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SOLUTIONS

Section 4.3 2. Θ(n2 )

3. Θ(n3 )

9. Θ(n lg n)

12. Θ(2n )

11. Θ(n)

18. Θ(n)

19. Θ(n2 )

25. Θ(n)

27. Θ(lg lg n)

31. (a) Even: 3n/2 − 2

6. Θ(n6 )

5. Θ(n lg n)

8. Θ(n2 )

15. Θ(n3 )

21. Θ(n2 )

16. Θ(n5/2 )

22. Θ(n3 )

24. Θ(n3 )

28. Θ(n)

Odd: (3n − 1)/2 − 1

(b) Θ(n)

33. Use Example 4.3.6.

35. 2n = |2 · 2{z· · · 2} = 2 · 2| · 2{z· · · 2} ≤ 2(2 · 3 · · · n) = 2n! n 2’s

n − 1 2’s

36. First note that

n X i=1

Therefore

i lg i ≤ n(n lg n) = n2 lg n. n X

i lg i = O(n2 lg n).

i=1

Now n X i lg i ≥ i=1

n X

i=d(n+1)/2e

i lg i ≥

n X

i=d(n+1)/2e

As in Example 4.3.9, if n ≥ 4,

n+1 n n+1 n+1 n 2 n n+1 lg = lg ≥ lg .

2 2 2 2 2 2 2 n 2

Therefore if n ≥ 4,

n X i=1

It follows that

i lg i ≥

n 2 2

n X

n 2

≥

n

≥

n n lg n

lg

lg

2

n lg n .

4

2

4

=

n2 lg n .

8

i lg i = Ω(n2 lg n).

i=1

Therefore

n X

i lg i = Θ(n2 lg n).

i=1

38. For sufficiently large n, nk ≥ max{2, c}. Therefore, for sufficiently large n,

lg(nk + c) ≤ lg(nk + nk ) = lg 2nk ≤ lg nk nk = lg n2k = 2k lg n.

Also, lg(nk + c) ≥ lg nk = k lg n for all n. Therefore, lg(nk + c) = Θ(lg n).

39.

k X

lg(n/2i )

=

(k + 1) lg n − (1 + 2 + · · · + k)

=

(k + 1)k −

i=0

= =

k(k + 1) 2

(k + 1)k 2 (1 + lg n) lg n = Θ(lg2 n) 2 Copyright c 2018 Pearson Education, Inc.

65

SOLUTIONS

41. We show that if f(n) = Ω(g(n)), and f(n) > 0 and g(n) ≥ 0 for all n ≥ 1, then, for some constant C, f(n) ≥ Cg(n) for all n ≥ 1.

Proof. Since f(n) = Ω(g(n)), g(n) = O(f(n)). By Exercise 40, for some constant C 0 , g(n) ≤ C 0 f(n) for all n ≥ 1. Taking C = 1/(1 + C 0 ), we have Cg(n) ≤ f(n) for all n ≥ 1.

42. Exercises 40 and 41 show that if f(n) = Θ(g(n)), and f(n) > 0 and g(n) > 0 for all n ≥ 1, then, for some constants C1 and C2 , C1 g(n) ≤ f(n) ≤ C2 g(n) for all n ≥ 1.

44. True: |2 + sin n| ≤ 3 ≤ 3|2 + cos n|.

45. True

47. True

50. False

51. True

53. False

59. True

61. True

63. False. A counterexample is f(n) = 1 for all n, and g(n) = 1 + 1/n.

64. False. A counterexample is f(n) = n, g(n) = n2 .

55. True

48. False

56. True

66. True

58. True

67. True

70. If f(n) 6= O(g(n)), then for every C > 0, |f(n)| > C|g(n)| for infinitely many n. If g(n) 6= O(f(n)), then for every C > 0, |g(n)| > C|f(n)| for infinitely many n. However, there is no guarantee that even one n for which |f(n)| > C|g(n)| is true also makes |g(n)| > C|f(n)| true.

1 if n is even 71. f(n) = g(n) = 1 − f(n) 0 if n is odd, 73. We prove the result by using induction on k.

Basis Step (k = 1). By Exercise 18, Section 2.4, 1 + nx ≤ (1 + x)n , for x ≥ −1 and n ≥ 1. Thus, for x > 0 and n ≥ 1, nx < (1 + x)n or 1 n < (1 + x)n . x Taking C = 1/x and c = 1 + x gives the desired result.

Inductive Step. Assume that if c > 1, there exists a constant C such that nk ≤ Ccn for all but finitely many n. Let c > 1. By the inductive assumption, there exists a constant C1 such that √ nk ≤ C1 ( c)n for all but finitely many n. By the Basis Step, there exists a constant C2 such that √ n ≤ C2 ( c)n for all but finitely many n. Multiplying these inequalities, we obtain √ √ nk+1 = nk n ≤ C1 C2 ( c)n ( c)n = C1 C2 cn for all but finitely many n. The Inductive Step is complete. Therefore nk = O(cn ) for all k ≥ 1 and c > 1.

74. f(n) = h(n) = t(n) = n, g(n) = 2n 75. The Θ-notation ignores constants that are present in the formula for the actual time.

77. Yes 79. By referring to a graph like that of Exercise 78, with y = 1/x replaced by y = xm , we find that Z n+1 (n + 1)m+1 − 1 (n + 1)m+1 1 m + 2 m + · · · + nm < xm dx = < . m+1 m+1 1 The other inequality is proved in a similar manner.

Copyright c 2018 Pearson Education, Inc.

66

SOLUTIONS

81. We rewrite the inequality of Exercise 80 as bn [b − (n + 1)(b − a)] < an+1 .

If we set a = 1 + 1/(n + 1) and b = 1 + 1/n, the term in brackets reduces to 1 and we have

n n+1 1 1 1+ < 1+ . n n+1 Therefore, the sequence {(1 + 1/n)n } is increasing.

82. We rewrite the inequality of Exercise 80 as bn [b − (n + 1)(b − a)] < an+1 .

If we set a = 1 and b = 1 + 1/(2n), the term in brackets reduces to 21 , and we have

1 1+ 2n

n

< 2.

Squaring both sides gives

2n 1 1+ < 4.

2n

By Exercise 81, {(1 + 1/n)n } is increasing; thus,

1+

1 n

n

<

2n 1 1+ < 4.

2n

84. Using Exercise 83, we have n X 1 i=1

i

≤

n X i=1

[lg(i + 1) − lg i] = lg(n + 1) ≤ lg 2n = 1 + lg n ≤ 2 lg n,

if n ≥ 2. Thus,

n X 1 i=1

i

= O(lg n).

Again, using Exercise 83, we have n X 1 i=1

i

n

>

1X 1 1 [lg(i + 1) − lg i] = lg(n + 1) ≥ lg n.

2 i=1 2 2

Thus, n X 1 i=1

i

= Ω(lg n).

Therefore, n X 1 i=1

i

= Θ(lg n).

Copyright c 2018 Pearson Education, Inc.

67

SOLUTIONS 85. We derive the following equivalent inequalities: (n + 1)1/(n+1)

<

n1/n

(n + 1)n

<

nn+1

<

n

<

n.

(n + 1)n n n n 1 1+ n

By Exercise 82, (1 + 1/n)n < n if n ≥ 4. It may be directly verified that (1 + 1/n)n < n for n = 3, and thus the last displayed inequality holds for all n ≥ 3. Since the displayed inequalities are equivalent, it follows that the first is also true for all n ≥ 3. Therefore {n1/n }∞ n=3 is decreasing.

87. Set a =1−

1 n

and

b=1−

1 n+1

to prove that {(1 − 1/n)n }∞ n=1 is increasing. Set 1 2n

a =1−

and

b=1

to prove that {(1 − 1/n)n }∞ n=1 is bounded above by 4/9.

88. Notice that

1−

1 n

n

=

n−1 n

n

.

If we replace n by n + 1, we find using Exercise 87 that the sequence whose general term is

n n+1

n+1

is increasing. It follows that the sequence whose general term is

1 n+1 =

n n+1

n+1 n

n+1

=

1+

1 n

n+1

is decreasing.

90. In the argument given, the constant is dependent on n.

92. False. A counterexample is f(n) = n and g(n) = n2 .

93. True. In fact, we can conclude that f(n) = Θ(g(n)) (see the hint to Exercise 94).

95. False. A counterexample is f(n) = 1 for all n, and g(n) =

1, if n is even 2, if n is odd.

96. False. For a counterexample, see the solution to Exercise 95.

Copyright c 2018 Pearson Education, Inc.

68

SOLUTIONS

97. Inductive Step. Assume that the inequality holds for n. Now lg(n + 1)! = lg(n + 1) + lg n! n n ≥ lg(n + 1) + lg .

2 2 If we can show that

n+1 n+1 n n lg ≥ lg , 2 2 2 2

lg(n + 1) + the inductive step will be complete.

This last inequality is equivalent, in turn, to

n n lg n − 2 2 n lg n 2 n lg n n lg n 1 + lg(n + 1)

lg(n + 1) +

≥ ≥ ≥ ≥ ≥

1 + lg(n + 1)

≥

lg(2(n + 1))

≥

2(n + 1)

≥

By Exercise 82,

n+1 n+1 lg(n + 1) − 2 2 n−1 1 lg(n + 1) − 2 2 (n − 1) lg(n + 1) − 1 n lg(n + 1) − lg(n + 1) − 1 n[lg(n + 1) − lg n]

n+1 n lg n

n n+1 lg n

n n n+1 1 = 1+ n n

n 1 < 4, 1+ n

for all n. Thus inequality (12) holds for all n.

98. In the following figure, y

6 y = ln x

1

2

3

···

-x

n−1 n

the area of the rectangles exceeds the area under the curve from 1 to n, so Z n 1

ln x dx ≤

n X

ln k.

k=1

Now n ln n − n < n ln n − n + 1 =

Z n

ln x dx.

1

Copyright c 2018 Pearson Education, Inc.

(12)

69

SOLUTIONS 100. The inequality may be rewritten

n [(lg n) − 1] ≤ lg n!.

2 Assuming the result of Exercise 99, it suffices to show that n [(lg n) − 1] ≤ n lg n − n lg e, 2

or, equivalently 1 [(lg n) − 1] ≤ lg n − lg e, 2 or, equivalently (lg e) − Since (lg e) −

1 2

≤

lg n .

2

(13)

1 = 1.44 . . . − 0.5 = 0.9 . . . , 2

inequality (13) is obviously true for n ≥ 4. (The given inequality is clearly true for n = 1, 2, 3.)

Problem-Solving Corner: Design and Analysis of an Algorithm 1.

Input: Output:

s1 , . . . , sn max, maximum sum of consecutive values begin ind, starting index of values that give the maximum sum, or 0 if every sum is negative end ind, ending index of values that give the maximum sum, or 0 if every sum is negative

max sum4 (s, n) { max = 0 sum = 0 begin ind = 0 end ind = 0 local begin ind = 1 for i = 1 to n { if (sum + si > 0) { local end ind = i sum = sum + si } else { local begin ind = i + 1 sum = 0 } if (sum > max) { begin ind = local begin ind end ind = local end ind max = sum } } } Copyright c 2018 Pearson Education, Inc.

70

SOLUTIONS

Section 4.4 2. Since n 6= 2, we proceed to line 6 where we divide the board into four 2 × 2 boards:

At line 8 we place one right tromino in the center:

At lines 9 through 12, we recursively tile the 2 × 2 boards:

3. Since n 6= 2, we proceed to line 6 where we divide the board into four 4 × 4 boards:

At line 8 we place one right tromino in the center (we do not show the board rotated):

At lines 9 through 12, we recursively tile the 4 × 4 boards:

5. Since n 6= 1 and n 6= 2, we execute the line return robot walk(n − 1) + robot walk(n − 2) with n = 4. When we compute robot walk(3), since n 6= 1 and n 6= 2, we execute the line Copyright c 2018 Pearson Education, Inc.

71

SOLUTIONS return robot walk(n − 1) + robot walk(n − 2)

with n = 3. The algorithm returns the value 2 for robot walk(2) and the value 1 for robot walk(1).

Therefore for n = 3, the algorithm returns the value robot walk(2) + robot walk(1) = 2 + 1 = 3.

Therefore for n = 4, the algorithm returns the value robot walk(3) + robot walk(2) = 3 + 2 = 5.

6. Since n 6= 1 and n 6= 2, we execute the line return robot walk(n − 1) + robot walk(n − 2)) with n = 5. Exercise 5 shows that the algorithm returns the value 5 for robot walk(4) and the value 3 for robot walk(3). Therefore for n = 5, the algorithm returns the value robot walk(4) + robot walk(3) = 5 + 3 = 8.

8. We use induction on n. The Basis Steps, which are readily verified, are n = 1, 2. For the Inductive Step, assume that walk(k) = fk+1 for all k < n. From the formula in this section, we have walk(n) = walk(n − 1) + walk(n − 2).

By the inductive assumption, walk(n − 1) = fn

and

walk(n − 2) = fn−1 .

Now walk(n) = walk(n − 1) + walk(n − 2) = fn + fn−1 = fn+1 .

9. (a)

Input: Output:

n 1+2+···+ n

sum(n) { if (n == 1) return 1 return n + sum(n − 1) } (b) Basis Step (n = 1). If n is equal to 1, we correctly output 1 and stop.

Inductive Step. Assume that the algorithm correctly computes the sum when the input is n − 1.

Now suppose that the input to this algorithm is n > 1. Since n 6= 1, we invoke this procedure with input n − 1. By the inductive assumption, the value v returned is equal to 1 + · · · + (n − 1).

We then return v + n = 1 + · · · + (n − 1) + n, which is the correct value.

11. (a)

Input: Output:

n The number of ways the robot can walk n meters Copyright c 2018 Pearson Education, Inc.

72

SOLUTIONS walk3 (n) { if (n == 1) return 1 if (n == 2) return 2 if (n == 3) return 4 return walk3 (n − 1) + walk3 (n − 2) + walk3 (n − 3) } (b) Basis Steps (n = 1, 2, 3). If n is equal to 1, 2, or 3, we correctly output the number of ways the robot can walk 1, 2, or 3 meters.

Inductive Step. Assume that the algorithm correctly computes the sum when the input is less than n. The robot’s first step is either 1, 2, or 3 meters. If the first step is 1 meter, the robot must finish its walk by walking n − 1 meters. By the inductive assumption the robot can complete the walk in walk3 (n − 1) ways. Similarly, if the first step is 2 meters, the robot can complete the walk in walk3 (n − 2) ways, and if the first step is 3 meters, the robot can complete the walk in walk3 (n − 3) ways. Thus the total number of ways the robot can walk n meters is walk3 (n − 1) + walk3 (n − 2) + walk3 (n − 3).

Since this is the value computed by the procedure, the algorithm is correct.

12.

Input: Output:

The sequence s1 , s2 , . . . , sn and the length n of the sequence The minimum value in the sequence

find min(s, n) { if (n == 1) return s1 x = find min(s, n − 1) if (x < sn ) return x else return sn } We prove that the algorithm is correct using induction on n. The Basis Step is n = 1. If n = 1, the only item in the sequence is s1 and the algorithm correctly returns it.

Assume that the algorithm computes the minimum for input of size n, and suppose that the algorithm receives input of size n + 1. By assumption, the recursive call x = find min(s, n) correctly computes x as the minimum value in the sequence s1 , . . . , sn . If x is less than sn+1 , the minimum value in the sequence s1 , . . . , sn+1 is x—the value returned by the algorithm. If x is not less than sn+1 , the minimum value in the sequence s1 , . . . , sn+1 is sn+1 —again, the value returned by the algorithm. In either case, the algorithm correctly computes the minimum value in the sequence. The Inductive Step is complete, and we have proved that the algorithm is correct.

14.

Input: Output:

The sequence si , . . . , sj , i, and j The sequence in reverse order Copyright c 2018 Pearson Education, Inc.

73

SOLUTIONS reverse(s, i, j) { if (i ≥ j) return swap(si , sj ) reverse(s, i + 1, j − 1) } 15.

Input: Output:

n n!

factorial(n) { fact = 1 for i = 2 to n fact = i ∗ fact return fact } 17. To list all of the ways that a robot can walk n meters, set s to the null string and invoke this algorithm.

Input: Output:

n, s (a string) All the ways the robot can walk n meters. Each method of walking n meters includes the extra string s in the list.

list walk2 (n, s) { if (n == 1) { println(s + “ take one step of length 1”) return } if (n == 2) { println(s + “ take two steps of length 1”) println(s + “ take one step of length 2”) return } if (n == 3) { println(s + “ take three steps of length 1”) println(s + “ take one step of length 1 and one step of length 2”) println(s + “ take one step of length 2 and one step of length 1”) println(s + “ take one step of length 3”) return } s0 = s + “ take one step of length 3” // + is concatenation list walk2 (n − 3, s0 ) s0 = s + “ take one step of length 2” list walk2 (n − 2, s0 ) s0 = s + “ take one step of length 1” list walk2 (n − 1, s0 ) } 19. 233 20. We use strong induction on n. The Basis Steps are n = 1, 2. If n = 1, there is 1 = f2 way to tile a 2 × 1 board with 1 × 2 rectangular pieces: Copyright c 2018 Pearson Education, Inc.

74

SOLUTIONS

2 1

If n = 2, there are 2 = f3 ways to tile a 2 × 2 board with 1 × 2 rectangular pieces: 2

2 2

2

Now suppose that n > 2 and if k < n, the number of ways to tile a 2 × k board with 1 × 2 rectangular pieces is fk+1 . Now the first two vertical 1 × 1 squares can be covered in two ways: using one 1 × 2 rectangular piece 2 × (n − 1)

2 1

or by using two 1 × 2 rectangular pieces 2 × (n − 2)

2 2

By the inductive assumption, in the first case the remaining 2 × (n − 1) board can be tiled in fn ways, and in the second case, the remaining 2 × (n − 2) board can be tiled in fn−1 ways. Thus the total number of ways to tile a 2 × n board with 1 × 2 rectangular pieces is fn + fn−1 = fn+1 .

The inductive step is complete.

22. Using Exercises 21 and 23 and the recurrence relation for the Fibonacci sequence, we have 2 2 fn−2 fn+2 = fn−2 (fn + fn+1 ) = fn fn−2 + fn−2 fn+1 = fn−1 + (−1)n+1 + fn2 − fn−1 .

The conclusion now follows immediately.

2 2 23. fn+2 − fn+1 = (fn+2 + fn+1 )(fn+2 − fn+1 ) = fn+3 fn

25. Basis Steps (n = 1, 2). f1 and f2 are both odd (f1 = f2 = 1) and neither 1 nor 2 is divisible by 3.

Therefore, the statement is true for n = 1, 2.

Inductive Step. Assume that the statement is true for all k < n. We must prove that the statement is true for n. We can assume that n > 2. We consider two cases: n is divisible by 3 and n is not divisible by 3.

If n is divisible by 3, then neither n − 1 nor n − 2 is divisible by 3. By the inductive assumption, both fn−1 and fn−2 are odd. Since fn = fn−1 + fn−2 , fn is even. Therefore, if n is divisible by 3, fn is even.

If n is not divisible by 3, then exactly one of n − 1 or n − 2 is divisible by 3. By the inductive assumption, one of fn−1 and fn−2 is odd and the other is even. Since fn = fn−1 + fn−2 , fn is odd. Therefore, if n is not divisible by 3, fn is odd.

We have shown that fn is even if and only if n is divisible by 3, so the Inductive Step is complete.

26. Basis Step (n = 2). f4 = 3 = 4 − 1 = f32 − f12 . f5 = 5 = 1 + 4 = f22 + f32 .

Inductive Step. Assume true for n. Now f2n+2

= =

f2n+1 + f2n 2 2 2 fn2 + fn+1 + fn+1 − fn−1

Copyright c 2018 Pearson Education, Inc.

75

SOLUTIONS = = = = = =

2 2 2fn+1 + fn2 − fn−1 2 2fn+1 + fn2 − (fn+1 − fn )2 2 2 2fn+1 + fn2 − fn+1 + 2fn fn+1 − fn2

2 fn+1 + 2fn fn+1 (fn+1 + fn )2 − fn2 2 fn+2 − fn2 .

We now use the just proved formula for f2n+2 to prove the formula for f2n+3 : 2 2 2 2 f2n+3 = f2n+2 + f2n+1 = fn+2 − fn2 + fn2 + fn+1 = fn+1 + fn+2 .

The inductive step is complete.

28. Basis Steps (n = 1, 2). f1 = 1 ≤ 1 = 20 ; f2 = 1 ≤ 2 = 21

Inductive Step. Assume that the equation is true for n − 2 and n − 1. Now fn = fn−1 + fn−2 ≤ 2n−2 + 2n−3 < 2n−2 + 2n−2 = 2n−1 .

29. Basis Steps (n = 1, 2). For n = 1, we have f2 = 1 = 2 − 1 = f3 − 1, and f1 = 1 = f2 .

For n = 2, we have f2 + f4 = 1 + 3 = 5 − 1 = f5 − 1, and f1 + f3 = 1 + 2 = 3 = f4 .

Inductive Step.

n+1 X

f2k

=

k=1

n+1 X

n X

f2k + f2n+2

n X

f2k−1 + f2n+1

k=1

=

f2k−1

=

k=1

f2n+1 − 1 + f2n+2 = f2n+3 − 1

k=1

=

f2n + f2n+1 = f2n+2

31. We show that the representation n=

m X

fki

(14)

i=1

given in the hint for Exercise 30 is unique.

By Exercise 29, the partial sum of Fibonacci numbers with even indexes is n X

k=1

f2k = f2n+1 − 1.

If we do not allow f1 as a summand, the partial sum of Fibonacci numbers with odd indexes becomes n X

f2k−1 =

k=2

n X

k=1

f2k−1 − f1 = f2n − 1,

where we have again used Exercise 29.

Suppose by way of contradiction that some integer has a representation as the sum of distinct Fibonacci numbers no two of which are consecutive different from (14). Let n denote the smallest such integer. Let n=

j X

fti ,

i=1

Copyright c 2018 Pearson Education, Inc.

76

SOLUTIONS where t1 > t2 > . . . , be another representation of n as the sum of distinct Fibonacci numbers no two of which are consecutive.

Since ft1 ≤ n and fk1 is the largest Fibonacci number less than or equal to n, ft1 ≤ fk1 . If ft1 = fk1 , n − fk1 has at least two representations as the sum of distinct Fibonacci numbers no two of which are consecutive, which contradicts the minimality of n. Therefore ft1 < fk1 . Thus ft1 ≤ fk1 −1 .

In the representation, no two Fibonacci numbers are consecutive, thus ft2 ≤ fk1 −3 , ft3 ≤ fk1 −5 , . . .

Therefore n

= ≤ ≤ = <

ft1 + ft2 + · · · + ftj fk1 −1 + fk1 −3 + · · · + fk1 −(2j−1) fk1 −1 + fk1 −3 + · · · + fp fk1 − 1 n,

where p = 2 or 3 by the preceding comments

which is a contradiction. Therefore the representation of an integer as the sum of distinct Fibonacci numbers no two of which are consecutive is unique if we do not allow f1 as a summand.

32. Exercise 21 shows that fn2 = fn−1 fn+1 + (−1)n+1 , n ≥ 2, so

fn2 = fn−1 (fn + fn−1 ) + (−1)n+1 , n ≥ 2

or 2 fn2 − fn−1 fn − fn−1 − (−1)n+1 = 0, n ≥ 2.

The quadratic formula gives fn

= =

fn−1 ±

q 2 2 fn−1 − 4[−fn−1 − (−1)n+1 ]

2 q 2 fn−1 ± 5fn−1 + 4(−1)n+1 2

, n ≥ 2.

The negative sign gives an extraneous root since if chosen, we would have, for n ≥ 3, fn <

fn−1 2

or 2(fn−1 + fn−2 ) < fn−1 or fn−1 + 2fn−2 < 0 which is impossible. The formula for n = 2 is correct by inspection.

34. Basis Steps (n = 3, 4). For n = 3, we have g1 f1 + g2 f2 = g1 + g2 = g3 .

For n = 4, we have g1 f2 + g2 f3 = g1 + 2g2 = g2 + (g1 + g2 ) = g2 + g3 = g4 .

Copyright c 2018 Pearson Education, Inc.

77

SOLUTIONS Inductive Step. Assume that n > 4. Now gn = gn−1 + gn−2

= =

(g1 fn−3 + g2 fn−2 ) + (g1 fn−4 + g2 fn−3 ) g1 (fn−3 + fn−4 ) + g2 (fn−2 + fn−3 )

=

g1 fn−2 + g2 fn−1 .

35. Basis Step (n = 2). −f1 + f2 = −1 + 1 = 0 = 1 − 1 = f1 − 1 Inductive Step. Assume true for n. Then n+1 X

(−1)k fk

=

k=1

n X

(−1)k fk + (−1)n+1 fn+1

k=1

= (−1)n fn−1 − 1 + (−1)n+1 fn+1 = (−1)n+1 (fn+1 − fn−1 ) − 1 = (−1)n+1 fn − 1.

dx = 1 = 1 · x0 dx Inductive Step. Assume true for n. Then

37. Basis Step (n = 1).

dxn+1 d(x · xn ) dxn dx = =x + xn = x(nxn−1) + xn · 1 = nxn + xn = (n + 1)xn . dx dx dx dx 38. The formula reduces the problem of integrating logn |x| to the problem of integrating logn−1 |x|, a simpler instance of the original problem. Eventually the problem is reduced to integrating log |x|, which is straightforward.

Another example of a recursive integration formula is Z Z sin2n−1 x cos x 2n − 1 + sin2n−2 x dx. sin2n x dx = − 2n 2n

Section 5.1 √ 2. Since b 209c = 14, the for loop in Algorithm 5.1.8 runs from d = 2 to 14. For d = 2 to 10, n mod d 6= 0.

When d = 11, n mod d = 0. Therefore the algorithm returns 11 to signal that 209 is composite and 11 is a divisor of 209. √ 3. Since b 47c = 6, the for loop in Algorithm 5.1.8 runs from d = 2 to 6. For each of these values, n mod d 6= 0. Therefore the algorithm returns 0 to signal that 47 is prime. √ 5. Since b 4141c = 64, the for loop in Algorithm 5.1.8 runs from d = 2 to 64. For d = 2 to 40, n mod d 6= 0.

When d = 41, n mod d = 0. Therefore the algorithm returns 41 to signal that 4141 is composite and 41 is a divisor of 4141. √ 6. Since b 1007c = 31, the for loop in Algorithm 5.1.8 runs from d = 2 to 31. For d = 2 to 18, n mod d 6= 0.

When d = 19, n mod d = 0. Therefore the algorithm returns 19 to signal that 1007 is composite and 19 is a divisor of 1007. √ 8. Since b 1050703c = 1025, the for loop in Algorithm 5.1.8 runs from d = 2 to 1025. For d = 2 to 100, n mod d 6= 0. When d = 101, n mod d = 0. Therefore the algorithm returns 101 to signal that 1050703 is composite and 101 is a divisor of 1050703.

10. (For Exercise 1) 9 = 3 · 3 Copyright c 2018 Pearson Education, Inc.

78

SOLUTIONS

11.

11!

13. 30

= = =

11 · 10 · 9 · 8 · 7 · 6 · 5 · 4 · 3 · 2 11 · (5 · 2) · (3 · 3) · (2 · 2 · 2) · 7 · (3 · 2) · 5 · (2 · 2) · 3 · 2 28 34 52 71 111 14. 5

16. 15

17. 20

19. 1

20. 331

22. 15

23. 7 26. (For Exercise 13) We have gcd(60, 90) = 30 and lcm(60, 90) = 180. Thus gcd(60, 90) · lcm(60, 90) = 30 · 180 = 60 · 90.

27. Since d | m, m = dq1 for some integer q1 . Since d | n, n = dq2 for some integer q2 . Now m − n = dq1 − dq2 = d(q1 − q2 ).

Therefore, d | (m − n).

29. Since d1 | m, m = d1 q1 for some integer q1 . Since d2 | n, n = d2 q2 for some integer q2 . Now mn = (d1 q1 )(d2 q2 ) = (d1 d2 )(q1 q2 ).

Therefore, d1 d2 | mn.

30. Since dc | nc, nc = dcq for some integer q. Since dc is a divisor of nc, c 6= 0. Therefore, we may cancel c in nc = dcq to obtain n = dq. Therefore, d | n.

32. After checking whether 2 is a divisor of n, we need not check whether 4, 6, 8, . . . divide n. Implementing this change cuts the time by about one-half: is prime(n) { if (n mod 2 == 0) return 2 d=3 √ while (d ≤ b nc) { if (n mod d == 0) return d d=d+2 } return 0 } √ Continuing this idea, if we store the primes less than or equal to b nc, we need only check whether any of these primes divides n.

33. 2 · 3 · 5 · 7 · 11 · 13 + 1 = 30031 = 509 · 59 Qn 34. Inductive Step. Assume that {a1 , . . . , an } is a ∗ -set. Let b0 = k=1 ak and bi = b0 + ai for 1 ≤ i ≤ n.

We prove that {b0 , b1 , . . . , bn } is a ∗ -set.

Qn We first show that (b0 − bj ) | b0 for all j ≥ 1. Since b0 = k=1 ak , b0 − bj = −aj , and j ≥ 1, it is clear that (b0 − bj ) | b0 .

We next show that (bj − Q b0 ) | bj for all j ≥ 1. Since bj − b0 = aj and bj = b0 + aj , we must show that n aj | (b0 + aj ). Since b0 = k=1 ak and j ≥ 1, aj | b0 . Trivially, aj | aj ; therefore, aj | (b0 + aj ).

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79

SOLUTIONS Finally, we show that for all i, j > 0, (bi − bj ) | bi . Since bi − bj = (b0 + ai ) − (b0 + aj ) = ai − aj and b i = b 0 + ai ,

Qn it suffices to show that (ai −aj ) | (b0 +ai ). By the inductive assumption, (ai −aj ) | ai . Since b0 = k=1 ak and i ≥ 1, ai | b0 . Thus (ai − aj ) | b0 . Therefore (ai − aj ) | (b0 + ai ) and the Inductive Step is complete.

35. {2, 3, 4}, {24, 26, 27, 28} 37. Suppose that d | Fm and d | Fn , 0 ≤ m < n. By Exercise 36, d | 2. So, d = 1 or 2. Since Fn is odd for all n, d 6= 2. Therefore d = 1 and gcd(Fm , Fn) = 1.

38. By Exercise 37, if m 6= n, Fm and Fn contain distinct primes in their prime factorizations. Since the Fermat numbers are infinite, so are the primes.

41. E-composite: 8 = 2 · 4.

42. E-prime: 10 is not the product of any elements in E.

44. 36 = 2 · 18 = 6 · 6 45. An integer in E is either of the form 2m, where m is an odd integer in Z+ , or 2k, where k is even. In the first case, the integer 2m is E-prime. In the second case, the integer 2k is E-composite. Therefore an integer n is E-prime if and only if it can be written in the form 2m, where m is an odd integer in Z+ .

47. The solution to Exercise 45 shows that an integer n is E-prime if and only if it can be written in the form 2m, where m is an odd integer in Z+ . In this case 2m + 2 = 2(m + 1). Since m + 1 is even, 2m + 2 is not E-prime.

48. The solution to Exercise 45 shows that an integer n is E-prime if and only if it can be written in the form 2m, where m is an odd integer in Z+ . Then n + 4 = 2m + 4 = 2(m + 2). Since m + 2 is odd, n + 4 is also E-prime. It follows that there are infinitely many pairs of E-primes that differ by 4.

Section 5.2 2. 6

3. 7

13. 219

5. 8

15. 255

6. 1001 16. 3547

21. 10000000000

22. 11000000110100

27. 1000100010

28. 100010100

34. 307322 39. 903

36. (For Exercise 17) 22 40. 565D

8. 3001

9. 763

18. 111101

19. 11011111

24. 101000 30. 489

12. 27

25. 1100011

31. 15996

33. 8349

37. (For Exercise 29) 111010

42. 130FF7

44. 1101010 represents a number in binary, decimal, and hexadecimal.

46. 4003

47. 4041

52. (For Exercise 17) 42

49. 519

50. 179889

53. (For Exercise 29) 72

55. Yes

56. 30470 does not represent a number in binary, but it does represent a number in octal, decimal, and hexadecimal.

Copyright c 2018 Pearson Education, Inc.

80

SOLUTIONS

58. Suppose that the base b representation of m is m=

k X

ci b i ,

i=0

ck 6= 0. Then

bk ≤ ck bk ≤ m,

and m≤ k

k X i=0

(b − 1)bi = (b − 1)

k X i=0

bi = (b − 1)

bk+1 − 1 = bk+1 − 1 < bk+1 . b−1

Since b ≤ m, taking logs to the base b, we obtain k ≤ logb m.

Since m < bk+1 , again taking logs, we obtain logb m < k + 1.

Combining these inequalities, we have k + 1 ≤ 1 + logb m < k + 2.

Therefore, the number of digits required to represent m is k + 1 = b1 + logb mc.

60. The algorithm begins by setting result to 1 and x to a. Since n = 15 > 0, the body of the while loop executes. Since n mod 2 is equal to 1, result becomes result ∗ x = 1 ∗ a = a. x becomes a2 , and n becomes 7.

Since n = 7 > 0, the body of the while loop executes. Since n mod 2 is equal to 1, result becomes result ∗ x = a ∗ a2 = a3 . x becomes a4 , and n becomes 3.

Since n = 3 > 0, the body of the while loop executes. Since n mod 2 is equal to 1, result becomes result ∗ x = a3 ∗ a4 = a7 . x becomes a8 , and n becomes 1.

Since n = 1 > 0, the body of the while loop executes. Since n mod 2 is equal to 1, result becomes result ∗ x = a7 ∗ a8 = a15 . x becomes a16 , and n becomes 0.

Since n = 0 is not greater than 0, the while loop terminates. The algorithm returns result , which is equal to a15 .

61. The algorithm begins by setting result to 1 and x to a. Since n = 80 > 0, the body of the while loop executes. Since n mod 2 is not equal to 1, result is not modified. x becomes a2 , and n becomes 40.

Since n = 40 > 0, the body of the while loop executes. Since n mod 2 is not equal to 1, result is not modified. x becomes a4 , and n becomes 20.

Since n = 20 > 0, the body of the while loop executes. Since n mod 2 is not equal to 1, result is not modified. x becomes a8 , and n becomes 10.

Since n = 10 > 0, the body of the while loop executes. Since n mod 2 is not equal to 1, result is not modified. x becomes a16 , and n becomes 5.

Since n = 5 > 0, the body of the while loop executes. Since n mod 2 is equal to 1, result becomes result ∗ x = 1 ∗ a16 = a16 . x becomes a32 , and n becomes 2.

Since n = 2 > 0, the body of the while loop executes. Since n mod 2 is not equal to 1, result is not modified. x becomes a64 , and n becomes 1.

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81

SOLUTIONS

Since n = 1 > 0, the body of the while loop executes. Since n mod 2 is equal to 1, result becomes result ∗ x = a16 ∗ a64 = a80 . x becomes a128 , and n becomes 0.

Since n = 0 is not greater than 0, the while loop terminates. The algorithm returns result , which is equal to a80 .

63. The algorithm begins by setting result to 1 and x to a mod z = 143 mod 230 = 143. Since n = 10 > 0, the body of the while loop executes. Since n mod 2 is not equal to 1, result is not modified. x is set to (x ∗ x) mod z = (143 ∗ 143) mod 230 = 20449 mod 230 = 209, and n is set to 5.

Since n = 5 > 0, the body of the while loop executes. Since n mod 2 is equal to 1, result is set to (result ∗ x) mod z = 209 mod 230 = 209. x is set to (x ∗ x) mod z = (209 ∗ 209) mod 230 = 43681 mod 230 = 211, and n is set to 2.

Since n = 2 > 0, the body of the while loop executes. Since n mod 2 is not equal to 1, result is not modified. x is set to (x ∗ x) mod z = (211 ∗ 211) mod 230 = 44521 mod 230 = 131, and n is set to 1.

Since n = 1 > 0, the body of the while loop executes. Since n mod 2 is equal to 1, result is set to (result ∗ x) mod z = (209 ∗ 131) mod 230 = 27379 mod 230 = 9. x is set to (x ∗ x) mod z = (131 ∗ 131) mod 230 = 17161 mod 230 = 141, and n is set to 0.

Since n = 0 is not greater than 0, the while loop terminates. The algorithm returns result , which is equal to an mod z = 14310 mod 230 = 9.

64. The algorithm begins by setting result to 1 and x to a mod z = 143 mod 230 = 143. Since n = 100 > 0, the body of the while loop executes. Since n mod 2 is not equal to 1, result is not modified. x is set to (x ∗ x) mod z = (143 ∗ 143) mod 230 = 20449 mod 230 = 209, and n is set to 50.

Since n = 50 > 0, the body of the while loop executes. Since n mod 2 is not equal to 1, result is not modified. x is set to (x ∗ x) mod z = (209 ∗ 209) mod 230 = 43681 mod 230 = 211, and n is set to 25.

Since n = 25 > 0, the body of the while loop executes. Since n mod 2 is equal to 1, result is set to (result ∗ x) mod z = 211 mod 230 = 211. x is set to (x ∗ x) mod z = (211 ∗ 211) mod 230 = 44521 mod 230 = 131, and n is set to 12.

Since n = 12 > 0, the body of the while loop executes. Since n mod 2 is not equal to 1, result is not modified. x is set to (x ∗ x) mod z = (131 ∗ 131) mod 230 = 17161 mod 230 = 141, and n is set to 6.

Since n = 6 > 0, the body of the while loop executes. Since n mod 2 is not equal to 1, result is not modified. x is set to (x ∗ x) mod z = (141 ∗ 141) mod 230 = 19881 mod 230 = 101, and n is set to 3.

Since n = 3 > 0, the body of the while loop executes. Since n mod 2 is equal to 1, result is set to (result ∗ x) mod z = (211 ∗ 101) mod 230 = 21311 mod 230 = 151. x is set to (x ∗ x) mod z = (101 ∗ 101) mod 230 = 10201 mod 230 = 81, and n is set to 1.

Since n = 1 > 0, the body of the while loop executes. Since n mod 2 is equal to 1, result is set to (result ∗ x) mod z = (151 ∗ 81) mod 230 = 12231 mod 230 = 41. x is set to (x ∗ x) mod z = (81 ∗ 81) mod 230 = 6561 mod 230 = 121, and n is set to 0.

Since n = 0 is not greater than 0, the while loop terminates. The algorithm returns result , which is equal to an mod z = 143100 mod 230 = 41.

66. Basis Step (n = 1). Since Tn! = 0 and Sn = 1, the result holds for n = 1.

Inductive Step. Assume true for n. Now T(n+1)! = T(n+1)n! = Tn+1 + Tn! = Tn+1 + n − Sn , where we have used the result of Exercise 62 and the inductive assumption. Thus the inductive step will be complete if we can show that Tn+1 = 1 + Sn − Sn+1 .

Suppose that n is even. Then n + 1 is odd, so Tn+1 = 0. Now n = . . . 0 in binary, so n + 1 = . . . 1.

Therefore Sn+1 = 1 + Sn . Thus Tn+1 = 1 + Sn − Sn+1 , if n is even.

Copyright c 2018 Pearson Education, Inc.

82

SOLUTIONS Suppose that n is odd, say, n = . . . 0 11 . . 11} . | .{z k 1’s

Then

n + 1 = . . . 1 00 . . 00} . | .{z k 0’s

Thus Sn − Sn+1 = k − 1 and Tn+1 = k. Therefore Tn+1 = 1 + Sn − Sn+1 , if n is odd.

67.

Input: Output:

b, m, b0, n prod, the binary product of b and b0

add(p + q, i, r) computes the sum of the binary numbers p and q, except that q is shifted i places left (effectively appending i zeros on the right), with the result in r. binary product (b, m, b0 , n) { prod = 0 for i = 1 to n if (b0i == 1) add (prod + b, i, prod) return prod } 68. The for loop runs in time O(n). Adding prod and b takes time m (the appended zeros are not added).

Thus the time is O(mn). Since m and n are the number of bits to represent the numbers that are multiplied, the result follows.

Section 5.3 2. 1 12. 425

3. 20

5. 20

6. 331

8. 495

9. 23

11. 27

14. 89, 55

15. Suppose that lines 3 and 4 in Algorithm 5.3.3 are deleted. If a ≥ b, clearly the result is the same as in the original form. If a < b, then b 6= 0 and the first iteration of the while loop swaps a and b. Thereafter, the algorithm proceeds as in the original form, and again the result is the same.

17. Let m be a common divisor of a and b. By Theorem 5.1.3(a), m divides a + b. Thus m is a common divisor of a and a + b.

Let m be a common divisor of a and a + b. By Theorem 5.1.3(b), m divides (a + b) − a = b. Thus m is a common divisor of a and b.

Since the set of common divisors of a and a + b is equal to the set of common divisors of a and b, gcd(a, b) = gcd(a, a + b).

18. Let m be a common divisor of a and b. By Theorem 5.1.3(b), m divides a − b. Thus m is a common divisor of b and a − b.

Let m be a common divisor of b and a − b. By Theorem 5.1.3(a), m divides (a − b) + b = a. Thus m is a common divisor of a and b.

Since the set of common divisors of b and a − b is equal to the set of common divisors of a and b, gcd(a, b) = gcd(a − b, b).

20. The algorithm given in the hint in the book to Exercise 19 does m subtractions in the worst case, which occurs when a = m and b = 1.

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83

SOLUTIONS 21.

0

1

2

3

4

5

6

7

8

9

10

11

12

13

14

15

16

17

18

19

20

21

a 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21

b

– 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0

0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1

0 1 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2 1 2

0 1 2 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1 2 3 1

0 1 1 2 1 2 2 3 1 2 2 3 1 2 2 3 1 2 2 3 1 2

0 1 2 3 2 1 2 3 4 3 1 2 3 4 3 1 2 3 4 3 1 2

0 1 1 1 2 2 1 2 2 2 3 3 1 2 2 2 3 3 1 2 2 2

0 1 2 2 3 3 2 1 2 3 3 4 4 3 1 2 3 3 4 4 3 1

0 1 1 3 1 4 2 2 1 2 2 4 2 5 3 3 1 2 2 4 2 5

0 1 2 1 2 3 2 3 2 1 2 3 2 3 4 3 4 3 1 2 3 2

0 1 1 2 2 1 3 3 2 2 1 2 2 3 3 2 4 4 3 3 1 2

0 1 2 3 3 2 3 4 4 3 2 1 2 3 4 4 3 4 5 5 4 3

0 1 1 1 1 3 1 4 2 2 2 2 1 2 2 2 2 4 2 5 3 3

0 1 2 2 2 4 2 3 5 3 3 3 2 1 2 3 3 3 5 3 4 6

0 1 1 3 2 3 2 1 3 4 3 4 2 2 1 2 2 4 3 4 3 2

0 1 2 1 3 1 2 2 3 3 2 4 2 3 2 1 2 3 2 4 2 3

0 1 1 2 1 2 3 3 1 4 4 3 2 3 2 2 1 2 2 3 2 3

0 1 2 3 2 3 3 3 2 3 4 4 4 3 4 3 2 1 2 3 4 3

0 1 1 1 2 4 1 4 2 1 3 5 2 5 3 2 2 2 1 2 2 2

0 1 2 2 3 3 2 4 4 2 3 5 5 3 4 4 3 3 2 1 2 3

0 1 1 3 1 1 2 3 2 3 1 4 3 4 3 2 2 4 2 2 1 2

0 1 2 1 2 2 2 1 5 2 2 3 3 6 2 3 3 3 2 3 2 1

a 1 2 3 5 8 13 21

b 0 1 2 3 5 8 13

n (= number of modulus operations) 0 1 2 3 4 5 6

23. Using induction on n, we prove that when the pair fn+2 , fn+1 is input to the Euclidean algorithm, exactly n modulus operations are required.

Basis Step (n = 1). Table 5.3.2 shows that when the pair f3 , f2 is input to the Euclidean algorithm, one modulus operation is required.

Inductive Step. Assume that when the pair fn+2 , fn+1 is input to the Euclidean algorithm, n modulus operations are required. We must show that when the pair fn+3 , fn+2 is input to the Euclidean algorithm, n + 1 modulus operations are required.

At line 6, since fn+3 = fn+2 + fn+1 , r = fn+1 . The algorithm then repeats using the values of fn+2 and fn+1 . By the inductive assumption, exactly n additional modulus operations are required. Thus a total of n + 1 modulus operations are required.

24. We prove this result by induction on max(a, b). We omit the Basis Step.

Without loss of generality, we assume that a ≥ b. If b = 0, zero modulus operations are required to compute gcd(a, b) and gcd(ka, kb). Suppose that b > 0. Suppose that when we divide a by b, we obtain a = bq + r, 0 ≤ r < b.

Multiplying these last inequalities by k, we obtain ka = (kb)q + kr, 0 ≤ kr < kb.

Thus when we divide ka by kb, we obtain the remainder kr. In computing gcd(a, b), we continue by computing gcd(b, r). In computing gcd(ka, kb), we continue by computing gcd(kb, kr). By the inductive Copyright c 2018 Pearson Education, Inc.

84

SOLUTIONS assumption, these computations require the same number of modulus operations. Thus the number of modulus operations required to compute gcd(a, b) is the same as the number of modulus operations required to compute gcd(ka, kb). The Inductive Step is complete.

26. By Theorem 5.3.7, there exist integers s and t such that gcd(a, b) = sa + tb. Since d | a and d | b, d | (sa + tb). Therefore d | gcd(a, b).

27. If p divides a, we are done; so suppose that p does not divide a. We must show that p divides b. Since p is prime, gcd(p, a) = 1. By Theorem 5.3.7, there are integers s and t such that 1 = sp + ta.

Multiplying both sides of this equation by b, we obtain b = spb + tab.

By Theorem 5.1.3(c), p divides spb and p divides tab. By Theorem 5.1.3(a), p divides spb + tab = b.

28. p = 4, a = 6, b = 10 31. X has a least element by the Well-Ordering Property.

32. Let g = sa + tb. If c | a and c | b, then c | sa + tb = g.

34. Exercise 33 shows that g is a common divisor. Exercise 32 shows that g is the greatest common divisor.

36. s = 1

37. s = 3

39. s = 134

40. s = 67

43. The subsection Computing an Inverse Modulo an Integer shows that if gcd(n, φ) = 1, then n has an inverse modulo φ.

Now suppose that n has an inverse s modulo φ. Then ns mod φ = 1.

Since 1 is the remainder, there exists q such that ns = φq + 1.

Now suppose that c is a positive common divisor of n and φ. Then c divides ns and φq and also ns − φq = 1.

Therefore c = 1 and gcd(n, φ) = 1.

Problem-Solving Corner: Making Postage 1. Let k = gcd(p, q). Suppose that we can make m cents postage using a p-cent and b q-cent stamps. Then ap + bq = m. Then k | m. Since k > 1, k 6 | m + 1. Therefore we cannot make m + 1 cents postage. The conclusion follows.

Section 5.4 2. DRINK YOUR OVALTINE

3. WEISERKYEIEFTKK

6. a = cs mod z = 411569 mod 713 = 500 10. φ = (p − 1)(q − 1) = 16 · 22 = 352 15. φ = (p − 1)(q − 1) = 58 · 100 = 5800

5. c = an mod z = 33329 mod 713 = 306

8. a = cs mod z = 387569 mod 713 = 60 11. s = 159

13. a = cs mod z = 250159 mod 391 = 10

16. s = 3961

18. a = cs mod z = 2503961 mod 5959 = 5648 Copyright c 2018 Pearson Education, Inc.

85

SOLUTIONS

Section 6.1 2. 3 · 3 · 5

5. 26 − 1

3. 2 · 3 · 5

6. 5 · 6 · 2 · 3 · 3 9. 263 102 , 26 · 25 · 24 · 10 · 9

8. Yes, the ad is correct; there are 5 · 14 · 3 = 210 dinners.

11. Choose F. We must next append a string of length 4 chosen from the letters ABCDEG that does not end EB. There are 4 · 3 such strings that do end EB. Since there are 6 · 5 · 4 · 3 strings of length 4 chosen from the letters ABCDEG without any restriction, there are 6 · 5 · 4 · 3 − 4 · 3 strings that begin with the letter F and do not end with EB in that order. with the letters CEG in that order and two others selected from 12. Consider filling five blanks ABDF. There are 3 ways to place CEG and, having placed these three letters, there are 4 ways to fill the first remaining blank and 3 ways to fill the last blank. Therefore there are 3 · 4 · 3 strings that contain CEG together in that order.

14. 2 + 3

15. 2 + 4

17. n + n + · · · + n = mn | {z } m n’s

18. 32 + 16

20. 3 + 6

21. 3 + 3

23. There are 5 · 4 · 3 strings that begin with AB and there are 6 · 5 · 4 · 3 that begin with D. Since there are 7 · 6 · 5 · 4 · 3 total strings, there are 7 · 6 · 5 · 4 · 3 − 5 · 4 · 3 − 6 · 5 · 4 · 3 that do not begin with AB or D. 24. Consider filling five blanks with the letters BD in that order and three others selected from ACEFG. There are 4 ways to place BD and, having placed these two letters, there are 5 ways to fill the first remaining blank, 4 ways to fill the next remaining blank, and 3 ways to fill the last blank. Thus there are 4 · 5 · 4 · 3 strings that contain BD together in that order. Similarly, there are 4 · 5 · 4 · 3 strings that contain DB together in that order. Therefore there are 2 · 4 · 5 · 4 · 3 strings that contain BD together in either order.

27. 6

28. 6 + 2

30. 10

36. 50 · 49

38. 26

46. 3 · 2 · 4

48. 5 · 4 + 5 · 4 · 3

54. 43 62. 200 − 72

49. 2 · 5 · 4

72. (5!)(5!)

51. 5 · 4 · 3

59. (200 − 4)/2 65. 196 − (9 + 2 + 18)

69. (a) 12 · 11 · 10 · 9 · 8

35. 502 42. 28 − 1

41. (8 · 7)/2

57. 3 · 4 · 3

63. 5 + 9 · 9 + 9 · 8

68. 2 + 3 + · · · + 9 = 44 71. (5!)(2!)(3!)

39. 3 · 26

55. 4 · 3 · 2

33. 2 · 32

31. 11

(b) 125

45. 4 · 3 · 2 52. 52

60. (200 − 4)/2 66. 1 + 1 · 9 · 9 − 2

(c) 125 − 12 · 11 · 10 · 9 · 8

74. 8! · 9 · 8

75. 26 + 26 · 36 + 26 · 362 + 26 · 363 + 26 · 364 + 26 · 365

76. mn

78. 2 · 3 · 3 · 2

79. A subset X has n elements or less if and only if X has more than n elements. Thus exactly half of the subsets have n elements or less. Therefore, the number of subsets is 12 22n+1 = 22n .

81. 221(28 − 2)

82. (27 − 1)(224 − 2) + 214 (216 − 2) + 221 (28 − 2)

84. We count the number of antisymmetric relations on {1, 2, . . . , n} by computing the number of ways to construct the n × n matrix of an antisymmetric relation.

Each element of the diagonal can be either 0 or 1. Thus there are 2n ways to assign values to the diagonal.

For i and j satisfying 1 ≤ i < j ≤ n, we can assign the entries in row i, column j and row j, column i in three ways: Copyright c 2018 Pearson Education, Inc.

86

SOLUTIONS Row i, Column j 0 1 0

Row j, Column i 0 0 1

Since there are (n2 − n)/2 values of i and j satisfying 1 ≤ i < j ≤ n, we can assign the off-diagonal values 2 2 in 3(n −n)/2 ways. Therefore, there are 2n · 3(n −n)/2 antisymmetric relations on an n-element set.

85. We count the number of reflexive and symmetric relations on {1, 2, . . . , n} by computing the number of ways to construct the n × n matrix of a reflexive and symmetric relation.

The diagonal must consist of all 1’s. The (n2 − n)/2 elements above the diagonal can be assigned in 2 2 2(n −n)/2 ways. The elements below the diagonal are then determined. Thus there are 2(n −n)/2 reflexive and symmetric relations on an n-element set.

87. We count the number of symmetric and antisymmetric relations on {1, 2, . . ., n} by computing the number of ways to construct the n × n matrix of a symmetric and antisymmetric relation.

The off-diagonal elements must all be 0’s. The n diagonal elements can be assigned in 2n ways. Thus there are 2n symmetric and antisymmetric relations on an n-element set.

88. Since the matrix of a reflexive, symmetric, and antisymmetric relation on an n-element set must have all diagonal elements 1 and all off-diagonal elements 0, there is one reflexive, symmetric, and antisymmetric relation on an n-element set.

90. nn

2

91. nn(n+1)/2

93. Let X = eight-bit strings that start with 1, and Y = eight-bit strings that end with 1. Then |X| = |Y | = 27 and |X ∩ Y | = 26 . Now X ∪ Y is the set of eight-bit strings that start with 1 or end with 1 or both and, by the Inclusion-Exclusion Principle, |X ∪ Y | = |X| + |Y | − |X ∩ Y | = 27 + 27 − 26 .

94. Let X = selections in which Ben is chairperson, and Y = selections in which Alice is secretary. Then |X| = |Y | = 5 · 4 and |X ∩ Y | = 4. Now X ∪ Y is the set of selections in which Ben is chairperson or Alice is secretary or both and, by the Inclusion-Exclusion Principle, |X ∪ Y | = |X| + |Y | − |X ∩ Y | = 20 + 20 − 4 = 36.

96. Let X = selections in which the blue die is 3, and Y = selections in which the sum is even. Then |X| = 6, |Y | = 18, and |X ∩ Y | = 3. Now X ∪ Y is the set of selections in which the blue die is 3 or the sum is even or both and, by the Inclusion-Exclusion Principle, |X ∪ Y | = |X| + |Y | − |X ∩ Y | = 6 + 18 − 3 = 21.

97. Let X = integers from 1 to 10,000 that are multiples of 5, and Y = integers from 1 to 10,000 that are multiples of 7. An integer n is a multiple of 5 if there exists an integer k such that n = 5k. Thus an integer n from 1 to 10,000 that is a multiple of 5 (i.e., n = 5k for some k) satisfies 1 ≤ 5k ≤ 10,000.

Thus k satisfies

1 ≤ k ≤ 2000.

5 Since there are 2000 such k, |X| = 2000. Similarly |Y | = 1428. Now X ∩ Y consists of integers from 1 to 10,000 that are multiples of both 5 and 7, that is multiples of 35. Arguing as before, we see that |X ∩ Y | = 285. Also X ∪ Y consists of integers from 1 to 10,000 that are multiples of 5 or 7 or both and, by the Inclusion-Exclusion Principle, |X ∪ Y | = |X| + |Y | − |X ∩ Y | = 2000 + 1428 − 285 = 3143.

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87

SOLUTIONS 98. Letting A = X and B = Y ∪ Z, we have |X ∪ Y ∪ Z|

= = =

|X| + |Y ∪ Z| − |X ∩ (Y ∪ Z)| |X| + |Y | + |Z| − |Y ∩ Z| − |X ∩ (Y ∪ Z)| |X| + |Y | + |Z| − |Y ∩ Z| − |(X ∩ Y ) ∪ (X ∩ Z)|

= =

|X| + |Y | + |Z| − |Y ∩ Z| − [|X ∩ Y | + |X ∩ Z| − |X ∩ Y ∩ Z|] |X| + |Y | + |Z| − |Y ∩ Z| − |X ∩ Y | − |X ∩ Z| + |X ∩ Y ∩ Z|.

100. Let M, P, C denote the sets of students taking calculus, psychology, and computer science, respectively.

We are given |M ∩ P ∩ C| = 8, |M ∩ P | = 20, |M ∩ C| = 33, |P ∩ C| = 24, |M | = 79, |P | = 83, |C| = 63.

Now |M ∪ P ∪ C| = =

|M | + |P | + |C| − |M ∩ P | − |M ∩ C| − |P ∩ C| + |M ∩ P ∩ C| 79 + 83 + 63 − 20 − 33 − 24 + 8 = 156.

Since there are 165 total students, 9 students are taking none of the three subjects.

101. Let L denote the set of persons who watched “Law and Disorder,” let N denote the set of persons who watched “25,” and let T denote the set of persons who watched “The Tenors.” We are given |L| = 68, |N | = 61, |T | = 52, |L ∩ N | = 16, |L ∩ T | = 25, |N ∩ T | = 19, |L ∪ N ∪ T | = 151 − 26 = 125.

Now |L ∩ N ∩ T | = =

|L ∪ N ∪ T | − |L| − |N | − |T | + |L ∩ N | + |L ∩ T | + |N ∩ T | 125 − 68 − 61 − 52 + 16 + 25 + 19 = 4.

Thus four persons watched all three shows.

103. Proceed as in the solution given to Exercise 98.

104.

|A ∪ B ∪ C ∪ D|

=

|A| + |B| + |C| + |D| − |A ∩ B| − |A ∩ C| − |A ∩ D| − |B ∩ C| − |B ∩ D| − |C ∩ D| + |A ∩ B ∩ C| + |A ∩ B ∩ D| + |A ∩ C ∩ D| + |B ∩ C ∩ D| − |A ∩ B ∩ C ∩ D|

105. Let A denote the set of integers between 1 and 10,000 that are multiples of 3, let B denote the set of integers between 1 and 10,000 that are multiples of 5, let C denote the set of integers between 1 and 10,000 that are multiples of 11, and let D denote the set of integers between 1 and 10,000 that are multiples of 13. Then |A| = 333, |B| = 2000, |C| = 909, |D| = 769.

(The solution to Exercise 97 shows how these numbers were computed.)

Now A ∩ B is the set of integers between 1 and 10,000 that are multiples of 3 and 5, that is, multiples of 15. Thus |A ∩ B| = 666.

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88

SOLUTIONS Similarly, |A ∩ C| = 303, |A ∩ D| = 256, |B ∩ C| = 181, |B ∩ D| = 153, |C ∩ D| = 69.

Now A ∩ B ∩ C is the set of integers between 1 and 10,000 that are multiples of 3, 5, and 11, that is, multiples of 165. Thus |A ∩ B ∩ C| = 60.

Similarly, |A ∩ B ∩ D| = 51, |A ∩ C ∩ D| = 23, |B ∩ C ∩ D| = 13.

Finally, A ∩ B ∩ C ∩ D is the set of integers between 1 and 10,000 that are multiples of 3, 5, 11, and 13, that is, multiples of 2145. Thus |A ∩ B ∩ C ∩ D| = 4.

Thus |A ∪ B ∪ C ∪ D| =

|A| + |B| + |C| + |D| − |A ∩ B| − |A ∩ C| − |A ∩ D| − |B ∩ C| − |B ∩ D| − |C ∩ D| + |A ∩ B ∩ C| + |A ∩ B ∩ D| + |A ∩ C ∩ D| + |B ∩ C ∩ D| − |A ∩ B ∩ C ∩ D|

=

333 + 2000 + 909 + 769 − 666 − 303 − 256 − 181 − 153 − 69 + 60 + 51 + 23 + 13 − 4 = 2526.

Section 6.2 2. abcd, abdc, acbd, acdb, adbc, adcb, bacd, badc, bcad, bcda, bdac, bdca, cabd, cadb, cbad, cbda, cdab, cdba, dabc, dacb, dbac, dbca, dcab, dcba 3. P (4, 3) = 4(3)(2) = 24 8. P (12, 4) = 12 · 11 · 10 · 9

5. 11!

6. P (11, 5) = 11 · 10 · 9 · 8 · 7

9. P (12, 3) = 12 · 11 · 10

11. 3! · 3!

12. Tokens labeled AE, C, and DB can be permuted in 3! ways.

14. 12 5! since half have A before D and half have A after D. 15. We first count the number of strings containing either AB or CD or both. To this end, let X Y

= =

strings that contain AB strings that contain CD.

By previous methods, we find that |X| = |Y | = 4!,

|X ∩ Y | = 3!.

Now X ∪ Y is the set of strings that contain either AB or CD or both. By the Inclusion-Exclusion Principle, |X ∪ Y | = |X| + |Y | − |X ∩ Y | = 4! + 4! − 3!.

Since there are 5! total strings, the number that contain neither AB nor CD is 5! − (4! + 4! − 3!).

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SOLUTIONS 17. C(5, 3) · 2!. Pick three slots for A, C, and E. Then place the two remaining letters.

18. Let X Y

= =

strings that contain DB strings that contain BE.

By previous methods, we find that |X| = |Y | = 4!,

|X ∩ Y | = 3!.

Now X ∪ Y is the set of strings that contain DB or BE or both. By the Inclusion-Exclusion Principle, |X ∪ Y | = |X| + |Y | − |X ∩ Y | = 4! + 4! − 3!.

20. First, line up the Vesuvians and the Jovians. This can be done in 18! ways. For each of these arrangements, we can place the Martians in 5 of the 19 in-between and end positions, which can be done in P (19, 5) ways. Thus there are 18! · P (19, 5) arrangements.

22. 9!

23. Seat the Martians (4! ways). Seat the Jovians in the in-between spots (5 · 4 · 3 · 2 · 1 ways). The answer is 4! · 5!.

26. {a, b, c}, {a, b, d}, {a, c, d}, {b, c, d} 27. {a, b, c}

Cc Sc ## # C Sc

c Cc S ## # C Sc

c Cc S ## # C Sc

{b, c, d}

c Cc S ## # C Sc

3-combinations

c b c a b a b c a c a b a a b b c c

d b d a b a b d a d a b a a b b d d

d c d a c a c d a d a c a a c c d d

d c d b c b c d b d b c b b c c d d

3-permutations

29. C(12, 4)

{a, b, d}

{a, c, d}

30. C(44, 6), C(48, 6)

32. The solution to Exercise 31 shows that there are 838 different pizzas. In selecting four pizzas, there are four possibilities: (a) Four kinds of pizzas. (b) Three kinds of pizzas. (c) Two kinds of pizzas. (d) One kind of pizza.

[Possibility (a).] Four kinds of pizzas can occur in C(838, 4) ways.

[Possibility (b).] Three distinct of pizzas can occur in C(838, 3) ways, and then the fourth pizza is the same as one of the three. Thus three kinds of pizzas can occur in 3C(838, 3) ways.

[Possibility (c).] Two distinct pizzas can occur in C(838, 2) ways. The two remaining pizzas can occur in three ways: the two can be the same as the first pizza, the two can be the same as the second pizza, one can be the same as the first and the other the same as the second. Thus two kinds of pizzas can occur in 3C(838, 2) ways.

[Possibility (d).] One kind of pizza can occur in C(838, 1) ways.

Thus the total number of ways that four pizzas can be selected is C(838, 4) + 3C(838, 3) + 3C(838, 2) + C(838, 1).

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34. C(6, 3)C(7, 4) 35. C(13, 4) − C(6, 4) [The number of possible committees is C(13, 4). The number that have no women is C(6, 4).]

37. C(13, 4) − [C(6, 4) + C(7, 4)] (The total number of minus the number with all men or all women.)

38. C(13, 4) − C(11, 2) [The number of possible committees is C(13, 4). The number in which Mabel and Ralph serve together is C(11, 2).]

40. C(8, 3)

41. Six: 00011111, 10001111, 11000111, 11100011, 11110001, 11111000

44. 13 · C(48, 1) (Choose the denomination and then the odd card.)

45. C(13, 5)

47. 4 · C(13, 2) · 133 (You must pick two of one suit and one of each of the three remaining suits. First choose the suit to have two cards. Then choose two cards. Then choose one of each of the remaining suits.)

48. 4 50. 9 · 45 (Pick the lowest card’s denomination. Then pick the suit of each of the denominations.)

51. C(13, 2)C(11, 1)C(4, 2)C(4, 2)C(4, 1) (Pick the two denominations that receive two cards; pick the denomination to receive one card; pick two cards from each of the chosen denominations; pick one card from the other denomination.)

53. 4 54. C(4, 2)[C(26, 13) − 2] [Pick the two suits. The number of hands containing cards from the chosen suits is C(26, 13). Subtract the two hands that contain only cards of one of the chosen suits.]

56. C(13, 5)C(13, 4)C(13, 3)C(13, 1) 57. 4!C(13, 5)C(13, 4)C(13, 3)C(13, 1) [Pick the suits (the order determines which gets 5, 4, 3, 1). Then pick the desired number of cards from the selected suits.]

59. C(32, 13) (Select 13 cards from among the 32 non-face cards.)

61. C(10, 3)

62. C(10, 3) + C(10, 2) + C(10, 1) + C(10, 0)

64. C(10, 5)

66. C(46, 4)

67. C(46, 2)C(4, 2) (Select 2 good and 2 defective.)

69. Represent each of the two 10’s by a star (“∗”). The remaining n − 4 bits can be placed in the three in-between and end positions with respect to the two stars. For each of these three groups of bits, if a 1 occurs, the bits to its right in that group must also be 1’s. We place a vertical bar “|” in between the 0’s and 1’s in each of these groups. Each string can be represented by 0a |1b ∗ 0c |1d ∗ 0e |1f , where 0 ≤ a, b, c, d, e, f ≤ n − 4 and a + b + c + d + e + f = n − 4. Note that the length of the string is n + 1. A particular string is determined by the choice of five slots from n + 1 slots for the pattern | ∗ | ∗ |. Hence there are C(n + 1, 5) such strings.

70. Fix n − k 1’s. The k 0’s must be assigned to the n − k + 1 positions between the 1’s or at either end.

This can be done in C(n − k + 1, k) ways.

71. Look at the formula for C(n, k).

74. Argue as in Example 6.2.23.

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SOLUTIONS 75. We first prove that if u = 0, the formula is correct. For u = 0, we have r−u+1 r+1 C(r + u, r) = C(r, r) = 1, r+1 r+1

(15)

which is correct.

Basis Step (n = 1). Since n = 1, u = 0 and r = 1. The formula is correct by (15).

Inductive Step Let N (r, u) be the number of ways that Wright receives r votes and Upshaw receives u votes. Assume that the formula is true for n − 1. If u = 0, the formula is correct by (15). So assume that 0 < u ≤ r. First suppose that u = r. The number of ways Wright can get the last vote is 0. (Prior to the last vote, Upshaw would have u − 1 votes and Wright would have u votes.) Using the inductive assumption, the number of ways that Upshaw can get the last vote is then N (r, u) = N (r, u − 1) = N (r, r − 1) =

r − (r − 1) + 1 C(2r − 1, r). r+1

Some algebraic manipulation shows that this latter term is equal to r−u+1 C(r + u, r) r+1 as desired.

Now suppose that r > u. The number of ways that the last vote can be for Upshaw is N (r, u − 1), and the number of ways that the last vote can be for Wright is N (r − 1, u). Then N (r, u) = N (r, u − 1) + N (r − 1, u).

By the inductive assumption, N (r − 1, u) =

(r − 1) − u + 1 C((r − 1) + u, r − 1), (r − 1) + 1

and N (r, u − 1) = Thus N (r, u) =

r − (u − 1) + 1 C(r + (u − 1), r). r+1

(r − 1) − u + 1 r − (u − 1) + 1 C((r − 1) + u, r − 1) + C(r + (u − 1), r). (r − 1) + 1 r+1

More algebraic manipulation shows that this latter sum is equal to

r − u + 1 (r + u)! r−u+1 = C(r + u, r), r+1 r!u! r+1 and the proof is complete.

76. Take n = r = u in the formula in Exercise 75, and use Exercise 73.

77. Note that the minimum number of votes for Wright is dn/2e. Thus, by Exercise 74, the number of ways the votes could be counted is 1+

n−1 X

[C(n, r) − C(n, r + 1)] = C(n, dn/2e).

r=dn/2e

(The first term, 1, is the number of ways Wright receives n votes and Upshaw receives 0 votes.)

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SOLUTIONS

79. By Exercise 77, k vertical steps can occur in C(k, dk/2e) ways, since, at any point, the number of up steps is greater than or equal to the number of down steps. Then, n − k horizontal steps can be inserted among the k vertical steps in C(n, k) ways. These n−k horizontal steps can occur in C(n−k, d(n−k)/2e) ways, since, at any point, the number of right steps is greater than or equal to the number of left steps.

Thus, the number of paths that stay in the first quadrant containing exactly k vertical steps is C(k, dk/2e)C(n, k)C(n − k, d(n − k)/2e).

Summing over all k, we find that the total number of paths is n X

k=0

C(k, dk/2e)C(n, k)C(n − k, d(n − k)/2e).

80. Fix a starting position, and move around the table. When a handshake begins, write R; when a handshake ends, write a U . The result is a sequence of n R’s and n U ’s in which the number of R’s is always greater than or equal to the number of U ’s. Furthermore, the correspondence between such sequences and handshakes is one-to-one and onto. Since the number of sequences of n R’s and n U ’s in which the number of R’s is always greater than or equal to the number of U ’s is Cn (see Example 6.2.23), the number of ways that 2n persons seated around a circular table can shake hands in pairs without any arms crossing is also Cn .

81. Number the students 1 to n, with i < j meaning that student i is shorter than student j. In the encoding of a good route in Example 6.2.32, using U for up and R for right, if the letter in position i (the first letter is in position 1) is U , assign student i to the bottom row; otherwise, assign student i to the top row. This gives a one-to-one correspondence between good routes and proper student arrangements.

82. We show a one-to-one, onto correspondence between the output i1 , i2 , . . . , in reversed and sequences of n R’s and n U ’s in which the number of R’s is always greater than or equal to the number of U ’s. Since the number of such sequences is Cn , the result then follows.

For each sequence of n R’s and n U ’s in which the number of R’s is always greater than or equal to the number of U ’s, under each R write the number of D’s that precede it. For example, for n = 3 we would have R 0

R 0

D

R 1

D

D

Then add one to each value of the resulting sequence. In our case, we obtain the sequence 112. For n = 3, the complete correspondence is RD Sequence RRRDDD RRDRDD RRDDRD RDRRDD RDRDRD

Numeric Sequence 111 112 113 122 123

Numeric Sequence Reversed (Output) 111 211 311 221 321

83. There are P (n, r) ways to place the r distinct objects. There is one way to place the identical objects in the remaining slots. Thus the total number of orderings is P (n, r).

There are C(n, n − r) = C(n, r) ways to choose positions for the n − r identical objects. After placing the identical objects in these positions, there are r! ways to place the distinct objects. Thus the number of total orderings is r!C(n, r). Therefore P (n, r) = r!C(n, r).

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SOLUTIONS

84. The Addition Principle must be applied to a family of pairwise disjoint sets. Here the sets involved are not pairwise disjoint. For example, if X is the set of hands containing clubs, diamonds, and spades and Y is the set of hands containing clubs, diamonds, and hearts, X ∩ Y contains hands that contain clubs and diamonds.

86. Double-counting occurs. For a specific example, consider X = {1, 2, 3, 4},

Y = {a, b, c}.

We first select a 3-permutation of X, say 2, 1, 4. We then assign 2 the value a, 1 the value b, and 4 the value c. We then assign the remaining element of X, namely 3, an arbitrary value in Y : Suppose that we assign 3 the value b. The function is {(1, b), (2, a), (3, b), (4, c)}.

Now suppose that we select the 3-permutation 2, 3, 4 of X. We then assign 2 the value a, 3 the value b, and 4 the value c. We then assign the remaining element of X, namely 1, an arbitrary value in Y : Suppose that we assign 1 the value b. The function is {(1, b), (2, a), (3, b), (4, c)}, which is the same as the first function.

88. Six, since there are six ways to choose five of the slots marked 0, after which we can fill the remaining slots with 0’s and 1’s to obtain the string 10001000.

89. C(8, 5), since there are C(8, 5) way to choose five slots to fill with 0’s, after which we can fill the remaining slots with 0’s to obtain the string 00000000.

90. (a) If there are more tables than people, it is impossible to seat at least one person at each table. (b) If there are equal numbers of tables and people and there is at least one person at each table, there will be exactly one person at each table. (c) See Example 6.2.7. (d) In this case, there are two people at one table and one person at each other table. The two people to sit at one table can be chosen in C(n, 2) ways. (e) We prove this equation by induction on n. The Basis Step, n = 2, is part (b).

Assume that the equation is true for n and that we have n + 1 people. Choose one person. Either this person sits alone or with others. If the person sits alone, the other persons can be seated at the other table in (n − 1)! ways. If this person is not alone, by the inductive hypothesis the remaining n persons can be seated in n−1 X1 (n − 1)! i i=1

ways. The (n + 1)st person can be added to this seating arrangement in n ways (to the right of any of the other n persons). Thus there are n · (n − 1)!

n−1 X i=1

n−1

X1 1 = n! i i i=1

ways to seat n + 1 people if the (n + 1)st person does not sit alone. Therefore the total number of seatings is n−1 n X X 1 1 (n − 1)! + n! = n! . i i i=1 i=1 The Inductive Step is complete.

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SOLUTIONS (f) Fix n. Each seating of n persons at k round tables, with at least one person at each table, determines a unique permutation of 1, . . . , n. If p(i, 1), . . . , p(i, ei ) are seated clockwise at table i, i = 1, . . . , k, in this order, we interpret this as the permutation defined by the mapping p(1, 1) → p(1, 2) p(1, 2) → p(1, 3) .. . p(1, e1 − 1) → p(1, e1 ) p(1, e1 ) → p(1, 1) .. . p(k, 1) → p(k, 2) p(k, 2) → p(k, 3) .. . p(k, ek − 1) → p(k, ek ) p(k, ek ) → p(k, 1) (This representation is called the decomposition of a permutation into its cycles.) Since all permutations are accounted for, the equation follows. (g) We show that s3,1 = 2 and sn,n−2 = 2C(n, n − 3) + 3C(n, n − 4) for n ≥ 4. s3,1 = 2 by part (c).

If n ≥ 4, there are two basic seating arrangements:

1. n − 3 tables of one person each and one table of three persons. There are 2C(n, n − 3) such seatings since we may select the n − 3 solitary persons in C(n, n − 3) ways, and then seat the remaining three persons at one table in 2! ways (using the formula from part (c)).

2. n − 4 tables of one person each and two tables of two persons each. Select the n − 4 solitary persons in C(n, n − 4) ways, and then seat the remaining four persons in three ways. In this case, there are 3C(n, n − 4) seatings.

91. (a) If k > n, an n-element set cannot be partitioned into k nonempty subsets. (b) There is one way to partition an n-element set into n nonempty subsets: Each subset must consist of one element. (c) There is one way to partition an n-element set into one nonempty subset: The subset is the n-element set itself. (d)–(f) See (g) and (h). (g) Let X be an n-element set and let x ∈ X. For each nonempty subset Y of X − {x}, {Y, X − Y } is a partition of X. Since these are also all the partitions, Sn,2 = 2n−1 − 1.

(h) A partition of an n-element set into n − 1 subsets consists of a subset containing two elements and n − 2 subsets each containing one element. The 2-element subset can be chosen in C(n, 2) ways.

Therefore Sn,n−1 = C(n, 2). (i) Sn,n−2 = C(n, 3) + 3C(n, 4).

If we partition an n-element set into n − 2 subsets, either there is a subset consisting of three elements with all other subsets consisting of one element [there are C(n, 3) of these], or there are Copyright c 2018 Pearson Education, Inc.

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SOLUTIONS

two subsets each consisting of two elements with all other subsets consisting of one element [there are C(n, 4) ways to choose the elements to be the doubletons and three ways to organize the four elements into doubletons].

93. P (m, n), since each n-permutation of Y determines a one-to-one function from X to Y and conversely.

If X = {x1 , . . . , xn } and y1 , . . . , yn is an n-permutation of Y , f(xi ) = yi is a one-to-one function from X to Y . Similarly, if f is a one-to-one function from X to Y , f(x1 ), . . . , f(xn ) is an n-permutation of Y .

94. n!. The solution to Exercise 93 shows that there are P (m, n) one-to-one functions from an n-element set to an m-element set. Taking m = n, we find that there are P (n, n) = n! one-to-one functions from an n-element set to an n-element set.

Problem-Solving Corner: Combinations 1. From the following figure, we derive the formula m X

k=0 n

z

C(k + p, p)C(m − k + n − p, n − p) = C(m + n, m)

{

m ff

> >

: k

8 > > <

2.

|

}|

{z p

Pmin{m,n} k=0

}

C(m, k)C(n, k) = C(m + n, m)

Section 6.3 2. 6!/2!

3. 12!/(4!2!)

5. 8!/(2!3!)

6. 8!/(3!2!)

8. We form strings in which no two S’s are consecutive by first placing the letters ALEPERON, which can be done in 8!

2! ways. We then place the four S’s in the nine in-between positions — A — L — E — P — E — R — O — N —, which can be done in C(9, 4) ways. Thus C(9, 4)8!

2! strings can be formed by ordering the letters SALESPERSONS if no two S’s are consecutive.

9. We count the number of strings of length zero, the number of strings of length one, and so on, and then sum these numbers. The number of strings of length zero is one, and the number of strings of length one is five.

There is one string of length two that uses the two O’s, and there are 5 · 4 strings of length two that do not use two O’s (formed by selecting a 2-permutation of SCHOL). Thus there are 21 strings of length two.

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SOLUTIONS 3!

There are four ways to choose three letters including two O’s. There are 2! = 3 ways to permute these letters. Thus there are 4 · 3 strings of length three that use the two O’s, and there are 5 · 4 · 3 strings of length three that do not use two O’s (formed by selecting a 3-permutation of SCHOL). Thus there are 72 strings of length three.

4!

There are C(4, 2) = 6 ways to choose four letters including two O’s. There are 2! = 6 ways to permute these letters. Thus there are 6 · 6 = 36 strings of length four that use the two O’s, and there are 5 · 4 · 3 · 2 strings of length four that do not use two O’s (formed by selecting a 4-permutation of SCHOL). Thus there are 156 strings of length four.

5! = 60 ways to permute There are C(4, 3) = 4 ways to choose five letters including two O’s. There are 2! these letters. Thus there are 4 · 60 = 240 strings of length five that use the two O’s, and there are 5 · 4 · 3 · 2 · 1 strings of length four that do not use two O’s (formed by selecting a 5-permutation of SCHOL). Thus there are 360 strings of length five.

6!

There are 2! = 360 strings of length six. Thus there are

1 + 5 + 21 + 72 + 156 + 360 + 360 = 975 strings that can be formed by ordering the letters SCHOOL using some or all of the letters.

11. C(10 + 6 − 1, 6 − 1)

12. C(4 + 6 − 1, 6 − 1)

14. Assign each problem five points, and let xi denote the number of additional points that can be assigned to problem i. Now the question is: How many solutions are there of 12 X

xi = 40?

i=1

Arguing as in Example 6.3.8, the answer is C(40 + 12 − 1, 12 − 1).

15. 4100

16. C(100 + 4 − 1, 4 − 1)

22. C(8 + 2 − 1, 8) 27. C(14 + 2 − 1, 14)

19. C(9 + 3 − 1, 9)

23. C(10 + 2 − 1, 10) + C(9 + 2 − 1, 9)

20. C(4 + 3 − 1, 4) 26. C(12 + 3 − 1, 12)

29. C(15 + 3 − 1, 15) − C(8 + 3 − 1, 8)

30. There are C(14 + 3 − 1, 14) solutions satisfying 0 ≤ x1 , 1 ≤ x2 , 0 ≤ x3 . Of these, C(8 + 3 − 1, 8) have x1 ≥ 6; C(6 + 3 − 1, 6) have x2 ≥ 9; and there is one with x1 ≥ 6 and x2 ≥ 9. Thus there are C(8 + 3 − 1, 8) + C(6 + 3 − 1, 6) − 1 solutions with x1 ≥ 6 or x2 ≥ 9. Therefore there are C(14 + 3 − 1, 14) − [C(8 + 3 − 1, 8) + C(6 + 3 − 1, 6) − 1] of the desired type.

32. The number of solutions to the given equation, where x1 , x2 , and x3 are positive integers, is the same as the number of solutions to x1 + x2 + x3 = n − 3, where x1 , x2 , and x3 are nonnegative integers (just add 1 to each of x1 , x2 , and x3 ). The number of solutions to the latter equation is C(n − 3 + 3 − 1, n − 3) = C(n − 1, n − 3) =

(n − 1)(n − 2) .

2

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97

SOLUTIONS 33. The problem is equivalent to solving x1 + x2 + · · · + xn + xn+1 = M since 0 ≤ xn+1 = M − x1 + x2 + · · · + xn .

Thus the number of solutions is C(M + (n + 1) − 1, (n + 1) − 1) = C(M + n, n).

34. We must count the number of solutions of x1 + x2 + x3 + x4 + x5 + x6 = 15

satisfying 0 ≤ xi ≤ 9, i = 1, . . . , 6. There are C(15 + 6 − 1, 15) solutions with xi ≥ 0, i = 1, . . . , 6.

There are C(5 + 6 − 1, 5) with x1 ≥ 10. There are 6C(5 + 6 − 1, 5) with some xi ≥ 10. (Note that there is no double counting, since we cannot have xi ≥ 10 and xj ≥ 10, i 6= j.) Thus the solution is C(15 + 6 − 1, 15) − 6C(5 + 6 − 1, 5).

35. C(20 + 6 − 1, 20) − [6C(10 + 6 − 1, 10) − C(6, 2)]

37. 8!/(4! · 2! · 2!)

40. C(5 + 3 − 1, 5)

43. C(20, 5)C(15, 5)

41. C(6, 2)C(6, 3)C(8, 2)

44. [C(20, 5) − C(14, 5)][C(14, 5) + 6C(14, 4)]

46. C(15 + 5 − 1, 15)C(10 + 5 − 1, 10)

38. C(7 + 2 − 1, 2) 47. C(12, 10)

49. Consider the number of orderings of kn objects where there are n identical objects of each of k types.

50. The number of times the print statement is executed is 1 + 2 + · · · + n.

Example 6.3.9 shows that this is the same as C(2 + n − 1, 2) = (n + 1)n/2.

52. list sols(n) { for x1 = 0 to n for x2 = 0 to n − x1 println(x1 , x2 , n − x1 − x2 ) } 53. Many partitions are not counted. For example, the partition {{x1, x3 }, {x2 }, {x4}, {x5}, {x6}, {x7 }, {x8}, {x9, x10 }} is not counted.

Section 6.4 2. 12456

3. 23456

5. 631245

6. 13245678

8. (For Exercise 5) After the while loop in lines 7–9 finishes, m is 2. After the while loop in lines 11–13 finishes, k is 5. At line 14, we swap s2 and s5 . Now the sequence is 635421. The while loop of lines 17–22 reverses s3 , . . . , s6 . The result is 631245.

10. 12, 13, 14, 15, 16, 23, 24, 25, 26, 34, 35, 36, 45, 46, 56 Copyright c 2018 Pearson Education, Inc.

98

SOLUTIONS

11. 12345, 12346, 12347, 12356, 12357, 12367, 12456, 12457, 12467, 12567, 13456, 13457, 13467, 13567, 14567, 23456, 23457, 23467, 23567, 24567, 34567 13. 123, 132, 213, 231, 312, 321 19. C(4, 3)

15. 34568

21. 865421397

16. C(7, 3)

18. None

22. 5!

25. Change line 5 to while (true) { Add the following lines after line 13 to the body of the while loop at lines 11–13 if (k == 0) return (Also, since there will now be multiple lines in the body of the while loop, enclose them in braces.)

26.

Input: Output:

r, n A list of all r-permutations of {1, 2, ..., n}

list r perms(r, n) { for i = 1 to r si = i r comb(r) for i = 2 to C(n, r) { find the rightmost sm not at its maximum value sm = sm + 1 for j = m + 1 to r sj = sj + 1 r comb(r) } } r comb(r) { for i = 1 to r ti = si println(t) for i = 2 to r! { find the largest index m satisfying tm < tm+1 find the largest index k satisfying tk > tm swap(tm , tk ) reverse the order of the elements tm+1 , . . . , tr println(t) } }

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99

SOLUTIONS 28.

Input: Output:

s1 , . . . , sn (a permutation of {1, . . . , n}) and n s1 , . . . , sn , the next permutation. (The first permutation follows the last permutation.)

next perm(s, n) { s0 = 0 // dummy value m=n−1 while (sm > sm+1 ) m = m−1 k=n while (sm > sk ) k = k−1 if (m > 0) swap(sm , sk ) p = m+1 q=n while (p < q) { swap(sp , sq ) p = p+1 q = q−1 } } 30.

Input: Output:

s1 , . . . , sn (a permutation of {1, . . . , n}) and n s1 , . . . , sn , the previous permutation. (The last permutation precedes the first permutation.)

prev perm(s, n) { s0 = n + 1 // dummy value // working from right, find first index where si > si+1 i=n−1 while (si < si+1 ) i = i−1 // reverse si+1 , . . . , sn j = i+1 k=n while (j < k) { swap(sj , sk ) j = j +1 k = k−1 } // if i > 0, swap si with the first s value after si that is less than si if (i > 0) { j = i+1 while (sj > si ) j = j +1 swap(si , sj ) } } Copyright c 2018 Pearson Education, Inc.

100

SOLUTIONS

32.

Input: Output:

s1 , . . . , sn , n, and a string α All permutations of s1 , . . . , sn , each prefixed by α. (To list all permutations of s1 , . . . , sn , invoke this procedure with α equal to the null string.)

perm recurs(s, n, α) { if (n == 1) { println(α + s1 ) return } for i = 1 to n { α0 = α + si perm recurs({s1 , . . . , si−1 , si+1 , . . . , sn }, n − 1, α0) } } 33. Imitate the proof of Theorem 6.4.11.

Section 6.5 2. (H,2), (H,4), (H,6)

3. (H,1), (H,3), (H,5), (T,1), (T,3), (T,5)

6. (1,1), (2,2), (3,3), (4,4), (5,5), (6,6) 7. (1,4), (2,4), (3,4), (4,4), (5,4), (6,4), (4,1), (4,2), (4,3), (4,5), (4,6) 9. Suppose that the experiment is: Roll three dice. An event is: The sum of the dice is 8.

10. Suppose that the experiment is: Roll three dice. The sample space is the set of all possible outcomes.

(There are 216 possible outcomes.)

12. 36

13. 56

4 15. 52

16. 13 52

18. Since an odd sum can be obtained in 18 ways, (1,2), (1,4), (1,6), (2,1), (2,3), (2,5), (3,2), (3,4), (3,6), (4,1), (4,3), (4,5), (5,2), (5,4), (5,6), (6,1), (6,3), (6,5), 18 the probability of obtaining an odd sum is 36 .

19. Since doubles can be obtained in six ways, (1,1), (2,2), (3,3), (4,4), (5,5), (6,6), the probability of obtaining 6 doubles is 36 .

21. Exactly one defective microprocessor can be obtained in 10 · C(90, 3) ways. (Choose one defective microprocessor and choose three good microprocessors.) Since four microprocessors can be chosen in C(100, 4) ways, the probability of obtaining exactly one defective microprocessor is 10 · C(90, 3) .

C(100, 4) 22. At most one defective microprocessor can be obtained in C(90, 4) + 10 · C(90, 3) ways. (Choose four good microprocessors, or choose one defective microprocessor and three good microprocessors.) Since four microprocessors can be chosen in C(100, 4) ways, the probability of obtaining at most one defective microprocessor is C(90, 4) + 10 · C(90, 3) .

C(100, 4) Copyright c 2018 Pearson Education, Inc.

101

SOLUTIONS

24.

3!

103

25.

32.

10 210

35. 1 −

1 C(49, 6)

27.

1 C(31, 7)

1 310

36.

1 310

29.

C(26, 13) C(52, 13)

31.

1 210

39. An equivalent problem is to count strings of three C’s and nine N ’s in which no two C’s are consecutive since we can regard the positions of the C’s as representing the chosen lockers. For example, the string NNCNNNCNNNNC represents the choice of lockers 3, 7, and 12, no two of which are consecutive. We can obtain such strings by placing the three C’s in the 10 in-between positions of the nine N ’s: —N—N—N—N—N—N—N—N—N— The number of ways of choosing the positions for the C’s is, thus, C(10, 3). Therefore the probability that no two lockers are consecutive is C(10, 3) .

C(12, 3) C(10, 3) 40. 1 − C(12, 3)

42.

18 38

2

1 43. 38

46. If you make a random decision, you are choosing randomly between two doors—one with a goat and one with the car; therefore, the probability of winning the car is 21 .

47. Suppose that behind the first two doors are goats, and behind the third door is the car. Consider your three initial choices. If you initially choose door one, your switch will move you to the door with the car, and you win. Similarly, if you initially choose door two, your switch will move you to the door with the car, and you win. However, if you initially choose door three, your switch will move you to a door with a goat, and you lose. Therefore the probability of winning the car is 23 .

48. One-third of the time the host picks a door that hides a car and the game is over. Consider the two-thirds of the time when the host picks a door that hides a goat. Either the contestant has picked a door with a car behind it or a door with a goat behind it, so the contestant wins one-third of the time regardless of which strategy was followed.

50. If you make a random decision, you are choosing randomly between three doors—two with goats and one with the car; therefore, the probability of winning the car is 31 .

51. Suppose that behind the first three doors are goats, and behind the fourth door is the car. There are eight equally probable outcomes. If you initially choose door one, you switch to either a door with a goat or a door with a car. Similarly, if you initially choose door two or three, you switch to either a door with a goat or a door with a car. If you initially choose door four, you switch to one of two doors, each of which hides a goat. Of the eight possibilities, three win a car. Therefore the probability of winning the car is 83 .

53. The reasoning is not correct. As a small example, suppose that the sample space consist of eight eggs: g1 , g2 , g3 , g4 , g5 , g6 , b1 , b2 , where gi denotes a good egg and bi denotes a bad egg. Then the probability of a bad egg is 1/4. However, the set {g1 , g2 , g3 , g4 } of four eggs contains no bad eggs.

54. There are four ways that the sequence of tosses can start: TT, HT, TH, HH. If the sequence starts TT, the second player wins. If the sequence starts HT, the first player wins. If the sequence begins TH or HH, the first T after the first two coin tosses must be preceded by H and the first player wins. Thus the probability that the first player wins is 3/4. It’s best to be the first player!

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102

SOLUTIONS

56. The 10 disks can be given to Mary, Ivan, and Juan in C(10 + 3 − 1, 3 − 1) ways. If Ivan receives exactly three disks, the remaining seven disks can be given to Mary and Juan in C(7 + 2 − 1, 2 − 1) ways. Thus the probability that Ivan receives exactly three disks is C(7 + 2 − 1, 2 − 1) .

C(10 + 3 − 1, 3 − 1)

Section 6.6 2. Since P (2) = P (4) = P (6) =

1 , 8

the probability of getting an even number is P (2) + P (4) + P (6) =

3 .

8

3. Since the probability of getting a 5 is 18 , the probability of not getting a 5 is 1 − 18 = 78 .

3 5. 10

1 6. 10

4 8. 10

5 9. 10

11. 14

12. 41

15. The sum of 7 is obtained in six ways: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1). Now P (1, 6) = P (1)P (6) =

1 1 1 · = .

4 12 48

Similarly, P (2, 5) = P (3, 4) = P (4, 3) = P (5, 2) = P (6, 1) =

1 .

48

Therefore the probability of the sum of 7 is 6

1 48

=

1 .

8

16. Doubles or a sum of 6 is obtained in 10 ways: (1,5), (2,4), (3,3), (4,2), (5,1), (1,1), (2,2), (4,4), (5,5), (6,6). Now 2 1 1 P (1, 5) = P (1)P (5) = = .

4 16 Similarly, P (3, 3) = P (5, 1) = P (1, 1) = P (5, 5) =

1 , 16

P (2, 4) = P (4, 2) = P (2, 2) = P (4, 4) = P (6, 6) = Therefore the probability of getting doubles or the sum of 6 is

1 1 5 + .

16 144 Copyright c 2018 Pearson Education, Inc.

1 .

144

103

SOLUTIONS

18. Let E1 be the event “sum of 6,” let E2 be the event “doubles,” and let E3 be the event “at least one 2.”

We want P ((E1 ∪ E2 ) ∩ E3 ) P (E1 ∪ E2 | E3) = .

P (E3 ) The event (E1 ∪ E2 ) ∩ E3 comprises (2,4), (4,2), (2,2); therefore, 2 1 P ((E1 ∪ E2 ) ∩ E3 ) = 3 .

12 The solution to Exercise 17 shows that P (E3 ) = Therefore P (E1 ∪ E2 | E3) =

23 .

144

3 P ((E1 ∪ E2 ) ∩ E3 ) 3 = 144 23 = 23 .

P (E3 ) 144

19. Let E1 be the event “sum of 6,” let E2 be the event “sum of 8,” and let E3 be the event “at least one 2.” We want P ((E1 ∪ E2 ) ∩ E3 ) P (E1 ∪ E2 | E3) = .

P (E3 ) The event (E1 ∪ E2 ) ∩ E3 comprises (2,4), (4,2), (2,6), (6,2); therefore, 2 1 .

P ((E1 ∪ E2 ) ∩ E3 ) = 4 12 The solution to Exercise 17 shows that P (E3 ) = Therefore P (E1 ∪ E2 | E3) = 21. (H,3) 27.

22. No

23 .

144

4 P ((E1 ∪ E2 ) ∩ E3 ) 4 = 144 23 = 23 .

P (E3 ) 144

24. No

26. 1 −

C(90, 10) C(100, 10)

C(10, 3)C(90, 3) + C(10, 4)C(90, 2) + C(10, 5)C(90, 1) + C(10, 6) C(100, 10)

29.

C(4, 2) 24

30. The probability of all boys or all girls is 224 , so the probability of at least one boy and at least one girl is 1−

2 .

24

32. Let E be the event “exactly two girls,” and let F be the event “at least one girl.” We want to compute P (E | F ) =

P (E ∩ F ) .

P (F )

Now P (E ∩ F ) = P (E) =

C(4, 2) , 24

and P (F ) = 1 − P (no girls) = 1 − Therefore

1 .

24

C(4,2)

P (E | F ) =

P (E ∩ F ) 4 = 2 1 .

P (F ) 1 − 24

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104

SOLUTIONS

33. Let E be the event “at least one boy,” and let F be the event “at least one girl.” We want to compute P (E ∩ F | F ) =

P ((E ∩ F ) ∩ F ) P (E ∩ F ) = .

P (F ) P (F )

Now

2 , 24

P (E ∩ F ) = 1 − and P (F ) = 1 − Therefore

1 .

24

1 − 224 P (E ∩ F ) = .

P (F ) 1 − 214

P (E ∩ F | F ) =

35. Let E be the event “at most one boy,” and let F be the event “at most one girl.” Then P (E) = P (F ) = 1 − 254 , and P (E ∩ F ) = 0. Since P (E ∩ F ) = 0 6=

1−

5 24

2

= P (E)P (F ),

E and F are not independent.

36. Let E be the event “children of both sexes,” and let F be the event “at most one girl.” Then P (E) = 1 −

2 n+1 n , P (F ) = , P (E ∩ F ) = P (exactly one girl) = n .

2n 2n 2

Now E and F are independent if and only if P (E ∩ F ) = P (E)P (F ), or

n = 2n

1−

1 2n−1

n+1 2n

.

This equation simplifies to 2n−1 = n + 1, whose only solution is n = 3. (By inspection, 2n−1 6= n + 1 if n = 1, 2. For n > 3, 2n−1 > n + 1.) Therefore E and F are independent if and only if n = 3.

38.

C(10, 5) 210

41.

C(10, 0) + C(10, 1) + C(10, 2) + C(10, 3) + C(10, 4) + C(10, 5) 210

39.

C(10, 4) + C(10, 5) + C(10, 6) 210

42. Let E be the event “exactly five heads,” and let F be the event “at least one head.” We want to compute P (E | F ) =

P (E ∩ F ) .

P (F )

Now P (E ∩ F ) = P (E) =

C(10, 5) , 210

and P (F ) = 1 − P (no heads) = 1 − Therefore

1 .

210

C(10,5)

P (E | F ) =

P (E ∩ F ) 10 = 2 1 .

P (F ) 1 − 210

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105

SOLUTIONS

44. Let E be the event “at least one head,” and let F be the event “at least one tail.” We want to compute P (E | F ) =

P (E ∩ F ) .

P (F )

Now P (E ∩ F ) = 1 − P (all heads or all tails) = 1 − and P (F ) = 1 − P (no tails) = 1 − Therefore P (E | F ) =

2 , 210

1 .

210

1 − 2210 P (E ∩ F ) = .

P (F ) 1 − 2110

45. Let E be the event “at most five heads,” and let F be the event “at least one head.” We want to compute P (E | F ) =

P (E ∩ F ) .

P (F )

Now P (E ∩ F ) = =

P (1 or 2 or 3 or 4 or 5 heads) C(10, 1) + C(10, 2) + C(10, 3) + C(10, 4) + C(10, 5) , 210

and P (F ) = 1 − P (no heads) = 1 − Therefore P (E | F ) = 48. C(68, 18)/C(70, 20)

P (E ∩ F ) = P (F )

1 .

210

C(10,1)+C(10,2)+C(10,3)+C(10,4)+C(10,5) 210 .

1 − 2110

49. 2C(68, 18)/C(70, 20)

51. The proof is by induction on n, with the Basis Step, n = 1, clear. So, assume that Algorithm 4.2.4 can potentially output any permutation of input of size n with equal probability. Consider the case n + 1. When i = 1, the random selection j, which is swapped with 1, can be any of 1, . . . , n + 1 with equal probability. If j = 1, for i = 2 to n + 1, the algorithm behaves exactly like the algorithm for n, except that 1, . . . , n are called 2, . . . , n + 1, respectively. By the inductive assumption, all permutations of 2, . . . , n + 1 can occur and are equally likely. If j 6= 1, for i = 2 to n + 1, the algorithm behaves exactly like the algorithm for n, except that 1, . . . , n are called 2, . . . , j − 1, 1, j + 1, . . . , n + 1, respectively. By the inductive assumption, all permutations of 2, . . . , j − 1, 1, j + 1, . . . , n + 1 can occur and are equally likely. The proof is complete.

52. The proof that all permutations are possible is similar to the proof in Exercise 51, with the “equally likely” part omitted.

Consider the case n = 3. When i = 1, the following permutations occur with equal probability: 123, 213, 321. When i = 2, the permutation 123 can be changed to 213, 123, 132; the permutation 213 can be changed to 123, 213, 231; and the permutation 321 can be changed to 231, 321, 312 each occurring with probability 1/9. In particular, P (123) = 2/9 and P (132) = 1/9.

54. 253 Copyright c 2018 Pearson Education, Inc.

106

SOLUTIONS

55. Let E1 denote the event “at least three people have birthdays on the same month and date,” let E2 denote the event “exactly two people have birthdays on the same month and date,” and let E3 denote the event “no two people have birthdays on the same month and date.” Then P (E1 ) = 1 − P (E1 ) = 1 − P (E2 or E3 ) = 1 − P (E2 ) − P (E3 ).

The event E2 can be constructed by choosing two people to have a common birthday [in C(n, 2) ways]; selecting the date on which they have the same birthday (365 ways); choosing non-common birthdays for the remaining n − 1 people [364 · 363 · · · (364 − n + 3) ways]. Thus P (E2 ) =

C(n, 2) · 365 · 364 · · · (364 − n + 3) .

365n

From Example 6.6.7, P (E3 ) =

365 · 364 · · · (365 − n + 1) .

365n

It follows that P (E1 ) = 1 −

C(n, 2) · 365 · 364 · · · (364 − n + 3) 365 · 364 · · · (365 − n + 1) − .

365n 365n

56. Let E1 denote the event “at least three people have birthdays on the same month and date.” Using the formula in the solution to Exercise 55, we find that for n = 35, P (E1) = 0.480722, and for n = 36, P (E1) = 0.511802.

Thus the answer is n = 36.

58. Let H be the event “has headache,” and let F be the event “has fever.” We are given P (H) = 0.01, P (F | H) = 0.4, P (F ) = 0.02.

Using Bayes’ Theorem, we have P (H | F ) =

(0.4)(0.01) P (F | H)P (H) = = 0.2.

P (F ) 0.02

60. P (B | A) = 0.01, P (B | D) = 0.03, P (B | N ) = 0.03 61.

P (A | B) = =

P (B | A)P (A) P (B | A)P (A) + P (B | D)P (D) + P (B | N )P (N )

(0.01)(0.55) = 0.289473684.

(0.01)(0.55) + (0.03)(0.1) + (0.03)(0.35)

Similarly, P (D | B) = 0.157894736, P (N | B) = 0.552631578.

64. By Theorem 6.6.9, Since

P (E1 ∩ E2 ) = P (E1) + P (E2 ) − P (E1 ∪ E2 ).

P (E1 ∪ E2 ) ≤ 1,

the result follows.

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107

SOLUTIONS 65. The Basis Step is P (E1 ) ≤ P (E1 ), which is clearly true.

Assume the statement is true for n. By Theorem 6.6.9, P (E1 ∪ E2 ∪ · · · ∪ En ∪ En+1 )

=

P (E1 ∪ E2 ∪ · · · ∪ En ) + P (En+1 ) − P ((E1 ∪ E2 ∪ · · · ∪ En ) ∩ En+1 ).

Since P ((E1 ∪ E2 ∪ · · · ∪ En ) ∩ En+1 ) ≥ 0, we have P (E1 ∪ E2 ∪ · · · ∪ En ∪ En+1 ) ≤ P (E1 ∪ E2 ∪ · · · ∪ En ) + P (En+1).

Using the inductive assumption, we have P (E1 ∪ E2 ∪ · · · ∪ En ∪ En+1 ) ≤ ≤

P (E1 ∪ E2 ∪ · · · ∪ En ) + P (En+1 ) n n+1 X X P (Ei ) + P (En+1 ) = P (Ei ). i=1

i=1

67. Yes. Since E and F are independent, P (E ∩ F ) = P (E)P (F ). Since E ∩ F and E ∩ F are mutually exclusive and E = (E ∩ F ) ∪ (E ∩ F ), P (E) = P (E ∩ F ) + P (E ∩ F ).

Now P (E ∩ F ) = P (E) − P (E ∩ F ) = P (E) − P (E)P (F ) = P (E)[1 − P (F )[= P (E)P (F ).

Therefore E and F are independent.

68. No. If the person carries a bomb on the plane, the probability of a bomb on the plane is 1. The probability of two bombs on the plane is then 1 · 0.000001 = 0.000001.

Section 6.7 2. 32c5 − 240c4 d + 720c3d2 − 1080c2 d3 + 810cd4 − 243d5 4. 59136s6t6

5. C(10, 2)C(8, 3) = 10!/(2!3!5!)

11. C(12 + 4 − 1, 12)

7. C(5, 2)

8. C(5, 2)

12. C(12 + 3 − 1, 12) + C(11 + 3 − 1, 11) + C(10 + 3 − 1, 10)

14. (a) C(n, k) < C(n, k + 1) if and only if n! n! < k!(n − k)! (k + 1)!(n − k − 1)! if and only if k + 1 < n − k if and only if k < (n − 1)/2.

15. The Basis Step is immediate, so we turn to the Inductive Step. Assume that the formula is true for m.

Then m+1 X

(−1)k C(n, k) =

k=0

m X

(−1)k C(n, k) + (−1)m+1 C(n, m + 1)

k=0

= =

(−1)m C(n − 1, m) + (−1)m+1 C(n, m + 1) (−1)m+1 [−C(n − 1, m) + C(n, m + 1)] = (−1)m+1 C(n − 1, m + 1).

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108

SOLUTIONS

16. Set a = 1 and b = −1 in the Binomial Theorem.

18.

C(n, k − 1) + C(n, k)

=

n! n! + (k − 1)!(n − k + 1)! k!(n − k)!

=

(n!)(n − k + 1) (n!)k + k!(n − k + 1)! k!(n − k + 1)!

=

(n!)(n + 1) = C(n + 1, k). k!(n − k + 1)!

19. Choosing a k-element set X also selects an (n − k)-element set X. 21. (n + 1)n(n − 1)/3 22. Use the fact that k 2 = 2C(k, 2) + C(k, 1).

24. Use Exercise 16 and equation (6.7.3).

25. Imitate the combinatorial proof of the Binomial Theorem.

27. Take a = b = c = 1 in Exercise 25.

28. Use the fact that C(2n − 1, n) = C(2n − 1, n − 1).

29. Let X and Y be disjoint sets where |X| = n − 1 and |Y | = n. Then C(n − 1, i) is the number of ways of picking an i-element subset of X, and C(n, n − i) is the number of ways of picking an (n − i)-element subset of Y . So C(n − 1, i)C(n, n − i) is the number of ways of picking an n-element subset of X ∪ Y with i elements from X and n − i elements from Y . Summing over all i is the total number of ways of picking n-element subsets of X ∪ Y . On the other hand, C(2n − 1, n) also is the total number of ways of picking n-element subsets of X ∪ Y . The conclusion follows.

30. Think of C(n, k)2 as C(n, k)C(n, n − k). Let X and Y be disjoint sets each having n elements. Now, C(2n, n) is the number of ways of picking n-element subsets of X ∪ Y . Picking an n-element subset of X ∪ Y is the same as picking a k-element subset of X and an (n − k)-element subset of Y .

31. To distribute the balls, we simply select q of them to put into the container of size q, and put the rest into the other container. Since there are C(n, q) ways to select q balls, we can place the balls into the containers in C(n, q) ways.

Now consider taking into account the colors of the balls. First suppose that k ≤ q. We can place i red balls in the container of size q in C(k, i) ways. We can then fill the container by choosing q − i blue balls, which can be done in C(n − k, q − i) ways. We then put the remaining balls in the other container. Thus, in this case, we can fill the containers in C(k, i)C(n − k, q − i) ways. Summing k X i=0

C(k, i)C(n − k, q − i) = C(n, q)

gives the result if k ≤ q. In fact we can extend the upper limit of the sum from k to q since if k < q, C(k, i) = 0 for i = k + 1, . . . , q: q X i=0

C(k, i)C(n − k, q − i) = C(n, q).

(16)

We claim that formula (16) is also correct when k > q. As i increases from 0 to q, q − i decreases from q to 0. If q ≤ n − k, the argument for the previous case is still valid. However if q > n − k, for initial values of i, we will have q − i > n − k so C(n − k, q − i) = 0, which is correct. Therefore, formula (16) is correct if k > q.

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109

SOLUTIONS 33. Set x = 1 in Exercise 32.

34. Inductive Step. n+1 X

kC(n + 1, k) =

k=1

n X

k=1

= = =

n+1 X

k=1 n+1 X

k[C(n, k − 1) + C(n, k)] + (n + 1)C(n + 1, n + 1) kC(n, k − 1) +

kC(n, k)

k=1

(k − 1)C(n, k − 1) +

k=1 n−1

n2

n X

n+1 X k=1

C(n, k − 1) +

+ 2n + n2n−1 = (n + 1)2n

n X

kC(n, k)

k=1

36. We count the number of ways to choose sets A and B with A ⊆ B ⊆ X. Fix an integer k with 0 ≤ k ≤ n.

There are C(n, k) ways to choose a subset A of X with k elements. After choosing such a set A, there are n − k elements not in A, and there are 2n−k ways to choose a subset of them to union with A to produce a set B that contains A. Thus there are C(n, k)2n−k ways to choose subsets A and B satisfying A ⊆ B ⊆ X in which A has k elements. Summing over all k we obtain the number of ordered pairs (A, B) satisfying A ⊆ B ⊆ X: n X C(n, k)2n−k . k=0

Taking a = 2 and b = 1 in the Binomial Theorem, we find that this sum is equal to (a + b)n = (2 + 1)n = 3n .

37. We use induction. The Basis Step is n = m:

C(m, m)Hm = Hm = C(m + 1, m + 1) Hm+1 −

1 m+1

.

Assume true for n. Then n+1 X

C(k, m)Hk

=

k=m

n X

C(k, m)Hk + C(n + 1, m)Hn+1

k=m

= =

=

= = =

1 C(n + 1, m + 1) Hn+1 − + C(n + 1, m)Hn+1 m+1

1 1 C(n + 1, m + 1) Hn+2 − − n+2 m+1

1 + C(n + 1, m) Hn+2 − n+2 [C(n + 1, m + 1) + C(n + 1, m)]Hn+2 C(n + 1, m + 1) + C(n + 1, m) C(n + 1, m + 1) − − n+2 m+1 C(n + 2, m + 1) C(n + 1, m + 1) C(n + 2, m + 1)Hn+2 − − n+2 m+1 C(n + 1, m) C(n + 1, m + 1) C(n + 2, m + 1)Hn+2 − − m+1 m+1 C(n + 2, m + 1) C(n + 2, m + 1)Hn+2 − m+1

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(17)

110

SOLUTIONS Equality (17) follows from the formula C(n + 2, m + 1) =

n+2 C(n + 1, m). m+1

39. The Basis Steps are n = 1, 2 and are omitted. We give the Inductive Step only for odd n > 2: fn = fn−2 + fn−1

=

b(n−2)/2c

C(n − 2 − k, k) +

X

C(n − 2 − k, k) + 1 +

k=0

=

b(n−2)/2c k=0

=

1+

b(n−2)/2c

X

k=0

=

b(n−1)/2c

X

1+

bn/2c

X

k=1

X

k=0

C(n − 1 − k, k)

b(n−2)/2c

X

k=0

C(n − 2 − k, k + 1)

C(n − 1 − k, k + 1)

C(n − k, k) =

bn/2c

X

k=0

C(n − k, k).

40. Let p be a prime. If i = 1, p clearly divides C(p, 1) = p so suppose that 3 ≤ p and 2 ≤ i ≤ p − 1. Since C(p, i) =

p! p(p − 1)! = i!(p − i)! [i(i − 1) · · · 2 · 1][(p − i)(p − i − 1) · · · 2 · 1]

is an integer, each of i, i − 1, . . . , 2 and p − i, p − i − 1, . . . , 2 divides p!. Since p is prime, in fact, each of i, i − 1, . . . , 2 and p − i, p − i − 1, . . . , 2 divides (p − 1)!. Thus M = (p − 1)!/[i!(p − i)!] is an integer.

Because C(p, i) = pM , p divides C(p, i) for all i, 1 ≤ i ≤ p − 1.

42. Let x = m/(m + n) and y = n/(m + n). The binomial theorem asserts that

m n + m+n m+n

m+n

=

m+n X

C(m + n, k)

k=0

m m+n

k

n m+n

m+n−k

.

(18)

Now the left-hand side of equation (18) is 1. The term in the right-hand side of equation (18) for k = m is

m n m n C(m + n, m) . m+n m+n Since the other terms on the right-hand side of equation (18) are positive, the inequality follows.

43. We call Ck nk + Ck−1 nk−1 + · · · + C1 n + C0 ,

Ck 6= 0, Pn k a polynomial in n of degree k. The exercise is, then, to show that i=1 i is a polynomial in n of degree k + 1, where the coefficient of nk+1 is 1/(k + 1). We use strong induction on k.

Basis Step (k = 0).

n X

ik =

i=1

Inductive Step. Assume that for p < k, coefficient of np+1 is 1/(p + 1).

n X

1=n

i=1

Pn

p i=1 i is a polynomial in n of degree p + 1, where the

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111

SOLUTIONS With ∆ defined as before Exercise 70, Section 2.4, by the binomial theorem ∆ik+1 = (i + 1)k+1 − ik+1

0 [ik+1 + (k + 1)ik + Ck−1 ik−1 + · · · + C10 i + C00 ] − ik+1

=

0 (k + 1)ik + Ck−1 ik−1 + · · · + C10 i + C00 .

=

By Exercise 70, Section 2.4, n X i=1

∆ik+1 = (n + 1)k+1 − 1 = (k + 1)

n X

0 ik + Ck−1

i=1

n X i=1

ik−1 + · · · + C10

n X i=1

i + C00

n X

1.

i=1

By the inductive assumption, (k + 1)

n X

0 ik + Ck−1

i=1

n X i=1

ik−1 + · · · + C10

n X

i + C00

i=1

(n + 1)k+1 − 1 = (k + 1)

(n + 1)k+1 − 1 = =

1 = (k + 1)

i=1

Therefore

Also, by the binomial theorem,

n X

n X

n X

ik + polynomial in n of degree k.

i=1

ik + polynomial in n of degree k.

i=1

(nk+1 + Ck00 nk + · · · + C100 n + 1) − 1

nk+1 + polynomial in n of degree k.

Therefore nk+1 + polynomial in n of degree k = (k + 1)

n X

ik + polynomial in n of degree k.

i=1

Solving for

Pn

k i=1 i , we have

n X

ik

=

nk+1 + polynomial in n of degree k − polynomial in n of degree k k+1

=

nk+1 + polynomial in n of degree ≤ k. k+1

i=1

The Inductive Step is complete.

Section 6.8 2. Let the six students be the pigeons and the five grades be the pigeonholes. Assign each student (pigeon) to his grade (pigeonhole). By the Pigeonhole Principle, some pigeonhole (grade) will contain at least two pigeons (students), that is, at least two students received the same grade.

3. Let the 32 people be the pigeons and the 31 days in January be the pigeonholes. Assign each person (pigeon) to the day (pigeonhole) he receives a check. By the Pigeonhole Principle, some pigeonhole (day) will contain at least two pigeons (people), that is, at least two people receive checks on the same day.

5. Let the elements of X be the pigeons and the elements of Y be the pigeonholes. Assign each x ∈ X (pigeon) to f(x) (pigeonhole). By the Pigeonhole Principle, some pigeonhole (value in Y ) will contain at least two pigeons (elements of X), that is, there exist x1 , x2 ∈ X, x1 6= x2 , such that f(x1 ) = f(x2 ); thus f is not one-to-one.

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SOLUTIONS

6. Let the six integers selected be the pigeons and the five sets {1, 10}, {2, 9}, {3, 8}, {4, 7}, {5, 6} be the pigeonholes. Assign each integer selected (pigeon) to the set (pigeonhole) to which it belongs. Notice that since the sets partition {1, 2, . . . , 10}, this assignment is unique. By the Pigeonhole Principle, some pigeonhole (set) will contain at least two pigeons (selected integers). Regardless of which set contains two of the selected integers, these integers sum to 11.

8. There are six possible combinations of first and last names. Each of the 18 persons is to be assigned a first and last name. By the Pigeonhole Principle, at least d18/6e = 3 of them will be assigned the same first and last names.

9. Professor Euclid is paid 26 times per year. Since there are 12 months, by the Pigeonhole Principle, at least d26/12e = 3 pay periods will occur in the same month.

11. Let A = {x1 , . . . , x60 } be the set of positions for the available items. Each xi assumes a distinct value in {1, . . . , 115}. Let B = {x1 + 4, . . . , x60 + 4}. The set X = {x1 , . . . , x60, x1 + 4, . . . , x60 + 4} of 120 numbers can take on values from 1 to 119. By the Pigeonhole Principle at least two of these 120 elements are identical. Since the elements in A are distinct, so are the elements in B. There is an element xi in A and an element in xj + 4 in B which are identical.

12. Let ai denote the position of the ith available item. The 110 numbers a1 , . . . , a55 ; a1 + 9, . . . , a55 + 9 have values between 1 and 109. By the second form of the Pigeonhole Principle, two must coincide. The conclusion follows.

14. If any pair (Pi , Pj ), (Pi , Pk ), (Pk , Pj ) is dissimilar, then the two dissimilar pictures together with P1 are three mutually dissimilar pictures. If none of these pairs are dissimilar, then Pi , Pj , and Pk are three mutually similar pictures.

15. No. Consider five pictures P1 , . . . , P5 in which P1 is similar to P2 ; P2 is similar to P3 ; P3 is similar to P4 ; P4 is similar to P5 ; P5 is similar to P1 .

16. Yes, since any subset of six pictures has the given property.

17. If a remainder when the n integers, {1, 11, 111, . . .} are divided by n is equal to 0, we have found a multiple of n consisting of at most n 1s. If no remainder is equal to 0, then the remainders are in the set {1, . . . , n − 1}. Since this set has n − 1 elements and there are n remainders, by the Pigeonhole Principle, at least two remainders are equal. The difference d between two of the original integers {1, 11, 111, . . .} with the same remainders is of the form 111 . . . 11000 . . . 000. Thus d is a multiple of n consisting of at most n 0s and 1s.

22. Let n = 2, and consider the subset {3, 4, 5} of {1, 2, 3, 4, 5}.

23. Let xi denote the longest increasing subsequence and yi denote the longest decreasing subsequence starting at ai . Consider (bi , ci ) and (bj , cj ). We may assume that i < j. If ai < aj , {ai , xj } is an increasing subsequence starting at ai which is longer than xj . Hence bi ≥ length of {ai , xj } > bj . If ai > aj , {ai , yj } is a decreasing subsequence starting at ai which is longer than yj . Hence ci ≥ length of {ai , yj } > cj . Since the ak ’s are distinct, the preceding cases are the only cases. For each, we have shown that (bi , ci ) and (bj , cj ) are distinct.

24. The number of ordered pairs (bi , ci ) is m = n2 + 1, one for each i = 1, . . . , m.

25. By assumption, every increasing or decreasing subsequence has length less than or equal to n. Thus 1 ≤ bi ≤ n and 1 ≤ ci ≤ n.

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SOLUTIONS

26. By Exercise 24, we have n2 + 1 pairs (bi , ci ). By Exercise 25, these pairs can take on only n2 values.

By the Pigeonhole Principle, at least two of these pairs must be identical. This contradicts the result of Exercise 23.

27. If ri ≤ 8, since ri ≥ 0 and si = ri , we have 0 ≤ si ≤ 8.

If ri > 8, since ri ≤ 15, we have 1 ≤ 16 − ri < 8. Thus 1 ≤ si < 8.

28. The set {s1 , . . . , s10 } is a subset of the 9-element set {0, . . . , 8}. By the second form of the Pigeonhole Principle, sj = sk for j 6= k.

29. Suppose that sj = rj and sk = rk . Then rj = rk , so aj mod 16 = ak mod 16. Therefore 16 divides aj − ak .

If sj = 16 − rj and sk = 16 − rk , we again find that rj = rk and the conclusion follows.

30. We may suppose that sj = rj and sk = 16 − rk . Thus rj = 16 − rk so rj + rk = 16. By definition, aj mod 16 = rj and ak mod 16 = rk so there are integers qj and qk satisfying aj = 16qj + rj and ak = 16qk + rk .

Now aj + ak

=

16(qj + qk ) + rj + rk

= =

16(qj + qk ) + 16 16(qj + qk + 1).

Therefore 16 divides aj + ak .

32. Suppose that the numbers around the circle are x1 , . . . , x12. We argue by contradiction. Suppose that x1 + x2 + x3 x2 + x3 + x4

≤ ≤ .. .

19 19

x10 + x11 + x12 x11 + x12 + x1 x12 + x1 + x2

≤ ≤ ≤

19 19 19.

Summing, we obtain the contradiction

12 · 13 234 = 3 = 3(x1 + · · · + x12 ) ≤ 12 · 19 = 228.

2 33. The sum of some four consecutive players’ numbers must be at least 26. Suppose that x1 + x2 + x3 + x4 x2 + x3 + x4 + x5 x12 + x1 + x2 + x3

≤ 25 ≤ 25 .. . ≤ 25

Summing, we obtain the contradiction

12 · 13 312 = 4 = 4(x1 + · · · + x12 ) ≤ 12 · 25 = 300.

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SOLUTIONS

34. Each of the n! + 1 functions

f, f 2 , f 3 , . . . , f n!+1

is a permutation of {1, . . . , n}. Since there are n! permutations of {1, . . . , n}, by the Pigeonhole Principle, fi = fj

(19)

for some distinct positive integers i and j.

Notice that if we compose each side of (19) with f −1 , we obtain f i−1 = f j−1 .

We may assume that i > j. If we compose each side of (19) with f −1 j times, we obtain f i−j = I, where I(x) = x for x = 1, . . . , n. We may take k = i − j to obtain the desired conclusion.

36. Suppose that the numbers around the circle are x1 , . . . , xp+q . We argue by contradiction. Suppose that each consecutive group of k numbers contains a zero. Then each consecutive group of k numbers sums to k − 1 or less. Thus x1 + · · · + xk x2 + · · · + xk+1 xp+q + x1 + · · · + xk−1

≤ ≤ .. . ≤

k−1 k−1 k−1

Summing, we obtain or

kp = k(x1 + · · · + xp+q ) ≤ (p + q)(k − 1) p ≤ (k − 1)q.

Since p ≥ kq > (k − 1)q, this is a contradiction.

37. See Section 6.11.1, pages 167–169, of U. Manber, Introduction to Algorithms, Addison-Wesley, Reading, Mass., 1989.

39. Let {a1 , a2 , . . . , an+1 } be an (n + 1)-element subset of {1, 2, . . . , 2n}. The 2n + 2 numbers a1 , . . . , an+1 , a1 + 1, . . . , an+1 + 1 cannot be distinct since each is in the (2n + 1)-element set {1, 2, . . . , 2n, 2n + 1}. Thus ai = aj + 1 for some i and j; hence, gcd(ai , aj ) = 1.

40. A counterexample is {2, 4, . . ., 2n}.

42. Basis Step (n = 1). The only 2-element subset of {1, 2} is {1, 2} and 1 divides 2.

Inductive Step Assume that the assertion is true for n. Let A be an (n + 2)-element subset of {1, 2, . . ., 2n, 2n + 1, 2n + 2}. If A does not contain both 2n + 1 and 2n + 2, we may delete an element from A (either 2n + 1 or 2n + 2 or, if neither is in A, an arbitrary element of A) and apply the inductive hypothesis to the altered set to derive the desired conclusion.

Suppose that both 2n + 1 and 2n + 2 are in A. If n + 1 ∈ A, then n + 1 divides 2n + 2 and we are finished.

If n + 1 ∈ / A, consider the (n + 1)-element set A0 obtained from A by deleting 2n + 1 and 2n + 2 and adding n + 1. By the inductive hypothesis, there exist a, b ∈ A0 such that a divides b. Now a 6= n + 1, for if a = n + 1, then b ≥ 2n + 2, which is not the case. If b 6= n + 1, then we have found a, b ∈ A such that a divides b. If b = n + 1, then a divides 2n + 2. In this case, a and 2n + 2 are the desired integers in A such that a divides 2n + 2.

43. A counterexample is {n + 1, n + 2, . . . , 2n}.

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SOLUTIONS

Section 7.1 2. an = an−1 + an−2 ; a1 = 3, a2 = 6 9. An = 1.10An−1 + 2000 12.

An

3. an = 2an−1an−2 ; a1 = a2 = 1

10. A0 = 2000

11. A1 = 4200, A2 = 6620, A3 = 9282

= = = = = .. .

1.10An−1 + 2000 1.10(1.10An−2 + 2000) + 2000 1.102An−2 + (1.10)2000 + 2000 1.102(1.10An−3 + 2000) + (1.10)2000 + 2000 1.103An−3 + (1.102 )2000 + (1.10)2000 + 2000

= =

1.10nA0 + (1.10n−1)2000 + (1.10n−2)2000 + · · · + (1.10)2000 + 2000 (1.10n )2000 + (1.10n−1)2000 + (1.10n−2)2000 + · · · + (1.10)2000 + 2000 (1.10n+1 )2000 − 2000 1.10 − 1 20000(1.10n+1 − 1)

= =

13. An = (1.03)4An−1

14. A0 = 3000 16. An = (1.03)4n3000

15. A1 = 3376.53, A2 = 3800.31, A3 = 4277.28

17. 5.86

20. An n-bit string that does not contain the pattern 00 either begins 1 and is followed by an (n − 1)-bit string that does not contain 00, or it begins 01 and is followed by an (n − 2)-bit string that does not contain 00. Thus we obtain the recurrence relation Sn = Sn−1 + Sn−2 , which is the same recurrence relation that the Fibonacci sequence f satisfies. The initial conditions for the sequence S are S1 = 2, S2 = 3.

Since f3 = 2 and f4 = 3, the result follows.

21. We count the number of n-bit strings with exactly i 0’s that do not contain the pattern 00. Such a string has n − i 1’s: — 1 — 1 — ... — 1 —.

To avoid the pattern 00, the 0’s must be placed in the n − i + 1 spaces. This can be done in C(n − i + 1, i) ways. Thus there are b(n+1)/2c X C(n + 1 − i, i) (20) i=0

n-bit strings that do not contain the pattern 00. By Exercise 20, (20) is equal to fn+2 .

23. We count the number of strings not containing 010 having i leading 0’s. For i = 0, there are Sn−1 such strings. For i = 1, the string begins 011, so there are Sn−3 such strings. Similarly, for i = 2, there are Sn−4 such strings; . . . for i = n − 3, there are S1 such strings. For i = n − 2, n − 1, or n, there is one such string. The equation now follows.

24. The formula for n − 1 is

Sn−1 = Sn−2 + Sn−4 + Sn−5 + · · · + S1 + 3.

Subtracting Sn−1 from Sn , we obtain

Sn − Sn−1 = Sn−1 + Sn−3 − Sn−2 .

Solving for Sn , we obtain the desired recurrence relation.

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116

SOLUTIONS

26. We use the explicit formula for the nth Catalan number derived in Example 6.2.22 to obtain (n + 2)Cn+1

=

(n + 2)C(2n + 2, n + 1) n+2

=

(2n + 2)! (n + 1)!(n + 1)!

=

2(2n + 1)(2n)!(2n + 2) (n + 1)n!n!(2n + 2)

=

(4n + 2)(2n)! (n + 1)n!n!

=

(4n + 2)C(2n, n) = (4n + 2)Cn . n+1

27. We use induction on n.

Basis Step (n = 4). n + 2 = 6 < 14 = C4 Inductive Step. Assume that n + 2 < Cn . Then (n + 2)/Cn < 1. Using Exercise 26, we have 4n + 2 n+2 = < 1.

Cn+1 Cn Therefore n + 3 < 4n + 2 < Cn+1 .

29. The proof is by induction on n with the Basis Step omitted.

Assume that the inequality holds for n. We use the formula from Exercise 26 to derive Cn+1 =

4n + 2 4n + 2 4n−1 4n Cn ≥ ≥ .

2 n+2 n+2 n (n + 1)2

The last inequality is successively equivalent to 4n + 2 (n + 2)n2

≥

(2n + 1)(n + 1)2 ≥ 2n3 + 5n2 + 4n + 1 ≥ n2 + 4n + 1 ≥

4 (n + 1)2 2(n + 2)n2 2n3 + 4n2 0,

which is clearly true for all n ≥ 1.

30. Let bn denote the number of ways to parenthesize the product a1 ∗ · · · ∗ an+1 .

Then b0 = b1 = 1.

Suppose that n > 1, 1 ≤ i ≤ n, and that the multiplication is carried out by multiplying a1 ∗ · · · ∗ ai ,

(21)

ai+1 ∗ · · · ∗ an+1 ,

(22)

parenthesized in some way, by

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SOLUTIONS

parenthesized in some way. There are bi−1 ways to parenthesize (21) and bn−i ways to parenthesize (22).

Therefore there are bi−1 bn−i ways to parenthesize the product a1 ∗ · · · ∗ an+1 . if the multiplication is carried out by multiplying (21), parenthesized in some way, by (22), parenthesized in some way. Summing over all i we obtain bn =

n X

bi−1 bn−i .

i=1

Since the sequence {bn } satisfies the same initial condition and recurrence relation as the Catalan sequence {Cn }, it follows that bn = Cn for all n.

32. The answer is (2n)!. We can order the variables in (n + 1)! ways. We can insert the operators in between the variables in n! ways, and, by Exercise 30, we can insert the parentheses in Cn ways. Thus the number of expressions is

1 1 (2n)! (n + 1)!n!Cn = (n + 1)!n!

C(2n, n) = (n + 1)!n! = (2n)!. n+1 n + 1 n!n!

33. We assume that the paths start at (0, 0) and end at (n, n). Suppose that a route first meets the diagonal at (k + 1, k + 1). It is either always below the diagonal [from (1, 0) to (k + 1, k)] or always above the diagonal [from (0, 1) to (k, k + 1)]. Thus there are 2Ck such paths from (0, 0) to (k + 1, k + 1). There are C(2(n+1 −(k +1)), n+1 −(k + 1)) = C(2(n−k), n−k) paths from (k +1, k +1) to (n+1, n+1) (with no restrictions). Thus the number of paths that first meet the diagonal at (k+1, k+1) is Ck C(2(n−k), n−k).

Since there are C(2(n + 1), n + 1) paths from (0, 0) to (n + 1, n + 1) (with no restrictions), C(2(n + 1), n + 1) =

n X

k=0

Thus C(2(n + 1), n + 1) = 2Cn +

2Ck C(2(n − k), n − k). n−1 X k=0

2Ck C(2(n − k), n − k).

Dividing by 2 and moving the last summation to the left side of the equation gives the desired result.

35. We assume that the paths start at (0, 0) and end at (n, n). Let Di denote the number of paths that first touch the diagonal at (i, i) after leaving (0, 0). Then Sn = D1 + D2 + · · · + Dn .

Since D1 is the product of the number of paths from (0, 0) to (1, 1) and the number of paths from (1, 1) to (n, n), D1 = 2Sn−1 .

If i ≥ 2, Di is the product of the number of paths from (1, 0) to (i, i − 1) and the number of paths from (i, i) to (n, n), so Di = Si−1 Sn−i .

Thus Sn = 2Sn−1 +

n X

Si−1 Sn−i .

i=2

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118

SOLUTIONS

37. price = ak/(k + b), quantity = a/(k + b) 39. We have

b b b b pn − a + pn−1 = (pn−1 − pn ) = |pn − pn−1 |. k k k k

|pn+1 − pn | = a −

Now b > k, so b/k > 1. Thus |pn+1 − pn | > |pn − pn−1 |.

41. Basis Step. A(1, 0) = A(0, 1) = 2 Inductive Step. A(1, n + 1) = A(0, A(1, n)) = A(0, n + 2) = n + 3 42. Basis Step. A(2, 0) = A(1, 1) = 3 by Exercise 41.

Inductive Step. Assume that the statement is true for n. Then A(2, n + 1)

= = = =

A(1, A(2, n)) A(1, 3 + 2n) 3 + 2n + 2 2n + 5.

by the inductive assumption

by Exercise 41

44. Basis Step (m = 0). A(0, n) = n + 1 > n for all n ≥ 0.

Inductive Step. Assume that A(m, n) > n for all n ≥ 0. We use induction on n to prove that A(m + 1, n) > n

for all n ≥ 0.

Basis Step (n = 0).

A(m + 1, n)

= = > >

A(m + 1, 0) A(m, 1) 1 0 = n.

= > >

A(m, A(m + 1, n)) A(m + 1, n) n

by the original inductive assumption

Inductive Step.

A(m + 1, n + 1)

by the original inductive assumption by the present inductive assumption.

Now A(m + 1, n + 1) > A(m + 1, n) ≥ n + 1,

which completes the current inductive step. Therefore

A(m + 1, n) > n for all n ≥ 0. Thus the original induction is complete.

45. If n = 0, A(m, 0) = A(m − 1, 1) > 1 by Exercise 44.

If n > 0, A(m, n) > n ≥ 1 by Exercise 44.

y

.y ..

for z > 0, where there are z y’s. We prove the 48. The formula is AO(4, y, 0) = 1, and AO(4, y, z) = y formula by induction on z. The Basis Step (z = 0) is obvious.

Inductive Step. Assume true for z. If z = 1, we have AO(4, y, 1) = AO(3, y, AO (4, y, 0)) = y1 = y, where we have used Exercise 47. .y .. y If z ≥ 1, we have AO (4, y, z + 1) = AO(3, y, AO (4, y, z)) = yy , where there are z + 1 y’s. We used Exercise 47 again. The proof is complete.

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119

SOLUTIONS 50. Basis Step (x = 2). AO(2, 2, 1) = 2 · 1 = 2.

Inductive Step. Assume that the statement is true for x. Now AO(x + 1, 2, 1) = AO(x, 2, AO(x + 1, 2, 0)) = AO(x, 2, 1) = 2.

51. Inductive Step.

AO(x + 1, 2, 2)

= = =

AO(x, 2, AO(x + 1, 2, 1)) AO(x, 2, 2) 4

by Exercise 50

53. (a) a2 = 1 because two nodes must establish one link to share files.

Suppose that we have three nodes A, B, and C. If the following successive links are established, A ↔ B, A ↔ C, A ↔ B, then all nodes know all files. Since three links suffice, a3 ≤ 3.

Suppose that we have four nodes A, B, C, and D. If the following successive links are established, A ↔ B, C ↔ D, A ↔ C, B ↔ D, then all nodes know all files. Since four links suffice, a4 ≤ 4.

(b) Suppose that we have n ≥ 3 nodes. Let A and B be nodes. First A and B share files. Next, all nodes except A share files (requiring an−1 links). Finally, A and B again share files. At this point all nodes know all files. Thus an ≤ an−1 + 2.

54. P1 = 1, Pn = nPn−1 56. Suppose that we have n dollars. If we buy tape the first day, there are Rn−1 ways to spend the remaining money. If we buy paper the first day, there are Rn−1 ways to spend the remaining money. If we buy pens the first day, there are Rn−2 ways to spend the remaining money. If we buy pencils the first day, there are Rn−2 ways to spend the remaining money. If we buy binders the first day, there are Rn−3 ways to spend the remaining money. Thus Rn = 2Rn−1 + 2Rn−2 + Rn−3 .

57. Because of the assumptions, when the (n + 1)st line L is added, it will intersect the other n lines. If we imagine traveling along L, each time we pass through one of the original regions, we divide it into two regions. Since we pass through n + 1 regions, Rn+1 = Rn + n + 1. h n−1 i 59. Sn = 32 1 − − 12

61. A string α of length n with C(α) ≤ 2 either begins with 1 (there are Sn−1 of these), 01 (there are Sn−2 of these), or 001 (there are Sn−3 of these). Thus Sn = Sn−1 + Sn−2 + Sn−3 .

62. Basis Steps (n = 1, 2).

2f2 − 1 = 2 − 1 = 1 = g1 , 2f3 − 1 = 4 − 1 = 3 = g2 Inductive Step. gn = gn−1 + gn−2 + 1 = (2fn − 1) + (2fn−1 − 1) + 1 = 2(fn + fn−1 ) − 1 = 2fn+1 − 1

64. The problem is that the Inductive Step assumes two previous cases, but the Basis Step proves only one.

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SOLUTIONS

65. C(n + 1, k) = C(n, k − 1) + C(n, k) 67. There are k n functions from X = {1, . . . , n} onto Y = {1, . . . , k}. We will count the number N of functions from X into, but not onto, Y . Then, the number of functions from X onto Y will be k n − N .

Let Z be a proper, nonempty subset of Y with i elements. There are S(n, i) functions from X onto Z. The number of subsets of Y having i elements is C(k, i). Then, there are C(k, i)S(n, i) functions from X onto some i-element subset of Y . If a function from X to Y is not onto Y , it is onto some proper nonempty subset of Y . The result follows.

68. (a) L3 = 4, L4 = 7, L5 = 11 (b) Basis Steps (n = 1, 2).

L3 L4

= =

4 = 1 + 3 = f2 + f4 7 = 2 + 5 = f3 + f5

Inductive Step. Assume that Ln+1 = fn + fn+2 and Ln+2 = fn+1 + fn+3 . Now Ln+3

= =

Ln+2 + Ln+1 = fn+1 + fn+3 + fn + fn+2 (fn + fn+1 ) + (fn+2 + fn+3 )

=

fn+2 + fn+4 .

70. Let X be an (n + 1)-element set and choose an element x ∈ X. We count the number of partitions of X containing k subsets in which x appears as a singleton and the number of partitions of X containing k subsets in which x appears in a subset with at least two elements.

A partition of X containing k subsets in which x appears as a singleton consists of {x} together with a partition of X − {x} containing k − 1 subsets. There are Sn,k−1 such partitions.

A partition of X containing k subsets in which x appears in a subset with at least two elements can be constructed in the following way. Select a partition of X − {x} containing k subsets. This can be done in Sn,k ways. Next add x to one of the subsets. This can be done in k ways. Thus there are kSn,k partitions of X containing k subsets in which x appears in a subset with at least two elements. The recurrence relation now follows.

71. We prove the formula by induction on n leaving the Basis Step (n = 1) to the reader.

Assume that the formula is true for n. We must prove the formula is true for n + 1. If k = 0, the formula is clearly true, so assume that k > 0. By Exercise 70, Sn+1,k

=

Sn,k−1 + kSn,k k−1

=

=

=

=

=

k

X k X 1 (−1)i (k − 1 − i)n C(k − 1, i) + (−1)i (k − i)n C(k, i) (k − 1)! k! i=0 i=0 "k−1 # k X X 1 i n i n (−1) (k − 1 − i) C(k − 1, i) + (−1) (k − i) C(k, i) (k − 1)! i=0 i=0 " # k X 1 n i−1 n i n k C(k, 0) + {(−1) (k − i) C(k − 1, i − 1) + (−1) (k − i) C(k, i)} (k − 1)! i=1 " # k X 1 n i n k C(k, 0) + (−1) (k − i) [−C(k − 1, i − 1) + C(k, i)] (k − 1)! i=1 " # k X 1 n i n k C(k, 0) + (−1) (k − i) C(k − 1, i) (k − 1)! i=1 Copyright c 2018 Pearson Education, Inc.

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SOLUTIONS

=

" # k 1 k n+1 C(k, 0) X i n C(k, i)(k − i) + (−1) (k − i) (k − 1)! k k i=1

k

=

1 X (−1)i (k − i)n+1 C(k, i). k! i=0

73. an = n(an−1 + 1) 75. 1,5,2,4,3; 1,5,3,4,2; 2,5,1,4,3; 2,5,3,4,1; 3,5,1,4,2; 3,5,2,4,1; 4,5,1,3,2; 4,5,2,3,1; 1,3,2,5,4; 2,3,1,5,4; 1,4,2,5,3; 2,4,1,5,3; 1,4,3,5,2; 3,4,1,5,2; 2,4,3,5,1; 3,4,2,5,1; E5 = 16 76. An item in the first, third, . . . position has a larger neighbor; therefore, n cannot be in any of these positions.

78. The solution is similar to that of Exercise 77, which is given in the book, except that the portion following 1 must be a “fall/rise” permutation. However, the number of “fall/rise” permutations of 1, . . . , n is equal to the number of “rise/fall” permutations of 1, . . . , n, so the argument proceeds as in the solution to Exercise 77.

79. Add the recurrence relations of Exercises 77 and 78.

Section 7.2 2. No

3. No

9. Yes; order 2 13.

an

5. No

6. Yes; order 3

8. Yes; order 2

12. an = 2n n!

=

an−1 + n = an−2 + (n − 1) + n = · · ·

=

a0 + 1 + 2 + · · · + n = 1 + 2 + · · · + n =

n(n + 1) 2

14. an = 2n an−1 = 2n 2n−1 an−2 = 2n 2n−1 2n−2 an−3 = · · · = 2n 2n−1 2n−2 · · · 21 a0 = 2n+(n−1)+···+1 = 2n(n+1)/2

16. The roots of t2 − 2t − 8 = 0 are 4 and −2. Thus the solution is an = b4n + d(−2)n . To satisfy the initial conditions, we must have 4 = 10 =

b+d 4b − 2d.

Solving, we find b = 3 and d = 1. Thus the solution is an = 3 · 4n + (−2)n .

17. Solving 0 = t2 − 7t + 10 = (t − 2)(t − 5), we obtain two roots t = 2 and t = 5. Thus there exist constants b and d such that an = b2n + d5n . The initial conditions require that 5 = b + d and 16 = 2b + 5d. Solving these two equations simultaneously for b and d, we obtain b = 3 and d = 2. Thus an = 3 · 2 n + 2 · 5 n .

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122

SOLUTIONS

19.

an

= = = =

an−1 + 1 + 2n−1 (an−2 + 1 + 2n−2 ) + 1 + 2n−1 an−2 + 2 + 2n−1 + 2n−2 = · · · a0 + n + 2n−1 + 2n−2 + · · · + 1 = n + 2n − 1

20. an = 3n − 2n3n−1 25. sn =

n(n + 1) +1 2

22. Similar to Example 7.2.13

23. an = 6

h n−1 i 26. Sn = 23 1 − − 12

1 n + 9n 3

1 n 3

27. Let p be the population of Utopia in 1970. Arguing as in Example 7.1.3, we find that n years after 1970, the population of Utopia is (1.05)n p. Therefore 10000 = (1.05)30 p. Solving for p, we find that the population of Utopia in 1970 was p = 2314.

30. 1/2 31. The equation follows since W and W are disjoint and W ∪ W exhausts the sample space.

33. 1

34. pn = 2pn−1 − pn−2

2 36. The recurrence relation becomes bn = bn−1 + 2bn−2. Solving gives an = b2n = 19 2n+1 + (−1)n .

37. Taking the logarithm to the base 2 of both sides of the equation, we obtain lg an =

1 (lg an−2 − lg an−1 ).

2

If we let bn = lg an , we obtain bn =

−bn−1 bn−2 + .

2 2

The quadratic equation t2 +

1 t − =0 2 2

has two roots, 12 and −1. Thus there exist constants p and q such that n 1 + q(−1)n . bn = p 2 Now b0 = lg a0 = lg 8 = 3 and a similar calculation shows that b1 = − 23 . Therefore 3 = 3 − = 2

p+q p − q.

2

Solving for p and q, we obtain p = 1 and q = 2. Thus n 1 bn = + 2(−1)n 2 and n

n

an = 2bn = 2[(1/2) +2(−1) ] .

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123

SOLUTIONS 39. Subtracting the given recurrence relation from the recurrence relation for n + 1 cn+1 = 2 +

n X

ci .

i=1

gives cn+1 − cn = cn so cn+1 = 2cn , n ≥ 2.

This last recurrence relation can be solved by iteration to yield cn+1 = 2cn = 22 cn−1 = · · · = 2n−1c2 = 3 · 2n−1 . for n ≥ 2. By inspection, the formula also holds for n = 1. Thus we obtain the formula cn = 3 · 2n−2 , for n ≥ 2.

40. Let S(n, m) = A(n, m) − C(n, m) + 1. [C(n, m) is the number of m-element subsets of an n-element set.]

Then S(n, n) = 1 = S(n, 0). Also S(n − 1, m − 1) + S(n − 1, m)

= =

A(n − 1, m − 1) − C(n − 1, m − 1) + 1 + A(n − 1, m) − C(n − 1, m) + 1

A(n, m) − C(n, m) + 1 = S(n, m).

Since {S(n, m)} and {C(n, m)} satisfy the same recurrence relation and have the same initial conditions, they are equal. Therefore A(n, m) = S(n, m) + C(n, m) − 1 = 2C(n, m) − 1.

42. Show that Un − g(n) satisfies

an = c1 an−1 + c2 an−2 .

44. Assume that g(n) = C1 n + C0 is a solution. We must have C1 n + C0 = 7[C1 (n − 1) + C0 ] − 10[C1(n − 2) + C0 ] + 16n.

The coefficient of n on the left must equal the coefficient of n on the right: C1 = 7C1 − 10C1 + 16.

Thus C1 = 4.

The constant on the left must equal the constant on the right: C0 = −7C1 + 7C0 + 20C1 − 10C0 .

Thus C0 = 13. Therefore g(n) = 4n + 13.

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124

SOLUTIONS The general solution of an = 7an−1 − 10an−2 is b2n + d5n .

Thus the general solution of the original recurrence relation is an = b2n + d5n + 4n + 13.

45. Assume that g(n) = C2 n2 + C1 n + C0 is a solution. We must have C2 n2 + C1 n + C0 = 2[C2 (n − 1)2 + C1 (n − 1) + C0 ] + 8[C2 (n − 2)2 + C1 (n − 2) + C0 ] + 81n2 .

The coefficient of n2 on the left must equal the coefficient of n2 on the right: C2 = 2C2 + 8C2 + 81.

Thus C2 = −9.

The coefficient of n on the left must equal the coefficient of n on the right: C1 = −4C2 + 2C1 − 32C2 + 8C1 .

Thus C1 = −36.

The constant on the left must equal the constant on the right: C0 = 2C2 − 2C1 + 2C0 + 32C2 − 16C1 + 8C0 .

Thus C0 = −38. Therefore

g(n) = −9n2 − 36n − 38.

The general solution of an = 2an−1 + 8an−2 is an = b4n + d(−2)n .

Thus the general solution of the original recurrence relation is an = b4n + d(−2)n − 9n2 − 36n − 38.

3 24 47. an = b4n + dn4n + 25 n + 125

48. an = b(1/3)n + dn(1/3)n + (5/4)n2 − (5/2)n + 25/8 50. We must have C0 C1

= a0 = b = a1 = b + d.

Set b = C0 and d = C1 /r − C0 .

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125

SOLUTIONS 51. By Exercise 53, Section 7.1, we have an ≤ an−1 + 2, n ≥ 4; a4 ≤ 4.

Now an

≤

≤ ≤ 53.

n 1 2 3 4 5 6 7 8 9 10

T (n) 1 3 5 9 13 17 25 33 41 49

an−1 + 2 ≤ an−2 + 2 + 2 < · · ·

an−i + 2i ≤ · · · a4 + 2(n − 4) ≤ 4 + 2(n − 4) = 2n − 4.

Opt Moves for 3-Peg Problem 1 3 7 15 31 63 127 255 511 1023

54. We show only the inductive step. Using the recurrence relation of Exercise 52 and the inductive assumption, we have T (n) = 2T (n − kn ) + 2kn − 1 = 2[(kn−kn + rn−kn − 1)2kn−kn + 1] + 2kn − 1.

First suppose that n − kn < Since

kn X i=1

it follows that

kX n −1 i=1

Therefore,

kn X

i.

i=1

i ≤ n,

i ≤ n − kn .

kn−kn = kn − 1.

Also, rn = n − Now T (n)

= = = =

kn X i=1

i = n − kn −

kX n −1

i = rn−kn .

i=1

2[(kn−kn + rn−kn − 1)2kn−kn + 1] + 2kn − 1 2[(kn − 1 + rn − 1)2kn−1 + 1] + 2kn − 1

(kn − 1 + rn − 1)2kn + 2 + 2kn − 1 (kn + rn − 1)2kn + 1.

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126

SOLUTIONS The case n − kn =

kn X

i

i=1

is treated similarly. (In this case, kn−kn = kn and rn−kn = 0.)

56. We call a stacking in which the disks are arranged from top to bottom in order from smallest to largest a proper stacking. We call a stacking in which the disks in arbitrary order except that the largest disk is on the bottom an arbitrary stacking.

Now consider the given problem at the point when the bottom disk first moves. There must be an empty peg for it to move to; thus, the remaining n − 1 disks must be optimally moved to a third peg in an arbitrary stacking. We first determine the minimum number of moves required to move disks from a proper stacking to another peg in an arbitrary stacking.

Our new problem is, given n disks in a proper stacking, find the minimum number of moves, which we denote sn , to move these n disks to another peg in an arbitrary stacking. Except for the original position, arbitrary stackings are allowed.

Clearly, s1 = 1. Suppose n > 1. Consider the point at which the bottom disk is moved. One peg must be empty (to receive the largest disk), and the n − 1 smaller disks must be moved to a third peg. By definition, this move requires sn−1 moves. Now the largest disk moves (which requires one additional move). Now the n − 1 smaller disks must be placed on top of the largest disk. We can simply peel them off one-by-one and place them on top of the largest disk. This requires n − 1 moves, which is surely optimal. Therefore, sn = sn−1 + 1 + n − 1.

Solving, we obtain sn = 1 + 2 + · · · + n =

n(n + 1) .

2

Now we can answer the original question. Consider the point at which the bottom disk first moves. There must be an empty peg for it to move to; thus, the remaining disks must be moved to a third peg, which requires n(n + 1)/2 moves. Now the largest disk moves (which requires one additional move). Finally, the smallest n − 1 disks must be optimally moved and properly stacked on top of the largest disk. This can be done by reversing the first n(n + 1)/2 moves. [If there were some way to move the smallest n − 1 disks and properly stack them on top of the largest disk in fewer than n(n + 1)/2, we could reverse this technique and obtain a method of moving from a proper stacking to an arbitrary stacking in fewer than n(n + 1)/2 moves.] Thus optimum number of moves is n(n + 1) n(n + 1) +1+ = n(n + 1) + 1.

2 2

Section 7.3 2. At line 2, since i > j (1 > 5) is false, we proceed to line 4 where we set k to 3. At line 5, since key (0 P 0 ) is not equal to s3 (0 J 0 ), we proceed to line 7. At line 7, key < sk (0 P 0 < 0 J 0 ) is false, so at line 10, we set i to 4. We then invoke this algorithm with i = 4, j = 5 to search for key in s4 = 0 M 0 , s5 = 0 X 0 .

At line 2, since i > j (4 > 5) is false, we proceed to line 4 where we set k to 4. At line 5, since key (0 P 0 ) is not equal to s4 (0 M 0 ), we proceed to line 7. At line 7, key < sk (0 P 0 < 0 M 0 ) is false, so at line 10, we set i to 5. We then invoke this algorithm with i = j = 5 to search for key in Copyright c 2018 Pearson Education, Inc.

127

SOLUTIONS s5 = 0 X 0 .

At line 2, since i > j (5 > 5) is false, we proceed to line 4 where we set k to 5. At line 5, since key (0 P 0 ) is not equal to s5 (0 X 0 ), we proceed to line 7. At line 7, key < xk (0 P 0 < 0 X 0 ) is true, so at line 8, we set j to 4. We then invoke this algorithm with i = 5, j = 4 to search for key in an empty list.

At line 2, since i > j (5 > 4) is true, we return 0 to signal an unsuccessful search.

3. At line 2, since i > j (1 > 5) is false, we proceed to line 4 where we set k to 3. At line 5, since key (0 C 0 ) is not equal to s3 (0 J 0 ), we proceed to line 7. At line 7, key < sk (0 C 0 < 0 J 0 ) is true, so at line 8, we set j to 2. We then invoke this algorithm with i = 1, j = 2 to search for key in s1 = 0 C 0 , s2 = 0 G0 .

At line 2, since i > j (1 > 2) is false, we proceed to line 4 where we set k to 1. At line 5, since key (0 C 0 ) is equal to s1 (0 C 0 ), we return 1, the index of key in the sequence s.

5. We give a proof using induction. We omit the Basis Step.

Assume that ai ≤ ai+1 for all i < n. We must prove the inequality for n.

Using (7.3.2) and the inductive assumption, we have

an = 1 + abn/2c ≤ 1 + ab(n+1)/2c = an+1 .

6. We use induction on n. The Basis Step (n = 1) is omitted.

Inductive Step. Suppose that n > 1 and ak = blg kc + 2 for all k < n.

If n is odd,

an

= = = = = =

1 + abn/2c = 1 + a(n−1)/2 1 + blg(n − 1)/2c + 2

3 + blg(n − 1) − 1c 3 + blg(n − 1)c − 1 2 + blg(n − 1)c 2 + blg nc

by the inductive assumption bx − 1c = bxc − 1 if j is odd, blg jc = blg(j − 1)c.

The case when n is even is treated similarly.

8. No. The only way that the algorithm finds key is when the condition key == sk evaluates to true.

9. This algorithm searches for key in the nondecreasing sequence si , . . . , sj , i ≥ 1. If key is found, the algorithm returns an index k such that sk equals key. If key is not found, the algorithm returns 0. bsearch nonrecurs(s, i, j, key) { while (i ≤ j) { k = b(i + j)/2c if (key == sk ) // found return k if (key < sk ) // search left half j =k−1 else // search right half i = k+1 } return 0 // not found } Copyright c 2018 Pearson Education, Inc.

128

SOLUTIONS

11. The idea is to run the algorithm of Exercise 10 twice. Suppose that the values returned are i and j.

Since only one incorrect response is allowed, either i or j is the correct value. If i = j, then i (or j) is the correct value. If i < j, we can compare sj with key. If sj = key, then j is the correct value; otherwise, i is the correct value. Similarly, If j < i, we can compare si with key. If si = key, then i is the correct value; otherwise, j is the correct value.

Running the algorithm of Exercise 10 twice uses at most 2(1 + dlg ne) = 2 + 2dlg ne comparisons. Since one additional comparison may be required, our algorithm uses and most 3 + 2dlg ne comparisons.

12. (a) We use induction on the size n of the sequence to prove that binary search2 is correct when the input is a sequence of size n.

The Basis Step is n = 0. In this case, the sequence is empty, i > j, and the algorithm correctly returns 0 to indicate that key is not found.

Now assume that if a sequence of length less than n is input to binary search2 , the algorithm returns the correct value. Suppose that a sequence of length n > 0 is input to binary search2 . Since i ≤ j, the algorithm proceeds to the line where it computes k. If key is at index k, the algorithm correctly returns k. If key is not at index k, since the sequence is sorted, key, if present, is either in si , . . . , sk−1 or sk+1 , . . . , sj , but not both. The algorithm then executes k1 = binary search2 (s, i, k − 1, key) If key is present in si , . . . , sk−1 , by the inductive assumption, binary search2 returns the index where it is located. If key is not present in si , . . . , sk−1 , by the inductive assumption, binary search2 returns 0. The value returned is stored in k1. The algorithm then executes k2 = binary search2 (s, k + 1, j, key) If key is present in sk+1 , . . . , sj , by the inductive assumption, binary search2 returns the index where it is located. If key is not present in sk+1 , . . . , sj , by the inductive assumption, binary search2 returns 0. The value returned is stored in k2. It follows that if key is present in si , . . . , sk−1 or sk+1 , . . . , sj , it is at index k1 + k2. If key is not present in si , . . . , sj , k1 + k2 = 0. Since the algorithm returns k1 + k2, it follows that the algorithm is correct. (b) We define the worst-case time required by the algorithm to be the number of times the algorithm is invoked in the worst case for a sequence containing n items. Let let an denote the worst-cast time.

Suppose that n is 0, that is, i > j. In this case, there is one invocation; so a0 = 1.

Now suppose that n > 1. In the worst case, the item will not be found at the line if (key == sk ) so the algorithm will be invoked twice more: k1 = binary search2 (s, i, k − 1, key) k2 = binary search2 (s, k + 1, j, key) By definition, the first invocation will require a total of ab(n−1)/2c invocations, and the second invocation will require a total of abn/2c invocations. Thus we obtain the recurrence relation an = 1 + ab(n−1)/2c + abn/2c.

We use strong induction to show that an ≤ 3n + 1 for all n ≥ 0, thus proving that an = O(n). The Base Case, n = 0, is readily verified. Now assume that n > 0. By the inductive assumption, ab(n−1)/2c ≤ 3b(n − 1)/2c + 1

and

abn/2c ≤ 3bn/2c + 1.

Copyright c 2018 Pearson Education, Inc.

129

SOLUTIONS Now an

= ≤ = =

1 + ab(n−1)/2c + abn/2c 1 + 3b(n − 1)/2c + 1 + 3bn/2c + 1 3 + 3(b(n − 1)/2c + bn/2c) 3 + 3(n − 1) = 3n < 3n + 1.

We conclude by using strong induction to show that an ≥ n for all n ≥ 0, thus proving that an = Ω(n) and, therefore, an = Θ(n). The Base Case, n = 0, is readily verified. Now assume that n > 0. By the inductive assumption, ab(n−1)/2c ≥ b(n − 1)/2c

and

abn/2c ≥ bn/2c.

Now an

= ≥ =

1 + ab(n−1)/2c + abn/2c 1 + b(n − 1)/2c + bn/2c 1 + (n − 1) = n.

14. The algorithm is correct. The worst-case time is Θ(n).

15. The algorithm is not correct. The input s1 = 10 and key = 8 results in infinite recursion.

17. The algorithm is not correct. If i = j, the index k is out of bounds.

19.

— 1 — — 9 — — 7 — — 3 — Merge one-element arrays

— 1

1

9 — — 3

3

7 — Merge two-element arrays

9

7

22. We give a recursive description. We assume that the input consists of the integers from 1 to n. Let m = b(n + 1)/2c.

Set

s1 = 1, s2 = 3, s3 = 5, . . . , sm = 2m − 1 and sm+1 = 2, sm+2 = 4, . . . , sn = 2bn/2c.

Now arrange s1 , . . . , sm to produce worst-case behavior for merge sort and arrange sm+1 , . . . , sn to produce worst-case behavior for merge sort.

Copyright c 2018 Pearson Education, Inc.

130

SOLUTIONS

23. Seven, which occurs for an already sorted array.

25. Basis Step. a1 = 0 < 2 = 2a1 + 2 = a2 .

Inductive Step. Assume that the inequality holds for k < n. Now an+1

= ≥

ab(n+1)/2c + ab(n+2)/2c + n abn/2c + ab(n+1)/2c + n − 1 = an .

26. The last inequality in the proof of Theorem 7.3.10 gives an ≤ 2n lg n + 2n + 1 for all n. If 3 ≤ lg n (or, equivalently, if 8 ≤ n), 2n + 1 ≤ 3n ≤ n lg n. Therefore if 8 ≤ n, an ≤ 2n lg n + 2n + 1 ≤ 3n lg n.

The cases 1 ≤ n < 7 can be checked directly.

27. Sequences of length m and n, where m ≤ n, require, at the minimum, m − 1 comparisons for merging.

Thus if bn denotes the least number of comparisons used by merge sort, bn satisfies jnk bn = bbn/2c + bb(n+1)/2c + − 1.

2 This recurrence relation can be estimated in the same way as the one for the worst-case time; therefore, the best-case time is Θ(n lg n).

29. 3, 2; 3, 1; 2, 1; 5, 4; 5, 1; 4, 1

30. There are none.

32. n(n − 1)/2

33. Swapping an adjacent pair removes exactly one inversion. In the worst case, by Exercise 32 there are n(n − 1)/2 inversions. Thus the worst case time is at least n(n − 1)/2 = Ω(n2 ).

36. See the solution to Exercise 37.

38. It is not necessary to check every pair to detect an inversion. For example, if a < b < c an algorithm can compare a and b to determine that a < b, and compare b and c to determine that b < c. Logic can then used by the algorithm to determine that a < c.

44. If n = 1, an = a; thus, a is returned. For n > 1, if n is even, m = bn/2c = n/2 and an = am am . If n is odd, m = (n − 1)/2 and an = am am a.

45. If n is odd, m = (n − 1)/2. If n > 1, after am is computed in line 5, line 6 is executed to compute am am and line 10 is then executed to compute am am a. Thus bn = b(n−1)/2 + 2. If n is even, m = n/2 and line 6 is executed to compute am am . In this case, line 10 is not executed, so bn = bn/2 + 1.

46. b1 = 0, b2 = 1, b3 = 2, b4 = 2 47. Assume that n = 2k . Now bn = b2k−1 + 1 = b2k−2 + 2 = · · · = b1 + k = 0 + k = lg n.

48. b7 = 4, b8 = 3 49. We use induction on n to prove that lg n ≤ bn ≤ 2 lg n, from which we deduce bn = Θ(lg n). The Basis Step (n = 1) is omitted.

Assume that n > 1. If n is even, we have bn = bn/2 + 1 ≤ 1 + 2 lg

n = 1 + 2[(lg n) − 1] ≤ 2 lg n 2

Copyright c 2018 Pearson Education, Inc.

131

SOLUTIONS and bn = bn/2 + 1 ≥ 1 + lg

n = 1 + (lg n) − 1 = lg n.

2

If n is odd, we have bn = b(n−1)/2 + 2 ≤ 2 + 2 lg and bn = b(n−1)/2 + 2 ≥ 2 + lg

n−1 = 2 + [2 lg(n − 1)] − 2 ≤ 2 lg n 2

n−1 = 2 + [lg(n − 1)] − 1 = 1 + lg(n − 1) ≥ lg n.

2

The last inequality is equivalent to 1 ≥ lg or 2≥ which is easily seen to be true for n > 1.

n n−1

n n−1

50. If i = j, there is only one element in the array, which is both largest and smallest. In this case, the algorithm simply returns these values.

If i < j, the algorithm divides the array into two nearly equal parts at line 7. At lines 8 and 9, the algorithm recursively finds that largest and smallest elements in each of the parts. The overall largest element is the larger of the largest in each of the parts (computed at line 11 or 13), and the overall smallest element is the smaller of the smallest in each of the parts (computed at line 15 or 17).

51. For input of size 1, i = j; no comparisons are made since the algorithm returns at line 5. Thus b1 = 0.

For input of size 2, no comparisons are made during the recursive calls at lines 8 and 9 (since each involves input of size 1). There is one comparison at line 10 and one at line 14. Thus b2 = 2.

52. 4 53. At lines 8 and 9, abn/2c + ab(n+1)/2c comparisons are made. At lines 10 and 14, two comparisons are made.

54. If n = 2k , the recurrence relation becomes b2k = 2b2k−1 + 2. Now bn

= b2k = 2b2k−1 + 2 = 2[2b2k−2 + 2] + 2 = 22 b2k−2 + 22 + 2 = · · ·

= 2k b20 + 2k + 2k−1 + · · · + 2 = 2k + 2k−1 + · · · + 2 = 2k+1 − 2 = 2 · 2k − 2 = 2n − 2 55. We give only the Inductive Step.

Assume that bk = 2k − 2, for k < n. The inductive assumption gives bbn/2c

bb(n+1)/2c

= =

2bn/2c − 2

2b(n + 1)/2c − 2.

If n is even, bn/2c = n/2 = b(n + 1)/2c, so bn = 2[2(n/2) − 2] + 2 = 2n − 2.

If n is odd, bn/2c = (n − 1)/2 and b(n + 1)/2c = (n + 1)/2, so bn = [2(n − 1)/2 − 2] + [2(n + 1)/2 − 2] + 2 = 2n − 2.

Copyright c 2018 Pearson Education, Inc.

132

SOLUTIONS

61. If n = 1, there is nothing to sort so the algorithm simply returns. If n > 1, the elements s1 , . . . , sn−1 are sorted as a result of the recursive call. To sort the entire array, the nth element sn , stored in the variable temp, is compared to each of the preceding elements which are moved up one position successively as long as they are greater than temp. The index i runs down the list. As soon as an element less than or equal to temp is found (pointed to by i), the loop is exited and temp is stored at position i + 1. If there is no element less than or equal to temp, i is 0 and temp is stored at position 1.

62. The worst-case behavior occurs when the items are in reverse order.

63. b1 = 0, b2 = 1, b3 = 3

64. bn = bn−1 + n − 1

66. bn = 1 + blg nc + bbn/2c , b1 = 1, b2 = 3, b3 = 3

n(n − 1) 2 67. bn = (1 + lg n)(2 + lg n)/2 65. bn =

68. An arbitrary value of n falls between two powers of 2, say 2k−1 < n ≤ 2k .

This inequality implies that k − 1 < lg n ≤ k. Since the sequence b is nondecreasing, b2k−1 ≤ bn ≤ b2k .

Now

(1 + k)(2 + k) (2 + lg n)(3 + lg n) ≤ = O((lg n)2 ).

2 2 Similarly, bn = Ω((lg n)2 ). Therefore bn = Θ((lg n)2 ). bn ≤ b2 k =

71. b1 = 1, b2 = 2, b3 = 3,

72. bn = bn−1 + 1

76. Similar to the proof for Exercise 68

73. bn = n

74. Θ(n)

78. bn = bb(1+n)/2c + bbn/2c + n

79. Let ck = b2k . Then ck = 2ck−1 + 3. If n = 2k , bn

= = =

ck = 2ck−1 + 3 = 2(2ck−2 + 3) + 3 = · · · 2k c0 + 3(2k−1 + 2k−2 + · · · + 1) = 2k · 0 + 3(2k − 1) 3(n − 1).

81. Use the method of Exercise 79.

82. bn = n(1 + lg n)

84. Use Exercise 81 to show that if n is a power of 2, bn = n lg n. Now let n be arbitrary. Choose k so that 2k < n ≤ 2k+1 . By Exercise 83, bn ≤ b2k +1 . Now b2k+1 = 2k+1 (k + 1) ≤ 2k+1 (k + k) = 4(2k k) ≤ 4n lg n.

86. an ≤ abn/2c + ab(n+1)/2c + 2 + lg(bn/2cb(n + 1)/2c). If n is even, bn/2cb(n + 1)/2c = n2 /4. If n is odd, bn/2cb(n + 1)/2c = [(n − 1)/2][(n + 1)/2] = (n2 − 1)/4 ≤ n2 /4. Thus we can write an ≤ abn/2c + ab(n+1)/2c + 2 + lg

n2 = abn/2c + ab(n+1)/2c + 2 lg n.

4

87. Let n = 2k and use induction on k.

Basis Step. If k = 0, b1 = 0 = 4 · 1 − 2 lg 1 − 4 Inductive Step. b2k+1

= =

2b2k + 2(k + 1) = 2[4 · 2k − 2k − 4] + 2(k + 1) 4 · 2k+1 − 2(k + 1) − 4

89. Inductive Step. bn = bbn/2c + bb(n+1)/2c + 2 lg n ≤ bb(n+1)/2c + bb(n+2)/2c + 2 lg(n + 1) = bn+1 90. Choose k with 2k < n ≤ 2k+1 . Then

an ≤ bn ≤ b2k+1 = 4 · 2k+1 − 2(k + 1) − 4 ≤ 8 · 2k ≤ 8n.

Copyright c 2018 Pearson Education, Inc.

133

SOLUTIONS

Section 7.4 2. Some pair is equal.

3. {(1, 1), (1, 1), (1, 1), (1, 1)}

5. Modify Algorithm 7.4.2 as follows.

Add the parameters x1 , y1 , x2 , y2 in which the x- and y-coordinates of a closest pair will be returned.

Add the lines x1 = x-coordinate of first member of closest pair y1 = y-coordinate of first member of closest pair x2 = x-coordinate of second member of closest pair y2 = y-coordinate of second member of closest pair after the line directly find the distance δ between a closest pair Replace the lines δL = rec cl pair(p, i, k) δR = rec cl pair(p, k + 1, j) by δL = rec cl pair(p, i, k, x1L, y1L, x2L, y2L) δR = rec cl pair(p, k + 1, j, x1R, y1R , x2R, y2R) if (δL < δR ) { x1 = x1L y1 = y1L x2 = x2L y2 = y2L δ = δL }

else { x1 = x1R y1 = y1R x2 = x2R y2 = y2R δ = δR }

Replace the line

δ = min{δ, dist(vk , vs )} by if (dist(vk , vs ) < δ) { δ = dist(vk , vs ) x1 = vk .x y1 = vk .y x2 = vs .x y2 = vs .y } Copyright c 2018 Pearson Education, Inc.

134

SOLUTIONS

6.

Input: Output:

x1 , . . . , xn , n the distance between a closest pair

one dim cl pair (x, n) { sort x1 , . . . , xn dist = ∞ for i = 2 to n dist = min{dist , |xi − xi−1 |} return dist } 8.

q q

q

q

q

l

q

9. No. There could be points q1 and q2 with q2 closer to p than q1 , yet the y-coordinate of q2 exceeds the y-coordinate of q1 .

11. find all(p, n) { δ = closest pair (p, n) // original function if (δ > 0) { sort p1 , . . . , pn by x-coordinate rec find all (p, 1, n, δ) } } rec find all (p, i, j, δ) { if (j − i < 3) { sort pi , . . . , pj by y-coordinate directly find and output all pairs within δ of each other return } k = b(i + j)/2c l = pk .x rec find all (p, i, k, δ) rec find all (p, k + 1, j, δ) merge pi , . . . , pk and pk+1 , . . . , pj by y-coordinate t=0 Copyright c 2018 Pearson Education, Inc.

135

SOLUTIONS for k = i to j if (pk .x > l − δ) ∧ pk .x < l + δ) { t = t+1 vt = pk } for k = 1 to t − 1 for s = k + 1 to min{t, k + 7} if (dist(vk , vs ) = δ) println(vk , vs ) } 12. find all 2 δ(p, n) { δ = closest pair (p, n) // original function if (δ > 0) { sort p1 , . . . , pn by x-coordinate rec find all 2delta(p, 1, n, δ) } } rec find all 2 δ(p, i, j, δ) if (j − i < 3) { sort pi , . . . , pj by y-coordinate directly find and output all pairs less than 2δ apart return } k = b(i + j)/2c l = pk .x rec find all 2 δ(p, i, k, δ) rec find all 2 δ(p, k + 1, j, δ) merge pi , . . . , pk and pk+1 , . . . , pj by y-coordinate t=0

for k = i to j if (pk .x > l − 2δ ∧ pk .x < l + 2δ) { t = t+1 vt = pk } for k = 1 to t − 1 for s = k + 1 to min{t, k + 31} if (dist(vk , vs ) < 2δ) println(vk , vs ) } 14. Consider points pi = (i, (i − 1)100) for i = 1 to 31

and

p32 = (1, 32).

The closest pair is p1 , p32 and δ = 32. After the closest-pair algorithm is called, the points are sorted by y-coordinate: p1 , p2 , . . . , p32. The only pair less than 2δ is p1 , p32, which is not found by algorithm exercise14 .

15. Replace the code that stores all points in the vertical strip for the initial value of δ with code that finds the first eight points in the strip (or less if there are less than eight points in the strip). After comparing Copyright c 2018 Pearson Education, Inc.

136

SOLUTIONS each point to the next seven points in the strip, store an additional point in the strip (if any) using the updated value of δ.

17. We show that for each point p, there are at most 64 distinct points whose distance to p is less than 2δ.

It will then follow that the number of pairs is less than or equal to 64n.

Consider a 4δ × 4δ square centered at p: r

p

Any point whose distance to p is less than 2δ is contained in the square. Since each of the δ × 2δ subrectangles contains at most eight points (see Figure 7.4.3), the 4δ × 4δ square contains at most 8 · 8 = 64 points.

19. Basis Steps (n = 2, 3). t2 = a2 ≤ max{a2 , a3 } = t3 .

t4 = t2 + t2 + (c + a3 )4 = 2a2 + 4a3 + 4c. Therefore t4 ≥ a2 and t4 ≥ a3 . It follows that t4 ≥ max{a2 , a3 } = t3 .

Inductive Step. Assume that n > 3. Then tn+1

= tb(n+1)/2c + tb(n+2)/2c + (c + a3 )(n + 1) ≥ tbn/2c + tb(n+1)/2c + (c + a3 )n = tn

20. See the proof of Theorem 7.3.10.

Section 8.1 2. The following simple, directed graph models the tournament: r Y

Snow

r

Skyscrapers

-? r

jr

Pheasants

Tuna

3. The following undirected graph models the tournament: r

Snow

r

Skyscrapers

r

Pheasants

r

Tuna

6. There are vertices of odd degree {b, d}.

7. There are vertices of odd degree {b, d}.

9. (a, b, c, e, b, d, e, f, c, g, h, i, f, h, e, g, d, a) 10. (a, c, f, e, c, b, e, d, b, a) Copyright c 2018 Pearson Education, Inc.

137

SOLUTIONS

12. V = {v1 , v2 , v3 }. E is empty. There are no parallel edges and no loops. G is simple. e1 does not exist.

All vertices are isolated.

13. V = {v1 , v2 , v3 , v4 , v5 }. E = {e1 , e2 , e3 , e4 , e5 , e6 , e7 , e8 }. There are no parallel edges, no loops, and no isolated vertices. G is simple. e1 is incident on v2 and v4 .

15. V1 = {v1 , v2 , v3 , v4 }, V2 = {v5 , v6 , v7 }.

16. n(n − 1)/2 v1 rP

PPrv5

v2 rP PPrv6 v3 rP PPrv7 v4 r

18. The graph is bipartite. V1 = {v1 , v3 , v4 , v6 , v8 , v9 , v10 }, V2 = {v2 , v5 , v7 }.

19. The graph is bipartite. V1 = {Gre, Buf , Sho, Dou, Mud}, V2 = {She, Wor , Cas, Gil, Lan}.

21. Not bipartite

22. Not bipartite

25. mn

26. Example 8.1.12 is bipartite, but Examples 8.1.13 and 8.1.14 are not bipartite. K1 is not bipartite because there is no way to partition the vertices into two nonempty subsets.

28. (a, c, d, e, b)

29. (c, a, b, e, d)

30. The vertices are mathematicians, and an edge connects two mathematicians if they co-authored a paper.

The Erdős number of mathematician m is the length of a shortest path from m to Erdős.

31. Yes 33. One class.

1

4

r @ r

@3r @

r2

@r 5

34. No. If similarity is defined solely by the dissimilarity function, in the graph of Figure 8.1.8, v1 is similar to v3 and v3 is similar to v5 , but v1 is not similar to v5 .

38.

10

10

11 11

00 00

010 01

01

000

110

111

011 100

101

001

41. n2n−1 (There are 2n vertices and each is incident on n edges. Since n2n counts each edge twice, the formula follows.)

42. n!2n (The result can be proved by induction on n.)

Copyright c 2018 Pearson Education, Inc.

138

SOLUTIONS

44.

01

11

r

r

r

r 00 r 10

10 r

r

11

r @ @r 00

01

r

10 r

10

r

11

r @ @r 11

r @ @r 00

01

r

r

01

r 00

10 r

10

r

11

r @ @r 11

r @ @r 00

01

r

r

01

r 00

010 r

110

r

011

r @ @r

111

47. 5 ar @ b @r j r rc @ g k ir r @r rd @ @ @ @r @r @r e h f - a= x+y P 51. x = 1 PP C PP PP C PP C q P - x= a+b+c y = 2 C- b = y + z C C

C CW - c = c+1 z = 3 - c= x+z 48.

52. x = 1

- w = x+5

y=2

- z =y+2

@ R x =z +w @

Section 8.2 2. Simple path–no; cycle–no; simple cycle–no 3. Simple path–yes; cycle–no; simple cycle–no 5. Simple path–no; cycle–yes; simple cycle–no 6. Simple path–no; cycle–no; simple cycle–no 8. Simple path–yes; cycle–no; simple cycle–no 9. Simple path–yes; cycle–no; simple cycle–no 11. r r r r

12. There is no such graph since the number of vertices of odd degree is always even.

14.

r @ @r

ir

r Copyright c 2018 Pearson Education, Inc.

r @ @r 000

001

r

r

101

r 100

139

SOLUTIONS 15. No such graph exists. One-half the sum of the degrees (= number of edges) is 5, not 4.

17. No such graph exists. Suppose, by way of contradiction, that there is such a graph with vertices a and b of degree 2 and c, d, and e of degree 4. Since c is of degree 4, it is incident on a, b, d, and e. Similarly, d is incident on a, b, c, and e, and e is incident on a, b, c, and d. But now a has degree at least 3.

Contradiction. r r

B A B A B A

r Ar B H

H B H

r HB

18.

20. (a, b, c, d, e), (a, b, c, d, f, e), (a, b, c, g, f, e), (a, b, c, g, f, d, e), (a, b, g, f, e), (a, b, g, c, d, e), (a, b, g, f, d, e), (a, b, g, c, d, f, e) r

21. r

r @ @r

r @ @r

r

r

r

r

r @

r

r

@r

r

r @

The second is a simple path. Neither is a cycle or a simple cycle.

r

@r

r @ r

@r

23. δ(v1 ) = 2, δ(v2 ) = 2, δ(v3 ) = 3, δ(v4 ) = 6, δ(v5 ) = 2, δ(v6 ) = 3, δ(v7 ) = 4, δ(v8 ) = 4, δ(v9 ) = 4, δ(v10) = 2 25. The following are the subgraphs with no edges: G1 = ({v1 }, ∅), G2 = ({v2 }, ∅), G3 = ({v3 }, ∅), G4 = ({v1 , v2 }, ∅), G5 = ({v1 , v3 }, ∅), G6 = ({v2 , v3 }, ∅), G7 = ({v1 , v2 , v3 }, ∅). Besides G itself, the other subgraphs are

v1 r v1 r v1 r v1 r

A

A A e2 e1 A e2 e1 A A

v2 r v2 r r v3 Ar v3 v2 r Ar v3 G11 G9 G10 G8 26. There are six subgraphs, the following five subgraphs and the original graph itself. e1 e2 '$ '$ r r

v1 v2

r

v1

r

v2

r

v1

r

v2

r

v1

r

v2

29. There is no Euler cycle since there are vertices of odd degree.

30. (v1 , v5 , v2 , v4 , v5 , v3 , v2 , v1 , v4 , v3 , v1 ) 32. (a, b, d, c, b, f, g, j, f, e, j, h, c, i, d, e, i, h, a) 33. (a, b, c, b, d, e, h, f, i, j, k, i, h, g, f, e, g, d, c, a) 35. When n is odd 36. Both m and n must be even.

38. When n is even 40. There are zero vertices of even degree, and zero is even.

Copyright c 2018 Pearson Education, Inc.

140

SOLUTIONS

41. Suppose that each team plays exactly nine other conference teams. Model the situation as a graph in which the vertices represent the teams and an edge between vertices (teams) represents a game played by these teams. Then every vertex has degree 9. But the number of vertices (11) of odd degree is even.

Contradiction.

42. (d, a, b, d, e, b, c, e, h, g, d, f, g, j, h, i, e) 44. (b, c, d, g, b, a, f, g, i, f); (c, h, e) 45. Let G be a connected graph with n vertices v1 , . . . , vn of odd degree. There are paths with no repeated edges from v1 to v2 , v3 to v4 , and so on, such that every edge in G is in exactly one of the paths.

47. False. Consider the cycle (a, b, c, a) for the graph rb

a r @ @r c

rd

49. The graph of Exercise 25 50. No. Let the vertices of a graph be the squares of the chessboard. Insert an edge between two vertices if a knight can make a move between the corresponding squares. The degrees of the vertices that correspond to the border squares next to the corner squares have degree 3. Since there are eight of these, there is no Euler cycle.

52. Let H be one of the connected subgraphs in the partition. Let v be a vertex in H. Let C be the component to which v belongs. We show that H = C. Let w be a vertex in H. Since H is connected, there is a path from v to w in H. Therefore w is in C. Let w be a vertex in C. Then there is a path from v to w (v0 , v1 , . . . , vn ), with v0 = v, vn = w, in C. The edge (v0 , v1 ) must belong to H since vertex v0 is in H. Thus vertex v1 is in H. Continuing in this way, we see that w is in H. Therefore the vertex sets of H and C are equal.

Similarly, the edges sets of H and C are equal. Therefore the subgraphs of the partition are components.

53. There is a path P from v to w. Change the orientation of each edge in P .

55. Let G be a simple, bipartite graph having the maximum number of edges with disjoint vertex sets having k and n − k vertices. Then G has k(n − k) edges. The maximum of the integer-valued function f(k) = k(n − k) occurs when k = n/2, if n is even, and when k = (n − 1)/2 or k = (n + 1)/2, if n is odd.

Thus the maximum number of edges is bn2 /4c = n2 /4, if n is even, and bn2 /4c = [(n − 1)/2][(n + 1)/2], if n is odd.

57. K6 61. If

(1) (1)

(1)

(2) (2)

(2)

(b1 b2 . . . bn−1 , b1 b2 . . . bn−1 , . . .) is a directed Euler cycle in G, (1) (1)

(1)

(2)

(3)

b1 b2 . . . bn−1 bn−1 bn−1 . . . is a de Bruijn sequence.

Copyright c 2018 Pearson Education, Inc.

141

SOLUTIONS 63. We first show that if G is a connected bipartite graph, then every closed path in G has even length.

Suppose that the disjoint vertex sets are V1 and V2 . Let P = (v0 , v1 , . . . , vn )

be a closed path from v0 to vn . Suppose that v0 ∈ V1 . Then v1 ∈ V2 , v2 ∈ V1 , . . .. Notice that if i is odd, vi ∈ V2 , and if i is even, vi ∈ V1 . Since vn ∈ V1 , it follows that n is even. Thus P has even length.

We conclude by showing that if every closed path in a connected graph G has even length, then G is bipartite.

Choose a vertex v in G. Let V1 denote the set of vertices w in G that are reachable from v on a path of even length. Let V2 denote the set of vertices w in G that are reachable from v on a path of odd length.

Notice that since G is connected, every vertex in G is either in V1 or V2 . We claim that V1 and V2 are disjoint. To show this, we argue by contradiction. Suppose that some vertex w belongs to both V1 and V2 . Then there is a path Po0 of odd length and a path Pe of even length from v to w. Let Po be the path from w to v obtained by reversing the order of the vertices. Then Pe followed by Po is a closed path of odd length from v to v. This contradiction shows that V1 and V2 are disjoint. Now let e be an edge incident on vertices x and y. Suppose that x belongs to V1 . Then there is a path P of even length from v to x. Now P followed by y is a path of odd length from v to y. Thus y is in V2 . Therefore G is bipartite.

64. n(n − 1)k [Choose (v0 , v1 , . . . , vk ) with vi−1 6= vi .]

66. (a) pm = (n − 1)m−1 − pm−1 . The first term counts the number of paths of length m − 1 that start with v, and the last term counts the number of paths of length m − 1 that start with v and end with w. A path of length m − 1 that starts with v and does not end with w can be extended to a path of length m that starts with v and ends with w. (b) pm

= = .. . = = = =

(n − 1)m−1 − [(n − 1)m−2 − pm−2 ] (n − 1)m−1 − (n − 1)m−2 + pm−2

(n − 1)m−1 − (n − 1)m−2 + · · · + (−1)m (n − 1) + (−1)m+1 p1

(n − 1)m−1 − (n − 1)m−2 + · · · + (−1)m (n − 1) + (−1)m+1 −(n − 1)m − (−1)m+1 −(n − 1) − 1 (n − 1)m + (−1)m+1 n

67. There is one path of length 1, (v, w), from v to w.

There are n − 2 paths of length 2, (v, x1 , w), from v to w since vertex x1 can be chosen in n − 2 ways.

(Vertex x1 must be different from v and w.)

There are (n − 2)(n − 3) paths of length 3, (v, x1 , x2, w), from v to w since vertex x1 can be chosen in n − 2 ways, and vertex x2 can be chosen in n − 3 ways. (Vertex x1 must be different from v and w, and vertex x2 must be different from v, w, and x1 .)

In general, there are (n − 2)(n − 3) · · · (n − k) paths of length k, (v, x1 , . . . , xk−1, w), from v to w since vertex x1 can be chosen in n − 2 ways, vertex x2 can be chosen in n − 3 ways, and so on. The result now follows.

Copyright c 2018 Pearson Education, Inc.

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SOLUTIONS

68. The number of simple paths of length 0 is n; the number of simple paths of length 1 is n(n − 1); the number of simple paths of length 2 is n(n − 1)(n − 2); and so on. Thus the number of simple paths is n + n(n − 1) + n(n − 1)(n − 2) + · · · + n(n − 1) · · · 1 = n!

n−1 X k=0

1 . k!

Now n!

n−1 X 1 k=0

=

k!

= Since, for n ≥ 2, n!

∞ X

k=n+1

1 k!

n! e −

∞ X 1

k=n

n!e − 1 − n!

k!

!

∞ X

k=n+1

1 . k!

=

1 1 + +··· n + 1 (n + 1)(n + 2)

<

1 1 1 1 + 2 +··· = n 1 = ≤ 1, n n n−1 1− n

the result follows. The result is true by inspection for n = 1.

70. A connected graph with one vertex, consists of the vertex, say v, and none or more loops incident on v.

An Euler cycle consists of a cycle that traverses each loop once.

A connected graph with two vertices, say v and w, each of which has even degree, consists of 2k edges incident on v and w, k ≥ 1, none or more loops incident on v, and none or more loops incident on w.

An Euler cycle consists of a path that begins at v, traverses all of the loops incident on v, traverses one edge from v to w, traverses all of the loops incident on w, traverses one edge from w to v, and traverses all remaining edges incident on v and w. This path will end at v since the number of edges incident on v and w is even.

72. diameter = n. The diameter is the maximum time for two processors to communicate.

73. 1, since dist(v, w) = 1 for every pair of distinct vertices in Kn 75. First we show that, if n mod 4 = 1, k≥

n−1 .

2

Suppose that n mod 4 = 1. In particular, n is odd. Since every vertex has degree k, k must be even. (If k is odd, we obtain a contradiction to the theorem that states that the number of vertices of odd degree is even). We show that (n − 3)/2 is an odd integer and, consequently, k≥

n−3 n−1 +1= .

2 2

Since n mod 4 = 1, we may write n = 4q + 1. Thus n − 3 = 4q − 2 and (n − 3)/2 = 2q − 1, which is an odd integer. Therefore n−1 k≥ , 2 regardless of the value of n mod 4.

Now suppose that G is not connected. Let C1 and C2 be components. Since every vertex has degree k, C1 and C2 each have at least k + 1 vertices. Thus G has at least 2(k + 1) ≥ n + 1 vertices, which is a contradiction.

Copyright c 2018 Pearson Education, Inc.

143

SOLUTIONS 77. We prove the result by induction on n. We omit the Basis Step (n = 1).

Assume that the result is true for n. Let G be an (n + 1)-vertex dag with the maximum number of edges.

By Exercise 75, G has a vertex v with no out edges. In fact, there must be edges of the form (w, v) for all w 6= v; otherwise, G would not have the maximum number of edges. This accounts for n edges.

Let G0 be the graph obtained from G by eliminated v and the n edges incident on v. G0 is an n-vertex dag and since G has the maximum number of vertices, G0 must also have the maximum number of vertices.

By the inductive assumption, G0 has n(n − 1)/2 vertices. Thus G has n(n − 1) (n + 1)n +n= 2 2 vertices.

78. Let I(Pn ) denote the number of independent sets in Pn . Note that I(P1 ) = 2 and I(P2 ) = 3. Now suppose that n > 2. Let v be a vertex of degree 1 in Pn . An independent set Pn that contains v consists of v and an independent set of Pn−2, and there are I(Pn−2 ) such independent sets. An independent set of Pn that does not contain v is an independent set of Pn−1 , and there are I(Pn−1 ) such independent sets. Therefore I(Pn ) = I(Pn−1 ) + I(Pn−2 ).

Since {I(Pn )} satisfies the same initial conditions and recurrence relation as {fn+2 }, I(Pn ) = fn+2 for all n.

79. (a) Let v and w be nonadjacent vertices in G. Let {x1 , . . . , xk } denote the vertices adjacent to v. Then the mapping N (xi ) = yi , where yi is adjacent to xi and w, is a bijection to the set of vertices adjacent to w. Therefore δ(v) = δ(w). (b) Let V1 denote the set of vertices of degree k. Suppose that V1 is nonempty. By part (a), every vertex in V1 is adjacent to every vertex in V1 . Since no vertex is adjacent to all other vertices, |V1 | ≤ 2 and |V1 | ≤ 2. Let v1 and w1 be distinct vertices in V1 , and let v2 and w2 be distinct vertices in |V1 |.

Now v1 adjacent to v2 and w2 and w1 adjacent to v2 and w2 , which is a contradiction since there are two vertices (v2 and w2 ) adjacent to both v1 and w1 .

Section 8.3 2. (a, b, c, d, e, f, n, p, m, l, k, j, o, i, h, g, a) 4. We would have to eliminate one edge at f, three edges at c, one edge at b, one edge at i, three edges at j, and three edges at m, leaving 15 edges. Since there are 16 vertices, a Hamiltonian cycle would have 16 edges.

5. Suppose that the graph has a Hamiltonian cycle. Since each vertex in a cycle has degree 2, we would have to include the edges (a, b), (a, f), (f, g), (b, c), (c, d), (d, e), (e, h), and (g, h). Since these edges already form a cycle, there is no Hamiltonian cycle.

7. (a, b, c, g, l, m, r, q, p, k, j, f, e, i, n, o, t, s, h, d, a) 8. There is no Hamiltonian cycle. We would have to eliminate two edges at c, three edges at e, and one edge at f, leaving six edges. Since there are seven vertices, a Hamiltonian cycle would have seven edges.

10. K3 Copyright c 2018 Pearson Education, Inc.

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SOLUTIONS

11. r r r

r r @ @r r

13. We begin the Hamiltonian cycle at row 1, column 1 and proceed along the first row to column m − 1.

Then we go down to row 2 and move back along row 2 to column 1. Then we go down to row 3 and along row 3 to column m − 1. Then we go down to row 4 and back to column 1 along row 4. We continue this serpentine path until we arrive at the last row. If n is odd, we finish in column 1. We then take the edge from row n, column 1 to row n, column m and proceed up column m to row 1. We then follow the edge from row 1, column m to row 1, column 1 finishing the cycle. If n is odd, we ended our path at row n, column m − 1. We then move along row n to column m, up column m to row 1, and along the edge from row 1, column m to row 1, column 1 to finish the cycle.

14. Choose any vertex v to start. After arriving at a vertex, move to a not-yet-visited vertex (except when returning to v for the nth and last move). Since the degree of every vertex is n − 1 and there are n moves, such moves are always possible.

16. The five edges of smallest weight have weights 3, 4, 4, 5, 5. Thus the shortest Hamiltonian cycle has a weight of at least 21. However, three of these edges (those with weights 3, 4, 4) are incident on vertex c.

Thus the edges of weight 3, 4, 4 cannot all be in a Hamiltonian cycle. If we replace an edge of weight 4 with an edge of minimum replacement weight 6, we can conclude that the shortest Hamiltonian cycle has weight at least 3 + 6 + 4 + 5 + 5 = 23.

Since the given Hamiltonian cycle has weight 23, we conclude that it is minimal.

17. (e, d, a, b, c, e) 19.

G1 : GR 1: G01 : G001 : G2 : GR 2: G02 : G002 : G3 : GR 3: G03 : G003 : G4 :

01 10 00 01 11 10 00 01 11 10 10 11 01 00 000 001 011 010 110 111 101 100 000 001 011 010 110 111 101 100 100 101 111 110 010 011 001 000 0000 0001 0011 0010 0110 0111 0101 0100 1100 1101 1111 1110 1010 1011 1001 1000 0000 0001 0011 0010 0110 0111 0101 0100 1100 1101 1111 1110 1010 1011 1001 1000

20. Let C be a Hamiltonian cycle in G. Consider a traversal of C. When we traverse an edge from a vertex v1 in V1 to a vertex v2 in V2 , this uniquely associates one vertex v2 with v1 . Since C traverses all vertices |V1 | = |V2 |.

22. Let each vertex of a graph represent a permutation. Put an edge between two vertices p and q if and only if pi 6= qi for all i = 1, 2, . . . , n.

23. (n = 1) 1 (n = 2) 12 21 Copyright c 2018 Pearson Education, Inc.

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SOLUTIONS

(n = 3) Consider a graph with six vertices representing the permutations and with an edge between two vertices if the permutations differ in each coordinate. A solution to the problem is a Hamiltonian path in the graph. There is no solution for n = 3 because the graph is not connected. (n = 4) 1234 3412 4321 1432 2341 4132 3241 4123 3214 1423 2314 1243 2134 3421 4312 2431 1342 4231 3142 4213 3124 2413 1324 2143 26. If Algorithm 8.3.10 starts with a and then f, the next choice must be e. From e, the next choice will be either c or d. We consider the two possibilities.

Possibility 1: If the next choice is c, the current path becomes (a, f, e, c). Then the next choice must be e. The current path becomes (a, f, e, c, d). Since we cannot extend the path from d, we revise the path to (a, f, e, d, c). From c, the next choice must be b. Since b is adjacent to a, we have found the Hamiltonian cycle (a, f, e, d, c, b, a), and the algorithm terminates.

Possibility 2: If the next choice is d, the current path becomes (a, f, e, d). Then the next choices must be c and then b. Since b is adjacent to a, we have again found the Hamiltonian cycle (a, f, e, d, c, b, a), and the algorithm terminates.

Thus the algorithm always finds the Hamiltonian cycle (a, f, e, d, c, b, a).

27. Suppose that Algorithm 8.3.10 starts with a. The next vertex selected will be either b or d. By symmetry, we may assume that b is chosen. After b, the algorithm may choose c or d. If the algorithm chooses c, the next choice must be d. Since d is adjacent to a, the algorithm has found the Hamiltonian cycle (a, b, c, d, a). If, after choosing a and then b the algorithm next chooses d, the next choice must be c. The current path becomes (a, b, d, c). Since this path cannot be extended, the path is revised to (a, b, c, d).

Since d is adjacent to a, the algorithm has again found the Hamiltonian cycle (a, b, c, d, a).

Suppose that Algorithm 8.3.10 starts with b. The next vertex selected will be either a, c, or d. By symmetry, we may limit the possible choices to a and d. If the algorithm chooses a, the next choices must be d and then c. Since c is adjacent to b, the algorithm has found the Hamiltonian cycle (b, a, d, c, b).

If, after choosing b and then d, the algorithm will choose either a or c. By symmetry, we may assume that the choice is a. The current path becomes (b, d, a). Since this path cannot be extended from a, the path is revised to (b, a, d) This path is extended to (b, a, d, c). Since c is adjacent to b, the algorithm has found the Hamiltonian cycle (b, a, d, c, b).

Since starting with c is symmetric with starting with a, the algorithm will always find a Hamiltonian cycle if it starts with c. Since starting with d is symmetric with starting with b, the algorithm will also always find a Hamiltonian cycle if it starts with d. Thus, regardless of which vertex is chosen as the starting vertex, the algorithm always finds a Hamiltonian cycle.

29. Suppose that a graph with no Hamiltonian cycle and all vertices of degree at least 2 is input to Algorithm 8.3.10. Let v be the last vertex on the current path examined by the algorithm. Since v extended to the first vertex is not a Hamiltonian cycle, the algorithm will continue by examining N (v), the set of vertices adjacent to v. If there is a vertex, say w, in N (v), that is not on the current path, the current path can be extended to w. In this case the algorithm executes again. Suppose that every vertex in N (v) is on the current path. Since v has degree at least 2, there are at least two vertices adjacent to v that are on the current path. One of them, say w, not the predecessor of v on the current path. Thus the path (v1 , . . . , w, vj , . . . , v) is revised to (v1 , . . . , w, v, . . . , vj ), and the algorithm executes again. Therefore, the algorithm will not terminate.

30. True. Examining the pseudocode of Algorithm 8.3.10 shows that a vertex on the path is never removed from the path.

31. Yes. Let v be the last vertex on the current path. If not all vertices are on the current path, since v adjacent to all the other vertices, there is a vertex not on the current path adjacent to v. In this case, the path can be extended, and the algorithm executes again. If all vertices are on the current path, then v is adjacent to the start vertex, so the algorithm has found a Hamiltonian cycle.

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SOLUTIONS

33. No. If m = 2 and n = 3, Km,n has no Hamiltonian cycle, so by Exercise 29, Algorithm 8.3.10 will not terminate.

34. Yes. Km,n has a Hamiltonian cycle if and only if m = n > 1. In this case, whatever the current path, there is a vertex adjacent to the last vertex that is not on the path (unless it is the start vertex, in which case the algorithm terminates having found a Hamiltonian cycle) so that algorithm can extend the path.

36. After the number of executions of the loop exceeds a value proportional to the number of vertices in the graph without finding a Hamiltonian cycle, restart the algorithm with a different start vertex. After trying half of the vertices as start vertices without finding a Hamiltonian cycle, terminate the algorithm.

38. No. Consider Figure 8.3.5.

39. Yes, (v1 , v2 , v5 , v4 , v3 ).

41. No. Suppose the graph has a Hamiltonian path. First note that the path must either start or end at either a or c. If not, we must use edges (a, b), (a, d), (b, c), and (c, d), which make a cycle. Similarly, the path must start or end at either j or l. Suppose that the path starts at a. The path begins with either (a, b) or (a, d). By symmetry, we may assume that the path begins with (a, d).

First suppose that the path ends at l. Since the path does not end at c, it must include edges (b, c) and (c, d). Also, since the path does not end at j, it must include edges (i, j) and (j, k). The path must include either (b, e) or (b, f). If (b, e) is in the path, then (f, e) and (f, i) must be in the path. Since the path ends at l, it must contain (l, k). If (b, f) is in the path, then (f, e) and (e, i) must be in the path.

Again, since the path ends at l, it must contain (l, k). In either case, h must have degree one, which is a contradiction. Therefore the path cannot end at l.

Now suppose that the path ends at j. Since the path does not end at c, it must include edges (b, c) and (c, d). Also, since the path does not end at l, it must include edges (i, l) and (k, l). Arguing as in the previous case, we find that either e or f connects to i. Since the path ends at j, it must include (j, k).

Again, h must have degree one, which is a contradiction. Therefore the path cannot end at j.

Now suppose that the path begins at c. By symmetry, we may assume that the path begins with (c, d).

Since the path does not end at a, it must include edges (d, a) and (a, b). Now the argument is exactly as in the preceding paragraphs; again a contradiction is reached. The proof is complete.

42. No. We would have to eliminate at least one edge at f, at least three edges at c, at least one edge at b, at least one edge at i, at least three edges at j, at least three edges at m, and at least one edge at p leaving 14 edges. Since there are 16 vertices, a Hamiltonian path would have 15 edges.

44. Yes, (a, b, c, j, i, m, k, d, e, f, l, g, h) 45. Yes, (a, b, c, g, l, m, r, q, p, k, j, f, e, i, n, o, t, s, h, d) 47. The graph contains a Hamiltonian path for all m and n. Start in the upper-left corner. Continue right until reaching the end of this row. Drop down to the next row. Continue left until reaching the end of this row. Drop down to the next row. Now repeat, continue right until reaching the end of this row . . .

Continue until all vertices have been visited.

48. For all n 49.

Input: Output:

A simple graph G = (V, E) with n vertices If the algorithm terminates, it returns true if it finds a Hamiltonian path and false otherwise.

randomized hamiltonian path(G, n) { if (n == 1) return true v1 = random vertex in G Copyright c 2018 Pearson Education, Inc.

147

SOLUTIONS if (n == 2) if (N (v1 ) 6= ∅) return true else return false i=1 while (i ¬ = n) { N = N (vi ) − {v1 , . . . , vi−1 } // if i is 1, {v1 , . . . , vi−1 } is ∅ // N contains the vertices adjacent to vi (the current last vertex // of the path) that are not already on the path if (N 6= ∅) { i = i+1 vi = random vertex in N } else if (vj ∈ N (vi ) for some j, 1 ≤ j < i − 1) (v1 , . . . , vi ) = (v1 , . . . , vj , vi , . . . , vj+1 ) else return false } return true }

Section 8.4 2. 11; (a, b, c, g)

3. 10; (a, b, c, d, z)

5. 10; (h, f, c, d)

7. Change line 8 of Algorithm 8.4.1 to while (T ¬ = ∅) { 8.

Input:

Output:

A connected, weighted graph with n vertices in which all weights are positive (if there is no edge between i and j, set w(i, j) = ∞) dist(i, j), the length of a shortest path from i to j for all i and j

all paths(w, n) { for j = 1 to n for k = 1 to n dist(j, k) = w(j, k) for i = 1 to n for j = 1 to n for k = 1 to n if (dist(j, i) + dist(i, k) < dist(j, k)) dist(j, k) = dist(j, i) + dist(i, k) } 10. False

a

b r @ 100 150@rz

r @ 40 @r 5

60

c

Copyright c 2018 Pearson Education, Inc.

148

SOLUTIONS

11. False

b r @ 1 −1 a r −1 rc @rz @ 1 1 @r −1

d

Section 8.5 2. Relative to the ordering a, b, c, d, e, f, g, the adjacency matrix is   0 2 0 0 0 0 0  2 0 1 0 1 0 0     0 1 2 0 1 0 0     0 0 0 2 1 0 1     0 1 1 1 0 1 0     0 0 0 0 1 0 1  0 0 0 1 0 1 0 2. Relative to the ordering a, b, c, d, e, the adjacency matrix is   0 1 0 0 0  1 0 0 0 0     0 0 0 1 1     0 0 1 0 1  0 0 1 1 0

3. Relative to the ordering a, b, c, d, e, f, g, the adjacency matrix is   0 2 0 0 0 0 0  2 0 1 0 1 0 0     0 1 2 0 1 0 0     0 0 0 2 1 0 1     0 1 1 1 0 1 0     0 0 0 0 1 0 1  0 0 0 1 0 1 0 5. If V1 = {a, b} and V2 = {c, d, e} are the vertex sets and the ordering is a, b, c, d, e, the adjacency matrix is   0 0 1 1 1  0 0 1 1 1     1 1 0 0 0     1 1 0 0 0  1 1 0 0 0 6. Every entry is one except along the main diagonal, which consists of zeros.

8. Relative to the orderings x1 , x2 , x3 , x4 and a, b, c, d, e the adjacency matrix is   1 0 0 0  1 0 0 0     0 1 1 0     0 0 1 1  0 1 0 1 Copyright c 2018 Pearson Education, Inc.

149

SOLUTIONS

9. Relative to the orderings x1 , x2 , x3, x4 , x5 , x6 , x7 , x8, x9 , x10, x11 and a, b, c, d, e, f, g, the adjacency matrix is   1 0 0 0 0 0 0 0 0 0 0  1 1 0 1 0 0 0 0 0 0 0     0 1 1 0 0 0 0 1 0 0 0     0 0 0 0 1 1 1 0 0 0 0     0 0 0 1 0 0 1 1 0 1 1     0 0 0 0 0 0 0 0 1 1 1  0 0 0 0 0 1 0 0 1 0 0 

1  0  11.   1  0 0

1 0 0 1 0

1 0 0 0 1

0 1 1 0 0

0 1 0 1 0



 0 1   0   0  1

1  1  12.   0  0 0

1 0 1 0 0

1 0 0 1 0

1 0 0 0 1

0 1 1 0 0

0 1 0 1 0

14. a

f

17.

r r

0 1 0 0 1

irb

rc

d

r

0 0 1 1 0

0 0 1 0 1

 0 0   0   1  1

re

15. a

b

i i i i r r r r @ @ r @r @r i i i

r r

c

d

r @ @er i r

18. (For Exercise 14) The first matrix is relative to the ordering a, c, f and the second is relative to the ordering b, d, e.     0 1 1 2 1 2  1 0 1   1 0 1  1 1 0 2 1 0 20. 93 25.

r

a

ri

b

21. The graph is not connected. r

c

r

d

r

e

26. The vertex corresponding to the row of zeros is an isolated vertex.

Copyright c 2018 Pearson Education, Inc.

150

SOLUTIONS

29. Use the fact that  dn+1 an+1  ·   ·   · an+1 an+1

  · · · an+1 dn  an ·    n+1 ·  = =A  ·  · ·  · · · dn+1 an

an dn

an

 · · · an 0  1 · · · an    ·   1   1 · · · · dn 1

1 0 1 1 1

1 1 0 1 1

30. We solve the second-order linear homogeneous recurrence relation (see Exercise 29)

1 1 1 0 1

 1 1   1  .

1  0

an = 3an−1 + 4an−2 by the method of Section 7.2.

Solving the equation t2 − 3t − 4 = 0

for t, we obtain t = 4 and t = −1. Thus the solution is of the form an = b4n + d(−1)n .

The initial conditions give the equations 1 = 3 =

a1 = 4b − d a2 = 16b + d.

Solving for b and d, we obtain b = 1/5 and d = −1/5. Therefore an =

4n (−1)n 1 − = [4n + (−1)n+1 ].

5 5 5

Section 8.6 2. Relative to the vertex orderings a, b, c, d, e for G1 , and 3, 4, 1, 5, 2 for G2 , the adjacency matrices of G1 and G2 are equal.

3. Relative to the vertex orderings a, b, c, d, e, f for G1 , and 4, 2, 1, 3, 6, 5 for G2 , the adjacency matrices of G1 and G2 are equal.

5. Relative to the vertex orderings a, b, c, d, e, f, g, h, i, j for G1 of Exercise 4, and 1, 2, 3, 4, 7, 6, 10, 9, 5, 8 for the given graph, the adjacency matrices are equal.

6. Replacing the vertex labels in the graph G1 in Exercise 4 with the sets in the table Original Vertex a b c d e f g h i j

Set {1, 2} {3, 5} {1, 4} {2, 5} {3, 4} {4, 5} {2, 4} {2, 3} {1, 3} {1, 5} Copyright c 2018 Pearson Education, Inc.

151

SOLUTIONS yields the graph G1 in Exercise 4.

8. The graphs are not isomorphic. G2 has a vertex of degree 4, but G1 has no vertex of degree 4.

9. Since G1 has eight edges and G2 has nine edges, the graphs are not isomorphic.

11. The graphs are not isomorphic. G1 has a vertex of degree 2, but G2 does not.

12. The graphs are isomorphic. Relative to the vertex orderings a, b, c, d, e, f, g, h, i, j, k, l for G1 , and 1, 5, 6, 2, 3, 7, 8, 4, 9, 10, 11, 12 for G2 , the adjacency matrices of G1 and G2 are equal.

14. The graphs are isomorphic. Relative to the vertex orderings a, b, c, d, e, f, g, h for G1 , and 5, 3, 1, 7, 4, 6, 2, 8 for G2 , the adjacency matrices of G1 and G2 are equal.

15. The graphs are not isomorphic. G1 has two simple cycles of length 3, but G2 has only one simple cycle of length 3 (see also Exercise 25).

16. Extend the definition in Example 8.6.3 as follows: f(v) = v1 v2 . . . vk , where vi is the ith coordinate determined as the members of a ti Gray code. Note that if (v, w) is an edge in M , the strings v1 v2 . . . vk and w1 w2 . . . wk will differ in exactly one bit. So, (v, w) is an edge of the (t1 + t2 + · · · + tk )-cube. Define g on the edges of M by g((v, w)) = (v1 v2 . . . vk w1 w2 . . . wk ). f and g define an isomorphism from M onto the subgraph (V, E) of the (t1 + t2 + · · · + tk )-cube where V E

= {f(v) | v is a vertex in M }, = {f(e) | e is an edge in M }.

18. Since the maximum number of edges in a simple n-vertex graph is C(n, 2) = n(n − 1)/2, the maximum number of edges in a 3-vertex simple graph is 3. Because the number of vertices of odd degree is even, no 3-vertex, regular graph has all vertices of odd degree. It follows that the only regular, simple 3-vertex graph is a triangle.

19. Since the maximum number of edges in a simple n-vertex graph is C(n, 2) = n(n − 1)/2, the maximum number of edges in a 4-vertex simple graph is 6. There are no 1-regular, simple, connected, 4-vertex graphs. The only 2-regular, simple, connected, 4-vertex graph is a square. The only 3-regular, simple, 4-vertex graph is K4 , which has 6 edges.

21. K3,3 and the following graph, which is not bipartite since it has odd-length cycles, are not isomorphic. r r

A r Ar A

Ar r

23. We use the notation of Definition 8.6.1. Suppose that G1 is connected. We must show that G2 is connected. Let v0 and w 0 be distinct vertices in G2 . Then there exist vertices v and w in G1 with f(v) = v0 and f(w) = w 0 . Since G1 is connected, there exists a path (v0 , v1 , . . . , vn ) in G1 with v0 = v and vn = w. Now (f(v0 ), f(v1 ), . . . , f(vn )) is a path in G2 from v0 to w 0 . Therefore G2 is connected.

24. Suppose that G1 and G2 are isomorphic. We use the notation of Definition 8.6.1. Suppose that G1 has n vertices v1 , . . . , vn of degree k and that G2 has m vertices of degree k. By Example 8.6.8, f(v1 ), . . . , f(vn ) each have degree k in G2 . Therefore m ≥ n. By symmetry, m ≤ n. Thus m = n.

26. Let (v, w) be an edge in G1 with δ(v) = i and δ(w) = j. Example 8.6.8 shows that δ(f(v)) = i and δ(f(w)) = j. Now the edge (f(v), f(w)) has the desired property in G2 .

29. Invariant Copyright c 2018 Pearson Education, Inc.

152

SOLUTIONS

30. Not an invariant

r

32.

r A rA

r r r r

Ar

r

r r r r

r r r r

A

r A

A A

r

Ar

r r r r

r r r r

r r r r @

r r r@r

r r r r

r r r r @

r r r r

r r r@r

33. r r r r r 37.

r r r r r

rH r A H H

A H

A HH

H r A Hr HH A

H

H A H

H A H

r HAr

r r r r r

38. Suppose that G is not connected. Let C be a component of G and let V1 be the set of vertices in G that belong to C. Let V2 be the set of vertices in G not in V1 . In G, for every v1 ∈ V1 and v2 ∈ V2 , there is an edge e incident on v1 and v2 . Thus, in G there is a path from v to w if v ∈ V1 and w ∈ V2 . Suppose that v and w are in V1 . Choose x ∈ V2 . Then (v, x, w) is a path from v to w. Similarly if v and w are in V2 , there is a path from v to w. Thus G is connected.

40. Suppose that G1 and G2 are isomorphic. We use the notation of Definition 8.6.1. We construct an isomorphism for G1 and G2 . The function f is unchanged. Let (v, w) be an edge in G1 . Set g((v, w)) = (f(v), f(w)).

It can be verified that the functions f and g provide an isomorphism of G1 and G2 .

If G1 and G2 are isomorphic, by the preceding result, G1 = G1 and G1 = G2 are isomorphic.

41. Yes 44. f(1) = w, f(2) = x, f(3) = y, f(4) = z, f(5) = y, f(6) = x 45. f(1) = a, f(2) = b, f(3) = c, f(4) = d, f(5) = c, f(6) = b 47. f(a) = 1, f(b) = 2, f(c) = 3, f(d) = 4, f(e) = 5, f(f) = 3, f(g) = 2 48. See [Hell].

Copyright c 2018 Pearson Education, Inc.

153

SOLUTIONS

Section 8.7 2. c

d

r

r

r

e

ra

b

r

rf

3.

a c r r r @ @ g ri @rh l r @r J @ J@fr rk JA JA re JA JAr b

rd

j

5. Remove (g, e) and (a, c) to obtain a graph homeomorphic to b f g rH r r @H @ H

@ H H@

@ H@ H

r @r Hr @ a c d 7. Planar a

r rb H @HH @ Hrc @ r @r

e

d

$ %

8. Not planar. The following graph is homeomorphic to K3,3 . a rH

r b @H

H

@ H @ HH

H r

@ Hr d h H H @ H

H @

H r H@ f H r H @r g e r c

11. Let G be a graph having four or fewer vertices. By Exercise 10, the planarity of G is not affected by deleting loops or parallel edges; so we can assume that G has neither loops nor parallel edges. Now G is a subgraph of K4 and, since K4 is planar, so is G. 13. Since every cycle has at least three edges, each face is bounded by at least three edges. Thus the number of edges that bound faces is at least 3f. In a planar graph, each edge belongs to at most two bounding cycles. Therefore 2e ≥ 3f = 3(e − v + 2). Thus 3v − 6 ≥ e.

14. K3,3 16. Suppose that G and G are both planar. Let v denote the number of vertices in G. Let e (respectively, e) denote the number of edges in G (respectively, G). If either G or G is not connected, add just enough edges, preserving planarity, to connect it. Let the connected graphs so obtained be denoted G∗ (with e∗ edges) and G∗ with e∗ edges). Using Exercise 13, we obtain v(v − 1) = e + e ≤ e∗ + e∗ ≤ 2(3v − 6).

2 Copyright c 2018 Pearson Education, Inc.

154

SOLUTIONS Thus v2 − 13v + 24 ≤ 0.

The roots of the equation obtained by replacing ≤ by = are √ 13 ± 73 x= , 2

so

√ 13 + 73 v≤ < 11.

2

17. See Amer. Math. Mo., April 1983, pages 287–288.

18. We use induction on the number e of edges. If e = 3, the graph is a triangle, so it can be colored using three colors. Assume that the statement is true for a graph having fewer than e edges. Choose a diagonal that connects two vertices v1 and v2 . Now edge (v1 , v2 ) divides the original graph into two subgraphs: one, say G1 containing edge (v1 , v2 ) and all vertices and edges on one side of (v1 , v2 ); the other subgraph, say G2 , containing edge (v1 , v2 ) and all vertices and edges on the other side of (v1 , v2 ). By the inductive assumption, we may color G1 with three colors, say C1 , C2, C3 , with v1 colored C1 and v2 colored C2 .

Also, by the inductive assumption, we may color G2 with three colors, C1 , C2, C3 , with v1 colored C1 and v2 colored C2 . Now the original graph is colored using three colors.

20. Pick a city in each country. Draw a line through the common border between two cities in countries sharing a common border. This can be done with no lines crossing.

21. Color D, say, red. Now B and C must be different colors and different from red.

22. A–red, B–green, C–blue, D–red, E–green, F –blue, G–green 24. Color L red. Now G needs a different color–say blue. Now K needs a color different from L and G–say yellow. Now J needs a fourth color.

25. A–blue, B–green, C–red, D–yellow, E–green, F –red, G–yellow, H–green, I–yellow, J–green, K–blue, L–red 27. Suppose that G0 can be colored with n colors. If we eliminate edges from G0 to obtain G, G is colored with n colors.

28. Each face is bounded by three edges and each edge is in a boundary for two faces.

30. Suppose that G has a vertex of degree 3. Then, we find the configuration v1 r r v2 Z

J Z

J Z v J Zr J J Jr v3

Consider the map G0 obtained from G by removing vertex v and the three edges incident on v. By assumption, G0 can be colored with four colors. Now v1 , v2 , and v3 require at most three colors. Color v with the fourth color. Now G is colored with four colors–a contradiction.

31. If G has a vertex v of degree 4, we find the configuration Copyright c 2018 Pearson Education, Inc.

155

SOLUTIONS v1 r

@

@ @

v4 r

@r @

r v2

v

@ @

@r v3

Consider the graph G0 obtained from G by removing vertex v and the four edges incident on v. By assumption, G0 can be colored with four colors. Show that if v1 , v2 , v3 , and v4 use three or fewer colors, we get an immediate contradiction.

Suppose that v1 , v2 , v3 , and v4 require four colors and that vi is colored Ci . Consider the subgraph G01 of G0 consisting of all simple paths starting at v1 whose vertices are alternately colored C1 and C3 . If G01 does not include v3 , we may change each C1 to C3 and each C3 to C1 in G01 and produce a coloring of G0 with four colors. If this is done, we can then color G with four colors.

Suppose that G01 includes v3 . Consider the subgraph G02 of G0 consisting of all simple paths starting at v2 whose vertices are alternately colored C2 and C4 . Show that G02 cannot include v4 . We may change each C2 to C4 and each C4 to C2 in G02 and produce a coloring of G0 with four colors. If this is done, we can then color G with four colors.

Deduce that G cannot have a vertex of degree 4.

33. Use the methods of Exercise 30–32.

Section 8.8 2.

R

r @

3

3.

@ 4 @

1

@ @

W r R

r

2

@ @

rB

R

2

2

@r G rB

3

4

W r

rG

1

r

W r

rB

1

4

rG

3

r @ @ 2 @ @ 1

R

W r

4

@ @

rB 3

@ @r G

Copyright c 2018 Pearson Education, Inc.

156

SOLUTIONS

5. R r i3

i rB

2

R

r

1 W r

1

3

W r

rG

4

rB

4

2

ir G

6. There is no solution.

8. (a)

r @

R

1

@ @ 2

@4 @ 1

ir

3

r

R

4

@ @

2 3

W

(b)

r i1

B

2

3

@r

G

4

r

r @ @

B

2

1

R

3

r

r

4

W

2

r

G

W

G1

(c)

r

B

2

r

4

r

G G3

r

R

3

r

W

@ @ 4 @ @ 3

r i1

@ @r

G

G2

R

1

B

2

r

B

1

4

r

W

3

r

G

G4

(d) Note that Gi and Gj have edges in common for i = 1, j = 3; i = 1, j = 4; i = 3, j = 4; i = 2, j = 3; and i = 2, j = 4. Thus the only solution is G1 , G2.

9. We cannot select the edge incident on R and B for, if we do, there is no way to make the degree of R = 2. Similarly, we cannot select the edge incident on B and G for, if we do, there is no way to make the degree of B = 2. Since we must have an edge labeled 4, we must select the loop incident on W . This means we cannot select any of the edges incident on W and G. Now, G cannot have degree 2. Thus, no subgraph satisfies (8.8.1) and (8.8.2).

Copyright c 2018 Pearson Education, Inc.

157

SOLUTIONS 10. We claim that the following graph has the desired properties.

3

R i r

4

1

r i3

B

1

2

4 3

ir i4 i

2

r i1

G

2 W

Note that each cube contains all four colors. The following subgraph satisfies properties (8.8.1) and (8.8.2). r

R

1

r

B

4

2 3

r

W

r

G

We show that there is no subgraph satisfying (8.8.1) and (8.8.2) that is edge-disjoint from the preceding subgraph. If there is, consider an edge labeled 1. We must choose either the loop labeled 1 on W or the loop labeled 1 on G. If we choose the loop labeled 1 on W , we cannot add an edge labeled 2. Suppose that we choose the loop labeled 1 on G. Now the only edge labeled 4 that we can choose is the loop on W . Now we cannot add an edge labeled 2. Therefore there is no subgraph satisfying (8.8.1) and (8.8.2) that is edge-disjoint from the preceding subgraph and, hence, the puzzle has no solution.

11. There are six choices for the top and, having chosen the top, there are four choices for the front for a total of 6 · 4 = 24 choices.

12. By Exercise 11, there are 24 orientations of one cube. Thus there are 244 = 331,776 stackings.

14. Let v1 r

1 2 3 4 H1

r v2

v3 r

1 2 3 4

r v4

H2

be subgraphs of the graph representing the four cubes in the puzzle such that the intersection of the edge sets and the intersection of the vertex sets are empty.

We can use H1 to construct front and back sides of the stack with the front having color v1 and back having color v2 . This is possible since the edges incident on v1 and v2 contain all the labels 1, 2, 3, and 4. Similarly, H2 can be used to construct the left and right sides of the stack with the left color v3 and the right color v4 .

Any solution is of this form, for if a solution exists, let v1 , v2 , v3 , and v4 be the colors of the front, back, left, and right faces of the solution stack. Then v1 and v2 appear on opposite faces of all four cubes, and Copyright c 2018 Pearson Education, Inc.

158

SOLUTIONS v3 and v4 appear on the other opposite faces of all four cubes. Thus H1 and H2 exist in the graph, as shown previously, representing the solution stack.

16.

4

R r

rG

3 2 1

W r

H1

17.

r

1 2

B

4

rB

3 2 1 H2

R

3

4

1

r

3 4

r

2

r

W

G

18. There is no solution.

20. Let H1 and H2 be a solution to the modified version as in, for example, the solution to Exercise 16. We construct subgraphs G1 and G2 as follows. We let the vertex set of G1 be {R, B, G, W }. We let the edge set of G1 be the set of edges labeled 1, 2 from H1 and the edges labeled 3, 4 from H2 . We let the vertex set of G2 be {R, B, G, W }. We let the edge set of G2 be the set of edges labeled 3, 4 from H1 and the edges labeled 1, 2 from H2 .

21. Yes. See the graph of Exercise 9.

Section 9.1 2. This graph is a not tree because, if v is the left, middle vertex and w is the left, bottom vertex, there is no simple path from v to w.

3. This graph is a not tree because, if v is the upper-left vertex and w is the bottom, middle vertex, there are two simple paths from v to w.

5. n = 1, 2 10. 5 c r

r

6. If either m or n, or both, equals 1

9. 4 Hospital ABC

a

12.

rb @

@r e @ rf @r g dr @ @r i hr

! !! @ ! ! @ !! @ ! !! @ Dr. A

Dr. B

A A

A A

a1 a2 · · · am

···

b1 b2 · · · bn

r

j

Copyright c 2018 Pearson Education, Inc.

Dr. Z

A

A

z1 z2 · · · zk

Patients

159

SOLUTIONS 15. DEAL

16. LAP

19. 0111000100111100010

20. 010000001111

22. Overhead in decoding, memory addressing capability, compatibility with other systems, amount of memory available 23. See G. Williams and R. Meyer, “The Panasonic and Quasar hand-held computers: beginning a new generation of consumer computers,” BYTE, 6 (January 1981), 34–45.

25.

r @0 r @r A 0 1 @0 @r Ar r P S 1 @0 r @r H 1 @0 r @r I 1 @0 r @r C 1 @0 r @r 1

r

U

1

B

28.

r

f

1

r @0 @r 1 @0 r @r e 1 @0 r @r d 1 @0 r @r c 1 @0 r @r a

b

M

29. The proposed code is ambiguous. For example, 01 could represent EA or C. 30. Since K3,3 and K5 contain cycles, a tree cannot contain a subgraph homeomorphic to either; thus, a tree is planar.

31. A terminal vertex has degree 1.

33. Consider the tree to be rooted. Color the vertices on even levels one color and those on odd levels another color.

34. a–5, b–4, c–5, d–4, e–3, f–4, g–3, h–4, i–4, j–5 36. In this solution, we call a simple path in a tree from the root to a terminal vertex a drop. Also, we let ecc(v) denote the eccentricity of the vertex v.

Let c be a center of a tree T . Root T at c. Notice that if ecc(c) = L, the height of T is L. We first show that no vertex on level 2 or greater can be a center. For suppose that there is a center c0 on level two or greater. Then ecc(c0 ) = L. A simple path starting at c0 of length L must pass through c. But now any simple path starting at the parent of c0 has length at most L − 1. This contradicts the definition of “center.”

Notice that no vertex different from c on a drop whose length is less than L can be a center. Thus the only possible centers besides c are the children of c which lie on drops of length L. It is easy to see that if c has at least two children each lying on a drop of length L, then c is the unique center. If c has a unique child c0 lying on a drop of length L, c and c0 are the only centers.

37. In the solution to Exercise 36, we showed that all centers are on level 0 or level 1. Therefore the centers are adjacent.

39. In the following tree, (a, b) and (a, b, a, b) are distinct paths from a to b. r r a

b

Copyright c 2018 Pearson Education, Inc.

160

SOLUTIONS

Section 9.2 2. Aphrodite, Kronos, Atlas, Prometheus 6. r Aphrodite r Eros

3. Aphrodite, Uranus

5. Zeus, Poseidon, Hades

8. Children of d: h, i. Child of e: j.

9. Ancestors of c: b, a. Ancestors of j: e, c, b, a.

11. Terminal vertices: j, f, g, h, i

12. Siblings of f: e, g. Sibling of h: i.

14. r e rj

r

15.

18. It is the root.

j

20. It is a terminal vertex.

21. One is the ancestor of the other.

23. No such graph exists. A terminal vertex has degree 1.

26.

19. They are siblings.

24. r

r Q

Q r r Qr A A A r Ar r Ar r Ar

r

r r

r r

28. The graph is not a tree since, according to Definition 9.1.1, a tree is a simple graph satisfying: If v and w are vertices, there is a unique simple path from v to w.

29. In this case, if the edge is (v, w), we would have the cycle (v, w, v).

31. n − m

32. No. Consider the paths (a, c, d, b) and (a, e, d, c, f, b) in the graph r A

a

d b r r r Q

Q A Q

Q A

Q Ar Qr c

e

f

34. First, suppose that T is a tree. By Theorem 9.2.3b, T is connected. Suppose that for some vertex pair v, w, when edge (v, w) is added, at least two cycles are created. Then there must be at least two distinct simple paths from v to w to account for the distinct cycles. But this contradicts the definition of a tree.

Now suppose that T is connected and when an edge is added between any two vertices, exactly one cycle is created. It follows that if v and w are vertices in T , there is a unique simple path from v to w. For if there were no simple path from v to w, inserting an edge between v and w would not create a cycle. If there were two or more simple paths between v and w, inserting an edge between v and w would create two or more cycles. Thus T is a tree by the definition of tree.

35. Let v be a vertex of degree at least 2 in a tree G and let P = (v0 , . . . , vn ) be a simple path of maximum length passing through v. Since G is a tree, P is not a cycle and, since v has degree at least 2, v 6= v0 and v 6= vn . If removing v and all edges incident on v leaves a connected graph, then there is a simple path, distinct from P , from v0 to vn . Since G is a tree, this is impossible. Therefore v is an articulation point.

Copyright c 2018 Pearson Education, Inc.

161

SOLUTIONS

Section 9.3 2.

r @ @r

r

r

r

3.

r @ @r

6. r @ @r

r

r

r

r

r

r

r

r

r

r

r

5.

r @ @r

r r

r r

r

r @

r @r

r @ @r

8. r @ @r

r

r r

H Hr r

r

r

9. The path (a, b, c, d, e, f, g, h, i, j, k, l) 11.

×

×

×

× 12.

×

×

× ×

14. 10

×

× × ×

×

× ×

×

× ×

×

15. The solution is given as the second part of the solution to Exercise 12.

16. False. Consider K4 . A breadth-first search spanning tree will produce a tree whose root has degree 3.

Thus it cannot produce the tree (a, b, c, d).

18. If T is a tree, every vertex ordering with the same initial vertex produces the same spanning tree, namely T itself.

19. If T is a tree, every vertex ordering with the same initial vertex produces the same spanning tree, namely T itself.

Copyright c 2018 Pearson Education, Inc.

162

SOLUTIONS

21. First show that the graph T constructed is a tree. Now use induction on the number of iterations of the loop to show that T contains all of the vertices of G. 22. If the edge is not contained in a cycle of G 24.

Input: Output:

A connected graph G with vertices ordered v1 , . . . , vn ; and d d(vi ) = length of a shortest path from v1 to vi

short paths(V, E, d) { S = (v1 ) V 0 = set consisting of v1 E0 = ∅ d(v1 ) = 0 while (true) { for each x ∈ S, in order for each y ∈ V − V 0 , in order if ((x, y) is an edge) { add edge (x, y) to E 0 and y to V 0 d(y) = d(x) + 1 } if (no edges were added) return T S = children of S ordered consistently with the original vertex ordering } } 25. Both algorithms find simple paths from v in increasing order of length.

28. The fundamental cycle matrix relative to the orderings (a, b, a), (b, d, c, b), (b, c, f, b), (d, e, c, d), (c, f, e, c), (c, e, g, d, c), (c, f, g, d, c) and e2 , e3 , e5 , e12 , e13 , e10 , e11 , e1 , e4 , e6 , e7 , e8 , e9 is   1 0 0 0 0 0 0 1 0 0 0 0 0  0 1 0 0 0 0 0 0 1 1 0 0 0     0 0 1 0 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 1 0 0 0 0 0 1 1 0     0 0 0 0 0 1 0 0 0 1 1 0 1  0 0 0 0 0 0 1 0 0 1 0 1 1 29. The fundamental cycle matrix relative to the orderings (a, b, d, e, a), (a, b, d, e, a)0, (b, c, d, b), (d, e, f, d), and e3 , e4 , e2 , e9 , e1 , e5 , e6 , e7 , e8 is   1 0 0 0 1 1 0 1 0  0 1 0 0 1 1 0 1 0     0 0 1 0 0 1 1 0 0  0 0 0 1 0 0 0 1 1 31. Modify Algorithm 9.3.7 as follows. Change the line return T to if (|V 0 | == n) return true else return false Copyright c 2018 Pearson Education, Inc.

163

SOLUTIONS If the graph is connected, the value true is returned; otherwise, the value false is returned.

32. Modify Algorithm 9.3.10 as follows. Change the line return true to print solution Delete the line return false.

35. dfs track parent (V, E, parent) { // set v1 ’s parent to 0 to indicate that v1 has no parent parent (v1 ) = 0 V 0 = {v1 } E0 = ∅ w = v1 while (true) { while (there is an edge (w, v) that when added to T does not create a cycle in T ) { choose the edge (w, vk ) with minimum k that when added to T does not create a cycle in T add (w, vk ) to E 0 parent (vk ) = w add vk to V 0 w = vk } if (w == v1 ) return T w = parent of w in T // backtrack } } 36. print parents(V, parent ) { for each v ∈ V println(v, parent (v)) }

38. The solution array x contains 0’s and 1’s. If x[k] = 0, k is not included in the subset. If x[k] = 1, k is included in the subset. all subsets(n) { rall subsets(n, 1) } rall subsets(n, k) { for x[k] = 0 to 1 if (k == n) { for i = 1 to n if (x[i] == 1) print(i + “ ”) println() } else rall subsets(n, k + 1) } Copyright c 2018 Pearson Education, Inc.

164

SOLUTIONS

39.

1

8

7

9

5

2

3

4

6

6

4

2

8

3

7

1

9

5

3

5

9

1

4

6

8

7

2

2

6

4

7

1

3

9

5

8

5

3

8

2

9

4

7

6

1

9

7

1

5

6

8

4

2

3

8

9

5

4

2

1

6

3

7

7

2

3

6

8

9

5

1

4

4

1

6

3

7

5

2

8

9

Section 9.4 2.

r r Q Q r Qr Q Q r Qr

3.

r r r

r r r

r r r

5.

r r r r

r r @ r @r A rA r A r Ar

r r r r

6. Suppose that the start vertex is 1, and that the vertices are added in the order 1, 2, . . . , n. When i is 1 and j is 1, the innermost line (line 11) of the nested for loops is executed n times. When i is 2, the innermost line of the nested for loops is executed n times for each of the values j = 1, 2. When i is 3, the innermost line of the nested for loops is executed n times for each of the values j = 1, 2, 3; and so on. Thus line 11 is executed n + 2n + 3n + · · · + (n − 1)n = n[1 + 2 + 3 + · · · + (n − 1)] =

n(n − 1)n = Θ(n3 ) 2

times. No input requires more than Θ(n3 ) time since the nested for loops take at most O(n3 ) time to execute. Therefore the worst-case time is Θ(n3 ).

8. The body of the last for loop executes n − 1 times the first time, n − 2 times the second time, and so on.

This time dominates, so the worst-case time is (n − 1) + (n − 2) + · · · + 1 =

(n − 1)n = Θ(n2 ).

2

9. The argument is similar to the proof of Theorem 9.4.5.

11. Yes Copyright c 2018 Pearson Education, Inc.

165

SOLUTIONS

12. Suppose that the weight of each edge in Kn is equal to 2. Suppose that some algorithm does not examine edge e. Let T denote the minimal spanning tree output by the algorithm. If e is in T , alter the input by changing the weight of e to 3. If e is not in T , alter the input by changing the weight of e to 1. Rerun the algorithm. Notice that since the algorithm does not examine e, it will still output T . However, for the modified input, T is not a minimal spanning tree. This is a contradiction. Therefore every minimal spanning tree algorithm examines every edge in Kn .

15. True 17. The proof is similar to the proof of Theorem 9.4.5. Let Gi be the graph produced at the ith iteration.

Use induction to show that Gi contains a minimal spanning tree.

18. In Algorithm 9.4.3, change ∞ in line 6 to −∞ and change < to > in line 10.

21. (For Exercise 1) If we break ties by picking the smallest vertices, Kruskal’s Algorithm picks, successively, (2, 3), (3, 5), (3, 4), (1, 2).

22. Argue as in the proof of Theorem 9.4.5.

24. The algorithm picks one 10-cent stamp and six 1-cent stamps to make 16 cents postage, but two 8-cent stamps is optimal.

25. We use induction on n to show that the greedy solution and any optimal solution to the n-cent problem are identical. The statement is clearly true for n = 1, 2, 3, 4, 5, 25.

Suppose that 5 < n < 25. Let S be an optimal solution to the n-cent problem. We must use a 5-cent stamp; for otherwise, we could replace five 1-cent stamps with one 5-cent stamp. Now S with a 5-cent stamp removed is an optimal solution to the (n − 5)-cent problem; for otherwise, an optimal solution to the (n − 5)-cent problem together with a 5-cent stamp would be smaller than S. By the inductive assumption, S, with a 5-cent stamp removed, is the greedy solution. Therefore S is the greedy solution.

Suppose that n > 25. Let S be an optimal solution to the n-cent problem. We must use a 25-cent stamp since we can make at most 24 cents postage optimally using only 5-cent and 1-cent stamps. Now S with a 25-cent stamp removed is an optimal solution to the (n − 25)-cent problem. By the inductive assumption, it is the greedy solution. Therefore S is the greedy solution.

26. a1 = 11, a2 = 5. For n = 15, the greedy method gives 11, 1, 1, 1, 1, but 5, 5, 5 is better.

28. The set {1, 5, 11} shows that the condition is not sufficient. For n = 15, the greedy algorithm gives one 11-cent stamp and four 1-cent stamps, but three 5-cent stamps is optimal.

The set {1, 5, 10, 20, 25, 40} shows that the condition is not necessary. (The example is due to Stephen B. Maurer, Amer. Math. Mo., 101 (5), 419.) The greedy algorithm is optimal for these denominations; however, the condition fails for i = 5: 25 ≥ 2 · 20 − 10 is false.

We can use induction to prove that the greedy algorithm is optimal for the set {1, 5, 10, 20, 25, 40}. We verify directly the cases 1 ≤ n ≤ 214. Now suppose that n > 214. Let S be an optimal solution for n. We claim that S contains a 40-cent stamp. If not, S contains at most four 1-cent stamps (since five 1-cent stamps could be replaced by one 5-cent stamp). For the same reason, S contains at most one 5-cent stamp, at most one 10-cent stamp, at most one 20-cent stamp, and at most seven 25-cent stamps.

But now S can make at most 4 · 1 + 1 · 5 + 1 · 10 + 1 · 20 + 7 · 25 = 214 cents postage. This contradiction shows that S contains a 40-cent stamp.

Now let Gn be the greedy solution for n-cents postage, and let S 0 be S with one 40-cent stamp removed.

Then S 0 is optimal for (n − 40)-cents postage. By the inductive assumption, |Gn−40| = |S 0 |. Therefore |Gn| = 1 + |Gn−40| = 1 + |S 0 | = |S|,

and the greedy solution is optimal for n-cents postage.

Copyright c 2018 Pearson Education, Inc.

166

SOLUTIONS

29. Suppose that such a k exists. Consider making postage for the amount ka2 . Since (k − 1)a2 < a3 ≤ k(a2 − 1), 1 < k < a2 . Thus, the greedy algorithm selects one a3 and ka2 − a3 ones. But, selecting k a2 ’s is optimal and 1 + ka2 − a3 > k, so the greedy algorithm is not optimal.

Suppose that the greedy algorithm is not optimal for all amount of postage. By Exercise 30, there is an amount A, with A < a2 + a3 , for which the greedy algorithm is not optimal. If the greedy algorithm uses no a3 ’s, then the greedy algorithm would be optimal. Therefore the greedy algorithm uses at least one a3 . Since A − a3 < a2 , in fact, the greedy algorithm uses exactly one a3 and A − a3 1’s. Thus for the greedy algorithm A = a3 + (A − a3 ). The optimal algorithm cannot use an a3 since then it would achieve the same result as the greedy algorithm. Thus the optimal algorithm uses k a2 ’s and r 1’s where A = ka2 + r, 0 ≤ r < a2 . Since ka2 ≤ A < a2 + a3 , (k − 1)a2 < a3 .

(23)

Since the greedy algorithm uses 1 + A − a3 stamps and the optimal algorithm uses k + r stamps, 1 + A − a3 > k + r or, replacing A by ka2 + r, 1 + k(a2 − 1) > a3 .

(24)

Combining inequalities (23) and (24), we have (k − 1)a2 < a3 < 1 + k(a2 − 1) and so (k − 1)a2 < a3 ≤ k(a2 − 1).

31. Let a1 = 1, a2 = 5, and a3 = 6. The greedy algorithm is optimal for n = 1, . . . , 9, but not optimal for n = 10.

32. There might be a non-greedy solution for n that does not use a 6-cent stamp. In this case, the greedy algorithm might be optimal for n − 6, but not for n (consider n = 10).

Section 9.5 2. 264, which has 1 + blog10 264 c = 20 digits.

6.

Input: Output:

3. 1/264 ≈ 5.4 × 10−20

root, the root of an n-vertex binary search tree; key, a value to find; and n The vertex containing key, or null if key is not in the tree

bst search(root , n, key) { ptr = root while (ptr ¬ = null ) if (ptr contains key) return ptr else if (ptr contains a value greater than key) ptr = left child of ptr else ptr = right child of ptr return null } Copyright c 2018 Pearson Education, Inc.

167

SOLUTIONS 7.

Input: Output:

s1 , . . . , sn , n A binary search tree T of minimum height that stores s1 , . . . , sn

optimal bst (s, n) { sort s1 , . . . , sn return o bst (s, 1, n) } o bst (s, i, j) { if (i > j) return null m = b(i + j)/2c T 0 = optimal bst (s, i, m − 1) T 00 = optimal bst (s, m + 1, j) let T be the tree whose root contains sm let the left subtree of T be T 0 let the right subtree of T be T 00 return T } 10.

r

r Q

Q r

Qr @ @ r r @r @r @ @ r @r r @r A A @ r Ar r Ar @r

11. There is no such graph. The existence of such a graph would contradict Theorem 9.5.6.

13.

Input: Output:

An integer n > 1 A full binary tree T with n terminal vertices

full binary tree(n) { T = a rooted tree with one vertex for i = 1 to n − 1 { let v be a terminal vertex give v two children } return T } 14.

Input: Output:

A word w to insert in a binary search tree T The updated binary search tree T

bst recurs(w, T ) if (T == null ) { let T be the tree with one vertex, root store w in root return T } s = word in T ’s root Copyright c 2018 Pearson Education, Inc.

168

SOLUTIONS if (w < s) if (T has no left child) give T a left child and store w in it else { left = left child of T bst recurs(w, left) } else if (T has no right child) give T a right child and store w in it else { right = right child of T bst recurs(w, right) } return T }

16.

Input: Output:

The root root of a nonempty binary tree in which data are stored true, if the binary tree is a binary search tree; false, if the binary tree is not a binary search tree. If the binary tree is a binary search tree, the algorithm sets small to the smallest value in the tree and large to the largest value in the tree.

is bst(root , small, large) { if (root has no children) { small = value of root large = value of root return true } lchild = left child of root rchild = right child of root if (is bst(lchild, small left, large left) ∧ is bst (rchild, small right, large right)) { val = value of root if (large left > val ∨ small right < val ) return false small = small left large = large right return true } else return false } 17. We prove the result using induction on n.

Basis Step (n = 1). In this case, the tree consists of three vertices–the root and its two children. Thus I = 0, E = 2, and E = 2 = 0 + 2 · 1 = I + 2n.

Inductive Step. Assume that the equation is true for n. Let T be a tree with n + 1 internal vertices.

Let T 0 be the tree obtained from T by deleting two sibling terminal vertices and the edges incident on them. Let p denote the (former) parent of the deleted siblings. The resulting tree T 0 has n internal vertices. Let I 0 and E 0 denote the internal and external path lengths for T 0 . By the inductive assumption, E 0 = I 0 + 2n.

Copyright c 2018 Pearson Education, Inc.

169

SOLUTIONS

If len denotes the length of the simple path from the root to p in T , the external path length in T is obtained from the external path length in T 0 by adding 2(len + 1), to account for the two new paths to the children of p, and by subtracting len, to account for the loss of the path to the former terminal p; thus, E = E 0 + 2(len + 1) − len = I 0 + 2n + len + 2.

The internal path length in T 0 is obtained from the internal path length in T by subtracting len to account for the loss of the path to p; thus, E = I 0 + 2n + len + 2 = I − len + 2n + len + 2 = I + 2(n + 1).

19. Not balanced

20. Balanced

23. If the balanced trees of heights h − 1 and h − 2 with the minimum number of vertices are found, the required tree of height h can be formed by attaching these two trees as right and left subtrees of a new root. Thus Nh = Nh−1 + Nh−2 + 1.

24. Let sh = Nh + 1. Then sh = Nh + 1 = 1 + Nh−1 + Nh−2 + 1 = sh−1 + sh−2 ,

by Exercise 23. Now s0 = N0 + 1 = 2, s1 = N1 + 1 = 3 (Exercise 22). Thus {sh } satisfies the same recurrence relation as the Fibonacci sequence. Since s0 = f3 and s1 = f4 , it follows that sh = fh+3 , h ≥ 0.

Therefore Nh = sh − 1 = fh+3 − 1.

26. We prove that n < 2h+1 using induction on n. Taking lg of both sides gives the desired result. We omit the Basis Step (n = 1).

Assume that the result is true for binary trees with less than n vertices. Let T be an n-vertex binary tree. Let nL be the number of vertices in T ’s left subtree, and let nR be the number of vertices in T ’s right subtree. Let hL be the height of T ’s left subtree, and let hR be the height of T ’s right subtree.

Note that 1 + hL ≤ h and 1 + hR ≤ h. By the inductive assumption, nL < 2hL +1 and nR < 2hR +1 . Now n = 1 + nL + nR < 1 + 2hL +1 + 2hR +1 ≤ 1 + 2h + 2h = 1 + 2 · 2h = 1 + 2h+1 .

Section 9.6 2.

Preorder: Inorder: Postorder:

ABHIKLM JCDEF G ILKM HJBADF EGC LM KIJHBF GEDCA

3.

Preorder: Inorder: Postorder:

ABCDEF CBEF DA CF EDBA

5.

Preorder: Inorder: Postorder:

ABCDEF G DCBAEF G DCBGF EA

7.

Prefix: Postfix:

−+*AB*CD−/AB+DE AB*CD*+AB/DE+− −

HH

H −

+

−

S

S ∗

A A A

S

S /

∗

+

A A A A

B C

D A

A A

B D

E

Copyright c 2018 Pearson Education, Inc.

170 8.

SOLUTIONS Prefix: Postfix:

/*−ACD+A−BD AC−D*ABD−+/

HH

H /

+

∗

S

S D

−

10.

S

S A

−

A A

A A

A

B

C

Prefix: Postfix:

D

++−*AB/CDE/− − −ABC*DD++ABC AB*CD/−E+AB−C−DD*−AB+C+/+ + PPP

P +

/

H

H

H

A −

S

S

−

A A

B C

A ∗

+

C

A A A A A A − C D D A B A A

D

A

12.

+

−

S

S

/

∗

A A A

E

Prefix: Usual Infix: Parened Infix:

B

−/A*B+CDE A/(B*(C+D))−E ((A/(B*(C+D)))−E)

−

A A /

E

A A A

∗

A A +

B

A A C

13.

D

Prefix: Usual Infix: Parened Infix:

−A+BC A−(B+C) (A−(B+C))

−

A A +

A

A A B

15.

C

Prefix: Usual Infix: Parened Infix:

*−+AB−*CD/EF A (A+B−(C*D−E/F ))*A (((A+B)−((C*D)−(E/F )))*A) Copyright c 2018 Pearson Education, Inc.

171

SOLUTIONS

∗

A A A H H

H −

+

−

S

S

A A A

∗

B

/

A A C

17. −16

A A

D E

18. 0

F

20. −6

21. 16

23. The tree is A

A A B

A A C

D

A A E

F

Because of the preorder listing, A is the root. If A had a left child, the inorder listing would not begin with A. Since A has no left child, the preorder listing tells us that the right child of A is B. The argument that the tree is correct continues in this way.

24.

Input: Output:

pr , the preorder list, and in, the inorder list root , the root of the binary tree with the given preorder and inorder lists

make tree(pr , in) { if (|pr | == 0) return null ch = first character in pr create a vertex v store ch in v root = v choose strings st1 and st2 such that in = st1 + ch + st2 // + is concatenation let pr 0 be the substring of pr obtained by omitting ch choose strings st1 0 and st2 0 such that pr 0 = st1 + st2 0 , where st1 0 (respectively, st2 0 ) is a permutation of st1 (respectively, st2) left subtree of root = make tree(st1 0 , st1 ) right subtree of root = make tree(st2 0 , st2 ) return root } 26. Not necessarily. Consider P1 = ABCDEF and P2 = DBCAEF .

27.

Input: Output:

pt , the root of a binary tree contents of the terminal vertices from left to right Copyright c 2018 Pearson Education, Inc.

172

SOLUTIONS print terminals(pt ) { if (pt == null) return if (pt is a terminal) { print contents of pt return } left = left child of pt print terminals(left ) right = right child of pt print terminals(right ) }

29.

Input: Output:

pt , the root of a binary tree the number of terminal nodes in pt

terminals(pt ) { if (pt == null ) { return 0 left = left child of pt right = right child of pt if (left == null ∧ right == null ) return 1 return terminals(left) + terminals(right) } 30.

Input: Output:

pt , the root of a binary tree initialize each vertex to the number of its descendants

descendants(pt ) { if (pt == null) return numb desc = 0 left = left child of pt if (left ¬ = null) { descendants (left) numb desc = 1 + contents(left ) } right = right child of pt if (right ¬ = null) { descendants (right) numb desc = numb desc + 1 + contents(right) } contents(pt ) = numb desc } 33.

Input: Output:

pt , the root of a binary tree that represents an expression the fully parenthesized infix form of the expression Copyright c 2018 Pearson Education, Inc.

173

SOLUTIONS print expression(pt ) if (pt == null) return if (pt is a terminal) { print (contents(pt )) return } print (“(”) left = left child of pt print expression(left ) print (contents(pt )) right = right child of pt print expression(right ) print (“)”) } 34.

Input: Output:

pt , the root of a Huffman coding tree, and a string α The characters and their codes. Each code is prefixed by α. To print just the codes, invoke this procedure with α set to the null string.

huffman(pt , α) if (pt is a terminal) { print (character stored in pt) print (α) return } left = left child of pt huffman (left, α + 1) // + is concatenation right = right child of pt huffman (right, α + 0) } 36. No. Exercise 37 shows that even if all edges are present, n − 1 vertices suffice for a cover.

37. First, note that any subset of n − 1 vertices is a vertex cover. Second, note that any subset V 0 of n − 2 edges is not a vertex cover. [If v and w are distinct vertices not in V 0 , then edge (v, w) violates the condition that either v or w is in V 0 .]

39. Let E 0 be an edge disjoint set for G, and let V 0 be a vertex cover of G. We define a function f from E 0 to V 0 in the following way: Let e = (v, w) ∈ E 0 . Then either v or w is in V 0 . Choose one of v or w that is in V 0 , but not both, and let f(e) denote the chosen vertex. The function f is one-to-one because the set E 0 is an edge disjoint set. Therefore |E 0 | ≤ |V 0 |.

40. The graph r

r @ @r

has the desired property since it is impossible to put more than one edge in an edge disjoint set and a single vertex is not a vertex cover.

Copyright c 2018 Pearson Education, Inc.

174

SOLUTIONS

Section 9.7 2. A tree of height one has at most three terminal nodes. Since four outcomes are possible, the decision tree must have height at least two. Thus at least two weighings are required to solve the problem of Exercise 1.

3. In the following figure, if the coins in the left pan weigh less than the coins in the right pan, we go to the left child. If the coins in the left pan weigh more than the coins in the right pan, we go to the right child. If the coins in the left pan weigh the same as the coins in the right pan, we go to the middle child.

C1 C2 C3 : C4 C5 C6

C1 C4 : C2 C5

Q

Q

C1 : C7

C3 : C7

C1 , L

C5 , H

C3 , L

C6 , H

PP PP PP PP PP PP

C7 : C1

Q Q

C1 C4 : C2 C5

S S

S

S

Q Q

C7 , L

C4 : C7

S S C2 , L

C8 : C1

C7 , H

S S

C4 , H C8 , L

C8 , H

C4 : C7

C4 , L

C2 , H

Q

Q Q

Q Q

Q

C3 : C7

C1 : C7

S S

S S

C6 , L

C3 , H

C5 , L

C1 , H

5. If we weigh four coins against four coins and they balance, the problem does not reduce to the problem of finding the bad coin from among four coins, but rather to the problem of finding the bad coin from among four coins and eight good coins. This latter problem can be solved in at most two weighings.

6. Four weighings are required in the worst case. We prove this result by considering several cases.

Suppose that we begin by weighing four coins against four coins. If they balance, in two additional weighings, we can account for at most nine outcomes. Since ten outcomes are possible with five coins, we cannot identify the bad coin in at most three weighings in this case.

Similarly, if we begin by weighing three coins against three coins, two coins against two coins, or one coin against one coin, at least four weighings are required in the worst case.

Suppose that we begin by weighing five coins against five coins. Consider one of the instances in which they do not balance: C1 C2 C3 C4 C5 : C6 C7 C8 C9 C10

@ r @@ @

In two more weighings, we can account for at most nine outcomes, but there are ten C1 , L, C2 , L, C3 , L, C4 , L, C5 , L, C6 , H, C7 , H, C8 , H, C9 , H, C10, H. Therefore, if we begin by weighing five coins against five coins, at least four weighings are required in the worst case.

Similarly, if we begin by weighing six coins against six coins, at least four weighings are required in the worst case.

Copyright c 2018 Pearson Education, Inc.

175

SOLUTIONS

We conclude that the 13-coins puzzle requires at least four weighings in the worst case. In fact, the puzzle can be solved in at most four weighings: Begin by weighing four coins against four coins. If they do not balance, proceed as in the solution to the 12-coins puzzle (see Exercise 4). If they balance, five coins remain. We can identify a bad coin among five in at most three weighings (see Example 9.7.1).

8. If there is an algorithm that solves the puzzle in k < n weighings, the algorithm can be described by a decision tree of height k. Every internal vertex of this tree has at most three children; thus, there can be at most 3k terminal vertices. But there are 2((3n − 3)/2) = 3n − 3 possible outcomes and 3n − 3 > 3k for n ≥ 2, k < n, which is a contradiction.

10. Pick a coin and call it x. Set variables l and s to 0. Compare x with each of the other n − 1 coins. In comparing x with a coin y, if x is heavier than y, increment l. If x weighs the same as y, increment s. If l 6= 0, the number of bad coins is equal to l. If l = 0, the number of bad coins is equal to s + 1.

11.

Input: Output:

a 1 , a2 , a3 , a4 a1 , a2 , a3 , a4 (in increasing order)

sort 4 (a1 , a2 , a3 , a4 ) // sort a1 and a2 if (a1 > a2 ) swap(a1 , a2 ) // sort a3 and a4 if (a3 > a4 ) swap(a3 , a4 ) // find largest if (a2 > a4 ) swap(a2 , a4 ) // find smallest if (a1 > a3 ) swap(a1 , a3 ) // sort a2 and a3 if (a2 > a3 ) swap(a2 , a3 ) } 13. There are 6! = 720 possible outcomes to the problem of sorting six items. To accommodate 720 vertices, we must have a tree of height at least 10 since 29 < 720 < 210 . Thus we need 10 comparisons in the worst case.

To sort six items using at most 10 comparisons, we first sort five items using at optimal sort (see Exercise 12). This requires at most seven comparisons. We then find the correct position for the sixth item using binary search. This last step requires at most three comparisons.

15.

35

Z Z Z

30

30

12

30

k

16. 2 − 1

1 12

9

0

3

Z J J

12

9

A A A A

A A 35

9

30

35

9

Z Z

J J

J J

A A A A

30

30

35

J J A A

0

30

1

12

0

0

A A A A 3

9

0

0

18. k Copyright c 2018 Pearson Education, Inc.

176

SOLUTIONS

Section 9.8 2. Isomorphic. f(v1 ) = w3 , f(v2 ) = w5 , f(v3 ) = w6 , f(v4 ) = w2 , f(v5 ) = w1 , f(v6 ) = w4 .

3. Not isomorphic. Tree T2 has a vertex of degree 4 (w4 ), but T1 has no vertex of degree 4.

5. Isomorphic. f(v1 ) = w7 , f(v2 ) = w4 , f(v3 ) = w6 , f(v4 ) = w10 , f(v5 ) = w3 , f(v6 ) = w2 , f(v7 ) = w9 , f(v8 ) = w11 , f(v9 ) = w1 , f(v10 ) = w8 , f(v11 ) = w5 , f(v12 ) = w12 .

They are isomorphic as free trees (see the solution to Exercise 3).

6. Not isomorphic. Vertex v10 in T1 must be mapped to vertex w4 in T2 since these are the only vertices of degree 4. The vertices adjacent to v10 have degree 1, 1, 2, 3, while the vertices adjacent to w4 have degree 1, 2, 3, 2.

8. Isomorphic. f(vi ) = wi , i = 1, . . . , 5. Also, they are isomorphic as free trees.

9. Not isomorphic as rooted trees. The root of T1 has degree 3, but the root of T1 has degree 1.

11. Not isomorphic. The root of T1 has no right child, but the root of T2 has a right child. They are not isomorphic as rooted trees, but they are isomorphic as free trees.

12. Isomorphic. f(vi ) = wi , i = 1, . . . , 6. Also, they are isomorphic as rooted trees and as free trees.

14. r

15. r r 17.

r r

r

r

r

r r r

r r r

r r

r

r Q A Q r r Ar Qr

r

r

r @ r @r

r

r

r

r r

r A

r

r r H

Hr r HHr r r

r

r

r

Ar

r @ r @r r

r

r

r A

r

r

r

r

r A

r Ar A r Ar r

Ar

r

r

r

r

r r

r A

Ar

r r r r

r

Copyright c 2018 Pearson Education, Inc.

r

r

r

r r r

r A

Ar

r

r

r

r

r

177

SOLUTIONS 18.

20.

21.

r

r

r @ @r

r A r Ar A r Ar A r Ar

r A

r @ r @r A A r Ar r Ar

r A r Ar A r Ar A r Ar A r Ar

r A

r A

r Ar A r Ar A r Ar

r A

r Ar A r Ar A r Ar A r Ar

Ar

r S

r Sr A A r Ar r Ar A r Ar

r S r Sr A A r Ar r Ar A r Ar

r A r Ar A r Ar A r Ar A r Ar

r A

r

r Ar A r Ar A r Ar

r S r Sr A A r Ar r Ar A r Ar r A

Ar A

Ar A

r

r Ar A r Ar

r A

r A

Ar A

r Ar A r Ar

r A r Ar S r Sr A A r Ar r Ar

r Ar A r Ar A r Ar A r Ar

r S r Sr A A r Ar r Ar A r Ar r A

r Ar A r Ar A r Ar A r Ar

r Ar A r Ar A r Ar

r

r A

r Ar A r Ar A r Ar A r Ar

r

r A

Ar S Sr r A A r Ar r Ar

23. For Exercise 9, there are 10 spanning trees obtained by replacing the left triangle by r @ r

r @

@r c

r

@r c

and by replacing the leftmost figure by

f

r

r

r

HHr r

f

r

r

r r

r

r HHr f r r

r

f

r

r

r r

r

r HHr

f r HHr

r

Copyright c 2018 Pearson Education, Inc.

178

SOLUTIONS

24. Let bk denote the number of comparisons when two k-vertex isomorphic binary trees are input to Algorithm 9.8.13. We use induction on k to show that bk = 6k + 2.

(25)

If k = 0, the trees are empty. In this case, there are two comparisons at line 1 after which the procedure returns. Thus (25) is correct for k = 0.

Assume that bi = 6i + 2 for i < k. There are four comparisons at lines 1 and 3. Let L denote the number of vertices in the left subtree (of either tree) and R denote the number of vertices in the right subtree (of either tree). By the inductive assumption, line 9 requires bL + bR = (6L + 2) + (6R + 2) comparisons. Thus the total number of comparisons is 4 + 6L + 2 + 6R + 2 = 6(1 + L + R) + 2 = 6k + 2.

The inductive step is complete.

26.

Input: Output:

n an n-vertex random binary tree

rand bin tree(n) { if (n == 0) return null let k be a random integer between 0 and n − 1 inclusive T1 = rand bin tree(k) T2 = rand bin tree(n − 1 − k) let T be the binary tree with left subtree T1 and right subtree T2 return T } 28. By Exercise 27, there are Cn nonisomorphic full binary trees with n+1 terminal vertices. We may choose a terminal vertex to mark in n + 1 ways. Therefore |X2 | = (n + 1)Cn .

31. Let T 0 ∈ X2 . Let t denote the marked terminal vertex, let v be t’s parent, and let t0 be v’s sibling. Delete t, delete edge (v, t0 ), and replace v by t0 . If T is the resulting tree, T 0 ∈ XT .

32. Each element of X1 gives rise to 2(2n − 1) elements of X2 . By Exercise 27, |X1 | = Cn−1 . By Exercise 31, all elements of X2 are included. Therefore |X2 | = Cn−1[2(2n − 1)]. By Exercise 28, |X2 | = (n + 1)Cn and the conclusion follows.

34. The hint provides a one-to-one correspondence between n-edge ordered trees and strings of n zeros and n ones in which, reading from the left, the number of ones is always greater than or equal to the number of zeros. The number of such strings is Cn , since these strings also encode paths in an n × n grid from the lower-left corner to the upper-right corner that never go above the diagonal from the lower-left corner to the upper-right corner (see Example 6.2.23). The encoding is obtained by interpreting a one as a move right and a zero as a move up.

Copyright c 2018 Pearson Education, Inc.

179

SOLUTIONS

Section 9.9 2. The second player always wins. If the first player leaves {1, 3} or {0, 3}, leave one. If the first player leaves {2, 3}, leave {2, 2}. After the first player moves, the second player can leave one. Part of the tree is 3 3

HH m 13m 03m

P

PP Q

Q P 2 3

1 3

0 3

2 2

P @ PP 0m 1m 1m 1m 3 2 1 0 PP

P

0 2

0 1

@ m 00m 00m

2 1

2 0

0 0

0 1

0 0

3. The tree is the same as Figure 9.9.1. The terminal vertices are assigned values as in Figure 9.9.2 with 0 and 1 interchanged. After applying the minimax procedure, the root receives the value 1; thus the first player will always win. The optimal strategy is to first leave {2, 2}. If the second player leaves only one pile, take it; otherwise, leave {1, 1}.

5. The tree is the same as in Exercise 1. The terminal vertices are assigned values as in the hint for Exercise 1 with 0 and 1 interchanged. After applying the minimax procedure, the root receives the value 1; thus the first player will always win. The optimal strategy is take 2. No matter how many player 2 chooses, player 1 can take the rest.

6. Figure 9.9.2 8. The strategy for winning play is: Play nim0 exactly like nim unless the move would leave an odd number of singleton piles and no other pile. In this case, leave an even number of piles.

10. The value of the root is 10.

11.

5

XXX

X m 4 5m HH @

H 10

4

15

5

13

@ @ @ m 10m 4m 15m 1m 5m 9m 13m

5

13. The value of the root is 9.

16. 1 − 1 = 0

17. 4 − 2 = 2

20. No. Assign a larger value to a winning position.

21.

Input:

Output:

the root pt of a game tree, the level pt level of pt , the maximum level n to which the search is to be conducted, and an evaluation function E the game tree with the values of the vertices stored in the vertices Copyright c 2018 Pearson Education, Inc.

180

SOLUTIONS minimax (pt , pt level , n, E) { if (pt level == n) { contents(pt ) = E(pt ) return } let c1 , . . . , ck be the children of pt for i = 1 to k { minimax (ci , pt level + 1, n, E) ei = contents(ci ) } if (pt is a box vertex) contents(pt ) = max{e1 , . . . , ek } else contents(pt ) = min{e1 , . . . , ek } }

25. We first obtain the values 9, 6, 7 for the children of the root. Thus we order the children of the root 9, 7, 6 and use alpha-beta pruning to obtain

8

7

8

9

7

9

9

8

1

13

3

1

7

9

7

4

8

16

8

8

12

4

12

4

4

28. m : n means player m plays player n. Thr left child of an m : n vertex means that m won that match, and the right child of an m : n vertex means that n won that match. The terminal vertices show who was the game winner.

Copyright c 2018 Pearson Education, Inc.

181

SOLUTIONS

1:2

1:3

2:3

1:4

1

4:3

4:2

4

2:4

2:4

2:3

2

1:2

3:1

3

1

2:3

4

2

1:2

1

2

3

4:3

4:1

4

2

1:4

1:3

1

1

2:1

3:2

3

2

1:3

4

1

2:1

2

3

1

2

29. Notice that the height of the game tree is the maximum number of rounds that can be played. We first show that the maximum number of rounds that can be played is at most n(n − 1)/2.

Consider the situation just before the last round. At this point, every player has at most n − 2 points.

Thus, if TP is the total number of points just before the last round, TP ≤ n(n − 2).

If x is the number of rounds after the first n − 1 rounds, but just before the last round, then TP = n − 1 + 2x.

Combining the last inequality and equation, we have n − 1 + 2x ≤ n(n − 2) or

n2 − 3n 1 + .

2 2 Noting that (n2 − 3n)/2 is always an integer, we, in fact, have x≤

x≤

n2 − 3n .

2

To see that this maximum can occur, first suppose that n = 2k is even. The situation in which 1 wins n − 2 times,  n wins (n − 2)/2 times, k wins (n − 2)/2 times,     n − 1 wins (n − 2)/2 times, k − 1 wins (n − 2)/2 times, 2(k − 1) = n − 2 pairs ..  .    n − (k − 2) wins (n − 2)/2 times, k − (k − 2) wins (n − 2)/2 times, Copyright c 2018 Pearson Education, Inc.

182

SOLUTIONS k + 1 wins (n − 2)/2 times, 1 wins, takes (n − 2) + [(n − 2) + 1]

n−2 n(n − 1) +1 = 2 2

rounds.

Now suppose that n is odd. Consider the sequence of wins  1345 . . . n2    1345 . . . n2  n − 1 lines ..  2 .    1345 . . . n2.

After the wins in the first line, players 1 and 3 through n each earn 1 point in the order shown and lastly, player 2 earns 2 points. After the wins in the second line, players 1 and 3 through n each earn 2 points in the order shown, for a total of 3 points each, and lastly, player 2 earns 2 points, for a total of 4 points.

After the wins in the last line, players 1 and 3 through n each earn 2 points in the order shown, for a total of n − 2 points each, and lastly, player 2 earns 2 points, for a total of n − 1 points, and player 2 wins the game. Thus, if the wins occur in the order shown, there are n(n − 1)/2 rounds.

32–33. It is possible to always force a draw in Mu Torere, see P. D. Straffin, Jr., “Position graphs for Pong Hau K’i and Mu Torere,” Math. Mag., 68 (1995), 382–386, and “Corrected figure for position graphs for Pong Hau K’i and Mu Torere,” Math. Mag., 69 (1996), 65.

Section 10.1 2. (a, d)–1, (b, d)–2, (e, c)–2, (e, z)–1. The value of the flow is 5.

3. (a, b)–3, (a, d)–1, (d, c)–0, (d, f)–1, (c, e)–1, (g, z)–2, (c, z)–2. The value of the flow is 6.

5.

r -3

r -4 Ar @ 8 @ ∞ 2 3 R @ R @ @ @ 4 w 2 d ∞ 2r r -6 @rB @rz a r @ Q QQ s Q R @ 7@ 5 4 2 3 ∞ 3

Q @ r - r - r - Qr w1

w3

5

r -4

b

e

2

c

f

6

C

6. Make the capacities of (A, z), (B, z), and (C, z), 4, 3, and 4, respectively.

8.

1000 A - 7:00 ∞

a

∞

∞

4000 B - 7:30

3000

C - 7:15

∞

B - 7:45

∞

1000 A - 7:15 ∞ ∞

3000 3000

A - 7:30

D - 7:45 ∞ 2000 2000 4000

∞ D - 7:30 ∞

C - 7:30 ∞

2000

C - 7:45

Copyright c 2018 Pearson Education, Inc.

D - 8:00

z

183

SOLUTIONS 9. Replace each undirected edge by two directed edges r r

each having weight equal to the weight of the undirected edge.

11. n2 − 3n + 3. To prove this, we sum the in and out degrees of the vertices and then divide by 2. The source has n − 1 out edges. The sink has n − 1 in edges. All other vertices have n − 2 out edges (since an edge cannot go to itself or the source) and n − 2 in edges (since an edge cannot come from itself or the sink). Since there are n − 2 vertices besides the sink and source, the sum of the in and out degrees is (n − 1) + (n − 1) + (n − 2)[(n − 2) + (n − 2)] = 2n2 − 6n + 6.

We obtain n2 − 3n + 3 by dividing by 2.

Section 10.2 2. 2 5.

3. 1

4000, 4000 r r P PP 3000, 2000 qP @ PPr @ ∞, 0 R ∞, 4000 ∞, 6000 ?

PP

2000, 2000 P qP @ ∞, 6000 ∞, 6000 P r r 4000, 4000 Pr @r P PP @ qP ? ∞, 0 P r ∞, 4000@ R ∞, 6000 ?3000, 2000 PP PP 2000, 2000 qP @ ∞, 0 PPr @r 4000, 4000

6.

r

3, 3 -

c

r

4, 3 f

r @ @ 2, 2@ R @

4, 3 a

1, 0

6 @ @r

r

w1

2, 2

a

r

-

b

8.

r @ @@ R 3, 3 @ @

d

2, 2

r

-

3, 2

b

@r z

5, 2

e

-

4, 4

3, 2

r r @ @ @ 8, 0 @ 4, 4 R@ @ R@ @ @ @ 6, 3 3, 3 @rB @r z c

4, 4

A

4, 4 2, 2 d r r r

H w 2 @

H

H @

HH

j 5, 3 4, 0 2, 2 3, 0 7, 5@ R *

H

H H @

H r r r Hr @ w3

5, 2

e

2, 2

f

6, 2

4, 4

C

Copyright c 2018 Pearson Education, Inc.

184

SOLUTIONS

9.

r

3, 2

2, 2

a

4, 2 r @ @ R 7, 7@

r

-

w1

r

r

2, 0 -

6, 5 -

d

5, 5

4, 0

6 r -

-

w3

4, 4

3, 2

w2

@ @r

r r @ @ @ 8, 0 @ 4, 4 R @ R @ @ @ @ @ d0 6, 3 3, 3 r @Br @r z @ @ 3, 0 4, 4 @ R 2, 2 @

@ r @r

-

b

5, 2

e

2, 2

c

4, 4

A

f

6, 2

C

11. r

r @

3, 3 -

c

b

5, 5

a

r @

-

6, 6

@ @ 6,R 4

?

6 5, 0

5, 0

d

r

4, 0

@ 6 @ @r

4, 0

R 6, 2

3, 0

6, 0

I 10, 6 10, 0

?

4, 4

2, 2

re 6, 6

6, 6 6, 0

@ R 10, 5 2, 0@ ?

6 @ @ 4, 4 rf @rz

5, 0 6, 0

R

I

4, 0

6

r

g

5, 0 8, 6 ?

h

12. (a, b)–10, (a, f)–11, (a, j)–1, (b, c)–6, (b, g)–9, (f, b)–5, (f, g)–6, (j, g)–1, (c, d)–8, (c, h)–2, (g, c)–4, (g, h)– 12, (d, e)–8, (h, i)–6, (h, n)–6, (h, m)–2, (m, n)–2, (e, z)–8, (i, z)–6, (n, z)–8. All other edges have flow equal to 0.

14. 8

15.

r

b

r @ @ 6, 4 R@ @ @

-

3, 3

c

3, 3

a

r @

?5, 1

@@ R @

2, 2

6

2, 2

@ @r d

5, 3

r

@r z

4, 1

e

Copyright c 2018 Pearson Education, Inc.

185

SOLUTIONS 17.

r

w1

∞, 6

a

r @

6, 6 -

-4, 4

b

3, 0

@ 2, 2@ R @

r @ @ R @ @∞, 5 @ @r z

A

2, 1

@ ∞, 0 - w2 r r @cr

@ 3, 3 @

*

@

@ d

r R R ∞, @ 3

4, 4@ @ 3, 3 @

@ *

@ r @r w3

∞, 4

B

18. (a, w1)–3, (a, w2 )–3, (a, w3 )–3, (w1 , b)–3, (w2 , b)–3, (w3 , d)–3, (b, A)–4, (b, c)–2, (d, c)–3, (c, A)–2, (c, B)– 3, (A, z)–6, (B, z)–3

Section 10.3 2. 9; not minimal

3. 15; not minimal

5. P = {a, w1, w2 , w3 , b, d}

6. P = {a, A−6:00,B−6:15,A−6:15,B−6:30,A−6:30} 8. P = {a}

9. P = {a, b}

11. P = {a}

12. P = {a, w1 , w2 , w3, b, c, d, e, f, A, B, C} 14. P = {a, A−7:00,C−7:15,A−7:15,C−7:30, A−7:30,C−7:45} 15. P = {a, b, d, e, g} 19. Use Exercise 18 and imitate the proofs of Theorems 10.3.7 and 10.3.9.

21. The argument is similar to that of Exercise 19.

22. Modify Algorithm 10.2.4.

Section 10.4 2. A minimal cut is P = {a}. r r Q S Q 1, 1 @ S Q s @ w S Q 1, 1 1, 0S @ 1, 1 Q R@

S QrP 1, 1@

P 1, 0 S PP qP@ 1, 1 1 S PP

r @r z a r

1, 0 @ @

S

1

R @ 1, 1

@ @ S r @

@ R @ 1, 1 1, 1 @ 1, 1 @ @

@ @ r @r 1, 0

4. Filling the maximum number of jobs

5. J1 − C, J2 − A, J5 − D

7. There is not a complete matching because there are fewer persons than jobs.

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186

SOLUTIONS

8. (a)(b) J1

A

J2

B

J3 a

z

C J4 D

J5

E

J6

C1

V

(c) Yes 11. (a)(d)

W C2 a

X

z

C3 Y

C4

Z

(b) The maximum number of committees that can be represented (c) All committees are represented. (e) Yes 12. Let V = {v1 , . . . , vm } and W = {w1 , . . . , wn} be the disjoint vertex sets. Order the vertices v1 , . . . , vm , w1 , . . . , wn .

14. Each row has exactly one label and each column has at most one label.

15. Among all labelings in which every column has at most one label, the maximum number of rows are labeled.

16. See A. Tucker, Applied Combinatorics, Wiley, New York, 1980, page 348.

18. See C. L. Liu, Introduction to Combinatorial Mathematics, McGraw-Hill, New York, 1968.

19. False. Consider r r H

H

H

r Hr Copyright c 2018 Pearson Education, Inc.

187

SOLUTIONS

Problem-Solving Corner: Matching 1.

r

r

*

- r r r -

r

Section 11.1 2.

x1 1 1 1 1 0 0 0 0

x2 1 1 0 0 1 1 0 0

x3 1 0 1 0 1 0 1 0

(x1 ∧ x2 ) ∨ x3 1 1 0 1 0 1 0 1

x1 1 1 0 0

x2 1 0 1 0

x1 ∨ x2 0 1 0 0

3.

H H c

x1 x2

5.

x1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0

x2 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0

x3 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0

x1

x4 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0

x1 ∨ x2

H H c

x1 ∨ x2

((x1 ∨ x2 ) ∧ (x3 ∨ x4 )) ∧ (x2 ∨ x4 ) 1 0 1 0 0 0 0 0 1 0 1 0 1 0 1 1 Copyright c 2018 Pearson Education, Inc.

188

SOLUTIONS

6. x1 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0

x2 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0

x3 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0

x4 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0

((x1 ∧ x2 ) ∧ (x3 ∨ x4 )) ∨ (x1 ∧ x2 ) 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0

8. If x = 0, the output of the AND gate is 0 regardless of the value of the other input. Thus the output of the NOT gate is 1. Therefore, y = 1.

9. Suppose that x = 1 and y = 0. Then the input to the AND gate is 1, 0. Thus the output of the AND gate is 0. Since this is then NOTed, y = 1. Contradiction. Similarly, if x = 1 and y = 1, we obtain a contradiction.

11. 1

12. 1

14. 1

15. (For Exercise 10) x1 and x2 are Boolean expressions by (11.1.2); x1 ∧ x2 is a Boolean expression by (11.1.3d); x1 ∧ x2 is a Boolean expression by (11.1.3b).

17. Is a Boolean expression

18. Is a Boolean expression

20. Is a Boolean expression

21. The circuit for Exercise 10 is Exercise 1. The solution to Exercise 1 gives the logic table.

23.

b

b

A

B

26. A ∧ (B ∨ (C ∧ B)) 28. A ∧ ((B ∨ D) ∨ (C ∧ (A ∨ D ∨ C))) ∧ B 30.

A 1 1 1 1 0 0 0 0

B 1 1 0 0 1 1 0 0

C 1 0 1 0 1 0 1 0

A ∨ (B ∧ C) 1 1 1 1 0 0 1 0

b

A

B

C

b

Copyright c 2018 Pearson Education, Inc.

189

SOLUTIONS 31.

A 1 1 1 1 0 0 0 0

B 1 1 0 0 1 1 0 0

C 1 0 1 0 1 0 1 0

(A ∧ B) ∨ (C ∧ A) 1 0 1 0 1 1 0 0

A 1 1 1 1 1 1 1 1 0 0 0 0 0 0 0 0

B 1 1 1 1 0 0 0 0 1 1 1 1 0 0 0 0

C 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0

D 1 0 1 0 1 0 1 0 1 0 1 0 1 0 1 0

b

A

B

C

A

b

33.

A ∧ ((B ∧ C ∧ D) ∨ ((B ∧ C) ∨ D) ∨ (B ∧ C ∧ D)) ∧ (B ∨ D) 1 1 1 0 0 1 0 0 0 0 0 0 0 0 0 0

C

b

B A

D B

C

D

b

Section 11.2 2. x1 1 1 1 1 0 0 0 0

x2 1 1 0 0 1 1 0 0

x3 1 0 1 0 1 0 1 0

(x1 ∨ x2 ) ∨ (x1 ∧ x3 ) 0 0 0 0 1 1 1 0

x1 ∧ (x2 ∨ x3 ) 0 0 0 0 1 1 1 0

Copyright c 2018 Pearson Education, Inc.

190

SOLUTIONS

3.

x1 1 1 0 0

x2 1 0 1 0

(x1 ∧ x2 ) ∨ x1 1 1 0 0

(x1 ∨ x2 ) ∧ x1 1 1 0 0

5. The Boolean expression for the second circuit can be transformed to the Boolean expression for the first circuit (x1 ∨ x3 ) ∧ (x2 ∨ x3 ) ∧ (x2 ∨ x4 ) ∧ (x1 ∨ x4 ) 7.

8. x1 1 1 0 0

x2 1 0 1 0

x1 ∧ x2 0 1 0 0

(x1 ∨ x2 ) 0 1 0 0

x1 1 1 0 0

= =

x2 1 0 1 0

(x3 ∨ (x1 ∧ x2 )) ∧ (x4 ∨ (x2 ∧ x1 )) (x1 ∧ x2 ) ∨ (x3 ∧ x4 )

x1 ∨ (x1 ∧ x2 ) 1 1 0 0

10. The left expression can be transformed into the right expression by successive applications of the distributive and commutative laws: (x1 ∨ x2 ) ∧ (x3 ∨ x4 )

= =

((x1 ∨ x2 ) ∧ x3 ) ∨ ((x1 ∨ x2 ) ∧ x4 ) (x1 ∧ x3 ) ∨ (x2 ∧ x3 ) ∨ (x1 ∧ x4 ) ∨ (x2 ∧ x4 )

=

(x3 ∧ x1 ) ∨ (x3 ∧ x2 ) ∨ (x4 ∧ x1 ) ∨ (x4 ∧ x2 )

12. False. Take x1 = 1, x2 = 0.

13. False. Take x1 = x2 = x3 = 1.

15. False. Take x1 = x3 = 1 and x2 = x4 = 0.

19. (For Exercise 27, Section 11.1) b

A

b

C D

B C

22.

23.

A A

b

C B D

F

@ A

b

b

B

@

F B

C

@

E A

b

D

Section 11.3 3. If (S, +, ·,0 , 1, 8) is a Boolean algebra, then x + x0 = 1. In this case, we must have lcm(x, 8/x) = 8, for x = 1, 2, 4, 8.

However, for x = 4, lcm(4, 8/4) = lcm(4, 2) = 4. Therefore, this system is not a Boolean algebra.

Copyright c 2018 Pearson Education, Inc.

191

SOLUTIONS 5. In this solution, we denote the 0 (respectively, 1) of a Boolean algebra by m (respectively, M ).

Every Boolean algebra has at least two elements since m and M are distinct. Thus n ≥ 2.

Suppose that n > 2 and that Sn is a Boolean algebra. Then n = n · M = min{n, M }. Thus M = n. For any x ∈ Sn , we have M = x + x0 = max{x, x0 }. It follows that if x 6= M , then x0 = M . Therefore, if x 6= M, x = (x0 )0 = M 0 . This says that M 0 is not unique. Contradiction.

If n = 2, we may take m = 1, M = 2, 10 = 2, and 20 = 1, and show that Sn is a Boolean algebra.

6. (a) X ∪ X = X, X ∩ X = X (b) X ∪ U = U, X ∩ ∅ = ∅ (c) X ∪ (X ∩ Y ) = X, X ∩ (X ∪ Y ) = X (d) X = X (e) ∅ = U, U = ∅ (f) X ∪ Y = X ∩ Y , X ∩ Y = X ∪ Y 9. (x0 y0 )0 = x + y 10. If xy = xz and x0 y = x0 z, then y = z.

12. x = 1 if and only if y = (x + y0 )(x0 + y) for all y.

13. If xy = 1, then x = 1 = y.

16. (For the dual of Exercise 8) xy + x0

= =

xy + 0 xy

by 11.3.6b by 11.3.1d

x(x + y)y

= = = = = = = =

(xx + xy)y (x + xy)y y(x + yx) yx + y(yx) yx + (yy)x yx + yx yx xy

by 11.3.1c by 11.3.6a by 11.3.1b by 11.3.1c by 11.3.1a by 11.3.6a by 11.3.6a by 11.3.1b

17. If xy = 0 and x + y = 1, then y = x0 . The dual of Theorem 11.3.4 is Theorem 11.3.4.

19. (For part c) x(x + y) = (x + 0)(x + y) = x + 0y = x + y0 = x + 0 = x 20. (For part d) We have x0 + x

= =

x + x0 1

by 11.3.1b by 11.3.1e.

Dually, x0 x = 0. By Theorem 11.3.4, (x0 )0 = x.

22. Since 11.3.1a–e hold for all sets, they hold for S. We need only note that ∅ and U are present and that S is closed under taking unions, intersections, and complements.

Copyright c 2018 Pearson Education, Inc.

192

SOLUTIONS

24. First notice that the proof of the absorption laws [Theorem 11.3.6(c)] does not use the associative laws.

Therefore, the absorption laws follow from the definition of Boolean algebra (Definition 11.3.1) without the associative laws.

We show that a + (b + c) = (a + b) + c by showing that each side of the preceding equation is equal to ((a + b) + c)(a + (b + c)).

Throughout the proof we make implicit use of the commutative laws, and we use the equations a(a + (b + c)) = a, b(a + (b + c)) = b, c(a + (b + c) = c, a((a + b) + c) = a, b((a + b) + c) = b, c((a + b) + c) = c.

The first equation follows immediately from the absorption laws. To prove the second equation, we first use the distributive law and then the absorption laws to obtain b(a + (b + c)) = ba + b(b + c) = ba + b = b.

The proof of the third equation is similar to the proof of the second equation, and the proofs of the last three equations are similar to the proofs of the first three equations.

We have ((a + b) + c)(a + (b + c))

= = = =

((a + b) + c)a + ((a + b) + c)(b + c) a + ((a + b) + c)(b + c) a + (((a + b) + c)b + ((a + b) + c)c) a + (b + c)

distributive law 4th equation distributive law 5th and 6th equations.

((a + b) + c)(a + (b + c))

= = = =

(a + b)(a + (b + c)) + c(a + (b + c)) (a + b)(a + (b + c)) + c (a(a + (b + c)) + b(a + (b + c))) + c (a + b) + c

distributive law 3rd equation distributive law 1st and 2nd equations.

Also

Therefore, the associative law holds for +. By duality, the associative law holds for ·.

Problem-Solving Corner: Boolean Algebras 1. If x ∈ R, then x ∈ [x, x + 1) ⊆ R, thus R is open.

If ∅ is not open, then ∃ x ∈ ∅ such that for all a, b ∈ R, with a < b, x ∈ [a, b) ⊆ ∅. But ∀ x ∈ R, x ∈ / ∅; therefore, ∅ is open.

Since both R and ∅ are open, R = ∅ and ∅ = R are open. Therefore R and ∅ are clopen.

2. Let X = [1, 2) ∪ [3, 5). If x ∈ X, either x ∈ [1, 2) or x ∈ [3, 5). If x ∈ [1, 2), we have x ∈ [1, 2) ⊆ X. If x ∈ [3, 5), we have x ∈ [3, 5) ⊆ X. Therefore X is open.

If x ∈ / X, either x < 1, x ∈ [2, 3), or x ≥ 5. If x < 1, then x ∈ [x, 1) ⊆ X. If x ∈ [2, 3), then x ∈ [2, 3) ⊆ X. If x ≥ 5, then x ∈ [5, x + 1) ⊆ X. Therefore X is open and therefore X ∈ A. Copyright c 2018 Pearson Education, Inc.

193

SOLUTIONS 3. To show that X is open, let x ∈ X. Then x ∈ [x, 10) ⊆ X. Therefore X is open.

Suppose, by way of contradiction, that X is open. Since 3 ∈ X, there exist a and b with a < b such that 3 ∈ [a, b) ⊆ X. Thus [3, b) ⊆ X. If b ≥ 10, then 9 ∈ [3, b) ∩ X, which is a contradiction. If b < 10, then (3 + b)/2 ∈ [3, b) ∩ X, which is also a contradiction. Therefore, X is not open.

4. By definition, if X ∈ A, then X and X are both open. Since X and X = X are both open, X ∈ A. 5. Suppose that X and Y are open. Let x ∈ X ∪ Y . If x ∈ X, since X is open, there exists a, b with a < b such that x ∈ [a, b) ⊆ X. Thus x ∈ [a, b) ⊆ X ∪ Y . Similarly, if x ∈ Y , there exists a, b with a < b such that x ∈ [a, b) ⊆ X ∪ Y . Therefore X ∪ Y is open.

6. Suppose that X and Y are open. Let x ∈ X ∩Y . Then x ∈ X and y ∈ Y . Thus there exist a, b with a < b such that x ∈ [a, b) ⊆ X and there exist c < d such that x ∈ [c, d) ⊆ Y . Thus x ∈ [a, b) ∩ [c, d) ⊆ X ∩ Y .

Noting that c < b and a < d {because [a, b) ∩ [c, d) 6= ∅}, if we let e = max{a, c} and f = min{b, d}, then x ∈ [e, f) ⊆ X ∩ Y and hence X ∩ Y is open.

7. Suppose that X, Y ∈ A. Then X and Y are open. By Exercise 6, X ∩ Y is open. Again, since X, Y ∈ A, X and Y are open. By Exercise 5, X ∪ Y is open. Since X ∪ Y = X ∩ Y , X ∩ Y is open. Therefore X ∩ Y ∈ A. 8. Example 11.3.3 shows that (P(R), ∪, ∩, , ∅, R) is a Boolean algebra. Since A is a subset of P(R), by the result of this Problem-Solving Corner, it is enough to show that R ∈ A and for all X, Y ∈ A, X ∩ Y ∈ A. Exercise 1 shows that R ∈ A. Let X, Y ∈ A. By Exercise 4, Y ∈ A. By Exercise 7, X ∩ Y ∈ A. Therefore (A, ∪, ∩, , ∅, R) is a Boolean algebra.

9. No; Exercises 1 and 5 remain true, but the others are false.

A counterexample for Exercise 4 is X = (−∞, 0) ∪ (0, ∞). A counterexample for Exercises 6 and 7 is X = [0, 1], Y = [1, 2]. A counterexample for Exercise 8 is X = [0, 1], Y = (−∞, 1) ∪ (2, ∞), since it is false that for all x ∈ X ∩ Y , there exists a, b with a < b such that x ∈ [a, b] ⊆ X ∩ Y .

Section 11.4 In the following hints, a ∧ b is written ab.

2. xy ∨ x y x

y

H Hc

H H c

3. xyz ∨ xyz ∨ xy z ∨ x yz ∨ x y z

5. xyz ∨ xyz ∨ xy z ∨ xyz ∨ xyz ∨ x yz 6. xyz ∨ xyz ∨ xyz ∨ xyz ∨ x yz ∨ x y z 8. xy z ∨ xyz ∨ xyz ∨ x yz 9. wxyz ∨ wxyz ∨ wx y z ∨ wxyz ∨ w xyz Copyright c 2018 Pearson Education, Inc.

194

SOLUTIONS

12. xy ∨ xy 13. xyz ∨ xyz ∨ xyz ∨ xy z ∨ xyz 15. xyz ∨ xyz ∨ xyz ∨ xy z ∨ x yz ∨ x y z 16. x yz ∨ x y z ∨ xyz 18. wxyz ∨ wxyz ∨ wxyz ∨ wxy z ∨ wxyz ∨ wxyz ∨ wxyz ∨ w xyz ∨ w x yz ∨ wxyz 19. x yz ∨ xy z ∨ x y z ∨ xyz 21. 22

n

24. Let f: Z2n → Z2 . Let A1 , . . . , , Ak denote the elements Ai of Z2n for which f(Ai ) = 0. For each Ai = (a1 , . . . , an ), set mi = y1 ∨ . . . ∨ yn where

xi if ai = 0 yi = xi if ai = 1.

Then f(x1 , . . . , xn ) = m1 ∧ m2 ∧ . . . ∧ mk .

26. (For Exercise 13) (x ∨ y ∨ z)(x ∨ y ∨ z)(x ∨ y ∨ z) 27. If f(x1 , . . . , xn ) = m1 ∨ . . . ∨ mk , then f(x1 , . . . , xn ) = m1 m2 · · · mk . Since each mi = y1 y2 · · · yn , where each yj is either xj or xj , mi = y1 ∨ y2 ∨ . . . ∨ yn . Thus f(x1 , . . . , xn ) is expressed as the conjunction of maxterms.

29. If j > k, then some term m0t does not occur in the expansion m1 ∨ . . . ∨ mk . Choose xi ∈ Z2 so that m0t = 1. Show that mi = 0 for i = 1, . . . , n. Conclude that j ≤ k. Similarly, j ≥ k. Therefore, j = k.

Give a similar argument to show that each mi is equal to some m0t .

Section 11.5 3. A combinatorial circuit consisting only of NOT gates would have as many outputs as inputs.

4. A combinatorial circuit consisting only of OR gates would output 0 when all inputs are 0.

7. xy = (x ↑ y) ↑ (x ↑ y) 8. False. Take x = 1 and y = z = 0.

10.

y1 y2 y3 y4

= = = =

x1 ∧ x2 (x1 ∧ x2 ) ∨ x3 ((x1 ∧ x2 ) ∨ x3 ) ∧ x1 ∧ x2 ∧ x3 ∧ x4 x3 ∧ x4

11. y1 = x1 , y2 = x1 ∨ x2 , y3 = x2 x3 (x1 ∨ x2 )x1 16. See the solution to Exercise 15.

18. x ↓ y = ((x ↑ x) ↑ (y ↑ y)) ↑ ((x ↑ x) ↑ (y ↑ y)) 19.

x1 1 1 0 0

x2 1 0 1 0

x1 ↓ x2 0 0 0 1 Copyright c 2018 Pearson Education, Inc.

195

SOLUTIONS 21. (For Exercise 3) Write xy ∨ xz ∨ x y = x(y ∨ z) ∨ x y = x ∨ y ∨ z ∨ x ∨ y, which gives c

x

y

z

c

c

c

c

c

c 24. See the solution to Exercise 23.

26. The standard multiplication algorithm gives x2 x1 y2 y1 x2 y1 x1 y1 x2 y2 x1 y2 z4 z3 z2 z1 This computation can be performed by the following circuit

x1 y1

x2

y2

29. 00101

z1

s

Half Adder c

z2

s

z3

Half Adder c

z4

30. 100101010

32. The simplest approach is to simply rewrite each of the properties in 11.3.1 using the star operator. Notice that in some cases the dual properties can be given in one star expression. For example, x ∗ ((y ∗ z) ∗ (y ∗ z)) = ((x ∗ y) ∗ (x ∗ y)) ∗ z Copyright c 2018 Pearson Education, Inc.

196

SOLUTIONS is sufficient to prove both associative laws.

J. G. P. Nicod in “A reduction in the number of primitive propositions of logic,” Proceedings of the Cambridge Philosophical Society, 19 (1916), 32–40, has shown that the following single axiom is sufficient to establish all of the properties of a Boolean algebra (a ∗ (b ∗ c)) ∗ ((d ∗ (d ∗ d)) ∗ ((e ∗ b) ∗ ((a ∗ e) ∗ (a ∗ e)))) = 1.

35. We use induction on the number n of occurrences of ↔ in B(x, y) to prove parts a and b simultaneously.

For the Basis Step, suppose that there are 0 occurrences of ↔ in B(x, y). If B(x, y) contains an even number of x’s, then either B(x, y) = 0, B(x, y) = 1, or B(x, y) = y. In all cases, B(x, y) = B(x, y) for all x and y. If B(x, y) contains an odd number of x’s, then B(x, y) = x. In this case, B(x, y) = B(x, y) for all x and y.

Now suppose that there are n occurrences of ↔ in B(x, y). Then B(x, y) = (B1 (x, y) ↔ B2 (x, y)), where each of B1 (x, y) and B2 (x, y) contains fewer than n occurrences of ↔.

Suppose first that B(x, y) contains an even number of x’s. If B1 (x, y) contains an even number of x’s, so does B2 (x, y). By the inductive assumption, B(x, y) = (B1 (x, y) ↔ B2 (x, y)) = (B1 (x, y) ↔ B2 (x, y)) = B(x, y).

If B1 (x, y) contains an odd number of x’s, so does B2 (x, y). By the inductive assumption, B(x, y) = (B1 (x, y) ↔ B2 (x, y)) = (B1 (x, y) ↔ B2 (x, y)) = (B1 (x, y) ↔ B2 (x, y)) = B(x, y).

Now suppose that B(x, y) contains an odd number of x’s. Then one of B1 (x, y) or B2 (x, y) contains an odd number of x’s. We may assume that B1 (x, y) contains an odd number of x’s. Then B2 (x, y) contains an even number of x’s. By the inductive assumption, B(x, y) = (B1 (x, y) ↔ B2 (x, y)) = (B1 (x, y) ↔ B2 (x, y)) = B(x, y).

The Inductive Step is complete, and parts a and b are proved.

Using parts a and b, we may show that there is no Boolean expression using only ↔ that computes the function x 1 1 0 0

y 1 0 1 0

f(x, y) 0 1 0 0

For suppose that there is a Boolean function B(x, y) using only the ↔ operator with f(x, y) = B(x, y). If B(x, y) contains an even number of x’s, part a tells us that we must have B(0, 0) = B(1, 0), which is not the case. If B(x, y) contains an odd number of x’s, part b tells us that we must have B(0, 1) = B(1, 1), which is not the case. Therefore {↔} is not functionally complete.

Section 12.1 2.

3. b/0 a/0 0

a/0 1

a/1 b/1

0

1

b/1 a/0

b/0

a/1

2

Copyright c 2018 Pearson Education, Inc.

b/1

197

SOLUTIONS 5. b/1 a/1 a/2

0

1

a/2 c/1

c/2

c/2

b/0

c/0 b/0

b/0 3

2

a/1

7. I = {a, b}; O = {0, 1, 2}; S = {σ0 , σ1 , σ2}; initial state = σ0 S\I σ0 σ1 σ2

a σ1 σ1 σ1

b σ0 σ0 σ0

a 0 2 0

b 2 1 1

8. I = {a, b}; O = {0, 1}, S = {A, B, C}; initial state = A S\I A B C

a A A C

b B C A

a 0 0 1

b 1 1 0

10. I = {a, b, c}; O = {0, 1, 2}; S = {A, B, C, D}; initial state = B S\I A B C D

a B A A D

b A D C C

12. 0100

c C D D A

a 1 2 0 2

b 0 0 1 2

c 2 0 2 0

13. 0101100

15. 121121

16. 011

18. 001110001

19. 010000000001 22.

23.

0/0

0/0

0/1

0/0 No 1's

1/0

One 1

1/1

Two or more 1's

0/0

1/0 0

1/1

1/0 1

1/1

25.

1/0

1/1

0/1 0

0/0

0/1 1

0/0

2

1/0

Copyright c 2018 Pearson Education, Inc.

2

198

SOLUTIONS

26.

1/0

1/0

1/1 1/1

0

1

2

0/0

3

0/1

0/0

0/0

29. Suppose that such a finite-state machine M exists and has m states. Let X and Y each be 1 followed by m + 2 0’s. Then the sequence 00, 00, . . ., 00, 00, 00, . . . , 00 {z }| {z } | m+2

m+3

is input to M . The product of X and Y is 1 followed by 2m + 4 0’s. The output is 0, 0, . . . , 0, 1, 0. | {z } 2m+4

After 11 is input, a sequence of m + 1 00’s is input and the output is 0 each time. Since there are only m states, we must return to a state that we previously visited. That is, the path in the transition diagram contains a cycle. Since the input is constant (00), we must remain on this cycle. Therefore, we continue outputting 0’s and we never output 1.

Section 12.2 2. All incoming edges to σ0 output 0, all incoming edges to σ1 output 1, and all incoming edges to σ2 output 0; hence, the finite-state machine is a finite-state automaton. b a a

0

b

1

2

b

a

3.

5. a

0

a

1

0

a/1

b/0 1

b/1

a

b

b

b

2

a

b

3

a/0

b/0

a/0

2

b/1

Copyright c 2018 Pearson Education, Inc.

a/1

3

199

SOLUTIONS b/0

6.

a/0

b/1

0

b

8.

1

2

a/0

0

b/0 c

b

a a

0

1

a

a/0

9.

a c

c

a

2

Correction: σ0 is b accepting; σ1 is not.

a c

b

3

11. 1,7,9

b 1

2

b

b a

12. The following must hold for all states σ: All entries in the output table corresponding to occurrences of σ in the next-state table must be identical.

14. Not accepted

15. Accepted

17. Accepted

19. If a string α, which ends bb is input, no matter which state we are in prior to bb, we will end at state σ2 , as can be seen by checking the three possibilities. Since σ2 is accepting, α is accepted.

Suppose that α is accepted by Figure 12.2.5. We end in state σ2 . Thus the last character in α is b. There is at least one character before b. If the last two characters are ab and we are in state α just before the a, by checking the three possibilities for σ, we can show that we will not end in σ2 . Therefore, the last two characters are bb.

22.

a

a

a

One or more b's

b

No b's

23.

a

0

b

1

a b

2

b b

25.

26.

a

0

1

b b

b

b

a

a

a

2

b

No a's

a

One a

a

Two or more a's b

Copyright c 2018 Pearson Education, Inc.

200 28.

b a

0

SOLUTIONS

a b

1

b

2

a

3

4

a a

b

b

29. a

a b

0

b

1

2

a

b

31. b a a

1

2

b

b

3

b

4

a

5

a

a

b

b

6

b

7

a

a

33. If a string consisting only of b’s is input to either finite-state automaton, it is accepted. If a string contains an a, in either finite-state automaton we move to a nonaccepting state. In neither finite-state automaton is there an edge from a nonaccepting state to an accepting state. Thus once an a is encountered, both finite-state automata reject the string. Therefore, the set of strings accepted by each finite-state automaton is the same–namely, the set of strings over {a, b} that do not contain an a.

34. If L is empty, the finite-state automaton a

b

accepts L. If L consists of the null string, the finite-state automaton a

b

a

b

accepts L. If L consists of one nonnull string x1 x2 . . . xk , the finite-state automaton Copyright c 2018 Pearson Education, Inc.

201

SOLUTIONS

x1

x2

xk y2

yk

a

y1 b

a

b

where yi =

a b

if xi = b if xi = a

accepts L. If L = {s1 , . . . , sn }, we have shown how to construct finite-state automata that accept Li = {si }, i = 1, . . . , n. Exercise 37 shows how to construct a finite-state automaton that accepts L = L1 ∪ L2 ∪ . . .∪ Ln .

36. The argument is similar to that given in Exercise 37.

37. Suppose that x1 . . . xn ∈ L1 . Then there exist states s10 , . . . , s1n satisfying s10 = f1 (s1,i−1 , xi ) = s1n ∈

σ1 ; s1i for i = 1, . . . , n; A1 .

Define s20 s2i si

= = =

σ2 ; f2 (s2,i−1 , xi ) (s1i , s2i )

for i = 1, . . . , n; for i = 0, . . . , n.

Now s0 f(si−1 , xi ) sn

= = =

σ; (s1i , s2i ) for i = 1, . . . , n; (s1n , s2n ) ∈ A.

Thus L1 ⊆ Ac(A). Similarly, L2 ⊆ Ac(A). Therefore, L1 ∪ L2 ⊆ Ac(A).

A similar kind of argument may be used to show that Ac(A) ⊆ L1 ∪ L2 .

39. Use Exercises 36 and 37.

Section 12.3 2. Context-sensitive

3. None

5. Regular, context-free, context-sensitive

8. σ ⇒ AAB ⇒ AaaB ⇒ ABaaB ⇒ ABaab ⇒ ABBaab ⇒ aaBBaab ⇒ aabBaab ⇒ aabbaab 9. σ ⇒ AB ⇒ aAB ⇒ aABb ⇒ aBAb ⇒ abAb ⇒ abab Copyright c 2018 Pearson Education, Inc.

6. None

202

SOLUTIONS

11. < S > ⇒ a < A > ⇒ ab < B > ⇒ aba < S > ⇒ abaa < A > ⇒ abaab < B > ⇒ abaabb < A > ⇒ abaabba < S > ⇒ abaabbab < S > ⇒ abaabbabb < S > ⇒ abaabbabba 13. The productions σ → bσ, A → bA, and σ → b generate any number of b’s. If these are omitted, the only derivations possible are σ ⇒ aA ⇒ aaσ ⇒ · · · ⇒ (aa)n σ ⇒ (aa)n aA ⇒ (aa)n+1 .

14. An accepted string is of the form x1 x2 . . . xn , where x1 = b∗ a and x2 , . . . , xn are any of ab∗ a, (bb)∗ , or bab∗a.

16. S → aS, S → bA, A → bA, A → aB, B → aB, B → bB, B → a, B → b, A → a 17. S → aS, S → bS, S → bA, A → a 19.

<digit> <nonzero digit> <integer> <signed integer> <unsigned integer> <digit string>

::= ::= ::= ::= ::= ::=

0|1|2|3|4|5|6|7|8|9 1|2|3|4|5|6|7|8|9 <signed integer> | <unsigned integer> +<unsigned integer> | −< unsigned integer> <digit> | <nonzero digit><digit string> <digit> | <digit><digit string>

20. <float number> ::= . <integer> | <integer> . | <integer> . <integer> 22. S → aS, S → bS, S → λ 23.

<BOOL>

::=

0 | 1 | X1 | X2 | . . . | Xn | (<BOOL>) | < BOOL > | <BOOL> ∨ <BOOL> | <BOOL> ∧ <BOOL>

26. This grammar does not generate L; aabb is a counterexample.

27. This grammar does generate L. Every string that the grammar generates is in L, since anytime an a is generated, a b is generated and vice versa.

To show that every string s ∈ L is generated, we argue by induction on the length of s. If the length of s is 0, s is generated. Suppose that s ∈ L and that s starts with a. (The argument is similar if s starts with b.)

Suppose first that s ends with b. Then s = atb, where t is a string shorter than s and t ∈ L. By the inductive assumption, S ⇒ t. But now S ⇒ aSb ⇒ atb = s is a derivation of s. Therefore s is in the language generated by the grammar.

Suppose that s ends with a. Then s = tu, where t and u are each shorter strings than s and t, u ∈ L. By the inductive assumption, S ⇒ t and S ⇒ u. But now S ⇒ SS ⇒ tu = s is a derivation of s. Again s is in the language generated by the grammar.

29. This grammar does not generate L; abba is a counterexample.

30. This grammar does not generate L; aabbbbaa is a counterexample.

34. The language is generated by the context-free language S → AC, C → cC, C → c, A → aAb, A → ab Copyright c 2018 Pearson Education, Inc.

203

SOLUTIONS 36.

S A D + −

→ → → → →

−AD+SDS+DA− | s +SD−ADA−DS+ | a D|d + −

a and s are ignored when drawing the figure.

37. In order, the notes (all quarter notes) are: C, D, F-sharp, E, F-sharp, D, C.

Section 12.4 2.

3.

b b

b 0

1

0

a

a

1

a

b b

b

b a b

2

3

5. b a

0

1

b

c

c a

b a c

2

b

3

7. I = {a, b}, S = {A, B, C}, A = {A, C}, initial state = A S\I A B C

a {A, C} {C} ∅

b {B} {B, C} ∅

8. I = {a, b}, S = {σ0 , σ1 , σ2 , σ3 }, A = {σ3 }, initial state = σ0 S\I σ0 σ1 σ2 σ3

a ∅ {σ2 } ∅ {σ3 }

2

b {σ1 } ∅ {σ3 } {σ3 } Copyright c 2018 Pearson Education, Inc.

204

SOLUTIONS

10. I = {a, b}, S = {σ0 , σ1 , σ2 , σ3 }, A = {σ3 }, initial state = σ0 S\I σ0 σ1 σ2 σ3

a {σ0 } {σ2 } ∅ ∅

b {σ0 , σ1} ∅ {σ3 } ∅

12. Yes. The path (σ, σ, σ, C, C, C, F ) represents bbabbb and ends at an accepting state.

13. (For Exercise 1, Section 12.3) b

b a

a

A

F

a b

15. No. For the first three characters, aaa, either we follow the path (σ, σ, σ, C) or the path (σ, σ, σ, σ). From C, on the first path, the next two moves are determined and we remain at C. But now we cannot move on the final a. From the last σ on the second path, the next move is determined and we move to D. But now we cannot move on the next b.

16. The string α is of the form bn abm , where n ≥ 0 and m ≥ 1. A path representing this string terminating at F is (σ n+1 C m F ). Any path starting at σ terminating at F is of the form (σ n+1 C m F ) and thus represents bn abm , where n ≥ 0 and m ≥ 1.

18. {an bm | n ≥ 1, m ≥ 0} ∪ {ban bm | n ≥ 1, m ≥ 0} 19. To reach σ3 on a path from σ0 , we must have ended bab. Any string that ends bab is accepted, since we can remain at σ0 until we encounter the final bab, at which time we move successively to σ1 , σ2 , and σ3 .

22. a

a b a

0

1

b

b

2

b

3

a

b

4

23. a

0

a

b

1

b

b

2

b

3

a 4

Copyright c 2018 Pearson Education, Inc.

205

SOLUTIONS 25. a a

0

b

1

b

2

a

3

b

4

5

b

26. b a

0

1

2

a a

28. a

b

0

b

1

2

29. Use Exercise 36, Section 12.2.

Section 12.5 3.

4. b a S

a

S

b B

A,F a

a

b

b

b

6. a

b

b b

b

a

a

A,B a S,F

a

B,C

b

b b

B,C,S,F a A,B,C,S,F

A a

a

b

b

a A,S,F

Copyright c 2018 Pearson Education, Inc.

a b

a b

a

S

A,B

S,F

B b a

206

SOLUTIONS

8. a

*

*

b

a

,C b

a

D

a

C,D

b

b

a a

b

C

b

9. Exercise 19, Section 12.2, shows that Figure 12.5.4 accepts precisely the strings over {a, b} that end bb.

We may now use Example 12.5.7 to conclude that Figure 12.5.5 accepts precisely the strings over {a, b} that start bb.

12. a b

b

0

1

a

2

b a

13. a

a

0

b

1

b a 4

a

b

a a

b

2

3

b b

15. To find the nondeterministic finite-state automaton that accepts Ac(A)+ , allow an edge in A that terminates on an accepting state to optionally return to the starting state: b b 0

a

b 1

a

2

b a

Copyright c 2018 Pearson Education, Inc.

207

SOLUTIONS 16. a b a

0

1

b

b b

a

a

b

b

2

18.

a

a

3

b a a

0

1

b a b

a b

0

1

2

a

b a

19. Allow an edge in A1 that terminates on an accepting state in A1 to terminate optionally on the starting state in A2 . The accepting states of the nondeterministic finite-state automaton are the accepting states of A2 : b a 0

b 1

a

2

b

a b a 0

a

b

a

1

2

b

a

b

3

b

21. See the hint for Exercise 19.

23. Allow any terminating production, alternatively, to return to the start: <S> <A> <B>

::= ::= ::=

b<S> | a<A> | a<S> | a a<S> | b<B> b<A> | a<S> | b<S> | b

24. Replace any terminating production of L1 with the starting symbol of L2 : Copyright c 2018 Pearson Education, Inc.

208

SOLUTIONS <S> <A> <B> <σ> <C>

::= ::= ::= ::= ::=

b<S> | a<A> | a<σ> a<S> | b<B> b<A> | a<S> | b<σ> b<σ> | a<C> b<C> | b

26. Use Exercises 35–37, Section 12.2, and the methods of Exercises 23 and 24.

27. Consider L1

=

L2

=

{an bn ck | n, k ∈ {1, 2, . . .}}

{ak bn cn | n, k ∈ {1, 2, . . .}}.

Appendix A

−9 −8 −2 4 3.

4.

6.

2 14 −12 −2  

46 −35 9 8 5 −11 −6   7.

10. −18 −20 11. −14 16 0 18 −8 23 −2   2a + 4c + e 2b + 4d + f 13.  6a + 9c + 3e 6b + 9d + 3f  15. x = 1, y = −2, z = 8 a − c + 6c b − d + 6f 5 7 7 −7 10 −1

−1 −6 0 −4

16. x = 38/5, y = 0, z = −81/5, w = −304/5

2 1 1 −1 1 −1 2 18. = I2 = 1 1 −1 2 −1 2 1 19. Let

1 1

a b c d

.

e g

f h

.

af + bh cf + dh

= AB =

A= Suppose that AB = I2 for some 2 × 2 matrix B= We have

ae + bg ce + dg

1 0 0 1

.

Now (ad − bc)(eh − fg)

= (ae + bg)(cf + dh) − (af + bh)(ce + dg) = 1 · 1 − 0 · 0 = 1.

Thus ad − bc 6= 0.

If s = ad − bc 6= 0, setting B=

d/s −b/s −c/s a/s

,

we obtain AB = I2 = BA. Copyright c 2018 Pearson Education, Inc.

209

SOLUTIONS 21. Let A = (aij ), B = (bpq ), AB = (crs ), B T AT = (duv ), AT = (a0ji ), B T = (b0qp ), (AB)T = (c0sr ).

Then duv =

k X

b0uxa0xv =

x=1

k X

avxbxu = c0uv .

x=1

Appendix B 2. 4y − 6a 6.

3. −2a + 14b

56x2 − 20xb − 4b2 9

5.

1 9. 81

8x − 4b 7x + b 16x − 8b 7x + b 9x − 9b − = − = 2 4 4 4 4 10. 81

12. 1

13. 1

15. 5n + 4 · 5n = 5n (1 + 4) = 5n · 5 = 5n+1 17. x2 + x − 12 23. x2 − a2 29. (x + 9)(x − 9) 33. (2x − 3)2 38.

18. 6x2 + x − 12 24. 4x2 − 9

26. (x − 5)(x + 2)

30. (x + 2b)(x − 2b)

21. 9x2 + 24x + 16 27. (x + 3)2

32. (3x + 5)(2x − 3)

35. (3a + 2b)(3a − 2b)

n(n + 1)(2n + 1) + (n + 1)2 6

= = = =

39.

20. x2 − 8x + 16

n 1 + 2n + 1 (2n + 1)(2n + 3)

36. 2(3x − 5)(2x − 5)

n(2n + 1) (n + 1) + (n + 1) 6

n(2n + 1) + 6(n + 1) (n + 1) 6 2

2n + 7n + 6 (n + 1) 6

(n + 2)(2n + 3) (n + 1) 6

=

(n + 1)(n + 2)(2n + 3) 6

=

n(2n + 3) + 1 2n2 + 3n + 1 = (2n + 1)(2n + 3) (2n + 1)(2n + 3)

=

(n + 1)(2n + 1) n+1 = (2n + 1)(2n + 3) 2n + 3

41. 2r(n − 1)r n−1 − r 2 (n − 2)r n−2 = 2(n − 1)r n − (n − 2)r n = [2(n − 1) − (n − 2)]r n = nr n 43. (3x − 2)(2x − 1) = 0, x = 23 , 12 p √ √ √ −(−4) ± (−4)2 − 4 · 2 · 1 4± 8 4±2 2 2± 2 44. x = = = = 2·2 4 4 2 46. −9 > x

47. − 12 11 < x Copyright c 2018 Pearson Education, Inc.

210

SOLUTIONS

49. We have (1 + ax)(1 + x)

=

1 + ax + x + ax2

= ≥

1 + (a + 1)x + ax2 1 + (a + 1)x

because ax2 ≥ 0.

50. We may multiply 5 = 2.5 > 2.25 = 2 by

2 3 2

n−2 3 2

to obtain

n−2 n−2 2 n 3 5 3 3 3 > = .

2 2 2 2 2

52. The given inequality is equivalent to 0 < 4n + 1, which is true for all n ≥ 1.

53. 6n2 + 4n + 1 ≤ 6n2 + 4n2 + n2 = 11n2 55. −7

56. 1

63. −4.736965594

58. 1000 65. −0.603845495

60. 4.906890596 66. 99.4300753

61. 15.84962501 69. 1.336810137

70. 2.069864223

Appendix C 2. First large is set to 8 and i is set to 2. Since i ≤ n is true, the body of the while loop executes. Since si > large is false, the value of large does not change. i is set to 3 and the while loop executes again.

Since si > large is false, the value of large does not change. i is set to 4 and the while loop executes again.

Since si > large is false, the value of large does not change. i is set to 5 and the while loop executes again.

Since i ≤ n is false, the while loop terminates. The value of large is 8, the largest element in the sequence.

3. First large is set to 1 and i is set to 2. Since i ≤ n is true, the body of the while loop executes. Since si > large is false, the value of large does not change. i is set to 3 and the while loop executes again.

Since si > large is false, the value of large does not change. i is set to 4 and the while loop executes again.

Since si > large is false, the value of large does not change. i is set to 5 and the while loop executes again.

Since i ≤ n is false, the while loop terminates. The value of large is 1, the largest element in the sequence.

5. First x is set to 4. Since b > x is false, x = b is not executed. Since c > x is false, x = c also is not executed. Thus x is the largest of the numbers a, b, and c.

6. First x is set to 8. Since b > x is false, x = b is not executed. Since c > x is false, x = c also is not executed. Thus x is the largest of the numbers a, b, and c.

Copyright c 2018 Pearson Education, Inc.

211

SOLUTIONS 8. max (a, b) { if (a > b) return a else return b } 9. swap(a, b) { temp = a a=b b = temp } 11. negatives(s, n) { for i = 1 to n if (si < 0) println(si ) } 12. find val (s, n, val) { for i = 1 to n if (si == val ) println(i) } 14. alternate(s, n) { i=1 while (i ≤ n) { println(si ) i = i+2 } }

Copyright c 2018 Pearson Education, Inc.

SOLUTIONS MANUAL

D ISCRETE M ATHEMATICS E IGHTH E DITION Richard Johnsonbaugh