SOLUTIONS MANUAL for Discrete Mathematics with Ducks 2nd Edition by Sarah Marie Belcastro

SOLUTIONS MANUAL Discrete Mathematics with Ducks 2nd Edition by Sarah Marie Belcastro Chapter 1

Counting and Proofs 1.2 Try This! Let’s Count 1. 2 + 6 + 1 + 1 = 10.

2. 3 · 5 = 15.

3. Six as there are five different ice-cream containers.

4. 5 · 5 · 5 = 125 orders because one might want all flavors the same.

5. This one is a pain, intentionally. First, 51 + 2 52 + 53 = 5+ 20+ 10 = 35 (the 2 52 is because it matters which flavor is ordered with 1 quart and which is ordered with 2 quarts), then just 5 4 4 4 4 = 10, then + + = 4 + 6 + 4 = 14 and = 4. Of course, students do not yet 3 1 2 3 3 know this notation, so they will likely write out solutions exhaustively and at length.

6. 3 + 2 + 1 = 6.

7. Three, as there are only two colors. To assure 7 monochromatic armbands, 13 armbands must be drawn (worst case: b-g-b-g-b-g-b-g-b-g-b-g-b/g).

8. 5 · 3 · 6 = 90.

9.

(a) n − 1.

(b) n − 1, then n − 2. (c) n − i.

(d) (n − 1) + (n − 2) + . . . 1 = n(n − 1)/2.

10.

(a) 3 · k.

(b) k + 1 as there are k containers.

11.

(a) The sum of their sizes. (b) The sum of their sizes! (c) Yes, the WEBS problem and the Frisbee tournament problems. (d) m + m + · · · + m = n · m.

12. m · k; yes, related to the ice-cream and snack-box problems.

2

1.9. Problems That Use Counting or Proofs

3

1.9 Problems That Use Counting or Proofs 1. 4 · 7 = 28.

2. 28 · 3 · 2 = 168.

3. An even number n can be written as 2k for some integer k. So, the sum of two even numbers is n1 + n2 = 2k1 + 2k2 = 2(k1 + k2 ) = 2k3 , which is even.

4. An odd number n can be written as 2k + 1 for some integer k. So, the sum of two odd numbers is n1 + n2 = 2k1 + 1 + 2k2 + 1 = 2k1 + 2k2 + 2 = 2(k1 + k2 + 1) = 2k3 , which is even.

5. 28 by the product principle using slots.

6. 5 · 4 + 3 · 3 = 29.

7. A binary number n that ends in 0 is the sum of some powers of 2, e.g., n = 2a + 2b + · · · + 2c where the lowest possible power of 2 is 21 = 2. Then we may write n = 2(2a−1 + 2b−1 + · · · + 2c−1 ) and so n is even.

8. An odd number n can be written as 2k + 1 for some integer k. To write n as a binary number, we must write it as a sum of powers of 2. This sum must have lowest power 0 (as in 20 ) or else every term of the sum will be divisible by 2. That means that the binary representation will end in a 1 = 20 .

9. There are 11 · 3 · 5 · k = 165k choices and 1,095 days, so k ≥ 6.63. But, of course, k must be an integer . . . so there are at least seven possible drinks.

10. Two consecutive perfect squares are n2 and (n + 1)2 . n2 − (n + 1)2 = n2 − n2 − 2n − 1 = −2n − 1 = −2n + 1 − 2 = 2(−n − 1) + 1, which is odd.

11. 11 = 10 + 1; and, 3 because there are only two different first digits (1, 2).

12. Counterexample: 32 + 42 = 25 is odd.

13. 15 · 8 = 120.

14. 15 · 6 = 90.

15. 5 + 6 + 3 + 4 = 18.

16. Nope. 2 + 3 = 5 is odd. (However, if we exclude 2, then the remaining primes are odd and so the sum of any two of them will be even.)

17. There are 1,440 minutes per day and so (1,440) · 23 = 960 90-second intervals. 1,042 > 960 so two flights must take off within the same interval.

18. Write n = 2k for some integer k. Then (−1)n = (−1)2k = ((−1)2 )k = 1k = 1.

19. 107 .

4

1. Counting and Proofs

20. 94 = 6561.

21. An odd number 2k + 1 times an odd number 2r + 1 is (2r + 1)(2k + 1) = 4kr + 2k + 2r + 1 = 2(2kr + k + r) + 1 and thus is odd.

22.

(a) Check 2,000,000 versus 220 = 1, 048, 576: not enough subsets. (b) By trial and error, see that 2,100,000 > 221 = 2,097,152 but 2,200,000 < 222 = 4,194,304.

23. For even n = 2k, 3(2k)3 + (2k) + 5 = 2(3 · 4k3 ) + 2k + 2 · 2 + 1 = 2(3 · 4k3 + k + 2) + 1, which is odd.

For odd n = 2k + 1, 3(2k + 1)3 + (2k + 1) + 5 = odd + odd + odd = odd.

24. 99 · 2 + 1 = 199; 99 · 3 + 1 = 298; 99 · k + 1.

25. Green and brown socks: three for the first pair, then another two for the next pair, and so forth and so on, so 50 · 2 + 1 = 101.

Three colors of socks: four for the first pair, then another two for the second pair (using the leftover two socks), and so forth and so on: 50 · 2 + 2 = 102.

k colors of socks: k + 1 for the first pair, then another two for the second pair (using the leftover k − 1 socks), and so forth and so on: 50 · 2 + k − 1 = 99 + k.

26. All the usual primes are still prime, and 1 is also prime.

If 1 were prime, we would no longer have unique factorization of natural numbers—for example, 6 = 2 · 3 = 2 · 1 · 1 · 1 · 3 = 1 · 2 · 3 . . . .

27. There are no prime numbers with this definition, because 1 is a positive divisor of every integer, and 1 is not included in this definition!

It would be silly to have a class of numbers with no members.

28. There are 3 · 2 = 6 regular options plus 2 · 1 Fancy Tips, so 6 + 2 = 8.

29. By the pigeonhole principle, if each bin had only 13 skeins then you’d only have 104 miniskeins.

30.

(a) Note that the numbers 1–6, paired with the numbers 12–7, form six sums of 13. If no pair of the 7 numbers sums to 13, then all the numbers must be in one of these two subsets—but that’s only 6 numbers, so the seventh number must come from the other set and pair to make a sum of 13. (b) Similarly, the numbers 12–7, paired with the numbers 6–1, form differences of 6. The same argument applies. (c) Now we look at the numbers 1–3 with 4–6, and 7–9 with 10–12 to get six pairs with differences of 3. The same argument as above applies.

31. We have 7 numbers in the range 1–20. There are 27 = 128 possible subsets, and the highest possible sum is 14 + 15 + 16 + 17 + 18 + 19 + 20 = 119. So there are at most 119 different subset sums, and more subsets than that—thus, by the pigeonhole principle, there are two subsets that have the same sum.

1.9. Problems That Use Counting or Proofs

5

32. We can subdivide the square mile into four half-mile squares. By the pigeonhole principle, we know that two of the five Dunkin Donutses must be in the same half-mile square. The furthest these could be apart is the diagonal of the square, or √12 ≈ .71 < .75 miles apart.

However, we could have four Dunkin Donutses at the corners of our square mile, and one in the center, and no two of these would be within half a mile of each other.

33. Well, only one at a time, but there are 2 + 2 · 6 + 2 + 1 + 3 = 20 shelves on which it might be found. And yes, the author does own this book.

34. 12 times 2 is 24.

35. If a(ab + 1) is odd, then both a and ab + 1 must be odd. Therefore a is odd, so a = 2k + 1.

Now (2k + 1)b + 1 = 2kb + b + 1 is odd, which means that b + 1 is odd (because 2kb is even) and therefore b is even.

Chapter 2

Sets and Logic 2.4 Try This! Problems on Sets and Logic 1. 4; {0, / {0}, {cat}, {{dog}}, {{2.1, 6}}, {0, cat}, {0, {dog}}, {0, {2.1, 6}}, {cat, {dog}}, {cat, {2.1, 6}}, {{dog}, {2.1, 6}}, {0, cat, {dog}}, {0, cat, {2.1, 6}}, {0, {dog}, {2.1, 6}}, {cat, {dog}, {2.1, 6}}, {0, cat, {dog}, {2.1, 6}}}. (24 = 16.)

2. First logic-language it. For all times t, and for all people p, p can be fooled at time t. (There are certainly other possible interpretations.) The negation will be, There exists a time t and a person p such that p cannot be fooled at time t.

3. For example, −16, 8, 12, . . . . These are the even integers.

4.

(a) A ∪ B = A ∩ B; A ∩ B = A ∪ B.

(b)

P T T F F

Q T F T F

P∨Q T T T F

¬(P ∨ Q) F F F T

P T T F F

Q T F T F

¬P F F T T

¬Q F T F T

(¬P) ∧ (¬Q) F F F T

P T T F F

Q T F T F

P∧Q T F F F

¬(P ∧ Q) F T T T

P T T F F

Q T F T F

¬P F F T T

¬Q F T F T

(¬P) ∨ (¬Q) F T T T

(c)

6

2.6. Try This! A Tricky Conundrum

7

(d) Let x ∈ A ∪ B. Then x 6∈ (A ∪ B), so x 6∈ A and x 6∈ B, so x ∈ A ∩ B. Similarly, let x ∈ A ∩ B. Then x 6∈ A and x 6∈ B, so x 6∈ (A ∪ B). Therefore, x ∈ A ∪ B. Now let x ∈ A ∩ B. Then x 6∈ (A ∩ B), so x 6∈ A or x 6∈ B, so x ∈ A ∪ B. Similarly let x ∈ A ∪ B. Then x 6∈ A or x 6∈ B, so x 6∈ (A ∩ B). Therefore x ∈ A ∩ B. (e) Sure.

n [

Ai =

i=1

n \

Ai and

i=1

n \

Ai =

i=1

n [

Ai .

i=1

(f) Yes; ¬(P1 ∨ · · · ∨ Pn ) is logically equivalent to ¬P1 ∧ · · · ∧ ¬Pn , and ¬(P1 ∧ · · · ∧ Pn ) is logically equivalent to ¬P1 ∨ · · · ∨ ¬Pn. 5.

(a) {−10, −9, −8, −7, −6, −4, −2, 0, 2, 4, 6, 7, 8, 9, 10}.

(b) {−10, −9, −8, −7, 7, 8, 9, 10}. (c) A.

6. Check by truth table to see that the answer is “yes”:

7.

P⇒Q T F T T

P T T F F

Q T F T F

n [

Ai = An ;

i=1

n \

i=0

¬(P ⇒ Q) F T F F

P T T F F

Q T F T F

¬Q F T F T

P ∧ ¬Q F T F F

Ai = {0}.

8.

9. No; n + 5 has a different value for each n, so such an m cannot have a single value.

2.6 Try This! A Tricky Conundrum 1. Solutions will vary.

2. NA. 3.

(a) Let learn about sets be S, let learn about logic be L, and let go on to the next chapter be C. (b) C ⇒ (S ∨ L).

(c) ¬S ∧ ¬C; (C ⇒ (S ∨ L)) ∧ (¬S ∧ ¬C) ⇒ ¬L.

8

2. Sets and Logic

(d)

C

S

L

S∨L

C ⇒ (S ∨ L)

¬S ∧ ¬C

T T T T F F F F

T T F F T T F F

T F T F T F T F

T T T F T T T F

T T T F T T T T

F F F F F F T T

(C ⇒ (S ∨ L))∧ (¬S ∧ ¬C) F F F F F F T T

(C ⇒ (S ∨ L))∧ (¬S ∧ ¬C) ⇒ ¬L T T T T T T F T

4. Solutions will vary.

5.

(a) Use the contrapositive. If n is odd, then n = 2k + 1 and n3 + 6n2 − 2n = (2k + 1)3 + 6(2k + 1)2 − 2(2k + 1) = 2(4k3 + 6k2 + 3k + 3(2k + 1)2 − (2k + 1)) + 1, which is odd.

(b) Use the contrapositive. Suppose x < 0. Then x5 + 7x3 + 5x < 0 and x4 + x2 + 8 > 0, so x4 + x2 + 8 > x5 + 7x3 + 5x.

(c) Suppose for the sake of contradiction that it can be tiled with dominoes. Then it has an even number of squares. However, 65 is odd; contradiction. (d) Tile the right-most column with n−1 2 vertical dominoes. Then tile the left-most n − 1 columns of horizontal dominoes. columns with n−1 2

2.9 Bonus: Truth Tellers Puzzle1: The simplest way to do this problem is by reasoning. Examine the three cases where Rachel, Tess, or Nicol is the liar. If Rachel is the liar, Tess and Nicol are truthful. However, Nicol and Tess contradict each other because Tess claims ze was out of town whereas Nicol says zesaw Tess in town. Thus, Rachel is not the liar. If Tess is the liar, Rachel and Nicol are truthful. Rachel’s and Nicol’s statements are not really related, so this situation is consistent. If Nicol is the liar, then Rachel and Tess are truthful. However, Rachel and Tess contradict each other because Rachel says Tess knows Amy whereas Tess says ze doesn’t know Amy. Therefore, Tess is the liar.

M D M true D false (M says) ∧ (D says) T T T F F Puzzle2: T F TF / T F F FT / F F T F F FT / TF / F There is no way that the parental statements can be consistent.

Project:

1.

D T T F F

berries are safe T F T F

D says T F F T

2.9. Bonus: Truth Tellers

9

If duck says yes, that could result from the berries being safe or unsafe; the same is true if the duck says no.

2. This will give no information about berries . . . .

3.

D T T F F

berries are safe T F T F

D∧ berries are safe T F FT / FT /

If duck says yes, that could result from the berries being safe or unsafe.

4.

D T T F F

berries are safe T F T F

D∨ berries are safe T T TF / FT /

If duck says yes, that could result from the berries being safe or unsafe.

5.

D T T F F

berries are safe T F T F

D ⇒ berries are safe T F /T F /T F

If duck says no, that could result from the berries being safe or unsafe. (If the duck says yes, we do learn that the berries are safe. However, we can’t depend on that answer.)

6.

D T T F F

berries are safe T F T F

D ⇐ berries are safe T T /F T /T F

If duck says yes, that could result from the berries being safe or unsafe.

7. Again, this tells us nothing about berries.

We need to design a question that will produce answers that are consistent with the safety of the berries. Here are four questions that work: Would a duck who answers oppositely from you say that the berries are edible?

Do you tell the truth if and only if the berries are edible?

Is it true that (if the berries are safe, then you tell the truth) and (if the berries are safe, then you lie)?

10

2. Sets and Logic

If I asked you whether the berries are safe, what would you say?

D T We give a truth table for this last question: T F F

berries are safe T F T F

D says T F /F T /T F

2.12 Problems about Sets and Logic 1. {2}.

2. The question is asking about non-urgent care. It excludes dental care and hospital care and urgent care. The question could have been worded, “In the last 12 months, how often did you get a prompt appointment for non-urgent health care at a doctor’s office or clinic?”

3. Here they are: (doctor or health provider), (discuss or provide), (methods and strategies), (smoking or using). All three of the ors are meant mathematically, but the and is used to mean union instead of intersection.

4. These are the negative integers {−9, . . . , −1}, so the cardinality of the set is 9.

5.

P T T F F

Q T F T F

¬P ∨ Q T F T T

P ∧ (¬P ∨ Q) T F F F

6. {2n | n ∈ W} or {n | n = 2k , k ∈ W}.

7.

8.

(a) {(2, 2), (2, 3), (3, 2), (3, 3)}.

(b) {(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3), (2, 4), (3, 4), (4, 4), (4, 3), (4, 2)}. (c) {(1, 1), (2, 1), (3, 1)}.

2.12. Problems about Sets and Logic

11

9.

P T T F F

Q T F T F

¬P F F T T

¬Q F T F T

P⇒Q T F T T

¬Q ⇒ ¬P T F T T

10.

P T T F F

Q T F T F

¬P F F T T

¬Q F T F T

Q⇒P T T F T

¬P ⇒ ¬Q T T F T

11. Let A = {1, 2, 3} and B = {2, 3, 4}.

12. This is total nonsense. The moon is not made of green cheese. Even if Moscow had existed as a city in the time of Aristotle, he certainly was not and is not its President. Every fiber of your being is probably screaming “False, False, FALSE!” But think again. Because the moon is not made of green cheese, it doesn’t matter what the statement is claiming this implies; the implication holds anyway, because a false statement implies, well, everything. See the appropriate truth table.

13.

14. Let A,C be disjoint. For example, let A = {1, 2},C = {3}, B = {4}, D = {5}. Then (A × B) ∪ (C ×D) = {(1, 4), (2, 4), (3, 5)} but (A∪C)×(B∪D) = {(1, 4), (1, 5), (2, 4), (2, 5), (3, 4), (3, 5)}.

15. Use the contrapositive. Let B = 0. / Then A \ B = A \ 0/ = A, which is not equal to 0/ unless A = 0. / 16. Use a truth table and trace backwards. We will have one value for R but two possibilities each for P, Q, S, so we have eight rows.

P T T T T F F F F

Q T T F F T T F F

S T F T F T F T F

R F F F F F F F F

P⇒Q T T F F T T T T

R∧S F F F F F F F F

(P ⇒ Q) ⇔ (R ∧ S) F F T T F F F F

12

2. Sets and Logic

We see that (P ⇒ Q) ⇔ (R ∧ S) is true when P, S are true and Q is false, or when P is true and Q, S are false. Thus, we conclude that P is true and Q is false.

17. In math-logic language, this expression is (c > 5 ∧ b = a) ∨ c ≥ 5. The crucial point is to notice that (c ≥ 5) is logically equivalent to (c > 5) ∨ (c = 5). So let c > 5 be P, let b = a be Q, and let c = 5 be R. 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 F F F F

P∨R T T T T T F T F

(P ∧ Q) ∨ (P ∨ R) T T T T T F T F

Now notice that P ∨ R would have the same truth table. Thus, we can rewrite this as c ≥ 5.

18. For every k that is a multiple of 3, there is a subset of the natural numbers with cardinality k. It’s not true because k could be negative, but no set has negative cardinality. The negation is ∃k ∈ 3Z, such that ∀S ⊆ N, |S| = 6 k. In English, that says there is a k evenly divisible by 3 such that no subset of N has cardinality k. And that’s true, because we can take k = −63.

19. This is a biconditional, so do it in two parts. (⇒) Direct proof. Suppose n is odd, so n = 2k + 1. Then n2 = (2k + 1)2 = 2(2k2 + 2k) + 1, which is odd. (⇐) Use the contrapositive. Suppose n2 is even, so n2 = 2k. Because n2 = n · n, we see that k = n·n 2 ; and because k is an integer, n must be divisile by 2 and so is even.

20. Here goes some double-inclusion provin’: Consider x ∈ {3k | k ∈ Z}; because k ∈ Z, then 3k ∈ Z so x ∈ Z. Similarly, we see that for y ∈ {3k + 1 | k ∈ Z}, y ∈ Z and for z ∈ {3k + 2 | k ∈ Z}, z ∈ Z. Now consider ℓ ∈ Z. If ℓ is divisible by 3, then ℓ ∈ {3k | k ∈ Z}. Otherwise, divide ℓ by 3; the remainder must be 1 or 2 (as if it were 0, ℓ would be divisible by 3). If the remainder is 1, then ℓ ∈ {3k + 1 | k ∈ Z}, and if the remainder is 2, then ℓ ∈ {3k + 2 | k ∈ Z}.

Therefore, ℓ ∈ {3k | k ∈ Z} ∪ {3k + 1 | k ∈ Z} ∪ {3k + 2 | k ∈ Z}.

21. Suppose there are only finitely many primes. List these k primes, p1 , p2 , . . . , pk . Multiply them all together to get a number larger than any of the primes, and add 1. It’s not divisible by any known prime (as otherwise 1 would also be divisible by that prime), so it is prime itself. This is a contradiction, because there are now k + 1 primes.

22. The binary representation of a natural number is a sum of powers of two, so notice that n is divisible by 4 exactly when (binary representation of n) = 22 ( sum of some powers of 2).

In other words, the binary representation must end in 00. The contrapositive of the original statement is n ∈ N is divisible by 4 if and only if the binary representation of n ends in 00.

2.12. Problems about Sets and Logic

23.

P T T F F

Q T F T F

¬P F F T T

13

¬P ∨ Q T F T T

24. Example 1.5.4, Example 1.5.5.

25. In logic language, the statement says, If you own a boat, then you wash it after each use. If you don’t own a boat, then you wash your non-boat every time you use it. The truth table for implies says that if the hypothesis is false, then the statement as a whole is true, which is the case here.

26.

(a) 3. (b) 4. (c) 1.

27.

(a) (2, 5), (−3, 1), (0, 1). (b) 4. (c) 0/ because no element of A has a 0 in the right-hand position.

28.

(a) {1, 6, 7, 8, 9, 10}.

(b) B \C = {5, 7}; (B \C) \ A = {7}.

(c) A ∩ B = {5}; (A ∩ B) ×C = {(5, 1), (5, 6), (5, 9)}.

29. x is not prime and x ≥ 52.

30.

p T T F F

q T F T F

¬p F F T T

¬q F T F T

((¬p) ∧ q) F F T F

(p ∨ (¬q)) F T F F

((¬p) ∧ q) ∧ (p ∨ (¬q)) F F F F

x−y ∈ Z. Thus the negation is ∃ x, y ∈ Z such 31. In logic notation, this statement is ∀ x, y ∈ Z, 5 x−y x−y that 6∈ Z. This reads as There exist integers x and y such that is not an integer.

5 5 32.

(a) 6 ∃ k ∈ N such that 2|k ∧ k is prime.

(b) 6 ∃ △ such that △ has four sides. (c) 6 ∃ a, b ∈ Z such that a2 /b2 = 2.

(d) 6 ∃ k ∈ N such that (∃ m ∈ Z such that m2 = k) ∧ (k = p + 1) for p a prime.

33. If y does not have a sister, then x2 < 100.

34. See solutions to Section 2.4 Problem 4.

14

2. Sets and Logic

35. See solutions to Section 2.4 Problem 4.

36.

(a) If n is even, then n = 2k. Therefore n − 1 = 2k − 1 = 2(k − 1) + 1, which is odd. (b) Suppose n − 1 is not odd. Then n − 1 is even, so n − 1 = 2k and thus n = 2k + 1, which is odd. . . and therefore n is not even. (c) Suppose n is even but n − 1 is not odd. Then n − 1 is even, so n − 1 = 2k and thus n = 2k + 1, which is odd. . . contradiction!

37. Suppose x is even. Then x − 2k. Therefore, 4x2 − 3x + 1 = 4(2k)2 − 3(2k) + 1 = 16k2 − 6k + 1 = 2(8k2 − 3k) + 1 is odd.

Now suppose 4x2 − 3x + 1 is odd. Then 4x2 − 3x + 1 = 2k + 1, so 4x2 − 3x = 2k. Factoring, we have x(4x − 3) = 2k. Thus either x is even or 4x − 3 is even. Note that 4x − 3 = 2(2x − 2) + 1 is odd and thus cannot be even, so x must be even.

38. Let’s do this in stages.

• If an organization (is primarily religious in purpose and falls under the regulations 940 CMR 2.00) or (does not raise or receive contributions from the public in excess of $5,000 during a calendar year or does not receive contributions from more than ten persons during a calendar year, if all of their functions, including fundraising activities, are performed by persons who are not paid for their services and if no part of their assets or income inures to the benefit of, or is paid to, any officer or members) then it is not required to have a “Certificate of Solicitation.”

• If an organization ((is primarily religious in purpose) and (falls under the regulations 940 CMR 2.00)) or ((does not raise or receive contributions from the public in excess of $5,000 during a calendar year or does not receive contributions from more than ten persons during a calendar year), if all of their functions, including fundraising activities, are performed by persons who are not paid for their services and if no part of their assets or income inures to the benefit of, or is paid to, any officer or members) then it is not required to have a “Certificate of Solicitation.”

• If an organization ((is primarily religious in purpose) and (falls under the regulations 940 CMR 2.00)) or (if all of their functions, including fundraising activities, are performed by persons who are not paid for their services and if no part of their assets or income inures to the benefit of, or is paid to, any officer or members and (does not raise or receive contributions from the public in excess of $5,000 during a calendar year or does not receive contributions from more than ten persons during a calendar year)) then it is not required to have a “Certificate of Solicitation.”

• If an organization ((is primarily religious in purpose) and (falls under the regulations 940 CMR 2.00)) or (if (all organizational or fundraising functions are performed by unpaid persons) and (no part of their assets or income inures to the benefit of, or is paid to, any officer or members) and (does not raise or receive contributions from the public in excess of $5,000 during a calendar year or does not receive contributions from more than ten persons during a calendar year)) then it is not required to have a “Certificate of Solicitation.”

• If an organization ((is primarily religious in purpose) and (falls under the regulations 940 CMR 2.00)) or (if (all organizational or fundraising functions are performed by unpaid persons) and (no part of organizational assets or income benefits any officer or members) and

2.12. Problems about Sets and Logic

15

((does not collect contributions from the public in excess of $5,000 during a calendar year) or (does not receive contributions from more than ten persons during a calendar year))) then it is not required to have a “Certificate of Solicitation.”

• If an organization ((is religious) and (falls under 940 CMR 2.00)) or (if (no one is paid for organizational work) and (neither officers nor members benefit from organizational assets or income) and ((public contributions are < $5, 000 during a calendar year) or (contributions are from ≤ 10 persons during a calendar year))) then it is not required to have a “Certificate of Solicitation.”

39. {x ∈ Z | x is divisible by 4} or {4x | x ∈ Z}.

40. Because A ∩ B ⊂ B, by definition (A ∩ B) ∩ (A \ B) = 0. / Therefore they cannot be equal unless both A ∩ B and A \ B are empty. That’s only true if A is the empty set.

Chapter 3

Graphs and Functions 3.3 Try This! Play with Functions and Graphs 3.3.1 Play with Functions 1. f3 (a) = c, f3 (b) = c; f4 (a) = d, f4 (b) = c; f5 (a) = e, f5 (b) = e; f6 (a) = d, f6 (b) = e; f7 (a) = e, f7 (b) = d; f8 (a) = c, f8 (b) = e; f9 (a) = e, f9 (b) = c. Nine.

2. 23 = 8.

3. 102 = 100.

4. 210 = 1024.

5. For each of those m elements, there are q choices of target element. Thus, qm .

3.3.2 Play with Graphs 1. Solutions will vary.

2. The degrees are 2, 3, 3, 2, 2, 4 with sum 16. There are 8 edges. The total degree is twice the number of edges.

3. The total degree is twice the number of edges. Proof: Think of the degree as counting the number of edges at each vertex. In this way, every edge is counted exactly twice, once for each incident vertex.

4. For the graphs in Figure 3.3 on page 73, the number of odd vertices is four, six, and two, respectively. All are even. Conjecture: The number of odd vertices is always even. Proof (by contradiction): Suppose not. Then the total degree involves an odd number of odds and so will be odd, which contradicts the proof from the previous problem.

3.3.3 A Dot Game 1. For five vertices, there can be a draw. For six vertices, there must always be a winner and loser. See the argument given in Section 3.9.

2. See previous answer.

3. If you start with fewer vertices, then there can be a draw. If you start with more vertices, you must have a winner and a loser (use the same argument as for six vertices).

16

3.8. Try This! More Graph Problems

17

3.8 Try This! More Graph Problems 1. Wn contains a cycle of size n− 1, which contributes n− 1 edges. Then, there is an edge joining the nth vertex to each of the other n − 1 vertices; this contributes another n − 1 vertices for a total of 2n − 2 edges.

2. No. The left-hand graph has no vertices of degree 2, but the right-hand graph has one of them.

3. Well, Kv has

v(v − 1) v(v − 1) edges, so G has − e edges.

2 2

4. The complement of C5 is the star inside K5 , which unravels to make another C5 . If a graph is isomorphic to its complement, then v(v−1) − e = e, which is equivalent to v(v−1) = 2e and 2 2 to v(v − 1) = 4e. Also, a cycle graph has the same number of vertices as edges, so 4e = 4v.

Thus, v − 1 = 4 and we must have v = 5.

5. They are isomorphic; move the center and upper vertices over to the right, and then move the leftmost vertex upwards. To exhibit an isomorphism, the graphs must be labeled first and labelings will vary from person to person.

6. The seven nonisomorphic subgraphs of K3 are shown in a row . . .

. . . above the 17 distinct subgraphs of K3 .

7. The two left-hand graphs are isomorphic; the top and bottom pairs of vertices of the octagonal graph can be squished inwards to form the inner square of the cube graph. The right-hand graph is also isomorphic to the other two; the leftmost and rightmost vertices can be untangled to form the outer square of the cube graph, and the remaining four vertices can be untangled to form the inner square of the cube graph. To exhibit an isomorphism, the graphs must be labeled first and labelings will vary from person to person.

8. No. The left-hand graph has two disjoint 3-cycles, but the right-hand graph has two incident 3-cycles.

18

3. Graphs and Functions

3.12 Bonus: Party Tricks Short activity:

1. The vertices represent people at the party, and two are adjacent if and only if they are friends.

2. Every vertex has degree at least 5.

3. True. There is at least one person in the garden and at most nine people in the garden. But because friendship is symmetric, i.e., the graph is undirected, we only need to consider the cases of one to five people in the garden as the six to nine cases are covered by symmetry with the house. If there is only one person in the garden, ze has at least five friends in the house.

If there are two people in the garden, each has at least four friends in the house. If there are three people in the garden, they may all be friends but each has at least three other friends in the house. If there are four people in the garden, each may be friends with the other three but still has two or more friends in the house. If there are five people in the garden, then each has at least one friend in the house.

4. The graphical model is the same, and each vertex has degree at least n. Thus, if there are k people in the garden, each is friends with at most the k − 1 others and so has at least n− (k − 1) friends in the house. If there are more than n people in the garden, consider the folks in the house to discover that they have friends in the garden.

3.13 Bonus 2: Counting with the Characteristic Function Short activity:

1. They are the same size.

2. First suppose that F(A) = F(B). Then χA = χB , so A = χA−1 (1) = χB−1 (1) = B. Now examine χA ∈ Fn . The subset A ∈ Pn has the property that F(A) = χA , as desired.

3. We know that there are 2n functions from {1, . . . , n} → {0, 1}. Because these functions are in one-to-one correspondence with members of Pn , there are also 2n elements in Pn , which means that a set with n elements has 2n subsets.

3.15 Problems about Graphs and Functions 1.

(a) (b) 10 vertices and 15 edges. (c)

W6 is shown in grey.

3.15. Problems about Graphs and Functions

19

2. There are none! Every graph must have an even number of vertices of odd degree, and a 3regular graph with five vertices would have five vertices of odd degree; that’s a contradiction.

3. By Theorem 3.2.8, f is one-to-one because its domain and target are the same size and f is onto.

4. Solutions will vary. b+1 a b 5. One-to-one: If ⌊ a+1 2 ⌋(−1) = ⌊ 2 ⌋(−1) then both sides have the same sign; then, a and b b+1 must have the same parity, as otherwise (−1)a 6= (−1)b . Now, ⌊ a+1 2 ⌋ = ⌊ 2 ⌋ and if a and b b+1 have the same parity, then a+1 2 = 2 , which implies that a = b.

Onto: Given z ∈ Z, if z is positive, then

1 2z + 1 2z (−1) = z + = z; f (2z) = 2 2

if z is negative, then

2|z| 2|z| − 1 + 1 2|z|−1 (−1) = (−1)−1 = −|z| = z. f (2|z| − 1) = 2 2

6.

(a) C3 is the only one; comparing numbers of edges, 3 = 3·2 2 . But this is the only n for which n(n−1) n= 2 . (b) There are none, as every vertex in a cycle has degree 2 and the central vertex of a wheel has degree at least 3. (c) W4 = K4 . Otherwise, the wheel does not have enough edges to be complete; 2 · 4 − 2 = n(n−1) 4·3 ⇒ n(n − 1) = 4n − 4 ⇒ n = 4.

2 but 2n − 2 = 2 (d) There are none, as a path has two vertices of degree 1 and all vertices of a cycle have degree 2. (e) K2 = P2 . Otherwise, no complete graph has vertices of degree 1.

7. There are 2n functions from S to Z2 . Unless n = 1 or 2, none are one-to-one. All but two of them are onto (the two that respectively send everything to 0 and send everything to 1).

8.

20

3. Graphs and Functions

9.

There are 34 of them.

10.

11. A; B; neither is one-to-one, and both are onto.

12. First notice that 144 = 12 · 12 = (22 · 3)2 = 24 · 32 . This means that every positive factor of 144 divides 24 · 32 , meaning that it has four or fewer factors of 2 and two or fewer factors of 3. Therefore, each factor of 144 is generated by a number in {0, 1, 2, 3, 4} of factors of 2 and a number in {0, 1, 2} of factors of 3. Thus, f is onto, and because A and B are finite and have the same size, f is also one-to-one.

13. Proof by contradiction: Suppose all n vertices have different degrees. The smallest degree possible is 0, and the largest possible is n − 1, which is a total of n different degrees. However, if a vertex has degree n − 1, there cannot also be a vertex of degree 0. So, there are at most n − 1 different degrees, and thus by the pigeonhole principle two of the n vertices must have the same degree.

3.15. Problems about Graphs and Functions

14.

21

Nope.

15. (1) There are n vertices. Each is connected to the other n − 1 vertices for a total of n(n − 1) edges, each of which was counted twice (once for each vertex).

(2) The first vertex has n − 1 edges incident to it. The second vertex already has one edge and there are n − 2 vertices to which it must connect; continuing, the jth vertex needs n − j more vertices, for the same total.

16.

(a) Some vertices are adjacent, so that gives us 1. Two vertices on opposite corners of a face have distance 2. Two vertices on opposite corners of the cube have distance 3. (b) Six. There are three choices of the edge on which to leave the initial vertex, and then two choices from each of those. Product principle! (c) The cube is symmetric, so we might as well consider the paths from a single vertex.

There are three length-1 paths, six length-2 paths, and six length-3 paths. This gives us 43 3+2·6+3·6 3+6+6 = 15 = 2.8667.

17.

(a) One part has m vertices and the other has n vertices, so by the sum principle there are m + n vertices total. (b) Every one of the m vertices is connected by an edge to each of the n vertices, and there are no other edges. Thus, by the product principle, there are a total of mn edges.

18. The parts have sizes m, n, and p, so there are a total of m + n + p vertices; each of the m vertices is connected to each of the n vertices and to each of the p vertices for m(n + p) edges, and each of the n vertices is also connected to each of the p vertices, so the total is mn + mp + np.

19. K2 ; K3 . Conjecture: Kn \ Km ∼ = Kn−m . This is true because n − m vertices remain, as do all edges between them (and no more).

20. n − 1 edges because that’s the number of adjacencies to vertex 1.

21. Let the map be f : 2Z → Z defined by f (k) = k/2. Match up the 0s in the labeling. To show this is an isomorphism, we need to show that the appropriate four properties hold: well-defined and one-to-one: a = b ⇐⇒ a/2 = b/2 ⇐⇒ f (a) = f (b). onto: given any z ∈ Z, f (2z) = 2z/2 = z. operation-preserving: In the all-even-labeled graph, the vertex labeled j is adjacent to vertices labeled j − 4, j − 2, j + 4 (if j is divisible by 4) or to vertices labeled j − 4, j + 2, j + 4 (if j is not divisible by 4). This should translate to the same adjacencies if everything is divided by 2, and that in fact holds.

22. Solutions will vary depending on how students approached the problems.

23. Solutions will, of course, vary here; the point is to help students become conscious of their proof techniques so as to link their study of formal proof techniques with their developing reasoning skills.

22

3. Graphs and Functions

24. K5 : {{1, 2, 3, 4, 5}, {{1, 2}, {1, 3}, {1, 4}, {1, 5}, {2, 3}, {2, 4}, {2, 5}, {3, 4}, {3, 5}, {4, 5}}}.  0 1  1  1 1

1 0 1 1 1

1 1 0 1 1

1 1 1 0 1

 1 1  1 .

1 0

{{1, 2, 3, 4, 5}, {2, 1, 3, 4, 5}, {3, 1, 2, 4, 5}, {4, 1, 2, 3, 5}, {5, 1, 2, 3, 4}}.

P5 : {{1, 2, 3, 4, 5}, {{1, 2}, {2, 3}, {3, 4}, {4, 5}}}.   0 1 0 0 0 1 0 1 0 0   0 1 0 1 0 .   0 0 1 0 1 0 0 0 1 0 {{1, 2}, {2, 1, 3}, {3, 2, 4}, {4, 3, 5}, {5, 4}}.

Everything takes more room for a graph with more edges. Comparing the results for the two graphs suggests that for sparse graphs, a list will be more compact than a matrix.

25.

26.

(a)

v1

v2

v3

v1

v2

v3

(b) v4 27. g cannot be one-to-one because the domain has more elements than the target. It can certainly be onto.

3.15. Problems about Graphs and Functions

23

28. Solutions will vary.

29.

(a) Kn−1 . (b) Pn .

30.

(a) There are 3 choices for the image of each of the 10 elements of A, so 310 functions.

None of them are one-to-one because |A| > |B|. (b) There are 10 choices for the image of each of the 3 elements of B, so 103 . For one-toone functions, we must assure that no two of the images are the same. So 10 · 9 · 8 = 720 functions because there are 10 choices of image for one element, then 9 for the second element, and finally 8 for the remaining element.

31. The target space is the whole numbers W. 32. The length of the shortest walk in any graph is 1. The longest path in the Petersen graph is 9 (uses all 10 vertices). Nope, K3 is not a subgraph of the Petersen graph. The smallest cycle in the Petersen graph is length 5.

33.

(a) 6 · 5 · 4 · 3 · 2 = 720 (b) None, as Y has greater cardinality than X . (c) There are 5! = 120 because the elements of X may be the images of the domain elements X in any order.

34. False. With degree sequence (1, 1, 1, 1, 2, 3, 3), the two degree-3 vertices may be adjacent or nonadjacent.

35. The ith tree has vi −1 edges. So the total number of edges is ∑ni=1 (vi −1) = ∑ni=1 vi −n = v−n.

36.

(a) The function f is not one-to-one because both n and n + 21 map to n.

The function f is onto because for every integer, there is a real number—itself!—that maps to it. (b) The function f is one-to-one because n = ⌊n⌋.

And, f is onto because for every integer, there is an integer—itself!—that maps to it.

37.

(a) The function c is not one-to-one because both n and n − 12 map to n − 2.

The function c is onto because for every integer, there is a real number—x + 2—that maps to it. (b) The function c is one-to-one because n + 2 = ⌈n + 2⌉.

And, c is onto because for every integer, there is an integer—n − 2—that maps to it.

38. No, it’s not onto, because even though there exists w ∈ W, we cannot always have f (w) = n.

For example, given 2 ∈ N, there is no w ∈ W with f (w) = 2 as the range of f is the odd numbers.

39. The zero map z is one-to-one exactly when |A| = 1. z is always onto because A must be nonempty. z is a bijection exactly when |A| = 1 40. A; B; A × B. Both are one-to-one, and neither is onto.

Chapter 4

Induction 4.3 Try This! Induction 1. Base case: S4 has 3 edges. Suppose that for n ≤ k, Sn has n − 1 edges, and consider Sk+1 .

Remove one of the leaves to get Sk , which has k − 1 edges. Replacing the leaf, we get one more edge, so Sk+1 has k edges. n

2.

(a) ∑ 2 j − 1. j=1

(b) n

(c) ∑ 2 j − 1 = n2 . j=1

2 j − 1 = ∑kj=1 2 j − 1 + (d) Suppose that for n ≤ k, ∑nj=1 2 j − 1 = n2 . Consider ∑k+1 j=1

2(k + 1) − 1 = ∑kj=1 2 j − 1 + 2k + 1. By the inductive hypothesis, this is equal to k2 + 2k + 1 = (k + 1)2 , as desired.

3. We’ll do this twice. The first thing we need to decide is what we’ll induct on, i.e., what plays the role of n ∈ N.

First: Let’s induct on the number of vertices in the graph. A graph with one vertex has no edges, and we get 0 = 0 as a base case; a graph with two vertices and one edge has total degree 2 and 2 = 2 · 1, as desired. Suppose that for a graph with n ≤ k vertices, ∑v∈V (G) d(v) = 2|E(G)|.

Now consider a graph with k + 1 vertices. Remove a vertex v0 with degree deg(v0 ). This leaves us with a graph G′ with k vertices and, by the inductive hypothesis, ∑v∈V (G′ ) d(v) = 2|E(G′ )|. Note that ∑v∈V (G) d(v) = ∑v∈V (G′ ) d(v) + 2deg(v0 ) because each edge incident to v0 also contributed one to the degree of (v0 ); and, |E(G)| = |E(G′ )| + deg(v0 ). Thus, ∑v∈V (G) d(v) = ∑v∈V (G′ ) d(v) + 2deg(v0 ) = 2|E(G′ )| + 2deg(v0 ) = 2|E(G)|.

Second: Let’s induct on the number of edges in the graph. A graph with one vertex has no edges, and we get 0 = 0 as a base case; a graph with two vertices and one edge has total degree 2 and 2 = 2 · 1, as desired. Suppose that for a graph with n ≤ k edges, ∑v∈V (G) d(v) = 2|E(G)|.

Consider a graph with k + 1 edges. Remove an edge. This leaves us with a graph G′ with k edges and, by the inductive hypothesis, ∑v∈V (G′ ) d(v) = 2|E(G′ )|. Note that ∑v∈V (G) d(v) = 24

4.6. Try This! More Problems about Induction

25

∑v∈V (G′ ) d(v)+2 because the edge had two ends; so, ∑v∈V (G) d(v) = 2|E(G′ )|+2 = 2(|E(G′ )|+ 1) = 2|E(G)|.

4. Postage using 3-cent and 5-cent stamps is p = 3x + 5y, where p is the postage, x is the number of 3-cent stamps, and y is the number of 5-cent stamps. We wish to show that if p > 7, then there exist x, y such that p = 3x + 5y. The base case is that x = y = 1 gives p = 3 + 5 = 8. Now suppose that if k ≥ p > 7, there exist x, y such that p = 3x + 5y. Consider k + 1 = (k) + 1; we may write k = 3x + 5y so that k + 1 = 3x + 5y + 1 = 3x + 5y + 3 + 3 − 5. Then xnew = x + 2, ynew = y − 1 satisfy k + 1 = 3xnew + 5ynew .

5.

(a) Solutions will vary. (b) Solutions will vary. (c) Solutions will vary. (d) Solutions will vary. (e) (A more formally written solution is given for Problem 17 in Section 4.13.) This is a challenge! Let us again induct once on vertices and once on edges. Take as a base case the tree with a single edge. Assign each of the two vertices to a part, and we see that this tree is bipartite. Now assume that a tree with n ≤ k vertices is bipartite, and consider a tree with k + 1 vertices. Hopefully you have proved that every tree has a leaf. In this case, remove a leaf to obtain a tree with k vertices; it is bipartite. Replace the leaf and assign it to the other part from its neighbor, and our tree with k + 1 vertices is bipartite.

Now assume that a tree with n ≤ k edges is bipartite, and consider a tree with k + 1 edges. Remove an edge to obtain two trees, each of which has fewer than k + 1 edges, so each is bipartite. Let the parts of one tree be A, B and the parts of the other tree be C, D. Replace the edge, which without loss of generality connects a vertex in A with a vertex in C. Then let the two parts be A ∪ D and B ∪C, and we are done.

4.6 Try This! More Problems about Induction 1. We get 2, 6, 14, 30, which is also 4 − 2, 8 − 2, 16 − 2, 32 − 2. Let us suppose that for n ≤ k, k j k+1 . By the inductive hypothesis, this j ∑nj=1 2 j = 2n+1 − 2 and consider ∑k+1 j=1 2 = ∑ j=1 2 + 2 equals 2k+1 − 2 + 2k+1 = 2 · 2k+1 − 2 = 2k+2 − 2, as desired.

2.

(a) It clearly works for n = 1. For n = 2, split the grid into four 2 × 2 grids. Imagine that each is missing an upper-right corner and tile it. Then rotate three of the four grids so that their missing corners form an L-shaped tile, and put such a tile there. (b) Let’s turn this observation into an inductive proof. Suppose that for any n ≤ k we can tile a 2n × 2n grid with the upper-right-hand square missing with L-shaped tiles. Consider a 2k+1 × 2k+1 grid. Split it into four 2k × 2k grids. Imagine that each is missing an upperright corner and tile it using the inductive hypothesis. Then rotate three of the four grids so that their missing corners form an L-shaped tile, and put such a tile there. Ta da!

3. Note that when n = 4, we have that 16 < 24 (and that when n = 3, we have 8 > 6). Assume that when 4 ≤ n ≤ k, 2n < n! and consider 2k+1 = 2 · 2k < 2 · k! by the inductive hypothesis.

Now, 2 < k + 1 so 2k+1 < 2 · k! < (k + 1)k! = (k + 1)!.

26

4. Induction

4.10 Bonus: Small Crooks Short Activity:

1. Yes! We just proved that every tree with at least one edge has at least two leaves. And we know that the tree with no edges has one vertex. So consider any other tree T ; among those that have a number of edges different from |V (T )| − 1, choose one with the least number of vertices. It has at least two leaves; remove one and obtain a smaller tree T ′ , which has |V (T ′ )| − 1 edges because it has fewer vertices than T . Now put back the leaf and get that T has |V (T ′ )| − 1 + 1 = |V (T )| − 1 edges. Contradiction!

2. Suppose not. Then there is a graph without this property; among those, select one (G) with the smallest number of edges. If the graph has a leaf ℓ with edge e, remove that leaf to obtain a graph with fewer edges and therefore the desired walk W . Replace the leaf and insert e–ℓ after the first instance of ℓ in W to produce a walk W ′ of G that begins and ends at the same vertex and crosses every edge twice. Contradiction! Now suppose the graph has no leaf. It must have an edge e with endpoints u, v whose removal does not disconnect the graph. Remove this edge to obtain a graph with fewer edges and therefore the desired walk W . Replace e and insert e–v–e after the first instance of u in W to produce a walk W ′ of G that begins and ends at the same vertex and crosses every edge twice. Contradiction!

3. Suppose not. Then not all vertices have the same degree; consider a vertex v of minimal degree. Then either all of its neighbors have the same degree as v, or they don’t. If they don’t, then some neighbor of v has larger degree and none have smaller degree. Then the average degree of the neighbors is higher than deg(v), which is a contradiction. If, on the other hand, all of v’s neighbors have the same degree as v, then look at all vertices with deg(v) and find one with a neighbor of different degree. (If none exists, then either G is not regular or not connected.) Return to the previous case for a contradiction.

4.13 Problems That Use Induction (k+1)−1

1. 3+ 3 = 6; if ∑n−1 j=0 3 = 3n for n ≤ k, then ∑ j=0

3 = ∑kj=0 3 = ∑k−1 j=0 3+ 3 = 3k + 3 = 3(k + 1).

2. 25 > 6; if 25n > 6n for n ≤ k, then 25k+1 = 25 · 25k > 6 · 25k > 6 · 6k = 6k+1 . (k+1)−5

3. 4 + 4 = 8; if ∑n−5 j=−5 4 = 4(n + 1) for n ≤ k, then ∑ j=−5 1) + 4 = 4(k + 2).

k−5 4 = ∑k−4 j=−5 4 = ∑ j=−5 4 + 4 = 4(k +

4. n(n+1) k n for n ≤ k and consider ∑k+1 1 + 2 = 2(2+1) j=1 j = ∑ j=1 j + (k + 1). By 2 ; suppose ∑ j=1 j = 2 the inductive hypothesis, this is

k(k + 1) k(k + 1) 2(k + 1) + (k + 1) = + 2 2 2 k(k + 1) + 2(k + 1) (2 + k)(k + 1) (k + 1)(k + 2) = = , = 2 2 2

4.13. Problems That Use Induction

27

as desired.

5. 10 · 11 = 110 < 121 = 112 ; if 10n < n2 when k ≥ n ≥ 11, then 10(k + 1) = 10k + 10 < k2 + 10 = k2 + 9 + 1 < k2 + 2k + 1 = (k + 1)2 .

6. A set with 2 elements has 4 = 22 subsets. Suppose that when n ≤ k, any set with n elements has 2n subsets. Consider a set S with k + 1 elements, and designate one of these elements as peach. Every subset either has the peach element or does not have the peach element. The subsets without the peach element are subsets of S \ peach, which has k elements so by the inductive hypothesis there are 2k subsets. These 2k subsets are in one-to-one correspondence with subsets of S that contain the peach element (just union each of them with the peach element), so there are a total of 2k + 2k = 2k+1 subsets of S. 7. Base cases: 30 = 1 = (31 − 1)/2 and 30 + 31 = 4 = (32 − 1)/2. Suppose that for any n ≤ r, n+1 ∑nk=0 3k = 3 2 −1 . r+1 k 3 = ∑rk=0 3k + 3r+1 . We can use the inductive hypothesis on Inductive step: Consider ∑k=0 the sum to obtain

3r+1 − 1 + 2 · 3r+1 3 · 3r+1 − 1 3r+2 − 1 3r+1 − 1 + 3r+1 = = = , 2 2 2 2 as desired.

8.

We have shown by Venn diagram that (A1 ∩ A2 ) = A1 ∪ A2 . Suppose that as long as n ≤ k, (A1 ∩ A2 ∩ · · · ∩ An ) = A1 ∪ A2 ∪ · · · ∪ An . Consider (A1 ∩ A2 ∩ · · · ∩ Ak+1 ). We can also express this as ((A1 ∩ A2 ∩ · · · ∩ Ak ) ∩ Ak+1 ). If we rename (A1 ∩ A2 ∩ · · · ∩ Ak ) = B, then our previous expression becomes (B ∪ Ak+1 ). That’s just two sets, so our Venn-diagram-proved base case tells us that (B ∩ Ak+1) = B ∪ Ak+1 . Let us resubstitute for B in order to obtain (A1 ∩ A2 ∩ · · · ∩ Ak ) ∪ Ak+1 . We can now use the inductive hypothesis on (A1 ∩ A2 ∩ · · · ∩ Ak ); therefore, (A1 ∩ A2 ∩ · · · ∩ Ak ) ∪ Ak+1 = (A1 ∪ A2 ∪ · · · ∪ Ak ) ∪ Ak+1 . Linking all of our statements together, we have that (A1 ∩ A2 ∩ · · · ∩ Ak+1 ) = A1 ∪ A2 ∪ · · · ∪ Ak ∪ Ak+1 . n(n−1) 9. K2 has 2·1 edges. Consider Kk+1 . Remove one 2 = 1 edge. Suppose that for n ≤ k, Kn has 2 vertex (and its incident k edges) to obtain Kk , which by the inductive hypothesis has k(k−1) 2 edges. Replacing the removed vertex adds k edges for a total of

k(k − 1) 2k 2k + k2 − k k2 + k (k + 1)(k) + = = = 2 2 2 2 2 edges, as desired.

28

4. Induction

10. For n = 3, 3(0·−1+1·0+2·1) = 3·2·1, and for n = 2, 3(0·−1+1·0) = 2·1·0. Suppose that (k+1)−1 for n ≤ k, 3 ∑n−1 j( j − 1) = 3 ∑k−1 j=0 j( j − 1) = n(n − 1)(n − 2), and consider 3 ∑ j=0 j=0 j( j − 1) + 3k(k − 1). By the inductive hypothesis, this equals k(k − 1)(k − 2) + 3k(k − 1) = k(k − 1)[(k − 2) + 3] = k(k − 1)(k + 1) = (k + 1)k(k − 1), as desired.

11. (2!)2 < 2! · 4!; suppose that for n ≤ k, we have ((n + 1)!)n ≤ 2! · 4! · · · (2n)!, and consider (((k + 1) + 1)!)k+1 = ((k + 2)!)k+1 = ((k + 2)!)k (k + 2)! = ((k + 1)!)k (k + 2)k (k + 2)!. By the inductive hypothesis, this is ≤ 2! · 4! · · · (2k)!(k + 2)k (k + 2)!. Now k + 2 ≤ m ≤ 2k + 2, so we can consider all k different m in the interval and say (k + 2)k ≤ (k + 3) · (k + 4) · · · (2k + 2).

Therefore, (((k + 1) + 1)!)k+1 ≤ 2! · 4! · · · (2k)!(k + 2)k (k + 2)! ≤ 2! · 4! · · · (2k)! · (k + 3) · (k + 4) · · · (2k + 2)(k + 2)! = 2! · 4! · · · (2k)! · (2k + 2)!.

1 1 n 1 = 1+1 . Suppose that for n ≤ k, we have ∑ni=1 i·(i+1) . Consider = n+1 12. Well, 1+1 k+1

k 1 1 1 = ∑ i · (i + 1) ∑ i · (i + 1) + (k + 1) · ((k + 1) + 1) . i=1 i=1

By the inductive hypothesis, this is 1 k(k + 2) 1 k + = + (k + 1) (k + 1)(k + 2) (k + 1)(k + 2) (k + 1)(k + 2) k2 + 2k + 1 (k + 1)2 k+1 k(k + 2) + 1 = = = , = (k + 1)(k + 2) (k + 1)(k + 2) (k + 1)(k + 2) k + 2 as desired.

13. 2 · 1 < 22 ; suppose that for n ≤ k, we have n! < nn , and consider (k + 1)! = k!(k + 1). By the inductive hypothesis, this is < kk (k + 1) < (k + 1)k (k + 1) = (k + 1)k+1 , as desired. (n+1)2

14. This is a toughie. The right-hand side is n3 + (n + 1)3 . The left-hand side is ∑ j=n2 +1 j. (The author admits she found the lower bound by using the OEIS.) Four base cases are given in the (n+1)2 problem statement, so suppose that for n ≤ k, we have ∑ j=n2 +1 j = n3 + (n + 1)3 . Consider ((k+1)+1)2

(k+2)2

∑ j=(k+1)2 +1 j, and note that we can write this as ∑ j=1 formula from above to see that this is

(k+1)2

j − ∑ j=1

j. Now we can use the

(k+2)2 ((k+2)2 +1) (k+1)2 ((k+1)2 +1) (k + 2)2 ((k + 2)2 +1) − (k + 1)2 ((k + 1)2 +1) − = .

2 2 2 What sort of feat of algebra is needed next? Yuck. The numerator is (k2 + 4k + 4)(k2 + 4k+ 5) − (k2 + 2k + 1)(k2 + 2k + 2) = (k4 + 8k3 + 25k2 + 36k + 20) − (k4 + 4k3 + 7k2 + 6k + 2) = 4k3 + 18k2 + 30k + 18. Now dividing by two, we get 2k3 + 9k2 + 15k + 9 = k3 + 3k2 + 3k + 1 + k3 + 6k2 + 12k + 8, which is (k + 1)3 + (k + 2)3 . Notice that induction was not necessary (except to prove a result we used in an intermediate step)! It’s all just algebra. We could have done it for the k case, and it would have been simpler than the k + 1 case.

15. Counting edges of Kn and ∑nj=1 j.

4.13. Problems That Use Induction

29

16. We know the base case, that |A1 × A2 | = |A1 | · |A2 |. Suppose that for n ≤ k, |A1 × A2 × · · · × An | = |A1 | · |A2 | · · · |An |. Now consider |A1 × A2 × · · · × Ak+1 | = |(A1 × A2 × · · · × Ak ) × Ak+1 |, which by the base case is |A1 × A2 × · · · × Ak | · |Ak+1 |. Now, by the inductive hypothesis, this is |A1 | · |A2 | · · · |Ak | · |Ak+1 |.

17. The solution is the same as given for Problem 5(e) in Section 4.3, but is written more formally here.

We will induct once on vertices and once on edges to provide two possible proofs. For each proof, take as a base case the tree with a single edge. Assign each of the two vertices to a part, and we see that this tree is bipartite.

Now assume that a tree with n ≤ k vertices is bipartite, and consider a tree with k + 1 vertices.

Hopefully you have proved that every tree has a leaf ℓ with neighbor v. (If not, go back and prove this or induct on edges instead.) Remove ℓ to obtain a tree with k vertices; it is bipartite, so each vertex belongs to part A or part B. Replace ℓ; if v belongs to A, assign ℓ to B, and if v belongs to B, assign ℓ to A. Our tree with k + 1 vertices is bipartite.

Now assume that a tree with n ≤ k edges is bipartite, and consider a tree T with k + 1 edges.

Remove an edge e to obtain two trees T1 and T2 , each of which has fewer than k + 1 edges, so each is bipartite. Let the parts of T1 be A, B and the parts of T2 be C, D. Replace e. If its endpoints are in . . . . . . A,C, then let the two parts be A ∪ D and B ∪C and we are done.

. . . A, D, then let the two parts be A ∪C and B ∪ D and we are done. . . . B,C, then let the two parts be A ∪C and B ∪ D and we are done.

. . . B, D, then let the two parts be A ∪ D and B ∪C and we are done.

18. As a base case, consider n = 2. We have 13 + 23 = 9 = (1 + 2)2 . Now suppose that for k 3 3 3 n ≤ k, we have ∑nj=1 j3 = (∑nm=1 m)2 , and consider ∑k+1 j=1 j = (∑ j=1 j ) + (k + 1) . Using 2 the inductive hypothesis, this equals (∑km=1 m)2 + (k + 1)3 . Now notice that (∑k+1 m=1 m) = ((∑km=1 m) + k + 1)2 . That’s of the form (a + b)2 = a2 + 2ab + b2 so equals (∑km=1 m)2 + k 2 2(k + 1) ∑km=1 m + (k + 1)2 . Because ∑km=1 m = k(k+1) 2 , this is also equal to (∑m=1 m) + 2(k + 1) k(k+1) + (k + 1)2 = (∑km=1 m)2 + (k + 1)2 (k + 1). But this is exactly what we just showed 2 3 that ∑k+1 j=1 j equals! So putting it all together, we’re done.

19. For n = 2, we have 1 · 1 + 2 · 2 = 5 = 3! − 1. Now assume that for n ≤ k, we have ∑nj=1 j · k j! = (n + 1)! − 1, and consider ∑k+1 j=1 j · j! = ∑ j=1 j · j! + (k + 1)(k + 1)!. By the inductive hypothesis, this equals (k + 1)! − 1 + (k + 1)(k + 1)! = (k + 1)!(1 + k + 1) − 1 = (k + 1)!(k + 2) − 1 = (k + 2)! − 1, as desired.

20. As a base case, note that the sum of the interior angles of a triangle is π . Now assume that for n ≤ k, the sum of the interior angles of any n-gon is π (n − 2). Consider a (k + 1)-gon. Choose a vertex v that is farther from the middle of the polygon than its two neighbors. (Such a vertex must exist as otherwise the polygon is open.) Remove v and join its neighbors u, w by an edge. This leaves us with a k-gon, and its interior angles sum to π (k − 2). The vertices u, v, w

30

4. Induction

form a triangle. Glue it on to the k-gon and erase the interior edge to obtain a (k + 1)-gon with π (k − 2) + π = π (k − 1) total interior angle, as desired.

21. ∑nj=1 j( j+1) = n(n+1)(n+2) . For n = 2 we have 1 + 3 = 2·3·4 2 6 6 . Now assume that for n ≤ k, n(n+1)(n+2) j( j+1) j( j+1) . Consider ∑k+1 = ∑kj=1 j( j+1) + (k+1)(k+2) , which by the in∑nj=1 2 = j=1 6 2 2 2 ductive hypothesis equals

k(k + 1)(k + 2) (k + 1)(k + 2) k(k + 1)(k + 2) 3(k + 1)(k + 2) (k + 3)(k + 1)(k + 2) + = + = , 6 2 6 6 6 as desired.

22. This is, of course, true for a puzzle of size n. Certainly a puzzle of size 2 is solvable by fitting the two pieces together. Assume that any puzzle of size n less than 500 (or k + 1) can be solved using exactly n − 1 fits. Solve the 500-piece puzzle and remove a piece. The remaining puzzle has only 499 pieces so required 498 fits by the inductive hypothesis. Then replace the piece to solve the 500-piece puzzle using 499 fits.

23. We will start with two base cases.

For n = 1, rotate the tiling of Figure 4.10 on page 117 so that it matches the missing square.

For n = 2, split the grid into four 2 × 2 grids. One of these four grids contains the missing square; tile it using a single L-shaped tile. Tile the other three grids by imagining that each is missing the square closest to the center of the 4 × 4 grid. Cover these three “missing” squares with an L-shaped tile, and tile the remainder of each grid with one L-shaped tile.

Now suppose that for any n ≤ k we can tile a 2n × 2n grid with any square missing with Lshaped tiles. Consider a 2k+1 × 2k+1 grid. Split it into four 2k × 2k grids. One of these four grids contains the missing square; tile it using the inductive hypothesis. Then imagine that the other three grids is missing the square closest to the center of the 2k+1 × 2k+1 grid. Cover these three “missing” squares with an L-shaped tile, and tile the remainder using the inductive hypothesis. Ta da!

24. One circle placed in the white plane can be colored grey. Two overlapping circles may have their outer lunes colored grey and their intersection colored white. Assume that for n ≤ k, n circles drawn in the plane can be two-colored. Consider k + 1 circles drawn in the plane.

Erase one of them and two-color the resulting k circles. Re-draw the (k + 1)st circle and switch the color of every region within it. Done.

25. It does not work for squares because two squares drawn so that they share part of an edge must be different colors, yet they must each be a different color than the background. Thus three colors are needed. A similar problem occurs with triangles. If we consider a trio of intersecting spheres, there are curves along which three different regions meet. So the proof will not work. It seems unlikely that changing the theorem to use surfaces instead of curves, or to use three colors instead of two, will easily yield a proof. .

26. ∑nj=1 j( j + 2) = n(n+1)(2n+7) 6 Check a base case of n = 1. (1 · 3) = 3 = 1(1+1)(2+7) .

6

4.13. Problems That Use Induction

31

n Assume that the statement holds for k ≤ n and consider ∑n+1 j=1 j( j + 2) = ∑ j=1 j( j + 2) +

+ (n + 1)(n + 2) = (n + 1)(n + 2). By the inductive hypothesis this expression n(n+1)(2n+7) 6

(n+1)(2n2 +7n) (n+1)(n+2)(2(n+1)+7) n+1 2 + (n+1)(6n+18) = n+1 .

6 6 6 (2n +13n+18) = 6 (n+2)(2n+9) = 6

27. 132 and 366, so that together with the last fitting we get 499 fittings total.

A puzzle of size n can be solved using exactly n − 1 fits. Certainly a puzzle of size 2 is solvable by fitting the two pieces together. Assume that any puzzle of size n less than k + 1 can be solved using exactly n − 1 fits. Solve the (k + 1)-piece puzzle and remove a piece. The remaining puzzle has only k pieces so required k − 1 fits by the inductive hypothesis. Then replace the piece to solve the (k + 1)-piece puzzle using k − 1 + 1 = k fits.

28. Base case: when n = 2, the sum is 1 = 22 .

2

k(k−1) Inductive step: ∑k+1−1 i = k + ∑k−1 = 2k+k2 −k = (k+1)k i=1 i=1 i = k + 2 2 .

29. Base cas: If n = 1, we have 1 · 2 · 3 = 6 = 2 · 3.

Inductive Hypothesis: For n ≤ k, n(n + 1)(n + 2) is a multiple of 3.

Inductive Step: Consider (k + 1)(k + 2)(k + 3). Rewrite as [(k + 1)(k + 2)](k + 3) to expand to k(k + 1)(k + 2) + 3(k + 1)(k + 2) so we can apply the inductive hypothesis. We now have 3t + 3(k + 1)(k + 2) which is a multiple of 3.

30. Base case: If n = 1, we have 1 · 2 · 3 = 6.

Inductive Hypothesis: For n ≤ k, n(n + 1)(n + 2) is a multiple of 6.

Inductive Step: Consider (k + 1)(k + 2)(k + 3). Rewrite as [(k + 1)(k + 2)](k + 3) to expand to k(k + 1)(k + 2) + 3(k + 1)(k + 2) so we can apply the inductive hypothesis. We now have 6t + 3(k + 1)(k + 2). We know that k is either even or odd. If k is even, then k + 2 is even; if k is odd, then k + 1 is even. Either way, (k + 1)(k + 2) is even. Thus 6t + 3(2s), which is a multiple of 6.

31. There is a hidden assumption: by adding a vertex to a generic k-vertex graph with all vertices of degree at least 1, we can obtain a generic (k + 1)-vertex graph with all vertices of degree at least 1. But notice that the counterexample graph in Figure 4.11 on page 123 cannot be constructed from a 3-vertex graph with all vertices of degree at least 1. This is the peril of trying to move from k to k + 1 instead of finding case k within case k + 1!

32. We will prove that an n-pin pinwheel graph has 3n edges. The base cases are 1-pin and 2pin pinwheels. Take a (k + 1)-pin pinwheel graph and remove a pin, which has 2 vertices and 3 edges. The resulting k-pin pinwheel graph has 3k edges. Replace the pin and get 3k + 3 = 3(k + 1) edges.

33. We will prove that an n-bubble bubblepath graph has 3n − 1 edges. The base cases are 1bubble and 2-bubble bubblepaths. Hang a (k + 1)-bubble bubblepath graph over a nail. A single edge hangs over the nail because a bubble is too curved to stay there. Remove that edge and get two smaller bubblepath graphs, of sizes j and (k + 1)− j. Together their edges number 3 j − 1+ 3(k + 1− j)− 1 = 3k + 3− 2, and if we restore the nail edge we get 3(k + 1)− 2+ 1 = 3(k + 1) − 1 edges as desired.

32

4. Induction

34. When we restore the removed vertex, it may be adjacent to vertices in both parts of the remaining bipartite graph. Thus the restored graph need not be bipartite.

35. When we remove the leaf, the remainder may no longer have a leaf, and so the inductive hypothesis may not apply.

36. Base case: 2 piles of 1 teal stone each have a product of 1.

Inductive step: We start with a pile of k +1 small teal stones. Make the first division. We know that each pile has k or fewer stones, so the inductive hypothesis applies to each pile. Thus j−1 the sum of the products of one pile is ∑i=1 i and the sum of the products of the other pile is k+1− j−1 j) and (k− j+1)(k− . The sum i. By Problem 28, these are equal (respectively) to j( j−1) ∑i=1 2 2 (k− j+1)(k− j) j( j−1) + j(k + 1− j) of the products for the pile of k + 1 small teal stones is then 2 + 2 2

2

2

2

2

j − j+2 jk+2 j−2 j = k 2+k = ∑ki=1 i (for the first pile division), which is equal to j − j+k − jk+k− jk+ 2 as desired.

37. Smallest case: 2 piles of 1 teal stone each have reciprocals 11 + 11 = 2.

Other small cases: a pile of 3 stones breaks into 2 piles (of 2 stones and 1 stone), and then into 3 piles of 1 stone each to give a product of sums ( 12 + 11 )( 11 + 11 ) = 3; a pile of 4 stones breaks into 2 piles of 2 stones each, or 2 piles one of which has 3 stones and the other of which has 1 stone. These give products of sums ( 21 + 21 )( 11 + 11 )( 11 + 11 ) = 4 and ( 31 + 11 )( 21 + 11 )( 11 + 11 ) = 4.

Conjecture: ps(n) = n. This is hilarious!

Inductive step: We start with a pile of k + 1 small teal stones. Make the first division. We know that each pile has k or fewer stones, so the inductive hypothesis applies to each pile.

Thus the product of the sums of the reciprocals of one pile is j and the product of the sums 1 of the reciprocals of the other pile is k + 1 − j. We then compute j(k + 1 − j)( 1j + k+1− j) = j+ j j(k + 1 − j) k+1− j(k+1− j) = k + 1 as desired.

38. One side of one line placed in the white plane can be colored teal.

Assume that for n ≤ k, the regions formed by n lines drawn in the plane can be two-colored.

Consider k + 1 lines drawn in the plane and the resulting regions. Erase one line and two-color the resulting configuration formed by the k lines. Re-draw the (k + 1)st line and switch the color of every region on one side of it. Done.

39. Base case: 8 − 1 = 7.

Inductive hypothesis: If n ≤ k, 8n − 1 is a multiple of 7.

Inductive step: Consider 8k+1 − 1 = 8 · 8k = 1. We may rewrite as (7 + 1)8k − 1 = 7 · 8 ∗ k + 1 · 8k − 1 = 7 · 8 ∗ k + (8k − 1) and apply the inductive hypothesis to obtain 7 · 8 ∗ k + 7t, which is a multiple of 7.

40. Base case: m = 2, where K2,2 \ e has 2 · 2 − 1 = 3 edges.

Inductive hypothesis: for m ≤ k, K2,m \ e has 2m − 1 edges.

Inductive step: Consider K2,k+1 . Each of the “k + 1” vertices has 2 incident edges. After removal of e, one of the vertices will have one incident edges and the others will have 2

4.13. Problems That Use Induction

33

incident edges each. Ditch a vertex with 2 incident edges to obtain K2,k \ e, which has by ind. hyp. 2k − 1 edges. Replace the vertex, and we see that K2,k+1 \ e has 2k − 1 + 2 = 2(k + 1) − 1 edges.

Chapter 5

Algorithms with Ciphers 5.5 Try This! Encryptoequivalent Modulalgorithmic Problems 1.

(a) Let’s check the three properties.

Symmetry: If α and β represent the same letter, then β and α represent the same letter, so symmetry holds.

Reflexivity: Certainly α represents the same letter as α does, so reflexivity holds.

Transitivity: If α and β represent the same letter, and β and γ represent the same letter, then α represents the same letter as γ does and so transitivity holds. Yes, ∼1 is an equivalence relation. The equivalence classes are [a], [b], . . . , [z] (or [A], [B], . . . , [Z]). (b) Let’s check the three properties.

Symmetry: If α and β have the same case, then β and α have the same case, so symmetry holds.

Reflexivity: Certainly α has the same case as α does, so reflexivity holds.

Transitivity: If α and β have the same case, and β and γ have the same case, then α has the same case as γ does and so transitivity holds. Yes, ∼2 is an equivalence relation.

The equivalence classes are [a] and [A]. (c) If we apply ∼1 , the corresponding equivalence relation in Z52 is arithmetic modulo 26. So the equivalence classes are [0], [1], . . . , [25]. If we apply ∼2 , the corresponding equivalence relation in Z52 is arithmetic modulo 2. So the equivalence classes are [0] and [1]. (d) Yes. Under ∼1 we have Z26 ; under ∼2 we have Z2 .

2.

(a) By definition of congruence modulo n, we know that a − b = kn and c − d = ℓn. We can rewrite these statements as a = b+kn and c = d +ℓn. Therefore, ac = (b+kn)(d +ℓn) = bd + dkn + bℓn + kℓn2 = bd + qn, where q = dk + bℓ + kℓn. Thus, ac − bd = qn and this means ac ≡ bd (mod n).

(b) Use the previous part with c = a, d = b to see that a2 ≡ b2 (mod n). Now proceed by induction, using the inductive hypothesis that ak−1 ≡ bk−1 (mod n) and in the inductive step applying the previous part (again). (c) 43 · 2 = 86 and 92 − 86 = 6. Therefore, 92 ≡ 6 (mod 43), so by the previous part with k = 57, 9257 ≡ 657 (mod 43).

3. Solutions will vary.

34

5.8. Bonus: Algorithms for Searching Graphs

35

4. Solutions will vary.

5. Solutions will vary.

6. Solutions will vary. But, for the last question there are only two possibilities, so trial and error on a few letters should suffice to separate plaintext from wacktext.

5.8 Bonus: Algorithms for Searching Graphs Practice: Solutions will vary.

5.9 Bonus 2: Pigeons and Divisibility There are at most ten different last digits of pk (actually, at most four last digits because it can’t end in anything even or in 5), so two powers of p must end in the same digit. Call these pk and pr . Then pk − pr = pr (pk−r − 1) is divisible by 10 and thus pk−r − 1 is divisible by 10 as pk is not divisible by 2 or by 5. So, pk−r − 1 ends in 0 and pk−r ends in 1. Similarly, there are finitely many different last-n-digits of pk , so two powers of p must end in the same n digits; then pk−r − 1 is divisible by 10n and so ends in n 0s and pk−r ends in (n zeros)1.

Related Problem.

5.11 Problems about Algorithms, Modular Arithmetic, and Ciphers 1. The instructions are designed to keep cheese from molding or drying out. What do we change twice weekly? Do we change the water or the foil/parchment/paper, or both? The instructions certainly fail on clarity.

2. lyicb kmfyui zyzw ayppmr hmaicwq.

3. sunset is beautiful. wish you were here.

4. If a is divisible by 3, then a · a is divisible by 3, so a ∼ a and ∼ is reflexive. If either a or b is divisible by 3, then a · b and b · a are divisible by 3, so ∼ is symmetric. However, if b is divisible by 3 but a, c are not divisible by 3 (e.g., a = 2, b = 6, c = 4) then a ∼ b and b ∼ c but a 6∼ c so ∼ is not transitive.

5. [0], [1], . . . , [n − 1]; {x ∈ Z|x ≡ 0 (mod n)}, {x ∈ Z|x ≡ 1 (mod n)}, . . . , {x ∈ Z|x ≡ n − 1 (mod n)} or {. . . , −2n, −n, 0, n, 2n, . . . }, {. . . , 1−2n, 1−n, 1, n+1, 2n+1, . . . }, . . . , {. . . , −1− 2n, −1 − n, −1, n − 1, 2n − 1, . . . }.

Note that the union of the sets is Z—every integer is in one of the sets because every integer has a remainder on division by n—and that no integer is in two of the sets because no integer has two different remainders on division by n, so any pair of the sets is disjoint. Thus the sets partition Z.

36

5. Algorithms with Ciphers

6. a ≡ b (mod n) means that a and b have the same remainder when divided by n. So, a = cn + k and b = dn + k. Now, a − b = cn + k − (dn + k) = cn − dn = (c − d)n, so n|(a − b).

Similarly, if n|(a − b) then a − b = qn. We can divide b by n to obtain b = dn + k. Then a − b + qn = (dn + k) + qn = (d + q)n + k, which shows that a and b have the same remainder when divided by n. Therefore, a ≡ b (mod n).

7. The inputs are two chopsticks, though this is not stated. The algorithm is not correct because it is possible to tuck a chopstick under one’s thumb in a way that will not bring it into contact with another chopstick. Also, surely the algorithm does not allow one to pick up a car. The algorithm is not very clear because there are many ways to tuck chopsticks under thumbs and to hold pencils. The role of food in the use of the chopsticks is also unclear, as one might just move the second chopstick up and down, seizing and dropping food without bringing the food to the mouth.

8. sdrawkcab si egassem siht.

9. If a ≡ b (mod n), then (a − b) = kn for some k ∈ Z; likewise, (b − c) = ℓn for some ℓ ∈ Z. Thus, a − c = (a − b) + (b − c) = kn + ℓn = (k + ℓ)n = qn for q ∈ Z, and therefore a ≡ c (mod n).

10. The algorithm does nothing, as 7 is not even. It terminates quite quickly.

11. fingerless flutes are fancy and fun.

12. The algorithm computes n!.

13. Direct proof.

14. a + b is even if and only if b + a is even, so symmetry holds. a + a = 2a is always even, so a ∼ a and reflexivity holds.

If a + b and b + c are even, then a, c are both even or both odd (if b is even, then both a, c are even and if b is odd, then both a, c are odd) and in either case, a + c is even.

So this is an equivalence relation.

The equivalence classes are [0] = {2k | k ∈ Z}, [1] = {2k + 1 | k ∈ Z} because any two odd numbers sum to an even number, but an odd plus an even is odd.

15. azzy zszd apgkclqlqylrw zsjkwa htrz bhtl hwhppk inegniep.

16. The remainder of 123 on division by 6 is 3, and 2 6≡ 3 (mod 6). So, no.

17.

1. Let k = 1.

2. If k > 9, then stop; otherwise, output 2k, replace k with k + 1, and go to step 1.

18. We know from the definition that a− b = kn and c− d = ℓn. Therefore, a− b+ c− d = kn+ ℓn or a + c − (b + d) = (k + ℓ)n = qn, so a + c ≡ b + d (mod n).

19.

(a) Solutions will vary, but one possible example is 20 ≡ 30 (mod 10) but 4 6≡ 12 (mod 10).

5.11. Problems about Algorithms, Modular Arithmetic, and Ciphers

37

(b) Solutions will vary, but one possible example is 60 ≡ 75 (mod 15) but 20 6≡ 25 (mod 15). (c) It holds when n is prime.

(d) Suppose ac ≡ bc (mod n). Then ac − bc = kn for some k ∈ Z. We can write this as (a − b)c = kn. Now notice that if c|k then we get a ≡ b (mod n). But if c6 | k then if we divide both sides by c, then the right-hand side is no longer divisible by n, so we do not have a ≡ b (mod n). Thus, in any counterexample to the converse of Theorem 5.3.3 c has a common factor with n. But c does not have to divide n, so we can construct an example: Let (a − b)c = kn, let n = αβ , and let c = β γ (so k = γ and a − b = α ). Now let α = 2, β = 3, γ = 5, and choose a, b odd. For example, let a = 5, b = 3, c = 15, k = 5, n = 6, so 75 ≡ 45 ≡ 3 (mod 6) but 5 6≡ 3 (mod 6).

20. No, because A ⊂ B does not imply B ⊂ A (e.g., {1, 2} ⊂ {1, 2, 3} but {1, 2, 3} 6⊂ {1, 2}); the symmetric property does not hold.

21. i can haz cheezburger.

22. stop calling stop calling I don’t want to think anymore.

Original Vigenère cipher was used.

23. 24 = 4!, which is the number of orders in which the charms might be fed into the machine.

In order to solve this problem we need to define “look the same.” There are two reasonable definitions.

(1) Two bracelets look the same if they are rotations of each other. In this case, the partition is {ABCF, BCFA,CFAB, FABC}, {ABFC, BFCA, FCAB,CABF}, {ACBF,CBFA, BFAC, FACB}, {AFCB, FCBA,CBAF, BAFC}, {AFBC, FBCA, BCAF,CAFB}, {ACFB,CFBA, FBAC, BACF}.

(2) Two bracelets look the same if they are rotations or reflections of each other. In this case, the partition is {ABCF, BCFA,CFAB, FABC, FCBA,CBAF, BAFC, AFCB}, {ABFC, BFCA, FCAB,CABF, ACFB,CFBA, FBAC, BACF}, {ACBF,CBFA, BFAC, FACB, AFBC, FBCA, BCAF,CAFB}.

24.

(a) Astep = 12, Bstep = 7.

Astep = 6, Bstep = 14.

Astep = 3, Bstep = 28; write down 28.

Astep = 1, Bstep = 56; write down 56.

Add 28 + 56 = 84. (b) Solutions will vary. (c) You write down Bstep when Astep is odd, and otherwise you don’t write down Bstep. (d) This is the tough part. The first time you write down Bstep, you have 2n B, where n is the number of powers of 2 by which A is divisible. Thereafter, Astep is odd, so is not divisible by 2. We take the floor after dividing by 2 and write down 2k B, where k is the number of powers of 2 by which Astep is divisible. So, we get a sum of powers of 2 of B in the end.

38

5. Algorithms with Ciphers

(e) Yes! Surreptitiously, the algorithm is working from the binary representation of A. When we write down Bstep, we have calculated where the next rightmost digit 1 is in A’s representation, and when we take the floor after dividing by 2, we are essentially truncating that rightmost 1. (f) Usual multiplication would return A·B. This algorithm essentially determines the binary representation of A right to left as it proceeds, one placeholder per iteration. If the next number in the binary representation is a 0, the algorithm increments the power of two by one, and if the next number is a 1, we write down the power of two in A’s binary representation. The result is that we get 2k1 B + 2k2 B + · · · + 2kr B = (sum of powers of 2)·B = A · B. 25. Solutions will vary. A lot.

26. a2 = a2 , so a ∼ a and ∼ is reflexive.

If a ∼ b then a2 = b2 , so b2 = a2 , and b ∼ a so ∼ is symmetric.

If a ∼ b and b ∼ c then a2 = b2 and b2 = c2 , so a2 = c2 , i.e. a ∼ c and so ∼ is transitive. Thus ∼ is an equivalence relation.

Note that a ≡ −a, and the equivalence classes are {{a, −a} | a ∈ . . . Z? R?}.

27. This translates to {1, 4, 7, 10, 13} ∪ {3, 8, 13} ∪ {2, 5, 8, 11, 14}, which is not a partition because two of the sets overlap (at 8) and not all numbers are present (see 0, 6, 12).

28. Solutions will vary, but one example is {0, 1, 4, 7, 9} ∪ {2, 3, 5, 6} ∪ {8, 10, 11}; {0, 5, 8} ∪ {7, 2, 4}; {0, 1, 2, 3, 4, 5, 6, 7, 8} ∪ {8, 9, 10, 11}; {6, 2, 8, 4} ∪ {6, 2, 7, 9, 11}. × 0 1 2 3 4

29.

+ 0 1 2 3 4

30.

(a) 107 − 77 = 30; 30 − 28 = 2; therefore, 2 ≡ 107 (mod 7).

0 0 1 2 3 4

1 1 2 3 4 0

2 2 3 4 0 1

3 3 4 0 1 2

4 4 0 1 2 3

0 0 0 0 0 0

1 0 1 2 3 4

2 0 2 4 1 3

3 0 3 1 4 2

4 0 4 3 2 1

(b) 483 − 450 = 33; 33 − 27 = 6. Thus, 6 is the smallest x. (c) 16 − 9 = 7.

(d) Yes, because 7 + 3 = 10. (e) No; 17 is already reduced mod 22 and 49 − 44 = 5 so 49 ≡ 5 (mod 22).

31. hsbdlmft bsf tijoz boe mjlf tijoz uijoht.

32. The algorithm computes. . . 0.

33. a + 2b is even if and only if a is even, so for b odd we have b + 2a odd. Thus symmetry does not hold. a + 2a = 3a is even exactly when a is even, so reflexivity does not hold.

5.11. Problems about Algorithms, Modular Arithmetic, and Ciphers

39

If a + 2b and b + 2c are even, then a is even so a + 2c is even. Thus transitivity holds.

So, this is not an equivalence relation.

34. stop the cat from eating the flowers.

35. The inputs of the algorithm are contact lenses and RepleniSH R solution. It does terminate after storing the lenses for at least 6 hours in a closed lens case.

36.

1. Let k = 1 and let sum = 0.

2. If k > n, then stop; otherwise, replace sum with sum + k2 .

3. Replace k with k + 1 and go to step 2.

37. 9 − 6 = 3, so yes; 389 − 184 = 205; 205 − 184 = 21 < 87, so no.

38. ex wtz e uikg afd vtfbml qazvk.

39. n + 1; k + n.

40. Using the standard Vigenère cipher, we get uuf ybm nwly ey zoj li yeg li Kefsgw Sgjywt.

Usng the original Vigenère cipher, we get uuf ybw trjz gx lzh fs nsp mc Yiltaw Wlrqil.

Theme I Supplement TI.2 Potential Practice Proof Problems TI.2.1 Problems pertaining to Chapter 1: 1. An odd number is 2k + 1 and an even number is 2ℓ. Note that k and ℓ must be different as otherwise a relationship is implied between numbers that are supposed to be generic! 2k + 1 + 2ℓ = 2k + 2ℓ + 1 = 2(k + ℓ) + 1 = (odd).

2. Counterexample: Let k = −2 and n = 45. Then kn = −90 = 2(−45) = (even).

3. An element of A × B × C is of the form (a, b, c). For each element of A, an ordered triple can be formed with each element of B and each element of C. There are |A| ways of placing an element a ∈ A in the first component of the triple, |B| ways of placing an element b ∈ B in the second component of the triple, and |C| ways of placing an element c ∈ C in the third component of the triple. Therefore, by the product principle, |A × B ×C| = |A| · |B| · |C|.

4. Let m be odd. It can be written as m = 2k + 1. Then 4m − 3 = 4(2k + 1) − 3 = 8k + 4 − 3 = 8k + 1 = 2(4k) + 1 = 2r + 1 = (odd).

5. Suppose n + 6 is even, so for some k we have n + 6 = 2k. Rewriting, we have n = 2k − 6 = 2(k − 3) = 2r = (even).

6. An even number is of the form n = 2k. We may write k in binary as a sum of powers of 2, e.g., k = 2a + 2b + · · · + 2c , where the lowest possible power of 2 is 20 = 1. Then we may write n = 2(2a + 2b + · · · + 2c ), and expansion shows that the lowest possible power of 2 is 2 · 20 = 2 · 1 = 2. Therefore, the coefficient of 20 is 0, and the binary representation ends in 0.

7. If a is even, then a = 2k. If b is odd, then b = 2m + 1. Then ab + 1 = (2k)(2m + 1) + 1 = 4km + 2k + 1 = 2(2km + k) + 1, which is odd.

TI.2.2 Problems pertaining to Chapter 2: 8. This is an if-and-only-if, so there are two parts. First show that A ⊆ B ⇒ A ∪ B = B. If A ⊆ B, then for a ∈ A, a ∈ B. Therefore, A ∪ B = {c | c ∈ A or c ∈ B} = {c | c ∈ B} = B. (One can also use double-inclusion, which is longer but perhaps easier to parse.) Now show that A ⊆ B ⇐ A ∪ B = B. If A ∪ B = B, then for a ∈ A, a ∈ B. Therefore, A ⊆ B. 9. Suppose that A ⊆ B, and consider an element q ∈ A ∩C. We know q ∈ A, q ∈ C and because A ⊆ B, q ∈ B. Therefore, q ∈ B, q ∈ C so q ∈ B ∩C, and therefore A ∩C ⊆ B ∩C. 40

TI.2. Potential Practice Proof Problems

41

10. Suppose A ∪ B ⊆ A ∩ B. Then for a ∈ A, a ∈ A ∩ B so a ∈ B. Therefore A ⊆ B. Likewise, for b ∈ B, b ∈ A ∩ B so b ∈ A. Therefore, B ⊆ A, and thus A = B. 11. Suppose that A ⊆ B and consider an element q ∈ A × C. The element q has the form (a, c).

Because A ⊆ B, a ∈ B so (a, c) ∈ B ×C. Therefore, A ×C ⊆ B ×C. 12. We will proceed by double-inclusion. (This could also be done by using a sequence of {element | condition} statements.) Consider q ∈ A × (B ∩ C). It has the form (a, r), where r ∈ B, r ∈ C. Therefore, (a, r) ∈ A × B and (a, r) ∈ A ×C, so A × (B ∩C) ⊆ (A × B) ∩ (A ×C).

Now consider s ∈ (A × B) ∩ (A ×C). It has the form (a,t), where a ∈ A,t ∈ B,t ∈ C. Therefore, (a,t) ∈ A × (B ∩ C). And therefore, A × (B ∩ C) ⊇ (A × B) ∩ (A × C). As a result, A × (B ∩C) = (A × B) ∩ (A ×C).

13. Use the contrapositive: prove that if m is odd, then m2 − 2m is odd. Suppose m is odd, so that m = 2k + 1. Then m2 − 2m = (2k + 1)2 − 2(2k + 1) = 4k2 + 4k + 1 − 4k − 2 = 4k2 − 2 + 1 = 2(2k2 − 1) + 1 = (odd).

14. Proof by contradiction: Suppose there is a c ∈ (A \ B) ∩ (B \ A). Then c ∈ A \ B, which means that c ∈ A, c 6∈ B. But also c ∈ B \ A, which means that c ∈ B, c 6∈ A. Therefore, c ∈ A and c 6∈ A, and c ∈ B and c 6∈ B. These are contradictions.

15. Suppose that z ∈ Z and z2 | z. Then z = kz2 for some k ∈ Z. It follows that 1 = kz unless z is 0. Because k, z ∈ Z, both k and z must be ±1. Therefore z ∈ {−1, 0, 1}.

16.

(a) Let n be even. Then n = 2k for some integer k. Next, n2 + n = (2k)2 + 2k = 4k2 + 2k = 2(2k2 + k) which is divisible by 2, and thus even. (b) If n2 + n is even, then n is even. (c) Let n = 3. Then n + n2 = 3 + 9 = 12. Now, 12 is even but 3 is odd, so the converse is false.

17. Suppose that A ∩ B = A. This implies that A ⊂ B because x ∈ A implies x ∈ A ∩ B implies x ∈ B. Now suppose that x ∈ U . If x 6∈ A, then x ∈ A and if x ∈ A then x ∈ B. Therefore, x ∈ A ∪ B. 18. Suppose that no child is at least 10 years old. Then each child is 9 years old or younger.

The maximum sum of the children’s ages is 36, so the average age is at most 9. That is a contradiction. Or we could conclude that the average age of the children is not 10 to complete the use of the contrapositive.

19.

(a) If n is even, then n = 2k. Therefore n + 2 = 2k + 2 = 2(k + 1), which is even. (b) Suppose n + 2 is not even. Then n + 2 is odd, so n + 2 = 2k + 1 and thus n = 2k − 1 = 2(k − 1) + 1, which is odd. . . and therefore n is not even. (c) Suppose n is even but n + 2 is not even. Then n + 2 is odd, so n + 2 = 2k + 1 and thus n = 2k − 1 = 2(k − 1) + 1, which is odd. . . contradiction!

42

Theme I Supplement

TI.2.3 Problems pertaining to Chapter 3: 20. Label the vertices of K4 as a, b, c, d and the vertices of W4 as 1, 2, 3, 4. We see that the map a 7→ 1, b 7→ 2, c 7→ 3, d 7→ 4 is a bijection, and the edge sets are identical, so this is a graph isomorphism.

21. Well-defined: Suppose z1 = z2 . Add 3 to both sides to obtain z1 + 3 = z2 + 3. This is f (z1 ) = f (z2 ).

One-to-one: Suppose f (z1 ) = f (z2 ). Then using the definition of f , z1 + 3 = z2 + 3. Subtract 3 from each side to obtain z1 = z2 .

Onto: Consider a z ∈ Z. f (z − 3) = (z − 3) + 3 = z, so we have produced an element of Z (namely z − 3) whose image is z ∈ Z. 22. Well-defined: Suppose (a1 , b1 ) = (a2 , b2 ). Then a1 = a2 and b1 = b2 . Therefore, −b1 = −b2 .

This means that (−b1 , a1 ) = (−b2 , a2 ), which also says that f ((a1 , b1 )) = f ((a2 , b2 )).

One-to-one: Suppose f ((a1 , b1 )) = f ((a2 , b2 )). By definition of f , this says (−b1 , a1 ) = (−b2 , a2 ). It follows that −b1 = −b2 and a1 = a2 , so b1 = b2 as well. Therefore (a1 , b1 ) = (a2 , b2 ).

Onto: Consider a (a, b) ∈ Z×Z. The element (b, −a) ∈ Z×Z has the property that f ((b, −a)) = (a, b).

23. Proceed by contradiction. Suppose there is such a graph. Then the handshaking lemma holds, so the sum of the degrees is even. However, 1 + 1 + 2 + 3 + 4 + 4 + 5 + 7 = 27, which is not even; contradiction.

24. Every vertex has degree 2, so there are 2n edges, except that we’ve counted each twice (once for each end), so there are n edges.

25. No! Suppose it were possible. The average degree of a vertex would be 7. However, a simple graph with 6 vertices can have maximal degree no more than 5. Contradiction!

26. The function f is not one-to-one because (1, 5) and (5, 1) both map to 26.

The function f is not onto because 1 has no preimage; the “smallest” element of N × N is (1, 1), which maps to 2.

27. f is not one-to-one because both (3, 2) and (2, 3) map to 5 (and so do (1, 4), . . . ). f is not onto because 1 has no preimage; the “smallest” element of N × N is (1, 1), which maps to 2.

TI.2.4 Problems pertaining to Chapter 4: 28. (1) Direct proof: 2n = n + n and n > 1 so n + n > n + 1.

(2) Induction: A base case of n = 2 gives 4 > 3. For the inductive step, 2(k + 1) = 2k + 2 > k + 1 + 2 > k + 2.

TI.2. Potential Practice Proof Problems

43

29. As base cases, we check n = 1 (2 + 3 = 5 = 1 + 4) and n = 2 (5 + 7 = 12 = 4 + 8). We then suppose that for n ≤ k, we have ∑nj=1 (2 j + 3) = n2 + 4n. Consider ∑k+1 j=1 (2 j + 3) = k 2 ∑ j=1 (2 j + 3) + 2(k + 1) + 3. By the inductive hypothesis, this is = k + 4k + 2(k + 1) + 3 = k2 + 6k + 5 = k2 + 2k + 1 + 4k + 4 = (k + 1)2 + 4(k + 1), as desired.

30. As base cases, we check n = 1 (3 + 1 = 4 = (2 + 3 + 3)/2) and n = 2 (4 + 12 + 1 = 17 = 3 2 +3n .

(16 + 12 + 6)/2). We then suppose that for n ≤ k, we have ∑nj=1 (3 j2 + 1) = 2n +3n 2 k+1 k 2 2 2 Consider ∑ j=1 (3 j + 1) = ∑ j=1 (3 j + 1) + (3(k + 1) + 1). By the inductive hypothesis, this is 2k3 + 3k2 + 3k + 2(3(k + 1)2 + 1) 2k3 + 3k2 + 3k + (3(k + 1)2 + 1) = 2 2 2k3 + 3k2 + 3k + 6k2 + 12k + 6 + 2 = 2 2k3 + 9k2 + 15k + 8 = 2 2k3 + 6k2 + 6k + 2 + 3k2 + 6k + 3 + 3k + 3 = 2 2(k + 1)3 + 3(k + 1)2 + 3(k + 1) = , 2 as desired.

31. As base cases, we check n = 1 (12 > 3) and n = 2 (144 > 9). We then suppose that for n ≤ k, we have 12n > 3n . Consider 12k+1 = 12k · 12. By the inductive hypothesis, this is greater than 3k · 12 = 3k+1 · 4 > 3k+1 , as desired.

2 n 2 = 31 . Suppose that for n ≤ k, we have ∑ni=1 (i+1)(i+2) . Consider = n+2 32. Well, 2·3 k+1

2

k

2

2

∑ (i + 1)(i + 2) = ∑ (i + 1)(i + 2) + (k + 1 + 1) · ((k + 2) + 1) .

i=1

i=1

By the inductive hypothesis, this is 2 k(k + 3) 2 k + = + (k + 2) (k + 2)(k + 3) (k + 2)(k + 3) (k + 2)(k + 3) k(k + 3) + 2 k2 + 3k + 2 (k + 1)(k + 2) k + 1 = = = = , (k + 2)(k + 3) (k + 2)(k + 3) (k + 2)(k + 3) k + 3 as desired.

TI.2.5 Problems pertaining to Chapter 5: 33. Suppose that [a] = [b]. Then by definition, {s ∈ S | s ∼ a} = {s ∈ S | s ∼ b}. Certainly a ∼ a and because these sets are equal, a ∈ {s ∈ S | s ∼ b} so a ∼ b. Now suppose that a ∼ b.

Consider [a] = {s ∈ S | s ∼ a}.

44

Theme I Supplement

34. By Theorem 5.3.3 we know that ac ≡ bc (mod n) (and that ad ≡ bd (mod n) and that ca ≡ da (mod n)) and that cb ≡ db (mod n). By the transitive property (which holds because ≡ is an equivalence relation), we have ac ≡ bc ≡ bd (mod n).

35. Suppose that 3|x. Then x = 3k for some integer k, so x2 = 9k2 = 3(3k2 ) = 3m so 3|x2 .

Suppose that 36 | x. Because 3 is prime, there are no two factors that can combine to make 3.

So 36 | x2 .

36. a + b is odd if and only if b + a is odd, so symmetry holds. a + a = 2a is never odd, so reflexivity does not hold.

If a + b and b + c are odd, then a, c are both even or both odd (if b is even, then both a, c are odd and if b is odd, then both a, c are even) and in either case, a + c is even—not odd. Thus transitivity does not hold.

So, this is not an equivalence relation.

37. |a| = |a|, so a ∼ a and ∼ is reflexive.

If a ∼ b then |a| = |b|, so |b| = |a|, and b ∼ a so ∼ is symmetric.

If a ∼ b and b ∼ c then |a| = |b| and |b| = |c|, so |a| = |c|, i.e. a ∼ c and so ∼ is transitive.

Thus ∼ is an equivalence relation.

Note that a ≡ −a, and the equivalence classes are {{a, −a} | a ∈ . . . Z? R?} 38. a ≤ a + 2, so a ∼ a and ∼ is reflexive.

Suppose a = 2 and b = 16. Then a ≤ b + 2 but b 6≤ a + 2, so ∼ is not symmetric.

If a ∼ b and b ∼ c then a ≤ b+2 and b ≤ c+2. Thus b+2 ≤ c+4. Suppose a = 5, b = 4, c = 2.

Then 5 ≤ 4 + 2 and 4 ≤ 2 + 2 but 5 6≤ 2 + 2, so ∼ is not transitive.

TI.3 Problems on the Theme of the Basics 1. We proceed by contradiction. Suppose that each victim worked on a different floor. Then there are at most 12 victims, one per floor. However, there were at least 20. Contradiction!

2. They are isomorphic; simply take the upper-right vertex of the left-hand graph and drag it below the center bottom of the graph. Then rotate the whole thing by π4 counterclockwise to obtain the graph on the right.

To exhibit an isomorphism, the graphs must be labeled first and labelings will vary from person to person.

3. We proceed by induction. Check first some base cases: When n = 1, 12 = 1(2)(3)/6. When . n = 2, 12 + 22 = 5 = 2(3)(5)/6. Now let us assume that when n ≤ k, ∑nj=1 j2 = n(n+1)(2n+1) 6

TI.3. Problems on the Theme of the Basics

45

k 2 2 2 Consider ∑k+1 j=1 j = ∑ j=1 j + (k + 1) . By the inductive hypothesis, this becomes

k(k + 1)(2k + 1) k(k + 1)(2k + 1) + (k + 1)2 = + k2 + 2k + 1 6 6 k(k + 1)(2k + 1) 6k2 + 12k + 6 = + 6 6 2 2 (k + k)(2k + 1) + 6k + 12k + 6 = 6 3 2 2 (2k + 2k + k + k) + 6k2 + 12k + 6 = 6 3 2 2k + 9k + 13k + 6 = 6 (k + 1)(2k2 + 7k + 6) = 6 (k + 1)(k + 2)(2k + 3) = , 6 as desired.

4. In order for a vertex to have five neighbors, there must be at least six vertices total. Many graphs on six vertices can be drawn that satisfy these conditions; here is one.

5. Yes. Two consecutive cubes are of the form n3 and (n + 1)3 . Note that (n + 1)3 = n3 + 3n2 + 3n + 1, so (n + 1)3 − n3 = 3n2 + 3n + 1. Now, n must either be odd or even. If n = 2k is even, then 3n2 + 3n + 1 = 3(2k2 ) + 3(2k) + 1, which is odd (or take the term modulo 2 and obtain 1). If n = 2k + 1 is odd, then 3n2 + 3n + 1 = 3((2k + 1)2 ) + 3(2k + 1) + 1, which reduces modulo 2 to 1 + 1 + 1 ≡ 1 (mod 2) (or odd + odd + odd = odd).

6.

(a) Let n = −2 and then n3 = −8.

(b) Let x = π so that cos(x) = −1.

(c) Let x = 32π so that sin(x) = −1.

(d) Let x = 0 in which case x2 = 0.

7. 2 · 3 = 6.

8. An m-digit number is less than 10m , so a length-k list has at most k · 10m possible sums. We need k · 10m < 2k or k · 5m · 2m−k < 1. This is equivalent to ln(k) + m ln(10) − k ln(2) < 0, which solves to m < (−k ln(2) − ln(k))/ ln(10).

9. If a graph G is not blue, then G has 30 vertices.

46

10.

Theme I Supplement

(a) As a base case, note that f (1) = q as given. Suppose that for k < n, we have f (k) = qk .

Now consider f (n). We know f (n) = f (1+ (n− 1)) = f (1)+ f (n− 1) = f (1)· f (n− 1).

Now, by the inductive hypothesis, f (n − 1) = qn−1 , so we have f (1) · f (n − 1) = q · qn−1 = qn . (b) f (n) = f (n + 0) = f (n) · f (0) = qn · f (0). And, if we have qn = qn · f (0), this implies that 1 = f (0) as long as q 6= 0.

11. If there are ten vertices, then the highest degree any vertex could have is 9. And because the graph is connected, the lowest possible degree for a vertex is 1. That means that there are nine possible degrees and ten vertices, so by the pigeonhole principle two vertices must have the same degree.

12. All n ≥ 4.

13. We have 26 letters and 10 numerals for a total of 36 characters. There are, therefore, 36 choices for each of the (6 to 12) characters. Thus, by combining the product and sum principles, we have 366 + 367 + 368 + 369 + 3610 + 3611 + 3612 = 4,873,763,662,211,469,312.

14. Solutions will vary.

15. The primes below 30 are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29. There are ten of them. So, 2 of our 11 numbers must have a prime divisor in common, by the pigeonhole principle. n

3 n = . Consider (i + 2)(i + 3) n + 3 i=1

3 = 41 . Suppose that for n ≤ k, we have ∑ 16. Well, 3·4 k+1

k 3 3 3 = ∑ (i + 2)(i + 3) ∑ (i + 2)(i + 3) + (k + 2 + 1) · ((k + 3) + 1) . i=1 i=1

By the inductive hypothesis, this is k 3 k(k + 4) 3 + = + (k + 3) (k + 3)(k + 4) (k + 3)(k + 4) (k + 3)(k + 4) k(k + 4) + 3 = (k + 3)(k + 4) k2 + 4k + 3 = (k + 3)(k + 4) (k + 1)(k + 3) = (k + 3)(k + 4) k+1 = , k+4 as desired.

17. Solutions will vary: for example, f (z) = 2z, f (z) = ⌊ 2z ⌋, and f (z) = 0.

TI.3. Problems on the Theme of the Basics

18.

47

(a) Proof: Let us induct on the number of vertices of the polygon (or, equivalently, sides of the polygon). The base case is a polygon with four sides. It can be decomposed into triangles exactly two ways (one corresponding to each diagonal of the quadrilateral), and in each of these decompositions there are two triangles; each of them has two sides in common with the original polygon. The inductive hypothesis is that for any convex polygon with no more than n sides, any decomposition of the polygon into triangles has two triangles that each have two sides in common with the original polygon. Consider a convex polygon P with n + 1 sides. Connect any two vertices with a chord C. This decomposes the original polygon into two polygons P1 and P2 , each with fewer vertices (and sides) than the original polygon. Apply the inductive hypothesis to P1 and P2 to obtain decompositions into triangles, where P1 (respectively P2 ) has two triangles that have two sides in common with P1 (respectively P2 ). At most one of those triangles can have a side in common with C, so when the triangle-decomposed P1 and P2 are combined to make a triangle-decomposed P, at least one triangle from P1 and one triangle from P2 have two sides in common with P. This completes the proof. (b) (i) It’s a tree. It’s acyclic because the triangle decomposition has no interior vertices, and it’s connected because all the triangles are contiguous in the polygon. (ii) No. An edge of the tree crosses a side of a triangle and a triangle has only three sides, so no vertex of the tree may have degree more than three (but there are lots of trees—for example, the star graphs—that have vertices with degrees higher than three). (iii) A triangle with two edges in common with the original polygon corresponds to a leaf in the tree. (iv) It shows that a tree with degree at most three must have at least two leaves!

19. Neat: 12 = 5 + 7, 18 = 7 + 11, 24 = 11 + 13, 36 = 17 + 19, 42 = 19 + 23, . . . . The proposition fails for 48: 48 = 23 + 25, but 23 + 29 = 52 and all smaller consecutive primes give smaller sums (as seen in the list above).

20. Well,

n r 1 n r = . Suppose that, for n ≤ k, we have ∑ = . r(r + 1) (r + 1) (i + r − 1)(i + r) n + r i=1 k r r r =∑ + . (k + r − 1 + 1) · ((k + r) + 1) i=1 (i + r − 1)(i + r) i=1 (i + r − 1)(i + r)

k+1

Consider ∑

By the inductive hypothesis, this is k r k(k + r + 1) r + = + (k + r) (k + r)(k + r + 1) (k + r)(k + r + 1) (k + r)(k + r + 1) k2 + rk + k + r = (k + r)(k + r + 1) k+1 , = k+r+1 as desired.

k(k + r + 1) + r (k + r)(k + r + 1) (k + 1)(k + r) = (k + r)(k + r + 1)

=

48

Theme I Supplement

21. If a ∼ b then a ≡ b (mod 3) or a ≡ b (mod 5). Because ≡ is an equivalence relation, this means that b ≡ a (mod 3) or b ≡ a (mod 5), so b ∼ a. Also, a ≡ a (mod 3) and a ≡ a (mod 5), so a ∼ a. It is not true that if a ∼ b and b ∼ c, then a ∼ c: Let a = 1, b = 4, and c = 9. We have that 1 ≡ 4 (mod 3) and 4 ≡ 9 (mod 5), but 9 ≡ 0 (mod 3) and 4 6≡ 1 (mod 5).

22. First suppose that G can be decomposed into copies of K1,3 . K1,3 has one vertex of degree 3 and three vertices of degree 1. Notice that each vertex of G either is contained in a single K1,3 or is contained in three separate K1,3 copies (because each vertex has degree exactly 3 and no two “fingers” of a claw lead to the same vertex). Take the vertex of degree 3 of each K1,3 (the “palm” or “knuckle” of the claw), and collect these into part 1. The remaining vertices may be placed in part 2. No two vertices in part 1 are adjacent because each has all three edges pointing into part 2. No two vertices in part 2 are adjacent because each of the three edges points to some “palm” in part 1.

Now suppose that G is bipartite. Notice that each vertex in part 1 is adjacent to three different vertices in part 2. Take all the vertices in part 1, and assign them to be “palms” of K1,3 copies, and for each of the three adjacent vertices in part 2, form the “fingers” of that K1,3 . Each vertex in part 2 is adjacent to three different vertices in part 1, so G has been decomposed.

23. At least two people want to get off at the same floor, by the pigeonhole principle.

Solutions will vary, but may include that someone made multiple mistakes in choosing a floor or that someone was playing a prank.

24.

(a) ∀n ∈ Z, 2n 6= 9.

(b) ∃ T such that (T is equilateral) ∧(T has perimeter 10). (c) ∀

,(

has diameter d ∈ Z) ∨ (

has area a ∈ Z).

(d) ∀a, b, ∈ N, ∃k ∈ Z such that (a < k < b) ∨ (a > k > b).

25. If there are four ways per algorithm, then 6 · 4 = 24.

26. Solutions will vary, but here’s one: A = {1, 2, 3, 4, 5, 6, 7, 8}, B = {1, 2, 3, 9, 10, 11}.

27. We proceed by induction. As base cases, we take n = 1, 2, 3 and note that 2−3 = −1, −1+4− 3 = 0 = 1− 1, and 0+ 6− 3 = 3 = 22 − 1. The inductive hypothesis is that for n ≤ k, ∑nj=1 2 j − k 3 = (n − 1)2 − 1. We examine ∑k+1 j=1 2 j − 3. This is equal to ∑ j=1 2 j − 3 + 2(k + 1) − 3 and 2 2 by the inductive hypothesis to (k − 1) − 1 + 2(k + 1) − 3 = k − 2k + 1 − 1 + 2k + 2 − 3 = k2 − 1 = (k + 1 − 1)2 − 1, as desired.

28. There are only k − 1 different residues modulo k − 1, so by the pigeonhole principle, for any k integers at least two must have the same residue mod (k − 1).

29.

(a) b ∧ ¬p.

(b) (s ∧ b) ∨ p.

(c) ¬b ∧ (p ∨ s).

TI.3. Problems on the Theme of the Basics

49

30. gjqqtg qpqn qjvvn qjbetgh.

31. We proceed by induction. As base cases we take n = 1, 2 and note that 2 = 4 − 2 and 2 + 4 = 6 = 8 − 2. The inductive hypothesis is that, for n ≤ k, ∑nj=1 2 j = 2n+1 − 2. We consider k j k+1 and by the inductive hypothesis equals 2k+1 − 2 + 2k+1 = j ∑k+1 j=1 2 . This equals ∑ j=1 2 + 2 2 · 2k+1 − 2 = 2k+2 − 2, as desired.

32. I left my head and my heart on the dance floor A standard Vigenère cipher was used.

33. This is an algorithm that outputs n cycling through values 2, 5, and 8; it does not terminate.

34.

(a) {{d, e}, f }.

(b) {d, e, f }.

35. The clue is that this is a shift cipher of shift 8! i 0 8 a

t 11 19 l

t 11 19 l

w 14 22 o

e 22 30/4 w

t 11 19 l

a 18 26/0 s

i 0 8 a

z 17 25 r

m 4 12 e

b 19 27/1 t

m 4 12 e

i 0 8 a

t 11 19 l

36. There are 9 different values modulo 9, but there are 10 numbers of taps, so by the pigeonhole principle two must be the same modulo 9.

37. We could write them out {3, 10, 17, . . . } and count the size of the set. Or we could note that 7 · 14 = 98 so there are 14 positive multiples of 7 less than 100; adding 3 to all of them takes the highest one out of range so only 13 are left, but including the number 3 = 3 + 0 · 7 brings the number back up to 14.

38.

(a) Amali is tall or dark or beautiful, and Amali is not tall and dark and beautiful all at the same time. So, Amali has one or two of the characteristics tall, dark, and beautiful. (b) Amali is neither tall nor dark. (c) Amali is dark and not both tall and not dark. So, Amali is dark. Alternatively, this is equivalent to d ∧ (¬t ∨ d), and thus we know she’s dark and may or may not be non-tall.

(d) Amali is both tall and dark, or ze is both not tall and dark. So, Amali is dark in either case. We can just say Amali is dark.

39. Yes!

Symmetry: If (a, b) ∼ (c, d), then a − c = b − d. Multiplying through by −1, we have c − a = d − b, which means that (c, d) ∼ (a, b).

Reflexivity: a − a = 0 = b − b, so (a, b) ∼ (a, b).

Transitivity: Suppose (a, b) ∼ (c, d) and (c, d) ∼ (e, f ). Then, a − c = b − d and c − e = d − f .

Add these together to obtain a − c + c − e = b − d + d − f , or a − e = b − f . Thus, (a, b) ∼ (e, f ).

50

Theme I Supplement

40. Let x ∈ B. Then x 6∈ B. Because A ⊆ B, this means that x 6∈ A. Therefore, x ∈ A, so B ⊆ A.

41.

(a) There are 45 = 1,024 because for each of the domain elements, there are four choices of target elements. (b) None, because |A| > |B|.

(c) There are 42 = 16. (d) There is only one target element not already hit. So, we must have either f (−2) = s or f (−1) = s. There are four functions that can be defined for each possibility, but this overcounts the function where f (−2) = s and f (−1) = s, so 4 + 4 − 1 = 7 onto functions.

42. We either need to show that there is an uncovered element of Z or that two of the sets overlap.

In this case, sets A and B both contain the element 10. And, the number 3 is not in any of the three sets!

43. This is double-inclusion. Suppose x ∈ B. Then, either x ∈ A or x 6∈ A. If x ∈ A, then x ∈ A ∪ (B \ A). If x 6∈ A, then x ∈ (B \ A) and so x ∈ A ∪ (B \ A). This shows that B ⊆ A ∪ (B \ A).

Now consider x ∈ A ∪ (B \ A). Either x ∈ A or x ∈ (B \ A). If x ∈ A then x ∈ B because A ⊆ B. If x ∈ (B \ A) then x ∈ B, so A ∪ (B \ A) ⊆ B. This completes the proof.

44.

g 6 1 7 h

r 17 0 17 r

a 0 19 19 t

p 15 1 16 q

h 7 0 7 h

t 19 19 38 m

h 7 1 8 i

e 4 0 4 e

o 14 19 33 h

r 17 1 18 s

45.

(a) 21 + 3 = 24, which is divisible by 12.

y 24 0 24 y

i 8 19 27 b

s 18 1 19 t

t 19 0 19 t

h 7 19 26 a

e 4 1 5 f

b 1 0 1 b

o 14 19 33 h

m 12 1 13 n

b 1 0 1 b

(b) 15 + 3 = 18, which is not evenly divisible by 12.

So, the first statement is true and the second is false.

46. f is one-to-one. Suppose f (c, d) = f (g, h). This means 2c · 3d = 2g · 3h . This implies that c = g, d = h, and so (c, d) = (g, h). f is not onto because 1 has no preimage; the “smallest” element of N × N is (1, 1), which maps to 6.

47. We know from the definition that a − b = kn and c − d = ℓn. Therefore, a − b − (c − d) = kn + ℓn, or a − c − (b − d) = (k + ℓ)n = qn, so a − c ≡ b − d (mod n).

48. Suppose that there are vertices x, y, z such that d(x, y) + d(y, z) < d(x, z). Then, there exists a path from x to z through y that is shorter than the shortest path from x to z. Contradiction!

TI.3. Problems on the Theme of the Basics

51

49. There are 3 colors of ducks and 2 colors of chickens. By the product principle, 3 · 2 = 6 is the number of color pairs.

50.

Yes; (A \ B) ∪ (B ∩C) is the same as (A ∪ B) \ (B \C).

Part II Theme: Combinatorics

Chapter 6

Binomial Coefficients and Pascal’s Triangle 6.3 Try This! Investigate a Triangle 1.

1 1

7

21

8

28

35 56

35 70

21 56

7 28

1 8

1

2. Yes; the triangular array is generated by the choice notation identity. We could number the rows, and n would be one less than the row number (or we could start counting rows with 0).

And we could number the entries in each row, and k would be one less than the entry number (or we could start counting entries with 0).

12 11 11 3. We want to calculate 12 .

Using the identity, = + to use the 3 3 2 3 . We now have 10 10 11 10 10 10 identity again, twice: 11 = + and = + .

We know = 10, so we 2 1 2 3 9 2 3

1 10 9 10 9 9 have two more uses of the identity: 2 = 1 + 2 and 3 = 2 + 3 . Again, we know

9 more work to do: 92 = 81 + 82 and 93 = 82 + 83 . Then 81 = 8, 1 = 9, and we have

so we continue: 82 = 71 + 72 and 83 = 72 + 73 . We continue with 72 = 61 + 62 and

7 6 6 6 5 5 6 5 5 5 = 53 = 10 3 = 2 + 3 ; and, 2 = 1 + 2 and 3 = 2 + 3 . Luckily, we know 2

from the start of Section 6.2. So, we now have 62 = 5 + 10 = 15 and 63 = 10 + 10 = 20,

which gives us 72 = 6 + 15 = 21 and 73 = 15 + 20 = 35, which in turn gives us 82 =

7 + 21 = 28 and 83 = 21 + 35 = 56, and then 92 = 8 + 28 = 36 and 93 = 28 + 56 = 84,

= 9 + 36 = 45 and 10 = 36 + 84 = 120 and then 11 = 10 + 45 = 55 which leads to 10 2 2 3

12 = 55 + 165 = 220. and 11 = 45 + 120 = 165, followed finally by 3 3

8 This time, we want to calculate 4 . We have written out extra rows in the array, so we simply look at the fifth entry in the eight row, which is 70.

Using the array to count Sushi Samplers means we have to generate four more rows of the array: 1 1 1 1

9 10

11 12

36 45

55 66

84 120

165 220

126 210

330 495

126 252

462 792

84 210

462 924

36 120

330 792

9 45

165 495

1 10

55 220

1 11

66

1 12

1

Then we simply read the fourth entry in the twelfth row; it’s 220, as we calculated earlier.

Using the identity to count cookie recipes is more of a pain: 84 = 73 + 74 ; we know 73 = 35 53

54

6. Binomial Coefficients and Pascal’s Triangle

from above, so we just need 74 = 63 + 64 . Again, we know 63 = 20 from above, so we just

need 64 = 53 + 54 . Then, we know 53 = 10 from above, so we just need 54 = 43 + 44 =

4 + 1 by inspection. Together, this gives us 64 = 10 + 5 = 15, so 74 = 20 + 15 = 35 and 8 4 = 35 + 35 = 70, as expected.

Really, they’re doing the same calculations, but the accounting method for the array makes them more algorithmic and thus less of a pain.

n 4. nk = n−k . If we choose k sugar-numbers from a bag of n sugar-numbers, then there are n−k sugar-numbers left in the bag. This creates a bijection between choices of k sugar-numbers and choices of n− k sugar-numbers (we could choose the sugar-numbers in our hands or those in the bag). Thus, the number of ways to choose k sugar-numbers is the same as the number of ways to choose n − k sugar-numbers.

5. The short answer is . . . they’re the same! Let us use the borification of nk as the number of k-subsets of an set of size n. So we will look at rows in Pascal’s triangle. Notice that the odd-n rows (such as 1 5 10 10 5 1) have an even number of entries (n + 1 entries, in fact). n We have n0 = nn , n1 = n−1 , etc. from the previous problem. Note that 0 is even and n is odd; 1 is odd and n − 1 is even; and, when k is even (odd), n − k is odd (even). Thus the even number of entries in the row is evenly split between odd and even, and the symmetry of Pascal’s triangle gives a one-to-one correspondence between the subsets of odd size and the subsets of even size.

6. The number of subsets of {1, 2, . . . , n} is 2n and that’s also the sum of the numbers across a row of Pascal’s triangle—each entry in the row counts the number of subsets of a particular size.

6.5 Overcounting Carefully and Reordering at Will Super-Quickie Questions:

1. 6, 3, 8, 12, respectively.

2. 12.

3. 216 24 = 9. . . . oh.

23 4. 47 24 = 1 24 on average. That means 23 students each have 2 ears and one student has 1 ear. So Vincent Van Gogh must be one of the students.

6.6 Try This! Play with Powers and Permutations 1.

(a) n. (b) n − 1.

(c) n · (n − 1) · (n − 2) · · · · · (n − (k − 1)).

6.7. Binomial Basics

55

(d) We’ve overcounted because the order in which we pulled out the k sugar-numbers doesn’t matter. So we need to divide n · (n − 1) · (n − 2) · · · · · (n − (k − 1)) by k!. (e) Multiply top and bottom by (n − k)!.

n 2. x3 +3x2 y+3xy2 +y3 . Conjecture: the coefficients of the expansion of (x+y) are the numbers n in the nth row of Pascal’s triangle (so the kth one is k ).

3. Write (x + y)n as (x + y)(x + y) · · · (x + y), i.e., n copies of (x + y) multiplied together. Every monomial in the expansion has a total degree of n. Now, to obtain a particular monomial xk yn−k , we choose k of the n copies of (x + y) from which to take xs, and then the other n − k n of the n copies of (x + y) will have ys taken from them.

There are ways to do this, so there k

will be nk copies of xk yn−k in the expansion. n n k n−k n 4. Yes, of course. (x + y) = ∑ xy . k=0 k 5. Apply the binomial theorem with x = 4, y = 1 to get (4 + 1)6 = 56 .

6. ∑nk=0 nk (−1)k . Using the binomial theorem, it’s more convenient to have k index powers of y to make sense of the exponents. So we have x = 1, y = −1 and obtain (1 − 1)n = 0.

6.7 Binomial Basics Mini-Activity (do it right now!):

1. 1 + 3 + 3 + 1 = 8 = 23 ; 1 + 5 + 10 + 10 + 5 + 1 = 32 = 25 .

6.9 Try This! Pancakes and Proofs 1.

(a) Six. (b) Four. (c) 6 · 4 = 24 edges, but each is counted twice, so 12 edges.

2.

(a) If we want three pancakes and there are 6 choices for each pancake, then we have 63 = 216 different pancake orders if we specify the stacking order for the pancakes. (b) 6 · 5 · 4 = 120 different pancake orders. (c) 3! = 6.

3.

(d) 120/6 = 20 = 63 ; 6. (a) n(n − 1)/2.

(b) They are connected by an edge.

(c) n2 = n(n − 1)/2 because both sides count the number of edges of Kn .

56

6. Binomial Coefficients and Pascal’s Triangle

4. We will mark m of p pens, chosen from a pile of w pens. On the left-hand side, we choose p pens from the pile, and then for each such choice there are mp ways of choosing m pens to mark. On the right-hand side, we choose m pens from the pile and mark them, and then choose another p − m pens from the remaining w − m pens.

5. The right-hand side is the number of ways to choose n objects from a pile of 2n objects. Let’s say that n of them are blue and n are purple. Then, the left-hand side adds up the number of ways to choose k from the blue pile and then n − k from the purple pile, for all choices of k. (By symmetry, there are the same number of ways to choose n − k from the purple pile as there are to choose k from the purple pile.)

6. Choose r things from a pile of m + n things on the left-hand side. For the right-hand side, make a pile of m things and a pile of n things. Then sum (over i) all possible ways to choose i things from the n-sized pile and the rest of the r things from the m-sized pile.

6.12 Bonus: Sorting Bubbles in Order of Size Activities:

1. 2.3 1.2 6.4 5.5 3.4 2.9 5.7 7.9 1.2 2.3 5.5 3.4 2.9 5.7 6.4 7.9 1.2 2.3 3.4 2.9 5.5 5.7 6.4 7.9 1.2 2.3 2.9 3.4 5.5 5.7 6.4 7.9 2. 9 8 7 6 5 4 3 2 1.

3. Solutions will vary.

4. Solutions will vary.

5. Solutions will vary.

6.13 Bonus 2: Mastermind 6.13.2 Mini-Project 1. There are no R, Y, G, or B in the sequence (from the first two guess/response pairs). Thus all colors are K or W. The third and fourth guesses have the first K in common but the second has no b, and so the other three letters must be correct. Thus we have –WKW. If K is not correct, then the first letter must be W, so the sequence is WWKW. 2. There are no B, K, or W in the sequence (from the first two guess/response pairs). From the last guess/response pair, we know that the only colors are R and Y and that there are two of each. Therefore, the two correctly placed pegs in the third guess are YY, and the correctly placed peg in the fourth guess is –R– –. This tells us the sequence is –RYY. Because there must be another R, the sequence is RRYY.

6.15. Problems Binomially Combinatorial in Nature

57

3. The last guess tells us that there are four colors, R, G, B, and W. However, the third response tells us that there is only one of R and B, not both. This is a contradiction.

4. From these guesses and responses, we know that the four pegs are R, Y, G, and G in some ordering. The first and third guesses have the final two Gs in common, so they must be in the wrong slots. However, that means that the first guess must have R and Y in the correct slots, in which case there are no slots left for the two Gs. This is a contradiction.

5. Here is a listing. One peg: b or w. Two pegs: bb, bw, or ww. Three pegs: bbb, bbw, bww, www. Four pegs: bbbb, bbww, bwww, wwww. That’s a total of 13. Note that bbbw is not possible because the guess peg corresponding to the white marker peg can’t be in a different slot.

6. (a) 42 = 6 for the different possible placements of the Ys. The remaining slots must be filled with Rs.

(b) 41 = 4 for the different possible placements of the Y. The remaining slots must be filled with Rs. (c) Just one.

7. It doesn’t—from the previous guesses we know how many slots are not Ws. The remaining slots must be Ws. 8. This tells us which slot(s) color C occupies.

9. Use the first five guesses on RRRR, YYYY, GGGG, BBBB, KKKK. This tells us which colors are present and with what frequency. Again using C for a color that is present in the sequence and N for a color that is not present in the sequence, use the next three guesses on CNNN, NCNN, and NNCN. We now know the position(s) of color C. In a worst-case scenario, it appears only once. Without loss of generality, assume that C is definitely in the first slot. Let C′ be a different color present in the sequence. Use the next two guesses on CC′ NN and CNC′ N. This tells us the position(s) of C′ . In a worst-case scenario, it appears only once. Without loss of generality, assume that C′ is definitely in the second slot. Let C′′ be a third color present in the sequence. Use the last guess on CC′C′′ N. This determines the position(s) of C′′ and of any 4th color peg.

6.15 Problems Binomially Combinatorial in Nature 1.

(a) 5 + 11 = 16. (b) 2 + 6 + 8 = 16.

= 10. (c) 1+4 2

(d) The same number of ways you can choose 1 spider from 10 spiders: 10.

2. Every handful of k candies either has a special pink piece candy or it doesn’t, so the total number of ways of grabbing handfuls of k candies is (number of different handfuls of k candies including the special pink one) + (number of different handfuls of k candies with no

58

6. Binomial Coefficients and Pascal’s Triangle

special pink one). Counting the number of handfuls of k candies containing the pink candy is the same as setting aside the pink candy to eat and counting the number of ways to grab k − 1 candies from the remaining n − 1 candies. Counting the number of handfuls of k candies without the special candy is the same as tossing the special candy in the garbage and counting the number of ways to grab all k candies from remaining n− 1 candies. And, that (in a nutshell or candy-bag) is the proof.

3. There are 41 2 handshakes because a handshake represents a choice of two Committee members.

Or, each handshake represents an edge in K41 and so there are 41·40 2 handshakes.

There are likely additional answers to this question. Few, if any, involve milkshakes.

n n! = 1.

4. There is only one way to not choose anything. = 0!n!

0 There

are n different elements in an n-element set, so n ways to choose 1 of them. n n! = = n.

1 1!(n − 1)! n n(n − 1) n! = . See Section 6.9 for a combinatorial justification. = 2 2 2!(n − 2)! n(n − 1)(n − 2) n n! = . = 3!(n − 3)!

6 3

p p!

5. Consider the factorial expression = . We know that this is an integer and that j j!(p − j)! all of the factors in the denominator are less than or equal to p. Thus, none divide p. Therefore p! the factor of p in the numerator remains after simplifying the fraction and = pk j!(p − j)! for some integer k. This satisfies the definition of being divisible by p.

n n 6. Using simplifications of the factorial expression to compute this we get 2 + = 2 1 2n(n − 1) + n = n(n − 1) + n = n2 − n + n = n2 .

2 7. Whenever we choose n objects from a pile of 2n objects, we are also not choosing the remaining n objects. But we could have also chosen the remaining n objects instead. Thus, every choice of n objects is in one-to-one correspondence with a second choice of n objects; they come in pairs. Therefore the total number of ways to choose n objects from 2n objects is even.

n−1 8. We know that nk = n−1 k−1 + k . And, when n is even n − 1 is odd so we can use the results of Problem 5 of Section 6.3. Moreover, when k is even (odd), k − 1 is odd (even). So let us consider nk for all 1 ≤ k ≤ n − 1. (When k = 0, n we have nk = 1.) From the first term of the

basic identity, we obtain n−1 1, which gives us all but the last ‘1’ in the k−1 for all 1 ≤ k ≤ n −

n previous row of Pascal’s triangle. If we stick n with these, we get a whole row. We know from Problem 5 of Section 6.3 that the sum of these for odd k − 1 equals the sum of them for even k − 1; thus, the contributions to the sum of nk for even k equals the contributions to

6.15. Problems Binomially Combinatorial in Nature

59

the sum of nk for odd k. From the second term of the basic identity, we obtain n−1 for all k 1 ≤ k ≤ n − 1, which gives us all but the first ‘1’ in the previous row of Pascal’s triangle. If we stick n0 with these, we get a whole row. Again, the contributions for even k equals the contributions for odd k. Thus, we discover that the number of subsets of {1, 2, . . . , n} of odd size equals the number of subsets of {1, 2, . . . , n} of even size.

9. Let x = 1, y = 3 and obtain 4m .

10. There are 49 6 = 13, 983, 816 sets of numbers with which to fill out the ticket (or win the lottery—despite the fact that the numbers are reported in the order in which they are drawn, winnings are determined by the number of ticket numbers that match draw numbers). But you can write them in any order on the ticket, so there are 49 6 · 6! = 10, 068, 347, 520 ways to fill out the ticket!

11.

15 31 127 511 , , , .

4 5 7 9 2n+1 − 1 . n+1

12. Let n = 2 and observe that 0 20 + 1 21 + 2 22 = 0 + 2 + 2 = 4 = 2 · 22−1 . This completes

j k kj . our base case. Now assume that ∑nk=0 k nk = n2n−1 holds for n < j. Consider ∑k=0

j j−1 + k−1 ) and rewrite as = Apply the basic choice notation identity to obtain ∑k=0 k( j−1 k

j j j j−1 j−1 j−1 ∑k=0 k k + ∑k=0 k k−1 . We consider these two sub-sums separately. ∑k=0 k k = j−1 j−1 j j−1 j−1 j−1 j−1 ∑k=0 k k + j j = ∑k=0 k k +0 = ( j −1)2 j−2 by inductive hypothesis. ∑k=0 k k−1 = j j j j−1 j−1 j−1 ∑k=1 k k−1 because the first term is 0. Rewrite as = ∑k=1 ((k − 1) k−1 + k−1 ) = ∑k=1 (k −

j j−1 j−1 j−1 j−1 j−1 j−1 1) k−1 + ∑k=1 .

Then by an index shift, we have = k + ∑ ∑ k=0 k=0 k−1 k k . By the inductive hypothesis and a known identity, this is ( j − 1)2 j−2 + 2 j−1 . In sum, we have ( j − 1)2 j−2 + ( j − 1)2 j−2 + 2 j−1 = 2( j − 1)2 j−2 + 2 j−1 = ( j − 1)2 j−1 + 2 j−1 = j2 j−1 , as desired.

13. (a) The possible pairings are 12 and 34, 13 and 24 , 14 and 23. One thinks 42 to choose the first pair, and the second pair is determined by the remaining two people; but the 2 (4) pairs could appear in either order, so 22 .

(b) There are 62 ways to make the first pair. Then there are 4 people left, and we know

there are 42 ways to make the remaining two pairs. However, the pairs could appear in (6)(4) any order, so 2 3! 2 .

m m−2 . . . 42 2 2 . (c) m 2!

14. This is completely different! Consider the case of 4 dancers. There are 4! orderings of the dancers, and these orderings are really each 2 pairs with leaders specified. The order of the pairs doesn’t matter, though, so we have 4!2 . For 6 dancers, we have 6! orderings of the dancers, which are really each a listing of 3 pairs with leaders specified. Again, the order of

60

6. Binomial Coefficients and Pascal’s Triangle m! the pairs doesn’t matter, so we have 6!

3! . In general, there are (m/2)! ways to pair-and-specify the dancers.

15. The facial information is a distractor. The important information is that each vertex has degree 4. Then the total degree is 60 · 4 = 240 and because each edge is counted twice (once for each end), there are 120 edges.

16. By the Binomial Theorem with x = −1, y = 2 we have (−1 + 2)q = 1q = 1.

17. Using factorials, we verify that

n−1 n (k + 1) · n! (k + 1) · n! n · (n − 1)! n = = = (k+1) = . k k+1 k!(n − 1 − k)! (k + 1)k!(n − 1 − k)! (k + 1)!(n − (1 + k))!

18. There are n ways to choose a single item from a pile of n items. Then for each choice there are n−1 ways to choose another k − 1 items, for a total of k items chosen with a distinguished k−1 first item. Or we could choose k items from a pile of n items, but we need a distinguished first item—and there are k different choices for that first item.

19. There are n ways to choose a single item from a pile of n items. And there are (n − 1) ways to choose a second item from the remaining (n − 1) items. Then for each choice of two items, there are n−2 k−2 ways to choose another k − 2 items, for a total of k items chosen with a distinguished first item and distinguished second item. Or we could choose k items from a pile of n items, but we need a distinguished first item—and there are k different choices for that first item—and we need a distinguished second item; there are (k − 1) choices remaining for that second distinction.

20. It’s equal to 2n−1 . Induction using the basic choice notation identity; observe that this gives you the sum across the previous row of Pascal’s triangle.

21. By the binomial theorem with x = 64, y = −64, we have (64 − 64)s = 0.

22. Prove by induction using the basic choice notation identity on the right-hand side.

23. From Problem 6 we know that for j ≥ 2, 2 2j + 1j = j2 . We will sum both sides to obtain j j ∑mj=2 2 2 + ∑mj=2 1 = ∑mj=2 j2 . Notice that the lower bound on the index is 2 because that’s a restriction on the identity we started with; for j = 1, j2 = 1, so we can just add 1 to both sides of the resulting identity desired identity. Now let’s try to simplify. We know

to get the

that ∑mj=2 2 2j = ∑mj=1 2 2j because 12 = 0. And, Problem 22 tells us this is equal to 2 m+1 3 . ! !

m m m m+1 j j Similarly, we have that ∑ − 1. The result = ∑ −1 = ∑ j −1 = 2 j=2 1 j=1 1 j=1

m m m+1 m+1 m+1 m+1 2 2 is that ∑ j = 2 + − 1 and therefore ∑ j = 2 + .

3 2 3 2 j=2 j=1 24. Here is one way to do it. For the first element of the target, there are m different elements of the domain that may map to it. We can choose ℓ1 of these. But we can’t reuse these elements,

6.15. Problems Binomially Combinatorial in Nature

61

so the second element − ℓ possible preimages, of which we can choose ℓ2 . In this

has m m m − ℓ1 m − ℓ1 − ℓ2 . . . . However, there are lots of ways to choose fashion, we get ℓ3 ℓ2 ℓ1 the ℓ j , so we will need to sum over the number of ways they can be determined. And, we have restrictions on them! We know that there must be n of the ℓ j , as there is one set of choices for each of the n elements of the target; and, we know that the ℓ j must sum to m, i.e., ℓ1 +ℓ2 +· · ·+ ℓn = m so that every element domain has a destination of the in the target space. Therefore we m m − ℓ1 m − ℓ1 − ℓ2 ℓn−1 + ℓn obtain ... . I’ve seen a calculation ∑ ℓ2 ℓ3 ℓn − 1 ℓ1 +ℓ2 +···+ℓn =m ℓ1

n n m claiming that we should get ∑ (−1)n−k k , but I don’t know what methodology produces k k=1 this expression.

25.

(a) n!. (b) k!. (c) (n − k)!.

(d) We have n! ways to lay out the sugar-numbers. Choosing k sugar-numbers is like taking the first k from a permutation of the sugar-numbers. There are as many ways of taking the first k as there are permutations, so we still have n!. However, order doesn’t matter, and there were k! ways of ordering any particular k sugar-numbers, so we’ve overcounted by a factor of k!. In fact, it also doesn’t matter what order the remaining sugar-numbers are in, so we’ve also overcounted by a factor of (n − k)!. Thus our n! !. formula is k!(n−k) 26.

(a) There are 6 vertices, and each touches 4 faces. Every face has 3 vertices. So, there are 6 · 4/3 = 8 faces.

(b) There are 12 edges, and each touches 2 faces. Every face has 3 edges. So there are 12 · 2/3 = 8 faces.

27.

(a) Here are the first 5 rows of this Pascal-ish triangle: 1 2 4 8 16

1 4

12 32

1 6

24

1 8

1

(b) If we count rows from 0 as we did in Pascal’s triangle, then the sum is 3n . (c) (2x + y)n = ∑nk=0 [n△k]xn−k yk .

(d) (2x + y)n = ∑nk=0 nk 2n−k xn−k yk . This tells us [n△k] = nk 2n−k .

(e) It ([n△k]) is the number of ways of finding 2k subsets of {1, . . . , n} that have exactly k elements in common. (The geometrically inclined may recognize this Pascal-ish triangle as encoding the number of k-faces of the n-cube, and thereby see this subset characterization via the binary-string representation of vertices of the n-cube.)

62

28.

6. Binomial Coefficients and Pascal’s Triangle 6 5 1 12 (−4) = 5, 971, 968.

29. There are 60 vertices, each of which meets 2 squares, 1 triangle, and 1 pentagon. Each square has 4 vertices, each triangle has 3 vertices, and each pentagon has 5 vertices. So there are 60 · 2/4 = 30 squares, 60 · 1/3 = 20 triangles, and 60 · 1/5 = 12 pentagons for a total of 52 faces.

6 30. (a) 12 4 · 4 .

6 (b) 12 6 · 2 .

6 12 6 (c) 12 4 · 4 · 8! and 6 · 2 · 8! respectively.

31. 7 · 6 · 5 · 4 = 74 · 4!.

32. Using the binomial theorem, ((−4) + 1)c = (−3)c .

33. There must be 11, one of whom is missing a leg or has it raised.

34. Let’s see. First we notice that the powers of b are decreasing by 2 each term, so we can say x is some multiple of b2 . Then we know that these terms involve (b2 )9 and (b2 )8 , so n ≥ 9.

Examine the coefficients: 495 = 32 · 5 · 11 and 4455 = 34 · 5 · 11. There are no factors of 495 missing from 4455, and definitely nothing is present to the 8th or 9th power, so x = 1 · b2 .

k+1 n We now have nk yk = 32 · 5 · 11 and k+1 y = 34 · 5 · 11. The possibilities for y are 1, 3, 9.

n = 55 but that doesn’t occur anywhere on Pascal’s triangle.

If y = 9 then we have nk = k+1

If y = 1 then from the factorial definition of nk we have 9(k + 1)!(n − k − 1)! = k!(n − k)! or 9k = n − k or n = 10k. But looking in Pascal’s Triangle at the multiple-of-10 rows, we see that 495 and 4455 don’t occur next to each other in any of them (checking 10 and the starts of 20, 30, 40 will show this). Therefore y = 3. This tells us k = 2 so k + 1 = 3, so n = 11.

35. 42 + 5 + 7 = 18.

Chapter 7

Balls and Boxes and PIE—Counting Techniques 7.3 Try This! Let’s Have Some PIE 1. Because there are 5 blueberry-peach pies, there are 10 − 5 = 5 more pies that are plain blueberry and 15 − 5 = 10 more pies that are plain peach, for a total of 5 + 5 + 10 = 20 fruit pies, and thus 5 chocolate mousse pies.

2. |A ∪ B| = (|A| − |A ∩ B|) + (|B| − |A ∩ B|) + |A ∩ B|. And |Q \ (A ∪ B)| = |Q| − |A ∪ B|.

3. Apple-strawberry with no rhubarb: 3 − 1 = 2. Strawberry-rhubarb with no apple: 4 − 1 = 3.

Plain rhubarb pies: 5 − (3 + 1) = 1. Plain apple pies: 7 − (2 + 1) = 4. Plain strawberry pies: 8 − (3 + 2 + 1) = 3. And that’s a total of 4 + 1 + 2 + 1 + 3 + 2 = 13 pies.

4. |A ∪ B ∪C| = |A| + |B| + |C| − |A ∩ B| − |A ∩C| − |B ∩C| + |A ∩ B ∩C|.

5. Yes, painful though it be. |A ∪ B ∪ C ∪ D ∪ F ∪ G ∪ H| = |A| + |B| + |C| + |D| + |F| + |G| + |H| − |A ∩ B| − |A ∩C| − |A ∩ D| − |A ∩ F| − |A ∩ G| − |A ∩ H| − |B ∩C| − |B ∩ D| − |B ∩ F| − |B ∩ G| − |B ∩ H| − |C ∩ D| − |C ∩ F| − |C ∩ G| − |C ∩ H| − |D ∩ F| − |D ∩ G| − |D ∩ H| − |F ∩ G| − |F ∩ H| − |G ∩ H| + |A ∩ B ∩C| + |A ∩ B ∩ D| + |A ∩ B ∩ F| + |A ∩ B ∩ G| + |A ∩ B ∩ H| . . . plus the rest of the 73 threefold intersections, minus the 74 fourfold intersections, plus the

7 5 fivefold intersections, minus the 7 sixfold intersections, plus |A ∩ B ∩C ∩ D ∩ F ∩ G ∩ H|.

7.6 Try This! What are the Balls and What are the Boxes? And Do You Want Some PIE? 1. First notice that we want ten boxes of fudge. There are eight kinds, so we have to have some repetition. In other words, we have 10 + (8 − 1) slots, and we need to choose which of them

possible orders. This is why get the (8 − 1) dividers between types of fudge, so there are 17 7 it was difficult to choose what to order. . .

2. All four letters are different, so 4! = 24 anagrams.

3. There are nine letters, of which three are identical, so 9!

3! = 60,480 anagrams.

4. Importantly, |N ∪ M| = |U | − |N ∩ M| is nummy. The formula can be seen by using a Venn diagram.

63

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7. Balls and Boxes and PIE—Counting Techniques

5. The line is 18 animals long, so 18 of placing the white cats and 3 = 816 is the number of ways

then the bluejays fill the remaining places (which equals 18 , the number of ways of placing 15 the bluejays and letting the white cats sit in the remaining spots).

One way to do the second problem is to start with a lineup of 15 bluejays and insert the 3 cats.

There are 14 spots between bluejays plus two end-spots so 16 3 = 560.

6. We seek x. We know 112 = I + (66 − x) + (37 − x) + x = I + 103 − x, 84 = L + (66 − x) + (30 − x) + x = L + 96 − x, 46 = C + (37 − x) + (30 − x) + x = C + 67 − x, and 184 = 45 + I + L + C + (66 − x) + (37 − x) + (30 − x) + x = 178 + I + L + C − x. From this we can deduce that I = 9 + x, L = x − 12, and C = x − 21, so 184 = 154 + 2x and thus x = 15.

10 7. First solution: 7! orderings of the 10 7 digits for 7! · 7 = 604,800.

Second solution: We have ten choices for the first digit, only nine for the second, only eight for the third, etc., so we have 10 · 9 · 8 · 7 · 6 · 5 · 4 = 604,800 all-digits-distinct phone numbers.

Third solution: Consider ten objects in all possible orders. We only need seven, so chop off the last three. But there are six different ways those three could be ordered, so each remaining phone number of length 7 has six possible (now-phantom) endings. This produces a total of 10!/6 = 604,800.

If we have ten-digit numbers, there’s only one choice for the digits but 10! orderings of those.

Including country codes means we have more than ten digits, so by the pigeonhole principle there are no such numbers with distinct digits.

8.

(a) 1,820 + 1,820 + 560 = 4,200. (b) 8,568 − 4,200 = 4,168.

(c) 560 have both and 4,168 have neither, so a total of 4,728.

(d) 1,820 + 1,820 = 3,640 have one or the other. But an additional 4,168 have neither, so a total of 7,808.

7.9 Bonus: Linear and Integer Programming Questions for LP/IP practice:

1. Solutions will vary depending on how students decide to interpret the objective and constraints, but here is one possibility. Let the set of foods {bananas, milk, oranges, frozen blueberries, protein powder, soy milk, frozen strawberries, honey} be indexed by j. Variables x j stand for the quantities of food and constants c j stand for their respective costs. We also have constants e j , p j , h j , f j , va j , vc j , vd j , ve j , etc. (calories, protein, carbohydrates, fats, vitamins) for the nutrients per unit volume of each food. Our objective function for cost would be ∑8j=1 c j x j ; for nutritive value, we would have ∑8j=1 (∑ v j )x j . Then we have constraints: ∑8j=1 e j x j ≥ k1 tells us we must have a certain number of calories, and we might also have ∑8j=1 e j x j ≤ k2 if we want to limit the calories we take in. To meet recommended amounts of nutrients, we have constraints of the form ∑8j=1 p j x j ≥ k3 . Hidden constraints are that x j ≥ 0.

7.11. Problems about Balls, Boxes, and PIEs

65

2. Solutions will vary depending on how students decide to interpret the objective and constraints, but here is one possibility. Let the set of foods {bananas, milk, oranges, frozen blueberries, protein bar, soy milk} be indexed by j. Variables x j stand for the quantities of food and constants c j stand for their respective costs. We also have constants e j , p j , h j , f j , va j , vc j , vd j , ve j , etc. (calories, protein, carbohydrates, fats, vitamins) for the nutrients per unit volume of each food. Our objective function for cost would be ∑6j=1 c j x j ; for nutritive value, we would have ∑6j=1 (∑ v j )x j . Then we have constraints: x j is an integer, ∑6j=1 e j x j ≥ k1 tells us we must have a certain number of calories, and we might also have ∑6j=1 e j x j ≤ k2 if we want to limit the calories we take in. To meet recommended amounts of nutrients, we have constraints of the form ∑6j=1 p j x j ≥ k3 . Hidden constraints are that x j ≥ 0.

3. Let’s call the additional variables yr so that we don’t get messed up by the numbering. For ∑6j=1 e j x j ≥ k1 , we introduce the variable y1 = ∑6j=1 e j x j − k1 so that y1 + ∑6j=1 e j x j = k1 and do likewise for the other inequalities. We need one slack variable per constraint.

4. We will convert x j into binary. Because x j ≤ 12 < 16, we know that the binary equivalent of x j can have no more than four digits. Thus x j = y j1 + 2y j2 + 4y j3 + 8y j4 , where the y ji can only take on the values 0 or 1. This also gives us the additional constraints y j1 + 2y j2 + 4y j3 + 8y j4 ≤ 12. (In turn, we would want to introduce slack variables for these constraints as well. . . )

7.11 Problems about Balls, Boxes, and PIEs 1. There are ℓw such words.

2. There are 93 · 104 such numbers.

3. False. For example, let A = {1, 2, 3}, B = {1, 3, 4},C = {2, 4, 5}. Then A ∩ B = {1, 3} = 6 0, / and |A ∪ B ∪C| = 5 6= 9.

4. The number of words of length w is a1 · a2 · · · · · aw .

5. 18 12 = 18,564 ways.

6. We want one, three, five, or seven even numbers.

One even number: there are 71 ways for that one slot to be chosen from our seven slots. For each of those choices of slot, there are five choices of digit for each of the seven slots (five even digits for the even slot, five odd digits for each of the odd slots), so there are 57 ways to fill in the slots.

Three even numbers: there are 73 ways for those three slots to be chosen from our seven slots. For each of those choices of slots, there are five choices of digit for each of the seven slots, so there are 57 ways to fill in the slots.

Five even numbers: there are 75 ways for those five slots to be chosen from our seven slots.

For each of those choices of slots, there are five choices of digit for each of the seven slots, so there are 57 ways to fill in the slots.

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7. Balls and Boxes and PIE—Counting Techniques

7.

Seven even numbers: All seven slots are even. There are five choices of digit for each of the seven slots, so there are 57 ways to fill in the slots.

In total, we have 57 ( 71 + 73 + 75 + 1) = 5,000,000. Cool! (a) A four-letter word containing ACE has only one other letter in it. That letter can occur at the beginning of the word or the end of the word, and there are six possibilities for that letter. Thus, there are 2 · 6 = 12 words containing ACE.

(b) If a word doesn’t begin with F, then there are five choices for the first letter and six choices for each of the other three letters, or 5 · 63 = 1,080 words in total.

If a word doesn’t end in E, there are six choices for each of the first three letters and five choices for the last letter, or 5 · 63 = 1,080 words in total.

It looks like we have 2,160 words, but we have overcounted those words that neither begin with F nor end in E. There are 5·62 ·5 = 900 of those, so our total is 2,160− 900 = 1,260 words. (c) We can have – –CAB, –CAB–, or CAB– –. For each of these three possibilities, we have six choices for each of the two remaining letters. Thus, the total number is 3 · 62 = 108.

(d) If a four-letter word begins with C, then there are six choices for each of the other three letters and so there are 63 = 216 such words.

If a four-letter word ends in two vowels, there are six choices of letter for the first two letters and two choices of letter (A or E) for the last two letters, for a total of 6 · 6 · 2 · 2 = 144 words.

This double-counts those words that begin with C and end in two vowels. There are 6 · 4 = 24 such words because there are one, six, two, and two choices for the first– fourth letters, respectively. Thus, there are 216 + 144 − 24 = 336 four-letter words that begin with C or end in two vowels.

8.

(a) There are six possibilities for the first letter, five for the second letter, etc., so there are 6! = 720 words in total. (b) A word can’t begin with both D and E, so we use the sum principle. If a word begins with D, there are five possibilities for the second letter, four for the third, etc., so there are 5! = 120 words beginning with D. Similarly, there are 120 words that begin with E, so there are 240 words beginning with D or E. (c) A word can’t end with both B and A, so we use the sum principle. If a word ends with B, there are five possibilities for the first letter, four for the second, etc., so there are 5! = 120 words ending with B. Similarly, there are 120 words that end with A, so there are 240 words ending with B or A. (d) There are two choices for the first letter, two choices for the last letter, and 4! ways to arrange the remaining four letters. Thus, there are a total of 4 · 4! = 96 such words.

(e) There are 720−240 = 480 words with first letter neither D nor E and also 480 words that have last letter neither B nor A. We will now use PIE. Let R be the set of words beginning with neither D nor E and S be the set of words ending with neither B nor A. We have |R ∪ S| = |R| + |S| − |R ∩ S|; we desire |R ∩ S|, so this is equal to |R| + |S| − |R ∪ S|. Now,

7.11. Problems about Balls, Boxes, and PIEs

67

notice that |R ∪ S| is the number of words that do not have first letter D or E and last letter B or A. This is 720 − 96 = 624. This allows us to calculate |R ∩ S| = 480 + 480 − 624 = 336.

9. There are 19 different chocolates. Place 20 slots in a line to indicate the 20 chocolates, and place 18 dividers to indicate switches between types of chocolates. That gives 20+18 = 18 33,578,000,610 possibilities.

10.

(a) We will associate the jth mini-pile location to the jth box. Then, the problems are the 1 same. Or, we could choose the first k1 items; for each choice, we have k−k ways to k2 choose the second pile, and so forth. This gives the same solution as to the balls-andboxes problem. (b) Symbolically,

(k − k1 )! k! k! k k − k1 k! = = . = k1 !(k − k1 )! k2 !(k − k1 − k2 )! k1 !k2 !0! k1 !(k − k1 )! k2 k1 Combinatorially, we could choose a pile of size k1 and put the remaining k2 objects in a second pile, or we could just choose k1 objects.

kn + kn−1 k − k1 − k2 k − k1 k ··· (c) kn−1 k3 k2 k1 =

(k − k1 )! (k − k1 − k2 )! kn + kn−1 k! ... k1 !(k − k1 )! k2 !(k − k1 − k2 )! k3 !(k − k1 − k2 − k3 )! kn−1 !kn !

=

k! . k1 !k2 !k3 ! . . . kn−1 !kn !

Recall that k1 + k2 + · · · + kn = k, so that k − (k1 + k2 + · · · + kn−1 ) = kn . (d) The number of ways of anagramming a word with k letters, the jth of which is repeated k j times, is k1 !k2k!!...kn ! . The previous problem shows that this is equal to the multinomial

coefficient k1 ,k2k,...,kn .

10 10! (e) = = 2,520.

3, 2, 5 3!2!5!

12 12! = 415,800. = 4, 4, 2, 1 4!4!2!

(f) Note first that k1 ,k2k,...,kn = k1 !k2k!!...kn ! . Also, multiplication of integers is commutative, so the order of the k j in the denominator of the expression does not matter.

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7. Balls and Boxes and PIE—Counting Techniques

(g)

(k − 1)! (k − 1)! (k − 1)! + + (k1 − 1)!k2 !k3 ! k1 !(k2 − 1)!k3 ! k1 !k2 (k3 − 1)! =

(k − 1)!(k1 !(k2 − 1)!(k3 − 1)! + k2 !(k1 − 1)!(k3 − 1)! + k3 !(k1 − 1)!(k2 − 1)!) k1 !k2 !k3 !(k1 − 1)!(k2 − 1)!(k3 − 1)!

=

(k − 1)!(k1 (k1 − 1)!(k2 − 1)!(k3 − 1)! + k2 (k2 − 1)!(k1 − 1)!(k3 − 1)! + k3 (k3 − 1)!(k1 − 1)!(k2 − 1)!)

=

(k − 1)!(k1 − 1)!(k2 − 1)!(k3 − 1)!(k1 + k2 + k3 ) k1 !k2 !k3 !(k1 − 1)!(k2 − 1)!(k3 − 1)!

=

(k − 1)!(k1 + k2 + k3 ) k1 !k2 !k3 !

=

k! (k − 1)!k = . k1 !k2 !k3 ! k1 !k2 !k3 !

k1 !k2 !k3 !(k1 − 1)!(k2 − 1)!(k3 − 1)!

On the left-hand side, we have chosen piles of size k1 , k2 , k3 from a pile of k objects. For the right-hand side, we will choose these piles differently. Suppose one of the objects is pink. Remove it and place it in one of the piles. It could be in pile 1 or pile 2 or pile 3. There are are k − 1 objects left to place, and if the pink object is in pile 1, there k−1 k−1 ways to finish the choosing; if the pink object is in pile 2, there are k1 −1,k2 ,k3 k1 ,k2 −1,k3

ways to finish the choosing, and if the pink object is in pile 3, there are k1 ,kk−1 ways 2 ,k3 −1 to finish the choosing. The sum principle completes the argument.

k k−1 k−1 k−1 (h) = + +· · ·+ . k1 , k2 , . . . , kn k1 − 1, k2 , . . . , kn k1 , k2 − 1, . . . , kn k1 , k2 , . . . , kn − 1 (i) A generic monomial of the expansion of (x1 + x2 + · · · + xn )k has the form Cxk11 xk22 . . . xknn , where k1 + k2 + · · · + kn = k. To obtain such a monomial while expanding, one chooses x1 from k1 of the copies of (x1 + x2 + · · · + xn ), chooses x2 from k2 of the copies of (x1 + x2 + · · · + xn ), and chooses xn from kn of the copies of (x1 + x2 + · · · + xn ). The number of ways of obtaining such a monomial is k1 ,k2k,...,kn , so C = k1 ,k2k,...,kn .

11. 263 · 103 = 17,576,000.

Well, it would be okay for some years after 2009. But unless we want to reuse licence plate codes relatively frequently, it would be better to have more options.

12.

(a) 8,568 + 1,260 + 56 = 9,884. (b) 65,780 − (24,480 + 22,848 + 8,568 + 1,260 + 56) = 8,568. (c) 8,568 + 24,480 + 22,848 = 55,896.

(d) There are 26 department members, of whom 8 are women. This can be discovered by x and the reasoning that the total number of committees in an x-person department is 5

number of all-female committees in a department with y women is 5y .

13. There are 11 different choices for the pillow the first strand will embellish, 10 for the second strand, and so forth to obtain 11 · 10 · 9 · 8 · 7 · 6 · 5 · 4 = 6,652,800 ways.

7.11. Problems about Balls, Boxes, and PIEs

69

14. Good grief! There are 34 letters! Okay . . . 3 Ss, 2 Us, 2 Ps, 2 Es, 2 Rs, 3 Cs, 3 As, 3 Ls, 7 Is, 2 Os, and one each of F, G, T, X, D. That gives 34! = 1,412,469,529,257,855,275,311,104,000,000 anagrams.

(3!)4 (2!)5 7!

15. By PIE, we should have 119 = 96 + 53 + 39 − 38 − 31 − 32 + 22. However, 96 + 53 + 39 − 38 − 31 − 32 + 22 = 109, and so some number must be incorrect. There are other PIE-like ways of reaching this same conclusion.

20 16. = 125,970.

8 17. We have four digits under our control. Were we to have no restrictions, there would be 104 such numbers. If there were no 9s, then there would be nine choices for each digit, so 94 such numbers. Thus, there are 104 − 94 = 3,439 seven-digit phone numbers that begin with 231and contain at least one 9.

18. First, give one chipmunk to every hobgoblin. This means there are n − h chipmunks (balls)

n−h+h+t −1 remaining to distribute among h+t creatures (boxes) and thus there are = n−h

n+t −1 n−h+h+t −1 n+t −1 or = ways to hand out chipmunk snax. n−h h+t −1 h+t −1 19. Consider each of the y j to be a box. Put a 1 in each of the boxes. Now we need to line up seven 1s and five plus signs (to figure out how to distribute the seven 1s into the six boxes).

There are 12 5 = 792 ways to do that.

20. Let us first see how many possibilities each individual component of a banana split yields.

We have 18 flavors and we need to select 3 scoops with repetition. So that’s 17 bars dividing

possibilities.

Then we have 5 toppings (4 bars) and 3 the 18 flavors and 3 stars, or 17+3 3 4+3 scoops to place them on (stars), so 3 possibilities. There is only one choice for whipped cream: yes, there will be some. There are two choices for or not having nuts and two having

20 7 for having or not having a cherry. In total, we have 3 3 2 · 2 = 159,600 possible banana split orders.

21.

(a) Let the p things represent balls. Then there are q + 1 boxes into which they may be placed (one to the left of all q things, one to the right of all q things, and q − 1 boxes representing between the q things). Then, the ball-and-box solution is that the spaces p+q+1−1 there are ways to distribute the balls into the boxes. p (b) Let the q things represent balls. Then there are p + 1 boxes into which they may be placed (one to the left of all p things, one to the right of all p things, and p − 1 boxes representing between the p things). Then the ball-and-box solution is that the spaces q+ p+1−1 there are ways to distribute the balls into the boxes. q

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7. Balls and Boxes and PIE—Counting Techniques

q+ p q+ p (c) Well, = . p q 22. We would first give each baby mouse two cupcakes, leaving 43 − 24 = 19 cupcakes to dis tribute to the baby mice. There are 19+12−1 = 54,627,300 ways to do that.

11

23. There are as many six-letter words equivalent to BLEFLA as there are anagrams of BLEFLA, 6! = 360. Now, is it an equivalence relation? We will need to check the three which is 2! conditions.

Is a word equivalent to itself? Yes. So this relation is reflexive.

If a word w1 is equivalent to a word w2 , is w2 also equivalent to w1 ? Yes, because the rearrangement can be reversed. The relation is symmetric.

If a word w1 is equivalent to a word w2 , and a word w2 is equivalent to a word w3 , then this means that w1 anagrams to w2 , which also anagrams to w3 , and thus the letters of w3 are a rearrangement of the letters of w1 . Thus, w1 is equivalent to w3 and the relation is transitive.

Yes; it’s an equivalence relation!

24. The criminal does not have any of the characteristics F, O, E, or T. Therefore, the number of residents who fit the murderer’s description should be (total population) − (number of residents with F or O or E or T). Now, (number of residents with F or O or E or T) equals, using PIE, 2400 + 3600 + 1200 + 800 − 1200 − 900 − 700 − 800 − 700 − 700 + 800 + 500 + 500 + 400 − 200 = 5,000. As this is exactly the number of residents of the town, the police chief says there are no potential suspects who live in the town.

The detective thinks it unlikely that there are no males between the ages of 16 and 35 who are under 6 feet tall and wear glasses in a town of 5,000 and so looks more closely at the DMV records. Ze notices that there are 200 F, O, E, T residents and 500 F, O, T residents, so there must be 500 − 200 = 300 residents who are F, O, T but not E. Similarly, there are 500 F, E, T residents, and so there must be 500 − 200 = 300 residents who are F, E, T but not O. Now, the records say there are 700 F, T residents. But just considering those who are F, O, E, T along with those who are F, E, T and not O, and those who are F, O, T, and not E (and not considering those who are F, T and neither O nor E) we have 200 + 300 + 300 = 800 > 700 F, T residents. Therefore the detective says the DMV records must be inaccurate. (This isn’t the only error in the data; a similar calculation on the O, E residents will produce a contradiction as well.)

25. The total number of six-character passwords that have no letter is 106 and the total number of six-character passwords that have no number is 266 . Thus, the number of six-character passwords that have no letter or no number is 106 + 266 , so that the number of six-character passwords that have at least one letter and at least one number is 366 −(106 +266 ) = 1,866,866,560.

We can repeat this argument for (7–12)-character passwords to obtain a total of 366 − (106 + 266 ) + 367 − (107 + 267 ) + 368 − (108 + 268 ) + 369 − (109 + 269 ) + 3610 − (1010 + 2610 ) + 3611 − (1011 + 2611 ) + 3612 − (1012 + 2612 ) = 4,774,516,436,184,676,480 passwords.

26. The total number of 8-character passwords that have no lower-case letter is 368 , the total number of 8-character passwords that have no upper-case letter is 368 , and the total number of

7.11. Problems about Balls, Boxes, and PIEs

71

8-character passwords that have no number is 528 ; the total number of 8-character passwords that have no letters is 108 , the total number with no upper-case letters or numbers is 268 , and the total number with no lower-case letters or numbers is also 268 (and there are no passwords with no lower-case letters, upper-case letters, or numbers). Thus by PIE, the number of 8-character passwords with at least one upper-case letter, lower-case letter, and number is 618 − (368 + 368 + 528 ) + (268 + 26 + 8 + 108 ) = 132,814,291,715,523. We can repeat this argument for (9–12)-character passwords to obtain a total of 618 − (368 + 368 + 528 ) + (268 + 26 + 8 + 108 ) + 619 − (369 + 369 + 529 ) + (269 + 269 + 109 ) + 6110 − (3610 + 3610 + 5210 ) + (2610 + 2610 + 1010 ) + 6111 − (3611 + 3611 + 5211 ) + (2611 + 2611 + 1011 ) + 6112 − (3612 + 3612 + 5212 ) + (2612 + 2612 + 1012 ) = 2,290,497,826,159,010,352,127 passwords.

This would be 2,290,497,826,159 seconds, or about 72,631 years. It seems like we’re safe (even if the speed of the password cracker is a few orders of magnitude faster).

27. There are 10 choices for the first duck, then 9 for the second, etc. so 10 · 9 · 8 · 7 · 6 · 5, or also 10 10 6 6!. If the ducks are new, then it’s just 6 .

52 49 45 40 34 27 19 10 28. 54 2 3 4 5 6 7 8 9 ( 10 ).

(a) There are 7 choices for each scoop, so 73 . (b) The order of the scoops doesn’t matter here, but repetition is allowed, so we use stars= 462. and-bars to get 7+4 5 7 (c) This is just 2 . (d) All the letters are different, so 6!. (e) There’s a double-O, so 5!/2.

17 17 30. If the textbooks and novels are indistinguishable, then 17 11 = 6 . If not, then 6 11!6!.

29.

31. 183 .

32.

33.

34.

(a)

8 3 .

= 120. (b) Stars-and-bars again for the win: 3+7 7 (c) There must be 2 plain apple, so 4 plain chocolate (by PIE).

2 (a) It’s 82 .

(b) BFF picks two and then you pick two from the remainder, so there are 82 62 ways to choose the order. To count the orders, note that there are 4 different flavors picked at once for 84 .

(a) 12 · 11 · 10 · 9.

10 5 (b) One solution is 13 3 5 2 . (c) 12!/(4!). (d) None. All anagrams contain 4 As.

35. Ha!

28! . Whew.

2!3!3!3!3!3!2!2!2!

Chapter 8

Recurrences 8.4 Try This! Sequences and Fibonacci Identities 1. 1, 2, 4, 8, 16, 32, . . . . Hey! That’s an = 2n ! (a) 2, 1, 5, 7, 17, 31, 65, . . . . It’s totally different. There’s no clear pattern. (b) 1, 1, 3, 5, 11, 21, 43, . . . . Also no clear pattern.

2.

(a) 1, 4, 7, 10, 13, 16, 19, . . . ; an = an−1 + 3; an − an−1 = 3n + 1 − (3(n − 1) + 1) = 3. (b) 7, 9, 11, 13, 15, 17, 19, . . . ; an = an−1 + 2; an − an−1 = 2n + 7 − (2(n − 1) + 7) = 2. (c) 0, 4, 14, 30, 52, 80, . . . ; an = an−1 + 6n − 2; an − an−1 = 3n2 + n − (3(n − 1)2 + (n − 1)) = 3n2 + n − 3n2 + 6n − 3 − n + 1 = 6n − 2.

2n−1 − 1 − (22(n−1)−1 − 1) = (d) −3 2 , 1, 7, 31, 127, 511, . . . ; an = 4an−1 + 3; an − an−1 = 2 22n−1 − 1 − 22n−3 + 1 = 22n−1 − 22n−3 = 22n−3 (4 − 1) = 3 · 22n−3 .

3.

(a) 3, 6, 12, 24, 48, 96, . . . ; an = 3 · 2n . We will proceed, abbreviatedly, by induction. The base case is covered by prior calculations. Consider an+1 . By the recurrence, this equals 2an−1+1 = 2an . This falls under the inductive hypothesis, so an+1 = 2 · 3 · 2n = 3 · 2n+1 , as desired. (b) 4, 8, 12, 16, 20, 24, 28, . . . ; an = 4n. We will proceed, abbreviatedly, by induction. The base case is covered by prior calculations. Consider an+1 . By the recurrence, this equals 4 + an−1+1 = 4 + an . This falls under the inductive hypothesis, so an+1 = 4 + 4n = 4(n + 1), as desired. (c) 1, 5, 9, 13, 17, 21, 25, 29, . . . ; an = 4(n − 1) + 1. We will proceed, abbreviatedly, by induction. The base case is covered by prior calculations. Consider an+1 . By the recurrence, this equals 4 + an−1+1 = 4 + an . This falls under the inductive hypothesis, so an+1 = 4 + (4(n − 1) + 1) = 4 + 4n − 4 + 1 = 4n + 1, as desired.

4. Solutions will vary.

5. They are Fibonacci numbers! Oh, we have to prove it? Let’s gather better data. Row 0 gives us 1. Row 1 gives us 1. Row 2 gives us 2. Row 3 gives us 3. Row gives 4 n−2 us 5. So0 it looks like the index of the Fibonacci number is n + 1. We get n0 + n−1 + 1 2 + · · · + n = Fn+1 .

Proof by induction, course.

Fn+1 = Fn + Fn−1 and by inductive hypothesis, the right-hand ofn−2

0 n−4 0 n−2 n−3 n−3 + · · · + . Rear+ + · · · + + + + side equals n−1 + 2 n−2 1 2 n−1 0 1 0 n−4 n−4 n−3 n−2 n−2 n−3 n−1 ranging terms, we have 0 + ( 0 + 1 ) + ( 1 + 2 ) + ( 2 + 3 ) + · · · +

0 0 n−2 ( n−1 + n−2 ) = n0 + n−1 1 + 2 + · · · + 0, as desired.

72

8.7. Try This! Recurrence Exercises

73

6. It’s false; try k = 2 as an example and expand the right-hand side.

7. We proved that Fn−1 Fn+1 − (Fn )2 = (−1)n−1 . Use the substitution that Fn−1 = Fn+1 − Fn and get the desired identity.

8.

(a) Fn+3 = Fn+2 + Fn+1 = Fn+1 + Fn + Fn+1 = 2Fn+1 + Fn = 2Fn+1 + F2 Fn . (b) Fn+4 = Fn+3 + Fn+2 . From the previous part, we have that this equal 2Fn+1 + Fn + Fn+2 = 2Fn+1 + Fn + Fn+1 + Fn = 3Fn+1 + 2Fn . (c) Fn+5 = Fn+4 + Fn+3 . From the previous two parts, we have that this equals 3Fn+1 + 2Fn + 2Fn+1 + Fn = 5Fn+1 + 3Fn . (d) Use induction on n; we have several base cases already, so let us assume that the statement holds for m ≤ n − 1. Then Fn+k = Fn−1+k + Fn−2+k and by the inductive hypothesis, this equals Fk Fn + Fk−1 Fn−1 + Fk Fn−1 + Fk−1 Fn−2 = Fk Fn + Fk Fn−1 + Fk−1 Fn−1 + Fk−1 Fn−2 = Fk Fn+1 + Fk−1 Fn .

9. Let k = n in the previous identity.

8.7 Try This! Recurrence Exercises 1.

(a) It is an = 3n . (b) Sequence: 1, 3, 9, 27, . . . hey! That looks like an = 3n again!

Let’s prove it. ak+1 = 2ak + 3ak−1 = 2 · 3k + 3 · 3k−1 = 2 · 3k + 3k = 3 · 3k = 3k+1 .

(c) Sequence: 1, −1, 1, −1, . . . Huh. That seems to be an = (−1)n .

Let’s prove it. ak+1 = 2ak + 3ak−1 = 2 · (−1)k + 3 · (−1)k−1 = 2 · (−1)k + (−3) · (−1)k = (2 − 3) · (−1)k = (−1)k+1 .

(d) We suspect an = rn .

Sequence: 2, 2, 10, 26, 82, . . . there’s no recognizable pattern of powers.

We might observe this is the sum of the previous two sequences, though! So the closed form should be an = 3n + (−1)n . Proof: ak+1 = 2ak + 3ak−1 = 2 · [3k + (−1)k ] + 3 · [3k−1 + (−1)k−1 ] = 2 · 3k + 3 · 3k−1 + 2 · (−1)k + 3 · (−1)k−1 = 3k+1 + (−1)k+1 . (e) Everything is 21 of the previous sequence, so we get a closed form of an = 21 · 3n + 1 n 2 (−1) .

2. Yes; kth differences should work.

1, 2, 5, 10, 17, 26, 37, . . . .

Yes; it is an = (n − 1)2 + 1.

Induction: an+1 = an + 2(n + 1) − 3 = (n − 1)2 + 1 + 2n + 2 − 3 = n2 − 2n + 1 + 1 + 2n − 1 = n2 + 1, as desired. The point is that sometimes applying a technique to obtain a closed form is slower than using intuition. Solving simultaneous equations for kth differences can be a pain.

74

8. Recurrences

3.

(a) Sequence is 1, 5, 13, 29, 61, . . . .

Closed form is an = 2n+1 − 3.

We will proceed, abbreviatedly, by induction. The base case is covered by prior calculations. Consider an+1 . By the recurrence, this equals an+1−1 + 2n+1 = an + 2n+1 .

This falls under the inductive hypothesis, so it equals 2n+1 − 3 + 2n+1 = 2 · 2n+1 − 3 = 2n+2 − 3, as desired.

(b) Sequence is 0, 2, 6, 12, 20, 30, . . . .

First differences are 2, 4, 6, 8, 10, . . . .

Second differences are 2, 2, 2, 2, . . . .

Therefore this is a quadratic, of the form an = c + dn + f n2 : a0 = 0 = c + 0 + 0, so c = 0; a1 = 2 = 0 + d + f and a2 = 6 = 0 + 2d + 4 f , so d = 2 − f ; 6 = 2(2 − f ) + 4 f = 4 + 2 f so 3 = 2 + f or f = 1; thus, d = 1. Thus, an = n + n2 .

We will proceed, abbreviatedly, by induction. The base case is covered by prior calculations. Consider an+1 . By the recurrence, this equals an+1−1 + 2(n + 1) = an + 2n + 2.

This falls under the inductive hypothesis, so it equals n + n2 + 2n + 2 = n2 + 3n + 2 = n + 1 + n2 + 2n + 1 = (n + 1) + (n + 1)2 , as desired. (c) Sequence is 4, 10, 16, 22, 28, . . . .

First differences are 6, 6, 6, 6, . . . .

Closed form is an = 4 + 6n.

We will proceed, abbreviatedly, by induction. The base case is covered by prior calculations. Consider an+1 . By the recurrence, this equals an+1−1 + 6 = an + 6. This falls under the inductive hypothesis, so it equals 4 + 6n + 6 = 4 + 6(n + 1), as desired. (d) Sequence is 2, 2, 4, 8, 32, 256, 8192, . . . .

Closed form is an = 2Fn .

We will proceed, abbreviatedly, by induction. The base case is covered by prior calculations. Consider an+1 . By the recurrence, this equals an+1−1 an+1−2 = an an−1 . This falls under the inductive hypothesis, so it equals 2Fn 2Fn−1 = 2Fn +Fn−1 = 2Fn+1 , as desired.

8.9 Try This! Find Closed Forms for These Recurrence Relations!

1. Looks like kth differences.

Sequence is 2, 10, 19, 29, 40, . . . .

First differences are 8, 9, 10, 11, . . . .

Second differences are 1, 1, 1, . . . .

Form is an = c + dn + f n2 ; need an a0 , so 2 = a1 = a0 + 1 + 6 and a0 = −5: a0 = c = −5; a1 = 2 = c + d + f ; a2 = 10 = c + 2d + 4 f ; 2 = −5 + d + f or d = 7 − f ; 10 = −5 + 2(7 − f ) + 4 f = 2 f + 9 or 1 = 2 f or f = 21 ; an = 12 n2 − 13 2 n − 5.

2. Looks like characteristic equation.

Sequence is 3, 10, 175, . . . .

8.9. Try This! Find Closed Forms for These Recurrence Relations!

75

√ √ 2 = 10x+25 or x2 −10x−25 = 0 = (x−(5 2+5))(x+(5 2− Characteristic polynomial is x √ √ n n ; a = 3 = q + q so q = 3 − q ; a = 10 = 5)): √an = q1 (5 2 + 5) 0 1 1 2 √1 √ + q2 (−5 2 + 5) √ √2 q1 (5 2 + 5) + q2 (−5 2 + 5) = (3 − q2 )(5 2 + 5) + q2 (−5√ 2 + 5) so 10 − 3(5 √2 + 5) = √ √ √ √ 2 and q1 = 3 − −5−15 √ 2 . Thus q2 (−5 2 + 5) − q2 (5 2 + 5) = q2 (−10 2) so q2 = −5−15 −10 2 −10 2 √ √ √ √ √ 2 )(5 2 + 5)n + ( −5−15 √ 2 )(−5 2 + 5)n . an = (3 − −5−15 −10 2 −10 2 3. Looks like characteristic equation.

Sequence is 62, −62, 62, −62, . . . .

Intuition works better: an = 62(−1)n .

Really, we don’t need to prove this one.

4. Looks like kth differences.

Sequence is −2, −8, −13, −17, −20, −22, . . . .

First differences are −6, −5, −4, −3, −2, . . . .

Second differences are 1, 1, 1, 1, . . . .

Form is an = c + dn + f n2 : a0 = c = −2; a1 = −8 = −2 + d + f so d = − f − 6; a2 = −13 = −2 + 2d + 4 f = −2 + 2(− f − 6) + 4 f = −2 − 2 f − 12 + 4 f so 1 = 2 f and f = 12 ; d = −13 2 .

Thus a + n = 21 n2 = 13 n − 2.

2 5. Looks like characteristic equation (characteristic polynomial roots are ±2).

Sequence is 1, 2, 4, 8, 16, 32, . . . .

Hey, looks like intuition will be better: an = 2n .

Abbreviated induction: an+1 = 4an−1 and, by inductive hypothesis, equals 4 · 2n−1 = 2n+1 , as desired.

6. Looks like kth differences.

Sequence is 0, 1, 7, 22, 40, . . . .

First differences: 1, 6, 15, 28, . . . .

Second differences: 5, 9, 13, . . . .

Third differences: 4, 4, . . . .

Form is an = c + dn + f n2 + gn3 : a0 = c = 0; a1 = 1 = d + f + g so d = 1 − f − g; a2 = 7 = 2d + 4 f + 8g = 2(1 − f − g) + 4 f + 8g = 2 + 2 f + 6g so 5 = 2 f + 6g or f = (5 − 6g)/2; a3 = 22 = 3d + 9 f + 27g = 3(1− f − g)+ 9 f + 27g = 3+ 6 f + 24g = 3+ 6(5− 6g)/2+ 24g = 2 3 1 2 1 18 + 6g; so 4 = 6g or g = 32 so f = 21 so d = −1 6 . So, an = 3 n + 2 n − 6 n.

7. Looks like characteristic equation.

Sequence is −3, 1, 117, . . . .

Characteristic polynomial is x2 = 12x − 35 or x2 − 12x + 35 = 0 = (x − 5)(x − 7), so an = q1 5n + q2 7n : a0 = −3 = q1 + q2 so q1 = −3 − q2 ; a1 = 1 = 5q1 + 7q2 = −15 − 5q2 + 7q2 = −15 + 2q2 so q2 = 8 and q1 = −11; an = (−11)5n + 8 · 7n .

76

8. Recurrences

8. Forced to use intuition.

Sequence is 1, 1, 0, −1, −1, 0, −1, −1, 0, . . . .

For n > 2, we have an = 0 when n is a multiple of 3 and −1 otherwise.

9. Looks like characteristic equation.

Sequence is 1, 2, 3, 10, 23, . . . .

Characteristic polynomial is x3 = 2x2 + 5x − 6 so x3 − 2x2 − 5x + 6 = 0.

This can be factored (try long division by (x − 1)). The poly roots are 1, −2, 3, so an = q1 1n + q2 (−2)n + q3 3n .

Now, a0 = 1 = q1 + q2 + q3 so q1 = 1 − q2 − q3 ; a1 = 2 = q1 − 2q2 + 3q3 so 2 = 1 − q2 − q3 − 2q2 + 3q3 = 1 − 3q2 + 2q3 so (1 + 3q2 )/2 = q3 ; a2 = 3 = q1 + 4q2 + 9q3 = 1 − q2 − q3 + 3 4q2 + 9q3 = 1 + 3q2 + 8q3 = 1 + 3q2 + 4(1 + 3q2 ) = 5 + 15q2 so q2 = −2 15 ; then q3 = 10 and 5 5 n 2 3 n n q1 = 6 . Thus, an = 6 1 − 15 (−2) + 10 3 .

8.12 Bonus: Recurring Stories Story problems:

1. There is one way to get from the first floor to itself, and two ways to get from the second floor to the first. From the third floor, there are two ways to get one floor down (using escalators 1a or 1b ), and three ways to get two floors down (using escalators 2a or 2b or 2c ), so one has a variety of options to get to the first floor: 1a −1a , 1a −1b , 1b −1a , 1b −1b , 2a , 2b , and 2c . Thus, the first three terms in our sequence en are 1, 2, 7. So, let us suppose we have descended a total of n floors. There were two ways we could have come from one floor above, and for each of those ways there were en−1 ways we could have gotten to the (n − 1)st floor. That’s a total of 2en−1 ways. Additionally, there were three ways we could have arrived from two floors above, so analogously that produces 3en−2 ways. Thus, our recurrence is en = 2en−1 + 3en−2 .

The closed form for this sequence is en = 41 3n − 41 (−1)n .

2.

(a) 1, 3, 5, 11, 21, . . . . (b) It is Tn = Tn−1 + 2Tn−2 because if the last tile is a square, then there are Tn−1 ways the rest of the trench could be filled, and if the last tile is a triangle, then there are two ways that the last two tiles could be filled with triangles and Tn−2 ways the rest of the trench could be filled for each of those two ways. (c) The characteristic equation is x2 = x + 2 so x2 − x − 2 = (x − 2)(x + 1) = 0. This gives us an = q1 (−1)n + q2 2n : a2 = a1 + 2a0 so 3 = 1 + 2a0 and a0 = 1; a0 = 1 = q1 + q2 so q2 = 1 − q1 ; a1 = 1 = 2q2 − q1 = 2 − 2q1 − q1 or −1 = −3q1 so q1 = 31 ; and q2 = 32 . So, an = 13 (−1)n + 32 2n .

3.

(a) 1, 2, 3, 5, . . . . (b) It is Rn = Rn−1 + Rn−2 because if the last tile is a square, then there are Tn−1 ways the rest of the trench could be filled, and if the last tile is a triangle, then there is only one

8.14. Recurring Problems

77

way that the last two tiles could be filled with triangles and Tn−2 ways the rest of the trench could be filled. (c) It’s the Fibonacci recurrence, so we obtain the Binet formula (with index shifted by 1).

4.

(a) 1, 2, 5, 13, 34, 89, 233, . . . . (b) There are (at least) two recurrences that work. In a brute force sort of way, bn = 1 + (n − 1)b1 · · · + 2bn−2 + bn−1 . For a row of n bricks, there can be nothing atop (1 stack); there are n − 1 ways to put one brick atop ((n − 1)b1 stacks); there are n − 2 ways to place a stack with two bricks-along-the-bottom atop ((n − 2)b2 stacks); two ways to place a stack with n − 2 bricks-along-the-bottom atop (2bn−2 stacks), and one way to place a stack with n − 1 bricks-along-the-bottom atop (bn−1 stacks). The figure (Figure 8.4, page 271) tries to suggest bn = 3bn−1 − bn−2 . There is one way to place a stack with n − 1 bricks-along-the-bottom atop a line of n bricks, or bn−1 . If there is no brick atop the leftmost (rightmost) brick in the line, that brick can be removed, leaving a stack with n − 1 bricks-along-the-bottom. This gives us an additional 2bn−1 . However, there are some stacks with no brick atop the leftmost brick and no brick atop the rightmost brick.

These stacks have been counted twice, and there are bn−2 of them because the removal of both leftmost and rightmost bricks leaves a stack with n − 2 bricks-along-the-bottom.

8.14 Recurring Problems 1. ∑nj=1 F2 j−1 .

Induction: As a base case, note that F1 + F3 = 1 + 2 = 3 = F4 . Now consider ∑n+1 j=1 F2 j−1 = F2(n+1)−1 + ∑nj=1 F2 j−1 = F2n+1 + ∑nj=1 F2 j−1 . By the (unstated) inductive hypothesis, this is F2n+1 + F2n = F2n+2 = F2(n+1) , as desired.

2. ∑nj=1 F2 j .

Induction: As a base case, note that F2 + F4 = 1 + 3 = 4 = F5 − 1. Now consider ∑n+1 j=1 F2 j = F2(n+1) + ∑nj=1 F2 j = F2n+2 + ∑nj=1 F2 j . By the (unstated) inductive hypothesis, this is F2n+2 + F2n+1 − 1 = F2n+3 − 1 = F2(n+1)+1 − 1, as desired.

3. Induction (yes!): As a base case, note that F3 = 2 and F6 = 8, both of which are even. For the inductive step, note that F3n = F3n−1 + F3n−2 = (F3n−2 + F3n−3 ) + F3n−2 = 2F3n−2 + F3n−3 .

The first term is even as it’s clearly divisible by 2, and the second term is even by inductive hypothesis. j+1 F .

4. ∑2n j j=1 (−1)

Induction: As a base case, note that F1 − F2 = 1 − 1 = 1 − F1 . Let’s do another: F1 − 2(n+1) F2 + F3 − F4 = 1 − 1 + 2 − 3 = −1 = 1 − F3 . Now consider ∑ j=1 (−1) j+1 Fj = F2n+1 − j+1 F . By the (unstated) inductive hypothesis, this equals F F2n+2 + ∑2n j 2n+1 − F2n+2 + j=1 (−1) 1 − F2n−1 = F2n+1 + F2n − F2n − F2n+2 + 1 − F2n−1 = F2n+1 + F2n − F2n+2 + 1 − F2n−1 − F2n = F2n+2 − F2n+2 + 1 − F2n+1 = 1 − F2(n+1)−1, as desired.

78

8. Recurrences

5. Using 2, 1 (for example) generates 2, 1, 3, 4, 7, 11, 18, 29, . . . .

Using 2, 3 (for example) generates 2, 3, 5, 8, 13, 21, . . . .

Any two consecutive Fibonacci numbers will generate more Fibonacci numbers.

6. First differences: 1, 4, 7, 10, 13, 16, . . . .

Second differences: 3, 3, 3, 3, . . . .

Form is an = c + dn + f n2 .

Let a0 = 0 = c, so a1 = 1 = d + f and d = 1 − f ; a2 = 5 = 2d + 4 f = 2 − 2 f + 4 f , so 3 = 2 f and f = 23 , d = − 12 ; an = 32 n2 − 12 n. Thus, the recurrence is an = an−1 + 3(n − 1) + 1.

7. Sequence: −1, 1, 3, 5, 7, 9, 11, . . . .

Uh, that looks like an = 2n − 1. Let’s prove it. Consider an+1 = 2an − an−1 and apply the inductive hypothesis to find 2(2n − 1) − 2(n − 1) − 1 = 4n − 2 − 2n + 2 − 1 = 2n − 1. Huh!

If we do things the hard way, we get characteristic equation x2 = 2x − 1 that becomes x2 − 2x + 1 = 0 and that factors as (x − 1)2 = 0. Because it has a double root, it is not covered by our algorithm and it’s therefore pretty lucky we can use intuition.

8. A binary string either ends in 0 or 1. If it ends in 1, then there are zn−1 binary substrings that could form the remainder of the digits. If it ends in 0, then really it either ends in 100 or 110 (because it can’t end in 000). Then, for each of those two endings, there are zn−3 binary substrings that could form the remainder of the digits. Thus we have zn = zn−1 + zn−3 .

Sequence: 2, 4, 7, 9, 13, 20, 29, . . . .

9. Sequence: 2, 6, 12, 20, 30, 42, . . . .

Closed form: an = n(n + 1).

Notice that an+1 = an + 2(n + 1) = n(n + 1) + 2(n + 1) = (n + 1)(n + 2), as desired.

10. Direct proof of a biconditional.

11. an = an−1 + an−2 + an−3 .

Just use different initial conditions. For example, start with 0, 1, 0, and obtain 0, 1, 0, 1, 2, 3, 6, 11, 20, 37, . . . .

12. Yes, though this may have changed by the time you read this sentence; the database is constantly updated. (For example, check 2, 4, 7, 9, 13, 20, 29, . . . from two problems ago.)

13. Induction!!

14. Sequence: −1, −2, −5, −4, −1, 8, 35, . . . .

Characteristic polynomial is x2 = 4x − 3 so x2 − 4x + 3 = (x − 1)(x − 3) = 0: an = q1 + q2 3n ; a0 = −1 = q1 + q2 so q1 = −1 − q2 ; a1 = −2 = q1 + 3q2 = −1 − q2 + 3q2 = −1 + 2q2 so q2 = − 21 and q1 = − 21 . Thus, an = − 12 − 21 3n .

8.14. Recurring Problems

79

15. an = an−1 + 2n−2 .

Just use different initial conditions. For example, start with 0, 1, and obtain 0, 1, 3, 7, 15, 31, . . . .

16. Prove each of the following identities. (a) Factor the last identity in Section 8.4. (b) See previous identity. (c) Recurse on Fn+1 in the second previous identity. (d) Factor the right-hand side and use that Fn−1 = Fn+1 − Fn .

(e) 2Fn+1 − Fn = 2Fn + 2Fn−1 − Fn = Fn + 2Fn−1 ; this reduces the problem to the identity two prior to this one. (f) Let k = n + 1 in Fn+k = Fk Fn+1 + Fk−1 Fn .

(g) Reverse the roles of n and k in the previously quoted identity, and then substitute k + 1 for k. (h) Substitute n − k for n in Fn+k = Fk Fn+1 + Fk−1 Fn .

(i) Reverse the roles of n and k in the previously quoted identity, and then substitute n − k for n. (j) Use k = 2n in the previously quoted identity to get F2n = (Fn+1 )2 − (Fn−1 )2 . Then recurse on Fn+1 and use k = n − 1 in the originally quoted identity.

(k) This is long. The summary is . . . recurse on Fn+1 , then recurse on one of the two Fn s, then recurse on the other Fn , and then rearrange terms and unrecurse to get an Fn as desired.

17. Sequence: 0, 1, 5, 21, 85, 341, 1365, 5461, . . . .

Characteristic polynomial is x2 = 5x − 4 so x2 − 5x + 4 = (x − 4)(x − 1) and the closed form is an = q1 + a2 4n : a0 = 0 = q1 + q2 so q1 = −q2 ; a1 = 1 = q1 + 2q2 = −q2 + 4q2 so q2 = 31 and q1 = 31 ; an = 13 − 31 4n .

18. x4 = 2x2 − 3x + 3 becomes xn = 2xn−2 − 3xn−3 + 3xn−4 and then an = 2an−2 − 3an−3 + 3an−4 .

19. Sequence: 1, 4, 7, 10, 13, 16, 19, 22 . . . .

Closed form: an = 3n − 2.

20. This tells us we have a characteristic polynomial with roots −3, 4. So (x + 3)(x − 4) = x2 − x − 12 = 0 or x2 = x + 12 or xn = xn−1 + 12xn−2 or an = an−1 + 12an−2 .

21. 1, 0, 1, 2, 1, 2, 3, 2, 3, 4, 3, 4, 5, 4, 5, 6, 5, 6, 7, 6, 7, 8, 7, 8, 9, 8, 9, 10, 9, 10.

The challenge is mostly notational: if n ≡ 0 (mod 3), then an = n3 + 1; if n ≡ 1 (mod 3), then n+1 an = n−1 3 ; and if n ≡ 2 (mod 3), then an = 3 .

22.

(a) (x + 1)(x − 2)(x + 3) = x3 + 2x2 − 5x − 6, so x3 = −2x2 + 5x + 6.

(b) an = −2an−1 + 5an−2 + 6an−3 .

80

8. Recurrences

(c) We need a0 , a1 , a2 . Let them be 1, −1, 1 in accordance with the closed form. Check the next two terms: −2 + 5(−1) + 6 = −1; −2(−1) + 5 + 6(−1) = 1. Yup.

(d) We need a0 , a1 , a2 . Let them be 2, 1, 5 in accordance with the closed form. Check the next three terms: −2(5) + 5(1) + 6(2) = 7; −2(7) + 5(5) + 6(1) = 17; −2(17) + 5(7) + 6(5) = 31. Yup. (e) We need a0 , a1 , a2 . Let them be 3, −2, 14 in accordance with the closed form. Check the next three terms: −2(14) + 5(−2) + 6(3) = −20; −2(−20) + 5(14) + 6(−2) = 98; −2(98) + 5(−20) + 6(14) = −212. Yup. 23. Intuition only here: the closed form is an = n2n , and to find a recurrence, compute an − an−1 to find that an = an−1 + n2n−1 + 2n−1 .

24. First differences: 2, 8, 18, 32, 50, 72, 98, 128, . . . .

Second differences: 6, 10, 14, 18, 22, 26, 30, . . . .

Third differences: 4, 4, 4, 4, 4, 4, . . . .

Criminy. an = c + dn + f n2 + gn3 : a0 = 0 = c and a1 = 2 = d + f + g so d = 2 − f − g; a2 = 10 = 2d + 4 f + 8g = 2(2 − f − g) + 4 f + 8g = 4 + 2 f + 6g so 6 = 2 f + 6g so f = 3 − 3g; a3 = 28 = 3d + 9 f + 27g = 6 − 3 f − 3g + 9 f + 27g = 6 + 6 f + 24g = 6 + 18 − 18g + 24g = 24 + 6g so 4 = 6g and g = 32 , f = 1, d = 31 . Thus, an = 13 n + n2 + 32 n3 .

25.

(a) x3 − x − 2x2 + 2 = (x + 1)(x − 1)(x − 2) so 1, −1, 2.

(b) Solutions will vary depending on the choice of assignment of 1, −1, 2 to r1 , r2 , r3 . Sample: We need a0 , a1 , a2 . Let them be 1, 1, 1 in accordance with the closed form. Check the next three terms: 2(1) + (1) − 2(1) = 1; 2(1) + (1) − 2(1) = 1; 2(1) + (1) − 2(1) = 1.

Yup. (c) Solutions will vary depending on the choice of assignment of 1, −1, 2 to r1 , r2 , r3 .

(d) Solutions will vary depending on the choice of assignment of 1, −1, 2 to r1 , r2 , r3 .

26. Sequence: 1, 0, −1, 0, 5, 16, 35, 64, . . . .

First differences: −1, −1, 1, 5, 11, 19, 29, . . . .

Second differences: 0, 2, 4, 6, 8, 10, . . . .

Third differences: 2, 2, 2, 2, 2, . . . .

Thus we have a closed form of the form an = c + dn + f n2 + gn3 : a0 = 1 = c + 0 + 0 + 0 so c = 1; a1 = 0 = 1 + d + f + g; a2 = −1 = 1 + 2d + 4 f + 8g; a3 = 0 = 1 + 3d + 9 f + 27g. Let d = −1 − f − g so that we obtain 0 = 2 + 2(−1 − f − g) + 4 f + 8g = 2 f + 6g and 0 = 1 + 3(−1 − f − g) + 9 f + 27g = −2 + 6 f + 24g. Next, let f = −3g to obtain 0 = −2 + 6(−3g) + 1 3 1 2 24g = −2 + 6g. Then, g = 13 , f = −1, d = −7 3 , so the closed form is an = 3 n − n − 3 n + 1.

27. The recurrence an = an−1 − 9n + 5, a0 = 1 is arithmetic with a first-degree polynomial, so we expect that second differences will be constant. The first five terms of the sequence are 1, −3, −16, −38, −69. We compute first and second differences.

8.14. Recurring Problems

81

1 −4

−3 −9

−13

−16 −9

−22

−38 −9

−31

−69

...

As expected, the second differences are constant. Therefore, the closed form will be an = c + dn + f n2 , and we need to determine c, d, and f .

When n = 0, we have a0 = 1 = 1 + d0 + f 02 , so we know c = 1.

When n = 1, we have a1 = −3 = 1 + d + f , or d + f = −4 or f = −4 − d.

When n = 2, we have a2 = −16 = 1 + 2d + 4 f , or 2d + 4 f = −15.

We now have two linear equations and two unknowns, and using high-school algebra, we can solve for d and f . Substituting in for f , we get 2d + 4(−4 − d) = 2d − 16 − 4d = −2d − 16 = −15, or −2d = 1, so d = 12 and f = −4 − 21 = − 92 .

Thus, our closed form is an = 1 + 12 n − 29 n2 . Let’s check and make sure this is correct: a1 = 1 + 12 − 29 = 1 − 4 = −3. Yup. a2 = 1+ 21 ·2− 29 ·4 = 1+ 1− 18 = −16. Yup. a3 = 1+ 21 ·3− 92 ·9 = 1+ 3−81 2 = 1− 39 = −38.

Looks like we’re good.

28. The recurrence an = 2an−1 + an−2 − 2an−3 , a0 = 1, a1 = 0, a2 = −1 is geometric. We begin by finding the characteristic polynomial: first convert as to xs and subscripts to superscripts, as in xn = 2xn−1 + xn−2 − 2xn−3 , then cancel extra factors of x (here, xn−3 ), as in x3 = 2x2 + x − 2. Rewriting so we can factor gives us x3 − 2x2 − x + 2 = 0, and factoring gives us (x − 1)(x + 1)(x − 2) = 0, so the roots are x = ±1, x = 2.

This corresponds to the closed-form equation an = q1 (1)n + q2 (−1)n + q3 (2)n , which generates the equations a0 = 1 = q1 + q2 + q3 , a1 = 0 = q1 − q2 + 2q3 (notice that this tells us q2 = q1 + 2q3 ), a2 = −1 = q1 + q2 + 4q3 . Plugging our equation for q2 into the first and third equations gives us 1 = q1 + q1 + 2q3 + q3 = 2q1 + 3q3 or 2q1 = 1 − 3q3 and −1 = q1 + q1 + 2q3 + 4q3 = 2q1 + 6q3 or 2q1 = −1 − 6q3 , from which we can conclude that 1 − 3q3 = −1 − 6q3 or 2 = −3q3 or q3 = −2 3 .

3 From this, we have 2q1 = 1 − 3 −2 = 3 so q 1 = 2.

3 1 Then, q2 = 23 + 2 −2 3 = 6 . Our closed form is then n an = 23 (1)n + 16 (−1)n + −2 3 (2) . As a quick check on our arithmetic, we note that a0 = 1 = q1 + q2 + q3 does indeed hold.

29.

(a) This has a variable coefficient and is not linear. (b) This fits the definition.

82

8. Recurrences

(c) This is not homogeneous. (d) This has a nonconstant coefficient and is not linear.

30. x2 − x + 6 = 0 which is (x + 3)(x − 2) = 0. an = (4/5)3n + (1/5)(−2)n .

31. Sequence: 4, 10, 22, 46, 94, 190, 382, . . .

This is 3 · 2n − 2. Proof: ak+1 = 2ak + 2 = 2(3 · 2k−1 − 2) + 2 = 3 · 2k − 4 + 2 = 3 · 2k − 2.

32. First differences: −2, −4, −6, −8, −10, −12 . . .

Second differences: −2, −2, −2, . . .

Therefore, the closed form will be an = c + dn + f n2 , and we need to determine c, d, and f . a0 = 111 so c = 111. a1 = 109 = 111 + d + f so d = −2 − f .

a2 = 105 = 111 + (−2 − f )2 + f 4 = 2 f + 107 so f = (105 − 107)/2 = −1.

an = −n2 − n + 111.

33. Should be an = −4n + 126.

34. First differences: 31, 31, 31, 31, . . . an = an−1 + 31, a1 = 12.

35. Sequence: −23, −18, −7, 10, 33, 62, 97, . . .

First differences: 5, 11, 17, 23, 29, . . .

Second differences: 6, 6, 6, 6, . . .

Therefore, the closed form will be an = c + dn + f n2 , and we need to determine c, d, and f . a0 = −23 so c = −23.

a1 = −18 = −23 + d + f so d = 5 − f .

a2 = −7 = −23 + (5 − f )2 + f 4 = 2 f − 13 so f = (−7 + 13)/2 = 3. an = 3n2 + 2n − 23.

Chapter 9

Cutting up Food (Counting and Geometry) 9.2 Try This! Slice Pizza (and a Yam) 1. Seven, as evidenced in

2. Eleven, as evidenced in

.

.

3. Sixteen, as evidenced in

4.

Number of cuts on a pizza Maximum number of pieces possible

.

0 1

1 2

2 4

3 7

4 11

5 16

6 22

7 29

8 37

5. Next terms: 46, 56, 67, 79, 92, 106, . . . .

Recurrence relation: pn = pn−1 + n.

There are all kinds of mildly wrong conjectures to be made, including pn = pn−1 + n − 1 and pn = pn−1 + n + 1.

+ 1 = n+1 Closed-form formula: pn = n(n+1) 2 2 + 1. n There are all kinds of mildly wrong conjectures to be made, including pn = n(n+1) 2 , pn = 2 , pn = 2n , or pn = 2n − 1.

6. No three cuts may intersect at a single point. Each cut must intersect all other cuts.

7. Solutions will vary. See Section 9.3.

83

84

9. Cutting up Food (Counting and Geometry)

8. Four chunks:

.

9. Eight chunks:

.

10. An upper bound is 24 = 16; a lower bound is 8 + 4 = 12 by making a parallel slice. Students will find 13 pieces easily and 14 with a bit more work, but they will be quite challenged to convincingly find 15 pieces.

9.4 Try This! Spaghetti, Yams, and More .

1. Five, as depicted in .

Six, as depicted in 2.

Number of cuts on a strand of spaghetti Maximum number of pieces possible

0 1

1 2

2 3

3 4

4 5

5 6

6 7

7 8

8 9

3. Terms: 9, 10, 11, 12, . . . .

Recurrence: sn = sn−1 + 1.

Closed form: sn = n + 1.

The recurrence is correct because if there are n − 1 cuts in a piece of spaghetti, only one additional piece can be obtained by making an additional cut; it cuts one of the existing pieces in two.

The closed form is correct by induction: the first few terms of the sequence provide base cases, and sn+1 = sn + 1 = n + 1 + 1 = (n + 1) + 1, as desired.

4.

Number of cuts on a yam Maximum number of pieces possible

0 1

1 2

2 4

3 8

4 15

5 26

6 42

7 64

8 93

5. No four cuts should meet in a line, no three cuts should meet in a line, and no two cuts should be parallel.

6. Students may notice that sn are the first differences of pn and that these are the first differences of yn . Or they may not.

7. Possible conjectures include yn = yn−1 + n2 + 1 and yn = yn−1 + pn−1 . As with pizzas, it is common for conjectures to be off by an index somewhere.

9.8. Bonus: Geometric Gems

8.

n sn pn yn ?n

0 1 1 1 1

1 2 2 2 2

2 3 4 4 4

3 4 7 8 8

85

4 5 11 15 16

5 6 16 26 31

6 7 22 42 57

7 8 29 64 99

8 9 37 93 163

It is now clearer that pn are the first differences of yn and that sn are the first differences of pn . Students may realize that they can fill in the next row of the table by using yn as first differences, though they are likely to have trouble conceptualizing the physical situation to n n which this corresponds. Students may start to put pieces together to see that sn = 1 + 0

and that pn = n2 + n1 + n0 . They may also get sort of stuck on pn = n+1 + 1 and thus 2 have trouble generalizing.

9. What you see is a section of plane with a bunch of cuts. As it happens, those cuts are cross sections of the n planar cuts made in the yam. Thus, every additional chunk created by that slice corresponds to a region on the section of plane. This explains why yn = yn−1 + pn−1 .

10. They need to be in general position.

11. What you see is a section of (k − 1)-dimensional plane with n − 1 cuts. As it happens, those cuts are cross sections of the n − 1 cuts made in the hyperbeet. Thus, every additional piece created by the nth cut corresponds to a region on the section of (k − 1)-dimensional plane.

12. hn,k = hn−1,k + hn−1,k−1 ; it is reminiscent of the fundamental recurrence for binomial coefficients.

n 13. hn,k = nk + k−1 + · · · + n2 + n1 + n0 .

9.8 Bonus: Geometric Gems Extension:

1. 26, because it’s one more than 25.

2. n2 + 1 dots tossed into a 1 × 1 square include some pair of dots that are less than apart.

√

2 n units

3. There are n2 smaller squares, so by the pigeonhole principle, if we have n2 + 1 dots at least two dots must be in the same smaller square. The farthest two √such dots can be from each other is the length of the diagonal of that small square, which is n2 .

9.10 Problems That Combine Combinatorial Topics 1. We need all the intersections to occur inside the pizza. Then each line extends twice out of the pizza, so there are 2n radii and thus 2n regions of the plane.

86

9. Cutting up Food (Counting and Geometry)

2. Sequence: 1, 2, 4, 8, 15, . . . .

First differences: 1, 2, 4, 7, 11, 16, . . . .

Second differences: 1, 2, 3, 4, 5, . . . .

Third differences: 1, 1, 1, 1, . . . .

Form is yn = c + dn + f n2 + gn3 : y0 = 1 = c; y1 = 2 = 1 + d + f + g so d = 1 − f − g; 1−6g y2 = 4 = 1 + (1 − f − g)2 + 4 f + 8g = 3 + 2 f + 6g so f = 1−6g 2 ; y3 = 8 = 1 + (1 − 2 − 1 5 g)3 + 1−6g 2 9 + 27g = 4 + 3(1 − 6g) − 3g + 27g = 7 + 6g so g = 6 ; f = 0 and d = 6 . Thus, 1 3 5 yn = 6 n + 6 n + 1.

3. Solutions will vary.

4. On a k-dimensional hyperbeet, whenever n ≤ k. This is because there are k different directions (or degrees of freedom) in a k-dimensional object, and so cuts can be made in k different directions at right angles to each other.

5. Consider sn , pn as our base cases. Suppose we have n − 1 hypercuts placed in general position in k-dimensional space. We will suppose that an nth hypercut has been added in general position, so that we obtain the largest possible number of additional regions and thus calculate hn,k . If we examine the cross section of this configuration that passes through the nth hypercut, we see cross sections of hyperregions and of hypercuts. A cross section of a hyperregion is (k − 1)-dimensional (because the dimension has been reduced by one), and a cross section of a hypercut is similarly (k − 2)-dimensional. The inductive hypothesis applies, so we have hn−1,k−1 hyperregions appearing; each of these hyperregions corresponds to a new region created by the nth hypercut cleaving an old region in two. So, we have the hn−1,k regions created by the first n − 1 hypercuts, and to this we add the hn−1,k−1 regions created by the nth hypercut. Thus, hn,k = hn−1,k + hn−1,k−1 .

n−1 6. We know that hn,k = hn−1,k + hn−1,k−1 , so by induction we have hn,k = ( n−1 k + · · · + 3 +

n−1 n−1 n−1 n−1 n−1 n−1 n−1 n−1 + · · · + n−1 + n−1 )+ 2 + 1 + 0 ) + ( k−1 3 + 2 1 + 0 ) = ( k + k−1

n n n n−1 n−1 n n−1 n−1 n−1 n−1 + . . . + + = ) + ( + ) + + · · · + ( n−1 + ) + ( 2 1 + 3 0 0 k 1 2 2 1 3 n n n n−1 n n = k + . . . 3 + 2 + 1 + 0 . (This argument is cleaner when done with summation 0 notation, but not as easy to follow.)

7. This is Pascal’s triangle—the recurrence is the same and so are the initial conditions.

8. Sequence: 0, 1, 3, 6, 10, 15, 21, . . . .

First differences: 1, 2, 3, 4, . . . .

Second differences: 1, 1, 1, 1, . . . .

This can be completed using the method of second differences, or one can notice by the

pattern that the sequence is an = n+1 .

2

n+1 Proof by induction that this is correct: Consider an+1 = an + n+1 = n+1 = n+2 1 2 + 1 2 by the basic recurrence.

9.10. Problems That Combine Combinatorial Topics

87

9. Sequence: 0, 0, 1, 4, 10, 20, 35, 56, . . . .

First differences: 0, 1, 3, 6, 10, 15, 21, . . . .

Notice that this is the sequence in the previous problem. This can be completed using the

method of third differences, or one can notice by the pattern that the sequence is an = n+1 3 .

n+1 n+1 n+1 n+2 Proof by induction that this is correct: Consider an+1 = an + 2 = 3 + 2 = 3 by the basic recurrence.

10. Sequence: 0, 0, 0, 1, 5, 15, 35, 70, 126, . . . .

First differences: 0, 0, 1, 4, 10, 20, 35, . . . .

Notice that this is the sequence in the previous problem. This can be completed using the

method of fourth differences, or one can notice by the pattern that the sequence is an = n+1 4 .

n+1 Proof by induction that this is correct: Consider an+1 = an + n+1 = n+1 = n+2 3 4 + 3 4 by the basic recurrence.

11. Sequence: 0, 0, 0, 0, 1, 6, 21, 56, 126, 252, . . . .

First differences: 0, 0, 0, 1, 5, 15, 35, 70, 126, . . . .

Notice that this is the sequence in the previous problem. This can be completed using the

. method of fifth differences, or one can notice by the pattern that the sequence is n+1 5

n+1 n+1 n+1 Proof by induction that this is correct: Consider an+1 = an + 4 = 5 + 4 = n+2 5 by the basic recurrence.

12. From the previous problems, we see that the closed form is an = n+1 j+1 .

13.

(a) No. He will always need to take an even number of steps to end up on an even number.

(b) It has the same parity as k does (i.e., if k is odd, so is n; if k is even, so is n). (c) He can do RRLL, LLRR, LRLR, RLRL, LRRL, RLLR, so there are six ways. (d) There have to be the same number of steps to the right and to the left. This means there are 2n steps to the right and the rest are to the left. So, we want to count the number of n meanders that have n steps, of which 2n are to the right. So we want n/2 .

(e) Every meander takes n steps, and choosing a particular sequence of steps means we need to know which of those steps are to the left and which are to the right. Every step that is not to the right is a step to the left, so we just need to know which of the steps are to the right.

The number of steps to the left plus the number of steps to the right gives n. And, a step to the left moves the location of Grover by −1 whereas a step to the right moves the location of Grover by +1. Therefore, the number of steps to the right minus the number of steps to the left is going to be k. Let the number of steps to the left be L and the number of steps to the right be R. Then our previous observations become L + R = n and R − L = k. Adding these equations together we have 2R = n + k or R = n+k 2 .

Now, we need to count the number of meanders that have n steps, of which n+k 2 are to

n the right. So we want (n+k)/2 .

88

9. Cutting up Food (Counting and Geometry)

14. Sequence: 0, 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, 176, . . . .

Recurrence: an = an−1 + 3(n − 1) + 1. (This can be seen as adding three sides, except for one ending dot, to the two existing top sides of the pentagon and then adding the missing ending dot.)

First differences: 1, 4, 7, 10, 13, 16, . . . .

Second differences: 3, 3, 3, 3, . . . .

Closed form is an = c + dn + f n2 . Let a0 = 0 = c, so a1 = 1 = d + f and d = 1 − f ; a2 = 5 = 2d + 4 f = 2 − 2 f + 4 f , so 3 = 2 f and f = 32 , d = − 12 . Thus, an = 23 n2 − 12 n.

15. Sequence 0, 1, 6, 15, 28, . . . .

Recurrence: an = an−1 + 4(n − 1) + 1. (This can be seen as adding four sides, except for one ending dot, to the two existing top sides of the hexagon and then adding the missing ending dot.)

First differences: 1, 5, 9, 13, . . . .

Second differences: 4, 4, 4, . . . .

Closed form is an = c + dn + f n2 . Let a0 = 0 = c, so a1 = 1 = d + f and d = 1 − f ; a2 = 6 = 2d + 4 f = 2 − 2 f + 4 f , so 4 = 2 f and f = 2, d = −1. Thus, an = 2n2 − n.

16.

(a) (b) (c) Recurrence: an = an−1 + 2(n − 1) + 1. (This can be seen as adding two sides, except for one ending dot, to the two existing sides of the square and then adding the missing ending dot.

Closed form: an = n2 . (d) Recurrence: an = an−1 + (n − 1) + 1 = an−1 + n. (This can be seen as adding one side, except for one ending dot, to the two existing sides of the triangle and then adding the missing ending dot.)

Closed form: an = n+1 2 .

17. Recurrence: an = an−1 + (k − 2)(n − 1) + 1. (This can be seen as adding k − 2 sides, except for one ending dot, to the two existing top sides of the k-gon and then adding the missing ending dot.)

Sequence: 0, 1, k, k + (k − 2)2 + 1 = 3k − 3, 3k − 3 + 3(k − 2) + 1 = 6k − 8, 6k − 8 + 4(k − 2) + 1 = 10k − 15, . . . .

First differences: 1, k − 1, 2k − 3, 3k − 5, . . . .

Second differences: (k − 2), (k − 2), (k − 2), . . . .

Form is an = c + dn + f n2 . Let a0 = 0 = c, so a1 = 1 = d + f and d = 1 − f ; a2 = k = 4−k k−2 2 4−k 2d + 4 f = 2 − 2 f + 4 f , so k = 2 + 2 f and f = k−2 2 , d = 2 . Thus, an = 2 n + 2 n.

18.

(a) One.

9.10. Problems That Combine Combinatorial Topics

89

(b) Two. (c) n − 2. This can be proved by induction; the base case is a triangle. Remove a triangle and be left with (a) one polygon with n− 2+ 1 = n− 1 sides or (b) two or three polygons with fewer sides.

19.

(a) Six. (b) Eight. (c) Ten.

20.

(d) If the larger polygon has n sides and the hole has k sides, then the triangulation has n + k triangles. This can be proved by induction. Use the triangle-in-a-triangle as a base case.

For the inductive step, either the hole or the larger polygon has more than three sides, so pick one of those sides; it has a triangle extending from it. Contract that triangle (eliminating one triangle from the triangulation and one side from the n + k sides), and the inductive hypothesis holds on the result.

(a) 42 = 6 lines. (b) Exactly one: cd.

(c) Four: ac, ad, bc, bd. (d) Three: ab and two of its intersecting lines (ac, bc or ad, bd or ac, ad). (e) Either ab, ac, bc or ab, ad, bd. (f)

21.

(a)

6 3

= 20 lines.

(b) Exactly one: de f .

(c) For each of a, b, c, we may choose two of the remaining points to form a line, so 3 · 32 = 9 lines intersect.

(d) There are 32 pairs of non-{a, b, c} points, and none of those pairs can be matched with more than one of a, b, c or else there will be two lines that intersect in two points. (e) Yes; abc, ade, bd f , ce f , for example. (f) abc, ade, bd f , ce f , for example.

(g)

22.

(a) (b)

k r lines.

k−r lines. r

90

23.

9. Cutting up Food (Counting and Geometry)

(c) For each of the r points in the given line, we may choose r − 1 of the remaining points k−r to form a line, so r · r−1 lines intersect.

(a) 73 = 35 lines.

(b) 43 = 4 lines.

(c) For each of a, b, c, we may choose two of the remaining points to form a line, so 3 · 42 = 18 lines intersect.

(d) There are 42 pairs of non-a, b, c points, and none of those pairs can be matched with more than one of a, b, c or else there will be two lines that intersect in two points. (e) Yes. abc, ade, a f g, bd f , beg, ce f , cdg for example. (f) For example, abc, ade, a f g, bd f , beg, ce f , cdg.

(g)

24.

(a) (b)

25.

8 3 = 56 lines.

5 3 = 10 lines.

(c) For each of a, b, c, we may choose two of the remaining points to form a line, so 3 · 52 = 30 lines intersect.

(d) There are 52 pairs of non-a, b, c points, and none of those pairs can be matched with more than one of a, b, c or else there will be two lines that intersect in two points. (e) Yes; abc, ade, a f g, bd f , beg, ce f , cdg, for example. But this doesn’t involve the eighth point! The possible lines are abc, xde (where x can be a, b, or c), x f g, xd f , xeg, xe f , xdg, xdh, xeh, x f h, xgh. Note that xde, x f g must have the same value of x to avoid parallel lines. Likewise, x f g, xdh must have the same value of x; xdh, xe f must have the same value of x; xe f , xgh and xgh, xd f and xd f , xeg and xeg, x f h and x f h, xdg and xdg, xeh must have the same value of x. So, all lines must have the same value of x in order to avoid parallel lines. However, we then have pairs of lines such as xd f and xde that intersect in more than one point. This produces a contradiction.

(a) 84 = 70 lines.

(b) 43 = 4 lines. (c) For each of a, b, c, d, we may choose three of the remaining points to form a line, so 4 4 · 3 = 16 lines intersect.

(d) There are 43 triples of non-a, b, c, d points, and none of those triples can be matched with more than one of a, b, c, d or else there will be two lines that intersect in two points.

However, because there are only four triples, namely e f g, e f h, egh, f gh, each pair of triples has two points in common! Therefore, we can only have one more line to intersect abcd.

9.10. Problems That Combine Combinatorial Topics

91

(e) Now there are 53 = 10 triples. They’re e f g, e f h, e f i, egh, egi, ehi, f gh, f gi, f hi, ghi.

But, if we have the least amount of overlap, e.g., e f g, ehi, we still need to place these with different letters. So at most we can have four additional lines. Can we achieve that? Let’s try: ae f g, behi, c f h . . . nope. We’ve gone all the way up to two additional lines. (f) It should, because then we can have enough points that we can pass two additional lines through each of a, b, c, d, e.g., ae f g, ahi j. However, when we try to construct additional lines, we run into problems. No triple from e, f , g, h, i, j shares only one point in common with e f g and only one point in common with hi j.

Theme II Supplement: Problems on the Theme of Combinatorics 1. By the principle of inclusion-exclusion, 3 + 5 − 2 = 6 cats like tuna or salmon, so seven cats were consulted in total.

2.

(a) There are three slots; a list that doesn’t begin with C may have four choices for the first letter, five choices for the second letter, and five choices for the third letter, so there are 100 possibilities. Similarly, there are 100 possibilities for three-letter lists that don’t end in P. Now, how many lists neither begin with C nor end in P? (If we add 100+100 = 200, we’ve overcounted those.) These have four choices for the first letter, five choices for the second letter, and four choices for the third letter, for a total of 80. This means there are 120 three-letter lists that don’t begin with C or don’t end with P. (b) Well, first notice that such a list must either begin with P, E, A or end with P, E, A. Thus, there are only two places the remaining letter can be, and there are five choices for that letter, so there are only ten such lists. (c) Again, there are only two places the remaining letter can be and five choices for that remaining letter, so there are ten such lists.

3.

(a) Five: red-red, red-green, green-red, green-green, and white-white. (b) Twelve: if all colors are red or green, there are 23 = 8 possibilities, and with white there are only red-white-white, green-white-white, white-white-red, and white-white-green. (c) If the rightmost gnat has a green or red hat, then there are two colors possible for that gnat and gn−1 possible hat-color strings prior to that gnat’s hat. If the rightmost gnat has a white hat, then there are gn−2 possible hat-color strings prior to that gnat’s—and its partner’s—hat.

So, the recurrence is gn = 2gn−1 + gn−2 . This is a linear homogeneous recurrence ren lation with constant coefficients, so we try the algorithm given √ 8.8: x = √ in Section n−1 n−2 2 2 2x + x √, so x = 2x + 1, and so x − 2x√ − 1 = 0 = (x − √2 − 1)(x + 2 − 1). The roots are ± 2 + 1. So, we know gn = q1 ( 2 + 1)n + q2 (− 2 + 1)n . Then, g1 = 2, g2 = 5, g3 = 12, . . . . Note that √ √ 2 = q1 ( 2 + 1) + q2 (− 2 + 1), √ √ √ √ 5 = q1 ( 2 + 1)2 + q2 (− 2 + 1)2 = q1 (3 + 2 2) + q2 (3 − 2 2), √ √ √ 1 + 2 = (3 + 2 2)/( 2 + 1).

92

Theme II Supplement: Problems on the Theme of Combinatorics

93

√ √ √ √ √ Therefore, 2(1 + 2) = q1 (3 + 2 2) + q2 (− 2 + 1)(1 + 2). Thus, 2(1 + 2) − 5√= √ √ √ √ √ + 2) − q (3 − 2 2) or 2 2 − 3 = −q − q (3 − 2 2) or q2 = 2−4 2 . q2 (− 2 + 1)(1 2 2 2 √

Then, q1 = 2+4 2 .

4. Base cases: 1 = 20 , 1 + 1 = 21 , 1 + 2 + 1 = 22 . n n Inductive hypothesis: For n ≤ k, ∑ = 2n . j j=0

k+1 k Consider ∑k+1 and remember that k+1 = kj + j−1 by the basic binomial recurrence. j=0 j j

k k k+1 k+1 k+1 k k+1 k k+1 k k Then ∑ j=0 j = ∑ j=0 j + j−1 = ∑ j=0 j + ∑ j=0 j−1 = ∑kj=0 kj + ∑k+1 j=1 j−1 = 2 +

k ∑kj=0 j = 2k + 2k = 2k+1 , as desired.

5. We are given the base case that b(1) = 1. Suppose that for n ≤ k, b(n) counts the number of 1s in the binary representation of n. Now consider b(k + 1). Either k is even or it is odd.

If k is even, k + 1 is odd and b(k + 1) = b( 2k ) + 1; this is the number of 1s to the left of the ones place of the binary representation of k + 1 plus the 1 in the ones place, which is equal to the number of 1s in the binary representation of k + 1. If k is odd, then k + 1 is even and b(k + 1) = b( k+1 2 ); all of the 1s in the binary representation of k + 1 are to the left of the ones place. Done.

6.

(a) −2, 3, 8, 13, 18, . . . ; an = an−1 + 5, with a0 = −2.

(b) 2, 4, 10, 28, 82, . . . ; an = 3(an−1 − 1) + 1 = 3an−1 − 2. (c) −6, 1, 8, 15, 22, . . . ; an = an−1 + 7, with a0 = −6.

7.

(a) One. (b) Two. (c) Three. (d) Five. (e) Fn+1 . Either the rightmost tile is a 2 × 1 rectangle, in which case we have the same number of ways of tiling the remainder as there are ways to tile a 2 × (n − 1) grid, or the rightmost tiles are two 1 × 2 rectangles, in which case we have the same number of ways of tiling the remainder as there are ways to tile a 2× n − 2 grid. So, we have a recurrence Tn = Tn−1 + Tn−2 . This is the same recurrence as defines the Fibonacci sequence and has the same starting terms, so it is the Fibonacci sequence.

8.

(a) 5, 25, 125, 625, 3125, . . . ; an = 5n . (b) 2, 10, 50, 250, 1250, . . . ; an = 2 · 5n−1 .

(c) 6, 9, 12, 15, 18, . . . ; an = 3n + 3 = 3(n + 1).

9. A passerby either takes a given brochure or does not. Thus, we are asking for the number of subsets of the n brochures, of which there are 2n .

94

Theme II Supplement: Problems on the Theme of Combinatorics

10. The characteristic polynomial is x3 = x2 + 4x − 4. Then, x3 − x2 − 4x + 4 = 0 = (x − 1)(x − 2)(x + 2), so the solution is an = q1 1n + q2 2n + q3 (−2)n . Thus, we have a0 = 4 = q1 + q2 + q + 3, a1 = 2 = q1 + 2q2 − 2q3 , and a2 = 10 = q1 + 4q2 + 4q3 . Solving these equations gives a1 = 2, q2 = 1, q3 = 1. The closed form is an = 2 + 2n + (−2)n .

11. sn = sn−1 + an . sn = sn−1 + an = sn−1 + 2n − 1 with s1 = 1.

12. This depends on the country; in the US there are pennies, dimes, nickels, quarters, halfdollars, and dollars. Thus, there are six options for each of nine slots: 69 ways.

13. Sequence: 1, 4, 8, 13, 19, 26, 34, 42, . . . .

First differences: 3, 4, 5, 6, 7, 8, 9, . . . .

Second differences: 1, 1, 1, 1, 1, . . . .

So, this has the form an = c + dn + f n2 : a1 = 1 = a0 + 1 + 1, so a0 = −1 = c; a1 = 1 = −1 + d + f , so d = 2 − f ; a2 = 4 = −1 + 2d + 4 f = −1 + 2(2 − f ) + 4 f = 3 + 2 f , so f = 21 and d = 23 . Thus, an = −1 + 23 n + 21 n2 .

14. a1 = 1, a2 = 2 · 1, a3 = 3 · 2, a4 = 4 · 6 = 24, . . . , so an = n!.

15. There are 14 choices for who gets the squash, then 13 for who gets the crown of broccoli, 12 for who gets the tomato, 11 for who gets the Swiss chard, and 10 for who gets the large radish, so there are a total of 14 · 13 · 12 · 11 · 10 = 240,240 ways to distribute the vegetables.

16. We have a line of six 1s and need to stick in three +s. Thus, there are 93 = 84 ways to line them up.

17. There are k edges per vertex, so there are kn edges, but we have double-counted because each edge is incident to two vertices; thus, the total is kn 2 edges.

18. This sounds delicious: 27 = (1 + 12 + 3 + 7) + (18 − 15 − x) + (12 − 10 − x) + x = 23 + 3 − x + 2 − x + x = 28 − x, so there is exactly one peanut-butter-banana-containing pie. Yum.

19. The number of ways to have two on-campus members is 92 , and there are 31 ways to pick

an off-campus member; thus, the total number of ways is 92 · 3 by the product principle.

20. This is a stars-and-bars argument. The stars are the dye doses and the bars are the 11 spaces between skeins. Thus, there are 28+12−1 = 1, 676, 056, 044 ways to recolor the yarn.

11 10!

9! = 3!10 21. That would be 10!/3!

7 = 3!106 = .06048 or 6% . . . hardly any.

107

22. Let the flavors be p, h, cm, g, s, and let a, b, c, d, e be the (nonnegative, integer) number of quarts of each flavor ordered. Then, a + b + c + d + e = 3, and the number of solutions to this equation is 73 = 35.

23.

(a) Three slots are fixed, and there are ten choices for each of the four remaining slots, so 104 = 10,000.

Theme II Supplement: Problems on the Theme of Combinatorics

95

(b) There are four slots to fill, with ten choices for the first slot, nine choices for the second slot, eight choices for the third slot, and seven choices for the fourth slot, for a total of 10 · 9 · 8 · 7 = 5,040 telephone numbers.

(c) This is slightly ambiguous, so here are two reasonable responses.

There is exactly one 9, exactly one 6, and exactly one 2. So, there are seven choices for the fourth digit, four ways to place that digit, and six ways to order the remaining three digits in the remaining three slots, so there are 7 · 4 · 6 = 168 telephone numbers.

There are at least one 9, at least one 6, and at least one 2. So, there are ten choices for the fourth digit, four ways to place that digit, and six ways to order the remaining three digits in the remaining three slots, so there are 10 · 4 · 6 = 240 telephone numbers.

24. 10 6 = 210 ways.

25. Try finite differences.

First differences: 32, 70, 115, 167, 226, . . . .

Second differences: 38, 45, 52, 59, . . . .

Third differences: 7, 7, 7, . . . .

The next third difference should be 7, so the next second difference should be 63, so that the next first difference should be 289, so that the next score should be 906.

26. Students who like gymnastics or the decathlon: 730 + 460 − 240 = 950.

All three: 1200 − 620 − 730 − 460 − 170 + 270 + 330 + 240 = 60.

27. First with Venn diagrams:

Now without Venn diagrams: Note that |A| = 80, |B| = 80, |C| = 80, |A∩B| = 30, |A∩C| = 30, |B ∩ C| = 20, and |A ∩ B ∩ C| = 20. Then, we want |A| − |A ∩ B| − |A ∩ C| + |A ∩ B ∩ C| = 80 − 30 − 30 + 20 = 40.

28. The facial information is a distraction. The important information is that each vertex has degree 4. Then, the total degree is 30 · 4 = 120, and because each edge is counted twice (once for each end), there are 60 edges.

29. The characteristic equation is xn − 5xn−1 + 6xn−2 = 0, so x2 − 5x + 6 = 0 is the characteristic polynomial, which factors to (x − 3)(x − 2) and thus has roots 2, 3. Thus, an = c · 2n + d · 3n : 0 = c · 2 + d · 3 = 2c + 3d, so −2c = 3d and c = − 23 d; 4 = c · 22 + d · 32 = 4c + 9d, so 4 = 4(− 23 d) + 9d = −6d + 9d = 3d; so d = 43 and c = −2. Thus, an = −2 · 2n + 34 · 3n = 4 · 3n−1 − 2n+1 .

30. Let x = 10, y = −1 in the binomial theorem to obtain (10 − 1)m = 9m .

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Theme II Supplement: Problems on the Theme of Combinatorics

8! because there are 8! arrangements of eight letters, but that overcounts by the 3! arrange31. 3!·2! ments of the three Es and the 2! arrangements of the two Rs.

n·(n−1)! n!

32. (a) Using factorials, we verify that nk = k!(n−k)! = k·(k−1)!(n−1−(k−1))! = nk n−1 k−1 .

(b) On the left-hand side, we are simply choosing k items from a pile of n items. On the right-hand side, we choose a single item from a pile of n items. Then, for each choice there are n−1 k−1 ways to choose another k − 1 items. However, we have distinguished the first item we chose, and any of our k items could have appeared as first, so we have overcounted by a factor of k.

33. 43 + 71 + 36 − 26 − 16 − 22 + x = 100 = 86 + x so x = 14 students.

34. r1 = 1, r2 = 2 (1, 1 or 2), r3 = 3 (1, 1, 1 or 2, 1 or 1, 2), and r4 = 5 (1, 1, 1, 1 or 1, 2, 1 or 1, 1, 2 or 2, 1, 1 or 2, 2). rn = rn−1 + rn−2 .

To obtain a sequence of steps that total n feet, the robot can take a sequence of steps totaling n − 1 feet and then take a 1-foot step, or it can take a sequence of steps totaling n − 2 feet and then take a 2-foot step.

n−1

n−1

35. Notice that − n1 = − n−1 and n2 = n−1 + 2 , so their sum is − n−1 + n−1 .

0 − 1 1 0 2 n n n n n−1 n−1 In this fashion, the entire sum telescopes: 0 − 1 + 2 · · · ± n = 1 − 0 − 1 + n−1 n−1 n−1 n−1 1 + 2 · · · ∓ n−2 ∓ n−1 ± 1 = 0.

36. Take 31 to obtain the Jeong representative and 8−3 for the non-Jeong representative. That’s 1 3 · 5 = 15 ways.

37.

(a) There are four slots for a word; so, the number of choices per slot is six, five, four, and three, respectively, and the total number of words is 6 · 5 · 4 · 3 = 360.

(b) This fixes the last two slots, so there are only 4 · 3 choices for the first two slots, or 12 possible words. (c) Such words could be of the form − − ef , −ef −, or ef − −, so 3 · 12 = 36 words.

(d) One way to compute this is to note that there are 12 words that end in ef and there are 4! ways to order the letters in each word, though this also counts each word twice (i.e., abef , baef ) so really there are 12 · 12 = 144 words.

38.

(e) Let the set of words containing an e be A and the set of words containing an f be B. Then, we seek |A ∪ B|. We know |A| = |B| = 4(5 · 4 · 3) = 240 because there are four ways to place the e (or the f ) and then, for the remaining three slots, there are five choices for the first slot, four for the second slot, and three for the third slot. We also know that |A ∩ B| = 144 from the previous problem. Now, |A ∪ B| = |A| + |B| − |A ∩ B|, so we compute 240 + 240 − 144 = 336 words.

(a) 10 4 = 210 sets.

(b) 74 = 35 sets.

Theme II Supplement: Problems on the Theme of Combinatorics

(c)

10 4

97

− 74 = 175 sets.

39. We have not specified how many cards each player receives. So for each card, we just need to decide which of the seven players gets the card. In other words, we have 52 slots and each has seven choices, so 752 = 88,124,787,089,723,195,184,393,736,687,912,818,113,311,201 is the number of ways.

40. There are multiple approaches; one is (1 · 9 · 9 · 9) · 4 = 2,916, for the four placements of the 9.

41. 20 5 = 15,504 ways.

42.

1. Write 1.

2. On a new line, write 1 1.

3. On a new line, write 1 and let j = 0.

4. If there is a ( j + 1)st entry in the previous line, add the jth and ( j + 1)st entries of the previous line and write this down; otherwise, write 1 and go to step 3.

5. Replace j with j + 1 and go to step 4.

Notice that this algorithm does not terminate. However, neither does Pascal’s triangle!

43. Sequence: 12, 5, −2, −9 . . . .

Relation: an = −7(n − 1) + 12 = −7n + 19.

44.

(a) There are 15 chips, and we need to choose which slots will be occupied by the 4 poisoned chips, so 15 4 = 1365 ways.

Note that these are distinctly different from poissoned chips.

(b) There are 15 2 = 105 ways of choosing two chips from the given lineup. The number of choices that contain at least one poisoned chip is the same as (the number of choices with one poisoned chip) + (the number of choices with two poisoned chips), or 105 − (the

= 55 ways to get two nonnumber of choices with no poisoned chips). There are 11 2 poisoned chips, so there are 105−55 = 50 choices that will contain at least one poisoned chip.

45. Check the base case: (x + y)1 = 10 x0 y1−0 + 11 x1 y1−1 . Consider (x + y)n+1 = (x + y)n (x+ y).

By the inductive hypothesis, this is

n i n−i n i+1 n−i n ∑ i x y (x + y) = ∑ i x y + ∑ i xi yn−i+1

n n i n−i+1 i−1+1 n−(i−1) =∑ x y +∑ xy i−1 i

n n i n−i+1 i n−i+1) =∑ xy +∑ xy i−1 i

n + 1 i n+1−i =∑ xy , i as desired.

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Theme II Supplement: Problems on the Theme of Combinatorics 6 3

= 20 choices.

47. We want 31 3 .

31!

Directly, 31 3 = 3!28! = 4,495. This is the number of permutations of the set of flavors divided by the overcounting of the number of permutations of the chosen flavors and of the nonchosen flavors.

30 30 Recursively, 31 3 = 3 + 2 , and we could break this down forever. This uses the basic choice identity.

48. This is Question D from Chapter 7, and so there are 12+3−1 = 91 ways.

2 46. Ze has

n−1 n 49. The correct statement is = ∑ i.

2 i=1

We can prove this by induction using the basic choice notation identity on the left-hand side n along with the fact that 1 = n.

50. The key is to see that we have 6 choices for each bar, so 62 = 36 total ways.

Part III Theme: Graph Theory

Chapter 10

Trees 10.3 Try This! Spanning Trees 1.

2.

3. Find a cycle, remove an edge. Repeat until the graph is acyclic.

Or, hide all the edges. Replace them one at a time, always choosing to replace an edge that is incident to some other replaced edge that does not complete any cycle. Continue until all vertices are included in the resulting tree.

4. See previous problem’s solution.

5. If we only remove edges, then we certainly include all the edges of the original graph. At each step, the graph remains connected; the end result is acyclic, so it is a tree. If we add edges so that we remain connected and do not make cycles, we have a tree; if we continue until all vertices are included, then we have a spanning tree. The only question is whether we can continue the algorithm at each step. Suppose we cannot add an edge. Then every edge of the graph that is not yet present either completes a cycle (in which case both of its vertices are already in the growing tree) or is not incident to the growing tree. In the latter case, consider one of the endpoints of the edge. The graph is connected, so there is a path from this vertex to a vertex of the growing tree. This path contains at least one edge that does not complete a cycle, and so the algorithm can continue.

6. Each connected component of the graph has a spanning tree. The union of these spanning trees is a spanning forest. (Yes, it’s possible for a forest to have only one tree. Silly, perhaps, but possible.)

7. The weights are, respectively, 17, 12, and 16.

The middle one has the smallest weight.

No, there’s a smaller weight spanning tree; here is a spanning tree with total weight 11:

100

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101

8. Remove an edge of highest weight that doesn’t disconnect the graph. Repeat until none exists (or until a tree is achieved). Then move on to an edge of second-highest weight and repeat.

Or, include as many lowest-weight edges as possible without creating cycles, and then move on to next-lowest weight edges, etc. Or, start with an edge of lowest weight and grow a tree from that edge, always choosing lowest-possible-weight edges from the set of edges incident to the growing tree.

9. See previous problem’s solution.

10. Solutions will vary. It is straightforward to show that (a) xor (b) holds, but getting both is pretty hard.

10.6 Try This! Binary Trees and Matchings 1.

(a)

(b)

2. There are lots of ways this could be done; we could use a color criterion, size criterion, edgeridges-or-not criterion, etc.

3. Only the middle one.

4. Add one edge joining the two uncovered vertices to the left-hand graph; the middle graph already shows a perfect matching; the right-hand graph has an odd number of vertices and so cannot have a perfect matching.

5. The odd path has no perfect matching but the even path has one. The other tree has no perfect matching because there are two leaves incident to the same vertex.

102

10. Trees

6. Odd number of vertices; multiple leaves incident to a vertex; leaves incident to vertices with an odd number of degree-2 vertices between them.

7. No. Consider a bipartite graph with one vertex in the first part and two vertices in the second part.

10.13 Tree Problems 1.

2. The sum of the degrees of a graph is two times the number of edges. Here, we know that the number of edges is one less than the number of vertices by Example 4.2.5. So, if v is the number of vertices and e is the number of edges, then 2e = 2(v − 1). The average degree of a graph is the total degree divided by the number of vertices, so here it’s 2 v−1 v < 2.

If a tree had no leaf, then every vertex would have degree at least 2, which would mean that the average degree would be at least 2. That’s a contradiction.

3.

w = 14

w=8

w = 10

w = 10

Exactly one.

4. Solutions will vary. Here is one possible spanning tree:

5. Solutions will vary. Here is one possible spanning tree:

w = 12

w = 12

w = 12

w = 10

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103

6. Solutions will vary. Here is one possible minimum-weight spanning tree:

7. Solutions will vary. Here is one possible minimum-weight spanning tree:

8. Solutions will vary. Here is one possible minimum-weight spanning tree:

9. We know that every tree has a leaf from Theorem 10.2.4. Start with this leaf and place it in part 1. Put the leaf’s neighbor in part 2. Put the neighbors of the leaf’s neighbor in part 1.

Continue in this fashion. Now, this will classify all of the vertices as belonging to part 1 or to part 2. How do we know that we do not classify any vertex twice? Suppose we did—then we must have hit the vertex a second time by going around a cycle, which is a contradiction as trees are acyclic. And, how do we know that we do not have edges between two vertices of part 1 (respectively part 2)? No vertex is in the same part as any of its neighbors.

This proof can also be done by induction in a couple of different ways. Consider a tree with n + 1 vertices and thus with n edges. Remove an edge. This gives us two trees, each of which is smaller than the original. So, each is bipartite. Label one of the trees as having parts 1 and 2 and the other as having parts A and B. Replace the edge. It joins a part-1 (respectively part-2) vertex to a vertex in either part A or part B. If the joined vertex is in part A, then relabel part B as part 1 (respectively part 2) and relabel part A as part 2 (respectively part 1).

If the joined vertex is in part B, then relabel part A as part 1 (respectively part 2) and relabel part B as part 2 (respectively part 1). Or, remove a leaf. This leaves us with a smaller tree.

Thus, it is bipartite. Let the leaf’s neighbor be in part 1 of this bipartition; put the leaf itself in part 2. This renders the original tree bipartite.

10. Yes. This graph has a perfect matching.

11. No. Let us prove it. If an edge is in a cycle, then its removal does not disconnect the graph as the desired paths can run through the rest of the cycle. And, if the removal of an edge does not disconnect a graph, it must be part of a cycle as there is more than one path from one endpoint of the edge to the other.

12. It’s more like an undoing of Kruskal because it uses the same verification step (cyclicity).

104

13.

10. Trees

1. Begin with a connected graph G. Order the edges arbitrarily as e1 , . . . , en .

2. Let e1 along with its vertices be called H and set j = 2.

3. If H ∪ e j has no cycles, then rename H ∪ e j as H; otherwise, do nothing.

4. If |E(H)| = |V (G)| − 1, output H as the desired tree; otherwise, do nothing.

5. Increment j by 1 and go to step 3.

14.

This is the only minimum spanning tree.

15. A tree that has a distinguished vertex called the root, has at most three edges growing “downwards” from each vertex, and has a designation of “left,” “center,” or “right” for each edge.

16. This requires finding the diameter of the coins! According to wikipedia.org, we have penny: 19.05 mm, nickel: 21.21 mm, dime: 17.9 mm, quarter: 24.26 mm, half-dollar: 30.6mm, and dollar: 26.5mm.

17. Solutions will vary depending on the choice of root. By the way, the non-real title is FowlWeather Friends.

18. Here are a graph and its matching:

10.13. Tree Problems

105

19. Here are a weighted graph and its minimum-weight matching:

20. Start with (1, ∗, ∗, ∗, 2, ∗, ∗, ∗, ∗). We then do (1, 1, ∗, ∗, 2, ∗, ∗, ∗, ∗), but that contradicts the row condition. So, we do (1, 2, ∗, ∗, 2, ∗, ∗, ∗, ∗), but that contradicts the column condition. So, we do (1, 3, ∗, ∗, 2, ∗, ∗, ∗, ∗). Then we do (1, 3, 1, ∗, 2, ∗, ∗, ∗, ∗) but that contradicts the row condition. So, we do (1, 3, 2, ∗, 2, ∗, ∗, ∗, ∗) (note that using 3 would violate row, so this is the only possible solution here). Then we do (1, 3, 2, 1, 2, ∗, ∗, ∗, ∗)—contradicts column—and then (1, 3, 2, 2, 2, ∗, ∗, ∗, ∗)—contradicts row—so finally

106

10. Trees

(1, 3, 2, 3, 2, ∗, ∗, ∗, ∗). Then we do (1, 3, 2, 3, 2, 1, ∗, ∗, ∗) (note that this is the only possible solution here from the row condition).

Then we do (1, 3, 2, 3, 2, 1, 1, ∗, ∗)—contradicts column—and so (1, 3, 2, 3, 2, 1, 2, ∗, ∗) (note that 3 would violate column, so this is the only possible solution).

Then we do (1, 3, 2, 3, 2, 1, 2, 1, ∗) (this was determined by constraints anyway). Lastly, we do (1, 3, 2, 3, 2, 1, 2, 1, 1)—contradicts column—and then (1, 3, 2, 3, 2, 1, 2, 1, 2)—contradicts column—so finally (1, 3, 2, 3, 2, 1, 2, 1, 3).

21. (∗, ∗, ∗, ∗, ∗, 2, . . . ), (∗, ∗, ∗, ∗, ∗, 3, . . . ), (∗, ∗, ∗, ∗, ∗, 4, . . . ), (∗, ∗, ∗, ∗, ∗, 1, ∗, ∗, ∗, ∗, ∗, ∗, 1 . . . ), (∗, ∗, ∗, ∗, ∗, 1, ∗, ∗, ∗, ∗, ∗, ∗, 3 . . . ), (∗, ∗, ∗, ∗, ∗, 1, ∗, ∗, ∗, ∗, ∗, ∗, 4 . . . ), (∗, ∗, ∗, ∗, ∗, 1, ∗, ∗, ∗, ∗, ∗, ∗, 2, 1 . . . ), (∗, ∗, ∗, ∗, ∗, 1, ∗, ∗, ∗, ∗, ∗, ∗, 2, 2 . . . ), (∗, ∗, ∗, ∗, ∗, 1, ∗, ∗, ∗, ∗, ∗, ∗, 2, 3 . . . ).

22. Try words in the order 1-down, 1-across, 2-down, 2-across. We proceed as follows: 1D blue blue blue blue blue blue blue blue blue blue blue teal teal teal teal teal teal teal teal teal teal teal rose rose

1A ocean sewer river storm cloud cloud cloud cloud cloud cloud cloud ocean sewer sewer sewer sewer sewer sewer sewer river storm cloud ocean sewer

2D

2A X X X X

sea tub tub tub tub bay

sea sea sea sea tub bay

X tank well lake

tank well lake

X X !!

X X

X !!

X X X X X X X X

1D rose rose rose grey grey grey grey grey ecru ecru ecru ecru ecru ecru ecru ecru ecru ecru ecru jade jade jade jade jade

1A river storm cloud ocean sewer river storm cloud ocean ocean ocean ocean ocean ocean ocean sewer river storm cloud ocean sewer river storm cloud

2D

2A X X X X X X X X

sea tub bay bay bay bay

X X tank well lake

X X X X X X X X X X X X

23. Suppose the graph is connected. This implies that it has a spanning tree by either algorithm given above. Now suppose that the graph G has a spanning tree T . Suppose also that it is

10.13. Tree Problems

107

not connected. Then there are two vertices x, y in G between which there is no path in G. However, T spans G and thus x, y ∈ T . Because T is a tree, there is some path between x, y in T and T ⊂ G, so this path is in G; this is a contradiction. Thus G is connected.

24. Five.

Fourteen (a left-descending edge with each three-datum tree beneath; a right-descending edge with each three-datum tree beneath; and two descendants from the root with four choices of additional edge).

25.

(a) 1 + 8 + 64 + 83 + 84 + 85 + 86 = 299,593 nodes. (b) There are three pairs for first numbers 2–7, so 18, and two pairs for first numbers 1 and 8, so 22 total. (c) 22 · 299,593 = 6,591,046. (d) 12,582,915 nodes. Not too shabby, but still not even as low as the 8! we got by preventing two coins from being set in the same row.

26. Solutions will vary, but three are shown here:

2

2 1

1

2

2

1

27. Solutions will vary, but two are shown here:

1

1

1 1

2

1 1

2

2 2

2 2

2

2 3

4

6

6

28. A minimum-weight spanning tree via Kruskal:

4

8 6

5 2

3

4 1

4

1

108

10. Trees 2

2 3

4

6

6

29. A minimum-weight spanning tree via Prim:

4

8 6

5 2

3

4 1

4 2

4

4

30. A non-minimum-weight spanning tree:

8 6 4 4

peapod

31. Here’s one:

pixie

lollipop

whisker

nudibranch edible?

yes

no

vegetable?

alive?

32. Here’s one:

no whisker

yes

no underwater?

no yes pixie nudibranch

lollipop

yes peapod

10.13. Tree Problems

109

33. Solutions will vary, but two different matchings of maximum size are shown here:

34. Solutions will vary, but two different perfect matchings are shown here: 2

2

2 3

4

3 6

6

4

4

4

8 6

5 2 4

1

6

6

8 6 3

2

4

5 2

3

4 1

4

35. This is like finding the ways to add numbers {1, 1, 1, 2, 2, 2, 3, 3, 3} to make 6. Abbreviatedly: 1 + 1 + 1 + 2 < 6. 1 + 1 + 1 + 2 + 2 > 6. Go back.

1 + 1 + 1 + 3 = 6. Keep and go back.

1 + 1 + 2 < 6. 1 + 1 + 2 + 2 = 6. Keep and go back.

1 + 1 + 3 < 6. 1 + 1 + 3 + 3 > 6. Go back.

1 + 2 + 2 < 6. 1 + 2 + 2 + 2 > 6. Go back.

1 + 2 + 3 = 6. Keep and go back.

2 + 2 < 6. 2 + 2 + 2 = 6. Keep and go back.

2 + 3 < 6. 2 + 3 + 3 > 6 Go back.

3 + 3 = 6. Done.

Chapter 11

Euler’s Formula and Applications 11.2 Try This! Planarity Explorations 1. Yes.

2. No; no proof is expected at this point.

3. Yes;

4. No; no proof is expected at this point;

5. Yes;

6. It is not planar; we will later be able to quickly show this by noting that e > 3v − 6 (27 > 24).

7.

Each drawing has three faces (just as did the original).

8.

110

11.7. Try This! Applications of Euler’s Formula

111

Each drawing has three faces (just as did the original).

9. Solutions will vary but here is one possibility:

Each drawing has four faces (just as did the original).

10. The sum of the face sizes is the same no matter how the graph is drawn. This is because each edge contributes twice to the sum, one for each “side” of the edge; if the edge is between two faces, it counts once for each face and if it is monofacial it counts twice on one face. Thus, the sum of the face sizes is twice the number of edges.

11.7 Try This! Applications of Euler’s Formula 1. For each component C considered alone, we have |V (C)| − |E(C)| + |F(C)| = 2. Now, all components share the same outside face. It is counted in |F(C)| for each component. So, if there are k components, we have overcounted the outside face k − 1 times. This tells us that |V (G)| − |E(G)| + |F(G)| = ∑C |V (C)| − ∑C |E(C)| + ∑C |F(C)| − (k − 1) = 2k − (k − 1) = 1 + k.

2. The smallest cycle length in the Petersen graph is 5. By similar reasoning to Theorems 11.6.2 and 11.6.3, if the Petersen graph were planar, then 5|F(G)| ≤ 2|E(G)|. Then, |V (G)| − |E(G)| + |F(G)| = 2 becomes |V (G)| − |E(G)| + 52 |E(G)| = |V (G)| − 35 |E(G)| ≥ 2. Now, for the Petersen graph, |V (G)| = 10 and |E(G)| = 15. However, 10 − 35 15 = 10 − 9 = 1 6≥ 2.

This is a contradiction, so the Petersen graph is nonplanar.

3. The condition means that 4|F(G)| = 2|E(G)|. Combining this with Euler’s formula, we get that 2 = |V (G)| − |E(G)| + 12 |E(G)| = |V (G)| − 21 |E(G)|, which rewrites to 4 = 2|V (G)| − |E(G)| and then the desired statement.

4. Conjecture: For n ≤ 4, Kn is planar and for n ≥ 5, Kn is nonplanar.

We know from Theorem 11.6.5 that K5 is nonplanar, and for n > 5 we have that K5 is a subgraph of Kn . Thus, for n ≥ 5 we know Kn is nonplanar. We exhibit planar drawings of K1 , K2 , K3 , and K4 to complete our proof:

5. Conjecture: For n ≤ 2 and any m, Km,n is planar and for n, m ≥ 3, Km,n is nonplanar.

112

11. Euler’s Formula and Applications

We know from Theorem 11.6.2 that K3,3 is nonplanar, and for n, m ≥ 3 we have that K3,3 is a subgraph of Km,n . Thus, for n, m ≥ 3 we know Km,n is nonplanar. We exhibit planar drawings of K1,m K2,m to complete our proof: 6.

(a) Solutions will vary. (b) Solutions will vary. (c) Euler’s formula does not hold. Instead, |V (G)| − |E(G)| + |F(G)| = 0. (d) This is really too challenging for students to write out rigorously, so here is a sketch of the proof. The simplest approach is by induction on the number of edges, using a single vertex graph with two loops as the base case (possibly with subdivisions, and with any number of trees glued on). Given a graph drawn on the torus with no edges crossing and both fundamental cycles “used,” select an edge that lies between two faces. One always exists or else we have the base case. Removing this edge gives G′ and eliminates a face, so |V (G)| − |E(G)| + |F(G)| = |V (G′ )| − (|E(G′ )| + 1) + (|F(G′ )| + 1) = 0.

11.10 Bonus: Topological Graph Theory Short activity:

1. Yes.

2. Yes.

3. Yes.

4. Yes.

5. Yes.

6. Yes.

11.12. Problems about Planar Graphs

113

11.12 Problems about Planar Graphs 1. Base case: one edge, two vertices, one face: 2 − 1 + 1 = 2. Consider a planar graph G with n edges. If there exists an edge whose removal does not disconnect the graph, then this edge separates two faces. Remove this edge to obtain G′ , so that two faces merge. Then |V (G′ )| − |E(G′ )| + |F(G′ )| = 2 by inductive hypothesis, and |V (G)| − |E(G)| + |F(G)| = |V (G′ )| − (|E(G′ )| + 1) + (|F(G′ )| + 1) = 2. If the removal of any edge would disconnect the graph, then G is a tree; it has one face, n vertices, and n − 1 edges so n − (n − 1) + 1 = 2.

2. The inequality arises from Theorem 11.6.2, so we want 3|F(G)| = 2|E(G)|. This means that every face has size 3.

3. The handshaking lemma says that the sum of the degrees of the vertices is twice the number of edges. In this case, we get 6k = 20. However, 20 is not evenly divisible by 6, and so no such graph can exist.

4. GL has 10 vertices and 15 edges, so it has 7 faces by Euler’s formula.

5. GR has 12 vertices and 20 edges, so it has 10 faces by Euler’s formula.

6. From the constraints, we have 3|V (G)| ≤ 2|E(G)| and 3|F(G)| ≤ 2|E(G)|. So 6 = 3|V (G)| − 3|E(G)| + 3|F(G)| ≤ 2|E(G)| − 3|E(G)| + 2|E(G)| = |E(G)|. So, no, P can’t have fewer than six edges.

7. From Euler’s formula, we have |V (G)| − 7 + |F(G)| = 2 or |V (G)| + |F(G)| = 9. Then, 2 3|V (G)| ≤ 2|E(G)| = 14, so |V (G)| ≤ 14 3 = 4 3 . Thus, there are at most four vertices and at most six edges. Contradiction!

8. This graph contains K5 , which is nonplanar.

9. Statement: Let G have girth g. If G is simple, planar, and connected, then g|F(G)| ≤ 2|E(G)|.

Proof: Because G is simple, no cycle can have fewer than g edges. Therefore, each face of G has at least g edges. As an inequality, this statement is g ≤ (# of edges bounding a face).

If we make such an inequality for each face of G, the sum of the left-hand sides is g|F(G)|.

When we sum the right-hand sides, we obtain 2|E(G)| because each edge borders two faces (or occurs twice on the boundary of one face). Thus, g|F(G)| ≤ 2|E(G)|, as desired.

10. Proof first: Because G is planar, we know by Theorem 11.5.1 that 2 = |V (G)| − |E(G)| + |F(G)|. From the previous problem it follows that |F(G)| ≤ 2g |E(G)|. Thus, 2 ≤ |V (G)| − |E(G)| + 2g |E(G)| = |V (G)| + 2−g g |E(G)|, which rewrites to 2g ≤ g|V (G)| + (2 − g)|E(G)|.

Statement: Let G have girth g. If G is simple, planar, and connected, then 2g ≤ g|V (G)| + (2 − g)|E(G)|.

11. K3,3 is nonplanar, so has thickness at least 2. Here is a decomposition of the edges of K3,3 into two internally non-intersecting sets:

114

11. Euler’s Formula and Applications

12. The Petersen graph is nonplanar, so has thickness at least 2. Here is a decomposition of the

edges of the Petersen graph into two internally non-intersecting sets:

13. Solutions will vary, but here are some examples:

14. This graph has girth 5. It has 12 vertices and 18 edges. If it were planar, it would have 8 faces

by Euler’s formula. But because it has no triangles, 4 · 8 ≤ 2 · 18, which is false.

15. K4,4 is nonplanar, so has thickness at least 2. Here is a decomposition of the edges of K4,4 into two internally nonintersecting sets:

16. Suppose that G has only two vertices of degree 5 or less and all the rest of the vertices are of degree 6 or more. Then, for each vertex (except those two), we may write 6 ≤ (the degree of the vertex). The remaining two vertices must have degree at least 1, because the graph is connected, so for each of those we get 1 ≤ (the degree of the vertex). If we add up all such statements, we get 6(|V (G)| − 2) + 2 on the left-hand side, and we get 2|E(G)| on the right-hand side. In total, we have 6(|V (G)| − 2) + 2 ≤ 2|E(G)| or 3(|V (G)| − 2) + 1 ≤ |E(G)|.

Now, if G has at least three vertices, then by Theorem 11.6.1, |E(G)| ≤ 3|V (G)| − 6. Putting these facts together, we have that 3|V (G)| − 6 + 1 ≤ |E(G)| ≤ 3|V (G)| − 6, which cannot be true. Contradiction!

17. No. Every vertex has degree more than 5, in contradiction to Theorem 11.6.6.

18.

(a) First notice that a regular polyhedron can be stretched and flattened into a regular graph.

So, we are really classifying regular planar graphs with every face of the same size. (b) Next we notice that by Theorem 11.6.6 every graph we seek has at least one vertex of degree at most 5. Because these graphs are regular, all vertices must have degree at most 5. (c) This tells us that we can have 2-regular, 3-regular, 4-regular, and 5-regular graphs only.

A connected 2-regular graph is a cycle, and that would correspond to a polygon rather than a polyhedron, so we’ll discard those. (d) Because all vertices have the same degree, we know that r|V (G)| = 2|E(G)|, for r = 3, 4, or 5.

11.12. Problems about Planar Graphs

115

(e) Because all faces are the same size, by the reasoning of Theorem 11.6.2, we have k|F(G)| = 2|E(G)|. (f) Now we will employ Euler’s formula. We know that |V (G)| − |E(G)| + |F(G)| = 2.

Translating everything into edges, we have 2r |E(G)| − |E(G)| + 2k |E(G)| = 2rk 2k − kr + 2r |E(G)| = 2. The number of edges must be a positive integer, so rk 2k − kr + 2r must be a positive integer. Let’s see what happens for r = 3, 4, or 5: 2 · 3k 6k r = 3: = , so k = 2, 3, 4, 5; 2k − k3 + 2 · 3 6 − k 8k 2 · 4k = , so k = 2, 3; r = 4: 2k − k4 + 2 · 4 8 − 2k 2 · 5k 10k r = 5: = , so k = 2, 3.

2k − k5 + 2 · 5 10 − 3k (g) We cannot have faces of size 2, so this gives us 3-regular graphs with face sizes 3, 4, and 5, a 4-regular graph with faces of size 3, and a 5-regular graph with faces of size 3.

Plugging back into our edge formula, we see that these have, respectively 6, 12, 30, 12, and 30 edges. We can draw the graphs to find, respectively, a tetrahedron, cube, dodecahedron, octahedron, and icosahedron.

19. Solutions will vary.

20. It has 13 edges. However, by Theorem 11.6.1, it could only have 3 · 6 − 6 = 12 edges if it were planar.

edges. Let’s proceed by 21. If G has n vertices, then between G and G there are n2 = n(n−1) 2 contradiction and suppose that both G and G are planar. We know from Theorem 11.6.1 that |E(G)| ≤ 3|V (G)|− 6, so |E(G)| ≤ 3n− 6 and n(n−1) − |E(G)| ≤ 3n− 6. That second equation 2 n(n−1) − 3n + 6 ≤ |E(G)| ≤ becomes 2 − 3n + 6 ≤ |E(G)|. Combining these, we have n(n−1) 2 n(n−1) 3n − 6 or 2 − 3n + 6 ≤ 3n − 6, and this becomes n2 − 13n + 24 ≤ 0. That equation holds when n ≤ 11.

22. Dude, what? Together, G and G have 55 edges. The most edges G can have is 3 · 11 = 6 = 27.

Then G has 55 − 27 = 28 edges . . . oh. Contradiction.

23. Yes. Solutions will vary, but one technique is to try to draw as much of K7 as possible without crossings, and then draw the remaining edges without crossings on any empty seven vertices.

24. Yes. 25. No; 28 − 8 = 20 6≤ 24 − 6 = 18.

26. The crucial insight is that Wn has a central vertex of degree n − 1, which is the highest degree of a vertex in Kn . Therefore in the complement, the central vertex will have degree 0, and only the Cn−1 has edges in the complement—indeed, the complement of Wn is the complement of Cn−1 plus an isolated vertex!

(There’s more; flip the page.)

116

11. Euler’s Formula and Applications

Thus we have reduced the problem to asking which cycles, if any, have planar complements. − n = n(n−3) edges, and using that e ≤ 3v − We note that the complement of a cycle has n(n−1) 2 2 n(n−3) 6 we see that if the complement of a cycle is planar, then 2 ≤ 3n − 6 or n2 − 9n + 12 ≤ 0.

The quadratic formula tells us that this is true when 1.62772 ≤ n ≤ 7.37228, so we only have to check n ∈ {3, 4, 5, 6, 7}.

A 3-cycle’s complement has no edges; a 4-cycle’s complement is just a matching; a 5-cycle’s complement is a 5-cycle. A 6-cycle’s complement is isomorphic to a triangular prism; a 7cycle’s complement contains a singly subdivided K3,3 and so is nonplanar. (That last step is pretty hard for students. They will conclude that the graph is nonplanar but will likely be unable to prove it.)

27. Here are two: K3,4 , and K3,3 with an edge added in one part.

28. If we consider two of the chords to be one planar graph, then the remainder can be made planar by pulling one of the two remaining chords to the outside. Thus the thickness is 2.

29. All three are planar.

30. Here are three examples: K4 with one edge subdivided; K4 with one edge subdivided and a leaf stuck onto the vertex of degree 2; a star graph.

31. One family: Wheels. Another family: Take any planar cubic graph and stick a vertex in the middle of one edge. A third family: Take any graph from the second family and add a leaf to the vertex of degree two. A fourth family: Star graphs.

32. It contains a K3,3 , so it is nonplanar.

33. By Euler’s formula, it must have 2 − v + e = 2 − 8 + 17 = 11 faces.

34. By Euler’s formula, it must have 2 − v + e = 2 − 12 + 32 = 22 faces. But wait! We should have e ≤ 3v − 6, and here we have 32 ≤ 36 − 6 = 30! That’s not possible! This graph cannot exist.

35. Let’s see. If every face is 5-sided or 6-sided, then 5 f5 + 6 f6 = 2e. We also know v − e + f = 2 and f5 + f6 = f and 3v ≤ 2e, or v ≤ 23 e. So 2 = v − e + f ≤ 32 e − e + f = 13 e + f , or 12 ≤ 6 f − 2e = 6( f5 + f6 ) − 5 f5 − 6 f6 = f5 . So the answer is 12.

Chapter 12

Graph Traversals 12.2 Try This! Euler Traversals 1. Solutions will vary. .

2. Solutions will vary. For example,

3. Solutions will vary. For example,

4. Solutions will vary. For example,

.

.

5. If a graph has an Euler circuit, then every vertex must have even degree.

6. If a graph has no odd vertices, then it has an Euler circuit. If a graph has all even vertices, then it has an Euler circuit.

7. If a graph has either two or zero odd vertices, then it has an Euler traversal.

8. Solutions will vary.

12.5 Try This!—Do This!—Try This!

1.

117

118

12. Graph Traversals

(14, a ) b

11

6 7 7 5 (11, e ) g (16, f ) 9 h 6 4 6 5e (25, g ) i (31, h ) 5 (7, j )f 5 6 4j 3 7 8 m 7 6 n k 7 l (4, S ) 7 4 p 4 (14, p ) (21, k ) (29, l ) o 6 (36, m ) (4, S ) (10, o ) 8 eend/gans 5 6 5 5 s (6, S ) 8 a

2.

9

(32, h ) d (36, i )

(22, g ) c

q

6

r (9, o ) (15, q ) (20, r )

3. 24 orderings.

4.

(a) They are the orderings of A, B, C, D. (24 orderings.) (b) Two vertices are adjacent if their labels differ by an adjacent transposition. (c) Each has degree 3, for switches first place/second place, second place/third place, and third place/fourth place. (d) A Hamilton circuit.

5.

12.8. Bonus: Digraphs, Euler Traversals, and RNA Chains

6.

12.8 Bonus: Digraphs, Euler Traversals, and RNA Chains Short activity:

1.

(a) AUG, G, CUCACUAUG, G, CAUACUCUAA. (b) AUG, UUG, UUCCAAAG, UUG, A. (c) AUG, CG, ACCCG, UUACCAAG, UAG.

2. AU, GGC, U, C, AC, U, AU, GGC, AU, AC, U, C, U, AA. AU, GU, U, GU, U, C, C, AAAGU, U, GA. AU, GC, GAC, C, C, GU, U, AC, C, AAGU, AG. 3.

(a) GA. (b) AUUAA. (c) UUAAG (can’t be CCAAG because that would produce AAG as a fragment). (d) UAG.

4.

Only the second has a unique reconstruction.

12.9 Bonus 2: Network Flows Some problems:

1. Follow the edges 5, 2, 4 or 5, 2, 3, 4. These are the only possibilities.

119

120

12. Graph Traversals

2. Yes. Use all edges at maximum capacity.

3.

(a) No, because the flow out of the next node is only 7.

(b) Solutions will vary; here is one example:

(c) Solutions will vary; here is an augmentation of the previous example that achieves the

maximum flow (as seen by the capacity flow into the sink):

4.

(a) 8 + 2 + 5 = 7 + 6 + 2 = 15.

(b)

(c) Here is one flow:

(d) It will return an error, as the maximum flow is 13 Mb.

12.10 Bonus 3: Two Hamiltonian Theorems 1. Just two places: showing that va has degree at least 2 and getting |Va | + |Vb | > n.

2. They cannot be adjacent, so neither Va nor Vb contains a.

3. For every two nonadjacent vertices, the sum of their degrees must be more than n.

4. We know that va , vb are not adjacent, so the sum of their degrees must be more than n. The degree of vb can be at most n − 2 because otherwise it would be adjacent to va . And, n − (n − 2) = 2, so the degree of va is more than two.

12.12. Problems with Traversing

121

5. Yes; there are only two places where the criterion is needed.

(1) Showing that va has degree at least 2: We know that va , vb are not adjacent, so the sum of their degrees must be at least n. The degree of vb can be at most n − 2 because otherwise it would be adjacent to va . And, n − (n − 2) = 2, so the degree of va is at least two.

(2) As before, |Va | = deg(va ) and |Vb | = deg(vb ). Neither Va nor Vb contains a because va is not adjacent to itself and {va , vb } is not in G′ . Therefore, |Va ∪Vb | ≤ n − 1 and so −|Va ∪Vb | ≥ 1 − n. Now PIE becomes |Va ∩ Vb | = |Va | + |Vb | − |Va ∪ Vb | ≥ n + 1 − n = 1, so |Va ∩ Vb | > 0 and Va ∩Vb is nonempty, as desired.

12.12 Problems with Traversing 1. Yes; it has two vertices of degree 3, so it has an Euler traversal but not a circuit.

2. Add an edge between the two vertices of odd degree, so that each now has even degree. Then tack that edge on to the start/end of the traversal.

3.

(a) Every vertex is adjacent to every other, so it is simply a matter of how many of the n vertices one travels through when leaving a (and one can choose the order, too!) before going to b.

(b) Solutions will very. Here is one example: (c) Adapt the previous example by adding more paths between a and b of longer and longer lengths.

4. No; all of the vertices have odd degree.

5. Yes, but only paths. Any other tree has more than two leaves and thus more than two vertices of odd degree.

6. Ze will announce that they are both implicated in the murder because they are liars, liars, liars! Here is a graph of the house:

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12. Graph Traversals

There are many vertices of odd degree, so the door inspector could not have gone to each door exactly once, and because there is a loop in the Best Bedroom (and because the Solarium of Plants either has a loop or is a leaf), there can be no Hamilton circuit for the cleaning service person to traverse.

7. No. Every wheel has all-but-one vertices of degree 3, and the smallest wheel has four vertices of degree 3.

8. Yes. It has exactly two odd vertices.

Nope. It has exactly two odd vertices.

Yes. Start in the upper-left corner, go around clockwise, and then into and out of the center.

9. Solutions may vary. Here is a planar drawing:

10.

(a) Yes. Every vertex of the network has even degree, so there is an Euler circuit. (b) Yes. OSdelta OSalpha

OSbeta

base

OSzeta

OSgamma

OSepsilon

11. The left-hand graph has a Hamilton circuit around the outer cycle. It has an Euler traversal but not an Euler circuit, as it has two vertices of odd degree (top and bottom).

The right-hand graph has a Hamilton circuit (start at lower right, go around counterclockwise, then up and down instead of completing the lower W). It is 3-regular, so it does not have an Euler traversal.

12.12. Problems with Traversing

123

12. The left-hand graph has four vertices of degree 3, so it has no Euler traversal. It has more than one Hamilton circuit.

The right-hand graph has four vertices of degree 3, so it has no Euler traversal. It has a Hamilton circuit: start at the upper right, go around counterclockwise, and then go in to the center and out again.

13. Graph F has an Euler traversal, formed by starting at g, circuiting G, going across the edge to h, and circuiting H.

14. Run Dijkstra and get 80 ft:

15. They all have Hamilton circuits;

.

16. The left-hand graph has four vertices of degree 3, so it has neither; the middle graph has all vertices of even degree, so it has an Euler circuit; the right-hand graph has two odd vertices, so it has only an Euler traversal.

17.

18.

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12. Graph Traversals

19. The left-hand graph has a Hamilton circuit, but the right-hand one does not; there’s a triangle to the left whose bottom vertex must be visited twice in a circuit. It does not even have a Hamilton traversal, though this must basically be shown exhaustively (look at the many vertices of degree 2).

20. The left-hand graph has two odd vertices, so it has only an Euler traversal; the right-hand graph has vertices all of even degree, so it has an Euler circuit.

21.

22. It should have all edges traversed by two trails. Ah. Add a vertex g and connect it to all four vertices of odd degree. Then, the new graph G′ has all even vertices, so it has an Euler circuit.

Now, when we remove the vertex and its four edges, we know that we are cutting two length-2 pieces out of the circuit (distinct because their endpoints are distinct). This leaves two trails that cover all the edges.

Or, just add an edge between two of the vertices of odd degree, find an Euler traversal, and break it by removing the edge.

23.

(a)

(b) Solutions will vary, but here is one solution:

12.12. Problems with Traversing

125

(c) Solutions will vary, but the idea is to find a minimum-weight tree that spans the hydrants.

Use Dijkstra (or equivalent) to find a bunch of shortest paths and meld them into a tree:

24. Yes.

25. Yes, because any path of length 4 on the dodecahedron can be mapped onto a segment of the previous solution and the circuit extended from there. Proving this can be done by brute force (I think).

26. This is a proof by induction. The base case is a simple closed loop, where one colors the inside a different color than the outside. We will induct on the number of regions. Because we have all even intersections, this loop is equivalent to a graph with all vertices of even degree. Now split one vertex into two vertices, one of which has two adjacent edges of the original vertex, and the other of which has the rest of the edges. (Every vertex has degree at least 4, so this is possible.) In this way, we have merged two incident, but not adjacent, regions. Note that the resulting graph still has all vertices of even degree, and thus has an Euler circuit. Therefore it is equivalent to a continuous closed loop with one fewer intersection. By the inductive hypothesis, we can 2-color the regions. Merging the vertices again retains the 2-coloring because the regions separated by this action are also separated by a single other region in the other color.

27. Yup, the same proof works if we take the union of the curves as our initial graph.

6

5 13 4

7

28. An Euler circuit. Yup:

1

2

14 12 8 11

3

10 9

29. There are no Euler circuits as all vertices have odd degree.

There is a Hamilton circuit found by following one side of the ladder, which twists into the other side and returns to the start point.

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12. Graph Traversals

30. Neither has a Hamilton circuit, but this is shown by brute force. Similarly, the right-hand graph does not have a Hamilton traversal. The left-hand graph does, though:

4 3

3

3

2

6

1

2

31. Dijkstra on the underworld:

5

4 2

2 2

9

5 10 3 3

2 1

2

2

3 5

7

2

32. Yes, all do. On the left-hand graph, use the outer cycle; on the middle graph, use three sides of the outer cycle, go in, and use three sides of the inner cycle; on the right-hand graph, zig-zag up-and-down going left to right and get back to the start using one of the long diagonal edges.

33. Yup, both have Hamilton circuits:

Neither has an Euler circuit or traversal because each has more than two vertices of degree three.

34. Yup, both have Hamilton circuits:

Neither has an Euler circuit or traversal because each has more than two vertices of degree three.

2

1

5

35. 9. See Dijkstra markings:

2 1 2 3

2

2

4

5 1

2 2 7 2 Flayed Finger

6

Hotel 2

4 1 5 2

3

7

2 2

2

4 9

3

2

3 3

4

Chapter 13

Graph Coloring 13.2 Try This! Coloring Vertices and Edges 13.2.1 Vertex Coloring

1. Three, three, and three:

2. Proof by contradiction: Each graph contains an odd cycle, and if we were to try to use only two colors on that cycle, we would have two adjacent vertices of one color.

3. Useless: We need at least one color.

Slightly better: Largest n such that there’s a copy of Kn in the graph.

13.2.2 Edge Coloring

4. Four, four, and four:

5. You bet! One. Less silly is the highest degree of any vertex.

13.2.3 More on Vertex Coloring

6.

These are 4-vertex colorings.

127

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13. Graph Coloring

7. These are nontrivial arguments!

For the left-hand graph, notice that each triangle needs three different colors and suppose that three colors suffice. Call these colors b, l, g (for black, light grey, and grey). Color the top vertex black and alternate l, g, l, g across the bottom of the three top triangles (all without loss of generality) so that no color conflicts arise. Then, both bottom vertices must be black, but they are adjacent and this creates a contradiction.

For the right-hand graph, again use colors black, light grey, and dark grey, and assume that three colors suffice. Suppose first that the top and bottom vertices are different colors (black and dark grey). Then the right-hand vertex and the center vertex must both be light grey.

Any choice of assigning dark grey and black to the four adjacent vertices either creates an immediate conflict or forces the two remaining vertices on this side of the graph to be the same color, which is a contradiction. If we instead suppose that the top and bottom vertices are the same color, a similar contradiction arises on the left-hand side of the graph.

13.2.4 More on Edge Coloring 8. The outer cycle of the Petersen graph requires at least three colors because it is an odd cycle, so let us assume that only three colors are needed. Assigning three colors to the outer cycle determines the color for each “spoke.” Spokes of different colors joined by an edge determine the color for that edge. This determines the colors for four of the five remaining edges, and two of those edges are incident but must have the same color; this is a contradiction. Each of the second and third graphs has a vertex of degree 4, so at least four colors are needed.

9. The number of edges in the graph!

Okay. Students are likely to either be stumped by the better upper bound, or come up with 2∆(G) − 1 or come up with the not-provable-in-this-class ∆(G) + 1. Encourage them to try to prove the former and treat the latter as dubious (for now).

13.4 Try This! Let’s Think about Coloring 1. χ ′ (K3 ) = 3, χ ′ (K4 ) = 3, χ ′ (K5 ) = 5, χ ′ (K6 ) = 5, and χ ′ (K7 ) = 6. (a) For n odd, χ ′ (Kn ) = n and for n even, χ ′ (Kn ) = n − 1.

(b) See Figure 13.15.

(c) See Theorem 13.5.1.

2.

(a) χ (K2,2 ) = 2, χ (K2,3 ) = 2, χ (K3,3 ) = 2, χ (K3,4 ) = 2, and χ (K3,5 ) = 2. (b) χ (Km,n ) = 2. Color the m-part teal and the n-part purple. (c) χ (B) = 2. There are two parts; color one part teal and the other part purple. (d) χ ′ (K2,2 ) = 2, χ ′ (K2,3 ) = 3, χ ′ (K3,3 ) = 3, χ ′ (K3,4 ) = 4, and χ ′ (K3,5 ) = 5. (e) It’s m or n, whichever is larger.

13.8. Bonus: The Four-Color Theorem

129

(f) One way to do this is to find an algorithm for edge coloring Km,n . For example, suppose that m ≥ n. Then assign m different colors, 1, 2, . . . , m to the edges incident to the first vertex in the n part. For the second vertex, assign the colors 2, 3, . . . , m, 1 so that no two edges of the same color hit the same vertex in the other part. Continue in this fashion, and a proper m-edge coloring results. Then, we know that χ ′ (Km,n ) ≥ ∆(Km.n ) = m, so we have achieved our lower bound.

3.

(a) Grab a vertex. Give it a color. Grab another vertex (probably a neighbor), and give it the lowest-numbered color not already in use by one of its neighbors. Lather, rinse, repeat. (b) No! For the left-hand diagram, it does; for the right-hand diagram, it does not. (c) Grab an edge. Give it a color. Grab another edge (probably incident), and give it the lowest-numbered color not already in use by an edge it touches. Lather, rinse, repeat. (d) No! For the right-hand diagram, it does; for the left-hand diagram, it does not.

4.

(a) There are two (the two different every-other-edge perfect matchings). (b) There are three disjoint perfect matchings. (c) Solutions will vary, but chances are that it will be 4-edge colorable. (d) If a k-regular graph can be edge-colored with k colors, then the color classes will form perfect matchings of the graph.

13.8 Bonus: The Four-Color Theorem Brain-enhancing activity:

1. The one labeled with a question mark.

2. Teal–1, purple–2, chartreuse–3, and black–4.

3. There is a teal–black–teal–black chain for the color-1–color-4 Kempe chain, and a purple– black–purple–black chain for the color-2–color-4 Kempe chain.

4. The color-3–color-2 chain is here chartreuse–purple. The color-3–color-1 chain is here chartreuse–teal. (Notice that there are longer chains of these colors on opposite sides—but they aren’t the chains that are cut off!)

5. Switch one of the chains you identified in Problem 3. What happens? The switches are among neighbors of v, so we have just permuted the colors. We still use all four colors around v.

6. Two adjacent vertices are now colored chartreuse!

7. There are cases in which the terminating/captured Kempe chains include two vertices adjacent to v as well as to each other, and thus the number of colors among v’s neighbors need not be reduced.

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13. Graph Coloring

13.10 Colorful Problems 1. Left: impossible because there’s a loop; center: 2; right: 3.

2. Left: impossible because there’s a loop; center: 3; right: 4.

3. Yes. Because adjacency is preserved, the set of colorings of the two graphs will be the same and so will have the same smallest number of colors.

4. χ (K2,2,3 ) = 3. χ (Kr,s,t ) = 3, and χ (Kr,s,t,u,v,w,x,y,z ) = 9.

5. The graph contains a K6 and has highest degree 13, so 6 ≤ χ (G) ≤ 14.

6. The graph has highest degree 13, so 13 ≤ χ (G) ≤ 25.

7. Three.

The same number of colors.

Depending on how you colored the graph to begin with, this may be possible. The idea is that with a protected left, lanes 1, 2, 5, and 6 can all go at once, whereas without the protected left, lanes 1 and 2 must always be separate from lanes 5 and 6.

8. Chromatic number is 3.

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131

9. Chromatic index is 4.

10. Solutions will vary.

11. For example, suppose that m ≥ n. Then assign m different colors, 1, 2, . . . , m, to the edges incident to the first vertex in the n part. For the second vertex, assign the colors 2, 3, . . . , m, 1 so that no two edges of the same color hit the same vertex in the other part. Continue in this fashion, and a proper m-edge coloring results. Then, we know that χ ′ (Km,n ) ≥ ∆(Km.n ) = m, so that’s that.

12. True; this follows by combining Theorems 13.5.2 and 13.5.3.

13. False! The Petersen graph is 3-vertex colorable but has some cycles of length 5.

14. No. An isomorphism preserves incidence because it preserves adjacency, so every proper edge coloring of G is also a proper edge coloring of H and vice versa. Therefore, G and H must have the same chromatic index.

15.

16. Chromatic numbers: 3, 3; chromatic indices: 4, 4.

17. The graph contains a K4 and has highest degree 24, so 4 ≤ χ (G) ≤ 25—but it’s also planar, so 4 ≤ χ (G) ≤ 6! And for those who read the bonus section, 4 ≤ χ (G) ≤ 5. Pretty tight bounds.

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13. Graph Coloring

18. If |V (G)| is even, then two colors; if it is odd, then three colors.

19. χ (G) = 3.

20. Solutions will vary.

21. G is Tutte’s uniquely edge-colorable cubic graph, the only nonplanar and triangle-free example of such that is known.

22. χ (D) = 3 and χ ′ (D) = 5.

23.

There is usually a choice of colors for a vertex of low degree. A vertex that can have more than one color can get a green light for more than one pattern of the cycle.

13.10. Colorful Problems

133

24. By the handshaking lemma, G must have an even number of vertices. Therefore, the Hamilton circuit has an even number of vertices. And therefore, the Hamilton circuit is 2-edge colorable. This leaves a perfect matching in G uncolored, so it can get a third color, rendering χ ′ (G) = 3.

25. First note our known bounds: clique size ≤ χ (G) ≤ ∆(G)+ 1 and ∆(G) ≤ χ ′ (G) ≤ 2∆(G)− 1.

The second can be rewritten as ∆(G) + 1 ≤ χ ′ (G) + 1 ≤ 2∆(G). This tells us that χ (G) ≤ χ ′ (G) + 1. Students will suspect, and they will be correct, that χ (G) ≤ χ ′ (G) (when G has more than one edge), but better bounds are needed to prove this.

26. Solutions will vary. Here’s one possibility: One root, two leaves from the root, one length-2 path from the root. Order the vertices as leaf 1, leaf 2, leaf 3, root, remaining vertex.

27. The start of a construction (or a construction for k = 5) is shown in Figure 13.1. The general 1 3 2 4 9

11

5

10 14

?

16

15 12

13

8

6 7

Figure 13.1. A weird start of a tree.

idea is to take ordered paths of lengths 1, 2, and 4 to produce vertices colored black, white, and grey, and add another vertex adjacent to those three vertices that must be colored BLACK. Use this 8-vertex gadget along with ordered paths of lengths 1, 2, and 4 to produce vertices that must be colored black, white, grey, and BLACK; add another vertex adjacent to those four vertices (it must be colored huh?). This process of using paths and increasingly large gadgets can be repeated as many times as necessary to force a vertex in a kth color.

28. For n even, there’s an odd cycle so 3 colors are needed (and suffice) for both vertex and edge coloring.

For n odd, 2 colors suffice for vertices and 3 for edges.

In both cases, for edges, just make all the rungs one color and alternate colors along the edges.

29. Five (has a vertex of degree 5):

.

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13. Graph Coloring

.

30. Four (has a vertex of degree 4):

Long Divsion Primes

Ordinality Projective

Knots

Game Theory

Combin

Strange

.

31. Yes; Geometry Random

Tropical

Fractals

Complex

Comb Opt Complexity

GenFunc

32. Four (there’s a K4 ) and five (there’s a vertex of degree 5) respectively:

13.10. Colorful Problems

135

.

33. Four (there’s a vertex of degree 4):

34. We give two vertices, connected by an edge, to each spy, because no one can be on shift with hirself. Then we create K4 s between agents who cannot work together. Here is a solution with 6 timeslots: 004 5

2

1

3

Natasha

Agent 99 4

2

1

5 Maxwell

Boris 6

6 4 Spy X 3

35. Three (there’s an odd cycle) and three (there’s a vertex of degree 3) respectively.

Theme III Supplement: Problems on the Theme of Graph Theory 1. Chromatic numbers: 3, 3; chromatic indices: 3, 4.

2. If n is odd, then the cycle around the central vertex is even and χ (Wn ) = 3 (alternate colors around the cycle, then add a third color for the central vertex); if n is even, then the outer cycle requires three colors and the central vertex is adjacent to all three colors, so χ (Wn ) = 4.

3. There are n − 1 vertices in the outer cycle, so the degree of the central vertex is n − 1 and χ ′ (Wn ) ≥ n − 1. Now, we can produce an (n − 1)-edge coloring because if two spokes have colors a, b, then the edge joining them can get color c, the edge to the “left” can get color b, and the edge to the “right” can get color a. We may then reuse a, b, around the remainder of the outer cycle.

4.

The smallest number of traffic patterns needed is three, as the graph is three-vertex colorable.

A vertex that can have more than one color can get a green light for more than one pattern of the cycle.

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Theme III Supplement: Problems on the Theme of Graph Theory

137

5. Either G ∪ e has all vertices of degree 2, in which case it is C5 and G is a path of length 5, or G ∪ e has all vertices of degree 4, in which case it is K5 and G is K5 /(an edge). One can show (quickly) by exhaustion that no graph with five vertices can have some vertices of degree 2 and others of degree 4.

6. Solutions will vary. The graph with one possible spanning tree:

7. Solutions will vary. The graph with one possible spanning tree:

8. Solutions will vary. Here is one possibility for spanning trees (graph at left, Kruskal at center, Prim at right):

9.

(a) False. Here is a counterexample with five vertices and four edges: (b) False. Here is a counterexample with three edges and four vertices:

(c) False. Here is a counterexample with 8 edges and 12 vertices: (d) False. Here is a counterexample with degree sequence (1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3):

10. Many are possible; here are three of the possibilities:

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Theme III Supplement: Problems on the Theme of Graph Theory

11. There’s the all-spokes matching, and then there are five other matchings that each have one spoke and the rest of the edges are determined.

12. If G has no Eulerian circuit, then k must be odd. From the handshaking lemma, it follows that G has an even number of vertices n. Now, the superposition of G and G is Kn , which has degree n − 1 for each vertex, so G is (n − 1 − k)-regular. We know n is even, so n − 1 is odd, and the difference between two odd numbers is even, so n − 1 − k is even and therefore G has an Eulerian circuit.

13. No. It has 16 vertices and 47 edges; to be planar, it can have at most 3 · 16 − 6 = 42 edges.

14. Yes; here is a planar drawing.

15. We know that such a graph must have all even vertices. Two examples are given below: one is not connected, so cannot have a Hamilton circuit, and the other forces a cycle before all vertices are visited.

16. Solutions will vary. Supposing that a standard deck is used (rather than Uno or Rook, etc.), one may (for example) have the top rank send red suits to the left and black suits to the right; then, the second rank can send hearts to the left of diamonds and clubs to the left of spades; then, the third rank can send values A–6 left and 7–K right; then, the fourth rank can send A–3 and 7–9 left and 4–6 and 10–K right; the fifth rank can send A–2, 7–8, 4–5, and 10–J left and 3, 9, 6, and Q–K right; the sixth and final rank can send A, 7, 4, 10, and Q left and 2, 8, 5, J, and K right.

17. This graph has girth 5. It has 12 vertices and 18 edges. If it were planar, it would have 8 faces

by Euler’s formula. But, because it has no triangles, 4 · 8 ≤ 2 · 18, which is false.

18.

A ⌋ $20 bills.

1. Let the amount of change needed be $A. Hand over ⌊ 20

A B 2. Compute B = A − ⌊ 20 ⌋. If B ≥ 10, then hand over ⌊ 10 ⌋ $10 bills; otherwise, continue.

B ⌋. If C ≥ 5, then hand over ⌊ C5 ⌋ $5 bills; otherwise, 3. Compute C = B − ⌊ 10 continue.

4. Compute D = C − ⌊ C5 ⌋. If D ≥ 1, then hand over ⌊D⌋ $1 bills; otherwise, continue.

5. Compute E = D − ⌊D⌋. If 100E ≥ 50, then hand over a 50-cent coin; otherwise, continue.

Theme III Supplement: Problems on the Theme of Graph Theory

139

100F 6. Compute F = 100E − 50⌊ 100E 50 ⌋. If 100F ≥ 25, then hand over ⌊ 25 ⌋ quarters; otherwise, continue.

100G 7. Compute G = 100F − 25⌊ 100F 25 ⌋. If 100G ≥ 10, then hand over ⌊ 10 ⌋ dimes; otherwise, continue.

100H 8. Compute H = 100G − 10⌊ 100G 10 ⌋. If 100H ≥ 5, then hand over ⌊ 5 ⌋ nickels; otherwise, continue.

9. Compute J = 100H − 5⌊ 100H 5 ⌋. If 100J ≥ 1, then hand over 100J pennies.

19. Any cycle with a chord added has a Hamilton circuit (just go around the cycle) but has two odd vertices and therefore no Euler circuit.

20. Yes, of course I can. (I am the solution!) We know that the “internal” canal intersections have degree 4, and so should be next to at most four other letters each. There are four such intersections. The four “corner” canal intersections have degree 2 and so will be next to exactly two other letters each. Then, the “edge” canal intersections have degree 3 and so will be next to at most three other letters each.

So, let us list all 16 letters along with the letters they appear next to in White Duck’s swim: a b c d e f g h

o, m g, m, f, o f, l, j, g l, j, k i, p p, c, e, b k, b, c, n n, k, j

i j k l m n o p

m, e, p d, c, h, k h, g, j, d c, d, p b, i, a o, h, g a, n, b e, f, l, i

We have only two intersections of degree 2 (a, e), and six intersections of degree 4 (b, c, f, g, j, k). We need to convert two of those six degree-4 intersections into degree-2 intersections.

If we switch the second instance of j with k, we get h–k–j–d and so j k h d

d, c, k h, g, j n, k l, j

Now we have the correct number of degree-4 intersections, and the correct number of degree2 intersections.

21. Every vertex has degree greater than 5. This contradicts Theorem 11.6.6.

22. It has 14 edges. However, by Theorem 11.6.1, it could only have 3 · 6 − 6 = 12 edges if it were planar.

23. The Petersen graph!

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Theme III Supplement: Problems on the Theme of Graph Theory

24. n = 1 works for any k. So does n = k + 1. Each rank of the tree has a multiple of k vertices in it (except for the root rank, which has one vertex). Thus, in total, the tree must have n = mk + 1 vertices, so we know that n ≡ 1 (mod k). Any such n is achievable: Consider a “fan” with a root vertex and k descendants. Start with a root vertex (there’s one of it), and glue on m fans by placing each fan’s root on an existing vertex of the tree-in-progress; this adds mk vertices as the fan-roots do not contribute vertices.

25. A star with four spokes has chromatic number 2 and largest degree 4.

Yes—a single edge has chromatic number 2 and largest degree 1. (This can be deduced in the following way: We know χ (G) ≤ ∆(G) + 1, so if 2∆(G) = χ (G) = ∆(G) + 1, then 2∆(G) = ∆(G) + 1 and ∆(G) = 1.)

26. First suppose that e is a bridge and there exists a spanning tree of G that does not contain e.

This means that e may be removed without disconnecting G (as the spanning tree contains all vertices of G and is connected), which is a contradiction to e being a bridge.

Now suppose that e is contained in every spanning tree of G but assume that e is not a bridge.

Therefore, its removal does not disconnect G and so we remove it; the remaining graph G \ e is connected and thus has a spanning tree. This tree also spans G, which contradicts our supposition that e is contained in every spanning tree of G. 27. If so, then it obeys Euler’s formula and 5 − 10 + 7 = 2. So far so good. But wait: if it is connected and simple, then it must obey Theorem 11.6.1, so |E(G)| ≤ 3|V (G)| − 6. Here that would be 10 ≤ 3 · 5 − 6 = 9, which is a contradiction. So, if such a graph exists, it must have multiple components or multiple edges. Here is one such graph: 28. G has 12 vertices with total degree 3(4 + 5 + 6 + 7) = 66 and so has 33 edges. By Theorem 11.6.1, if G is planar then |E(G)| ≤ 3|V (G)| − 6. Here that would be 33 ≤ 3 · 12 − 6 = 30, which is a contradiction.

29. Rania can use a binary tree approach. Hir first question will be “Is your number less than 8?”

This cuts the number of possibilities down to eight. Depending on Reza’s answer, Rania will either ask “Is your number less than 4?” or “Is your number greater than 11?” as hir second question. This cuts the number of possibilities down to four. With hir third question, Rania can cut the number of possibilities down to two, and a fourth question seals the deal and determines Reza’s number. So, Rania needs to ask only four questions.

30.

(a) Solutions will vary. One idea is to ask each question twice, so that the lie is detected immediately. (b) Yes! This references Section 2.10. (See this section for a full solution.) The possibilities include “If I asked you whether your number is less than 8, what would you say?” and “Are you telling the truth if and only if your number is less than 8?”

′ 31. Suppose that ∆(G) = 2χ ′ (G). We know that χ ′ (G) ≥ ∆(G), so we have ∆(G) 2 = χ (G) ≥ ∆(G), which reduces to ∆(G) ≥ 2∆(G). That’s a contradiction. So, nope, no such graph exists.

Theme III Supplement: Problems on the Theme of Graph Theory

141

32. A counterexample is that K100 contains a K3 subgraph: K100 is nonplanar, but K3 is planar.

33.

(a) Suppose that d1 = 2. Then, all di ≥ 2 and the sum of the di ≥ 2n, which is a contradiction.

Thus, d1 = 1. Similarly, if d2 = 2, then the sum of the di ≥ 2n − 1, which is also a contradiction. Thus, d2 = 1. (b) Suppose that dn = 1. Then, all di ≤ 1 and the sum of the di ≤ n, which is a contradiction.

However, it could be that d1 = d2 = 1 and the rest of the di = 2, so 2 is the smallest value that dn can have. (c) No. The total degree of G is 2n − 2, and so the number of edges is n − 1. However, there could be a disconnected graph with a single edge as one component and a cycle as the other component, and it has the degree sequence (1, 1, 2, . . . , 2). (d) Yes. Here’s how to construct such a tree. Start with a path; it has (1, 1, 2, . . . , 2). The two vertices of degree 1 are at the left and right ends of the path. If this is not the degree sequence we started with, it is related in the following way: we can remove 1 from some number of the 2s on the left and add those 1s to some number of the 2s on the right. (This keeps the sum of the di constant at 2n − 2.) Let us say that dn = k. We remove a vertex (and an edge) from the left end of the path and glue it on to the second-from-the-right vertex; repeat this process a total of k − 1 times. Now examine dn−1 and again move as many leaf-twigs as needed. Because n is finite, this process will terminate in the desired tree.

34. We will induct on the number of vertices. Use as a base case a graph with two vertices and one edge. It is its own spanning tree. Assume that every graph with k ≤ n vertices has a spanning tree. Consider a graph G with n + 1 vertices. Pick one (call it v) and remove it. Now, G \ v has a spanning tree by the inductive hypothesis. Now restore v and notice that G has a tree that spans all vertices but v. Certainly v is connected by some edge to another vertex, so add this edge and v to the almost-spanning tree to obtain a spanning tree. This is spanning because it covers all vertices by construction, and it is a tree because v only has degree 1 in the spanning subgraph and thus adding v and an edge did not complete a cycle.

35. Solutions will vary; the graph needs to have all vertices of even degree except two of odd degree. A square with a hair will do it.

36. For G, v = 7, e = (3 + 12 + 5 + 12)/2 = 16, and it follows from Euler’s formula that e = 3v − 6 = 21 − 6 = 15. Contradiction.

Equally, 7 − 16 + 11 = 2 so f = 11, and 33 is not less than 32. Contradiction.

37. Solutions will vary; one consists of the three Y-shapes that meet at the center.

38.

(a) χ (Sn ) = 2. Color the central vertex red and the rest of the vertices blue. (b) χ ′ (Sn ) = n. There are n edges, all incident to the central vertex, so each must be a different color.

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Theme III Supplement: Problems on the Theme of Graph Theory

39.

40. χ (tree) = 2 as every tree is bipartite. χ ′ (tree) ≥ ∆(tree) so unless the tree is just a path, ∆(tree) > 2. So, generally, χ ′ (tree) is larger than χ (tree).

41. Solutions will vary, but one such tree (of weight 36) is shown below. Kruskal’s algorithm was used.

42. Yes. This is equivalent to asking whether there is a Hamilton path in the graph of the cube.

The cube is shown with part of its graph, and then the full graph and Hamilton path are shown.

43. It is bipartite, so χ (G) = 2 and χ ′ (G) = 3, as that’s the highest degree of a vertex.

44.

(a) The degree sequence is (n, n, n, . . . n) where there are 2n entries. Each vertex has n coordinates and therefore n positions that could differ, and there are 2n binary strings of length n. (b) 1, 4, 12, 32. (c) en = 2en−1 + 2n−1 . There are en−1 edges in the (n − 1)-cube, and if we append either a 0 or a 1 to the end of the coordinates for those edges, we get an edge in the n-cube.

Then, there are edges that connect a vertex v of the (n − 1)-cube with a 0 appended to the same vertex with a 1 appended; there are 2n−1 such vertices. (d) Every vertex has degree n, and there are 2n vertices, so the total degree would be n2n but this counts every edge twice, so there are n2n−1 edges in the n-cube.

Theme III Supplement: Problems on the Theme of Graph Theory

143

(e) For n ≤ 3 the n-cube has a planar graph: Let us now prove that for n ≥ 4, the n-cube is nonplanar. First note that the n-cube contains the (n− 1)-cube, so it will suffice to show that the 4-cube is nonplanar. Suppose that the 4-cube is planar. Its shortest cycle length is 4, so we know that 4|F| ≤ 2|E|, or |F| ≤ 21 |E|. Check Euler’s formula: 2 = |V | − |E| + |F| ≤ 24 − 4(23 ) + 12 4(23 ) = 16 − 32 + 16 = 0. But 2 6≤ 0, so this is a contradiction and the 4-cube is nonplanar.

45.

46. Below is a graph of the reactions alongside its coloring; two bays are needed.

47. 1 + 1 < 6. 1 + 1 + 2 < 6. 1 + 1 + 2 + 2 = 6. Excellent. Keep that, go back.

1 + 1 + 2 + 3 > 6. Go back.

1 + 2 < 6. 1 + 2 + 2 < 6. 1 + 2 + 2 + 3 > 6. Go back.

1 + 2 + 3 = 6. Keep that, go back.

2 + 2 < 6. 2 + 2 + 3 > 6. Done.

48. We know that e = 23 v or v = 32 e from 3-regularity, and v − e + f = 2 from Euler’s Formula.

Also, 3 f3 + 4 f4 + 5 f5 + · · · = 2e and f = f3 + f4 + f5 + · · · = f .

Now, putting the first couple of equations together we get 23 e − e + f = 2 or 6 f − 2e = 12.

Substituting in all the subscripty stuff from the second pair of equations, we have 6( f3 + f4 + f5 +. . . )−(3 f3 +4 f4 +5 f5 +. . . ) = 12 which simplifies to 3 f3 +2 f4 + f5 − f7 −2 f8 −· · · = 12 as desired.

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Theme III Supplement: Problems on the Theme of Graph Theory

6

5 13 4

7

49.

1

2

14 12 8 3

11 10

9

50. The printers form a K6 missing some edges (where an edge means two printers cannot be run at the same time). We will color its vertices where colors represent timeslots for simultaneous running. The graph contains a triangle, so at least 3 hours are needed; they also suffice, as shown below.

c

p u

a

l

r

Part IV Other Material

Chapter 14

Probability and Expectation 14.4 You Are Probably Expected to Try This!

1.

(a) {(roll 1, roll 1), (roll 2, roll 1), (roll 3, roll 1), (roll 4, roll 1), (roll 5, roll 1), (roll 6, roll 1), (roll 7, roll 1), (roll 8, roll 1), (roll 9, roll 1), (roll 10, roll 1), (roll 11, roll 1), (roll 12, roll 1), (roll 1, roll 2), (roll 2, roll 2), (roll 3, roll 2), (roll 4, roll 2), (roll 5, roll 2), (roll 6, roll 2), (roll 7, roll 2), (roll 8, roll 2), (roll 9, roll 2), (roll 10, roll 2), (roll 11, roll 2), (roll 12, roll 2), (roll 1, roll 3), (roll 2, roll 3), (roll 3, roll 3), (roll 4, roll 3), (roll 5, roll 3), (roll 6, roll 3), (roll 7, roll 3), (roll 8, roll 3), (roll 9, roll 3), (roll 10, roll 3), (roll 11, roll 3), (roll 12, roll 3), (roll 1, roll 4), (roll 2, roll 4), (roll 3, roll 4), (roll 4, roll 4), (roll 5, roll 4), (roll 6, roll 4), (roll 7, roll 4), (roll 8, roll 4), (roll 9, roll 4), (roll 10, roll 4), (roll 11, roll 4), (roll 12, roll 4)}.

1 . (b) P((11, 3)) = 48 1 1 1 + 48 = 24 . (c) These are exclusive states, so the probability is 48

(d) One can roll a total of 14 by rolling (12, 2), (11, 3), or (10, 4). These are exclusive, so 3 the total probability is 48 .

45 3 = 48 . (e) By the probability axioms, P(roll a total 6= 14) = 1−P(roll a total of 14) = 1− 48

(f) Let F : S → {0, 1} be defined as F(roll (a, b)) = 1 if a + b = 14 and F(roll (a, b)) = 0 if a + b 6= 14.

3 . (g) E[F] = 0 · P(roll a total 6= 14) + 1 · P(roll a total of 14) = 48

2.

(a) {(1, 1), (2, 1), (3, 1), (4, 1), (5, 1), (6, 1), (1, 2), (2, 2), (3, 2), (4, 2), (5, 2), (6, 2), (1, 3), (2, 3), (3, 3), (4, 3), (5, 3), (6, 3), (1, 4), (2, 4), (3, 4), (4, 4), (5, 4), (6, 4), (1, 5), (2, 5), (3, 5), (4, 5), (5, 5), (6, 5), (1, 6), (2, 6), (3, 6), (4, 6), (5, 6), (6, 6)}. (b) Let E(roll (a, b)) = 1 if a + b is even and E(roll (a, b)) = 0 if a + b is odd.

1 · (the number of possible even rolls). In (c) E[E] = 0P(E = 0) + 1P(E = 1) Or, E[E] = 36 either case, 18 of the 36 rolls have even sums. So, E[E] = 21 . Interpretation: half the time we expect to roll an even sum.

(d) Let O(roll (a, b)) = 2 if a and b are odd, O(roll (a, b)) = 1 if exactly one of a or b is odd, and O(roll (a, b)) = 0 if both a and b are even.

1 · (the number of rolls with (e) E[O] = 0P(O = 0) + 1P(O = 1) + 2P(O = 2). Or, E[O] = 36 1 exactly one odd number) + 2 · 36 · (the number of rolls with two odd numbers). In either 2·9 case, E[O] = 18 36 + 36 = 1 odd number rolled.

146

14.6. Try This! . . . , Probably, Under Certain Conditions

147

(f) Nothing! While you can’t tell the difference between the two dice after rolling them, they are still distinct. The colors of the dice were presented in this problem to help students avoid conflating rolls of ( j, k) with rolls of (k, j).

3.

(a) {left/red, right/red, left/orange, right/orange, left/yellow, right/yellow, left/lime green, right/lime green, left/forest green, right/forest green, left/robin′ s egg blue, right/robin′ s egg blue, left/indigo, right/indigo, left/black, right/black}. (b) It should be 12 because half of the outcomes include left mittens. (c) It should be 21 because half of the outcomes include cool colors. (d) There are 12 outcomes with a left mitten or a cool color, so the probability is 34 . (e) There are 4 outcomes with a left mitten and a cool color, so the probability is 41 . (f) Solutions will vary, but here are examples.

Let R be the number of left mittens grabbed, with E[R] = 0P(R = 0) + 1P(R = 1) = 12 .

Let R be the number of cool colors grabbed, with E[R] = 0P(R = 0) + 1P(R = 1) = 12 .

Let R be the number of left mittens and/or cool colors grabbed, with E[R] = 0P(R = 0) 4 8 + 2 · 16 = 1 left or cool-color mitten. + 1P(R = 1) + 2P(R = 2) = 1 · 16

14.6 Try This! . . . , Probably, Under Certain Conditions 1.

(a) We assign random variables Wo, R, Wa, and F with value 1 if a Minbari is worker caste, is religious caste, is warrior caste, and enjoys eating flarn, respectively, and 0 otherwise.

Then, our question asks for P(R = 1 and F = 1). (b) Ah . . . this is tricksy. One might think this is the same question as the previous question, because we are talking about Minbari who are both religious caste and like to eat flarn, but actually we are looking at a religious-caste Minbari and wondering whether ze likes to eat flarn. So, this is a conditional probability question! We want P(F = 1|R = 1). (c) This is the same as the previous question: P(F = 1|R = 1). (Some students may only realize that the previous question is not an and after reading this one.) (d) This is the probability P(R = 1|F = 1). (e) This is the same as the previous question, so P(R = 1|F = 1).

2.

(a) 34 = 81 cards. (b) (i) One-third of the cards have three symbols. (ii) One-third of the cards are shaded. (iii) There are nine cards that are shaded and have three symbols, so 91 . (iv) Using PIE, we have 13 + 31 − 19 = 95 .

1 (c) Yes, because 31 · 13 = 91 . Also, 1/9 1/3 = 3 , which is the conditional probability of each event given the other, as well as the probability of each event.

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(d) We need to compute the probability that the first card is solid and the second card is solid. The probability that the first card drawn is solid is 13 . Now, the second card is drawn from a deck of only 80 cards—so it does not have the same probability of being solid! There were originally 27 solid-symbol cards. So, the probability that the second card is solid, if the first one is solid, is 26 80 . Now, let the event of the first card being solid be denoted as F and the event that the second card is solid be denoted by S. We know 1 13 P(F and S) = P(S|F)P(F), and so we have 26 80 · 3 = 120 . (e) This is where things get tricky. It could be that the first card is green and the second isn’t, or it could be that the first card is not green, or it could be that both cards are green.

So, we make three calculations this time. We already know that the probability that 13 by the previous problem (because greenness is an equivalent both cards are green is 120 property to solidness). The probability that the first card is green but the second isn’t 54 1 18 9 is 80 3 = 80 = 40 by the same reasoning as in the previous problem, and likewise the 2 9 probability that the first card is not green but the second is green is 27 80 3 = 40 . These 9 13 67 9 + 40 + 120 = 120 . three states are exclusive, so we sum the probabilities: 40 (f) First we need to know the probability that a single card has two diamonds: it’s 91 because 9 of the 81 cards have both two symbols and diamonds. The probability that two cards 8 1 1 drawn both have diamonds is 80 9 = 90 by previous reasoning. The probability that the 1 1 first card has two diamonds but the second does not is 72 80 9 = 10 . The probability that 1 9 8 the second card has two diamonds but the first does not is 80 9 = 10 . These three states 1 1 1 19 are exclusive, so we sum the probabilities: 10 + 10 + 90 = 90 . (g) Let the event of the first card being green be denoted by FG and the event of the second card having two diamonds be denoted by S2D. First compute P(S2D|FG). Now, we don’t know whether the second card is green or not. So, we need to compute P((S2D and G)|FG) + P((S2D and not G)|FG). Worse yet, we don’t know whether the first card has two diamonds or not! So, we actually need to split each of these— and compute P((S2D and G)|(FG and 2D)) + P((S2D and G)|(FG and not 2D)) + P((S2D and not G)|(FG and 2D)) + P((S2D and not G)|(FG and not 2D)). That’s 3 6 5 16 1 2 80 + 80 + 80 + 80 = 80 = 5 .

Now compute P(FG)P(S2D) and see if it’s the same as P(FG and S2D): P(FG) = 13 ; 1 1 1 P(S2D) = P(S2D|F2D) + P(S2D|not F2D) = 90 + 10 = 91 ; so, P(FG)P(S2D) = 27 .

1 11 And, P(FG and S2D) = P(S2D|FG)P(FG) = 5 3 = 15 . So, these events are not independent. (h) Let the event of both cards having solid symbols be denoted by S and the event of at least one card being green be denoted by G. We will compute P(G)P(S) and see if 13 67 and P(S) = 120 . Now, it’s the same as P(G and S). From earlier problems, P(G) = 120 18 9 13 P(G and S) = P( f irst G and S)+ P(second G and S)+ P(both G and S) = ( 26 )( 27 120 )+ 9 18 13 8 9 13 6 6 8 13 44 13 22 13 ( 26 27 )( 120 ) + ( 26 27 )( 120 ) = ( 26 + 26 + 78 )( 120 ) = ( 78 )( 120 ) = ( 39 )( 120 ). You can think of this, per probability, as computing, e.g., P( f irst G and S) = P( f irst G|S)P(S), 13 22 )( 120 ) 6= where conditioning on S means that the deck is reduced to 27 cards. And, ( 39 67 13 120 120 . So, they are not independent.

14.12. Expect Problems, Probably

3.

149

(a) There are six raised doughnuts out of 11 doughnuts total. The probability of getting a 6 , the probability of getting a second raised doughnut (given that first raised doughnut is 11 5 the first is raised) is 10 , and the probability of getting a third raised doughnut (given that the first two are raised) is 49 . We want P(d1 is raised and d2 is raised and d3 is raised) = P(d2 is raised and d3 is raised|d1 is raised)P(d1 is raised) = P(d3 is raised|d1 is raised 4 6·5·4 = 33 ≈ .12. and d2 is raised)P(d2 is raised|d1 is raised)P(d1 is raised) = 11·10·9

(b) You could get raised-raised-cake, raised-cake-raised, or cake-raised-raised. These three probabilities must be summed. Using the argument given in the previous part, these are 6 5 5 6 5 5 5 6 5 5 11 10 9 , 11 10 9 , and 11 10 9 . Their sum is 11 . (c) One way of computing this is 1 − (probability of getting three doughnuts without sprinkles). There are three doughnuts with sprinkles, so eight without. Then, by previous reasoning (see the first part of this question), the probability of getting three 56 8 7 6 doughnuts without sprinkles is 11 10 9 = 165 . Therefore the probability we seek is 56 = 109 1 − 165 165 ≈ .66.

14.12 Expect Problems, Probably 1. Let us denote each lightbulb being on by “1” and being off by “0.” Then, the states of the bulbs are the binary strings {0000, 0001, 0010, 0100, 1000, 0011, 0101, 1001, 0110, 1010, 1100, 1110, 1101, 1011, 0111, 0000}. The corresponding probabilities (for 4, 3, 2, 1, or 0 bulbs burnt out) are (.02)4 , (.02)3 (.98), (.02)2 (.98)2 , (.02)(.98)3 , and (.98)4 (respectively).

2. Define a random variable corresponding to the number of heads that show when the coins land. What is the probability that this random variable has value 3? The state space is all possible seven-word strings with words heads and tails. There are 27 of them. Let H be the number of heads that show when the coins land. Then, P(H = 3) be the number of states with

exactly three heads. And, there are 73 = 35 ways for three of the coins to show heads.

3. The state space is all 64 ordered pairs of numbers, where your chosen number is first and your friend’s chosen number is second. There are eight possibilities for states (k, k), so the probability that you and your friend choose the same number is 18 .

The possibilities of numbers one apart are (1, 2), (2, 3), (3, 4), (4, 5), (5, 6), (6, 7), (7, 8), and 7 (8, 7), (7, 6), (6, 5), (5, 4), (4, 3), (3, 2), (2, 1). Thus, the probability is 32 .

4.

11 13 ) = 40 . (a) E[R] = 1P(R = 1) + 12 P(R = 12 ) + 0P(R = 0) = 18 + 21 ( 18 + 40 11 ) = 13 (b) E[Y ] = 1P(Y = 1) + 21 P(Y = 12 ) + 0P(Y = 0) = 18 + 21 ( 81 + 40 40 .

13 13 (c) R +Y = 13 40 + 40 = 20 .

5.

11 9 (a) Using PIE, we have that 1 − 20 = 20 of the Folks own pets. That’s also the fraction of 9 3 5 Folks who own fish or ferrets. So, 20 = X + 25 − 20 = X + 20 , and X = 15 .

2 . (b) Hair type and pet ownership are (presumably) independent, so 15 · 52 = 25

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(c) Ah . . . this is tricksy. One might think this is the same question as the previous question, because we are talking about people who are both curly-haired and own ferrets, but actually we are looking at a person with curly hair and wondering whether ze owns a ferret. So, this is a conditional probability question! See next item for the answer. (d) So, this is the probability of having curly hair and a ferret divided by the probability of 2 divided by 25 , or 15 . (Some students may only realize that having curly hair. Thus, it’s 25 there is an issue with the previous question after reading this one.) (e) This is the probability of having curly hair and a ferret divided by the probability of 2 owning a ferret, or 25 divided by 15 , or 25 . (f) This is the same as the previous question, so 52 . (g) We assign random variables Fi, Fe, and C with value 1 if a Folk owns fish, owns ferrets, or has curly hair (respectively) and 0 otherwise. Then, our questions are really asking P(Fe = 1), P(Fe = 1 and C = 1), P(Fe = 1|C = 1), and P(C = 1|Fe = 1). (h) We presume again that pet ownership and hair type are independent, so we compute 1 1 )/ 10 = 52 ( 25 10 (i) The same independence assumption tells us that this is just the probability of having kinky hair, or 15 .

11 11 = 100 . (j) Again we presume independence, so 51 · 20

6. The average value of a random variable is (∑s∈S X (s))/|S|.

The expected value of a random variable is ∑k∈N kP(X = k). Let us rewrite the second expression to obtain the first: P(X = k) = |{s | X(s)=k}| . So kP(X = k) = k|{s | X(s)=k}| . For the s |S| |S|

under consideration, we have X (s) = k so we obtain ∑k∈N kP(X = k) = ∑k∈N k|{s | X(s)=k}| = |S| X(s)

∑s∈S |S| .

7. The probabilities of the two events in S1 are a and 1− a, and the probabilities of the two events in S2 are b and 1 − b. The product probabilities are ab, a(1 − b), b(1 − a), and (1 − a)(1 − b).

These correspond to states, which are exclusive, so P(C) = ab + a(1 − b) + b(1 − a) + (1 − a)(1 − b) = ab + a − ab + b − ab + 1 − a − b + ab = 1.

8.

(a) The representations of carp are photos of carp in the wild, origami carp, crocheted carp, and bead-knitted carp. These represent four of the nine possible items in a Bag of Carp.

Then, E[R] = P(R = 1); a straightforward way of computing P(R = 1) is to sum the probabilities of the kinds of states that contain representations of carp. There is one type of state with all three items representing carp; its probability is ( 94 )3 . There are three types of state with two items representing carp, each with probability ( 94 )2 ( 59 ), and three types of state with one item representing carp, each with probability ( 94 )( 59 )2 . The 604 total probability is 729 ≈ .83.

(b) The items that can be used by a pet carp are aquatic insect pieces (for feeding carp) and a CD of carp music (made by carp for carp). These represent two of the nine possible items in a Bag of Carp. Thus, E[C] = P(C = 1); a straightforward way of computing

14.12. Expect Problems, Probably

151

P(C = 1) is to sum the probabilities of the kinds of states that contain items that can be used by a pet carp. There is one type of state with all three items that can be used by a pet carp; its probability is ( 29 )3 . There are three types of state with two items that can be used by a pet carp, each with probability ( 29 )2 ( 97 ), and three types of state with one item that can be used by a pet carp, each with probability ( 92 )( 79 )2 . The total probability 386 ≈ .53. is 729

(c) First note that no item both represents carp and can be used by a pet carp (because paper doesn’t do well under water and pet carp do not have fins that can hug stuffed representations of carp). We will perform a similar calculation as above. Here, our possibilities are that two items represent carp and one can be used by carp (three types of state) and that one item represents carp and two can be used by carp (again, three types of state), and that one item represents carp, one can be used by carp, and one item falls into neither category (six types of state). Thus, P(R = 1 and C = 1) = 3( 49 )2 ( 92 ) + 3( 49 )( 92 )2 + 6( 94 )( 92 )( 31 ) = 32 81 . To determine P(R = 1 or C = 1), we will use PIE: 604 32 26 P(R = 1 or C = 1) = P(R = 1) + P(C = 1) − P(R = 1 and C = 1) = 729 + 386 729 − 81 = 27 .

9. Our solution follows that of Example 14.5.7.

The maximum of P(E1 or E2 ) occurs when E1 and E2 are exclusive. Then, P(E1 or E2 ) = 83 P(E1 ) + P(E2 ) = 38 + 97 = 83 72 . However, 72 > 1, so we know that E1 and E2 cannot be exclusive! Thus, the maximum of P(E1 or E2 ) = 1.

The minimum value of P(E1 or E2 ) = max{P(E1 ), P(E2 )} = P(E2 ) = 79 .

Now, the minimum of P(E1 and E2 ) occurs when P(E1 or E2 ) is maximum (by PIE). So, the 11 . The maximum of P(E1 and E2 ) = minimum of P(E1 and E2 ) = P(E1 ) + P(E2 ) − 1 = 72 3 min{P(E1 ), P(E2 )} = 8 by PIE. 10.

(a) Let L = 1 if the two letters are the same and L = 0 if they are different. Then, E[L] = 1P(L = 1) = 12 because there are two of four possible outcomes that have two of the same letter. (b) Let R = 1 if the two letters are placed in the same row and R = 0 if they are placed in different rows. Then, E[R] = 1P(R = 1) = 31 because there are three rows, a probability of 31 of placing either letter in a given row, and thus a probability of 3 · 31 · 13 of placing both letters in the same row (any one of the three). (c) Let C = 1 if the two letters are placed in the same column and C = 0 if they are placed in different columns. Then, E[C] = 1P(C = 1) = 13 because there are three columns, a probability of 31 of placing either letter in a given column, and thus a probability of 3 · 31 · 13 of placing both letters in the same column (any one of the three).

(d) This is P(C = 1 and R = 1). More pertinently, the only way that two letters can be in the same column and in the same row is if they occupy the same square! There are nine ways that can happen (one for each square), and for each of them, the probability is 19 · 91 .

Thus, P(C = 1 and R = 1) = 19 . (e) This is P(L = 1 and R = 1). One way to do this is using conditional probability, so that P(L = 1 and R = 1) = P(L = 1)P(R = 1|L = 1). Our computation of P(R = 1) did

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not use anything about the identity of the letters, so P(R = 1|L = 1) = P(R = 1). Thus, P(L = 1 and R = 1) = 12 · 31 = 61 . (f) Yes! Any two of the three events are independent; their probabilities are multiplicative.

11.

(a) Let L = 1 if the two letters are the same and L = 0 if they are different. Then, E[L] = 1P(L = 1). The probability of getting the same letter twice is equal to the probability of getting an X first and an X second plus the probability of getting an O first and an O second. Each of those probabilities is equal to the conditional probability of getting an X second given getting an X first, times the probability of getting an X first. That’s 9 4 8 8 8 17 · 18 = 17 each, so 17 total. Thus, E[L] = 17 . (b) There are three possible rows. For each row, we follow the reasoning of the previous part to obtain the probability 28 · 93 , and three times this is 14 .

12.

(a) The probabilities add to 1, so it is assumed that every prickly ball actually lands on the board! (b) That’s the probability of hitting the upper-left-most region and hitting a non-white region divided by the probability of hitting a non-white region. The probability of hitting the upper-left-most region and hitting a non-white region is just the probability of hit1 . The probability of hitting a non-white region is ting the upper-left-most region, or 16 8 1 7 3 16 + 12 = 12 . Thus, the desired probability is 28 . (c) We will define a random variable B = 1 if the prickly ball lands in a grey region, B = 3 if the prickly ball lands in a white region, and B = 5 if the prickly ball lands in the center 5 1 1 + 3 · 4 · 48 + 5 · 12 = 13 oval. Then, E[B] = P(B = 1) + 3P(B = 3) + 5P(B = 5) = 8 · 16 6 .

13. We will add a value to B, namely B = 0 if the ball does not hit the board. And, we will define a new random variable R denoting the number of points earned if the prickly ball sticks. We have R = 0 if the ball does not stick, R = 1 if the prickly ball lands in a grey region, R = 3 if the prickly ball lands in a white region, and R = 5 if the prickly ball lands in the center oval. We want to compute E[R] = P(R = 1) + 3P(R = 3) + 5P(R = 5). The event of the ball sticking/falling is independent of where on the board the ball hits. Let us assign a random variable S = 1 when the ball sticks and S = 0 when the ball falls. The events of hitting and missing the board are exclusive, so the event of where on the board the ball hits is dependent on the event of the ball hitting the board (let’s say H = 1). Thus, P(R = k) = P(B = k and H = 1 and S = 1) = P(B = k and H = 1)P(S = 1). Moreover, P(B = k and H = 1) = P(B = k|H = 1)P(H = 1). All of our old probabilities for B were determined assuming that the board was hit, so P(B = k|H = 1) = P(B = k). This all comes 4 is together to give us P(R = k) = P(B = k)P(H = 1)P(S = 1) = P(B = k) · 31 · 54 . The 15 4 26 common to all terms, so E[R] = 15 E[B] = 45 .

14. Because the two coins are identical and flipped at the same time, their individual flips are independent. Therefore, P(two heads) = P(heads)P(heads) = 31 , and so P(heads) = √13 .

Similarly, P(tails) = √13 . However, √23 6= 1.

15. Start with P(head 1 )P(head 2 ) = 13 = P(tail1 )P(tail2 ) and P(head 1 tail2 ) + P(head 2 tail1 ) = 13 .

So, P(taili ) = 1−P(head i ) and plugging this in to the previous equation, we have P(head 1 )(1−

14.12. Expect Problems, Probably

153

P(head 2 )) + P(head 2 )(1 − P(head 1 )) = 31 . This reduces to P(head 1 ) + P(head 2 ) = 1. Ah. More plugging in produces P(head 1 )P(head 2 ) = (1 − P(tail1 ))(1 − P(tail2 )) = 1 − P(tail1 ) − P(tail2 ) + P(tail1 )P(tail2 ) = 1 − P(tail1 ) − P(tail2 ) + 31 = 34 − P(tail1 ) − P(tail2 ). From the start of the expression, this is 13 , so we get 1 = P(tail1 ) + P(tail2 ). That tells us P(tail2 ) = P(head 1 ) and P(head 2 ) = P(tail1 ). Now, P(head 1 tail2 )+P(head2 tail1 ) = P(tail2 )P(head 1 )+ P(head 2 )P(tail1 ) = P(head 1 )P(head 1 ) + P(head 2 )P(head 2 ) = P(head 1 )P(head 1 ) + ((1 − P(head 1 ))((1 − P(head 1 )) = 31 . Now, there are no real roots to this last equation, so there is no possible assignment of probabilities that will work.

16. For your opponent, P(R) = P(P) = P(S) = 31 . So, the probability you win is 31 . If both of you draw at random, then your actions are independent. Thus, each combination (RR, RP, RS, PR, PP, PS, SR, SP, SS) has probability 19 , and three of those are winning combinations for you. So, your probability of winning is still 13 .

17. There is a 13 probability of a draw in the first game and a 13 probability of a win in the second game. The probability of both happening is 13 · 31 = 19 . The conditional probability of winning 1/9 the second game given a draw in the first game is 1/3 = 13 .

18. By Theorem 14.7.1, we just need to compute E[B] = E[H] + E[F]; E[H] = 1P(H = 1) + 0P(H = 0) = 1· 21 + 0· 12 = 12 and, similarly, E[F] = 1P(F = 1)+ 0P(F = 0) = 1· 41 + 0· 34 = 41 .

Thus, E[B] = 34 .

One way to decide whether the events H = 1 and F = 1 are independent is to compute P(H = 1|F = 1) and compare to P(H = 1). There are four of 16 states where F = 1. Of these, two have H = 1. Thus, P(H = 1|F = 1) = 21 . Similarly, we compute P(F = 1|H = 1) and compare to P(F = 1). There are eight of 16 states where H = 1. Of these, two have F = 1. Thus, P(F = 1|H = 1) = 41 . Therefore, these events are independent.

19. Let us define a random variable Bk = 1 if the kth box is delivered on the given day and Bk = 0 10 if the kth box is not delivered on the given day. Then, B = ∑10 k=1 Bk . Thus, E[B] = ∑k=1 E[Bk ].

Note that P(Bk = 1) = 61 because a box might arrive on any of the six days on which mail is 10 1 5 delivered, so E[Bk ] = 1P(Bk = 1) + 0P(Bk = 0) = 16 and E[B] = ∑10 k=1 E[Bk ] = ∑k=1 6 = 3 .

So, you expect to receive 53 packages on any given day.

20. By Theorem 14.7.1, we just need to compute E[X ] + E[Y ]; E[X ] = P(X = 1) + 2P(X = 2) and E[Y ] = P(Y = 1) + 2P(Y = 2). We have P(X = 1) = P(X = 2) = 21 , so E[X ] = 12 + 2 12 = 23 .

Then, P(Y = 1) = P(Y = 1 and X = 1) + P(Y = 1 and X = 2) = P(Y = 1|X = 1)P(X = 1) + P(Y = 1|X = 2)P(X = 1). If the first card drawn has value 1, then the remaining deck has three cards of which one has value 1, and if the first card drawn has value 2, then the remaining deck has three cards of which two have value 1. Thus, P(Y = 1) = 31 · 12 + 32 · 21 = 21 .

Similarly and symmetrically, P(Y = 2) = 21 . Therefore, E[Y ] = 12 + 2 · 21 = 32 . It then follows that E[X +Y ] = 32 + 23 = 3.

21. No, because P(Y = 1|X = 1) = 13 but P(Y = 1) = 21 ; or, because P(X = 1)P(Y = 1) = 41 but P(Y = 1 and X = 1) = 16 from above.

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22. We will use as our base case that ES [X1 + X2 ] = ES [X1 ] + ES [X2 ]. As our inductive hypothesis, we will assume that for k ≤ n, ES [∑kj=1 X j ] = ∑kj=1 ES [X j ]. Now consider ES [∑n+1 j=1 X j ]. This is equal to ES [Xn+1 + ∑nj=1 X j ]. By our base case, this equals ES [Xn+1 ] + ES [∑nj=1 X j ]. By our inductive hypothesis, that latter term equals ∑nj=1 ES [X j ], and so we now have ES [∑n+1 j=1 X j ] = n+1 n ES [Xn+1 ] + ∑ j=1 ES [X j ] = ∑ j=1 ES [X j ], as desired.

23.

(a) This restricts our sample space to the six raised doughnuts. Three of these doughnuts are glazed. Using the reasoning given in the first part of the previous problem, we have 1 3 2 1 6 · 5 · 4 = 20 .

(b) This is the probability that the first doughnut is maple plus the probability that the second doughnut is maple plus the probability that the third doughnut is maple. That’s 61 · 55 · 4 5 1 4 5 4 1 1 4 + 6 · 5 · 4 + 6 · 5 · 4 = 2.

24.

(a) We have restricted our sample space to the five cake doughnuts, of which two are double-chocolate. We wish to measure the probability that the first doughnut is double1 by previous chocolate and the second doughnut is double-chocolate, namely 52 · 41 = 10 reasoning (see the first part of the first doughnut problem). (b) This is the probability that the first doughnut is blueberry plus the probability that the second doughnut is blueberry, or 15 · 44 + 54 · 41 = 25 .

25. The probability of getting at least one raised doughnut and getting at least one doughnut with sprinkles is 1 − (probability of getting no raised doughnuts or getting no sprinkley doughnuts). And, that’s 1 − (probability of getting no raised doughuts) + (probability of getting no sprinkley doughnuts) − (probability of getting three doughnuts that are neither raised nor sprinkled). For that last item, there are three doughnuts that fit the criteria, so the probability 1 5 4 3 2 3 2 1 · 10 · 9 = 165 . The probability of getting no raised doughuts is 11 · 10 · 9 = 33 . So, we have is 11 2 56 1 20 1 − ( 33 + 165 − 165 ) = 33 . Now, the product of the two probabilities for getting at least one 109 3379 20 raised doughnut and getting at least one doughnut with sprinkles is ( 31 33 )( 165 ) = 5445 6= 33 .

Thus, these events are not independent.

26. First note that every state is also an event. (a) Event: there may be other cats present, so this represents many states. (b) State (and thus also event): we know exactly what’s happening in the yard. (c) Neither: we don’t know whether Tubbs is present or not. (d) State (and thus also event): we know exactly what’s happening in the yard. (e) Event: there may be other cats present, so this represents many states.

27. This is conditional probability; the probability of (rolling less than 5) and (rolling even) is 82 ; the probability of (rolling even) is /1 2, so P((< 5)|(even)) = 21 . The probability of (rolling less than 5) is also 12 , so yes, these events are independent.

28. Let’s use PIE. P(heads or prime) = P(heads)+ P(prime)− P(heads and prime). P(heads) = 1 4 3 1 2 ; P(prime) = 2 ; P(heads and prime) = 16 . Thus P(heads or prime) = 4 .

14.12. Expect Problems, Probably

155

29. Using conditional probability, we compute P(CRC and Cologuard+)/P(Cologuard+).

P(CRC and Cologuard+) = .002 · .923 = 0.001846.

P(Cologuard+) = .002 · .923 + (1 − .866)(1 − .002)) = 0.135578.

0.001846/0.135578 = 0.0136158 or 1.4%.

30. Using conditional probability, we compute P(CRC and Cologuard− )/P(Cologuard− ).

P(CRC and Cologuard− ) = .002(1 − .923) = 0.000154.

P(Cologuard− ) = .002(1 − .923) + (1 − .002).866 = 0.864422.

0.000154/0.864422 = 0.000178154 or 0.02%.

31. Using conditional probability, we compute P(CRC and FIT+)/P(FIT+).

P(CRC and FIT+) = .002 · .783 = 0.001476.

P(FIT+) = .002 · .783 + (1 − .949)(1 − .002)) = 0.052374.

0.001476/0.052374 = 0.0281819 or 2.8%.

32. Using conditional probability, we compute P(CRC and Epi proColon+)/P(Epi proColon+).

P(CRC and Epi proColon+) = .002 · .482 = 0.000964.

P(Epi proColon+) = .002 · .482 + (1 − .915)(1 − .002)) = 0.085794.

0.000964/0.085794 = 0.0112362 or 1.12%.

Using conditional probability, we compute P(CRC and Epi proColon− )/P(Epi proColon− ).

P(CRC and Epi proColon− ) = .002(1 − .482) = 0.001036.

P(Epi proColon− ) = .002(1 − .482) + (1 − .002).915 = 0.914206.

0.001036/0.914206 = 0.00113322 or 1.13%.

Given that the percentage chance of having CRC is the same whether the test gives a positive or negative result, the test tells us nothing!

3 33. Conditional probability. P((squid or star) and face) = 13 · 13 because two of the 6 suits are 3 present and 3 of the 13 cards in each suit are face cards. P(face) = 13 .

1 1 So P((squid or star)| face) = 3 . We could also say that 3 of the face cards are squid or star.

34. There’s one skull on each die, so the probability of rolling it is 16 .

25 For P(Sk = 0) we have 65 · 56 = 36 as neither die rolls a skull.

5 as either die could roll a skull.

For P(Sk = 1) we have 2 · 61 · 61 = 18

1 For P(Sk = 2) we have 61 · 16 = 36 as both dies must roll a skull.

The sum is 1, as it should be because some number of skulls are rolled.

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35. The dice rolls are independent, so we just need to know the probability of rolling a magnifying glass for each die. For a green die, P(M = 1) = 21 ; for the yellow die, P(M = 1) = 23 ; for the red die, P(M = 1) = 21 .

Now, for the eight dice, P(M = 0) means none of the dice roll magnifying glasses. Thus we 1 . have ( 12 )6 ( 13 )( 21 ) = 384 P(M = 1) means exactly one die has a magnifying glass. So if that’s one of the green dice,

we have 61 ( 21 )6 ( 13 )( 12 ); if it’s the yellow die, we have ( 12 )6 ( 32 )( 12 ); if it’s the red die, we have 3 ( 21 )6 ( 31 )( 12 ). The sum is 128 .

P(M = 2) can have two green dice, for 62 ( 21 )6 ( 13 )( 21 ), one green and the yellow for 61 ( 12 )6 ( 23 )( 21 ),

one green and the red for 61 ( 21 )6 ( 13 )( 21 ), or the yellow and the red, for ( 21 )6 ( 32 )( 12 ). This gives 35 a total of 384 .

P(M = 3) can have three green dice, for 63 ( 12 )6 ( 13 )( 21 ), two green and the yellow for 62 ( 12 )6 ( 32 )( 12 ),

two green and the red for 62 ( 21 )6 ( 13 )( 21 ), or one green and the red and the yellow for 61 ( 12 )6 ( 23 )( 21 ).

77 This gives a total of 384 .

P(M = 4) can have 4 green dice for 64 ( 21 )6 ( 13 )( 21 ), three green dice and the yellow for

6 1 6 2 1 dice and the red for 63 ( 12 )6 ( 13 )( 21 ), or two green dice and the 3 ( 2 ) ( 3 )( 2 ), three green

35 . yellow and the red for 62 ( 21 )6 ( 32 )( 12 ). This gives a total of 128

P(M = 5) can have 5 green dice for 65 ( 21 )6 ( 13 )( 21 ), 4 green dice and the yellow for 64 ( 12 )6 ( 23 )( 21 ),

4 green dice and the red for 64 ( 21 )6 ( 31 )( 12 ), or three green dice and the yellow and the red for 6 1 6 2 1 91 3 ( 2 ) ( 3 )( 2 ). This gives a total of 384 .

P(M = 6) can have all green dice for ( 21 )6 ( 13 )( 21 ), 5 green dice and the yellow for 65 ( 12 )6 ( 23 )( 21 ),

5 green dice and the red for 65 ( 12 )6 ( 13 )( 21 ), or 4 green dice and the yellow and the red for

6 1 6 2 1 49 4 ( 2 ) ( 3 )( 2 ). This gives a total of 384 .

P(M = 7) can have all green dice and the yellow for ( 12 )6 ( 23 )( 21 ), all green dice and the red for ( 21 )6 ( 13 )( 21 ), or 5 green dice and the yellow and the red for 6( 12 )6 ( 31 )( 12 ). This gives a total 5 of 128 .

1 .

There’s only one way to get eight magnifying glasses, so P(M = 8) = ( 21 )6 ( 23 )( 12 ) = 192

Luckily, these numbers do actually sum to 1.

Chapter 15

Fun with Cardinality 15.4 Try This: Investigating the Play 15.4.1 Questions about Sample Storage [⋆1 ].

(a) The natural numbers N. (b) For n < 59, f (n) = n; for n ≥ 59, f (n) = n + 1; f : N → N ∪ {sample}.

(c) Yes. We will show it is injective: if f (n) = f (m), then either n = m or n + 1 = m + 1, so n = m. We will show it is surjective: given n ∈ N, if n < 59, f (n) = n; if n = 59, f (sample) = 59; if n > 59, f (n − 1) = n.

(d) There are infinitely many samples when we arrive at the storage area, and we add one sample. But the samples are in one-to-one correspondence with N, and our map gives a one-to-one correspondence between N and N ∪ {sample}, so these two sets have the same size. Therefore, ∞ = ∞ + 1. If ∞ < ∞ + 1, then we wouldn’t have the bijection defined above.

[⋆2 ]. It seems that N and N ∪ (some set with eight elements) have the same cardinality. Solutions may vary as to the definition of the function, but here is one example. Define f : N ∪ (some set with eight elements) → N by sending the eight elements to 1, . . . , 8 and sending n to n + 8.

This is a bijection; there is a one-to-one correspondence between the eight elements and 1, . . . , 8; if n + 8 = m + 8 then n = m; and, for m ∈ N, m > 8, m − 8 gets mapped to m. [⋆3 ].

(a) f : N → N is defined by f (n) = 2n.

(b) Yes! There are at least two conjectures to be made here, actually: |N| = |2N| and |N| = |N| + |N|.

To prove that |N| = |2N|, note that we can modify f (n) = 2n to be g : N → 2N defined by g(n) = 2n. This is definitely a bijection because if g(n) = g(m), then 2n = 2m, so n = m, and if we consider n ∈ 2N, then 2n ∈ N, so g( 2n ) = 2 n2 = n.

To prove that |N| = |N| + |N|, consider that N = 2N ∪ O (where O is the odd natural numbers), and then use the bijections g : N → 2N defined by g(n) = 2n and h : N → O defined by h(n) = 2n − 1. [⋆4 ].

(a) f : N → N is defined by f (n) = n2 .

(b) Here is data one might generate to get a start on answering this question.

1 x

2 3 1 2

4 5 x 3

6 7 4 1

8 9 2 x

10 11 12 13 14 15 16 17 18 19 20 . . .

5 6 3 4 1 2 x 7 8 5 6 ...

157

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15. Fun with Cardinality

Let’s see how many spaces are between old samples that have been moved to new drawers. Notice that (n + 1)2 − n2 = (n2 + 2n + 1) − n2 = 2n + 1, which means that there are 2n + 1 drawers available after the previously moved sample—including the drawer for the next moved sample. So, there are 2n drawers available between moved samples.

Between drawer n2 and drawer (n + 1)2 , the Storage Coordinator places two samples from each of the first n field biologists. (c) It’s difficult for beginning students to explain why this means there will be enough room for all the samples. They are unlikely to come up with the following explanation: There are |N| field biologists. Given any particular field biologist k, hir samples will begin being placed after drawer k2 and before drawer (k + 1)2 . And, given any particular sample n from field biologist k, that sample will be placed somewhere between drawer (k + ⌈ 2n ⌉ − 1)2 and drawer (k + ⌈ 2n ⌉)2 .

(d) This is another difficult question for students to answer. Students may want to say things along the lines of |N| · |N| = |N| or |N| + |N| . . . |N| = |N|. The corresponding sets are N and ∪k∈N Nk —though note that students do not have the latter notation yet. (And, in fact, there is a subtlety whereby the index set is really N ∪ (a single-element set), but because that’s in bijection with N, it’s okay.) Essentially any justification will be that these two sets are in one-to-one correspondence because the movement and placement algorithms produce injections from each of the Nk to subsets of N, and collectively those injections are surjections onto N. (e) One seems infinitely larger than the other, so it’s often surprising that they have the same cardinality.

15.4.2 More Questions about Sample Storage [⋆5 ]. If the starting tag number was 59, then when eight more specimens arrived, it would become 59+8 = 67. Then, when the first (single) batch of infinitely many specimens was processed, it would become 67 · 2 = 134. Therefore, the storage coordinator has at least started processing the single batch of infinitely many specimens and has not yet gotten to the 134th drawer in processing the infinitely many sets of infinitely many specimens. [⋆6 ].

(a) This suggests a bijection between N and N × N, which suggests that |N × N| = |N|. (b) An excellent surjection is f (x, y) = xy . This is not injective because (for example) f (x, y) = f (2x, 2y) = yx . This tells us that |N × N| ≥ |Q| and thereby suggests that |N| = |Q|. (c) One way to interpret the statement that there are more samples than slides is to say (slides) ⊂ (specimens), which is true because there exist specimens of which the lab does not store slides. But, the sets can still have the same cardinality—consider, for example, 2N ⊂ N, yet |2N| = |N|.

15.4.3 Questions about Café Conversations [⋆7 ].

(a) Z. (b) Let Lab Assistant be at stool n (cup of coffee n) and LA2 be at stool n − 1. For k ≤ n − 2, f (k) = k + 2 and for k ≥ n − 1, f (k) = k.

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159

[⋆8 ]. Suppose this person is at stool k. Then, just considering cookies from higher-numbered stools, set up the map for k ≤ n ≤ k + 6, f (n) = n− 8 and for n ≥ k + 7, f (n) = k. Then, seven cookies will be passed to the left of k and infinitely many cookies will arrive at k. [⋆9 ].

(a) The function sends a group of lab assistants to the project it discusses. So, f (group) = (group discussion topic). Better yet, let the set of groups of lab assistants be G and the set of projects be P. Let the project discussed by group g be pg . Then f : G → P is defined by f (g) = pg . (b) It sends a group of lab assistants to the lab assistant whose primary project the group discusses. Here, f (g) = ℓg , where ℓg ∈ L and we use L to denote the set of lab assistants. (c) g = {ℓ | ℓ 6∈ g′ , where g′ is such that f (g′ ) = ℓg′ }.

(d) By definition, f (g) = ℓg .

(e) Aaaa! If f (g) ∈ g, then ℓg 6∈ g except that ℓg = f (g) ∈ g. Contradiction! The output f (g) is neither an element of g nor is it not an element of g . . . this is a paradox. (f) If there were a bijection, then the group of lab assistants who are not in the group that discusses their primary project would have to map to some assistant. But, the function can’t map to any lab assistant, so there is no bijection. (g) The set of groups of lab assistants is the power set of the set of lab assistants. (h) The power set of N does not have the same cardinality as N. It has to have a larger cardinality. Therefore, not all infinite sets have the same cardinality. [⋆10 ]. This is 1342 , so the Storage Coordinator must be dealing with the infinitely many field biologists with infinitely many samples.

15.4.4 Indiscrete Questions [⋆11 ].

(a) One minute. (b) Infinitely many. ∞ 1 1 1 1 (c) + + · · · = 1 = ∑ j .

2 4 8 j=1 2

[⋆12 ].

(a) [0, 1] (b) |[0, 1]| > |N|. Suppose there was a bijection between [0, 1] and N. Call it q : N → [0, 1].

This would create a numbered list of elements of [0, 1]. However, we can create an element d of [0, 1] not on that list using the Cataloguer’s technique. Therefore, there is no k ∈ N such that q(k) = d, and so q is not onto; that’s a contradiction.

[⋆13 ]. Yes, and no. There were infinitely many lab assistants missing earlier, and only finitely many missing now; certainly there are infinitely many lab assistants whose names begin with “J” present now who were not present earlier, so in this sense, yes, there are more lab assistants in the café now than there were earlier. The cardinality of the lab assistants in the café now is the same as the cardinality of the lab assistants in the café earlier (both are |N|). So, in that sense, no, there are not more lab assistants in the café now than there were earlier.

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15. Fun with Cardinality

15.10 Infinitely Large Problems 1. We will exhibit a bijection between N and N ∪ {thing}. Let f (1) = thing and f (k) = k − 1.

This is injective: if f (k) = f (r), then either k − 1 = r − 1 so k = r or thing = thing. It is also surjective: for n ∈ N, f (n + 1) = n and f (1) = thing.

2. We will exhibit a bijection between N and N ∪ Z100,000,000 . For k ≤ 100,000,000, let f (k) = k − 1 ∈ Z100,000,000 , and for k > 100,000,000, let f (k) = k − 100,000,000 ∈ N. This is injective: Note that if f (k) = f (r), then f (k), f (r) must be both in N or both in Z100,000,000 . If f (k) = f (r), then either k − 100,000,000 = r − 100,000,000 so k = r, or k − 1 = r − 1 so k = r.

It is also surjective: for n ∈ N, f (n + 100,000,000) = n, and for n ∈ Z100,000,000 , f (n + 1) = n.

3. We will exhibit a bijection between N and 3N. Let f (k) = 3k. This is injective: if f (k) = f (r), then 3k = 3r so k = r. It is also surjective: for n ∈ 3N, f ( 3n ) = n.

4. We will exhibit a bijection between N and W. Let f (k) = k − 1. This is injective: if f (k) = f (r), then k − 1 = r − 1, so k = r. It is also surjective: for n ∈ W, f (n + 1) = n.

5. We will exhibit a bijection between N and Z. Let f (2k + 1) = k + 1 and f (2k) = −k. This is injective: if f (k) = f (r), then either both values are nonnegative, so that k + 1 = r + 1 so k = r, or both values are negative, so that −k = −r so k = r. It is also surjective: for n ∈ Z, if n ≥ 0 then f (2(n − 1) + 1) = n and if n < 0 then f (−2n) = n.

6. We will exhibit a bijection between Z and 2Z. Let f (k) = 2k. This is injective: if f (k) = f (r), then 2k = 2r, so k = r. It is also surjective: for n ∈ 2Z, f ( n2 ) = n.

7. This will be a visual proof. Draw Z × Z as a lattice on the Cartesian plane. Then, draw a spiral beginning at (0, 0) and winding outwards through (0, 1), (−1, 1), (−1, 0), (−1, −1), (0, −1), (1, −1), (1, 0), (1, 1), (0, 2), etc. Then, label these same points as 1, 2, 3, . . . to place the lattice in one-to-one correspondence with N. 8. Instead of 59 → 59+8 = 67 → 67·2 = 134 → 1342 , we would have 59 → 59·2 = 118 → 1182 .

9. Send the current samples along using n 7→ 3n, and then use 1, 2 (mod 3) to place the remaining samples.

10. No, because two different people map their seats/coffee onto LA2 (and also onto Lab Assistant); therefore, the map is not injective.

11. There are probably lots of answers to this. Here is one. We will define a map f : N → N where f (n) = the person who gets coffee from person n. In order for everyone to get two cups of coffee, we will pick some person k to be the focus of the map. The person to the left of k, person k − 1, hands k a cup of coffee so that k has two cups. Then k − 1 gets coffee from k − 2, k − 3. Then k − 2 gets coffee from k − 4, k − 5. To extend this, note that k − 2 = k − 1 · 2, k − 3 = k − 1 · 2 − 1, k − 4 = k − 2 · 2, and k − 5 = k − 2 · 2 − 1. More generally, these are of the n+k form k − 2r, k − 2r − 1. Thus, for n ≤ k − 2 and n the same parity as k, f (n) = n + k−n 2 = 2 , and for n ≤ k − 2 and n the opposite parity as k, f (n) = n + k−n+1 = n+k+1 2 2 .

15.10. Infinitely Large Problems

161

12. Proof by contradiction: Suppose that [1, 2) is countable. Write each element as a decimal 1.k1 k2 k3 . . . , where ki ∈ {0, 1, . . . 9} are decimal-place digits, and list them in one-to-one correspondence with N. Now create the number 1.k1 k2 k3 . . . where digit k1 is different from k1 in the first number of the list, k2 is different from k2 in the second number of the list, and so forth and so on. This number is not on the list but is in [1, 2), which is a contradiction.

13. Proof by contradiction: Suppose that there is a bijection f between P(Z) and Z. Consider the set A ∈ P(Z) defined as A = {a | a 6∈ A′ , where A′ is such that f (A′ ) = a′ }. What is f (A)?

This leads to a contradiction.

14. This follows from the first problem in this section, noting that ℵ0 = |N|.

15. This follows from the fact that 2|N| = |2N| = |N|.

16. This follows from the fact that |N × N| = |N| and also from the fact that the union of countably many copies of N has the same cardinality as N itself.

17. It should be the smallest number strictly larger than ℵ1 .

18. There are 2ℵ0 such numbers; they are in one-to-one correspondence with binary decimal expansions.

19. There is a bijection between this set and N, defined by f ( 1k ) = k. Therefore, the cardinality is ℵ0 .

20. There is a bijection between this set and N × N, defined by f ( qp ) = (p, q). Therefore, the cardinality is ℵ0 .

21. |W | = ℵ0 because we can put W in one-to-one correspondence with N as follows: list all 26 words of length 1, then all 262 words of length 2, and so forth. (Number the entries in the list.)

22. L is uncountable because the slopes of the lines are in one-to-one correspondence with R. 23. Countable, because we can list all finite graphs with one vertex, then all finite graphs with two vertices, and so forth; there are finitely many graphs at each stage and making a list gives an automatic one-to-one correspondence with N. 24.

(a) There are ℵ0 of these polynomials; use the bijection f (ax) = a. (b) There are ℵ0 of these polynomials; use the bijection f (ax + b) = (a, b) and the bijection between N and N × N.

(c) There are ℵ0 of these polynomials; use the bijection f (ax2 + bx + c) = (a, b, c) and the bijection between N and N × N × N.

(d) There are ℵ0 of these polynomials; use the bijection f (an xn + an−1 xn−1 + · · · + a1 x+ a0 ) = (an , an−1 , . . . , a1 , a0 ) and the bijection between N and N × N × · · · × N. 25. It’s countable. There are countably many possible monomials and countably many possible coefficients for each; therefore, the cardinality of this set is ℵ0 · ℵ0 = ℵ0 .

162

26.

15. Fun with Cardinality

(a) Injective: Let f (a) = f (b) so that 5a = 5b. Then a = b.

Surjective: Consider b ∈ 5Z. Then b5 ∈ Z and f ( 5b ) = 5 b5 = b. (b) Injective: Let g(a) = g(b) so that 10a − 7 = 10b − 7. Now 10a = 10b and because 1 10 ∈ R, we can divide by 10 on both sides to get a = b.

b 7 b 7 b 7 Surjective: Consider b ∈ R. Then 10 + 10 ∈ R and g( 10 + 10 ) = 10( 10 + 10 )−7 = b + 7 − 7 = b as desired.

(c) This map fails surjectivity. Note that 10(1) − 7 = 3 and 10(2) − 7 = 13. There is no a ∈ Z such that h(a) = 8 because 23 6∈ Z. 27.

(a) There cannot be a 1-1 map because two elements of A would need to map to one element of B by the pigeonhole principle. There is an onto map, by taking two elements of A and mapping them to one element of B, and mapping each of the remaining elements of A to each of the remaining elements of B. (b) There is an injection B → A by mapping each element of B to a different element of A. There is no surjective function because only three elements of A can be hit by the elements of B in a well-defined gipo. If we consider a surjection to not necessarily be a function (and thus relax the well-defined condition) we can create a surjective gipo by sending one element of B to two different elements of A. (c) Generally S2 has larger cardinality, by the pigeonhole principle. (d) Generally, S1 has larger cardinality, by the pigeonhole principle.

28. At the end of Scene 3, we’re told the drawer number is 17,956.

First eight samples arrive, so 59 → 59 + 8 = 67.

Then, every sample is moved to a drawer with twice the label, so 67 → 2 · 67 = 134.

Finally, every sample is moved to a drawer with the square of its label, so 1342 = 17956.

1: 2: 3: 4: 29.

5: 6: 7: 8: 9: 10:

first specimen from the original field biologist with infinitely many samples first specimen from first of infinitely many field biologists second specimen from first of infinitely many field biologists specimen that was in drawer 2 before—that’s the first specimen from the biologist who gave 8 samples third specimen from first of infinitely many field biologists fourth specimen from first of infinitely many field biologists first specimen from second of infinitely many field biologists second specimen from second of infinitely many field biologists specimen that was in drawer 3 before—that’s the second specimen from the original field biologist with infinitely many samples fifth specimen from first of infinitely many field biologists

15.10. Infinitely Large Problems

11: 12: 13: 14: 15: 16: 17: 18: 19: 20:

163

sixth specimen from first of infinitely many field biologists third specimen from second of infinitely many field biologists fourth specimen from second of infinitely many field biologists first specimen from third of infinitely many field biologists second specimen from third of infinitely many field biologists specimen that was in drawer 4 before—that’s the second specimen from the biologist who gave 8 samples seventh specimen from first of infinitely many field biologists eighth specimen from first of infinitely many field biologists fifth specimen from second of infinitely many field biologists sixth specimen from second of infinitely many field biologists

30. This is the 5th (2 · 1 − 1, so n = 2) element of the 10th (kth) copy of N, so the drawer number is (10 + 2 − 1)2 + 2(10 − 1) + 1 = 140. . .

. . . and the 10th (2 · 5 so n = 5) element of the 5th (kth) copy of N, so the drawer number is (5 + 5 − 1)2 + 2(5 − 1) + 2 = 91.

31. For each trip (except maybe the last) the dolly crosses the hall twice, once with boxes and once empty. Because the entire sequence of trips takes a minute, the dolly is full of boxes for 30 seconds and empty for 30 seconds. If one wishes to argue that these numbers cannot be equal because the dolly doesn’t return empty at the end (which is questionable—the script doesn’t specify), then this is equivalent to claiming that in binary .1111 · · · = 6 1, which is false.

32. Note that S if we add an element of [0, 1) to each element of Z, we get a copy of Z within R. In this way, α ∈[0,1) Zα ⊂ R.

33. Show that | 10 i=1 Ni | = |N| by mimicking Example 15.5.2, then use our usual bijection between N and N × N. S

34. This set is actually just Z as every element is an integer and all integers are present. So the cardinality is ℵ0 .

35. There are lots! Here are a few: P(N), P(W), P(Z), P(Z × Z), P(Q), P(N × N). . .

Chapter 16

Number Theory 16.2 Try This! Divisors and Congruences 16.2.1 Phi, Phi, Pho, Phum 1. ϕ (p) = p − 1. This is true because every natural number less than p is relatively prime to p.

2. ϕ (4) = 2, ϕ (9) = 6, ϕ (25) = 20, and ϕ (49) = 42. The conjecture is that ϕ (p2 ) = p2 − p when p is a prime number. This is because every number not a multiple of p is relatively prime to p2 , and there are exactly p multiples of p less than or equal to p2 , namely {p, 2p, . . . , (p − 1)p, p · p}.

Similarly, ϕ (pk ) = pk − pk−1 for k ≥ 2 because the multiples of p less than or equal to pk are {p, 2p, . . . , (p − 1)p, p2 , (p + 1)p, . . . (pk−1 − 1)p, pk−1 p}.

3. ϕ (10) = 4, ϕ (14) = 6, ϕ (15) = 8, ϕ (21) = 12, ϕ (33) = 20, ϕ (33) = 24, and ϕ (5) = 40.

Each of these numbers is the product of two primes. It appears that ϕ (pq) = ϕ (p)ϕ (q).

16.2.2 Getting More out of Prime Factorizations 1. GCD(5, 7) = 1, GCD(6, 12) = 6, GCD(3, 10) = 1, and GCD(20, 30) = 10.

2. LCM(5, 7) = 35, LCM(6, 12) = 12, LCM(3, 10) = 30, and LCM(20, 30) = 60.

3. Conjecture: GCD(a, b)·LCM(a, b) = ab.

Proof: Consider the prime factorization of ab, and think of it as a pile of prime factors. We claim we can split this pile into two (possibly) smaller piles in such a way that one pile multiplies to GCD(a, b) and the other pile multiplies to LCM(a, b). Note that for each prime factor that a and b have in common, there are two copies of that prime factor in ab. And note that for each prime factor in a but not in b, and for each prime factor in b but not in a, there is one copy of that prime factor in ab. So make a pile containing one copy of each prime factor that a and b have in common. This multiplies, by definition, to the greatest common divisor of a and b because its product divides a and divides b and no larger number divides both. The product of the remaining prime factors (they form the other pile) is a multiple of a because it contains all prime factors of a, and a multiple of b because it contains all prime factors of b, and is the least possible multiple of a and b because it contains no additional prime factors.

4. GCD(a, b) divides a, so a = GCD(a, b)c, and GCD(a, b) divides b, so b = GCD(a, b)d.

Therefore a − kb = GCD(a, b)c − kGCD(a, b)d = GCD(a, b)(c − kd), so GCD(a, b) divides a − kb.

164

16.4. Try This! Congruence Experiments

165

5. b (mod a) = b − ka for some integer k. Thus GCD(a, b) divides b (mod a). By definition, it also divides a, and thus GCD(a, b) divides GCD(b (mod a), a). Now, any divisor of both b − ka and a is also a divisor of b (by the same argument as in the previous question). Thus GCD(b (mod a), a) divides both a and b and thus divides GCD(a, b). So we have GCD(a, b) = ℓ1 GCD(b (mod a), a) and GCD(b (mod a), a) = ℓ2 GCD(a, b), so that GCD(a, b) = ℓ1 ℓ2 GCD(a, b), and ℓ1 ℓ2 = 1; because ℓ1 , ℓ2 are integers and GCDs are positive, ℓ1 = ℓ2 = 1 and GCD(a, b) = GCD(b (mod a), a).

6. Here a = 20, b = 30. So 30 (mod 20) = 10 and we have GCD(20, 30) =GCD(10, 20) = 10 because 10 divides 20.

7. Note that 382 (mod 230) = 152. Then GCD(230, 382) =GCD(152, 230). Now 230 (mod 152) = 78 and GCD(230, 382) = GCD(78, 152). As it happens, 152 (mod 78) = 74 so GCD(230, 382) = GCD(74, 78). But 78 (mod 74) = 4, so GCD(230, 382) = GCD(4, 74), and so we compute 74 (mod 4) = 2. This tells us GCD(230, 382) =GCD(2, 4) = 2.

16.2.3 Back to ϕ : More Phi, Less Pho and Phum 1. Well, ϕ (20) = 8; ϕ (2) = 1, ϕ (4) = 2, ϕ (5) = 4, and ϕ (10) = 4. Also 20 = 2 · 10 = 4 · 5. We can’t really combine 1 and 4 to get 8, but 2 · 4 = 8.

2. ϕ (12) = 6, ϕ (18) = 6, ϕ (24) = 8, ϕ (30) = 8, and ϕ (36) = 12.

12 = 3 · 4 = 2 · 6; 18 = 2 · 9 = 3 · 6; 24 = 6 · 4 = 8 · 3; 30 = 2 · 3 · 5 = 3 · 10 = 2 · 15 = 5 · 6; 36 = 6 · 6 = 4 · 9. It looks like when we collect like prime terms we can use those factorizations multiplicatively. k1 k1 −1 km 1 ) · · · (pkmm − pkmm −1 ).

3. Let n = prk 1 . . . pm . We conjecture that ϕ (n) = (p1 − p1

4. We need to prove that ϕ (n) = ϕ (pk11 ) · · · ϕ (pkmm ) and that ϕ (pki i ) = pki 1 − piki −1 . We proved the second statement earlier.

5. ϕ (n) = ϕ (pk11 ) · · · ϕ (pkmm ). We can prove this by induction if we know that ϕ (ab) = ϕ (a)ϕ (b) when a and b are relatively prime.

6. ϕ (20) = ϕ (4)ϕ (5), or |{1, 3, 7, 9, 11, 13, 17, 19}| = |{1, 3} × {1, 2, 3, 4}|. More generally, we have a single set on the left-hand side, and a Cartesian product of sets on the right-hand side.

This suggests we look for a bijection between the single set and the Cartesian product of sets.

7. Now we have ϕ (20) = ϕ (4)ϕ (5) as |{m ∈ N | m < 20, m is relatively prime to 20}| = |{q ∈ N | q < 4, q is relatively prime to 4}| · |{t ∈ N | t < 5,t is relatively prime to 5}|.

16.4 Try This! Congruence Experiments 1. Those that satisfy x ≡ 2 − 9 (mod 4) ≡ −7 (mod 4) ≡ −3 (mod 4) ≡ 1 (mod 4).

2. x ≡ s − r (mod n).

3. None, because 2x is always even and 1 (mod 4) is always odd (it’s 4k + 1).

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16. Number Theory

4. Even numbers, because 3 · 2 = 6 and so 3 · 2k = (3 · 2)k = 6k.

5. If 2x ≡ 1 (mod 3) then 2x = 3k + 1. So 2x is even numbers that are one more than a multiple of 3, or {4, 10, 16, 22, . . . } which means that x ∈ {2, 5, 8, 11, . . . }, or x ≡ 2 (mod 3).

6.

(a) 2 · 2 ≡ 1 (mod 3).

(b) We multiply through by 2 to obtain 4a ≡ 2 (mod 3) or a ≡ 2 (mod 3). Note that if a ≡ 2 (mod 3), then 2a ≡ 2 · 2 ≡ 1 (mod 3), so these give the numbers that satisfy our original equivalence. (c) Yes, 2 · 3 = 6 ≡ 1 (mod 5). Therefore we get x ≡ 4 (mod 5).

(d) No. If we look at 3 · {0, 1, 2, 3, 4, 5} (mod 6) we get {0, 3, 0, 3, 0, 3}, so we can never get a product of 1. (e) Sure: 7 · 3 = 21 ≡ 1 (mod 10). And 5 · 5 = 25 ≡ 1 (mod 6). But the multiples of 6 only produce 6, 3, and 0 (mod 9). (f) Such a w exists when k and n are relatively prime.

7. x = 2k = 3n, so x must be a multiple of 6.

8. x = 2k + 1 = 3k + 2, so x is odd and x ∈ {2, 5, 8, 11, 14, 17, . . . }. Thus x ∈ {5, 11, 17, . . . } or x ≡ 5 (mod 6).

Also x = 2k + 1, and plugging in gives 2k + 1 ≡ 2 (mod 3) or 2k ≡ 1 (mod 3), which we solved earlier to give us k ≡ 2 (mod 3). Then we have x = 2 · 2 + 1 = 5 as one solution, and x = 5 · 2 + 1 = 11 as the next solution; we again get x ≡ 5 (mod 6).

9. The first condition means x is odd, and every x ≡ 3 (mod 4) is odd, so all x ≡ 3 (mod 4).

10. Solutions will vary, but might include: writing out two sets and finding their intersection; substituting one equation into the other.

16.6 Try This! Investigate Freaky Fraction Lists 1.

0 1

0 1

1 4

1 3

2 5

1 2

3 5

2 3

3 4

1 .

1

3 7

1 2

4 7

3 5

5 8

2 3

5 7

4 11

3 8

5 13

2 5

5 12

3 7

4 9

1 ...

2

7 9

4 5

5 6

1 .

1

0 1

1 5

1 4

2 7

1 3

3 8

2 5

1 6

1 5

2 9

1 4

3 11

2 7

3 10

5 9

4 7

7 12

...

2. Solutions will vary.

3 5

8 13

5 8

7 11

1 3 2 3

7 10

5 7

8 11

3 4

3 4

4 5

1 .

1

16.6. Try This! Investigate Freaky Fraction Lists

167

3. We can prove this by induction. The base case is our first FFL: 1 · 1 − 0 · 1 = 1. The inductive hypothesis is that for n < k, all adjacent pairs of fractions ba and dc in the nth FFL have the property bc − ad = 1. For the inductive step, consider the kth FFL. Adjacent fractions either a+c a+c have the form ba and b+d , or the form b+d and dc , where ab and dc are from the (k − 1)st FFL. For this reason, we know that bc − ad = 1. In order to complete the proof, we need to show that b(a + c) − a(b + d) = 1 and that (b + d)c − (a + c)d = 1.

Now, b(a + c) − a(b + d) = ab + bc − ab − ad = bc − ad = 1, and (b + d)c − (a + c)d = bc + cd − ad − cd = bc − ad = 1. Done!

a+c , GCD(a + c, b + d) = 1. Check out this slickness: bc − ad = 1 4. We need to show that for b+d can be rewritten as ab + bc − ab − ad = b(a + c) + (−a)(b + d) = 1. Therefore we have found k = b and ℓ = −a such that k(a + c) + ℓ(b + d) = 1, which means by Theorem 16.3.2 that GCD(a + c, b + d) = 1.

5. Again, we use that bc − ad = 1 (see why it was useful to show that first?). It follows that bc = 1 + ad, so bc > ad. Now a little bit of algebra gives: bc + ab > ad + ab

bc + cd > ad + cd

b(a + c) > a(b + d) a+c a > b+d b

c(b + d) > d(a + c) c a+c > . d b+d

a+c < dc , each FFL is in increasing order. Therefore, if a fraction 6. Notice that because ab < b+d a+c so appears more than once, it must appear next to itself. This is not possible because ba < b+d a+c a no two adjacent fractions can be equal. (Or, b = b+d implies that bc = ad which is false.)

7. Conjecture: They all appear in the freaky fraction lists.

Proof: Suppose the fraction mn does not appear on any FFL. Pick some FFL (any one will do, it turns out) and find the place where mn is between two consecutive fractions, so ba < mn < dc .

Now, ba < mn ⇒ an < bm ⇒ bm − an > 0, and mn < dc ⇒ md < cn ⇒ cn − md > 0. Because a, b, c, d, m, n are all integers, this actually means that bm − an ≥ 1 and cn − md ≥ 1. Now, using these two inequalities, we have that (a+b)(cn−md)+(c+d)(bm−an) ≥ a+b+c+d.

Check out the left-hand side: it expands to acn − amd + bcn − bmd + bcm − acn + bdm − adn, which then reduces to and factors as bc(m + n) − ad(m + n) = (bc − ad)(m + n). Using our old reliable relationship bc − ad = 1, the inequality becomes m + n ≥ a + b + c + d. a+c a+c Now, if we look at the next FFL, we have ab < b+d < dc , with either ba < mn < b+d < dc or a+c m c a b < b+d < n < d . Repeating the argument from the previous paragraph gives us m + n ≥ a+b+(a+c)+(b+d) = 2a+2b+c+d > a+b+c+d or m+n ≥ (a+c)+(b+d)+c+d = a + b + 2c + 2d > a + b + c + d. That is, with every new FFL we examine, we increase the right-hand side of the inequality. So after looking at (no more than) m + n more FFLs, we get a contradiction as the right-hand side will now be larger than the left-hand side!

Therefore mn appears in some FFL.

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16. Number Theory

16.10 Bonus: The RSA Cryptosystem 1. The prime factorization of 556 · 1452 is 24 · 3 · 112 · 139 so e = 5 will work but e = 11 will not.

The relative primeness and lack thereof could also be checked using the Euclidean algorithm.

2. Because P is 7 digits, the best block size is 6 digits so we know no block will exceed P in value.

The message converts to 150400, 021402, 100504, 001907, 041718, 120010, 040617, 040019, 020019, 191424, 18.

Raising each number to the 13th power and reducing mod 3314221 produces 2907752, 868886, 1512383, 1571359, 482910, 109135, 167472, 1271400, 664765, 2255049, 1892926.

In case anyone wants to decode: 3314221 = 4027 · 823 and the multiplicative inverse of 13 mod 3309372 is 3054805.

3. First, we need to know what number w has the property that w · 23 (mod (1867 − 1)(2017 − 1)) = 1. This is the tough part! Luckily Wolfram—Alpha understands “solve 23 x mod(1867− 1)(2017 − 1) = 1” and returns 163559.

Now we raise each number in the encrypted message to the 163559th power and reduce mod 1867 · 2017 = 3765739.

That returns 190700, 190620, 122414, 201108, 100408, 180614, 81306, 191402, 141204, 10002, 100813, 181924, 1104.

The message is that gum you like is going to come back in style, a quotation from Twin Peaks that was used to announce the 2017 season of the show.

4. We compute w such that w · 62989 (mod (919 − 1)(1907 − 1)) = 1 and find w = 760117.

Now we raise each number in the encrypted message to the 760117th power and reduce mod 919 · 1907 = 1752533.

This gives us 191412, 141717, 142201, 41114, 130618, 191419, 71418, 41405, 201822, 71402, 141302, 40821, 41405, 81900, 180104, 111413, 60813, 61914, 42104, 172414, 1304.

The message is tomorrow belongs to those of us who conceive of it as belonging to everyone, a quotation from Audre Lorde.

16.12 Problems on Number Theory Topics 1.

(a) Consider n. If it is not prime, then it is composite and can be factored as n = ab where both a and b are smaller than n. Repeat this process for a and b. Because we get smaller numbers at each step, eventually the process terminates and we have a prime factorization. (b) If there are two different prime factorizations of n, we have n = pk11 · · · pkmm = q1j1 · · · qℓjℓ .

Assume that we have arranged the primes to be increasing, so that p1 < p2 < · · · < pm .

Because p1 divides n, we must have that p1 divides q1j1 · · · qℓjℓ . The only positive divisors

16.12. Problems on Number Theory Topics

169

of qi are itself and 1, so p1 = qi for some i. Therefore we can cancel p1 from the lefthand side and cancel qi from the right-hand side. We can repeat this process until we’ve ditched all the pi and are left with 1 on the left-hand side. But then we have 1 equal to a product of things that are bigger than 1, which is a contradiction.

2. None of the ki in n = pk11 . . . pkmm is greater than 1.

3. Consider the eight ball numbers b1 , . . . b8 and compute b1 (mod 7), b2 (mod 7), . . . , b8 (mod 7).

These values cannot all be different (by the pigeonhole principle) so two of them must be the same. And if bi ≡ b j (mod 7), then bi = b j + 7q or bi − b j = 7q.

4. Note that 2, 4, . . . , 2⌊ n2 ⌋ are all less than n, whereas 2⌈ 2n ⌉, . . . , 2n are all greater than n. And 2⌊ 2n ⌋ = n − 1 whereas 2⌈ 2n ⌉ = n + 1. So modulo n, 2⌈ 2n ⌉, . . . , 2n become 1, . . . , 0. Thus all residues are covered and they must be distinct.

5. By the pigeonhole principle, there are at least 17 balls with the same tens digit, because there are only 10 different possible digits. And because there are 175 balls, there must be at least one digit represented more than 17 times—18 times at least. (This is like dividing each ball number by 10, rounding to an integer, and reducing modulo 10.)

6. Using x = 0 and all whole y, we get multiples of 6 for k; using x = 1 and all whole y, we get all numbers 3 (mod 6) (except 3) for k. Together, this gives us all multiples of 3 greater than 3 itself.

7. Using y = 0 and all whole x, we get all non-negative multiples of 3 for k; using y = 1 and all whole x, we get all numbers 1 (mod 3) and larger than 6 for k; using y = 2 and all whole x, we get all numbers 2 (mod 3) and larger than 12 for k. Together, this gives us all numbers greater than 11 as well as 0, 3, 6, 7, 9.

8. Solutions will vary, but: We conjecture that when GCD(a, b) = 1, there’s some natural number above which all k are achievable. And when GCD(a, b) = g > 1, we can only get k that are high enough multiples of g. This is because when we don’t restrict x and y to whole numbers, we can get ax1 + by1 = GCD(a, b). Given a solution in whole numbers axw + byw = m with xw ≥ |x1 |, yw ≥ |y1 |, we can add the two equations together to increment m by GCD(a, b) (and repeat).

9. (⇐) Suppose m is divisible by 4. Then m = 10n−1 an−1 + · · · + 10a1 + a0 ≡ 0 (mod 3). Because 10 ≡ 1 (mod 3), m ≡ an−1 + · · · + a1 + a0 (mod 3). Thus the sum of m’s digits is also ≡ 0 (mod 3) and therefore is divisible by 3. (⇒) Suppose the sum of m’s digits is divisible by 3. Then an−1 + · · · + a1 + a0 ≡ 0 (mod 3).

Because 10 ≡ 1 (mod 3), we can substitute powers of 10 into this expression without changing its value. Then m = 10n−1 an−1 + · · · + 10a1 + a0 ≡ 0 (mod 3), which means m is divisible by 3.

10. 10k = 2k 5k so ϕ (10k ) = 2k−1 5k−1 4 = 2k+1 5k−1 .

11. ϕ (2k 3) = 2k−1 2 = 2k .

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16. Number Theory

12. All n except n = 1, 2. If an odd prime p divides n, then ϕ (n) is divisible by p − 1.

13. Any n divisible by 5 has a factor of 5 − 1 = 4 in ϕ (n). Any n divisible by 2k for k > 2 has ϕ (n) divisible by 2k−1 and k − 1 > 1. Any n divisible by 12 is divisible by both 3 and 4, so has factors of 3 − 1 = 2 and 2 in ϕ (n). The only ways to get a factor of 4 are from something of the form p − 1 = 4, pk = 4, or (p − 1 = 2 and pk = 2).

14. 20 = 22 5. Every divisor of ϕ (n) is either a prime power or p − 1 for some prime p. So the possible primes are 2, 3 (because 3 − 1 = 2), 5, and 11 (because 11 − 1 = 10). And we could have n = 33, 25. These are the only possibilities.

15. Only 5 could divide n, because none of 5, 25, 125 are one less than another prime. And any power of 5 has a factor of 4 in ϕ , so there are no possible n with ϕ (n) = 125!

16. Well, 24 = 23 · 3. The primes that could divide n are 2, 3, 5, 7, and 13. Because none of 5, 7, 13 divides 24, we know there is no more than one factor of each of those in n. If 13 divides n, then 24 = 12 · 2 and n = 13 · 3. If 7 divides n, then 24 = 6 · 4. Because 4 = 23−1 (2 − 1) = 5 − 1 = 22−1 (1)(3 − 1), we have that ϕ (5) = ϕ (8) = ϕ (12). Then n = 35, 56, 84. If 5 divides n, then 24 = 4 · 6 and 6 = 7 − 1 = 32−1 (3 − 1) so the new value we get is n = 5 · 9 = 45. Now we can suppose that only 2 and/or 3 divide n. In fact, both must divide n; ϕ (2k ) = 2k−1 and ϕ (3k ) = 3k−1 2. There must be exactly 2 factors of 3 for a single 3 to show up in ϕ (n), so we have 24 = 3(2)(22 ). Thus n = 9 · 8 = 72. In total, we have n = 35, 39, 45, 56, 72, 84.

17. Recall that for each prime p dividing n, there is a factor of p − 1 dividing ϕ (n). Aside from the case p = 2 where p − 1 = 1, the primes dividing n and mn must be the same. Moreover, because for each prime power pk dividing n, there is a factor of pk−1 dividing ϕ (n). These prime powers dividing n and mn must be the same except for powers p1 . Therefore m = 21 and n must be odd.

18.

(a)

Divisors of 5: 1, 5.

Divisors of 6: 1, 2, 3, 6.

Divisors of 8: 1, 2, 4, 8.

Divisors of 9: 1, 3, 9.

Divisors of 11: 1, 11.

Divisors of 12: 1, 2, 3, 4, 6, 12.

ϕ of these divisors: 1, 4. ϕ of these divisors: 1, 1, 2, 2. ϕ of these divisors: 1, 1, 2, 4. ϕ of these divisors: 1, 2, 6. ϕ of these divisors: 1, 10. ϕ of these divisors: 1,1, 2, 2, 2, 4.

Sum of the ϕ s: 5.

Sum of the ϕ s: 6.

Sum of the ϕ s: 8.

Sum of the ϕ s: 9.

Sum of the ϕ s: 11.

Sum of the ϕ s: 12.

(b) ∑i ϕ (di ) = n. (c) First we show that the conjecture holds for numbers of the form pk . The divisors of pk are 1, p, . . . , pk−1 , pk . Their associated ϕ are 1, p − 1, . . . , pk−2 (p − 1), pk−1 (p − 1) or, more conveniently for this purpose, 1, p − 1, . . . , pk−1 − pk−2 , pk − pk−1 . The sum is 1 + (p − 1) + p2 − p + · · · + pk−1 − pk−2 + pk − pk−1 = pk .

Now consider ∑i ϕ (di ) for any n. Each of the di can be written as its prime factorization, and ϕ (di ) can be factored corresponding to the primes in di . By reordering and grouping terms, we can write this sum as the product of ∑i ϕ (di ) for each of the primes dividing n. Thus we have ∑i ϕ (di ) = (∑i p ϕ (di p1 ) · · · (∑i pm ϕ (di pm ) = pk11 · · · pkmm = n.

1

(d) For each divisor di of n, consider the set Sdi = {k | GCD(k, n) = di }. Notice that the Sdi partition the set {1, . . . , n} because every number j ∈ {1, . . . , n} has GCD( j, n) = some

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divisor of n, and no j has two different GCDs with n. Now notice that k ∈ Sdi if and only if k ∈ Tdi = {k | k = qdi , GCD(q, dni ) = 1, q ≤ dni }. By definition, |Tdi | = ϕ ( dni ). Therefore n = |{1, . . . , n}| = ∑i ϕ ( dni ). But this is just a reordering of the terms in ∑i ϕ (di ), so we’re done.

19. 7644 (mod 4620) = 3024; 4620 (mod 3024) = 1596; 3024 (mod 1596) = 1428; 1596 (mod 1428) = 168; 1428 (mod 168) = 84; 168 (mod 84) = 0.

Aha! GCD(4620, 7644) = GCD(84, 168) = 84. So. . . 84 = 1·84 = 1428−8·168. Then 168 = 1596− 1428 so 84 = 1428− 8·1596+ 8 · 1428 = 9·1428− 8·1596. And 1428 = 3024− 1596 so 84 = 9·3024− 17·1596. Then 1596 = 4620− 3024 so 84 = 9·3024− 17·(4620− 3024) = 26 · 3024 − 17 · 4620. Finally, 3024 = 7644 − 4620, so 84 = 26 · 7644 − 43 · 4620.

20. Clearly kGCD(m, n) divides km and divides kn, so also divides GCD(km, kn). On the other hand, if any divisor of both km and kn does not divide k, then it must divide both m and n. And k divides GCD(km, kn), so GCD(km, kn)/k divides GCD(m, n). Each of the two expressions divides the other, so they must be equal.

21. 2 · 3 = 6 ≡ 1 (mod 5); the equivalence becomes x ≡ 2 · 4 ≡ 3 (mod 5), so the desired set is all x ≡ 3 (mod 5).

22. 3 · 4 = 12 ≡ 1 (mod 11); the equivalence becomes x ≡ 3 · 2 ≡ 6 (mod 11), so the desired set is all x ≡ 6 (mod 11).

23. 5 · 13 = 65 = 16 · 4 + 1 ≡ 1 (mod 16); the equivalence becomes x ≡ 5 · 4 ≡ 4 (mod 16), so the desired set is all x ≡ 4 (mod 16).

24. 5 · 9 = 45 ≡ 1 (mod 22); the equivalence becomes x ≡ 5 · 8 ≡ 18 (mod 22), so the desired set is all x ≡ 18 (mod 22).

25. 2x is even numbers that are 4 more than a multiple of 6, or {4, 10, 16, 22 . . . }, and thus x ∈ {2, 5, 8, 11 . . . } . . . Wait. That’s the same solution set! What’s going on here? First notice that a number y ≡ 1 (mod 3) is either equivalent to 1 (mod 6) or 4 (mod 6). Then, the solution set for y ≡ 1 (mod 3) is the union of the solution sets to z ≡ 1 (mod 6) and w ≡ 4 (mod 6).

However, 2x is even and z ≡ 1 (mod 6) is always odd. Therefore the solution set for y ≡ 1 (mod 3) is only the solution set to w ≡ 4 (mod 6).

26. x = 5 j + 3 so 5 j ≡ 3 (mod 7). Then 3 · 5 ≡ 1 (mod 7) so j ≡ 3 · 3 ≡ 2 (mod 7). Because j = 7q + 2, we have that x = 5(7q + 2) + 3 so x ≡ 13 (mod 35).

27. x = 3 j + 2 so 3 j ≡ 2 (mod 5). Then 2·3 ≡ 1 (mod 5) so j ≡ 4 (mod 5). Because j = 5q+ 4, we have that x = 3(5q + 4) + 2 so x ≡ 14 (mod 15).

28. x = 5 j + 2 so 5 j ≡ 2 (mod 6). Then 5 · 5 ≡ 1 (mod 6) so j ≡ 10 ≡ 4 (mod 6). Because j = 6q + 4, we have that x = 5(6q + 4) + 2 so x ≡ 22 (mod 30).

29. x = 11 j + 3 so 11 j ≡ 9 (mod 13). Then 6 · 11 ≡ 1 (mod 1)3 so j ≡ 54 ≡ 2 (mod 13).

Because j = 13q + 2, we have that x = 11(13q + 2) + 3 so x ≡ 25 (mod 143).

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16. Number Theory

30. We start by solving the first two and try to substitute that into the third. So. . . x = 3 j + 1 so 3 j ≡ 2 (mod 5). Then 2·3 ≡ 1 (mod 5) so j ≡ 4 (mod 5). Because j = 5q+ 4, we have that x = 3(5q + 4) + 1 so x ≡ 13 (mod 15).

Now we have x = 15v + 13 and 15v ≡ 5 − 13 ≡ −1 ≡ 6 (mod 7). Also 15 ≡ 1 (mod 7) so really we have v ≡ 6 (mod 7). That becomes v = 7r + 6 so x = 15(7r + 6) + 13 or x ≡ 103 (mod 105).

Let’s double-check: does this work? 103 ≡ 1 (mod 3); 103 ≡ 3 (mod 5); 103 ≡ 5 (mod 7); and, the LCM(3, 5, 7) is 105, so adding multiples of 105 does not change these equivalences.

Aces!

31. First, because x ≡ s (mod a), we know x = ja + s. Substituting into x ≡ s (mod b) we have that ja + s ≡ s (mod b) or ja ≡ 0 (mod b) or ja = kb. Therefore, because a and b are relatively prime, j = qb for some integer q. We can then rewrite x = ja + s as x = qba + s and conclude x ≡ s (mod ab).

Now, consider x ≡ 5 (mod 12). This is equivalent to x ≡ 5 (mod 4) ≡ 1 (mod 4) and x ≡ 5 (mod 3) ≡ 2 (mod 3). Similarly, x ≡ 38 (mod 86) is equivalent to x ≡ 38 (mod 2) ≡ 0 (mod 2) and x ≡ 38 (mod 43). But a number cannot be ≡ 1 (mod 4) and ≡ 0 (mod 2), so we have a contradiction.

32.

0 1

1 5

0 1 1 4

0 1 1 4 1 3

0 1 1 3 1 3 2 5

1 2 1 2 1 2 1 2

1 .

1 2 3 2 3 3 5

1 .

1 3 4 2 3

1 .

1 3 4 4 5

1 .

1

Notice that the fractions we need to add all have denominator n and have numerators relatively prime to n. Therefore we add ϕ (n) fractions to Fn−1 to get Fn —in total, we have 2 + ∑nk=2 ϕ (k).

33. Notice that we can split up this sum by denominator, so we have ∑i (fractions with denominator i). The fractions with denominator i have numerators all relatively prime to i. Now notice a special case of an earlier rule: if GCD(a, i) = 1 then GCD(i − a, i) = 1 also. Thus these fractions come in pairs, and each pair sums to 1. (There can be no fraction paired with itself as then a = i − a so a = i/2 and ai = 12 , which is a contradiction.) It then follows that for each i > 1, the contribution to the total sum is ϕ2(i) . Therefore the whole sum is 1 + 12 ∑nk=2 ϕ (k).

34. Sequence is 2, 3, 5, 9, 17, 33, 65, . . .

Every FFLk has all the fractions from FFLk−1 along with new fractions placed between them.

So ak = ak−1 + ak−1 − 1. We might also observe that the total number is 1 + 2k−1 . By induction, we have ak = 2ak−1 − 1 = 2(1 + 2k−2 ) − 1 = 2 + 2k−1 + 1 = 1 + 2k−1 . Done.

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173

35. The sequence of numerator sums is 1, 2, 5, 14, 41,. . . .

Each numerator sum is the previous sum, plus the new numerators. Each new numerator is the sum of two previous numerators, and each previous numerator is used in two sums (the ones to the left and to the right of it) except for the ends. And one end is always 0 and the other is always 1. So we have ak = ak−1 + 2ak−1 − 1 = 3ak−1 − 1. This corresponds to a closed form of (3n + 1)/2.

The sequence of denominator sums is 2, 4, 10, 28, 82, . . . Wait. 3n + 1? Really?

Again we have that each is the previous sum, plus the new denominators. And each new denominator is the sum of two previous denominators, and each previous denominator is used in two sums (the ones to the left and to the right of it) except for the ends. And each end is always 1. So we have ak = ak−1 + 2ak−1 − 2 = 3ak−1 − 2.

Chapter 17

Computational Complexity 17.2 Try This! Count the Operations 1. First algorithm: The first three lines of code take 1 operation each. There are 2 operations in the 4th line. Then, for values k = 2, 3, 4, we have another instance of line 3 and line 4, so 3 · (1 + 2) = 9 more operations. That’s a total of 3 + 2 + 9 = 14.

Second algorithm: The first three lines of code take 1 operation each. There are 2 operations in the 4th line. Then, for values k = 2, 3, 4, 5, 6, 7, we have another instance of line 3 and line 4, so 6 · 3 = 18 more operations. For k = 8 we have another line 3, but only operation in line 4, so 2 more operations. Lines 5, 6, 7 are 1 operation each. There are 2 operations in the 8th line. Then, for values m = 6, 7 we have another line 7 and another line 8, so 2 · 3 = 6 more operations. For m = 8 we have another line 7 but line 8 has only one operation, so 2 more operations. Line 9 is an operation. In total, we have 3 + 2 + 18 + 2 + 3 + 2 + 6 + 2 + 1 = 39.

2. First algorithm: The first three lines of code take 1 operation each. There are 2 operations in the 4th line. Then, for values k = 2, 3, 4, 5, 6, 7, we have another instance of line 3 and line 4, so 6 · (1 + 2) = 18 more operations. That’s a total of 3 + 2 + 18 = 23.

Second algorithm: The first three lines of code take 1 operation each. There are 2 operations in the 4th line. Then, for values k = 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, we have another instance of line 3 and line 4, so 12 · 3 = 36 more operations. For k = 14 we have another line 3, but only operation in line 4, so 2 more operations. Lines 5, 6, 7 are 1 operation each. There are 2 operations in the 8th line. Then, for values m = 9, 10, 11, 12, 13 we have another line 7 and another line 8, so 5 · 3 = 15 more operations. For m = 14 we have another line 7 but line 8 has only one operation, so 2 more operations. Line 9 is a operation. In total, we have 3 + 2 + 36 + 2 + 3 + 2 + 15 + 2 + 1 = 66.

3. First algorithm: We have three operations (for the first three lines of code). Then step 4 is done n times, and each time includes a comparison and another action. For each of those n times except the last (output) we do another step 3, so there are n − 1 of those. The total is 3 + 2n + n − 1 = 3n + 2.

Second algorithm: We have three operations (for the first three lines). Step 4 is done 2n times.

Each time there’s a comparison, and for all but the last time, there’s a replacement. So we get 2n · 2 − 1 operations there. For all except the last iteration, we also do another step 3 so that gives us 2n − 1 more operations. Steps 5–7 take another three operations. For step 8 there are n executions; all but the last include a comparison and a replacement. So there are 2n − 1 operations there. For each of those except the last, there is an additional step 7 (so n − 1 more 174

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175

operations), and then step 9 is a operation. Our total is 3 + (4n − 1) + (2n − 1) + 3 + (2n − 1) + (n − 1) + 1 = 9n + 3.

(Problems about sorting algorithms) 1. Solutions will vary depending on what intermediate steps students write down.

Insertion sort: {3, 6, 7, 2, 5, 4, 1} {2, 3, 6, 7, 5, 4, 1} {2, 3, 5, 6, 7, 4, 1} {2, 3, 4, 5, 6, 7, 1} {1, 2, 3, 4, 5, 6, 7} Select sort: {1, 3, 7, 2, 5, 4, 6} {1, 2, 7, 3, 5, 4, 6} {1, 2, 3, 7, 5, 4, 6} {1, 2, 3, 4, 5, 7, 6} {1, 2, 3, 4, 5, 6, 7} 2. Insertion sort: Nothing changes. There are 37 operations. There is one setup operation, then 6 operations for each value of i from 2–7 inclusive.

Select sort: Nothing changes. There are 100 operations. There are 4 setup operations, then for i = 1–6, {24, 21, 18, 15, 12, 5} operations respectively, then one final operation.

3. Insertion sort: {7, 6, 5, 4, 3, 2, 1} {6, 7, 5, 4, 3, 2, 1} {5, 6, 7, 4, 3, 2, 1} {4, 5, 6, 7, 3, 2, 1} {3, 4, 5, 6, 7, 2, 1} {2, 3, 4, 5, 6, 7, 1} {1, 2, 3, 4, 5, 6, 7} This takes 109 operations: 1 for setup, then {8, 12, 16, 20, 24, 28} operations for i = 2–7.

Select sort: {7, 6, 5, 4, 3, 2, 1} {1, 6, 5, 4, 3, 2, 7} {1, 2, 5, 4, 3, 6, 7} {1, 2, 3, 4, 5, 6, 7} {1, 2, 3, 4, 5, 6, 7} {1, 2, 3, 4, 5, 6, 7} This takes 125 operations: 4 for setup, then for i = 1–6, {36, 29, 23, 15, 12, 5} operations, then one final operation.

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17. Computational Complexity

4. Of course, both algorithms need more operations for anti-sorted input than sorted input. Insertion sort takes about 3 times as many operations for anti-sorted input than it does for sorted input. Select sort takes about 1.25 times as many operations for anti-sorted input than it does for sorted input. But for sorted input, select sort takes about 2.5 times as many operations as insertion sort, and for anti-sorted input, select sort and insertion sort take almost the same amount of operations!

5. Insertion sort takes 1 operation at the beginning, and then 6 operations for each value of i, 2–n inclusive. This is a total of 6(n − 1) + 1 = 6n − 5 operations.

Select sort takes 4 operations at the beginning, and 1 operation at the end, and 5 operations for i = n − 1. For i = 1 to n − 2, select sort takes 3 fewer operations with each iteration, ending with 12 operations. This gives the same number as taking 3 operations more with each iteration, starting with 12 operations. So we compute ∑n−2 i=1 (3i + 9) and add 10, to get 1 2 + 9n) − 5.

(3n 2

6. Insertion sort takes 1 operation at the beginning, and then 4 more operations with each iteration i = 2 through n, starting with 8. So we have 1 + ∑ni=2 (4i), which comes out to 2n2 + 2n − 3.

Select sort takes 4 operations at the beginning, and then. . . for i = 1 it takes 5n + 1 operations, for i = 2 it takes 5(n − 2) + 4 operations, and continues like this to i = ⌊ 2n ⌋. Then for the next iterations through n − 2 the list is sorted so it takes 3 fewer operations with each iteration. We earlier calculated that this was some quadratic function. So we get something approximately 5n · 2n +more quadratic function, which comes out as some other quadratic function.

7. For sorted input, insertion sort is faster because its number of operations needed is a linear function of n and select sort is slower because its number of operations needed is a quadratic function of n.

For anti-sorted input, both sorts have a number of operations needed that’s a quadratic function.

So select sort uses about the same number of operations no matter what the input is, but insertion sort is faster on sorted input than on anti-sorted input.

17.4 Try This! Compute Runtime Functions 1. Insertion sort: ri (n) = 2n2 + 2n − 3.

Select sort: rs (n) ≈ 5n · n2 + 21 (3n2 + 9n) − 5, where this is an overestimation.

Both of these follow from the earlier work; both functions are quadratic.

2.

(a) It’s constant, possibly r(n) = 1. (b) This has a linear runtime function. Roughly, r(n) = 2n + 1. For example, we might have 1. Let i = 1.

2. If listi =′ \0′ , output i; otherwise continue.

3. Replace i with i + 1 and go to the previous step.

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177

(c) Runtime function: 2 + n + 2 + 2n + 2n = 5n + 4. This is linear. (d) Runtime function: 2 + n + 1 + n + 2n = 4n + 3. This is also linear (and smaller!).

3.

(a) A worst-case input would have no subsets with sum 0, which we can achieve by having all positive entries in ints. (b) There are 2n − 1 nonempty subsets of ints. For each of these, we do the construction, which involves putting the elements in a list—at worst there are n of these. Then we sum the elements, which is at worst n − 1 additions. We do a comparison. (We don’t do a replacement because we know none of the subsets sum to 0! But it doesn’t matter, because that would just be one operation.) So that means there are 2n operations for each subset. Thus the runtime function is r(n) ≈ 2n · 2n . (c) This is more likely an underestimate than an overestimate, because there may be additional complexity in the “construct a new subset” step.

4. Suppose we’re looking for the first word in the dictionary. Then we have the sequence i = 500, 250, 125, 63, 32, 16, 8, 4, 2, 1. If we’re looking for the last word, we have the sequence i = 500, 750, 875, 938, 969, 985, 993, 997, 999, 1000. For each value of i, we have 2 comparisons and 3 replacements. So there are about 50 operations in either case.

The input size is actually the size of the dictionary! It’s 1000. So the generalization of the algorithm is to a dictionary of size n. The number of iterations is basically the closest power of 2 to n, or log2 (n). So we get r(n) ≈ 5 log2 (n).

17.6 Try This! Determine Complexity Classes 1. Solutions will vary according to how students did the problems, but...

Early examples: Θ(n).

Insertion sort: Θ(n2 ).

Select sort: O(n2 ), because we had to overestimate at one point.

Lists: Θ(1), Θ(n) for finding length, and Θ(n) for both list-insertion algorithms.

Subset sums: Ω(n2n ).

Dictionary: O(log(n)).

2.

(a) It’s m + n. (b) Step 1: integers are broken into digits. One read/storage operation for each digit, so m + n here.

Step 2: m multiplications.

Step 3: m more multiplications. There are also m storage operations to write everything in the correct columns.

Step 4: m ·(n− 2) multiplications because there are n− 2 more digits of B, and m ·(n− 2) storage operations.

Step 5: The numbers are in columns of sizes 1, 2, . . . , m, (m − 1) . . . , 2, 1.

So to add it all up we have a copy, an addition, 2 additions, etc. so that we have 1 + i+m−1 = 2+ 1 + · · · + (m − 1) + (m − 2) · · · + 1 + 1 operations which is 2 + 2 ∑m−2 1 (m−1)(m−2) + m − 1 operations.

2

178

17. Computational Complexity

(c) The total of the above-listed operations is m + n + mn + m(n − 1) + m + 1 + (m−1)(m−2) 2 operations. (d) It’s O(mn), or Θ(mn) for more precision. (e) We assume m ≥ n. (f) The roles of m and n swap, which means the non-mn terms are mostly smaller! But it’s the same function class because the highest term is still mn.

We could say that in a worst-case scenario, m = n so we have O(m2 ).

3. Essentially this algorithm works by repeatedly substituting the recursive definition of the Fibonacci numbers into fib(n). It computes each Fibonacci number in terms of the first two Fibonacci numbers, and recomputes each term on the way: it is sort of creating a big binary tree of values for fib(k) and at the end adds up the relevant number of them. Because at each stage there is a computational split (into fib(k − 1) and fib(k − 2)) this is of exponential complexity.

17.10 Computation Problems 1. The input size is the number of chips in the bag, n. We assume this number is nonzero.

Throw-away algorithm: There are 4 steps, each of which uses one operation per chip. There are two additional operations when the bag is empty (from step 1). The function is 4n + 2.

Potato-chip eating algorithm: How many iterations happen depend on how hungry the executor is. In a worst-case scenario, the executor is hungry after eating the entire bag. Then the algorithm has exactly the same number of operations as the throw-away algorithm, so the function is again 4n + 2. (We could write a function of the hungriness level h, measured in chips: it would be 4h.)

Kale-chip eating algorithm: This is functionally identical to the potato-chip eating algorithm.

2. A worst-case input has no entries equal to 61.

Runtime function: 2 + 3n + 2n = 5n + 2. This is linear with r(n) ∈ O(n).

3. This algorithm checks to see whether any entry is banana.

A worst-case input has no banana entries.

Runtime function: 1 + 3n. This is linear with r(n) ∈ O(n).

2 4. First procedure: Examine each of the nn ways to place n coins on the n × n grid. Ignoring 2 the checking-for-correctness part of the procedure, we have Ω( nn ).

Second procedure: List the row-position of each coin in each column, thus assuring no two coins are in one column. That gives nn lists to check for correctness. This is horrible and so we have Ω(nn ).

Third procedure: Don’t allow any of the two row-positions to be the same, thus assuring no two coins are in the same row or column. That gives n! lists to check for correctness. This gives us Ω(n!).

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5. Pretty much these functions need to be plotted in the [1, 5] range so they can be distinguished.

All three are much larger than 2n , which is larger than n2 . That is, they are all worse than exponential—so Ω(exponential).

6. This algorithm tells us whether a list has any duplicate elements.

Runtime: r(n) ∈ Θ(n2 ).

7. In the earlier multiplication algorithm the input size was the number of digits in each number.

Here we will see that isn’t the right choice for this algorithm.

Step 1: 2 operations.

Step 2: 1 operation.

Step 3: 1 or 2 operations per iteration (worst case, 2).

Step 4: 3 operations per iteration (divide, floor, replace).

Step 5: 2 operations per iteration (add, replace).

Step 6: 1 or 2 operations per iteration (1 until last iteration). Note that the number of iterations is the same as the number of 1s in the binary representation of A. (See the solution for Problem 24 from Section 5.11 for an explanation.) In a worst-case scenario, A’s binary representation is all 1s, and the number of those is the number of binary digits in A, which is log2 (A).

Step 7: The number of nonzero things in the column is the number of 1s in the binary representation of A, so that (minus 1) is the number of additions.

In total, we have 3 + (2 + 3 + 2 + 1) log2 (A) + 1 + log2 (A) − 1 = 9 log2 (A) + 3. Interestingly, our input size is A! So we have a total of 3 + (binary digits of A − 1)3 + 2(number of 1s in binary rep of A) where that 2 comes from 1 write in the column, 1 add operation. Worst-casing it, we have log2 (A) for each of those measures, right? Well, ⌊log2 (A)⌋ gives the number of binary digits of A. (And at worst they’re all 1.) What is that in terms of the number of decimal digits? The number of decimal digits is log10 (A) + 1 (because log10 (10) = 1).

8. Challenge: Why is this a challenge? Because US-style multiplication has input of size m + n, where m and n are the number of decimal digits of inputs A and B respectively, whereas Russian-style multiplication has input of size A. The US-style multiplication has runtime of order mn. For comparison purposes, let’s say that m = n so that our runtime is of order m2 .

The Russian-style multiplication has runtime of order log2 (A), which is approximately the number of digits in the binary representation of A. A = c · 10m , so log2 (A) = log2 (c · 10m ) = log2 (c) + log2 (10m ) = log2 (c) + m log2 (10). This is a linear function of m! So Russian-style multiplication is faster.

9. Step 1: 3 operations.

Step 2: Unclear how many operations are involved in this write, but let’s say 1 (per iteration).

Step 3: 3 operations for the first n − 1 out of every n iterations, and 1 operation for the nth.

Step 4: 4 operations for the first n − 1 iterations, and 1 operation for the nth.

Okay. So how many times is each step called? Step 4 is called n times. Step 3 is called n

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17. Computational Complexity

times for each value of k, or n · n = n2 times. Step 2 is called with every value of k and every value of ℓ, which is to say n2 times. So our total is 3 + n2 + n2 + n = 2n2 + n + 3.

10. Step 1: 1.

Step 2: 1.

Step 3: 2 each time; happens n times.

Step 4: 2 each time except last; n − 1 of those, and 1 for the last compare.

Step 5: 1.

So r(n) = 3 + 2n + 2(n − 1) + 1 = 4n + 2. This is in O(n).

11. Depth-first search crawls the tree in a backtracking pattern, traversing every edge twice and marking each vertex once. Because we’re working with a tree, we know the number of edges is one fewer than the number of vertices. So for V vertices, we have V + 2(V − 1) ≈ 3V as the runtime. The input size is the number of vertices, and the complexity class is linear.

12. Breadth-first search crawls the tree rank by rank, in each rank traversing every edge twice and marking each vertex once. Depending on how the graph is stored as data, the edges between the root and the first rank may be traversed once (in each direction) for each rank of the tree; those between the first and second ranks one less time, etc. If there are V vertices, there are roughly log(V ) ranks.

Because we’re working with a tree, we know the number of edges is one fewer than the number of vertices. So for V vertices, we have V + 2(V − 1) for the general search, and log(V ) log(V ) ) i≈ accounting for edge re-traversals, we get log(V V −1 edges per rank on average so V −1 ∑i log(V )3 < V . In total, we get ≈ 4V as the runtime. The input size is the number of vertices, V

and the complexity class is linear.

13. The algorithm does 4 operations (replace, add, replace, subtract) for each i, so the algorithm is linear in n.

14. The algorithm does 3 operations (compare, add, append) for each i, so the algorithm is linear in n.

15. For the Problem 13 algorithm, there are two additions per iteration; for the Problem 14 algorithm, there is one addition per iteration. So these two algorithms should be similarly affected.

We could note that certainly the number of operations per addition is bounded above by kn, so that the algorithms are in O(n2 ).

To be more precise, we can note that the number of operations per addition depends on the value of i (and is bounded by ki). Then we have ∑i ki = k ∑i ≈ kn2 , which tells us the algorithms are in Θ(n2 ).

16. Whether or not students have done the associated challenge problem, the solution is that the algorithms of Problems 13 and 14 are polynomial, and thereby much faster than the exponential-time algorithm given in Section 17.6 Problem 3.

17.10. Computation Problems

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17. Proof by contradiction. Suppose n2 ∈ O(n). Then there exists some c and some k such that for all n > k, n2 ≤ cn. Consider such a c. Then for n > c, we have that n2 > cn, which is a contradiction.

18. Well, 45n5 − 89n4 + 40789n3 − 67n2 + 57840578n − 2 ≤ 45n5 + 40789n3 + 57840578n ≤ 45n5 + 40789n5 + 57840578n5 = 57881412n5 , so we can choose k = 1 and c = 57881412 to fulfil the definition of 45n5 − 89n4 + 40789n3 − 67n2 + 57840578n − 2 ∈ O(n5 ).

19. By definition, r1 (n) ∈ O(r2 (n)) means there exists c1 , k1 such that for n > k1 , r1 (n) ≤ c1 r2 (n).

Similarly, r2 (n) ∈ O(r3 (n)) means there exists c2 , k2 such that for n > k2 , r2 (n) ≤ c2 r3 (n).

Therefore, for n > max(k1 , k2 ), we have r1 (n) ≤ c1 r2 (n) ≤ c1 c2 r3 (n). Thus choosing k = max(k1 , k2 ) and c = c1 c2 , we see that r1 (n) ∈ O(r3 (n)).

20. For c1 = 1, c2 = 1011 , the definition is satisfied for all n.

21. Solutions will vary depending on the algorithm suggested, but here’s one possibility: Suppose the input list has length n. Loop from i = 1 to n − 1, and if listi > listi+1 keep going; otherwise, exit and say “no.” If the loop is completed, say “yes.”

This is linear in n.

22. The algorithm checks whether a given string of length n contains the substring matcha. Let’s look at a couple of sample inputs. If the string has no ms, then the algorithm takes about n − 6 operations. If the string is matchmatchmatch . . . match then the algorithm takes about 6(n/5) + 4n/5 = 2n operations. So this seems to be linear in n.

23. The algorithm does 10 operations per iteration if it’s before noon and 2 operations per iteration if it’s after noon. The complexity is linear.

24. The algorithm does a lot of 2-operation iterations (outputting 7 after adding 3 and 4). There are 3 nested loops, so this is of cubic complexity. (The two inner loops together do ∑ni i iterations, which is quadratic, and these are done for each outer-loop iteration; there are n of those.)

25. This produces (a list of) the list elements that are less than the average value of the list elements.

There are two parts to the algorithm. The first computes the average, which takes 4n operations.

The second does the comparisons and makes the lists, which takes at most 4n operations. So in total, the complexity is linear.

26. This finds a pair of elements from an integer list whose difference is maximal. It does ∑ni n − i step 2s, each of which has at most 6 operations, and for each of the instances of step 2 there are 2–4 other operations in other steps. . . so the complexity is quadratic.

182

17. Computational Complexity

27. If we examine r(n) and 236t(n) on the range [1, 500], then we see that r(n) ∈ O(t(n)).

If we examine r(n) and t(n) on the range [1, 500], then we see that r(n) ∈ Ω(t(n)).

Thus, we suspect r(n) ∈ Θ(t(n)).

28. Plotting r(n) and t(n) over subranges of [1, 25], it is clear that as n gets large, t(n) becomes larger than r(n). Thus r(n) ∈ O(t(n)), but r(n) 6∈ Ω(t(n)) or Θ(t(n)).

29. Algorithm A does m multiplications in the first line, and then there are 2m iterations that have one multiplication (and two replacements) each. So the complexity is exponential in m.

Algorithm B does m iterations, each of which has 2 operations, so the complexity is linear in m.

However, if we consider the size of the numbers we’re multiplying, a la Problem 15, then we should get quadratic complexity instead.

30. False, because for n > the logarithm’s base, log(n) > 1 so n log(n) > n; more generally, given m ∈ N, there exists k such that when n > k, log(n) > m, and so we have n log(n) > mn and cannot have n log(n) < cn for any c. Thus n log n ∈ Ω(n). √ √ 31. True, because for natural n, n > n, n > log(n). Thus n + log(n) < 2n and the definition is satisfied.

32.

1. Look in the top folder.

2. If there are no folders present, delete all the files, return one directory-level higher (to the place that contains the folder we’re in), and go to step 4. Otherwise, continue.

3. Look in the first folder you see and go to step 2.

4. If the first folder you see is the top folder, delete it and exit; otherwise, delete the first folder you see and go to step 2.

Complexity: It’s linear in the number of files plus the number of folders.

33.

1. Create an empty folder in location B and give it the same name as the top folder in location A. 2. Let pathstart be the address of the top folder in location B. 3. Look in the top folder in location A. 4. If there are folders present, identify the first folder that does not share a name with a folder in pathstart, make an empty folder (with the same name) in pathstart, add this folder name to pathstart, enter this folder in location A, and go to the beginning of this step. Otherwise, continue.

5. Copy all the files you see (keeping the names the same) to pathstart.

6. Go up one directory level and remove the ending directory of pathstart.

7. If pathstart is the address of the top folder in location B, exit.

8. Go to step 4.

Complexity: It’s linear in the number of files plus the number of folders.

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34. Solutions will vary, but here is one possibility: 1. Let i = 2 and let great = list1 .

2. If listi > great, replace great with listi and continue; otherwise, continue.

3. If i < n, replace i with i + 1 and go to the previous step; otherwise, output great.

The complexity is linear in the length of the list.

35. Solutions will vary, but here is one possibility: Compute 50 − d. Find, using a binary/phonebook search, the latest entry catpricei with catpricei,1 ≤ remain. Define imax as the index of this entry. If catpricei,1 = remain, output catpricei and continue. Otherwise, decrement i and continue. Compute temp = remain − catpricei,1 and look to see whether anything prior in the list has value temp. If so, output a list containing that item and the other items whose costs sum to remain. Otherwise, find the rightmost item with price less than temp, replace temp with the amended remaining cost, and continue searching. When the start of the list is reached, remove the last two items of the list and start checking again just to the left of the more expensive item.

Complexity: log(m) operations to find the starting place. In a worst-case scenario, there are no combinations of items that exactly total prices to remain. And. . . it seems to be polynomial.

But seriously, if you assign this problem, presumably your students will come up with better algorithms and better complexity estimates. Also, you can simply refuse to give feedback on an insufficiently clearly written algorithm/complexity estimate.

Discrete Mathematics with Ducks Solutions Manual sarah-marie belcastro