CS 173: Discrete Structures, Spring 2010 Homework 9 Solutions

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1 CS 173: Discrete Structures, Spring 010 Homework 9 Solutions This homework contains 5 prolems worth a total of 5 regular points. 1. Off to the races! [10 points] You decide to take in a day of races at the local horse racing track and are particularly interested in the action involving races numer 1 and for the day. In the first race there are 1 horses and in the second race there are 8. In oth races the horses are given numers, 1 through 1 in the first case and 1 through 8 in the second. At the conclusion of the race, the horses will e placed in the order they have finished. With photo finishes and other snazzy gadgets we assume that there is never a tie. In order to strategically use your money, you ve decided to do a couple of calculations to see just how many different ways certain things can happen for some of the various ets. (a) How many different outcomes (i.e. orderings of all the horses) are there for each of the two races? How many possiilities for the sequence of two races? 1st race: it s a permutation of 1 items, or 1! outcomes. nd race: it s a permutation of 8 items, or 8! outcomes. For each outcome of race 1, any outcome of race will work, and vice versa. Thus in sequence, there are 1!8! outcomes considering oth races. () A trifecta is a particularly well paying et where you pick the first place, second place, and third place horses for a single race. Convinced that horse 4 will win first, horse will win second and horse 10 will win third in the first race, you decide to uy a trifecta ticket to this effect. In how many different outcomes of the first race will you win money? There are 1 horses in the first race, and we have fixed the positions of three of them. The numer of matching outcomes is the numer of ways we can order the remaining horses: 9! (c) If you pay twice as much money, you can ox a trifecta and then you only have to pick the first three finishing horses without placing them in their specific order (this does not pay quite as much as a regular trifecta, unfortunately). If you are still convinced that the top 3 will e horses 4,, and 10, in how many different outcomes of the first race will you win money if you uy a trifecta ox with those horses? From the previous prolem, we see that there are 9! matching outcomes whenever we set the positions of horse4, horse, and horse10. There are 3! ways to order horse4/horse/horse10 in the first three positions, so there are 3!9! matching outcomes. 1

2 (d) An In the Money et on a given horse is a et that the horse will come in either first, second, or third place. In the first race, you like horse numer 4 and in the second race, horse numer 8 looks particularly interesting. If you were to make an In the Money et on horse 4 in race 1 and horse 8 in race, calculate the numer of possile outcomes for the first races in which you win at least one of the ets? How many outcomes where you win oth of the ets? The numer of outcomes in race 1 with horse4 in the top 3 positions is 3 11!, or the numer of ways to choose the position of horse4 and then permute the remaining 11. Similarly, the numer of outcomes in race with horse8 in the top 3 is 3 7!. We will win oth of the ets in (3 11!)(3 7!) = 3 11!7! outcomes. Considering oth races, we will win the race 1 et (and may or may not win the race et) in (3 11!)(8!) in outcomes. We will win the race et (and may or may not win the race 1 et) in (1!)(3 7!) outcomes. We will win at least one et in (3 11!)(8!)+(1!)(3 7!) 3 11!7! outcomes. Notice that we sutracted off 3 11!7! outcomes, since we otherwise would doule-count the outcomes where we win oth ets.

3 . Bracketology [1 points] Professor J has recently ecome a fan of college asketall. He s a gamling man, and he wants to et on the ACES Basketall Tournament ( Mulch Madness ). The numer of teams in the tournament varies from year to year, ut there are always n = k participating teams for some i Z +. Prof J will predict the tournament s outcome y making a racket that specifies the winners of every game. A racket is a alanced, complete, full laeled inary tree that stores the results of all of the games in the tournament. Every node in a racket is laeled with a team name: There are n = k leaf nodes, each laeled with one of the participating teams. Each interior node is laeled with the winner of the game etween the two children. For example, the following picture shows a racket with 4 teams. There are three games: T1 eats T, T4 eats T3, and T4 eats T1. T4 T1 T4 T1 T T3 T4 A team will continue to play games as long as it keeps winning, and the root of the racket is the tournament winner. Any particular racket can e seen as a prediction of the outcome of every game in the entire tournament. Say that the laels on the n = k leaves have een fixed (i.e. they are the same on every racket and Prof. J does not have to guess them). (a) How many games will e played in the tournament? (i.e. numer of interior nodes) For a inary tree with n = k leaves, there are n 1 = k 1 interior nodes. () How many games must a team win to ecome the champion of the tournament? (i.e. height of the tree) For a inary tree with k leaves, the height is k. (c) His department is low on funds, so Prof J will only et on the tournament if he is guaranteed to make a profit. It costs $1 for Prof J to sumit a racket as a guess, and he s allowed to sumit multiple rackets. Say that he will win prize money only if one of his rackets exactly guesses the entire tournament (i.e. every game s winner is predicted correctly y that racket). In terms of k, what s the minimum prize money that must e offered for Prof J to play? Give an exact formula and also state approximately how large a numer it is. To e guaranteed to win, Prof J must sumit every possile racket. Starting from the ottom (leaves) of a racket, there are k 1 ways to chose the winners of the 3

4 first set of games. Then there are k ways to choose the next level of winners, k 3 ways to chose the next level, etc. At the final game, there are 1 ways to choose the winner. Thus there are k 1 k... 0 = P k 1 i=0 i = k 1 rackets. Note that there are k 1 games (interior nodes) in a racket, so our answer is equivalent to just picking etween two options at each game, and we could have derived it as such. Surprisingly, it doesn t really matter which order we pick these in (e.g. going top to ottom, ottom to top, or filling them in randomly); this is ecause a racket with fixed leaf laels is uniquely defined y the choices etween left and right child at each interior node. Thus the minimum prize money must e greater than k 1 dollars. For n = 3, this is already 31, or over two illion dollars! (d) Now say that he will win money if one of his rackets guesses the tournament winner (even if he hasn t correctly predicted the results in all the intermediate games). In terms of k, what s the minimum prize money that must e offered for Prof J to play? To e guaranteed to win, Prof J must sumit one racket with each team winning. Thus the prize must e greater than k dollars. (e) Prof J has a favorite team in the tournament. How many different rackets can he make where his favorite team is the tournament winner? In a racket where his favorite team is the tournament winner, he knows the outcome of any game that this team plays. Thus there is one game at every level that is fixed and the rest can vary. We can perform the same calculation as in part (c), ut with one game already decided at each level: ( k 1 1 ) ( k 1 )... ( 0 1 ) = (P k 1 i=0 i ) k = k 1 k Similar to the second approach in part (c), this is the same as picking from two options for all the games minus the k games that we know the favorite team must win. 4

5 3. Tree induction [10 points] Fragrant trees are a special type of laelled tree, found only in Illinois in the early spring. Fragrant trees have laels on their nodes. Nodes whose laels aren t specifically mentioned elow have the placeholder lael. A fragrant tree is defined recursively as follows: A single node with lael is a fragrant tree. If S is a fragrant tree, then so is the tree consisting of a root whose left sutree is a node with lael and whose right sutree is S. If S and T are fragrant trees, then so is the tree consisting of a root node whose four child sutrees are (left to right) a single node with lael A, S, a single node with lael a, and T. The second and third cases look like: S a S a T (a) Draw two examples of fragrant trees with more than 4 nodes. Also two examples of trees with more than 4 nodes which have laels a and on the leaves and on the interior nodes, ut which are not fragrant. Here are two possile fragrant trees with more than 4 nodes: a a 5

6 a a And here are two that are not fragrant trees: a a a a () Prove y induction that a fragrant tree always has an even numer of leaves with lael a. By induction on the structure of fragrant trees: Base: Consider the fragrant tree with one node with lael. This tree has 0 leaves with lael a, so the claim holds. Induction: Say that the claim holds for fragrant trees T 1 and T. That is, T 1 has k 1 leaves with lael a, and T has k leaves with lael a, for k 1, k N. Consider tree U which is a root with a single node as its left sutree and T 1 as its right sutree. U is a fragrant tree, y definition. The numer of lael a leaves in U is exactly the numer in T 1, since the left sutree has none. Thus U has k 1 lael a leaves, which is even, so the claim holds for U. Now consider tree V which is a root with four children: lael a single node, sutree T 1, lael a single node, and sutree T. The numer of lael a leaves in V is two plus the numer in T 1 and T. Thus V has + k 1 + k = (1 + k 1 + k ) lael a leaves, which is even, so the claim holds for V. 6

7 4. Structural induction [10 points] Suppose that the set S containing -D points is defined as Rule 1: (1, 1) and (1, 3) are in S Rule : If (x, y) S then (kx, ky) S for any positive integer k. Rule 3: If (x, y) S and (p, q) S then (xp, yq) S. Use structural induction to prove that, for any point (x, y) in S, y = 3 m x for some integer m. By induction on S: Base: For (1, 1), choose m = 0: (1, 1) = (1, 3 0 1) For (1, 3), choose m = 1: (1, 3) = (1, 3 1 1) Induction: Say that the claim holds for pairs (a, ), (c, d) S. That is, = 3 m 1 a for some integer m 1, and d = 3 m c for some integer m Consider (ka, k) S for positive integer k: (ka, k) = (ka, k(3 m 1 a) = (ka, 3 m 1 ka). This is of the form (x, 3 m x), so the claim holds for (ka, k). Now consider (ac, d) S: (ac, d) = (ac, 3 m 1 a 3 m c) = (ac, 3 m 1+m ac). This is of the form (x, 3 m x), so the claim holds for (ac, d). 5. More astract induction [10 points] Let S e any non-empty set. Define the set E(S) inductively as: If S = 1 then E(S) = If S > 1, choose some x S and set A = S {x}. Then E(S) = E(A) {{x, y} y A } (a) List the elements of E(S) for S = {a, }, S = {a,, c}, and S = {a,, c, d}. E({a, }) = {{a, }} E({a,, c}) = {{a, }, {a, c}, {, c}} E({a,, c, d}) = {{a, }, {a, c}, {a, d}, {, c}, {, d}, {c, d}} () Use induction to prove that, for any set S containing n elements, where n, ( ) n E(S) = By induction on n: Base: If S =, then say S = {x 1, x }. E(S) = {{x 1, x }} = 1 = ( ) Induction: Assume the claim for some n = k. That is, for S with S = k, E(S) = ( k ). 7

8 Consider a set S with S = k + 1. By definition, E(S ) = E(A) {{x, y} : y A} for some x S and set A = S {x}. Note that for any A, A = S 1 = k. Thus y our inductive hypothesis, E(A) = ( k ) for any A. Now consider {{x, y} : y A}. There are k elements in this set, one for each element in A. This set is disjoint from E(A) ecause E(A) is a set of element sets, none of which contain x. (We really should e showing this fact y induction as well, ut we are glossing over that for this proof.) Thus E(S ) can e calculated as follows: E(S ) = E(A) + {{x, y} : y A} E(A) {{x, y} : y A} = ( ) k + k 0 = k! k(k 1) + k = + k (k )!! = k k + k = k + k = (k + 1)! (k 1)!! = ( ) k + 1 8