6.8 The Pigeonhole Principle

Transcription

1 6.8 The Pigeonhole Principle

2 Getting Started Are there two leaf-bearing trees on Earth with the same number of leaves if we only consider the number of leaves on a tree at full bloom?

3 Getting Started Are there two leaf-bearing trees on Earth with the same number of leaves if we only consider the number of leaves on a tree at full bloom? There are 37,493 seats available for a night game at Fenway park. Does there have to be a day that at least 100 of those people share as a birthday?

4 Stating the Pigeonhole Principle Theorem If n pigeons fly into k pigeonholes and k < n, some pigeonhole contains at least 2 pigeons.

5 Stating the Pigeonhole Principle Theorem If n pigeons fly into k pigeonholes and k < n, some pigeonhole contains at least 2 pigeons. Proof. Suppose each pigeonhole contains at most one pigeon. Then at most k pigeons have been assigned.

6 Stating the Pigeonhole Principle Theorem If n pigeons fly into k pigeonholes and k < n, some pigeonhole contains at least 2 pigeons. Proof. Suppose each pigeonhole contains at most one pigeon. Then at most k pigeons have been assigned. But since k < n, we have not assigned all of the pigeons to pigeonholes, contradicting that each pigeonhole had at most one pigeon. Therefore there must exist at least one pigeonhole with at least two pigeons.

7 Basic s How many people do we need to have in order to be guaranteed that two were born on the same day of the week?

8 Basic s How many people do we need to have in order to be guaranteed that two were born on the same day of the week? How many cards do you have to select from a standard deck to guarantee that two will be of the same suit?

9 Basic s How many people do we need to have in order to be guaranteed that two were born on the same day of the week? How many cards do you have to select from a standard deck to guarantee that two will be of the same suit? How many people must we have in order to have 3 people with the same birthday?

10 Another Way To Say It Theorem If f is a function from a finite set X to a finite set Y and X > Y, then f (x 1 ) = f (x 2 ) for some x 1, x 2 X, x 1 x 2.

11 Another Way To Say It Theorem If f is a function from a finite set X to a finite set Y and X > Y, then f (x 1 ) = f (x 2 ) for some x 1, x 2 X, x 1 x 2. Theorem If n objects are placed in into k boxes, then there is at least one box containing at least n k objects.

12 On Thanksgiving, do two or more of the consumed turkeys have to have the same weight when rounded to the nearest millionth?

13 On Thanksgiving, do two or more of the consumed turkeys have to have the same weight when rounded to the nearest millionth? What do you need to know to answer this?

14 On Thanksgiving, do two or more of the consumed turkeys have to have the same weight when rounded to the nearest millionth? What do you need to know to answer this? The largest recorded turkey weighed 37 pounds.

15 On Thanksgiving, do two or more of the consumed turkeys have to have the same weight when rounded to the nearest millionth? What do you need to know to answer this? The largest recorded turkey weighed 37 pounds. There are an estimated 46 million turkeys consumed on Thanksgiving.

16 On Thanksgiving, do two or more of the consumed turkeys have to have the same weight when rounded to the nearest millionth? What do you need to know to answer this? The largest recorded turkey weighed 37 pounds. There are an estimated 46 million turkeys consumed on Thanksgiving. There are 37,000,000 possible weights (when taken to the nearest millionth). Since there are more turkeys than possible weights, by the Pigeonhole Principle, there must be at least 2 turkeys with the same weight.

17 Suppose there are 6,000 American students at Salem State and at least one from each of the 50 states. Does there need to be at least 120 from the same state?

18 Suppose there are 6,000 American students at Salem State and at least one from each of the 50 states. Does there need to be at least 120 from the same state? The maximum must be at least the average...

19 Suppose there are 6,000 American students at Salem State and at least one from each of the 50 states. Does there need to be at least 120 from the same state? The maximum must be at least the average = So, yes.

20 Suppose there are 6,000 American students at Salem State and at least one from each of the 50 states. Does there need to be at least 120 from the same state? The maximum must be at least the average = So, yes. To see this is true, consider the alternative...

21 If you pick five numbers from the integers 1 to 8, then two of them must add up to nine.

22 If you pick five numbers from the integers 1 to 8, then two of them must add up to nine. The key is that every number can be paired with another to sum to 9.

23 If you pick five numbers from the integers 1 to 8, then two of them must add up to nine. The key is that every number can be paired with another to sum to 9. Pair up the integers: {1, 8}, {2, 7}, {3, 6}, {4, 5}. Now, select one from each pairing. No matter which 5 th one we select, it will be the second from one of these pairs.

24 Suppose you coach a soccer team. Over a 30 day period, you want to hold 45 practices with at least one per day. Then, there will be a period of consecutive days where there will be exactly 14 practices.

25 Suppose you coach a soccer team. Over a 30 day period, you want to hold 45 practices with at least one per day. Then, there will be a period of consecutive days where there will be exactly 14 practices. Let S i be the number of practices held by day i. Since there is at least one practice per day and a total of 45 practices, we know that 1 S 1 < S 2 <... < S 30 = 45

26 Suppose you coach a soccer team. Over a 30 day period, you want to hold 45 practices with at least one per day. Then, there will be a period of consecutive days where there will be exactly 14 practices. Let S i be the number of practices held by day i. Since there is at least one practice per day and a total of 45 practices, we know that 1 S 1 < S 2 <... < S 30 = 45 We want to show there is some i < j such that S i + 14 = S j.

27 Suppose you coach a soccer team. Over a 30 day period, you want to hold 45 practices with at least one per day. Then, there will be a period of consecutive days where there will be exactly 14 practices. Let S i be the number of practices held by day i. Since there is at least one practice per day and a total of 45 practices, we know that 1 S 1 < S 2 <... < S 30 = 45 We want to show there is some i < j such that S i + 14 = S j. So let s add 14 to each value in the inequality: 15 S < S <... < S = 59

28 So, now we have 60 numbers, S 1, S 2,..., S 30, S , S ,..., S

29 So, now we have 60 numbers, S 1, S 2,..., S 30, S , S ,..., S By the Pigeonhole Principle, two of these must be the same.

30 So, now we have 60 numbers, S 1, S 2,..., S 30, S , S ,..., S By the Pigeonhole Principle, two of these must be the same. Since the first 30 are distinct and the second 30 are likewise distinct, there must be one from the first set and one from the second set that are equal. So, for some i j, S i + 14 = S j.

31 In any cocktail party with two or more people, there must be at least two people who have the same number of friends.

32 In any cocktail party with two or more people, there must be at least two people who have the same number of friends. Suppose there are n people at the party. Then, each person can be friends with 0 to n 1 other people. There are two cases to consider.

33 In any cocktail party with two or more people, there must be at least two people who have the same number of friends. Suppose there are n people at the party. Then, each person can be friends with 0 to n 1 other people. There are two cases to consider. Case 1: Everyone has at least one friend.

34 In any cocktail party with two or more people, there must be at least two people who have the same number of friends. Suppose there are n people at the party. Then, each person can be friends with 0 to n 1 other people. There are two cases to consider. Case 1: Everyone has at least one friend. Then, each of the n people must be categorized with one of 1 to n 1 and so by the Pigeonhole Principle, at least two must have the same number of friends.

35 Case 2: Someone has no friends.

36 Case 2: Someone has no friends. Then the people with the most friends can have at most n 2 friends.

37 Case 2: Someone has no friends. Then the people with the most friends can have at most n 2 friends. So, each of the partygoers must be classified with 0 through n 2, which is n 1 choices. So by the Pigeonhole Principle, at least two must be the same.

38 Case 2: Someone has no friends. Then the people with the most friends can have at most n 2 friends. So, each of the partygoers must be classified with 0 through n 2, which is n 1 choices. So by the Pigeonhole Principle, at least two must be the same. So, in either case, each of the n partygoers can be classified in one of n 1 ways, meaning that there are at least two people with the same number of friends.