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The Pigeonhole Principle Rosen 4.2 Pigeonhole Principle If k+1 or more objects are placed into k boxes, then there is at least one box containing two.

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Presentation on theme: "The Pigeonhole Principle Rosen 4.2 Pigeonhole Principle If k+1 or more objects are placed into k boxes, then there is at least one box containing two."— Presentation transcript:

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2 The Pigeonhole Principle Rosen 4.2

3 Pigeonhole Principle If k+1 or more objects are placed into k boxes, then there is at least one box containing two or more objects.

4 Generalized Pigeonhole Principle If N objects are placed into k boxes, then there is at least one box containing at least  N/k  objects Examples –Among any 100 people there must be at least  100/12  = 9 who were born in the same month. –What is the minimum number of students needed in a class to be sure that at least 6 to get the same grade? (5 choices for grades:A,B,C,D,F) Smallest integer N such that  N/5  = 6, 5*5+1 = 26

5 Example What’s the minimum number of students, each of whom comes from one of the 50 states must be enrolled in a university to guarantee that there are at least 100 who come from the same state? 50*99 + 1 = 4951  4951/50  = 100

6 There are 38 different time periods during which classes at a university can be scheduled. If there are 677 different classes, how many different rooms will be needed?  677/38  = 18 A computer network consists of six computers. Each computer is directly connected to zero or more of the other computers. Show that there are at least two computers that are directly connected to the same number of other computers. Solution: Each computer can be directly connected to 0,1,2,3,4,5. But there are really only five choices, not six, since if one computer is connected to zero other computers, then no computer can be connected to five others. Six computers, 5 choices. Pigeonhole principle says that at least two must have the same number of direct connections.

7 Let (x i, y i, z i ), i = 1,2,3,..,9 be a set of nine distinct points with integer coordinates in xyz space. Show that the midpoint of at least one pair of these points has integer coordinates. For points (x j, y j, z j ) and (x k, y k, z k ) we compute the midpoint by ((x i +x j )/2, (y i +y j )/2, (z i +z j )/2 ). (1,1,2), (1,2,2), (3,2,7), (10,5,8), (3,1,4), (3,7,2), (2,1,1), (1,2,1), (0,0,0) The midpoint between (1,1,2) and (3,1,4) = (2,1,3) Remember from number theory that when we add an odd number to an odd number, or an even number to an even number, we get an even number. So the question becomes, does there exist two sets of coordinates that have the same parity (i.e., their odd/even order is the same)? From the product rule there are 2*2*2 = 8 possible parities. There are nine points, so by the pigeonhole principle two of them must be the same. Therefore at least one midpoint must have integer coordinates.

8 During a month with 30 days a baseball team plays at least 1 game a day, but no more than 45 games. Show that there must be a period of some number of consecutive days during which the team must play exactly 14 games. Proof: Let a j be the number of games played on or before the j th day of the month. Then a 1, a 2, …, a 30 must be an increasing sequence of distinct positive integers, with 1  a j  45. Day of MonthGames Played 1a11a1 2a22a2 3a33a3… 30a 30

9 Moreover, a 1 +14, a 2 +14,..., a 30 +14 is also an increasing sequence of distinct positive integers with 15  a j + 14  59. Together the two sequences, each containing 30 integers, contain 60 positive integers, all of which are less than or equal to 59. By the pigeonhole principle, at least two of these integers are equal. Since the integers a j, j = 1 to 30, are all distinct and the integers a j +14, j = 1 to 30 are all distinct, there must be indices i and j with a i = a j +14. This means that exactly 14 games were played from day j+1 to day i.

10 Some Definitions Suppose that a 1,a 2, … a n is a sequence of real numbers. A subsequence of this sequence is a sequence of the form a i 1, a i 2, …, a i m, where 1  i 1 < i 2 <... < i m  n A sequence is called strictly increasing if each term is larger than the term that precedes it. A sequence is called strictly decreasing if each term is smaller than the one that precedes it. –Example: {1, 5, 6, 2, 3, 9} is a sequence. –{5,6,9} is a subsequence that is strictly increasing

11 Theorem: Every sequence of n 2 +1 distinct real numbers contains a subsequence of (at least) length n+1 that is strictly increasing or strictly decreasing. Example: 8, 11, 9, 1, 4, 6, 12, 10, 5, 7 10 = 3 2 +1 terms so must be a subsequence of length 4 that is either strictly increasing or strictly decreasing. 1,4,6,12 1,4,6,7 11,9,6,5 ……

12 Theorem: Every sequence of n 2 +1 distinct real numbers contains a subsequence of at least length n+1 that is strictly increasing or strictly decreasing. Let a 1, a 2, …, a n 2 +1 be a sequence of n 2 +1 distinct numbers. Associate an ordered pair (i k,d k ) with each term of the sequence where i k is the length of the longest increasing subsequence starting at a k and d k is length of the longest decreasing subsequence starting at a k. Example: 8, 11, 9, 1, 4, 6, 12, 10, 5, 7 a 2 = 11, (2,4) a 4 = 1, (4,1) Proof by contradiction: Now suppose that there are no increasing or decreasing subsequences of length n+1 or greater. Then i k and d k are both positive integers  n, for k=1 to n 2 +1.

13 By the product rule, there are n 2 possible ordered pairs for (i k,d k ). Why? Because each has the range from 1 to n. By the pigeonhole principle, since we have n 2 +1 ordered pairs (one for each element in the sequence) two of them must be identical. Formally  terms a s and a t in the sequence, with s a t, it can be shown that d s must be greater than d t, which is also a contradiction.


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