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1 Section 4.2 The Pigeonhole Principle

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2 … states that, if there are more pigeons than there are pigeonholes, there must be at least one pigeonhole with 2 pigeons in it More formally: if k+1 objects are placed in k boxes, there is at least one box containing 2 or more objects

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3 Applying the pigeonhole principle to counting problems Among any group of 367 people, at least 2 must have the same birthday In any group of 27 English words, at least 2 must start with the same letter

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4 Example 1 A drawer contains 12 black socks and 12 blue socks. Your professor takes socks out of the drawer at random, in the dark. –How many must she remove to guarantee at least one matched pair? –How many to guarantee at least 2 black socks? –The answers are 3 and 14, respectively

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5 Example 2 During a month with 30 days, a baseball team plays at least 1 game a day, but no more than 45 games in the month. Show that there must be a period of some number of consecutive days during which the team must play exactly 14 games

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6 Example 2 Let a j = number of games played on or before day j (of the 30-day month) Then a 1, a 2, … a 30 is an increasing sequence of distinct positive integers with 1<=a j <=45 (sequence 1) And a 1 +14, a 2 +14, a 3 +14 and so forth is also an increasing sequence of positive integers, with 15<=a j +14<=59 (sequence 2)

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7 Example 2 The 60 integers from a 1, a 2, …, a 30 and from a 1 +14, a 2 +14, …, a 30 +14 are all less than or equal to 59 (since a 30 <= 45) Hence, by the pigeonhole principle, at least 2 of these integers are equal (59 pigeonholes, 60 “pigeons” Since the integers in the 2 sequences are all distinct, there must be indices i and j with a i =a j +14 - so there are exactly 14 games played from day j+1 to day i

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8 Example 3 Show that among n+1 positive integers less than or equal to 2n, there must be an integer that divides one of the other integers The n+1 integers are: a 1, a 2, … an, a n+1 Each of these integers can be written as a power of 2 times an odd integer (odd*2 is even; odd*2 0 is odd)

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9 Example 3 So, let a j = 2 k j q j for: –j=1, 2, …, n+1 –k j is a non-negative integer –q j is an odd integer The integers q 1, q 2, … q n+1 are all odd positive integers <= 2n (this must be true if a j is a product of q j and some power of 2)

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10 Example 3 Since there are only n odd positive integers less than 2n, the pigeonhole principle holds that 2 of the q’s must be equal - in other words, there are two integers i and j such that q i = q j If we represent this common value as q, then a i = 2 k i q and a j = 2 k j q It follows then that if k i k j, a j divides a i

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11 Pigeonholes & subsequences Can use the pigeonhole principle to show the existence of a subsequence of a certain length within a sequence of distinct integers For a sequence of the form: a 1, a 2, …, a N a subsequence is a sequence of the form: a i 1,a i 2, …,a i m where 1<=i 1 < i 2 < … < i m <=N So if we have this sequence: 2, 4, 6, 8; some subsequences are: 2,4; 4,6,8; 2,8 (we include terms in original order, but can skip terms)

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12 Subsequences A subsequence is strictly increasing if each term is larger than the preceding term, and strictly decreasing if each is smaller than its predecessor There is a theorem which states: every sequence of n 2 +1 distinct real numbers contains a subsequence of length n+1 that is either strictly increasing or strictly decreasing

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13 Example 4 For example, the sequence: 11, 2, 23, 14, 15, 81, 90, 4, 3, 5 has length 10, so n=3 (10 = 3 2 +1) According to the theorem, there is a subsequence of length 4 that is either strictly increasing or decreasing There are several examples of both, including: 11, 23, 81, 90 (increasing) 23, 15, 4, 3 (decreasing)

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14 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 of application of the principle: –Among 100 people there are at least 100/12 , or 9 people who share a birth month –Minimum number of students needed to guarantee that at least 3 receive the same grade (A,B,C,D or F) would be the smallest number N such that N/5 = 3; that would be 11

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15 Example 5 What is the least number of area codes needed to guarantee that 25 million phones in a state have distinct 10-digit numbers under the NXX-NXX-XXXX scheme? –Where N = 2..9 –And X = 0..9

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16 Example 5 There are 8*10*10*10*10*10*10=8,000,000 different 7-digit numbers of the form NXX- XXXX So, for 25,000,000 telephones, at least 25,000,000/8,000,000 = 4 will have identical 7-digit numbers - therefore, 4 area codes are needed

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17 Generalized pigeonhole principle & Ramsey Theory - example 6 Ramsey theory: in a group of 6 people, in which each pair consists of 2 friends or 2 enemies, there must be 3 mutual friends or 3 mutual enemies in the group (assuming anyone who is not a friend is an enemy) We can use the generalized pigeonhole principle to prove this theory

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18 Example 6 Let A be one of the 6 - of the other 5, 3 or more are either friends or enemies of A (because by the generalized pigeonhole principle, when 5 objects are divided into 2 sets, one set has at least 5/2 = 3 elements) Suppose that B, C and D are friends of A –If any 2 of these 3 are friends, then that pair + A make 3 mutual friends –If they are not friends, then they are mutual enemies - so theory is proven

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19 Section 4.2 The Pigeonhole Principle - ends -

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