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Section 4.2

The Pigeonhole Principle

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The Pigeonhole Principle

• … 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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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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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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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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Example 2

• Let aj = number of games played on or before day j (of the 30-day month)

• Then a1, a2, … a30 is an increasing sequence of distinct positive integers with 1<=aj<=45 (sequence 1)

• And a1+14, a2+14, a3+14 and so forth is also an increasing sequence of positive integers, with 15<=aj+14<=59 (sequence 2)

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Example 2

• The 60 integers from a1, a2, … , a30 and from a1+14, a2+14, … , a30+14 are all less than or equal to 59 (since a30 <= 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 ai=aj+14 - so there are exactly 14 games played from day j+1 to day i

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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: a1, a2, … an, an+1

• Each of these integers can be written as a power of 2 times an odd integer (odd*2 is even; odd*20 is odd)

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Example 3

• So, let aj = 2kjqj for:

– j=1, 2, …, n+1

– kj is a non-negative integer

– qj is an odd integer

• The integers q1, q2, … qn+1 are all odd positive integers <= 2n (this must be true if aj is a product of qj and some power of 2)

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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 qi = qj

• If we represent this common value as q, then ai = 2kiq and aj = 2kjq

• It follows then that if ki<kj, ai divides aj, and if ki>kj, aj divides ai

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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: a1, a2, … , aN a subsequence is a sequence of the form: ai1

,ai2, … ,aim

where 1<=i1<

i2< … < im<=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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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 n2+1 distinct real numbers contains a subsequence of length n+1 that is either strictly increasing or strictly decreasing

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Example 4

• For example, the sequence:11, 2, 23, 14, 15, 81, 90, 4, 3, 5

has length 10, so n=3 (10 = 32+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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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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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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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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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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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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Section 4.2

The Pigeonhole Principle

- ends -