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## Section 4.2

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**Section 4.2**The Pigeonhole Principle**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**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**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**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**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)**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**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)**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)**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**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)**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**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)**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**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**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**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**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**Section 4.2**The Pigeonhole Principle - ends -