The Pigeonhole Principle Pigeonhole principle The pigeonhole principle

The Pigeonhole Principle

Pigeonhole principle The pigeonhole principle : If k is a positive integer and k+1 or more objects are placed into k boxes, then there is at least one box containing two or more of the objects. A common way to illustrate this principle is by assuming that k+1 pigeons fly to k pigeonholes. Then, there must be at least one pigeonhole containing at least two pigeons. The pigeonhole principle is a trivial observation, which however can be used to prove many results.

Example 1 In a room of 13 people, 2 or more people have their birthday in the same month. Proof: (by contradiction) 1. Assume the room has 13 people and no 2 people have their birthday in the same month. 2. There must be at least 13 different months. 3. Statement 2. is false, so the assumption is false.

The pigeonhole principle • Let m 1, m 2, … , mn be positive integers. • If m 1 + m 2 +. . . + mn - n + 1 objects are put into n boxes, • Then either • the 1 st box has at least m 1, or • the 2 nd box has at least m 2, or …, or • the nth box has at least mn objects.

Proof by Contradiction • Assume m 1 + m 2 +. . . + mn - n + 1 objects are put into n boxes, and • the 1 st box has at most m 1 - 1, and • the 2 nd box has at most m 2 - 1, and …, and • the nth box has at most mn - 1 objects. • Then, at most m 1 + m 2 +. . . + mn - n objects are in the boxes, a contradiction.

Another Form of Pigeonhole Principle If A is the average number of pigeons/hole, then some hole contains at least A pigeons and some hole contains at most A pigeons.

Intuition A A A Cannot have all holes contain less than the average. Cannot have all holes contain more than the average.

Proof of Alternate Principle By contradiction: 1. Assume A is the average number of pigeons/hole. 2. Assume every hole contains at most A - 1 pigeons or every hole contains at least A + 1 pigeons. 3. Let n denote the number of holes. 4. Assume every hole contains at most A - 1 pigeons. 5. All holes contain at most n( A - 1 ) < n. A pigeons, a contradiction.

5. Assume every hole contains at least A + 1 pigeons. 6. All holes contain at least n( A + 1) > n. A pigeons, a contradiction. 7. Therefore, some hole contains at least A pigeons and some hole contains at most A pigeons.

Applications of pigeonhole principle • If n + 1 pigeons are distributed among n holes, then some hole contains at least 2 pigeons. • If 2 n + 1 pigeons are distributed among n holes, then some hole contains at least 3 pigeons. • If kn + 1 pigeons are distributed among n holes, then some hole contains at least k + 1 pigeons. The average number of pigeons/hole = k + 1/n and k + 1/n = k + 1.

Applications. . . • In any group of 367 people, there must be at least 1 pair with the same birthday. • If 4 different pairs of socks are scrambled in the drawer, only 5 socks need to be selected to guarantee finding a matching pair. • In a group of 61 people, at least 6 were born in the same month.

Applications. . . • If 401 letters were delivered to 50 houses, then some house received at most 8 letters. • If x 1, x 2, …, x 8 are distinct integers, then some pair of these have the same remainder when divided by 7.

Applications. . . • Given a set of 7 distinct integers, there are 2 whose sum or difference is divisible by 10. • Set this up so that there are 6 pigeon holes. • Partitioning the integers into equivalence classes according to their remainder when divided by 10 yields too many classes. • Consider: • {[0]}, {[1], [9]}, {[2], [8]}, {[3], [7]}, {[4], [6]}, {[5]}. • If 2 integers are in the same set either their difference is divisible by 10 or their sum is divisible by 10.

Applications. . . • Suppose • 50 chairs are arranged in a rectangular array of 5 rows and 10 columns. • 41 students are seated randomly in the chairs (1 student/chair). • Then, • • some row contains at least 9 students some row contains at most 8 students some column contains at least 5 students some column contains at most 4 students.

Applications. . . • A patient has 45 pills, with instructions to take at least 1 pill/day for 30 days. • Prove: there is a period of consecutive days in which the patient takes a total of 14 pills. 1. Let ai be the number of pills taken through the end of the ith day. 2. 1 a 1 < a 2 <. . . < a 30 45. 3. a 1 + 14 < a 2 + 14 <. . . < a 30 + 14 45 + 14 = 59

4. We have: • 60 integers: a 1, a 2 , . . . , a 30 , a 1 +14, a 2 +14 , . . . , a 30 +14 • 59 holes. 5. 2 of these integers must be the same. 6. They cannot both be in a 1, a 2 , . . . , a 30. 7. They cannot both be in a 1 +14, a 2 +14 , . . . , a 30 +14. 8. One is in each: ai = aj + 14, for some i and j. 9. For that i and j, ai - aj = 14. 10. That is, aj+1+aj+2 +. . . + ai = 14.