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The Pigeonhole Principle. 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.
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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.
Example 2 • If 41 balls are chosen from a set of red, white, blue, garnet, and gold colored balls, then at least 12 red balls, 15 white balls, 4 blue, 10 garnet, or 4 gold balls chosen. • Proof by contradiction: (use DeMorgan’s law) 1. Assume that 41 balls are chosen from this set and that at most 11 red, 14 white, 3 blue, 9 garnet, and 3 gold balls are chosen. 2. At most 40 balls were chosen, a contradiction.
The pigeonhole principle • Let m1, m2, … , mn be positive integers. • If m1 + m2 + . . . + mn - n + 1 objects are put into n boxes, • Then either • the 1st box has at least m1, or • the 2nd box has at least m2, or …, or • the nth box has at least mn objects.
Proof by Contradiction • Assume m1 + m2 + . . . + mn - n + 1 objects are put into n boxes, and • the 1st box has at most m1 - 1, and • the 2nd box has at most m2 - 1, and …, and • the nth box has at most mn - 1 objects. • Then, at most m1 + m2 + . . . + 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: • Assume A is the average number of pigeons/hole. • 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 ) < nA pigeons, a contradiction.
5. Assume every hole contains at least A + 1 pigeons. 6. All holes contain at least n(A + 1) > nA 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 2n + 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 leastk + 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 x1, x2, …, x8 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 a1 < a2 < . . . < a30 45. 3. a1 + 14 < a2 + 14 < . . . < a30 + 14 45 + 14 = 59
4. We have: • 60 integers: a1, a2 , . . . , a30 , a1 +14, a2 +14 , . . . , a30 +14 • 59 holes. 5. 2 of these integers must be the same. 6. They cannot both be in a1, a2 , . . . , a30 . 7. They cannot both be in a1 +14, a2 +14 , . . . , a30 +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.
