pigeon-hole principle ( php )

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Pigeon-Hole Principle( PHP )

A Simple Idea in Combinatorics

ButVery Powerfull and

Usefull

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Ilustration

Theorem

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Let n and k be positif integer, and n > k. Suppose we have to place n identical balls into k identical boxes. Then there will be at least one box in which we place at least two balls.

For warmup we begin with simple problems

without solution

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1.Among three persons, there are two of the same sex.

2.Among 13 persons, there are two born in the same month.

3.Among 400 persons, there are a least two have the same birthday.

Example 1

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In tournament with n players , everybody plays with everybody else exactly once. Prove that during the game there are always two players who have played the same number of game.

Proof

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The difficulties in proving with pigeon-hole principle is the identification of the ball and the box .

Then the question is . . .What is the box?

What is the ball?

Proof

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The number of players is n. And the number of possibilities for the number of game finished by any one player are

0, 1, 2, . . . , n-1.But the fact , however 0 and n-1 cannot both occur among the numbers of games finished by the players at any one time.

Proof

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Then, takeThe number of players ( n ) as balls

andThe number of game finished by one player ( n-1 ) as boxes. By PHP we get, there are two players that have the same number of the game.

Example 2

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(ONMIPA 2010) Let S is a set of n positive integer. Prove that there exist an element in S or sum of elements in S that divisible by n.

Proof

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Suppose the elements in S are{ a1, a2, a3, . . ., an } not necessary distinct

If there is an element in S divisible by n, we are done.Otherwise, none of them is divisible by n.Consider the Set T = { s1, s2, s3, . . ., sn }Where,

sn = a1 + a2 + . . . + an

Proof

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If there is i, 1≤ i ≤ n such that si is divisible by n, we are finished. So, assum all of si is not divisible by n. Then we get, the remainder if si divided by n are 1, 2, 3, . . . , n-1There are n-1 possibilities but as we have n numbers, by PHP there are two numbers that have the same remainder.

Proof

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Without loss of generality, assum these numbers are si and sj where i < j.Then

sj – si = ai + ai+1 + . . . + aj

is divisible by n as desired.

Example 3

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(ONMIPA 2009) We select n+1 different integers from the set { 1, 2, 3, … 2n }. Prove that there always be two among the selected integers such that one is divisible by other.

Hint.

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Write the numbers 1, 2, 3, . . . , 2nin the form

2p.awhere,a is odd number andp ≥ 0, p integer

Then used PHP to prove the statement.

Problem Solving can be learn ONLY by Solving

Problems

Thanks15

REMEMBER

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This presentation is wrote by Tutur Widodo,

In April 2010

Hi…hi…hi…!