pigeon-hole principle ( php )
DESCRIPTION
Pigeon-Hole Principle ( PHP ). A Simple Idea in Combinatorics But Very Powerfull and Usefull. Ilustration. 1. 2. 3. 4. 5. Theorem. - PowerPoint PPT PresentationTRANSCRIPT
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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
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