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Pigeonhole Principle

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Generalized Pigeonhole PrincipleExample

Proof of Pigeonhole Principle

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Pigeonhole PrinciplePigeonhole Principle

Sanjay Jain, Lecturer, Sanjay Jain, Lecturer,

School of ComputingSchool of Computing

Pigeonhole PrinciplePigeonhole Principle

If we place m balls in n boxes, m > n, then at least one box gets 2 balls.

Another formulation:

Consider a function f from A to B, where

#(A) > #(B), and

A, B are finite.Then f cannot be 1—1.

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ExampleExample

In a group of 13 people, there are at least 2 who are born in the same month.

S = set of people (13 elements)M = months of the year (12 elements)

f: S —> M

f(x)= the month x was born.Since #(S) > #(M), by pigeonhole principle, f cannot be

1—1.That is, there exist x, y in S, x y, such that

f(x)=f(y) In other words, x and y are born in the same month.

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ExampleExample

Let A={1,2,3,…,8}

If we pick 5 integers from A, then there exist 2 integers among the selected integers, which add up to 9.

Proof:

A1={1,8}; A2={2,7}; A3={3,6}; A4={4,5}B: set of numbers picked.C={1,2,3,4}f: B --> C,

if xB, is a member of Ai then f(x)=i.

Since #(B) > #(C), by pigeonhole principle, f cannot be 1 —1.Thus, there exist a, b, in B such that f(a) = f(b).But then a+b = 9

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ExampleExample

In a set of four numbers, two are same mod 3.

Proof:A = Set of 4 numbers.B={0,1,2}f: A —> B

where f(x)=x mod 3.Now #(A) > #(B). So f cannot be 1 —1.Thus, there exists distinct x, y in A, such that f(x) = f(y).In other words, x mod 3 = y mod 3.

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ExampleExample

There are 600 students. Each of them takes 3 compulsory modules, and 1 of 2 electives.

In each module they can get a grade of A, B, C, or D. Show that there are 2 students with identical grade sheet.

ExampleExampleProof:

A= set of students (600)

B= set of grade sheets.

How many grade sheets are there?T1: select the elective.

T2: select the grade for module 1

T3: select the grade for module 2

T4: select the grade for module 3

T5: select the grade for module 4

T1 can be done in two ways. Each of T2 to T5 can be done in four ways.

Using multiplication rule, the number of elements of B = 2*4*4*4*4 = 512

ExampleExampleA= set of students (600)B= set of grade sheets. (512)

f: A —> Bwhere f(x)=grade sheet of x.

Since #(A) > #(B), f cannot be 1 — 1Thus, there exist distinct x, y in A such that f(x) = f(y)In other words, x and y have the same grade sheets.

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ExampleExample

Suppose there are 19 people in a party. Suppose friendship relation is mutual.Show that there are 2 persons in the party with same number of friends

ExampleExampleA= set of people in the party. B={0,1,2…,18}f: A —> B

where f(x) = number of friends of x#(A) = 19, #(B) = 19.We cannot apply pigeonhole principle yet.Note that, either

for all x, f(x)0, orfor all x, f(x)18

Why? Suppose for all x, f(x) 0 is false.Then there is an a such that f(a)=0.Thus a has no friends.Then, a is not a friend of anyone.Thus, for all x, f(x)18

ExampleExampleA= set of people in the party. B={0,1,2…,18}f: A —> B

where f(x) = number of friends of x#(A) = 19, #(B) = 19.We cannot apply pigeonhole principle yet.Note that, either

for all x, f(x)0, orfor all x, f(x)18

Let B’=B - {w}, where w = 0 or 18, based on which of above cases holds.Note that #(B’)=18.Thus f: A—> B’ cannot be 1—1.Therefore, there exist distinct x and y in A, such that f(x)=f(y).In other words, there are two people in A, with the same number of friends.

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ExampleExample

Let X={1,2,….,2n}Let S be a subset of X containing n+1 elements.Then, there exist distinct x and y in S, such that x divides y.

ExampleExampleSuppose the elements of S are s1, s2, …,sn+1

Let si=2riwi, where wi is odd.Let B= set of odd numbers 2n. Note that #(B)=nLet f:S—> B

where f(si)=wi

Thus, f cannot be1— 1 (by PH principle)Thus, there exist distinct i and j, such that

f(si)= f(sj)

In other words, wi= wj.

si=2riwi, sj=2rjwj

Now if, ri>rj then, sj divides si

otherwise, si divides sj.

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Floors and CeilingsFloors and Ceilings

w denotes the largest integer w.For example: 6.9 = 6; -9.2 = -10; 9 = 9

w denotes the smallest integer w.For example: 6.9 =7; -9.2 = -9; 9 = 9

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Generalized Pigeonhole PrincipleGeneralized Pigeonhole PrincipleIf I place m balls in n boxes, then at least one box will get m/n balls.

If I place m balls in n boxes, then at least one box will get m/n balls.

For any function f from a finite set X to a finite set Y, if #(X) > k* #(Y), thenthere exists a y in Y such that y is the image of at least k+1 distinct elements of X.

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ExampleExampleSuppose I place 26 letters of English alphabet in a circle. Then there exists a consecutive sequence of 5 consonants.

B1

B4

B5

B3

B2

21 consonants ---> balls.B1,…, B5 boxes.At least one box will get 21/5 =5 balls. Thus there are 5 consecutive consonants.

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Proof of Pigeonhole PrincipleProof of Pigeonhole PrincipleRecall the pigeonhole principle:For any function f from a finite set X to a finite set Y, if #(X) > #(Y), then f is not 1—1.Proof: Suppose f is 1—1.

Let Y={y1, y2,…, ym}f -1(y)={x X | f(x)=y}

f -1(y1), f -1(y2), …,f -1(ym) are pairwise disjoint, whose union is X.Therefore by Addition Rule,

#(X) = #(f -1(y1)) + #(f -1(y2)) + … + #(f -1(ym) ) 1+ 1+……+1 = m#(X) #(Y)a contradiction. Thus f is not 1—1.

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