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Pigeonhole Principle
Example
Example
Example
Example
Example
Example
Generalized Pigeonhole Principle
Example
Proof of Pigeonhole Principle
Pigeonhole Principle
Sanjay Jain, Lecturer,
School of Computing
Pigeonhole 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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Example
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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Example
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 xB, 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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Example
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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Example
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.
Example
Proof:
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
Example
A= set of students (600)
B= set of grade sheets. (512)
f: A —> B
where f(x)=grade sheet of x.
Since #(A) > #(B), f cannot be 1 — 1
Thus, 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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Example
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
Example
A= 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, or
for 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
Example
A= 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, or
for 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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Example
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.
Example
Suppose 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)=n
Let 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 Ceilings
w denotes the largest integer w.
For example: 6.9 = 6; -9.2 = -10;
w denotes the smallest integer  w.
For example: 6.9 =7;
-9.2 = -9;
9 = 9
9 = 9
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Generalized Pigeonhole Principle
If 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), then
there exists a y in Y such that y is the image of at least k+1
distinct elements of X.
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Example
Suppose I place 26 letters of English alphabet in a circle.
Then there exists a consecutive sequence of 5
consonants.
B1
B5
B4
B2
B3
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 Principle
Recall 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.
END OF SEGMENT