Transcript The Pigeonhole Principle

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.
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:
1. Assume A is the average number of pigeons/hole.
2. 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 least k + 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.