The%20Pigeonhole%20Principle

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The Pigeonhole Principle
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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.

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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 DeMorgans 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.

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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.

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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.

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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.

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Intuition
?A?
A
?A?
Cannot have all holes contain less than the
average. Cannot have all holes contain more than
the average.
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Proof of Alternate Principle

• By contradiction
• Assume A is the average number of pigeons/hole.
• 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 ) lt nA
  pigeons, a contradiction.

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• 5. Assume every hole contains at least ?A? 1
  pigeons.
• 6. All holes contain at least n(?A? 1) gt nA
  pigeons, a contradiction.
• 7. Therefore, some hole contains at least ?A?
  pigeons and some hole contains at most ?A?
  pigeons.

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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.

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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.

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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.

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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.

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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.

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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 lt a2 lt . . . lt a30 ? 45.
• 3. a1 14 lt a2 14 lt . . . lt a30 14 ? 45 14
  59

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• 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, aj1aj2 . . . ai 14.