Functions: Part 3

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Functions Part 3

• Section 7.3
• The Pigeonhole Principle

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The Pigeonhole Principle

• If n pigeons fly into m pigeonholes and ngtm, then
  at least one pigeonhole must contain two or more
  pigeons

? It can not be one-to-one!
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Example (1)

• In a group of 6 people, must there be at least
  two who were born in the same month?

No! people lt months

• In a group of 13 people?

Yes! People gt months
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Example (2)

• Let A 1, 2, 3, 4, 5, 6, 7, 8
• A) If 5 integers are selected from A, must at
  least one pair of integers have a sum of 9?
• B) If 4 integers are selected?
• The set A can be partitioned into 4 subsets based
  on sums that equal 9
• 1,8 2, 7 3, 6 4,5
• If 5 integers are selected from A, then by the
  Pigeonhole Principle at least two must be from
  the same subset

Ints
Sums
A) Ints gt Sums, so Yes!
B) Ints Sums, so No!
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Generalized Pigeonhole Principle

• From a finite set X to a finite set Y, if N(X) gt
  kN(Y), then there is some y?Y that is the image
  of at least k1 distinct elements of X

1 2 3 7 8 9
N(X) 9 N(Y) 4
1 2 3 4
Using the definition above N(X) gt k N(Y) 9 gt 24
Pigeons
Pigeonholes
So ?y?Y that is the image of at least k1
elements in X or there is a y?Y that is the image
of at least 3 elements in X
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GPHP Example

• Out of 85 people, how many people have the same
  last initial?

People 85
Initials 26
N(X) 85 N(Y) 26
Using the definition above N(X) gt k N(Y) 85 gt
326 or 85 gt 78

• y is the image of at least k1 elements of X or
  at least 4 people have the same last initial

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Related Theorems

• Contrapositive
• For any function f from X ? Y, if for each y?Y,
  f-1(y) has at most k elements, then X has at most
  kN(Y) elements (Proof in book)
• Theorem 7.3.1
• For any function f from a finite set X to a
  finite set Y, if N(X) gt N(Y), then f is not
  one-to-one
• Theorem 7.3.2
• Let X and Y be finite sets with the same number
  of elements and suppose f is a function from X to
  Y. Then f is one-to-one if and only if f is onto