The Pigeonhole Principle

1
Section 7.4

• The Pigeonhole Principle

2
A Simple Idea

• If n pigeons fly into m pigeonholes and n gt m,
  then some pigeonhole must have two or more
  pigeons inside.

3
Pigeonhole Principle (formally)

• A function from one finite set to a smaller
  finite set cannot be one-to-one. There must be
  at least two elements in the domain that have the
  same image in the co-domain.

4
Application of this Rule

• There are 78 people in this class. Must there be
  two whose birthdays fall in the same month?
• YES there are twelve months in a year, and
  more people than months.
• In the same week?
• YES there are 52 weeks in a year, and more
  people than weeks.
• On the same day?
• NO there are more days in the year than people
  in the class.

5
Finding the Number to Pick to Ensure a Result

• Suppose six similar-looking pairs of boots are
  together in a big pile in a dark room. How many
  boots must you pick to be sure of getting a
  matched pair?
• Answer You must pick seven boots to be sure of
  getting a matched pair.
• This ensures more boots than pairs, at which
  point the pigeonhole principle tells us two boots
  must be from the same pair.

6
Application to Finite-State Automata

• It seems tempting to say that we can solve all
  our problems with computers, bt that is simply
  not true.
• Our finite-state automata model is not as
  sophisticated as a real computer, but even now we
  can show there is a language not accepted by any
  finite-state automaton.

7
Finite vs. Infinite Sets

• A set is called finite if, and only if, it is the
  empty set or there is a one-to-one
  correspondence from 1, 2, , n to it, where n
  is a positive integer.
• In the first case, the number of elements in the
  set is said to be 0, and in the second case it is
  said to be n.
• A set that is not finite is called infinite.

8
Theorem Let L be the language over the alphabet
? a, b defined by L s??s akbk where k
? Z.There is no finite-state automaton that
accepts L.

• Proof By contradiction.
• Suppose that there is a finite-state automaton A
  that accepts L. Since A has only a finite number
  of states, they can be denoted s1, s2, , sn,
  where n is a positive integer.
• Consider all input strings that consist entirely
  of as a, a2, a3, .
• There are infinitely many such strings and only
  finitely many states, so by the pigeonhole
  principle, there is a state sm and strings a j
  and ak such with j ? k such that when either a j
  or ak is input, A goes to state sm.

9
Theorem Let L be the language over the alphabet
? a, b defined by L s??s akbk where k
? Z.There is no finite-state automaton that
accepts L.

• By our definition of A, A accepts L, so it
  accepts both the strings a jb j and akbk. Thus,
  input of either j or k bs upon encountering
  state sm will lead to an accept state. As
  acceptance of L means that if a string is not in
  L, it will not be accepted.
• But this means that A accepts a jbk, which is not
  in L because j ? k. This is a contradiction.
• Therefore, our supposition is false, so there is
  no finite-state automaton that accepts L.

10
Theorem 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.

• Proof Suppose f is a function from X to Y, where
  X and Y are finite sets each with k elements.
• If f is one-to-one, then f is onto.
• Suppose f is one-to-one. Then f (x1), f (x2), f
  (x3), , f (xk) are all different.
• Consider the set S of elements of Y that are not
  the image of any element of X.
• The sets f (x1), f (x2), , f (xk), and S
  are mutually disjoint. Therefore,
• n(Y) n(f (x1)) n(f (x2)) n(f
  (xk)) n(S)

11
Theorem 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.

• n(Y) n(f (x1)) n(f (x2)) n(f
  (xk)) n(S)
• 1 1 1 n(S)
• k n(S)
• But n(Y) k, so n(S) k k 0. Thus S is
  empty, so there is no element of Y that is not
  the image of some element of X.
• Therefore, f is onto.

12
Theorem 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.

• If f is onto, then f is one-to-one.
• Suppose that f is onto. Then f -1(yi ) ? ? and
  thus n(f -1(yi )) ? 1 for all i 1, 2, , k.
• But X must be the union of the mutually disjoint
  sets f -1(y1), f -1(y2), , f -1(yk). Therefore,
• n(X) n(f -1(y1)) n(f -1(y2)) n(f
  -1(yk))
• n(X) ? k.
• Now if any of the sets f -1(yi) has more than one
  element, n(X) gt k. But we know this is not the
  case because n(X) k.
• Therefore, each set f -1(yi) has exactly one
  element, so f is one-to-one.

13
Wednesdays class

• Turn in homework.
• New homework.
• Section 7.6 countability and uncountability