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Unformatted text preview: Lecture 5: Pumping Lemma and Closure Properties David Dill Department of Computer Science 1 Outline Pumping Lemma Closure properties. 2 Pumping Lemma Primary application: Proving that certain languages are not regular. Lemma: For every regular language L , there exists a positive constant n such that for every string w in L such that  w  n , there exist strings x,y,z such that w = xyz with the following properties: y negationslash =  xy  n xy k z L for all k . 3 Understanding the Pumping Lemma The pumping lemma is a little difficult to understand because of all the alternating quantifiers ( L n w x,y,z k ... ) With such theorems, it is often helpful to think of a game with two players. Player is trying to make the theorem true, player is trying to falsify it. For applications of the Pumping Lemma, its most useful to treat L as the game board, which is given, and take the game from there. Player : Chooses the constant n . player : Chooses w L such that  w  n . Player : Chooses xyz = w such that y negationslash = and  xy  n . player : Chooses k . player wins if xy k z negationslash L . The pumping lemma says: If L is regular, Player has a winning strategy. (Important note: If L is not regular, Player may still have a winning strategy. This version of the Pumping Lemma is only necessary for regularity, not sufficient.) 4 Example Application of Pumping Lemma Theorem: L = All strings over = { , 1 } with the same number of 0s and 1s is not regular. Proof: We prove by contradiction, using the PL. Suppose L were regular. Let n be the pumping constant of the PL. [This is Player s move. player has no idea what n is, but gives a general rule for the next move for any n .] Consider the string w = 0 n 1 n L . [This is player s move, which works for any n .] By the PL, there must be xyz = 0 n 1 n , such that  xy  n and y negationslash = , and xy k z L for all k . [Player s move.] So xy will consist of all s and y will be a string of at least one . [We dont know what xyz are, but we know these properties hold if Player made a legal move.] If k negationslash = 1 , then xy k z will have a different number of s than xyz and the same number of 1...
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This note was uploaded on 01/12/2010 for the course CS 154 at Stanford.
 '08
 Motwani,R
 Computer Science

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