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Unformatted text preview: CS 373: Theory of Computation Gul Agha Mahesh Viswanathan Fall 2010 1 1 Expressiveness 1.1 Finite Languages Finite Languages Definition 1. A language is finite if it has finitely many strings. Example 2 . { , 1 , 00 , 10 } is a finite language, however, (00 11) * is not. Finiteness and Regularity Proposition 3. If L is finite then L is regular. Proof. Let L = { w 1 ,w 2 ,...w n } . Then R = w 1 w 2 w n is a regular expression defining L . 1.2 NonRegular Languages Are all languages regular? Proposition 4. The language L eq = { w { , 1 } *  w has an equal number of 0s and 1s } is not regular. Proof? No DFA has enough states to keep track of the number of 0s and 1s it might see. Above is a weak argument because E = { w { , 1 } *  w has an equal number of 01 and 10 substrings } is regular! Proving NonRegularity Proposition 5. The language L eq = { w { , 1 } *  w has an equal number of 0s and 1s } is not regular. Proof. Suppose (for contradiction) L eq is recognized by DFA M = ( Q, { , 1 } ,,q ,F ), where  Q  = n . There must be j < k n such that ( q , j ) = ( q , k ) (= q say). Let x = 0 j , y = 0 k j , and z = 0 n k 1 n ; so xyz = 0 n 1 n . Proving NonRegularity 2 Proof (contd). q q x = 0 j y = 0 k j q q q x = 0 j y = 0 k j z = 0 n k 1 n We have ( q , j ) = ( q , k ) = q Since 0 n 1 n L eq , ( q , n 1 n ) F . ( q , n 1 n ) = ( ( q , k ) , n k 1 n ) (since ( q,uv ) = ( ( q,u ) ,v )) = ( ( q , j ) , n k 1 n ) ( ( q , j ) = ( q , k )) = ( q , n k + j 1 n ) (since ( q,uv ) = ( ( q,u ) ,v )) So M accepts 0 n k + j 1 n as well. But, 0 n k + j 1 n 6 L eq ! 2 Pumping Lemma 2.1 Statement and Proof Pumping Lemma: Overview Pumping Lemma The lemma generalizes this argument. Gives the template of an argument that can be used to easily prove that many languages are nonregular....
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This note was uploaded on 10/04/2011 for the course CS 373 taught by Professor Viswanathan,m during the Fall '08 term at University of Illinois, Urbana Champaign.
 Fall '08
 Viswanathan,M

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