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Unformatted text preview: CS 373: Theory of Computation Gul Agha Mahesh Viswanathan Fall 2010 1 1 Closure Properties 1.1 Decidable Languages Boolean Operators Proposition 1. Decidable languages are closed under union, intersection, and complementation. Proof. Given TMs M 1 , M 2 that decide languages L 1 , and L 2 • A TM that decides L 1 ∪ L 2 : on input x , run M 1 and M 2 on x , and accept iff either accepts. (Similarly for intersection.) • A TM that decides L 1 : On input x , run M 1 on x , and accept if M 1 rejects, and reject if M 1 accepts. Regular Operators Proposition 2. Decidable languages are closed under concatenation and Kleene Closure. Proof. Given TMs M 1 and M 2 that decide languages L 1 and L 2 . • A TM to decide L 1 L 2 : On input x , for each of the  x  + 1 ways to divide x as yz : run M 1 on y and M 2 on z , and accept if both accept. Else reject. • A TM to decide L * 1 : On input x , if x = accept. Else, for each of the 2  x  1 ways to divide x as w 1 ...w k ( w i 6 = ): run M 1 on each w i and accept if M 1 accepts all. Else reject. Inverse Homomorphisms Proposition 3. Decidable languages are closed under inverse homomorphisms. Proof. Given TM M 1 that decides L 1 , a TM to decide h 1 ( L 1 ) is: On input x , compute h ( x ) and run M 1 on h ( x ); accept iff M 1 accepts. Homomorphisms Proposition 4. Decidable languages are not closed under homomorphism Proof. We will show a decidable language L and a homomorphism h such that h ( L ) is undecidable • Let L = { xy  x ∈ { , 1 } * ,y ∈ { a,b } * ,x = h M,w i , and y encodes an integer n such that the TM M on input w will halt in n steps } • L is decidable: can simply simulate M on input w for n steps • Consider homomorphism h : h (0) = 0, h (1) = 1, h ( a ) = h ( b ) = . • h ( L ) = HALT which is undecidable....
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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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