HW8Soln

HW8Soln - CSCI 2670 HW 8 11/2/05 Solutions 1. Show T =...

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CSCI 2670 HW 8 11/2/05 Solutions 1. Show T = {<M> | M is a TM with L(M) R = L(M)} is an undecidable language. Proof: Assume T is decidable and let D be a decider for T. Consider the following Turing machine S: S = “On input <M,w>, where M is a Turing machine 1. Create the following Turing machine M 1 M 1 = “On input x 1. If x = 0 n 1 n for some n, accept 2. If M accepts w, accept. 2. Run D on input <M 1 > 3. If D accepts, accept 4. If D rejects, reject S is a decider since step 1 creates a Turing machine and moves on, step 2 calls a decider, and steps 3 and 4 halt. Note that if M accepts w, then L(M 1 ) = Σ*, which is in T. If M does not accept w, then L(M 1 ) = 0 n 1 n , which is not in T. Therefore, D accepts <M 1 > iff M accepts w, and reject otherwise. By construction, S accepts if D accepts (i.e., if M accepts w) and rejects if D rejects (i.e., if M does not accept w). Therefore, S decides A
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HW8Soln - CSCI 2670 HW 8 11/2/05 Solutions 1. Show T =...

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