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Unformatted text preview: Week 3: Some undecidable 1. A TM = {<M,w>  M is a TM and M accepts w} is undecidable. 2. A TM is Turing recognizable (recursively enumerable). Diagonalization Method Countable, uncountable Z is countable, Z × Z is countable. Theorem : A is a subset of B, and B is countable, then A is also countable. R × R has the same size as R. R is uncountable. The set of all languages is uncountable. Theorem: The set L of all TMs, that is L = {<M> M is a TM}, is countable. Note that L is a subset of Σ * Theorem. There exist a language not decidable. Theorem : A TM = {<M,w>  M is a TM and M accepts w} is undecidable. (proof) Diagnalization . Suppose it is decidable . Then Lu = {<M>  M rejects <M> }. L is also decidable . Let M’ be the Turing machine which decides Lu. Then consider Mu with input <Mu> . If <Mu> is in Lu, then Mu must accepts <Mu>(Mu accepts Lu), also Mu must reject <Mu>(def of x in Lu) , If <Mu> is not in Lu, Mu must reject <Mu>, and if Mu reject <Mu>, then <Mu> must be in Lu,...
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This note was uploaded on 07/25/2008 for the course CSE 860 taught by Professor Chung during the Spring '04 term at Michigan State University.
 Spring '04
 CHUNG

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