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Unformatted text preview: CS103 HO#54 SlidesMore on Reductions 5/27/11 1 CS103 Mathematical Foundations of Computing 5/27/11 FINAL EXAM: Monday, June 6, 12:15 3:15 Braun Auditorium NOHALT TM = { M, w  Turing machine M does not halt on w } HALT TM = { M, w  Turing machine M halts on w } We know from PS8 that NOHALT TM is not Turingrecognizable . Problems that are not Turingrecognizable generally involve infinite search, or knowing that a TM will loop infinitely, or both. To solve NOHALT TM by simulation, we would have to run M forever. To solve { M  L(M) = * } by simulation, we would have to try all strings in *. To solve { M  there is no string on which M halts } by simulation, we would have to try all strings in * and show that M always fails to halt. We know from PS8 that NOHALT TM is not Turingrecognizable. Problems that are not Turingrecognizable generally involve infinite search, or knowing that a TM will loop infinitely, or both. To solve NOHALT TM by simulation, we would have to run M forever. To solve { M  L(M) = * } by simulation, we would have to try all strings in *. To solve { M  there is no string on which M halts } by simulation, we would have to try all strings in * and show that M always fails to halt. We most often show a language to be not Turingrecognizable by reduction from NOHALT TM , or by using Theorem 4.22: a language is decidable if and only if it is Turingrecognizable and coTuringrecognizable. Theorem 4.22 : A language is decidable if and only if it is Turingrecognizable and coTuringrecognizable. Suppose we know that: At least one of L or L is not decidable. L is recognizable We can conclude that L is not recognizable. Example . Show that H ANY = { M  there is no string on which TM M halts } is not recognizable. H ANY = H ANY . So we know from PS8 that H ANY is recognizable and undecidable. So H ANY is not recognizable, because if it were, H ANY would be decidable. Example: Reduction Proof to show a language is not recognizable. Show that H ANY = { M  there is no string on which TM M halts } is not recognizable. The reduction has to start with a language we know is not recognizable, and we hypothesize the existence of a recognizer rather than a decider. NOHALT TM is not recognizable and is often a good place to start. Here we show that NOHALT TM m H ANY . Suppose that HNA recognizes H ANY . Map the input M, w for NOHALT TM to M' that operates as follows: M' = "On input x: 1. Run M on w. 2. Accept." M' halts on everything if M halts on w, or nothing if M does not halt on W. So HNA accepts M' if M, w NOHALT TM , and does not accept if M, w NOHALT TM This would give us a recognizer for NOHALT TM , so HNA must not exist. CS103 HO#54 SlidesMore on Reductions 5/27/11 2 Problem 6(d). A Al Even = { M  M accepts all even length strings } Problem 6(d). A Al Even = {...
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 Spring '09

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