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# Lecture10x - 15-453 FORMAL LANGUAGES AUTOMATA AND...

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FORMAL LANGUAGES, AUTOMATA AND COMPUTABILITY 15-453

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UNDECIDABILITY II: REDUCTIONS
A TM = { (M,w) | M is a TM that accepts string w } A TM is undecidable: Assume machine H semi-decides A TM H( (M,w) ) = Accept if M accepts w Rejects or loops otherwise Construct a new TM D H as follows: on input M, run H on (M,M) and output the “ opposite ” of H whenever possible.

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D H ( M ) = Reject if M accepts M (i.e. if H( M , M ) = Accept) Accept if M rejects M (i.e. if H( M , M ) = Reject) loops if M loops on M (i.e. if H( M , M ) loops) D H D H D H D H D H D H D H D H D H D H D H D H D H Note: There is no contradiction here! D H loops on D H We can effectively construct an instance which does not belong to A TM (namely, ( D H , D H ) ) but H fails to tell us that.
That is: Given any semi-decision machine H for A TM (and thus a potential decision machine for A TM ), we can effectively construct an instance which does not belong to A TM (namely, ( D H , D H )) but H fails to tell us that. So H cannot be a decision machine for A TM

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HALT TM = { (M,w) | M is a TM that halts on string w } Theorem: HALT TM is undecidable THE HALTING PROBLEM Proof: Assume, for a contradiction, that TM H decides HALT TM We use H to construct a TM D that decides A TM On input (M,w), D runs H on (M,w) If H rejects then reject If H accepts, run M on w until it halts: Accept if M accepts and Reject if M rejects
H (M,w) (M,w) M w If M doesn’t halt: REJECT If M halts Does M halt on w? D

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In most cases, we will show that a language L is undecidable by showing that if it is decidable, then so is A TM We reduce deciding A TM to deciding the language in question A TM “<” L So, A TM “<” Halt TM Is Halt TM “<” A TM ?
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## This note was uploaded on 02/29/2012 for the course CS 15-453 taught by Professor Edmundm.clarke during the Spring '09 term at Carnegie Mellon.

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Lecture10x - 15-453 FORMAL LANGUAGES AUTOMATA AND...

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