Lecture15_Part3

# Lecture15_Part3 - Demonstration Consider HALTTM = M w | M...

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Unformatted text preview: Demonstration Consider HALTTM = M , w | M is a TM that halts on w The "Real" Halting Problem { } Theorem HALTTM is undecidable. 3. We assume that B is decidable and use this assumption to prove that A is decidable. In the following slides we assume (towards a contradiction) that HALTTM is decidable and use this assumption to prove that ATM is decidable. Proof By reducing HALTTM to ATM . 10 4. We conclude that B is undecidable. 9 Discussion Discussion Recall the definition of ATM : ATM = M , w | M is a TM that accepts w Assume by way of contradiction that HALTTM is decidable. { } Why is it impossible to decide ATM ? Because as long as M runs, we cannot determine whether it will eventually halt. Well, now we can, using the decider R for HALTTM . 12 Recall that a decidable set has a decider R: A TM that halts on every input and either accepts or rejects, but never loops!. We will use the assumed decider of HALTTM to devise a decider for ATM . 11 ...
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## This note was uploaded on 04/30/2009 for the course CSE 105 taught by Professor Paturi during the Winter '99 term at UCSD.

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