Lecture15_Part3

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

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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 ...
View Full Document

Ask a homework question - tutors are online