Turing machine halting problem

Question

I have a question about turing machines and halting problem.

Suppose that we have Atm = {(M,w) where M is a turing machine and w is an input} and
HALTtm = {(M,w) where M is a turing machine halts with an input w}

I want to prove that HALTtm <=m Atm

I've tried some methods but I think they're far from the solution. Anyone can give some clues ??

This question would have been perfect for the upcoming [Computer Science Stack Exchange](http://area51.stackexchange.com/proposals/35636/computer-science-non-programming?referrer=pdx8p7tVWqozXN85c5ibxQ2). So, if you like to have a place for questions like this one, please go ahead and help this proposal to take off! — Raphael, Dec 06 '11 at 12:36
What exactly is <=m supposed to mean? I read it as `\leq_m`, but how is that defined? — Raphael, Dec 06 '11 at 12:37

score 2 · Answer 1 · answered Mar 24 '10 at 18:55

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Well, observe that for all (M,w) in HALTtm, it must be that (M,w) is in Atm. Then show there exists some (M',w') which is a member of Atm but which does not halt, and so is not in HALTtm.

answered Mar 24 '10 at 18:55

Fred

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The second step isn't necessary. Note that he used "<=" which is presumably intended to mean "subset of *or equal to*". – sepp2k Mar 24 '10 at 18:57
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I suspect <=m is intended to mean the existence of a [many-one reduction](http://en.wikipedia.org/wiki/Many-one_reduction). In that case, what the OP is trying to prove is not true. – Prateek Dec 06 '11 at 14:06

score 0 · Answer 2 · answered Dec 14 '18 at 17:04

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we can do reduction from ATM to HALTTM let M2 is a new machine like On input x When run M2 on x if M2 accepts x then halt and acccept if M2 rejects then M2 goes for an infinite loop

so there exists a x that not halts M2 so ATm is not in HALTTM

answered Dec 14 '18 at 17:04

fkaftanci

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score 0 · Answer 3 · edited Nov 22 '12 at 23:48

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For each of the following languages, draw a transition diagram for a Turing Machine that accepts that language.

{aibj | i≠j}

edited Nov 22 '12 at 23:48

David G

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answered May 06 '11 at 03:51

rabea

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score 0 · Answer 4 · answered Dec 06 '11 at 14:00

What is <=m? I think you mean "many-one reduces to"? In that case, what you're asking for is a total computable function f such that for all strings x,

x belongs to HALTtm if and only if f(x) belongs to Atm

If such an f existed, we could decide the halting problem : given x, compute f(x) and check if f(x) belongs to Atm (Atm is easily recursive/decidable). But since the halting problem is not decidable, such an f cannot exist. So HALTtm does not many-one reduce to Atm.

Turing machine halting problem

4 Answers4