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Is there a difference between the end state of a Turing machine and the halt state? Especially, for example the Busy Beaver 3-state Problem. It is said that the TM is with 3 states but there is also a halt. Is the end state Q2 or the halt?

Dimitar
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    At close-voter: How is a question about Turing machines more suitable for SuperUser than for StackOverflow? This *might* be suitable for [CS SE](http://cs.stackexchange.com/), or possibly [Theoretical CS SE](http://cstheory.stackexchange.com/). However, comparing the results for search term "Turing Machine" on SO and SU quickly reveals that any concrete questions about Turing machines are more suitable here than on SU. – O. R. Mapper Jan 25 '15 at 12:47
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    I'm voting to close this question as off-topic because it belongs on Theoretical CS SE (can it be migrated?) – ali_m Jan 25 '15 at 12:57
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    @ali_m I know that CS SE is a better place, but unfortunately, there the comunity is not so big and I have not received an answer, having seen that here in SO are to find some questions about TM, I posted mine. – Dimitar Jan 25 '15 at 12:59
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    If this question should be migrated, what does that say about the particular tags that were applied to it here at SO. If those tags are allowed as tags here, I can't see any way this question is off-topic. If the question is off-topic, I can't see any way those tags should exist for this site. – ely Jan 25 '15 at 13:00
  • @prpl.mnky.dshwshr Unfortunately there are lots of tags that probably shouldn't exist on SO... See here: http://meta.stackoverflow.com/a/254764/1461210 – ali_m Jan 25 '15 at 13:06
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    @user4325010 I can definitely sympathize with you. The problem, as I see it, is that if people habitually ask their theoretical CS questions on SO then not only does it clutter SO, but it also undermines the Theoretical CS SE community. I have no objection to questions on SO that relate to a *specific software implementation* of an algorithm, but your question is purely theoretical, and is therefore off-topic as per the CS tag description (see the Meta discussion I linked to as well). – ali_m Jan 25 '15 at 15:27

1 Answers1

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These notes give one possible answer, found in the section titled Notes on "Turing Acceptable".

Some books define "acceptance by Turing machine" slightly differently. That is, in the Turing machines those books define, there are two halt states: "accept halt" and "reject halt". A Turing machine thus may accept a string and halt, reject a string and halt, or loop. With this definition, a string is accepted by a Turing machine if given the string, the Turing machine eventually goes into the accept halt state. As far as the material discussed in this class note, there is no difference between these two definitions of "accept".

So the likely answer is no, there is no material difference between "accept state" and "halt state".

Do note, however, that for a single instance of a problem running on some specific Turing machine, this question about halting is not going to be very relatable to the Halting Problem in general. The Halting Problem in general is about an algorithm that could, for any input program and starting state, tell you if that program, starting from that state, will halt.

That's quite different than being given merely one single instance of a problem, like Busy Beaver 3, and asked to reason about just that lone instance's possible start-state-to-halt-state paths.

ely
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