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I want to design a Turing machine that accepts at most 3 0s. Now, I have designed one, which goes to accept state overtime it sees 1, 2 and 3 0s and rejects any further 0s. I wanted to know if it is okay for TM to go to accepting state from 3 different states?

user9179677
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  • You'd be better off asking such a question on [the Computer Science site](https://cs.stackexchange.com/) – Jonathon Jul 02 '18 at 06:05
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    Cross-posted: https://cs.stackexchange.com/q/93768/755, https://stackoverflow.com/q/51129854/781723. Please [do not post the same question on multiple sites](https://meta.stackexchange.com/q/64068). Each community should have an honest shot at answering without anybody's time being wasted. – D.W. Jul 06 '18 at 20:57
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    @Jonathon, may I make a request for the future? In the future, if you suggest another site, can you let the poster know that they should not cross-post? You can suggest they delete the copy here before posting elsewhere. Hopefully this will provide a better experience for all. Thanks for listening! – D.W. Jul 06 '18 at 20:58

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Yes, this is perfectly fine. Even if you want a deterministic machine, several transitions going into the same state are fine. If several outgoing transitions read the same symbol the machine is not deterministic anymore. But even this is not a problem for nondeterministic TMs.

Peter Leupold
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