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This is what is known about the halting problem and semi-decidability :-

The halting problem says that for a given input x and a machine H, we can't say whether the machine H halts or not on input x.

A language is said to be Semi-decidable if there exists a Turing machine which halts if a word belongs to the language (YES cases) and may reject or go into infinite loop if the word doesn't belong to the language (NO case).

Now, in the halting problem we can't say whether the machine will halt even if the input belongs to the language (YES cases). Then how is it Semi-decidable? I think it should be non-recursively enumerable or undecidable.

Joshua Schwartz
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Zephyr
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1 Answers1

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Tl;dr: "(say) whether or not it halts" and "(say) if it halts" are not the same thing. Use mathematics to avoid confusion induced by language ambiguity.

Halting problem says that for a given input x and a machine H, we can't say whether the machine H halts or not on input x.

No, that's not what it says. The halting problem is the computational problem of deciding whether $H$ halts on $x$, given $x$ and $H$ as input. It is important to note that "decide" here means "say yes if it is so, and say no if it is not".

The undecidability of the halting problem states that there is no single algorithm (Turing machine) that solves the halting problem for all $H$ and $x$.

Now, in halting problem we can't say whether the machine will halt even if the input belongs to the language (YES cases). Then how is it Semi-decidable?

After the above clarification, your confusion here should be clear. It doesn't matter what "we can say". The "semi halting problem" is relaxed: the algorithm still has to say "yes" if $H$ halts on $x$, but it can do whatever it please if it does not (except answer "yes").

This is trivial to implement: just run $H$ on $x$. If it halts, answer "yes". If not, it doesn't matter since we're allowed to loop.

user207421
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Raphael
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