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I currently have a problem with the following question:

Let $L = \{ \langle M \rangle \mid \exists w: \text{$M$ halts for $w$ in at most $|w|^3$ steps} \}$. Construct an NTM (non-deterministic Turing machine) that decides $L$.

My idea was to simulate every possible input of length $\le|w|^3$ on a given TM $M$ by using a separate band in my NTM for every single input word. If there is one band where $M$ halts, then the NTM accepts.

Is this the right way to go about it? My problem is that I don't see why I would need a NTM to decide $L$. A standard TM would be able to do the same thing.

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  • $\begingroup$ "simulate every possible input"... Are you sure you can do this in finite time? For every possible input? Remember deciders need to halt to reject too! $\endgroup$ – dkaeae Jan 7 at 15:10
  • $\begingroup$ Also, are you sure you mean "at least $|w|^3$ steps" and not "at most $|w|^3$ steps"? $\endgroup$ – dkaeae Jan 7 at 15:13
  • $\begingroup$ sorry you are right, I meant at most |w|^3 steps. If it doesn't halt until then, I know that it is the wrong input. $\endgroup$ – freak14 Jan 7 at 15:38
  • $\begingroup$ maybe it helps to know that I use a NTM to show that L ist semi-decidable $\endgroup$ – freak14 Jan 7 at 15:47
  • $\begingroup$ Wait... Are you supposed to prove $L$ is decidable or semi-decidable? I do not think the former is correct... $\endgroup$ – dkaeae Jan 8 at 12:29
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$L$ is semi-decidable (or, synonymously, recursively enumerable). This can be proven (arguably) more easily by using an NTM which non-deterministically picks an input $w \in \{ 0, 1 \}^\ast$, simulates $M$ on it for at most $|w|^3$ steps, and accepts if and only if $M$ does. The NTM accepts exactly the language $L$ because, if there is no accepting branch of the NTM, then there is also no input $w$ for which $M$ halts in at most $|w|^3$ steps. For a properly written proof, it should also be said the semi-decidability of $L$ follows from the equivalence of DTM and NTM acceptors (remember (semi-)decidability is defined in terms of DTMs, not NTMs!).

Because of the equivalence of DTM and NTM acceptors, the above reasoning indirectly gives you a DTM which accepts $L$. The explicit construction could be picking an enumeration of inputs (e.g., lexicographical order), simulating $M$ on each input $w$ for at most $|w|^3$ steps, and accepting if and only if $M$ does. (Having to argue about the enumeration of inputs might be perceived as "hard" by some, hence why proving with an NTM could be considered "easier".)

A warning, however: $L$ is not decidable. This is because its complement is $\{ \langle M \rangle \mid \forall w: \text{$M$ does not halt for $w$ in at most $|w|^3$ steps} \}$ and, thus, a variant of the complement of the halting problem, which is notorious for not being semi-decidable.

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