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$L=\{\langle M,k\rangle \mid\exists w\in L(M) \text{ such that $M$ passes at least $k>2$ distinct states before accepting $w$}\}$

I try to think of reduction to prove that this language is neither RE nor coRE. How to approach this problem? Is there a hint, or intuition?

I usually check whether Rice can be used, but the question here is not about the language itself

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Clearly $L$ is acceptable (just simulate $M$ and keep track of the number of distinct states encountered during the simulation). We now show that it is not decidable.

If $L$ were decidable, you would be able to solve the Halting problem as follows: given a TM $T$ and an input $x \in \Sigma^*$, construct a TM $M$ that ignores its input, simulates $T$ with input $x$ and, when the simulation is complete, accepts. You can further ensure that, if $M$ accepts, then it also traverses at least $3$ distinct states by just transitioning from the initial state to another (distinct) state before starting the simulation of $T$.

Now check whether $\langle M, 3 \rangle \in L$. If the answer is affirmative, then there is some $w \in \Sigma^*$ for which $M(w)$ accepts, showing that $T(x)$ halts. If the answer is negative then $M$ never halts, showing that $T(x)$ does not halt.

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  • $\begingroup$ How can I ensure that M visits at least three states before accepting? It doesn't depend on T? Or do I have to build M so that I add 3 states, first going through them without changing anything, and then running T? $\endgroup$ – Ella Jun 23 at 10:55
  • $\begingroup$ Just build $M$ so that it goes form its initial state to another state, and then simulates $T$. This is already ensures that $M$ traverses at least $2$ distinct states. When (if) the simulation of $T$ completes, $M$ will move to the accepting state. Therefore if $T$ halts, $M$ visits at least $3$ distinct states. $\endgroup$ – Steven Jun 23 at 16:10
  • $\begingroup$ got it. thank you very much! $\endgroup$ – Ella Jun 24 at 7:17

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