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I was checking this question and wondered how to apply it to the following language to prove it is in $RE$: $$ L = \{ \langle M \rangle : \exists P \text{ such that } M \text{ halts on every input } w \text{ within } P(|w|) \text{ steps} \},$$

where $P(x)$ is a polynomial with coefficients $a_1,\ldots,a_n \in \mathbb{N}$.

Those are the two approaches I am thinking of so far, in order to build a Turing Machine $M_L$ which accepts $L$ above:

  1. Let $d$ be the degree of the polynomial $P$. For each $d=1,2,3,\ldots$, iterate over all possible $d$ coefficients, and check whether the current $P$ satisfies $L$'s condition above. I think it might work because if there is no such $P$, then the Turing machine $M$ won't stop and this is OK.
  2. Here I am taking $P$ to be a constant (i.e. $P(x)=k \in \mathbb{N}$), and then iterate over all $k$ and apply the Turing machine from the question I linked at the beginning, where the language at the question above is in $R$ then there is a Turing machine $M_k$ which decides it, thus, while we get Rejection from $M_k$ for the $k$ values we're iterating on, $M_L$ does not stop.

I am not sure even where I made a mistake above, since the concept of iterating over all $k \in \mathbb{N}$ does not seems robust.

Does the approach of building a Turing machine work? Since I could not get to a reduction which proves that $L \in RE$.

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ThunderWiring
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1 Answers1

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The halting problem reduces to $\overline{L}$: given a machine $T$, create a new machine that on input $n$ simulates $T$ for $n$ steps, and if $T$ halts then it enters an infinite loop. (I leave you the details.)

Since the halting problem is complete for $\mathsf{RE}$, it cannot be the case that $\overline{L} \in \mathsf{coRE}$, and so it cannot be the case that $L \in \mathsf{RE}$.

Yuval Filmus
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  • I might be wrong, but the way I understand 'complete for $RE$' is that $L \in RE$ then , any $L' \in RE$ yields $L' \leq L$, so i don't see how $\overline{L} \leq L$ and then $L$ cannot be in $RE$ (because $\overline{L} \notin coRE \rightarrow \overline{L} \in RE$)? could you please clarify that point a bit more? – ThunderWiring Dec 17 '16 at 14:12
  • I showed a reduction from the halting problem to $\overline{L}$. This shows that every problem in $\mathsf{RE}$ can be reduced to $\overline{L}$. If $\overline{L}$ were in $\mathsf{coRE}$, then it follows that $\mathsf{RE} \subseteq \mathsf{coRE}$, which we know to be false. – Yuval Filmus Dec 17 '16 at 14:50
  • to make sure I understood, basically we have this scenario: $HP \leq \overline{L} \rightarrow HP \notin coRE \text{ implies } \overline{L} \notin coRE \rightarrow L \notin RE$ where I used the reduction rule. correct? – ThunderWiring Dec 17 '16 at 15:21
  • No. The point isn't that the halting problem is not in coRE. It's that the halting problem is RE-complete, and that RE$\neq$coRE. – Yuval Filmus Dec 17 '16 at 16:03
  • i don't understand what part of my last correct is incorrect? Also it is not clear to me how the fact that HP is RE-complete is related to deduce something about L? – ThunderWiring Dec 18 '16 at 12:10
  • In short, an RE-complete problem cannot belong to coRE, since that would imply that RE=coRE, which we know doesn't hold. More explicitly, if the halting problem had a coRE algorithm, then by combining this with an RE algorithm you will be able to decide the halting problem, which we know isn't possible. – Yuval Filmus Dec 18 '16 at 12:12