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  • By "$T$ runs in polynomial time", I mean that $T$ halts for every input of length $n$ in $O(n^k)$ steps for some $k$.
  • By Turing-recognizable, I mean that there exists a Turing machine that halts in the accept state iff the input $w = \langle T\rangle \in L$.

I'm not really sure how to approach this problem. It seems like I should reduce from $A_{\text{TM}}$ (or $HALT_{\text{TM}}$ depending on convenience). (Edit: wrong, both of these are Turing-recognizable.)

I also tried to directly do a running time variant of the diagonalization proof that $A_{\text{TM}}$ is undecidable. I began by assuming that $L$ is Turing-recognizable by some $M$. Then we can create a TM $D$ that reads $\langle T\rangle$ and simulates $M$ on it. Towards contradiction we then make $D$ run in a loop (or do $2^{|\langle T\rangle|}$ additional steps) if $\langle T\rangle \in L$, or halt immediately otherwise. But this doesn't work as we don't know the complexity of $M$. More precisely I can get a contradiction only if $\langle M\rangle \in L$.

Rice's theorem doesn't seem to be of much help either. If I could prove that (1) $L$ is undecidable and (2) $L^C$ is recognizable, then I would obtain $L$ is unrecognizable. But Rice doesn't help with (1) (because for $L(T_1) = L(T_2)$ says nothing about the running times), and (2) seems to be straight-up false.

jcora
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2 Answers2

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The simplest example of a non-Turing-recognizable language is the complement of $A_{\text{TM}}$, $$\overline{A_\text{TM}}=\{\langle M,w\rangle\mid M\text{ is a TM and }M\text{ does not accepts }w\}.$$

Let us reduce $\overline{A_\text{TM}}$ to $L$.

Given a TM $M$ and a string $w$, construct the following TM.

$M'$= "On input $s$:

  1. Simulate $M$ on input $w$ for $|s|$ steps using less than $|s|^3+159759$ steps.
  2. If the simulated $M$ has accepted, possibly before $|s|$ steps, run forever. Else halt."

Here $3$ and $159759$ just mean some large-enough numbers independent of $M$ so that the simulation for $|s|$ steps can be done.

Let us verify that $\langle M,w\rangle\in \overline{A_\text{TM}}$ iff $\langle M'\rangle\in L$.

  • If $\langle M,w\rangle\in \overline{A_\text{TM}}$, i.e., $M$ does not accept $w$, then on input $s$, $M'$ will always halt in $|s|^3+159759$ steps. Hence $\langle M'\rangle\in L$.
  • If $\langle M,w\rangle\notin \overline{A_\text{TM}}$, i.e., $M$ accepts $w$, then for input $s$ that is long enough, the simulated $M$ will have accepted after step 1. Hence $M'$ will run forever on those input. Hence $\langle M'\rangle\notin L$.

Since $\overline{A_{\text{TM}}}$ is not Turing recognizable, neither is $L$.

John L.
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Suppose $L$ is decidable with a Turing machine $T$. Now we will show $\sim HALT\leq L$.

Create a Turing machine $\mathcal{M}$ from $\langle M,x\rangle$ which on any input runs $M$ on $x$ for $|x|$ many steps. If $M$ halts within $|x|$ many steps then $\mathcal{M}$ goes on a loop and halts after $2^{|x|^c}$ many steps for some constant $c>1$ and if $M$ does not halt within that many steps $\mathcal{M}$ halts.

Now it is easy to see that $\mathcal{M}$ is a polynomial time Turing Machine if $M$ halts within $|x|$ steps. Therefore $$\langle M,x\rangle\in\ \sim HALT\iff \mathcal{M}\in L$$Since $T$ can decide $L$, $T$ can decide if $\mathcal{M}$ is in $L$ or not, i.e. $T$ can decide the set $\sim HALT$. Which is not decidable. Hence contradiction. Hence $L$ is not decidable

Sassy Math
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