Let $L = \{ <M> | M$ halts on every input $x$ in at most $200 * |x|$ steps $\}$.

Is $L$ decidable? Recognizable?

Given that membership in $L$ asserts something about $M$'s behavior on an infinite set of strings, it seems extremely unlikely to me that $L$ could be either. I have shown that co-$L$ is Turing-recognizable (I think): you can make an enumerator that tests each $M_1, M_2, \dots, $ for each $s_1, s_2, \dots$ and emits $M$ if it does not accept some $s$ in at most $200|x|$ steps.

Since co-$L$ is recognizable, either $L$ is not recognizable, or it is decidable. I can't imagine that $L$ is decidable. However, it definitely cannot be reduced to the halting problem, nor can Rice's theorem be applied to it (since the quality in question is the quality of halting in a particular number of steps, being able to decide it doesn't let us decide other arbitrary properties).

It seems to me that the best way to go will be to show that $L$ lets me recognize something that is unrecognizable, since the only problems it solves are ones which require me to run over infinite sets of strings. But I can't think of what this could be. I thought that maybe co-HALT would work, but I can't ever prove that a TM will never halt on some input.

I'm stuck. What direction should I go in?

  • $\begingroup$ & wonder if there is some way to relate this to std language classes eg CSLs etc...? $\endgroup$
    – vzn
    May 16, 2014 at 18:47

2 Answers 2


The language is actually decidable, but the proof is quite involved, and uses crossing-sequence arguments. See this paper for details.

If you change $200|x|$ to a faster growing function, such as $|x|\log |x|$ (plus some constant), then the language becomes undecidable by standard reduction arguments.


This answer is wrong. I'm leaving it here since this sort of argument is the natural way to go, and does work for faster growing functions, as mentioned in Shaull's answer.

Hint: Let $M$ be a machine having no input. Define a new machine $M'$ as follows: on input $x$, $M'$ simulates $M$ for $f(|x|)$ steps, where $f(t)$ is a function tending to infinity such that the simulation takes less than $200 t$ steps. If $M$ ever halts, then $M'$ switches to an infinite loop. When is $\langle M' \rangle \in L$?

  • 1
    $\begingroup$ You can't necessarily perform the simulation in time $f(t)$. Specifically, we don't know how to simulate with linear overhead. $\endgroup$
    – Shaull
    May 16, 2014 at 15:58

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