It is known that the halting problem is decidable for every fixed $M_0$ Turing machine and every fixed $w_0$ input.

My related question would be the following: is it true that for every fixed $M_0$ Turing machine and every fixed $w_0$ input, an $M_{M_0,w_0}$ Turing machine can be constructed for which the possible inputs are $(M, w)$ machine-input pairs, and for the $(M_0, w_0)$ pair, the output is "1" if $M_0$ will halt on $w_0$ and "0" if $M_0$ will not halt on $w_0$? ($M_{M_0, w_0}$ can give false answers for other pairs, it is not demanded that it has to run correctly for every $(M, w)$ pair.)

  • $\begingroup$ What do you mean by "can be constructed"? Do you want an algorithm that constructs $M_{M_0, w_0}$ or do you want to know whether $M_{M_0, w_0}$ exists? $\endgroup$
    – Steven
    Commented Apr 2, 2020 at 23:15
  • $\begingroup$ @Steven An algorithm would be preferred, but knowing if it exists or not would be also really helpful. Edit: I see from the conversation with 6005 on my previous question that either or machine that always accepts or another one which always rejects will be good. I don't know though that can we decide for every fixed $(M_0, w_0)$ pairs which is going to be the right choice? $\endgroup$
    – Heisenberg
    Commented Apr 2, 2020 at 23:22

1 Answer 1


Since $M_0$ and $w_0$ are fixed parameters of the problem, the answer is yes: for every fixed $M_0$ and $w_0$, there exists a Turing machine $M_{M_0, w_0}$ (depending on $M_0$ and $w_0$) such that, for the input $(M_0, w_0)$, $M_{M_0, w_0}$ returns $1$ if $M_0(w_0)$ halts and 0 otherwise.

In particular one such Turing machine $M_{M_0, w_0}$ must be one of the following two machines:

  • $M'_1$: Write $1$. Halt.
  • $M'_0$: Write $0$. Halt.

If, instead, you are looking for an algorithm that takes $M_0$ and $w_0$ as input and outputs a machine $M_{M_0, w_0}$ with the above property, then you are out of luck: there is not such algorithm in general (it might exist if you restrict the set of input machines $M_0$). Suppose that such an algorithm (i.e., Turing machine) $A$ existed, then it would allow to solve the halting problem:

  • Given $M_0$ and $w_0$, compute $M_{M_0, w_0}$ by simulating $A$ on input $(M_0, w_0)$.
  • Simulate $M_{M_0, w_0}$ on input $(M_0, w_0)$. By definition of $M_{M_0, w_0}$ this step requires finite time.
  • Return "yes" if the output of $M_{M_0, w_0}( (M_0, w_0) )$ was $1$, otherwise return "no".
  • $\begingroup$ I'm a little confused: does this mean that if we have a fixed Turing machine $M_0$ and a fixed $w_0$ input, we cannot predict how $M_0$ will work on $w_0$, i.e. it will halt or not? $\endgroup$
    – Heisenberg
    Commented Apr 2, 2020 at 23:45
  • 1
    $\begingroup$ It depends on $M_0$ and $w_0$. For some of them you might be able to predict whether $M_0(w_0)$ halts or not (think for example, of the trivial machine that always halts on the first step). But you have no general method that, given $M_0$ and $w_0$, is able to predict whether $M_0(w_0)$ halts. $\endgroup$
    – Steven
    Commented Apr 2, 2020 at 23:49
  • $\begingroup$ You say I "might be able to predict whether $M_0(w_0)$ halts or not". Does this mean that there exists a $(M_0, w_0)$ pair for which it is known that it is impossible to predict whether $M_0(w_0)$ halts or not? If yes, could you give me please an example? $\endgroup$
    – Heisenberg
    Commented Apr 2, 2020 at 23:55
  • 1
    $\begingroup$ The proposition: $P_T$="This Turing machine $T$ halts on the empty string $\epsilon$" cannot be proven true or false for all $T$ (as otherwise you'd be able to solve the halting problem by generating, in parallel, all tentative proofs for $T$'s termination and non-termination until, eventually, a correct proof is found). Let $T$ be a Turing machine for which $P_T$ can't be proven true of false. Pick $M_0 = T$ and $w_0 = \epsilon$. I don't know what $T$ is explicitly. $\endgroup$
    – Steven
    Commented Apr 3, 2020 at 0:51
  • $\begingroup$ @T.Christopher Perhaps what you ask in your 2nd comment deserves a new question/post. It's not easy to answer it in comments. $\endgroup$
    – frabala
    Commented Apr 3, 2020 at 7:48

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