Consider the reduction $A_{TM} \le_m \overline{E}_{TM}$, where

$$A_{TM} = \{\langle M, w \rangle \mid \text{TM $M$ accepts $w$}\}\text{, and}$$

$$\overline{E}_{TM} = \{\langle M \rangle \mid \text{TM $M$ accepts some string}\}$$

Then define the mapping reduction, $f(\langle M, w \rangle)$ as follows:

Construct a new TM $M^\prime$ that on input $x$

  • rejects if $x \ne w$;
  • otherwise, runs $M(w)$ and accepts if it accepts.

Finally, it outputs $\langle M^\prime \rangle$. This construction ensures that $\langle M, w \rangle \in A_{TM} \Leftrightarrow \langle M^\prime \rangle \in \overline{E}_{TM}$.

Question. The idea makes sense, except that I am having a hard time wrapping my head around infinite loops and the computability of mapping reductions. In particular, what ensures that we can indeed write $\langle M^\prime \rangle$ even if $M$ does not halt on $w$? It seems like the answer should be obvious — that there is some generic way to encode $\langle M^\prime \rangle$, but I don't know how.

Secondly, is it correct to say that $f$ is computable because it is fully determined by its inputs?

  • $\begingroup$ Please ask only one question per post. $\endgroup$
    – D.W.
    Sep 24, 2022 at 5:20

1 Answer 1


"fully determined by its inputs" does not imply "computable". The halting function is fully determined by its inputs, but not computable.

  • $\begingroup$ Then (I think) both my questions are really asking the same thing. $\endgroup$
    – big-Oaf
    Sep 24, 2022 at 5:39

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