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I'm trying to prove the undecidability of the following language.

$$L=\{\langle M \rangle\mid M\text{ is a Turing machine and there is a string }w\\\text{ s.t. }M\text{ accepts }w\text{ and }M\text{ rejects }w'\},$$ where $w'$ is the mirrored version of $w$.

I know that my first steps should be to find a reduction from $A_{\text{TM}}$, which is undecidable but the rejection part of the mirrored string is proving troublesome.

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  • $\begingroup$ Be careful with the direction of the reduction. If $P$ is reduced to $Q$, then informally, $P$ is not "harder" than $Q$ and $Q$ is not "easier" than $P$. A reduction from $A_\text{TM}$ to $L$ will be useful if you wanted to show $L$ is undecidable. $\endgroup$
    – John L.
    Apr 2, 2019 at 22:39
  • $\begingroup$ Thank you for pointing that out. However, even though I can see the reduction with the language first part's description, I can't quiet understand how to take into account the part with the mirrored string. $\endgroup$
    – Celestius
    Apr 3, 2019 at 8:38
  • $\begingroup$ It looks like you accepted an answer that has an important typo. "Accepts your favourite string and its reverse" is different from "accepts $w$ and rejects its reverse $w'$". I would recommend that you should take some time to verify an answer before accepting it. Note $w=w'$ for some string. $\endgroup$
    – John L.
    Apr 3, 2019 at 14:46

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It's the standard reduction from the Halting problem: given a machine $M$ and an input $w$ to it, construct a new machine $M$ that accepts your favourite string and rejects its reverse if $M(w)$ halts, and accepts nothing if $M(w)$ does not halt.

This technique applies to any question of the form "Prove that it's undecidable whether a Turing machine accepts a string or language of some specific kind."

Or just use Rice's theorem, if you've been taught that.

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  • $\begingroup$ Thanks a lot ! I'll look into Rice's Theorem if I can prove the non-triviality as well as the semantic property. But knowing I indeed haven't been taught it, I'll go with the first proof. $\endgroup$
    – Celestius
    Apr 3, 2019 at 12:00

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