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Why is the $A_{ETM}$ for a variant of a Turing machine (an erasing Turing machine), where changing a tape symbol to a nonblank symbol is prohibited, decidable? Why does the following diagonalization argument not work:

Assume $A_{ETM}$ is decidable, and $R$ decides $A_{ETM}$. Then, $R$ accepts $\langle M, w \rangle$ when ETM $M$ accepts $w$. Let $D$ be an ETM that accepts exactly when $M$ does not accept $w$. Then, $D$ rejects $\langle D \rangle$ when $D$ accepts $\langle D \rangle$, a contradiction.

Why does this not work?

Alternatively, if $A_{ETM}$ is decidable, why is $E_{ETM}$ undecidable? The conventional contradiction argument would not work in that case, and I don't know how to reduce it any other way.

Thanks in advance!

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  • $\begingroup$ I have point out a flaw in your proof (the part that is in bold). What is $M$ so that you can define $D$ like that. $\endgroup$
    – Thinh D. Nguyen
    Oct 18 '18 at 12:24
  • $\begingroup$ Hmm, I'm not sure what you mean - $M$ is an ETM such that $L(M)$ is decidable? I don't see how that would help, since the conventional $A_{TM}$ argument works for all TMs $M$? $\endgroup$
    – Barış Ekim
    Oct 18 '18 at 12:47
  • $\begingroup$ Shouldn't the last step be: $D$ accepts $\langle D \rangle$ when $M$ does not accept $\langle D \rangle$? Why is that a contradiction? Why was $R$ never used? I can't see any diagonalization argument above. $\endgroup$
    – chi
    Oct 18 '18 at 13:53
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An ETM (Erasing Turing machine) can be safely simulated due to its erasing-only nature.

It can only pass the right-end of the input to a finite limit or else would plunge off to infinity. Adding this to the input itself, we can conclude that the languages accepted by an ETM must be context-sensitive.

So, an ETM can only accept some context-sensitive language. As a result, $A_{ETM}$ is decidable.

Where is the source that says $E_{ETM}$ is undecidable?

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  • $\begingroup$ Thanks for the response! If for any ETM $M$, $L(M)$ is a CSL, then we can say $E_{ETM}$ is undecidable, since $E_{LBA}$ is undecidable, can't we? $\endgroup$
    – Barış Ekim
    Oct 18 '18 at 15:00
  • $\begingroup$ No, because $E_{ETM}$ is a sub-problem of $E_{LBA}$. It might be easier. $\endgroup$
    – Thinh D. Nguyen
    Oct 18 '18 at 15:02

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