0

Past year paper question:

Let $M_i$ denote the Turing machine with code $i$ using the alphabet $\Sigma=\{a,b\}$.

Show that the following language is not recursively enumerable:

$L = \{a^i \;:\; (\exists x \in \mathrm{Lang}(M_i))\;[ xx \notin \mathrm{Lang}(M_i) ] \}$

I tried to show that we can reduce $L_e$, the set of Turing machines that accept no strings, to $L$. To do this, I need to find a transformation $f$ such that is $\mathrm{Lang}(M)$ is empty, then the machine represented by $f(M)$ has some string $x$ but not $xx$, and if $\mathrm{Lang}(M)$ is not empty, then the machine represented by $f(M)$ contains $xx$ if it contains $x$.

But I am not sure how to find such a transformation. Can someone show a transformation to solve this problem?

eatfood
  • 175
  • 6
  • Welcome to Computer Science! The title you have chosen is not well suited to representing your question. Please take some time to improve it; we have collected some advice here. Thank you! – Raphael Nov 21 '18 at 16:28
  • Hi @Raphael, I have edited the question I think it is more precise now – eatfood Nov 22 '18 at 14:18

1 Answers1

0

First, recall that the problem "$L(M)\neq\emptyset$" is semidecidable.

Define $f(M)$ as $a^i$ where $M_i$ is a TM such that

  • it accepts $aa$ iff $L(M)\neq\emptyset$,
  • it accepts any other string (including $a$).

For the first step: when $M_i$ receives $aa$ as input, it semi-decides $L(M)\neq\emptyset$. If that holds, the semi-decider halts in finite time, so $M_i$ can accept. If that does not hold, the semi-decider diverges, making $M_i$ diverge. This is fine, since $aa$ is not accepted.

Note that we never reject $aa$, because we do not have a full decider for $L(M)\neq\emptyset$. This however does not matter: we do not have to reject $aa$, we only have to avoid accepting it, and diverging suffices.

chi
  • 14,564
  • 1
  • 30
  • 40
  • Hi, should the 3rd point be "it accepts every other string"? – eatfood Nov 22 '18 at 14:55
  • @eatfood No, why? If you do that, then $f(M)$ would accept both $aaa$ and $aaaaaa$, always satisfying the condition in $L$ (since $x$ can be $aaa$). In that case the reduction would no longer work. – chi Nov 22 '18 at 15:09
  • But in that case, if $L(M)$ empty, then $f(M)$ would accept $a$ and not $aa$, so encoding of $f(M)$ in $L$. And if $L(M)$ is not empty, then $f(M)$ would accept $aa$ and not $aaaa$, and so encoding of $f(M)$ would also be in $L$. – eatfood Nov 22 '18 at 15:16
  • I was thinking that if the 3rd point was changed to "accept all other strings", then:We reduce $L_e$ to $L$. Then $L_e$ not RE implies $L$ not RE.

    Let $M$ be given.

    We define $M'$ as follows:

    if input is $a$, accept.

    if input is $aa$, if $L(M)=\emptyset$, we accept, else reject.

    accept all other inputs.

    Then define $f(M)=a^i$ where $i$ is the encoding of $M'$.

    Then if $L(M)=\emptyset$, $M'$ accepts every string except $aa$, so $f(M) \in L$.

    If $L(M) \neq \emptyset$, then $M'$ accepts all strings, so $f(M) \notin L$.

    So if $L_e$ is reduced to $L$, hence $L$ is not RE.

     – eatfood Nov 22 '18 at 15:17
  • @eatfood I think you are right, let me check. I think I missed one negation at one point. – chi Nov 22 '18 at 15:17
  • @eatfood Your argument looks essentially correct, I edited to reflect it. There are only two mistakes: we can't write "if $L(M)=\emptyset$" since that condition is not RE. We can't also write "if $L(M)\neq\emptyset$, we accept, else reject", we can only write "..., else diverge" since the "if" guard is not decidable, but is only RE. But we don't need to reject, diverging suffices for the reduction. – chi Nov 22 '18 at 15:28
  • Ok! Thanks for the answer and also for pointing that out! :) – eatfood Nov 22 '18 at 15:44