Prove that the language

$$ L=\{\langle M \rangle \mid \exists M_1, M_2 : L(M_1)=L(M_2)=L(M) \text{ and } |\langle M_1 \rangle| < |\langle M \rangle| < |\langle M_2\rangle| \}$$

is not recursive.

  • 1
    $\begingroup$ What do you think? What have you tried, and where did you get stuck? $\endgroup$ Jan 19, 2017 at 14:38
  • $\begingroup$ I have tried the recrusion theory but I can not get the answer $\endgroup$
    – Sikelef
    Jan 19, 2017 at 15:20
  • $\begingroup$ Hint: If you could decide $L$, you could compute (more or less) the Kolmogorov complexity of a string, which you shouldn't be able to do. $\endgroup$ Jan 19, 2017 at 16:58
  • $\begingroup$ We're happy to help you understand the concepts but just solving exercises for you is unlikely to achieve that. You might find this page helpful in improving your question. $\endgroup$
    – D.W.
    Jan 19, 2017 at 17:21
  • $\begingroup$ @YuvalFilmus I think I can define Kolmogorov complexity of $w$ as the shortest encoding of a TM that accepts $\{ w \}$. So if I could decide $L$, then I would be able to tell whether a specific TM that accepts some $\{ w \}$ is the shortest TM that accepts $\{ w \}$. But if it isn't the shortest, then even if I test whether all shorter TMs are in $L$, I wouldn't know which of them accept $\{ w \}$, and so it wouldn't help me in finding the Kolmogorov complexity of $w$... Am I misinterpreting your hint? $\endgroup$ May 22, 2019 at 12:16

1 Answer 1


First, let us notice that in the definition of $L$, a machine $M_2$ accepting the same language as $M$ but having a longer description always exists, so only the condition on $M_1$ is pertinent.

Suppose that $L$ were computable. Consider the following Turing machine $T_\ell$. The machine enumerates all Turing machines of size at least $2^\ell$, until it finds a machine $M$ such that $\langle M \rangle \notin L$ (such a machine must exist, since there are infinitely many computable languages but only finitely many Turing machines of size less than $2^\ell$). Then it transfers control to the machine $M$, i.e., it runs $M$ on the input.

By construction, $L(T_\ell) = L(M)$. Since $\langle M \rangle \notin L$, necessarily $|\langle T_\ell \rangle| \geq |\langle M \rangle| \geq 2^\ell$. However, by hardcoding $\ell$ into a Turing machine in which $\ell$ is an additional input, we see that $|\langle T_\ell \rangle| = O(\ell)$. We obtain a contradiction for large enough $\ell$.


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