Given a language $A \in RE$, is the following language also in $RE$? $$ L_{10}^{A} = \{ \langle M \rangle : \lvert A \cap L(M) \rvert \geq 10 \} $$ Where $L(M) = \{x \in \{0, 1 \}^* \mid M \text{ accepts } x \}$.

(I've managed to prove that if $A \in R$ then $L_{10}^A \notin R$ necessarily with a reduction from $HALTING$.)

While my intuition says it is in $RE$, I haven't been able to construct a turing machine that recognizes $L_{10}^{A}$ nor define a reduction from it to another language in $RE$.

And another challenge:

For every language $A \in RE$, show that there exists a language $B_A \in R$ such that for all $x \in \{0, 1 \}^*$ we have $$x \in A \iff \exists y \in \{0, 1\}^* : (x, y) \in B_A$$

All help with these two problems will be greatly appreciated.



1 Answer 1


$L^A_{10}$ is recursively enumerable. In particular, since $A$ is recursively enumerable, there is a Turing Machine $M_A$ that accepts $A$.

Let $M^t$ (resp. $M_A^t$) denote a Turing machine that simulates the first $t$ steps of $M$ (resp. $M_A$). If $M$ (resp. $M_A$) does not halt within $t$ steps, then $M^t$ (resp. $M_A^t$) halts and rejects.

For every $M$, $|A \cap L(M)| \ge 10$ if and only if there is some $T$ such that $|L(M_A^T) \cap L(M^T)| \ge 10$. Then you can accept $L^A_{10}$ as follows:

  1. Start with $t=1$ and $c=0$;
  2. Enumerate all words of length at most $t$ and, for each such word $x$:
  • Run $M_A^t$ on $x$;
  • Run $M^t$ on $x$;
  • If $x$ is accepted by both $M_A^t$ and $M^t$, increment $c$ by $1$.
  • If $c \ge 10$ halt and accept.
  1. Double $t$, set $c=0$, and repeat from point 2.

In particular, the above Turing machine accepts $M$ when $t \ge \max\{T, \ell\}$, were $\ell$ is the length of the the tenth shortest word in $L(M_A^t) \cap L(M^t)$.

Regarding your second problem: let $M$ be a Turing machine that accepts $A$ and, for any word $x \in A$, let $t_x$ be the number of step executed by $M$ on input $x$.

Define $B_A = \{ (x, 0^{t_x}) \mid x \in A \}$, where $0^{t_x}$ denotes the word consisting of $t_x$ zeros. To decide whether $(x, y) \in B_A$ it suffices to simulate $M$ for (up to) $|y|$ steps. If $M$ takes exactly $y$ steps to accept, then accept. Otherwise reject.

  • $\begingroup$ That is beautiful. Any ideas on the second problem? $\endgroup$ Commented Mar 23, 2022 at 14:05
  • 1
    $\begingroup$ See my edited answer. $\endgroup$
    – Steven
    Commented Mar 23, 2022 at 15:44

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