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How could I prove that the following language is decidable?

$\{\langle G\rangle \mid G\ \text{is a CFG over}\ \{0,1\}\ \text{and}\ 1^* \subseteq L(G)\}$

P.S. It's the problem 4.15 of the third edition of the "Introduction to the Thoery of Computation" by Michael Sipser.

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You can find the solutions to the exercises of that book here. For completeness of the answer, I rewrite the solution (with some more details) here.

Let $p$ be an upper bound of the pumping length of $G$, you only need to check whether $1^0, 1^1,\ldots,1^{p+p!} \in G$. If not, we can certainly reject $\langle G\rangle$. Otherwise, applying pumping lemma on $1^k$ ($p\le k\le p + p!$), we can conclude that there exist substrings $u,v,w,x,y$ such that

  • $1^k=uvwxy$, which means $u,v,w,x,y$ contain only 1s,
  • $|vwx|\le \text{pumping length} \le p$,
  • $|vx|\ge1$ and
  • $uv^nwx^ny\in G$, i.e. $1^{|uwy|+n|vx|}=1^{k+(n-1)|vx|}\in G$, for all $n\ge 0$, which means $1^{k+n(p!)}\in G$ for all $n\ge0$.

Note $\{1^{k+n(p!)}\mid n\ge0, p\le k\le p + p!\}$ covers all $1^m$ for large $m$, so $1^0, 1^1,\ldots,1^{p+p!} \in G$ indeed implies $1^*\subseteq G$.

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  • $\begingroup$ Thanks, but the only thing I cannot realize that is the p+p!. Could you explain more about that? $\endgroup$ – javadr Jun 5 '18 at 22:45
  • $\begingroup$ @javadr $p!$ is a multiple of all possible $|vx|$, so to make $1^{k+(n-1)|vx|}$ cover all $1^m$ for large $m$, it is natural to let $k$ range from $p$ to $p+p!$. $\endgroup$ – xskxzr Jun 6 '18 at 2:32

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