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How to generate CFG for this language?

$ L = \{ w \mid w \in \{ (, [, ], ) \}^* \text{ s.t. } $

  • In any prefix of $w$, no. of ( is more than no. of ), and
  • no. of [ is more than no. of ].

$\}$

Thus, ([(), [()[] etc. are valid.

I have tried,

$ S \to (S \mid (S)S \mid [S \mid [S]S \mid \epsilon $

But, this does not accept, ([)].

It seems possible to do with two stacks, by keeping counts of ( and [. Thus it seems it is not Context Free. Any help in proving it is not CF or a CFG exists?

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  • $\begingroup$ Word [()[] is not in $L$ since the no. of ( is not more than no. of ) in the prefix [(). $\endgroup$ – John L. Feb 22 '19 at 3:49
  • $\begingroup$ $S\to (S)S\mid\epsilon$ generates (), which is not in $L$. $\endgroup$ – John L. Feb 22 '19 at 3:50
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The language is not context-free, as you have suspected.

Intuitively, a PDA that accepts the language has to keep track of the difference of the number of (s and the number of )s as well as the difference of number of [s and the number of ]s. Since these two differences vary to arbitrary largeness independently to each other, one pushdown stack is not able to track them. However, this understanding is not a rigorous proof.

One standard way to disprove context-free-ness is to tap into the power of the pumping lemma. A bit of care should be taken to construct the witness word. For example, a word of the form $(^{p+1})^p[^{p+1}]^p$ or $(^{p+1}[^{p+1}]^p)^p$ can be pumped without any problem. However, that word is not far from the right word we need.

In case a more explicit hint is needed, here it is.

Check $(^{p+1}[^{p+1})^p]^p$ assuming the pumping length is $p$.

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