Example Context free grammar

Question

Is there a nice way to give context free grammar for $$\{a^nb^ma^kb^l:n+m=k+l\}?$$

From PDA point of view it seems we just push + on stack if we see a, push + on stack if we see b, pop + from stack if we see a, pop + from stack if we see b and accept if you have empty stack.

Every decent automata theory textbook gives a description of how to convert a PDA to a CFG. So, simply apply that construction and you'll get a CFG for that language. There's little point in us repeating the description that construction here, and a question that says "please run standard algorithm X on my particular input Y for me" seems of unclear value (to me) -- better for you to run it yourself. — D.W., Nov 07 '15 at 10:30

score 3 · Answer 1 · answered Nov 05 '15 at 20:09

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Here is a simple grammar: $$ \begin{align*} & S \to aSb|U|V \\ & U \to aUa|T \\ & V \to bVb|T \\ & T \to bTa|\epsilon \end{align*} $$

answered Nov 05 '15 at 20:09

Yuval Filmus

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And why would that be correct? – Raphael Nov 05 '15 at 20:46
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Just a feeling. The OP didn't ask for a proof, just for a grammar. – Yuval Filmus Nov 05 '15 at 20:56
Of course it is indeed correct. Very nice job, Yuval. – Rick Decker Nov 06 '15 at 14:38
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Looks like something Arul can use to practice structural induction on. – G. Bach Nov 06 '15 at 14:52
no I am just looking for a verbatim way to see if PDA has a canonical transfer to CFG – Nov 07 '15 at 20:10
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@Arul There is an algorithm for converting PDA to CFG. You could run it. – Yuval Filmus Nov 07 '15 at 23:04
@YuvalFilmus it seems very intuitive to think of pads rather than cfgs I don't know why this is – Turbo Nov 08 '15 at 01:02
Also is there a notion of non-ambiguous PDAs? – Turbo Nov 08 '15 at 01:03

Example Context free grammar

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