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I stumbled on this ambiguous grammar and I've been trying to make it unambiguous but it's still ambiguous.

Given the ambiguous CFG :
$S \to A\mid B$
$A \to aAb\mid ab$
$B \to abB\mid \epsilon$

My closet try was:

Given the ambiguous CFG :
$S \to A$
$A \to aAb\mid C$
$B \to b$
$C \to abC\mid\epsilon$

But the string "$ab$" is ambiguous

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  • $\begingroup$ What have you tried? Where did you get stuck? Here is a good series of questions to get you started: cs.stackexchange.com/questions/tagged/… $\endgroup$ – dkaeae Jan 17 '19 at 12:53
  • $\begingroup$ In fact, the original CFG does not generate generate $aababb$, which is generated by your CFG as follows, $S\Rightarrow A\Rightarrow aAb\Rightarrow aCb\Rightarrow aabCb\Rightarrow aababCb\Rightarrow aababb$. $\endgroup$ – John L. Jan 17 '19 at 18:34
  • $\begingroup$ Oh true, those that mean there isn't a solution to this? $\endgroup$ – Owonubi Job Sunday Jan 17 '19 at 19:00
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As you have observed, the only problem in the original CFG is that the string $ab$ can be generated by both $A$ and $B$.

Here is a hint. Instead of generating $ab, a^2b^2, a^3b^3,\cdots$, can you let $A$ generate $a^2b^2, a^3b^3,\cdots$ without generating $ab$?

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  • $\begingroup$ Any idea on how to implement that? $\endgroup$ – Owonubi Job Sunday Jan 17 '19 at 15:54
  • $\begingroup$ How did you avoid an ambiguous parse of the empty string? $\endgroup$ – rici Jan 17 '19 at 16:37
  • $\begingroup$ @OwonubiJobSunday Instead of $A\to aAb\mid ab$, how about $A\to aAb\mid a^2b^2$? $\endgroup$ – John L. Jan 17 '19 at 16:41
  • $\begingroup$ @rici can the empty string be parsed in two ways in the original CFG? $\endgroup$ – John L. Jan 17 '19 at 16:43
  • $\begingroup$ @apass.jack. no, it can't. Why not? Because the base productions differ. $\endgroup$ – rici Jan 17 '19 at 16:47

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