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The pumping lemma for regular languages can be used to prove that certain languages are not regular, and the pumping lemma for context-free languages (along with Ogden's lemma) can be used to prove that certain languages are not context-free.

Is there a pumping lemma for deterministic context-free languages? That is, is there a lemma akin to the pumping lemma that can be used to show that a language is not a DCFL? I'm curious because almost all of the proof techniques I know to show that a language is not a DCFL are really complicated, and I was hoping that there was an easier technique.

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    $\begingroup$ There are some related questions that may or may not be relevant. $\endgroup$ – Raphael Apr 2 '13 at 15:54
  • $\begingroup$ Computer scientists may be sadists, but they aren't are all masochists who use over-complicated proof techniques where simpler ones are known... $\endgroup$ – vonbrand Apr 2 '13 at 16:39
  • $\begingroup$ vonbrand: But any mathematician or computer scientist might use over-complicated proof techniques if simpler ones are not yet known or not known to him. $\endgroup$ – Blaisorblade May 13 '13 at 20:56
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There is a Pumping Lemma specifically for DCFL, under the title "A Pumping Lemma for Deterministic Context-Free Languages", by Sheng Yu; Information Processing Letters 31 (1989) 47-51, doi 10.1016/0020-0190(89)90108-7. With this explicit title I must apologize that I missed it!

The online copy unfortunately has a blank spot in one of the formula, so I hope I reconstructed the result properly. Below ${}^{(1)}y$ is the first symbol of $y$ (when it exists) or $\varepsilon$ (if $y=\varepsilon$).

Lemma 1 (Pumping Lemma). Let $L$ be a DCFL. Then there exists a constant $C$ for $L$ such that for any pair of words $w,w'\in $ if

(1) $w=xy$ [?] and $w'=xz$, $|x|>C$ and

(2) ${}^{(1)}y = {}^{(1)}z $, [?]

then either (3) or (4) is true:

(3) there is a factorization $x=x_1x_2x_3x_4x_5$, $|x_2x_4|\ge 1$ and $|x_2x_3x_4|\le C$, such that for all $i\ge 0$ $x_1x^i_2x_3x^i_4x_5y$ and $x_1x^i_2x_3x^i_4x_5z$ are in $L$;

(4) there exist factorizations $x=x_1x_2x_3$, $y=y_1y_2y_3$ and $z=z_1z_2z_3$, $|x_2|\ge 1$ and $|x_2x_3|\le C$, such that for all $i\ge 0$ $x_1x^i_2x_3y_1y^i_2y_3$ and $x_1x^i_2x_3z_1z^i_2z_3$ are in $L$.

Two applications of the Lemma are given: $\{ a^ib^i \mid i\ge 0 \} \cup \{ a^ib^{2i} \mid i\ge 0 \}$ as well as $\{ w\in\{a,b\}^* \mid w=uv, |u|=|v|, \mbox{ and } v \mbox{ contains an } a \}$ are not DCFL. The proof uses the fact that each DCFL has an LR(1) grammar in Greibach normal form.

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  • $\begingroup$ I hope you can use it. It is even more complicated to state than the known pumping lemma. $\endgroup$ – Hendrik Jan Apr 12 '13 at 21:04

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