How are these Context-Free Pumping Lemma Approaches differ? Maybe this might help understand pumping lemma better

$(a^{i}b^{i}c^{j}d^{j} \mid i, j \geq 0$}

$(a^{i}b^{j}c^{i}d^{j} \mid i, j \geq 0$}

$(a^{i}b^{j}c^{j}d^{i} \mid i, j \geq 0$}

I understand we use contradiction with these conditions

  1. $|vwx| \leq p$
  2. $|vx| \geq 1$
  3. for every $i \geq 0$, $uv^{i}wx^{i}y \in L$.
  • $\begingroup$ Welcome to Computer Science! Your question is a very basic one. Since you did not include much of an attempt to solve it on your own, we have litte to work with. Let me direct you towards our reference questions which cover your problem in detail. Please work through the related questions listed there, try to solve your problem again and edit to include your attempts along with the specific problems you encountered. Your question may then be reopened. Good luck! $\endgroup$ – Raphael Mar 22 '13 at 14:00
  • $\begingroup$ So, what is your question? You've listed some languages, and you mention the context-free pumping lemma. You seem to be interested in applying that lemma to each language. So, have you tried to do it? What part of the proof gave you trouble? $\endgroup$ – Gilles 'SO- stop being evil' Mar 22 '13 at 19:46
  • $\begingroup$ Dear Moderators. It is clear that not each of the three languages here is non-context-free. After that hint the user has correctly identified these. For the pumping lemma I have directed user to a standard answer that deals with that topic. It seems that the question has been answered to the satisfraction of the user. $\endgroup$ – Hendrik Jan Mar 22 '13 at 22:20

Not all three languages are in fact non-context-free. Try to spot those that have a CF-grammar, and apply pumping to the remaining example(s). That is the difference you mean, I presume?

(added) The grammars you give in a comment seem correct to me. General hints for applying the pumping lemma are given in various answers at this site, see in particular How to prove that a language is not context-free?

  • 1
    $\begingroup$ It seems like like only the second doesn't have a grammar, and the rest do. Is that right? a. $S \rightarrow aSd\mid A$ ; $A \rightarrow bSc \mid \epsilon$ and c. $S \rightarrow AB $ ; $A \rightarrow aAb \mid \epsilon$ ; $B \rightarrow cBd \mid \epsilon$ $\endgroup$ – Iancovici Mar 22 '13 at 11:37
  • $\begingroup$ @echadromani Your observation is correct (and so are the grammars you give) $\endgroup$ – Hendrik Jan Mar 22 '13 at 13:29

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