# Context free languages [closed]

I have stumbled on this question:

Which of the following languages over the alphabet ${a,b,c,d}$ are context-free and which not ?

a) $L_{1} = \{wa^{3n+1}b^nw^{R} \mid w\in \{c,d\}^*,\ n>0\}$;

b) $L_{2} = \{a^{3n+1}wb^nw^{R} \mid w\in \{c,d\}^*,\ n>0\}$.

For a) I think this grammar solves it :

\begin{align*} S&\to cMc \mid dMd \mid M\\ M&\to aN\\ N&\to aaaTb \mid\varepsilon\\ T&\to \varepsilon \end{align*}

b) doesn't look so nice so I think that we might prove it with the pumping lemma,any suggestions what word to pick?

## closed as unclear what you're asking by David Richerby, Evil, Discrete lizard♦, Yuval Filmus, Thomas KlimpelJun 10 '18 at 21:07

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• Pay attention to the difference between {c,d}* and {c,d}. Your proposed solution for (a) restricts w to {c,d}. – Draconis May 30 '18 at 19:32
• Thank you ,i think i have fixed it any suggestions for the second part? – Karmen May 30 '18 at 20:01
• Still not quite: now it'll accept {c,d}? = {c,d,ε} rather than {c,d}*. – Draconis May 30 '18 at 20:26
• Do you have any question about your answer to a)? This site isn't well-suited to "please check my answer" because it's only interesting to you and tends to degenerate into "OK, I fixed that problem. Now is it right?" and Stack Exchange can't handle discussions. – David Richerby May 30 '18 at 21:06

I think you could use the Pumping Lemma for Contextfree languages to prove that the language (b) is not contextfree. Have a look at the word $a^{3n+1}cdb^ndc$, given n as the Pumping Length.