Is this language context free?

$L = \{a^kb^lb^ka^l \ | \ k,l \in \mathbb{N}\}$

Using Pumping Lemma and $z = a^nb^nb^na^n$ I find it contradicting PL.

If $z = uvwxy$ and $|vwx| \leq n$, follows:

  1. $vwx$ is in either $a^n$ or $b^n$, second $b^n$, or second $a^n$. After pumping form of the word changes and thus pumped word not anymore in the language. Done

  2. $vwx$ lies between $a^nb^n$ or $b^nb^n$ or $b^na^n$. When we pump down, we get $a^pb^qb^na^n$ or $a^nb^pb^qa^n$ or $a^nb^nb^pa^q$ with $p,q \leq n$. Which shows that pumped word not in the language.

Where is the error? The language should be context free. Why PL shows that language is not context free?


1 Answer 1


The language is indeed context-free. Since it can be defined as $L = \{a^kb^kb^la^l\mid k,l\geqslant 0\}$, the following grammar can generate it:

$S \rightarrow XY$

$X\rightarrow AXB\mid \varepsilon$

$Y\rightarrow BYA\mid \varepsilon$

In your tentative of proof, the second case is wrongly considered. You could have $vwx \in a^*b^*$ without any problem. For example, if $u=a^{n-1}$, $v =a$, $w=\varepsilon$, $x=b$ and $y = b^{2n-1}a^n$, then for any $k\geqslant 0$, $uv^kwx^ky = a^{n-1 +k}b^{2n-1+k}a^{n}\in L$.

  • $\begingroup$ You wrote down wrong language. It is not $a^kb^k$. It is $a^kb^l$. And your grammar is wrong too. $\endgroup$
    – cs_student
    Sep 23, 2022 at 10:00
  • $\begingroup$ And why do write $a^*b^*$. Here we work with particluar word of the language $a^nb^nb^na^n$, not with language. $\endgroup$
    – cs_student
    Sep 23, 2022 at 10:06
  • $\begingroup$ @cs_student a word of the language $a^kb^lb^ka^l = a^kb^{l+k}a^l = a^kb^kb^la^l$. Also I wrote $a^*b^*$, to express "$vwx$ lies between $a^nb^n$" more formally. $\endgroup$
    – Nathaniel
    Sep 23, 2022 at 11:52
  • $\begingroup$ Shouldn't PL work for every combination? It seems in your case it only works for particular configuration. When there are different number of a's and b's in v and x, when pumping we get result, that doesn't sum up as elegantly as in your example. Or am I doing something wrong? $\endgroup$
    – cs_student
    Sep 23, 2022 at 16:46
  • $\begingroup$ Pumping lemma states that there exists a configuration that can be pumped. To prove that a language is not CFL, you have to prove that there isn't any configuration that can be pumped. That is why your proof failed: you didn't consider all configurations, and the one I proposed prove that $a^nb^nb^na^n$ cannot be a counter-example. $\endgroup$
    – Nathaniel
    Sep 24, 2022 at 8:37

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