I can see that $L$ has to be context-free but not regular here as regular languages are closed under concatenation and intersection. But $L\cap L^2$ looks too weird. I couldn't think of any $L$ that gives rise to meaningful $L\cap L^2$.

Any hint would be appreciated!


2 Answers 2


You can take $$ L = \{ a^n b^n : n \geq 1 \} \cup \{ a^k b^n a^n b^\ell : n,k,\ell \geq 1 \}. $$ You can check that $$ L \cap L^2 = \{ a^n b^n a^n b^n : n \geq 1 \}. $$


The intersection of the form $L\cap L^2$ actually turns out to be quite powerful. In fact languages like this can code Turing machine computations.

Let $L_1, L_2 \subseteq\{a,b\}^*$ and let $\#$ be a third "special" symbol. Consider $L = L_1\# \cup L_2\#\#$. The only strings in $L\cap L^2$ must end in $\#\#$, and before that must be a string in $L_1\cap L_2$.

Now one can do weird things indeed. Let $L = \{a^nb^{2n}\mid n\ge 1\}^*a^*\# \cup a^1\{b^na^n\}^*\#\#$. Then $L\cap L^2$ consists of strings of the form $a^1b^2a^2b^4a^4\dots b^{2^k}a^{2^k}\#\#$.


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