# The operator $A(L)= \{w \mid ww \in L\}$

Consider the operator $A(L)= \{w \mid ww \in L\}$. Apparently, the class of context free languages is not closed against $A$. Still, after a lot of thinking, I can't find any CFL for which $A(L)$ wouldn't be CFL.

Does anyone have an idea for such a language?

• To be more spesific, I thought of context free languages in a form similar to $L=\{0^n1^n0^m1^m|n,m >=0 \}$, which is CFL, but than A(L) is CFL as well, so it doesn't help me.. it seems to me that every CFL L that I would write down, its A(L) would be CFL as well, apperantly it doesn't correct. – Jozef May 3 '12 at 14:14
• Jozef: Think of some well-known noncontextfree language $L$, and then try to find $L'$ such that $A(L')=L$. – sdcvvc May 3 '12 at 15:07

As an amplification of sdcvvc's hint: You can find a context-free language $L$ such that $A(L) = \{a^m b^m c^m \mid m\geq 0\}$, which as you probably know is not context-free.
So we want to find $L$ such that if $ww\in L$, then $w$ has the form $a^m b^m c^m$. We might as well have $L$ be a sublanguage of $a^*b^*c^*a^*b^*c^*$. Now we just need to put enough restrictions on the words in $L$ so that words of the form $ww$ in $L$ will have the form we require, while ensuring $L$ is still context-free.
A word $a^{m_1} b^{m_2} c^{m_3} a^{m_4} b^{m_5} c^{m_6}$ has the form $ww$ if and only if $m_1 = m_4$, $m_2 = m_5$ and $m_3 = m_6$. Now we can add some restrictions on the $m_i$ for words in $L$ to force them to all be equal in words of the form $ww$. (Hint: we can require $m_1 = m_2$, but then we can't require $m_1 = m_3$ or $m_2 = m_3$, because then $L$ wouldn't be context-free anymore. This isn't a problem though, because we can make use of some of the equalities we already have between the $m_i$.)
• If I got you and the hint right, I am supposed to get that L is of some form of $\{a^mb^mc^ma^mb^mc^m \mid m \ge 0\}$, still it doesn't a CFL. – Jozef May 3 '12 at 16:51
• @Jozef: No, you just need to find $L$ such that the words in $L$ of the form $ww$ are $a^m b^m c^m a^m b^m c^m$ $(m\geq 0)$. – Tara B May 3 '12 at 16:55