Let $L$ be the language of all even-length strings whose first half is a palindrome.

Let $L$ be the language of all even length strings whose first half is imbalanced—with an unequal number of $\mathtt{a}$'s and $\mathtt{b}$'s. I believe $L$ is not context-free; the intuition is that the equipment of a PDA can either be used to keep both halves of the string the same length, or to construct complex internal structure like a palindrome or an imbalanced string, but not both simultaneously.

I would like to formalize this intuition as a general conjecture, beyond this particular example $L$. I suggest that, letting $A\bowtie B$ denote the language $\{xy : x\in A,\;y\in B,\; |x|=|y|\}$,

Conjecture: $A\bowtie \Sigma^*$ is context free if and only if $A$ is regular.

This conjecture would then establish that $L\equiv \mathsf{imbalanced}\bowtie \Sigma^*$ isn't context-free. One direction $(\Leftarrow)$ is easy, using regular grammars. But I wonder if anyone knows or can see a proof of the other direction?

Is a proof of this statement already well known or easily constructible?

Maybe it would be easier to prove this more broadly:

Conjecture: If $A\bowtie B$ is context free and $A$ and $B$ have the same word sizes $\{|x| : x\in A\} = \{|y| : y \in B\}$, then $A$ and $B$ are regular.

So far I have tried arguing about the derivation tree for $A\bowtie \Sigma^*$, considering cases of which parentheses are allowed on which halves of the string in the Chomsky Schuetzenberger theorem, limiting to the case where $A$ and $\Sigma^*$ have different(-colored) alphabets, and similar. I have proved it for the case where the alphabets of $A$ and $B$ are disjoint (e.g. color-coded), by arguing about the cases of colors of strings that the nonterminals of a grammar for $A\bowtie B$ can produce in CNF.

Edit: Having looked at generating functions, I know that if $A(x)$ is the generating function for strings of each length $n$ in $A$, then $A(2x^2)$ is the generating function for $A\bowtie \Sigma^*$, and I know that regular grammars correspond to rational generating functions, unambiguous context-free grammars have algebraic generating functions, and $\bowtie$ corresponds to the Hadamard product, which has certain closure properties relative to rational and algebraic functions—but since I don't know a characterization of what kind of context-free languages $A\bowtie B$ can be, I don't see what to uncover next.

  • 2
    $\begingroup$ Maybe I'm mistaken, but it seems to me that the first language can be proven not CFL using the pumping lemma. Take the words $s=uvwxy=a^Nba^Nb^{2N+1}$. Since $|vwx|\leq N$. By pumping, the palindrome is always destroyed. $\endgroup$
    – Lamine
    Apr 29, 2022 at 14:56
  • $\begingroup$ Thanks! I think you're right. I will edit the question with a better example, and hopefully still be able to prove the theorem. $\endgroup$
    – user326210
    Apr 30, 2022 at 7:42
  • $\begingroup$ cs.stackexchange.com/q/151275/755 $\endgroup$
    – D.W.
    May 7, 2022 at 1:44
  • $\begingroup$ @D.W. Useful reference to show that the conjecture for $A\bowtie B$ doesn't hold for arbitrary $B$. I wonder if there's a way to apply that reasoning to the original conjecture about $A\bowtie \Sigma^*$. $\endgroup$
    – user326210
    May 7, 2022 at 21:57
  • 3
    $\begingroup$ On alphabet $\Sigma = \{0, 1\}$, isn't the language $\{0^n 1 (0|1)^{3n+1} | n > 0\}$ context-free, for the same reason that $\{0^n 1^{3n} | n > 0\}$ from the CS.SE question is? If so, doesn't this refute your claim because $A = \{0^n 1 (0|1)^n | n > 0\}$ is context-free but not regular? $\endgroup$
    – a3nm
    May 10, 2022 at 9:12

1 Answer 1


This answer is inspired by this question. The conjecture is false, for the following reason. Consider the alphabet $\Sigma=\{0,1\}$, and the language $A=\{0^n1(0|1)^n|n>0\}$. This language is context-free but not regular.

Now, $A \bowtie \Sigma^*$ is the language $\{0^n 1 (0|1)^{3n+1}|n>0\}$. This language is context-free, for the same reason as the language $\{0^n 1^{3n}|n>0\}$ from the question mentioned before.

Thus, this refutes the conjecture.


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