|
I remember coming across the following question about a language that supposedly is context-free, but I was unable to find a proof of the fact. Have I perhaps misremembered the question? Anyway, here's the question:
|
|||||
|
|
Claim: $L$ is context-free. Proof Idea: There has to be at least one difference between the first and second half; we give a grammar that makes sure to generate one and leaves the rest arbitrary. Proof: For sake of simplicity, assume a binary alphabet $\Sigma = \{a,b\}$. The proof readily extends to other sizes. Consider the grammar $G$: $\qquad\begin{align} S &\to AB \mid BA \\ A &\to a \mid aAa \mid aAb \mid bAa \mid bAb \\ B &\to b \mid aBa \mid aBb \mid bBa \mid bBb \end{align}$ It is quite clear that it generates $\qquad \mathcal{L}(G) = \{ \underbrace{w_1}_k x \underbrace{w_2v_1}_{k+l}y\underbrace{v_2}_l \mid |w_1|=|w_2|=k, |v_1|=|v_2|=l, x\neq y \} \subseteq \Sigma^*;$ the suspicious may perform a nested induction over $k$ and $l$ with case distinction over pairs $(x,y)$. Now, $w_2$ and $v_1$ commute (intuitively speaking, $w_2$ and $v_1$ can exchange symbols because both contain symbols chosen independently from the rest of the word). Therefore, $x$ and $y$ have the same position (in their respective half), which implies $\mathcal{L}(G) = L$ because $G$ imposes no other restrictions on its language. The interested reader may enjoy two follow-up problems: Exercise 1: Come up with a PDA for $L$! Exercise 2: What about $\{xyz \mid |x|=|y|=|z|, x\neq y \lor y \neq z \lor x \neq z\}$? |
|||||||||||
|
