Context-free grammar for complement of $\{ab \mid b=\mathrm{complement}(a)\}$

Question

I want to construct a context-free grammar for $$L=\Sigma^*-\{ab \mid b=\mathrm{complement} (a) , a,b \in \{0, 1\}^*\}$$ and prove the correctness of answer. The complement of a string is obtained by switching $0$s and $1$s. For example, $\mathrm{complement}(001) = 110$.

I'm trying to decompose the language but I'm not sure about the different cases I should consider. For example, I know that $L$ contains all strings with odd length. But what else do I have to consider? Also, I can't make sure that the language of constructed grammar equals the given language, so a proof is needed.

Answer 1

As you have noticed, you can decompose between words of odd length and words of even length. Since the odd part is quite easy, I will focus on the even one, that is $L = \{uv \mid u, v\in \{0,1\}^*\wedge u\neq \mathrm{complement}(v)\}$.

Given $u, v\in\{0,1\}^*, |u|=|v|$, if we write $u = u_1…u_k$ and $v = v_1…v_k$, then $uv \in L$ if and only if $k \geq 1$ and $\exists i \in [\![1,k]\!], u_i = v_i$. In that case, we can write $uv = wu_ixv_iy$ where $|w| = u_1…u_{i-1}$ and $y = v_{i+1}…v_k$. We also have $|x| = k-i + i -1= k-1$, so $x$ can be written as $x = zt$ with $|z| = i-1 = |w|$ and $|t| = k-i = |y|$.

We will now write a grammar to generate these words. Let's define:

$S\to AA\mid BB$

$A \to 0 \mid 0A0 \mid 0A1 \mid 1A0 \mid 1A1$

$B\to 1\mid 0B0\mid 0B1 \mid 1B0 \mid 1B1$

It is clear that $A$ rules will generate a word $w0x$ with $|w| = |x|$. That means that $AA$ will generate $w0zt0y$, with $|w| = |z|$ and $|t| = |y|$. The same thing can be done with $BB$.

(This answer was greatly inspired by this one.)