How does this context-free grammar generate even length strings on either side?

Question

I came across this context-free grammar for the language L = {xy||x|=|y|, x≠y}, but I can't seem to see how it can generate all lengths for x and y. Could someone illustrate this? For example, how would you derive the string "abba" or "aaabbaaa"?

Answer 1

Suppose that $|x| = |y| = n$ and $x_i \neq y_i$, say $x_i = a$ and $y_i = b$. We can therefore write $x = r a s$ and $y = ubv$, where $|r| = |u| = i-1$ and $|s| = |v| = n-i$. Therefore $$xy = rasubv, $$ where $|r| = i$, $|su| = (n-i) + (i-1)$, $|v| = n-i$. Now let $p$ be the first $i-1$ letters of $su$, and let $q$ be the final $n-i$ letters of $su$, so that $su = pq$. Then $$ xy = rapqbv. $$ Now $A$ generates $rap$ (since $|r| = |p| = i-1$) and $B$ generates $qbv$ (since $|q| = |v| = n-i$), and so the grammar generates $xy$.

Now let's consider your two examples, starting with $abba$. In this case, $x = ab$ and $y = ba$, and so there are two choices for $i$: $i = 1$ and $i = 2$.

If we choose $i = 1$ then $x = ras$ for $r = \epsilon$ and $s = b$, and $y = ubv$ for $u = \epsilon$ and $v = a$. We have $su = b = pq$, and so $p = \epsilon$ and $q = b$. We now generate $xy$ as follows. First, $A \Rightarrow^* rap = a$. Second, $B \Rightarrow^* qbv = bba$. Finally, $S \Rightarrow AB \Rightarrow^* (a)(bba)$.

If we choose $i = 2$ then $x = rbs$ for $r = a$ and $s = \epsilon$, and $y = uav$ for $u = b$ and $v = \epsilon$. We have $su = b = pq$ and so $p = b$ and $q = \epsilon$. We can generate $rbp = abb$ using $B$ and $qav = a$ using $A$, and put them together via $S \Rightarrow BA$ to generate $(abb)(a)$.

In your second example $aaabbaaa$ we similarly have two options for $i$: $i = 1$ and $i = 4$. If we choose $i = 1$ then $x=ras$ for $r = \epsilon$ and $s = aab$, and $y = ubv$ for $u = \epsilon$ and $v = aaa$. Then $su = aab = pq$, and so $p = \epsilon$ and $q = aab$. We generate $rap = a$ using $A$ and $qbv = (aab)b(aaa)$ using $B$, and put them together via $S \Rightarrow AB$ to generate $(a)(aabbaaa)$.

Finally, if $i = 4$ then $x = rbs$ for $r = aaa$ and $s = \epsilon$, and $y = uav$ for $u = baa$ and $v = \epsilon$. Then $su = baa = pq$, and so $p = baa$ and $q = \epsilon$. We generate $rbp = (aaa)b(baa)$ using $B$ and $qav = a$ using $A$, and put them together via $S \Rightarrow BA$ to generate $(aaabbaa)(a)$.