Design a grammar for this context-free language

Question

I am doing an exercise from Models Of Computation - Ch - 5, Q-1(r).

Design a grammar that generates this context-free language

$\{ x\space\$\space y^R \,|\, x, y \in\{0, 1\}^* \text{ and } x \ne y\}$

Any hint will be nice.

(My try: I can't seem to come up with a grammar. The pushdown automaton that can accept this language seems easy though. First push all of x into the stack, which can be identified on seeing the $. Keep matching y's characters one by one with top of stack character and pop until the first mismatch. If no mismatch is encountered and the stack is empty as well as y is exhausted, then reject. If a mismatch is encountered, or if the stack becomes empty before y is exhausted, or y is exhausted and stack is still not empty, then accept.)

Answer 1

Hint: Break this into four different cases:

1. Words of the form $x\$y$ where $|x|>|y|$.
2. Words of the form $x\$y$ where $|x|<|y|$.
3. Words of the form $\Sigma^* 0 \Sigma^n \$ \Sigma^n 1 \Sigma^*$.
4. Words of the form $\Sigma^* 1 \Sigma^n \$ \Sigma^n 0 \Sigma^*$.

Answer 2

Since I have been told to somehow prove that the grammar indeed generates the language describe, I am adding that in the form of an answer.

Breaking the language into 6 subsets (not mutually exclusive) and giving rules for generating each. The first two cases generate all those words such that $|x| \ne |y|$

• Case 1: $|x| > |y|$

  $\begin{align*} X_{Longer} &\to (0 \mid 1) \space X_{Longer} \space (0 \mid 1) \\ X_{Longer} &\to (0 \mid 1)^+ \space \$ \end{align*}$

• Case 2: $|x| < |y|$

  $\begin{align*} Y_{Longer} &\to (0 \mid 1) \space Y_{Longer} \space (0 \mid 1) \\ Y_{Longer} &\to \$\space(0 \mid 1)^+ \end{align*}$

The next 4 cases generate words where the first mismatch occurs at the $(|x| - n + 1)$-th character of string $x$ and $n$-th character of string $y$ for all $n > 0$. (Note, $(|x| - n + 1)$-th character of string $x$ is the same as saying $n$-th character of $x^R$)

The rules have names of the form $M_{c, l, r}$ where $c$ is the character which repeats consecutively $n$ times toward the front and back of the $\$$ before the mismatch. $l$ and $r$ are the mismatching characters in $x$ and $y$ respectively.

• Case 3: Words of the form $(0\mid1)^*\space 0 \space 1^n\space\$ \space 1^n \space 1 \space (0\mid1)^*, \forall n \geq 0$

  $M_{1, 0, 1} \to (0\mid1)^*\space 0 \space O \space 1 \space (0\mid1)^*$

• Case 4: Words of the form $(0\mid1)^*\space 1 \space 1^n\space\$ \space 1^n \space 0 \space (0\mid1)^*, \forall n \geq 0$

  $M_{1, 1, 0} \to (0\mid1)^*\space 1 \space O \space 0 \space (0\mid1)^*$

• Case 5: Words of the form $(0\mid1)^*\space 0 \space 0^n\space\$ \space 0^n \space 1 \space (0\mid1)^*, \forall n \geq 0$

  $M_{0, 0, 1} \to (0\mid1)^*\space 0 \space Z \space 1 \space (0\mid1)^*$

• Case 6: Words of the form $(0\mid1)^*\space 1 \space 0^n\space\$ \space 0^n \space 0 \space (0\mid1)^*, \forall n \geq 0$

  $M_{0, 1, 0} \to (0\mid1)^*\space 1 \space Z \space 0 \space (0\mid1)^*$

And to complete, there's the $O$ and $Z$ rules

$\begin{align*} O &\to 1\space O \space 1 \\ O &\to \$ \\ Z &\to 0\space Z \space 0\\ Z &\to \$ \end{align*}$

So the starting non-terminal is:

$S \to X_{Longer} \mid Y_{Longer} \mid M_{1, 0, 1} \mid M_{1, 1, 0} \mid M_{0, 1, 0} \mid M_{0, 0, 1}$