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Give a context-free grammar for the following language: $\{ a^i b^n a^n \mid i \ge 0, n \ge 0 \}$

So far, this is the solution that I have been able to come up with, though I am not sure how accurate it is:

$$\begin{align*} S &\to aSa \mid X \\ X &\to bX \mid a \mid \varepsilon \end{align*}$$

I worked this out from outside first, which gives me the first and last a's and then used the second rule to get the b's or a's after.

The reason I came up with that answer, I thought the grammer need to start and end with "a" and then have b's in between those a's.

So for each letter with different power we need to have a rule? That is, for $a^i$ we need a rule and for the $n$ power we need another one ?

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  • $\begingroup$ If you are given a PDA that accepts the language, there is a recipe to convert that PDA to a CFG, which is not easy to follow by hand usually, though. $\endgroup$ – Apass.Jack Dec 17 '18 at 15:08
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You could break down the language into simpler parts, e.g., you can generate languages separately for $a^i$ and $b^na^n$, then concatenate them in the end (we know that CFG's are closed under concatenation).

In your case, you can do

A $\rightarrow$ aA | $\epsilon$

to generate $a^i$, and

B $\rightarrow$ bBa | $\epsilon$

to generate $b^na^n$, then concatenate by doing:

S $\rightarrow$ AB

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You have created a grammar for $\{ a^i b^k a^j \mid i, k \in \mathbb{N}_0, j \in \{ i, i + 1 \} \}$, which is not what you wanted.

Hint: Try to generate the a's and b's for the $b^n a^n$ part simultaneously.

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