What is the language of following CFG? A CFG G is given with the following productions where S is the start symbol, A is a non-terminal and a and b are terminals.
$$S → aS \mid A \\
A → aAb \mid bAa \mid \epsilon$$
Which of the following strings is generated by the grammar above?


*

*aabbaba 

*aabaaba 

*abababb 

*aabbaab



AFAIK :
Option $(4)$ is true.
$S→aS→aA→aaAb→aabAab→aabbAaab→aabbaab$.

Can you explain please, what is the language of this grammar?

I'd tried for :


*

*More or equal number of a's than b's : False

*Starting and ending with same symbol : False

*Starting with a and more number of a's than b's : False

 A: A derivation may start by using the production $S\to aS$ any number of times, but eventually it must use the production $S\to A$, since the only terminal production is $A\to\epsilon$. Say it uses $S\to aS$ $n$ times, where $n\ge 0$, and then $S\to A$; at that point we have $a^nA$.
Now consider what kinds of strings can be produced starting with $A$. The only available non-terminal productions are $A\to aAb$ and $A\to bAa$, each of which produces one $a$ and one $b$. Suppose that we apply these productions a total of $m$ times; then at that point we have
$$a^nx_1x_2\ldots x_{m-1}x_mA\overline{x}_m\overline{x}_{m-1}\ldots\overline{x}_2\overline{x}_1\;,$$
where each $x_k$ is either $a$ or $b$, and 
$$\overline{x}_k=\begin{cases}
b,&\text{if }x_k=a\\
a,&\text{if }x_k=b\;.
\end{cases}$$
If we now apply $A\to\epsilon$, we end up with something of the form
$$a^nx_1x_2\ldots x_{m-1}x_m\overline{x}_m\overline{x}_{m-1}\ldots\overline{x}_2\overline{x}_1\;.$$
For any word $w\in\{a,b\}^*$ let $\widehat{w}$ be the word obtained by reversing the order of $w$, replacing each $a$ by a $b$, and replacing each $b$ by an $a$. For instance, if $w=abb$, then $\widehat{w}=aab$. Then the language of this grammar is the set of all words of the form $a^nw\widehat w$, where $n\ge 0$, and $w\in\{a,b\}^*$:
$$L(G)=\big\{a^nw\widehat w:n\ge 0\text{ and }w\in\{a,b\}^*\big\}\;.$$
Now just test your four strings against this description; you’ll find that only one of them fits. For instance, $aabbaba$ doesn’t fit. To see this, observe that the $w\widehat w$ part must have even length, so it must be $abbaba$, $baba$, $ba$, or $\epsilon$. There can be no $b$s before it, so it must be $abbaba$. But then $w=abb$, so $\widehat w=aab\ne aba$, and $abbaba$ doesn’t have the right form.
A: The language of this grammar is the one given by any string that has the following structure. The string $s = s_+ s_-$ is divided into the prefix $s_+$ and the postfix $s_-$. The prefix consists of an arbitrary number (possibly zero) of $a$'s. Let $n$ be the length of $s_-$, so that we can write
$$s_- = s_1 s_2 \dots s_n$$
and note that $n$ must be an even number. Then the postfix $s_-$ must be so that if $s_i$ is $a$, then $s_{(n+1)-i}$ must be $b$ and if $s_i$ is $b$ then $s_{(n+1)-i}$ must be $a$. In other words, when there is an $a$ in the first half of the postfix, there must be a $b$ in the corresponding place in the second half of the postfix and vice versa.
With this in mind, you can also clearly see that none of the three first strings are generated by the grammar, whereas the fourth is.
