GATE CS 2017 Set 2 Q.16 (CS Q.16): Context-Free Grammars, Pushdown Automata and CFLs

Identify the language generated by the following grammar, where $S$ is the start variable.

\begin{aligned} S &\to XY \\ X &\to aX \mid a \\ Y &\to aYb \mid \epsilon \end{aligned}

$\{a^m b^n \mid m \ge n, n > 0\}$

$\{a^m b^n \mid m \ge n, n \ge 0\}$

$\{a^m b^n \mid m > n, n \ge 0\}$

$\{a^m b^n \mid m > n, n > 0\}$

Official answer and solution

Solution

Find the language generated by each non-terminal and concatenate their languages.

Analyze the production rules of the grammar:

Non-terminal $X$ has productions $X \to aX \mid a$ . This generates the regular language $a^+$ (i.e., $a^k$ with $k \ge 1$ ).
Non-terminal $Y$ has productions $Y \to aYb \mid \epsilon$ . This generates equal numbers of $a$ 's and $b$ 's: $\{a^n b^n \mid n \ge 0\}$ .
The start production is $S \to XY$ . Therefore, any string generated by $S$ is of the form:

w = a^k \cdot a^n b^n = a^{k+n} b^n \quad \text{where } k \ge 1, n \ge 0

Let $m = k + n$ . Since $k \ge 1$ and $n \ge 0$ , we have $m \ge n + 1$ , which is equivalent to $m > n$ , with $n \ge 0$ .

Thus, the generated language is $\{a^m b^n \mid m > n, n \ge 0\}$ , which matches option C.

Final answer

Common trap

Overlooking that $Y \to \epsilon$ allows $n = 0$ , which yields strings like $a^1, a^2, \dots$ where $n = 0$ .

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