GATE CS 2025 Set 1 Q.19: Context-Free Grammars, Pushdown Automata and CFLs

Consider the following context-free grammar $G$ , where $S$ , $A$ , and $B$ are the variables (non-terminals), $a$ and $b$ are the terminal symbols, $S$ is the start variable, and the rules of $G$ are described as:

S \to aaB \mid Abb

A \to a \mid aA

B \to b \mid bB

Which ONE of the languages $L(G)$ is accepted by $G$ ?

$L(G) = \{a^2 b^n \mid n \ge 1\} \cup \{a^n b^2 \mid n \ge 1\}$

$L(G) = \{a^n b^{2n} \mid n \ge 1\} \cup \{a^{2n} b^n \mid n \ge 1\}$

$L(G) = \{a^n b^n \mid n \ge 1\}$

$L(G) = \{a^{2n} b^{2n} \mid n \ge 1\}$

Official answer and solution

Official answer: A
From: Official GATE 2025 answer key

Solution

Find the languages generated by non-terminals $A$ and $B$ and take the union of the two productions of $S$ .

Non-terminal $A \to a \mid aA$ generates all non-empty strings of $a$ 's: $L(A) = \{a^n \mid n \ge 1\}$ .
Non-terminal $B \to b \mid bB$ generates all non-empty strings of $b$ 's: $L(B) = \{b^n \mid n \ge 1\}$ .
Substituting into $S \to aaB \mid Abb$ gives:

L(G) = \{a^2 b^n \mid n \ge 1\} \cup \{a^n b^2 \mid n \ge 1\}

Final answer

Option (A)

The answer is the one in the official GATE answer key. The solution is this site's own explanation.

Source

Paper: GATE 2025 Computer Science and Information Technology - Session 1 (CS1)
Question: 19 of 65
Section: Computer Science and Information Technology

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