GATE CS 2025 Set 2 Q.52: Regular Expressions, Finite Automata and Regular Languages

Let $\Sigma = \{a, b, c\}$ . For $x \in \Sigma^*$ , and $\alpha \in \Sigma$ , let $\#_\alpha(x)$ denote the number of occurrences of $\alpha$ in $x$ .

Which one or more of the following option(s) define(s) regular language(s)?

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

$\{a, b\}^* \cap \{a^m b^n c^{m-n} \mid m \ge n \ge 0\}$

$\{w \mid w \in \{a, b\}^*, \; \#_a(w) \equiv 2 \pmod 7, \text{ and } \#_b(w) \equiv 3 \pmod 9\}$

$\{w \mid w \in \{a, b\}^*, \; \#_a(w) \equiv 2 \pmod 7, \text{ and } \#_a(w) = \#_b(w)\}$

Official answer and solution

Official answer: A and C
From: Official GATE 2025 answer key

Solution

Test each language for finite-state recognizability versus unbounded comparison of counts.

(A) is Regular: $\{a^m b^n \mid m, n \ge 0\}$ is generated by the regular expression $a^* b^*$ .
(B) is NOT Regular: Any string in $\{a, b\}^*$ contains zero $c$ 's, so intersecting with $\{a^m b^n c^{m-n} \mid m \ge n \ge 0\}$ forces $m - n = 0 \implies m = n$ , yielding $\{a^n b^n \mid n \ge 0\}$ , which is a non-regular CFL.
(C) is Regular: Tracking $\#_a(w) \bmod 7$ requires $7$ states and tracking $\#_b(w) \bmod 9$ requires $9$ states; their product DFA has $7 \times 9 = 63$ states.
(D) is NOT Regular: Enforcing $\#_a(w) = \#_b(w)$ requires unbounded memory and cannot be done by a finite automaton.

Final answer

Options (A), (C)

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 2 (CS2)
Question: 52 of 65
Section: Computer Science and Information Technology

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