GATE CS 2000 Q2.8 — Regular and Context-free Languages

To find the correct option, let us analyze the possible minimal Deterministic Finite Automata (DFA) with exactly $2$ states over the alphabet $\Sigma = \{a\}$.

Let the two states be $q_0$ (start state) and $q_1$. Since the DFA is minimal, both states must be reachable, meaning there must be a transition $q_0 \xrightarrow{a} q_1$. There are two possibilities for the transition from $q_1$ on symbol $a$:

1. Case 1: $q_1 \xrightarrow{a} q_0$ (Loop of length 2)
   • If $q_0$ is the only accepting state, $L = \{a^n \mid n \text{ is even}\}$.
   • If $q_1$ is the only accepting state, $L = \{a^n \mid n \text{ is odd}\}$.
2. Case 2: $q_1 \xrightarrow{a} q_1$ (Self-loop on $q_1$)
   • If $q_1$ is the only accepting state, $L = \{a^n \mid n \geq 1\}$.
   • If $q_0$ is the only accepting state, $L = \{\epsilon\}$.

Since $L$ can also be $\{a^n \mid n \geq 1\}$ or $\{\epsilon\}$, none of the options (A), (B), (C), or (D) are correct. Thus, this question has no correct option among the choices provided, which is why the official key is marked as X (Marks to All).