Consider the DFAs M and N given above. The number of states in a minimal DFA that accepts the language $L(M) \cap L(N)$ is ____________

Let us analyze the languages accepted by the two DFAs:

1. DFA $M$:

   • Starts at state $s_0$ (non-accepting).

   • Any input ending with the symbol 'a' leads to state $s_1$ (accepting).

   • Any subsequent 'b' transitions back to $s_0$.

   • Hence, $L(M) = (a+b)^*a$ (all strings ending in 'a').

2. DFA $N$:

   • Starts at state $t_0$ (non-accepting).

   • Any input ending with the symbol 'b' leads to state $t_1$ (accepting).

   • Any subsequent 'a' transitions back to $t_0$.

   • Hence, $L(N) = (a+b)^*b$ (all strings ending in 'b').

3. Intersection $L(M) \cap L(N)$:

   • A string cannot simultaneously end in 'a' and end in 'b'.

   • Therefore, $L(M) \cap L(N) = \emptyset$.

4. A minimal DFA that accepts the empty language $\emptyset$ over the alphabet $\{a, b\}$ consists of a single non-accepting state with self-loops on both 'a' and 'b'.

Thus, the number of states in the minimal DFA is 1.