Which of the regular expressions given below represent the following DFA? I) $0^*1(1+00^*1)^*$ II) $0^*1^*1+11^*0^*1$ III) $(0+1)^*1$

Let the start state be $q_0$ and the final (accepting) state be $q_1$.

The transitions of the DFA are:

• $\delta(q_0, 0) = q_0$

• $\delta(q_0, 1) = q_1$

• $\delta(q_1, 0) = q_0$

• $\delta(q_1, 1) = q_1$

From the transitions, notice that whenever the automaton receives the symbol $1$, it enters the accept state $q_1$. Whenever it receives $0$, it moves to the non-accepting state $q_0$. Therefore, the automaton accepts any string over $\{0, 1\}$ that ends with the symbol $1$.

Evaluating each regular expression:

• **Expression III: $(0+1)^*1$** Directly generates the set of all strings ending in $1$. Hence, it represents the language of the DFA.

• **Expression I: $0^*1(1+00^*1)^*$** To reach $q_1$ from $q_0$, the path is $0^*1$. From $q_1$, any loop back to $q_1$ is either:

  • Reading $1$ directly: $1$

  • Reading $0$, looping on $0$ in $q_0$, and returning to $q_1$ on $1$: $00^*1$

  Combining these gives the loop expression $(1 + 00^*1)^*$. Thus, the overall regular expression from the start state to the final state is $0^*1(1+00^*1)^*$, which matches the DFA.

• **Expression II: $0^*1^*1+11^*0^*1$** Consider the string $0101$, which ends with $1$ and is accepted by the DFA. It cannot be generated by $0^*1^*1$ (which cannot have $0$ after $1$) nor by $11^*0^*1$ (which must start with $1$). Hence, II is incorrect.

Therefore, only I and III represent the DFA, which corresponds to option (B).