Let $M$ be the 5-state NFA with $\epsilon$-transitions shown in the diagram below. Which one of the following regular expressions represents the language accepted by $M$ ?

To determine the regular expression for the language accepted by the NFA $M$, we analyze the paths from the start state $1$ to each of the accepting states $2$ and $5$ [1]:

1. Paths reaching state $2$:

   • The start state $1$ reaches state $2$ via an $\epsilon$-transition ($1 \xrightarrow{\epsilon} 2$).

   • From state $2$, we can alternate between state $2$ and state $3$ via $2 \xrightarrow{0} 3 \xrightarrow{0} 2$, which corresponds to the loop $(00)^*$.

   • Thus, the set of strings accepted at state $2$ is $(00)^*$.

2. Paths reaching state $5$:

   • Via state $4$: $1 \xrightarrow{\epsilon} 4 \xrightarrow{1} 5$, giving string $1$. Once at state $5$, there is a loop $5 \xrightarrow{1} 4 \xrightarrow{1} 5$, giving $(11)^*$. The strings reaching state $5$ along this path are represented by $1(11)^*$.

   • Via state $2$ and state $3$: $1 \xrightarrow{\epsilon} 2 \xrightarrow{(00)^*} 2 \xrightarrow{0} 3 \xrightarrow{\epsilon} 5$. This produces strings of the form $0(00)^*$. From state $5$, looping gives $(11)^*$, so strings along this path are represented by $0(00)^*(11)^*$.

   • Combining both paths to state $5$, we get $(1 + 0(00)^*)(11)^*$.

3. Total language accepted by $M$: $$\begin{aligned} L(M) &= L(\text{at state } 2) + L(\text{at state } 5) \\ &= (00)^* + (1 + 0(00)^*)(11)^* \\ &= (00)^* + 1(11)^* + 0(00)^*(11)^* \\ &= (00)^* + 0(00)^* + 1(11)^* + 0(00)^*(11)^+ \\ &= 0^* + (1 + 0(00)^*)(11)^* \end{aligned}$$

Hence, the correct regular expression is **$0^* + (1 + 0(00)^*)(11)^*$, which corresponds to option (B)** [1].