Convert the regular expression to a NFA

[mathmari]

Hey!

I have to convert the following regular expressions to a NFA:

1. $$(0 \cup 1)^{\star} 000 (0 \cup 1)^{\star}$$
2. $$(((00)^{\star} (11)) \cup 01)^{\star}$$
3. $$\emptyset^{\star}$$
4. $$a(abb)^{\star} \cup b$$
5. $$a^+ \cup (ab)^{\star}$$
6. $$(a \cup b^+)a^+b^+$$

For the regular expressions $1-3$, $\Sigma=\{0,1\}$, and for the expressions $4-6$, $\Sigma=\{a, b\}$.

I have done the following:

View attachment 4165

Is this correct?? (Wondering)

How is the NFA for the regular expression $3.$ ?? (Wondering)

 

[Valkarie]

1. NFA: States: q0, q1, q2, q3 Alphabet: {0, 1} Transitions: q0 → q1 on 0 or 1 q1 → q2 on 0 q2 → q3 on 0 or 1 q3 → q3 on 0 or 1 Start state: q0 Accepting states: q3 2. NFA: States: q0, q1, q2 Alphabet: {0, 1} Transitions: q0 → q1 on 0 q1 → q1 on 0 q1 → q2 on 1 q2 → q0 on 0 Start state: q0 Accepting states: q2 3. NFA: States: q0 Alphabet: {ε} Transitions: q0 → q0 on ε Start state: q0 Accepting states: q0 4. NFA: States: q0, q1, q2, q3 Alphabet: {a, b} Transitions: q0 → q1 on a q1 → q2 on b q2 → q3 on b q3 → q3 on a or b Start state: q0 Accepting states: q3 5. NFA: States: q0, q1, q2 Alphabet: {a, b} Transitions: q0 → q1 on a q1 → q2 on b q2 → q2 on a or b Start state: q0 Accepting states: q2 6. NFA: States: q0, q1, q2, q3 Alphabet: {a, b} Trans