Question

Find the regular Expression for the FA (Finite Automata) using Arden's Theorem.

Answer 1

R = (a + b(b + aa)a) b(b + aa)* a

Answer 2

To find the regular expression for the given Finite Automata (FA) using Arden's Theorem, we first write down the equations for each state.

1. Define States and Transitions: Let $q_1, q_2, q_3$ be the regular expressions representing the set of all strings that can take the FA from the initial state to states $q_1, q_2, q_3$ respectively.

• $q_1$ is the initial state, so it has an $\epsilon$ transition.
• $q_3$ is the final state.

2. Write Equations for Each State: Based on the transitions in the FA:

• For state $q_1$:

  • It's the initial state, so it includes $\epsilon$.
  • There's a self-loop on $q_1$ with 'a'.
  • There's a transition from $q_2$ to $q_1$ with 'a'. Equation for $q_1$: $q_1 = \epsilon + q_1a + q_2a$ (Equation 1)

• For state $q_2$:

  • There's a transition from $q_1$ to $q_2$ with 'b'.
  • There's a self-loop on $q_2$ with 'b'.
  • There's a transition from $q_3$ to $q_2$ with 'a'. Equation for $q_2$: $q_2 = q_1b + q_2b + q_3a$ (Equation 2)

• For state $q_3$:

  • There's a transition from $q_2$ to $q_3$ with 'a'. Equation for $q_3$: $q_3 = q_2a$ (Equation 3)

3. Solve the Equations using Arden's Theorem: Arden's Theorem states that if an equation is of the form $R = Q + RP$, where $Q$ and $P$ are regular expressions and $R$ does not contain $R$ in $Q$, then the solution is $R = QP^*$.

Substitute Equation 3 into Equation 2: $q_2 = q_1b + q_2b + (q_2a)a$ $q_2 = q_1b + q_2b + q_2aa$ $q_2 = q_1b + q_2(b + aa)$

Now, apply Arden's Theorem to solve for $q_2$. Here, $R = q_2$, $Q = q_1b$, and $P = (b + aa)$. $q_2 = q_1b(b + aa)^*$ (Equation 4)

Now substitute Equation 4 into Equation 1: $q_1 = \epsilon + q_1a + (q_1b(b + aa)^*)a$ $q_1 = \epsilon + q_1a + q_1b(b + aa)^*a$ $q_1 = \epsilon + q_1(a + b(b + aa)^*a)$

Apply Arden's Theorem to solve for $q_1$. Here, $R = q_1$, $Q = \epsilon$, and $P = (a + b(b + aa)^*a)$. $q_1 = \epsilon(a + b(b + aa)^*a)^*$ Since $\epsilon R = R$, $q_1 = (a + b(b + aa)^*a)^*$ (Equation 5)

Finally, we need the regular expression for the language accepted by the FA, which is the expression for the final state $q_3$. Substitute Equation 5 into Equation 4, and then the result into Equation 3:

Substitute $q_1$ (Equation 5) into $q_2$ (Equation 4): $q_2 = (a + b(b + aa)^*a)^* b(b + aa)^*$ (Equation 6)

Now, substitute $q_2$ (Equation 6) into $q_3$ (Equation 3): $q_3 = q_2a$ $q_3 = ((a + b(b + aa)^*a)^* b(b + aa)^*) a$

This is the regular expression for the given FA.

Explanation of the solution:

1. Define regular expressions for each state ($q_1, q_2, q_3$) representing strings from the initial state to that state.
2. Formulate a system of linear equations based on the FA's transitions. $q_1$ is initial, so it gets $\epsilon$.
3. Solve the system of equations using Arden's Theorem ($R = Q + RP \implies R = QP^*$) by substituting equations backwards from the final state.
4. The regular expression for the FA is the expression for the final state ($q_3$).