Let's solve the problem step-by-step using the **Generalized Transition Graph (GTG) method** (also known as the state elimination method): --- ### **Step 1: Add New Start and Final States** Add a new start state `S` (with a λ transition to `q0`), and a new final state `F` (with λ transitions from the original final state(s) to `F`). Assume `q2` is the only accepting state, so add λ from `q2` to `F`. --- #### **Diagram with New States** ``` S --λ--> q0 --(a,λ)--> q1 | ^ | b | a | | | v | | q3 --a--> q0 | | ^ | | | | b | | | | | | +------v----+ | q2 --a--->q2 | | +--λ----->F ``` --- ### **Step 2: Build the GTG Table** Write the transitions as regular expressions between each state pair. | From | To | RegExp | |------|-----|----------------| | S | q0 | λ | | q0 | q1 | a + λ | | q0 | q3 | b | | q1 | q1 | a | | q1 | q2 | b | | q2 | q2 | a | | q2 | F | λ | | q3 | q0 | a | | q3 | q2 | b | Other transitions are empty (∅). --- ### **Step 3: Eliminate States One by One** #### **Eliminate q1:** Update all pairs (i, j) where i ≠ q1 and j ≠ q1. - From q0 to q2 via q1: q0 --(a+λ)--> q1 --[a]*-- (b) --> q2 - q0 to q2: (a+λ)a\*b = a\*b Now, q0 to q2: - Direct: ∅ - Via q1: a\*b So, q0 to q2 = a\*b --- #### **Update Table after Eliminating q1** | From | To | RegExp | |------|-----|------------| | S | q0 | λ | | q0 | q2 | a\*b | | q0 | q3 | b | | q3 | q0 | a | | q3 | q2 | b | | q2 | q2 | a | | q2 | F | λ | --- #### **Eliminate q3:** Update transitions: - q0 to q0 via q3: q0 --b--> q3 --a--> q0 So, q0 to q0: ba - q0 to q2 via q3: q0 --b--> q3 --b--> q2 So, q0 to q2: old(q0 to q2) + (b b) = a\*b + bb --- | From | To | RegExp | |------|-----|------------| | S | q0 | λ | | q0 | q0 | ba | | q0 | q2 | a\*b + bb | | q2 | q2 | a | | q2 | F | λ | --- #### **Eliminate q0:** - S to q2 via q0: - S --λ--> q0 --[ba]*-- (a\*b + bb) --> q2 - (ba)\* (a\*b + bb) So, S to q2: (ba)\* (a\*b + bb) --- ### **Final Step: q2 to F** - q2 --a--> q2 (loop) - q2 --λ--> F So, q2 to F: a\* --- ### **Final Regular Expression** From S to F (the language accepted by the original NFA): \[ (ba)^* (a^*b + bb) a^* \] --- ## **Final Answer** \[ \boxed{(ba)^* (a^*b + bb) a^*} \] This is the regular expression equivalent to the given NFA.