Question: Convert the nfa defined by delta (q_0, a) = {q_0, q_1} delta (q_1, b) = {q_1, q_2} delta (q_2, a) = {q_2} delta (q_1, lambda)

$Convert the nfa defined by delta (q_0, a) = {q_0, q_1}$

Convert the nfa defined by delta (q_0, a) = {q_0, q_1} delta (q_1, b) = {q_1, q_2} delta (q_2, a) = {q_2} delta (q_1, lambda) = {q_1, q_2} with initial state q_0 and final state q_2 into an equivalent dfa

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