Convertion from NFA to DFA I have converted nfa to dfa. But when I checked with some examples, it looks like it is wrong.

This is the full question.
Convert the nfa defined by
δ(q0, a) = {q0, q1}
δ(q1, b) = {q1, q2}
δ(q2, a) = {q2}
δ(q1, λ) = {q1, q2}
with initial state q0 and final state q2 into an equivalent dfa.


First, I made a table.

             a          b
ɸ            ɸ          ɸ
{q0}         {q0,q1}    ɸ
{q1}         ɸ          {q1,q2}
{q2}         {q2}       ɸ
{q0,q1}      {q0,q1}    {q1,q2}
{q1,q2}      {q2}       {q1,q2}
{q0,q2}      {q0,q1,q2} ɸ
{q0,q1,q2}   {q0,q1,q2} {q1,q2}


Finally, I have drawn dfa according to table above.

The bold line is accepting state.

Thanks for your help/
It looks like you have 3 symbols in your language: a, b, and \lambda.  But your transition table only has the symbols a and b.  You need to add \lambda to your transition table and integrate that into your diagram.  That should help some.