## NFA to DFA using Indirect method

Step 01 : First convert NFA with ∈ moves to NFA without ∈ moves.
Step 01 :Then convert NFA without ∈ to the DFA.

Let's take an example,

We have a NFA with epsilon moves, now convert it to the NFA without epsilon moves. NFA with epsilon moves

First convert above NFA to without epsilon moves,

Find ∈-closure of (q0), (q1) and (q2).
1. ∈-closure of (q0) = {q0, q1, q2}
2. ∈-closure of (q1) = {q1, q2}
3. ∈-closure of (q2) = {q2}

 State 0 1 2 ->q1 {q1,q2,q3} {q2,q3} {q3} q2 φ {q2,q3} {q3} q3 φ φ {q3}

From the above transition table, draw the transition diagram, NFA without  epsilon

From the question diagram, it is clear that only with ∈ input q1 and q2 state can reach to the final state.
So, now without ∈ input, q1 and q2 is also treated as final states.
As shown in diagram beabove.

Now, NFA without epsilon to DFA

Using Subset construction method.

Again I am going to use information from this transition table.

 State 0 1 2 ->q1 {q1,q2,q3} {q2,q3} {q3} q2 φ {q2,q3} {q3} q3 φ φ {q3}

Subsets are,
1. {q1}
2. {q2}
3. {q3}
4. {q1,q2,q3}
5. {q2,q3}
Treat each subset as state in DFA.
If there is no transition use φ symbol.

 State Input = 0 Input = 1 Input = 2 {q1} {q1,q2,q3} {q2,q3} {q3} {q2} φ {q2,q3} {q3} {q3} φ φ {q3} {q1,q2,q3} {q1,q2,q3} {q2,q3} {q3} {q2,q3} φ {q2,q3} {q3} φ φ φ φ

In NFA withoud epsilon, q1, q2, q3 was final states. In subset {q1,q2,q3} and {q2,q3} final states exist.
So these subsets will also treated as final states. DFA from NFA

Now from the transition table of above DFA,

 State Input = 0 Input = 1 Input = 2 {q1} {q1,q2,q3} {q2,q3} {q3} {q2} φ {q2,q3} {q3} {q3} φ φ {q3} {q1,q2,q3} {q1,q2,q3} {q2,q3} {q3} {q2,q3} φ {q2,q3} {q3} φ φ φ φ

We find state {q1,q2,q3} is equivalent to state {q1}, so replce state {q1,q2,q3} by state {q1}.
Simillary, replace state {q2,q3} by state {q2}.

So, new DFA transition table become,

 State Input = 0 Input = 1 Input = 2 {q1} {q1} {q2} {q3} {q2} φ {q2} {q3} {q3} φ φ {q3} φ φ φ φ

And DFA transition diagram,

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