## Conversion of NFA to DFA

Conversion of NFA to DFA

Let M2 = < Q2 , , q2,0 , 2 , A2 > be an NFA that recognizes a language L. Then the DFA M = < Q, , q0 , , A > that satisfies the following conditions recognizes L:

Q = 2Q2 , that is the set of all subsets of Q2 ,
q0 = { q2,0 } ,
( q, a ) = for each state q in Q and each symbol a in   and
A = { q Q | q A2 }

To obtain a DFA M = < Q, , q0 , , A > which accepts the same language as the given NFA M2 = < Q2 , , q2,0 , 2 , A2 > does, you may proceed as follows:

Initially Q = .
First put { q2,0 } into Q. { q2,0 } is the initial state of the DFA M.
Then for each state q in Q do the following:
add the set , where here is that of NFA M2, as a state to Q if it is not already in Q for each symbol a in .
For this new state, add ( q, a ) = to , where the on the right hand side is that of NFA M2.

When no more new states can be added to Q, the process terminates. All the states of Q that contain accepting states of M2 are accepting states of M.

Note: The states that are not reached from the initial state are not included in Q obtained by this procedure. Thus the set of states Q thus obtained is not necessarily equal to 2Q2 .

Example 1: Let us convert the following NFA to DFA.

Initially Q is empty. Then since the initial state of the DFA is {0} , {0} is added to Q.
Since 2( 0 , a ) = { 1 , 2 } , { 1 , 2 } is added to Q and ( { 0 } , a ) = { 1 , 2 } .
Since 2( 0 , b ) = , is added to Q and ( { 0 } , b ) = .
At this point Q = { {0} , { 1 , 2 }, } .
Then since { 1 , 2 } is now in Q, the transitions from { 1 , 2 } on symbols a and b are computed. Since 2( 1 , a ) = { 1 , 2 } , and 2( 2 , a ) = , ( { 1 , 2 } , a ) = { 1 , 2 } . Similarly ( { 1 , 2 } , b ) = { 1 , 3 } . Thus { 1 , 3 } is added to Q .
Similarly ( { 1 , 3 } , a ) = { 1 , 2 } and ( { 1 , 3 } , b ) = . Thus no new states are added to Q . Since the transitions from all states of Q have been computed and no more states are added to Q, the conversion process stops here.

Note that there are no states of Q2 in . Hence there are no states that M2 can go to from . Hence ( , a ) = ( , b ) = .

For the accepting states of M, since states 0 and 1 are the accepting states of the NFA, all the states of Q that contain 0 and/or 1 are accepting states. Hence { 0 }, { 1 , 2 } and { 1 , 3 } are the accepting states of M.

The DFA thus obtained is shown below.

Example 2: Similarly the NFA

is converted to the following DFA:

### Test Your Understanding of Conversion of NFA to DFA

Indicate which of the following statements are correct and which are not.
Click Yes or No , then Submit.
There are two sets of questions.

Answer the questions below on converting the following NFA to DFA.

The following notation is used in the questions:

: \delta

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