Conversion of an NFA to a DFAusing

Subset Construction Algorithm

Algorithm of Subset construction`

Operations on NFA States

By using subset construction algorithm convert the following NFA

N for (alb) * abb to DFA

Example:

we need to remove :

1. transition That require to construct closure (s)

2. A multiple transition on input symbol from some state s in T:Mov(T,a)

Example: To apply subset construction

{0 , 1, 2, 4, 7}{0} =

{1, 2, 4}{1} =

{2}{2} =

{1, 2, 3, 4, 6, 7}{3} =

{4}{4} =

{1, 2, 4, 5, 6, 7}{5} =

{1, 2, 4, 6, 7}{6} =

{7}{7} =

{8}{8} =

{9}{9} =

{10}{10} =

Example: First step: construct closure (s)

The start state A of the equivalent DFA is closure (0)

The closure (0) = {0, 1, 2, 4, 7}

Then the starting state A = {0, 1, 2, 4, 7}

Example: Second step: Looking for the start state for the DFA

{0, 1, 2, 4, 7}A

a

b

First determine the input alphabet

here input alphapet = {a,b} Second compute:

1. Dtran[A,a] = -closure(move(A,a))

2. Dtran[A,b] = -closure(move(A,b))

Example: Third step: Compute U = -closure(move(T,a)

Remember:

A = {0, 1, 2, 4, 7}

1. Dtran[A,a] = -closure(move(A,a))

Among the state 0, 1, 2, 4, 7, only 2 and 7 have transitions on a to 3 and 8, respectively.

Thus move(A,a) = {3, 8}

Also, - closure ({3,8} = {1, 2, 3, 4, 6, 7, 8}

So we conclude:

Dtran[A,a] = -closure(move(A,a)) = - closure ({3,8} = {1, 2, 3, 4, 6, 7, 8}

Let us call this set B so:

Dtran[A,a] = B

{1, 2, 3, 4, 6, 7, 8}

B

aA

2. Dtran[A,b] = -closure(move(A,b))

Among the states in A, only 4 has a transition on b, and it goes to 5

Thus:

Dtran[A,b] = -closure(move(A,b)) = - closure ({5} = {1, 2, 4, 5, 6, 7 }

Let us call this set C so:

Dtran[A,b] = C

{1, 2, 4, 5, 6, 7 }

C

bA

Third compute:

3. Dtran[B,a] = -closure(move(B,a))

4. Dtran[B,b] = -closure(move(B,b))

3. Dtran[B,a] = -closure(move(B,a))

Among the states in B, only 2, 7 has a transition on a, and it goes to {3, 8} respectively

Thus:

Dtran[B,a] = -closure(move(B,a)) = - closure ({3,8} = {1, 2, 3, 4, 6, 7, 8 }

Dtran[B,a] = B

{1, 2, 3, 4, 6, 7, 8 }

B

aA

a

4. Dtran[B,b] = -closure(move(B,b))

Among the states in B, only 4, 8 has a transition on b, and it goes to {5, 9} respectively

Thus:

Dtran[B,b] = -closure(move(B,b)) = - closure ({5,9} = {1, 2, 4, 5, 6, 7,9}

Dtran[B,b] = D

{1, 2, 4, 5, 6, 7,9}

B

aA

a

bB

D

Fourth compute:

5. Dtran[C,a] = -closure(move(C,a))

6. Dtran[C,b] = -closure(move(C,b))

5. Dtran[C,a] = -closure(move(C,a))

Among the states in C, only 2, 7 has a transition on a, and it goes to {3, 8} respectively

Thus:

Dtran[C,a] = -closure(move(C,a)) = - closure ({3,8} = {1, 2, 3, 4, 6, 7, 8 }

Dtran[C,a] = B

{1, 2, 4, 5, 6, 7 }

CA Bab

6. Dtran[C,b] = -closure(move(C,b))

Among the states in C, only 4 has a transition on b, and it goes to {5} respectively

Thus:

Dtran[C,b] = -closure(move(C,b)) = - closure ({5} = {1, 2, 4, 5, 6, 7}

Dtran[C,b] = C

{1, 2, 4, 5, 6, 7}b

AC

b

Fifth compute:

7. Dtran[D,a] = -closure(move(D,a))

8. Dtran[D,b] = -closure(move(D,b))

7. Dtran[D,a] = -closure(move(D,a))

Among the states in D, only 2, 7 has a transition on a, and it goes to {3, 8} respectively

Thus:

Dtran[D,a] = -closure(move(D,a)) = - closure ({3,8} = {1, 2, 3, 4, 6, 7, 8 }

Dtran[D,a] = B

{1, 2, 4, 5, 6, 7,9}

DBb

a

8. Dtran[D,b] = -closure(move(D,b))

Among the states in D, only 4 and 9 has a transition on b, and it goes to {5, 10} respectively

Thus:

Dtran[D,b] = -closure(move(D,b)) = - closure ({5,10} = {1,2, 4, 5, 6, 7, 10}

Dtran[D,b] = E

{1, 2, 4, 5,6, 7, 10}

bB

b

{1, 2, 4, 5, 6, 7, 9}Da E

Finally compute:

9. Dtran[E,a] = -closure(move(E,a))

10. Dtran[E,b] = -closure(move(E,b))

9. Dtran[E,a] = -closure(move(E,a))

Among the states in E, only 2, 7 has a transition on a, and it goes to {3, 8} respectively

Thus:

Dtran[E,a] = -closure(move(E,a)) = - closure ({3,8} = {1, 2, 3, 4, 6, 7, 8 }

Dtran[E,a] = B

{1, 2, 4, 5, 6, 7,9}

EDb Ba

10. Dtran[E,b] = -closure(move(E,b))

Among the states in E, only 4 has a transition on b, and it goes to {5} respectively

Thus:

Dtran[E,b] = -closure(move(E,b)) = - closure ({5} = {1,2, 4, 5, 6, 7}

Dtran[E,b] = C

{1, 2, 4, 5,6, 7, 10}b

DE

bC

Transition table Dtran for DFA D

Result of applying the subset Construction NFA for (alb) * abb