KEMBAR78
Automata theory - NFA ε to DFA Conversion | PPTX
Convert the given NFA-ε into its equivalent DFA.
• Solution:
Let us obtain the ε-closure of each state.
ε-closure(q0) = {q0, q1, q2}
ε-closure(q1) = {q1, q2}
ε-closure(q2) = {q2}
Let the initial state of DFA be ε-closure(q0) = {q0, q1, q2}
= [q0, q1, q2] ----- A.
Find transitions of A on inputs 0,1,2
δ'(A, 0) = ε-closure{δ((q0, q1, q2), 0)}
= ε-closure{δ(q0, 0) ∪ δ(q1, 0) ∪ δ(q2, 0)}
= ε-closure{q0}
= {q0, q1, q2}
=[q0, q1, q2] ----- A
δ'(A, 1) = ε-closure{δ((q0, q1, q2), 1)}
= ε-closure{δ(q0, 1) ∪ δ(q1, 1) ∪ δ(q2, 1)}
= ε-closure{q1}
= {q1, q2}
= [q1, q2] ------- B
δ'(A, 2) = ε-closure{δ((q0, q1, q2), 2)}
= ε-closure{δ(q0, 2) ∪ δ(q1, 2) ∪ δ(q2, 2)}
= ε-closure{q2}
= {q2}
= [q2] ---------C
Find transitions of B on inputs 0,1,2
δ'(B, 0) = ε-closure{δ((q1, q2), 0)}
= ε-closure{δ(q1, 0) ∪ δ(q2, 0)}
= ε-closure{ϕ}
= ϕ ------- ϕ
δ'(B, 1) = ε-closure{δ((q1, q2), 1)}
= ε-closure{δ(q1, 1) ∪ δ(q2, 1)}
= ε-closure{q1}
= {q1, q2}
= [q1, q2] ------- B
δ'(B, 2) = ε-closure{δ((q1, q2), 2)}
= ε-closure{δ(q1, 2) ∪ δ(q2, 2)}
= ε-closure{q2}
= {q2}
= [q2] ------- C
Find transition of C on inputs 0,1,2
δ'(C, 0) = ε-closure{δ(q2, 0)}
= ε-closure{ϕ}
= ϕ ------- ϕ
δ '(C, 1) = ε-closure{δ(q2, 1)}
= ε-closure{ϕ}
= ϕ ------- ϕ
δ'(C, 2) = ε-closure{δ(q2, 2)}
= {q2}
= [q2] ------- C
Resultant DFA :
input
State
0 1 2
*A A B C
• B ϕ B C
*C ϕ ϕ C
Note: Final state given nfa is {q2}
Since q2 lies in the states A = {q0, q1, q2}
B = {q1, q2}
C = {q2}
A,B, C becomes final states of resultant DFA.
DFA - Transition table
DFA - Transition diagram
Example :2 Convert it to DFA
δ’(Z, 1)=ε-closure{δ((C), 1)} = [C] ----------Z
δ’(X, 0)=ε-closure{δ((A,B,C), 0)} = [B,C] -------- Y
δ’(X, 1)=ε-closure{δ((A,B,C), 1)} = [A,B,C] ----- X
δ’(Y, 0)=ε-closure{δ((B,C), 0)} = [C] ---------Z
δ’(Y, 1)=ε-closure{δ((B,C), 1)} = [B,C] ------Y
δ’(Z, 0)=ε-closure{δ((C), 0)} = [C] -----------Z
Let the initial state of DFA be ε-closure(A) = {A,B,C}
= [A,B,C] ----- X
Automata theory - NFA ε to DFA Conversion

Automata theory - NFA ε to DFA Conversion

  • 1.
    Convert the givenNFA-ε into its equivalent DFA. • Solution: Let us obtain the ε-closure of each state. ε-closure(q0) = {q0, q1, q2} ε-closure(q1) = {q1, q2} ε-closure(q2) = {q2} Let the initial state of DFA be ε-closure(q0) = {q0, q1, q2} = [q0, q1, q2] ----- A.
  • 2.
    Find transitions ofA on inputs 0,1,2 δ'(A, 0) = ε-closure{δ((q0, q1, q2), 0)} = ε-closure{δ(q0, 0) ∪ δ(q1, 0) ∪ δ(q2, 0)} = ε-closure{q0} = {q0, q1, q2} =[q0, q1, q2] ----- A δ'(A, 1) = ε-closure{δ((q0, q1, q2), 1)} = ε-closure{δ(q0, 1) ∪ δ(q1, 1) ∪ δ(q2, 1)} = ε-closure{q1} = {q1, q2} = [q1, q2] ------- B δ'(A, 2) = ε-closure{δ((q0, q1, q2), 2)} = ε-closure{δ(q0, 2) ∪ δ(q1, 2) ∪ δ(q2, 2)} = ε-closure{q2} = {q2} = [q2] ---------C
  • 3.
    Find transitions ofB on inputs 0,1,2 δ'(B, 0) = ε-closure{δ((q1, q2), 0)} = ε-closure{δ(q1, 0) ∪ δ(q2, 0)} = ε-closure{ϕ} = ϕ ------- ϕ δ'(B, 1) = ε-closure{δ((q1, q2), 1)} = ε-closure{δ(q1, 1) ∪ δ(q2, 1)} = ε-closure{q1} = {q1, q2} = [q1, q2] ------- B δ'(B, 2) = ε-closure{δ((q1, q2), 2)} = ε-closure{δ(q1, 2) ∪ δ(q2, 2)} = ε-closure{q2} = {q2} = [q2] ------- C
  • 4.
    Find transition ofC on inputs 0,1,2 δ'(C, 0) = ε-closure{δ(q2, 0)} = ε-closure{ϕ} = ϕ ------- ϕ δ '(C, 1) = ε-closure{δ(q2, 1)} = ε-closure{ϕ} = ϕ ------- ϕ δ'(C, 2) = ε-closure{δ(q2, 2)} = {q2} = [q2] ------- C
  • 5.
    Resultant DFA : input State 01 2 *A A B C • B ϕ B C *C ϕ ϕ C Note: Final state given nfa is {q2} Since q2 lies in the states A = {q0, q1, q2} B = {q1, q2} C = {q2} A,B, C becomes final states of resultant DFA. DFA - Transition table DFA - Transition diagram
  • 6.
    Example :2 Convertit to DFA δ’(Z, 1)=ε-closure{δ((C), 1)} = [C] ----------Z δ’(X, 0)=ε-closure{δ((A,B,C), 0)} = [B,C] -------- Y δ’(X, 1)=ε-closure{δ((A,B,C), 1)} = [A,B,C] ----- X δ’(Y, 0)=ε-closure{δ((B,C), 0)} = [C] ---------Z δ’(Y, 1)=ε-closure{δ((B,C), 1)} = [B,C] ------Y δ’(Z, 0)=ε-closure{δ((C), 0)} = [C] -----------Z Let the initial state of DFA be ε-closure(A) = {A,B,C} = [A,B,C] ----- X