1.7. eqivalence of nfa and dfa

Equivalence of
NFA and DFA
-Sampath Kumar S,
AP/CSE, SECE

Equivalence of NFA and DFA
 The NFA with ε can be converted to NFA
without ε and this NFA without ε can be
converted to DFA.
11/21/2017
Sampath Kumar S, AP/CSE, SECE
2
ε-NFA
NFA
DFA

Conversion of NFA to DFA:
The conversion of NFA to DFA involves following steps:
1. QD = 2Q
N. If NFA has n states. DFA at most have 2n
states.
2. Σn=ΣD
3. [q0]=[q0]
4. FD = A set of all the states of QD that contains at
least one of the final states of FN .
5. δD((q1,q2,q3),a)=
δn(q1,a) u δn(q2,a) u δn(q3,a) ={p1,p2,p3}
Add state [P1,P2,P3] to QD if it is not there
11/21/20173

Problem to discuss:
38. Convert the following NFA to DFA
11/21/20174

Problem to discuss:
39. Construct DFA equivalent to NFA
M = ( {p,q,r,s}, {0,1}, δ, P, {q,s} )
where δ is defined in the following table.
11/21/20175
δ 0 1
->p {q, s} {q}
q* {r} {q,r}
r {s} {p}
s* - {p}

Problem to discuss:
40. Convert the given NFA to its equivalent DFA
11/21/20176

Conversion of ε-NFA to NFA:
The conversion of ε-NFA to NFA involves following
steps:
1. Find out all the ε-transition from each state from Q.
That will be called as ε-closure {qi} where qi ∈ Q.
2. Then δ’ transition can be obtained. The δ’ transition
means and ε-closure on δ moves.
3. Step 2 is repeated for each input symbol and for
each state of given NFA.
4. Using the resultant states the transition table for
equivalent NFA without ε can be built
11/21/20177

Rule for conversion:
δ’(q,a) = ε-closure (δ (δ (q, ε ), a)
Where
δ (q, ε ) = ε-closure (q)
11/21/20178

Problem to discuss:
41. Convert the following ε-NFA to NFA
11/21/20179

Solution:
Step 1: Find ε-closure for all states:
ε-closure (q)={q, r, s}
ε-closure (r)={ r, s}
ε-closure (s)={s}
11/21/201710

Solution:
δ’(q,a) = ε-closure (δ (q,r,s), a)
= ε-closure (δ(q,a) U δ(r,a) U δ(s,a) )
= ε-closure (q U Φ U Φ)
= ε-closure (q) = {q, r, s}
δ’(q,b) = ε-closure (δ (δ (q, ε ), b)
= ε-closure (δ (q,r,s), b)
= ε-closure (δ(q,b) U δ(r,b) U δ(s,b) )
= ε-closure (Φ U r U Φ)
= ε-closure (r) = {r, s}
11/21/201711

Solution:
δ’(q,c) = ε-closure (δ (δ (q, ε ), c)
= ε-closure (δ (q,r,s), c)
= ε-closure (δ(q,c) U δ(r,c) U δ(s,c) )
= ε-closure (Φ U Φ U s)
= ε-closure (s) = {s}
11/21/201712

Final Solution:
δ’(q,a) = {q, r, s}
δ’(q,b) = {r, s}
δ’(q,c) = {s}
δ’(r,a) = {Φ}
δ’(r,b) = {r, s}
δ’(r,c) = {s}
δ’(s,a) = {Φ}
δ’(s,b) = {Φ}
δ’(s,c) = {s}
11/21/201713

Problem to discuss:
42. Convert the following ε-NFA to NFA
11/21/201714

Conversion of ε-NFA to DFA:
The conversion of NFA to DFA involves following steps:
1. Compute ε* for the current state, resulting in a set
of the states S.
2. δ(S, a) is computed for all a Σ by
1. Let S ={p1,p2, ….. pK}
2. Compute R = δ(S, a) as
R = pi S δ(pi, a) ={r1,r2, ….. rm}
This set is achieved by the following input a, not by
following any ε transitions.
3. Add the ε transitions in by computing ε-closure(R)
3. Make a state, an accepting state if it includes any
final states in the NFA
11/21/201715

Problem to discuss:
43. Convert the following ε-NFA to DFA
11/21/201717

Problem to discuss:
44. Convert the given ε-NFA to its equivalent DFA
11/21/201718

11/21/2017
19

நன்றி
11/21/2017
20

1.7. eqivalence of nfa and dfa

More Related Content

What's hot (20)

Similar to 1.7. eqivalence of nfa and dfa (20)

More from Sampath Kumar S (19)

Recently uploaded (20)

1.7. eqivalence of nfa and dfa

Editor's Notes