epsilon Non-deterministic and non deterministic equivalence of languages

1. Given:
Conversion Of ε-NFA To NFA
Note:
1.No Change In Initial State
2.No Change In The Total No. Of States
3.May Be Change In Final States
4. Changes in Transitions

2. ε-Closure(q0)={q0,q1}
ε-Closure(q1)={q1}
• Let NFA –ε be M=(Q,Σ,δ,q0,F)
NFA be M1= (Q1,Σ,δ1,q0
1,F1)
Initial state of given nfa-ε is {q0}
Now, initial state of nfa will be {q0}
Now find the transition of q0 on inputs 0,1
δ1=(q0,0)=ε-Closure(δ(ε-Closure(q0),0))
=ε-Closure(δ({q0,q1},0))
=ε-Closure(q0)
={q0,q1}
δ1=(q0,1)=ε-Closure(δ(ε-Closure(q1),1))
=ε-Closure(δ({q0,q1},1))
=ε-Closure(q1)
={q1}
δ1=(q1,0)=ε-Closure(δ(ε-Closure(q1),0))
=ε-Closure(δ({q1},0))
=ε-Closure(Φ)
=Φ
δ1=(q1,1)=ε-Closure(δ(ε-Closure(q1)1))
=ε-Closure(δ({q1},1))
=ε-Closure(q1)
={q1}

3. Resultant NFA is
input
State
0 1
*q0 {q0,q1} {q1}
* q1 Φ {q1}
Final State?
q0 & q1 Both are final states because
ε-Closure(q0)={q0,q1}
ε-Closure(q1)={q1}
NFA - Transition diagram
NFA - Transition table

4. Example :2
Find δ=(q,a)
Find δ=(q,b)
Find δ=(q,c)

5. Find δ=(r,a)
Find δ=(r,b)
Find δ=(r,c)

6. Find δ=(s,a)
Find δ=(s,b)
Find δ=(s,c)