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### Conversion Of E-NFA To DFA

Conversion Of E-NFA To DFA Example:- ε-Closure (q0) --> {q0,q1} q01 = [q0q1]

### Conversion Of E-NFA To DFA

Conversion Of E-NFA To DFA Let M=(Q,Σ,δ,q0,F) Be The NFA M1=(Q1,Σ,δ1,q01,F1) Be The Equivalent DFA   1) Initial State:-  q01 - ε-Closure(q0) 2) Construction Of δ1 δ1(q,x)=ε-Closure(δ(q,x)) Start The Construction Of δ1 With Initial State And Continue For Every New State. 3) Final State:- Every Subset Which Contain The Final State Of ε-NFA Is Final State Of DFA   Example:- 1)  Initial State:- q01=ε-Closure(q0) ={q0,q1} q0=q0,q1 2) DFA

### NFA With ε-Moves Or ε-NFA

NFA With ε-Moves Or ε-NFA The NFA Which Has Transition Even For Empty String ε Is The Called An ε-nfa ε-NFA Has 5-Tuple M=(Q,Σ,δ,q0,F) Where Q=Set Of States Σ=I/p Alphabets δ=Transition Function(Q×Σ∪{ε}--> 2Q) q0=Initial State F= Final State Q×Σ--> Q--> DFA Q×Σ--> 2Q-->NFA Q×Σ∪{ε}--> 2Q--> ε-NFA   Example L={am,bm/m,n≥0} L={a,b,ε,ab,abb...} 2.L={anbUbna/n≥0} 3.L={(ab)n/n≥0}

### NFA Example: Consider The NFA M Given Below

Consider The NFA M Given Below Let 'M' Be The NFA Obtain From M By Converting All The Non-Final States Into Final States If L & L1 Be The Language Accepted  By M And M1 Then Which Of The Following Statement Is Correct. a) L=L1 b) L⊂L1 c) L1⊂L d) L∩L1=Φ e) L∪L1=Σ* f) None   Before Solving This Problem I'll Tell Yo

### NFA Problem

NFA Problem - Consider The NFA M Given Below. Let 'M' Be The NFA Obtain From By Interchanging Final And Non- Final States. Let L And L1 Be The Language Accepted By M And M1 Respectively Then Which Of The Following Statement Is True. L=L1 L⊆L1 L1⊆L L∩L1=Φ L∩L1=Σ* L∪L1=Φ L∪L1=Σ* Solution:- Given NFA 'M' Is He