NFA Example: Construct The NFA That Accepts All The String Of a & b Where No. Of a Are Exactly 2

Construct The NFA That Accepts All The String Of a & b Where No. Of a (i) Exactly 2 (ii) At Least 2 (iii) At-Most 2 (iv) ≅ 0(Mod 3) (v)  ≅ 1(Mod 5) (vi)  ≅ 0(Mod 2) (vii)  ≅ 1(Mod 2)   (i) Exactly 2 Σ=[a,b} L={aa,....} (ii) At Least 2   (iii) At-Most 2 (Not More Than 2) (iv) ≅ 0(Mod 3) (v)  ≅ 1(Mod 5) (vi)  ≅ 0(Mod 2) (vii)  ≅ 1(Mod 2)  

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NFA Example : Construct The NFA That Accepts All The Strings Of a & b Where Starts With aa Or bb

Construct The NFA That Accepts All The Strings Of a & b Where (i) Starts With aa Or bb (ii) End With aa Or bb (iii) Start With aa And End With bb Or Start With bb & End With aa   (i) Starts With aa Or bb L={aa or bb} NFA Transition Table (ii) End With aa Or bb L={......aa or

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NFA Example : Construct The NFA That Accepts All The Strings Of 0 & 1 Where Every String Start And End With 0

Construct The NFA That Accepts All The Strings Of 0 & 1 (i) Where Every String Start And End With 0 (ii) Every String Start & End With Same Symbol (iii) Every String Start & End With Different Symbol   (i) Where Every String Start And End With 0 L={0,00,010,0110,01010....}   (ii) Every String Start & End With

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NFA: Introduction About Non-Deterministic Finite Automata(NFA)

Introduction About Non-Deterministic Finite Automata(NFA) - Non-Deterministic Finite Automata Is A Incomplete System. The Non-Deterministic Finite Automata Which Has Zero, One Or More Than One Transition From Any State For Any Input Symbol Is Called Non-Deterministic Automata.   Non-Deterministic Finite Automata Is 5 Tuple Machine.   M={Q.Σ,δ,q0,F} Q-Set Of States Σ-Set Of Input Alphabets δ-Transition Function Q×Σ-2Q q0-Initial State F-

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DFA Example : Minimization Of DFA Example 3

Minimization Of DFA Example 3   Solution:- Step 1- Delete Unreachable State From (q0) q4,q5,q6,q7 Are Unreachable Because If You Observe From q3. q4,q5,q6,q7  Are Outward Arrows To q3. There Is No Inward Arrow Which Is Going From q3.   Delete q4,q5,q6,q7 Step 2:- Transition Table  Step 3 - State Equivalence Method 0-Equivalence - [q0,q1,q2][q3] 1-Equivalence- [q0,q1][q2][q3] 2-Equivalence- [q0][q1][q2][q3]      

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DFA Example : Minimization Of DFA Example 2

Minimization Of DFA Example 2 Solution:- Step 1- Delete Unreachable States From Start State(q0) If You Observe q3, It Has Outward Arrows, I Can't Reach q3, From q1, So I'm Deleting q3 Step 2:- Transition Table   Step 3:- State Equivalence Method 0-Equivalence [q0][q1,q2]  // q0  Is Non-Final State, q1,q2 Are Final States. 1-Equivalence [q0][q1,q2] 0-Equivalence & 1-Equivalence Are Same I'll Stop Here. I'll Merge q1,q2 Because Both

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DFA Type Example: Minimization Of DFA Example

Minimization Of DFA Example Step 1:- Delete Unreachable State From Starting State(q0)  q3 Is Unreachable  From q0 Because You Can See Outward Arrow For q3, It Is Not Inward Arrow From q0 I Can't Reach q3 From Starting State q0   -> I'm Removing q3 And I'm Removing Transitions To q3 Now There Are No Unreachable States From Starting State(q0)   Step

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DFA : Minimization Of DFA

Minimization Of DFA Is Reducing The Size Of DFA.You Have To Follow Certain Rules While Reducing The Size, Minimizing The DFA.   Reducing The Number Of States In DFA If Any Two States In DFA Is Equivalent Means Same. I'll Merge Those Two Same States And It Becomes Single State.   But How Can We Know

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Complement Of Finite Automata(DFA Example): Construct The Minimal Final Automata That Accepts All The Strings Of a & b Where No Of a’s In String Is Not Divisible By 4

Construct The Minimal Final Automata That Accepts All The Strings Of a & b Where No Of a's In String Is Not Divisible By 4 Finite Automata (Or) Machine (M) Divisible By 4 Complement Of Finite Automata (Or) Complement Of M1 Not Divisible By 4 Just Make All Finite States To Non-Final States And Make

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Complement Of Finite Automata : Construct The Minimal Finite Automata That Accept All The Strings Of a & b Where Every String Does Not End With ‘baa’

Construct The Minimal Finite Automata That Accept All The Strings Of a & b Where Every String Does Not End With 'baa'     States & Transitions Will Not Change While Constructing Complement Of Finite Automata (M1) From Machine (M)

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