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DFA Example : Construct Minimal Finite Automata That Accepts All Strings Of 0’s & 1’s Such That a) |w|≅0 (mod 2), b) |w|≅1 (mod 2), c) |w|≅2 (mod 3)

Construct Minimal Finite Automata That Accepts All Strings Of 0's & 1's Such That a) |w|≅0 (mod 2), b) |w|≅1 (mod 2), c) |w|≅2 (mod 3) |w|= Length Of 'w' String a) |w|≅0 (mod 2) Means If I Divide Length Of String With '2' I Should Get Remainder '1' 'ε' Is Length '0' String,

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Construct Minimal Finite Automata That Accepts All Strings Of 0’s And 1’s Such That i) η0|w|≅0 (mod 3) ii) η0|w|≅0 (mod 4)

Construct Minimal FA That Accepts All Strings Of 0's And 1's Such That i) η0|w|≅0 (mod 3) ii) η0|w|≅0 (mod 4) Σ={0,1} i) η0|w|≅0 (mod 3) // η0 Is No. Of Zero's, |w| Is In The String, If I Divide With '3', I'll Get Remainder '0'. Means, The No. Of 0's In The String,If I Divide

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DFA Example : Construct Minimal Finite Automata That Accept All The Strings Of 0’s And 1’s Such That Exactly Two 0’s

Construct Minimal Finite Automata That Accept All The Strings Of 0's And 1's Such That i) Exactly Two 0's ii) Almost Two 0's iii) Atleast Two 0's   Solution There Σ={0,1} // Here In The Place Of a,b The Input Alphabets Are a,b   i) Exactly Two 0's Possible Strings :- {00,001,01011,100,1001....} // String Should Contain Exactly Two 0's Small String Is 00 Here

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DFA Example : Construct The Minimal Finite Automata All The Strings Of a & b Where Second Symbol From Right End Is a

Construct The Minimal Finite Automata All The Strings Of a & b Where Second Symbol From Right End Is a Solution  In This Problem, The Second Symbol From Right 'a'   You Can Easily Solve Problems If The Question Is Asked Fixed Left Side Symbol. If They Ask You Fixed Right Side Symbol You Have

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DFA Example : Construct Minimal Deterministic Finite Automata Ends With aa Or Ends With bb

Construct Minimal Deterministic Finite Automata Ends With aa Or Ends With bb Condition Given Is The String Must End With aa Or Can End End With bb. L={aa,bb,aaa,bbb,abaa,babb} // Infinite String   Lets's Create NFA (Smallest String Is 'aa' Or 'bb') DFA Remember Dead State Concept Will Not Comes In The 'End With ' Problems. If I

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DFA Example : Construct Deterministic Finite Automata (DFA) Start And End With Same Symbol

Construct Minimal Deterministic Finite Automata9DFA) Start & End With Same Symbol Where Σ={a,b}   Conditions - If Your String Is Starting With 'a', It Should End With 'a' Only. If Your String Is Starting With 'b', It Should End With 'b' Only.   " 'Dead State' Concept Comes When I Say About Start Or End Symbol

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Finite Automata Example: Construct Minimal Deterministic Finite Automata (DFA) Start And End With Different Symbol

Construct Minimal Deterministic Finite Automata (DFA) Start And End With Different Symbol Where Σ={a,b}   Condition Given Is If Your String Is Starting With 'a' It Should End With 'b' If Your String Is Starting With 'b' Then It Should End With 'a'   L={ab,ba,abbb,abab,baba,bbaa.........} // Infinite Language  

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String Contain 'abb' As Substring ||String End's With 'abb': Finite Automata Example’s

Construct Minimal Deterministic Finite Automata (DFA) String Contain 'abb' Σ={a,b} L={abb,abbab,aabba,bbabb....} ababba n+1 are possible states for the string 3,4,5....... String Lengths For Any Substring Problems.   Construct Minimal DFA Σ={a,b} a) String End's With abb L={abb,aabb,aaabb,ababb.....} // Instring Language   We Will Convert DFA From NFA If Question Is ' ends with ' Dont Use ' Dead State(qd)  

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Length Of String || Exact || At Most || At Least : Finite Automata Example

Construct Finite Automata Whose Length Of String Is Exact 2 Σ={a,b} // Input Alphabets   L={aa,ab,ba,bb} // Finite String abc is invalid string // the length of abc is 3   {Q,Σ,δ,q0,F}   Q={q0,q1,q2,qd} Σ={a,b} q0={q0} F={q2}   (ab) // String   δ=(QXΣ->Q) δ(q0,a)->q1 δ(q1,b)->q2 δ(q0,a)->qd δ(q0,b)->qd δ(qd,a)->qd δ(qd,a)->qd     b) at most 2 (Maximum 2)  // ('0','1','2') L={ε,a,b,aa,ab,bb,ba}         {Q,Σ,δ,q0,F}   Q={q0,q1,q2,qd} Σ={a,b} q0={q0} F={q0,q1,q2} δ=(QXΣ->Q)   c) at least ' 2 ' // should be 2 and more than two L={aa,ab,ba,bb,aba,bab......}   0,1 are rejected in this   Example ababb(Valid String)   ababb Is

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String Start With | End With | Contains : Finite Automata Example

We Will Understand How The Problems Ar Designed On Finite Automata And How These Problems Are Solved.   I'll Design Problem And I'll Show You Three Variations Of The Problem And I'll Show You How Finite Automata Changes With " Small Changes". First Point For You Is, You Have To Read The Problem And

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