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Construct Minimal DFA That Accepts All The Strings Of 0’s & 1’s Where The Length Of The String Is a) Divisible By 2 And Divisible By 3 b) Divisible By 2 Or Divisible By 3

Construct Minimal DFA That Accepts All The Strings Of 0's & 1's Where The Length Of The String Is a) Divisible By 2 And Divisible By 3 b) Divisible By 2 Or Divisible By 3 a) Divisible By 2 And Divisible By 3 Divisible By 2 And Divisible By 3 // 2

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DFA Example : Construct Minimal Deterministic Finite Automata(FA) Σ={0,1} Such That a) String Start With 0 And Length Is Divisible By 3 b) String Start With 010 And Length Is Equal To 3(Mod 5)

Construct Minimal Deterministic Finite Automata(FA) Σ={0,1} Such That a) String Start With 0 And Length Is Divisible By 3 b) String Start With 010 And Length Is Equal To 3(Mod 5) a) String Start With 0 And Length Is Divisible By 3 Given Σ={0,1} String Start With 0 And Length Is Divisible By 3. L={011,001,010,010000,010101010.....}

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Construct Minimal DFA That Accepts All Strings Of 0’s & 1’s Where a) No Of 0’s Is Even b) No Of 1’s Is Odd

Construct Minimal DFA That Accepts All Strings Of 0's & 1's Where a) No Of 0's Is Even b) No Of 1's Is Odd In This Problem, He Asked About The Characters In The String, He Didn't Ask The Length Of String.   a) No 0s 0's Is Even In The String, The Number Of 0's

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DFA Example : Construct The Minimal FA(DFA) That Accepts All Strings Of a & b Where 3rd Symbol From Right End Is b

Construct The Minimal FA(DFA) That Accepts All Strings Of a & b Where 3rd Symbol From Right End Is b   Solution Means _ _ _ _ _ _ 3 _ _ 2n--> States(23=8) // 2 Is a,b(input alphabets), n Is 3rd Symbol. 2n-1=Final States(23-1=4)   Transition Table

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DFA Example : Construct Minimal FA(MEANS DFA) That Accept All The Strings Of a & b Where Second Symbol From Left End Is a

Construct Minimal FA(MEANS DFA) That Accept All The Strings Of a & b Where Second Symbol From Left End Is a, Third Symbol From Left End Is b, Fifth Symbol From Left End Is a. 2nd From Left End Is a 3rd Symbol From Left End Is b 5th Symbol From Left End Is a   Conditions Every

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DFA Example : Construct DFA Starts aa Or bb Where Σ={a,b}

Construct DFA Starts aa Or bb Where Σ={a,b} , Means 'aa' Or 'bb' Anyone Of The Condition Should Be Satisfied Let's Create Language.   L={aa,bb,aab,bba,aaa,bbb,.....}// Infinite String The Only Possibilities aa,bb,ab,ba DFA Final Minimal DFA Remember My Starting String Should Be 'aa' or 'bb' I'll Reject If I Get ab (or) ba In Starting Of My String. If

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DFA Example : Construct Minimal Deterministic Finite Automata(DFA) Starts And Ends With ab Where Input Symbols Are Σ={a,b}

Construct Minimal Deterministic Finite Automata(DFA) Starts And Ends With ab Where Input Symbols Are Σ={a,b}   This Is The Problem In Which The String Must Satisfy More Than One Condition.   It Is Easy To Solve One Condition Problem In DFA, It Is Little Difficult To Solve Problems With More Than One Condition.   We Have

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