#finiteautomata

Complement Of Finite Automata Means The Finite Automata Which Is Obtained By Interchanging Final And Non-Final States Is Known As Complement Of Finite Automata.

In This Concept, I'll Change Final States To Non-Final To Final State And Final To Non-Final State. By Doing This The Language Changes. M-Automata(DFA) L-Language M Supports L Language   The String Which Comes In Language L Are Accepted By Machine M L⊆Σ* L Belongs To Sigma Star Σ* Means Universal Set Of Strings(All The Strings) L⊆Σ* Means The Strings Which Are Accepted By Language Comes Under 'L' Not All Strings Comes Under 'L'   M→L⊆Σ* Means Machine M Accepts Strings Which Are Belongs To 'L' And L Is The Language(Set Of Strings) Which Are Subset Of Universal Set Of Strings i.e Σ* (Sigma Star)

Then What Is Complement Of Finite Automata?

If I Give Complement For Machine 'M' That Becomes M1 M1→L1= Σ*- L M1 Is The Machine Which Accepts The Language L1 L1= Σ*- L   This Machine M1 Accepts The Language L1 And L1  Is Universal Set Of Strings Language. M→L⊆Σ* M1→L1= Σ*- L   I Hope You Understood The Difference Between Machine M And Machine M1 Example

Σ={a,b}

L=String Starts With 'a'

This Is (M) Machine Example.  

     2. Σ={a,b}

L=String Does Not Start With 'a'  

This Is M1 (Machine Complement) Example In M1 The Non-Final States Becomes Final State Including Dead State And Final States Becomes Non-Final State.   Cross-Check baa  

Remember

Construct The Minimum Finite Automata That Accept All The Strings Of a & b Such That

i) String Contains At least 2 a's & ηb|w|≅0mod3

Solution:- (Check Previous Two Problems For Better Understanding) CONDITION GIVEN:- String Should Contain At least 2 a's And ηb|w|≅0mod3(Means No Of b's In String Is Approximately Equal To O Mod 3(If I Divide The No.Of b's With 3, I Should Get Remainder As 0)  

FINITE Automata 1:-

Σ={a,b} L={aa,aaa,aaaa,aaaaa....}

FINITE AUTOMATA 2:-

ηb|w|≅0mod3 Means No Of b's In String Is Approximately Equal To O Mod 3(If I Divide The No.Of b's With 3, I Should Get Remainder As 0

Possible Remainders For 3 Are 0,1,2   Mean If I Divide No. Of b's In A String With 3 I Should Get Remainder As 0   Imagine-{b,bbb,babbab,bbabbbbba....} If No Of b's Are '0' In A String And If I Divide 3 With 0 I'll Get Remainder '0'   0 Is Divisible By 3 0 Means ε Means L={ε,1,2,3}={ε,b,bb,bbbb...}

CROSS PRODUCT FINITE AUTOMATA 1 × FINITE AUTOMATA 2 => {(q0,q3),(q0,q4),(q0,q5), (q1,q3),(q1,q4),(q1,q5), (q2,q3),(q2,q4),(q2,q5)}  

FINITE AUTOMATA

Construct The Minimal Finite Automata That Accept All The String Of a & b Such That i) There Is Even No. Of a And Even Number Of b   Solution:- Question Is Final Automata Should Contain Even Number Of 'a's And Even Number Of 'b's Even Numbers Are 0,2,4,6,8,.....  

CREATE TWO SEPARATE AUTOMATAS WITH GIVEN TWO CONDITIONS

Finite Automata 1:- Even Number Of 'a's Σ={a,b}

q0 Is My Initial And Final State Because Of ε q0 Goes To q1 And q1 Goes To q0(aa) I Dont Bother About Number Of b's On q0q1 Finite Automata 2:- Even Number Of b's(ε,bb,bbbb,bbbbbb....)

I Have Drawn Finite Automata 2 Using The Same Procedure Used In Finite Automata 1.   Now Cross Product (Or) Cross Multiplication Finite Automata 1 × Finite Automata 2

=> {(q0,q2),(q0,q3) (q1,q2),(q1,q3)}  NOS - 4   Now DRAW FINAL DFA Using These States By Observing Finite Automata 1 And Finite Automata 2

Final State Will Be The Combination Of Finite Automata 1 And Finite Automata 2 Only. Because The Condition Given Is Two Conditions Should Be Satisfied. No Of Even 'a's And No. Of Even b's In A String.   Cross Check abaaba

Construct The Minimal Finite Automata That Accepts All Base 8 Numbers Which Are Divisible By '6' (Binary Means Base 2, i.e Σ(0,1) Input Symbols Will Be 2 Only 0 & 1) (Integer Means Base 10,i.e (0,1,2,3,4,5,6,7,8,9) Input Symbols Will Be 10 Only 0&1&2&3&4&5&6&7&8&9)   Here Given Input Symbols Are Base '8', It Means OCTAT. My Input Symbols Are Σ={0,1,2,3,4,5,6,7}   Example :- (22)8 = (2×81+2×80)10 =(16+2)10 =(18)   After Conversion Into Decimal (22)8 Is Divisible By '6' Possible Remainders For '6'

Question Is Divisible By '6' Means Remainder Should Be '0' For '0' We Gave 'q0'As State ,  'q0' Is Final State.  

q1= q4 q2=q5   q1& q4 Are Same, I'll Merge Both. q1= q4 q2 & q5  Are Same, I'll Merge Both. q2=q5  

No Commons States In This Table. MAKE DFA USING THIS TABLE.