#finiteautomataexamplewithsolution

[smartcherryposts]

Complement Of Finite Automata

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

L(FA)∩l(FA1)=Φ L(FA)∪(FA1)=Σ* M→n states k final states M1→ n states n-k final states       Read the full article

[smartcherryposts]

Construct The Minimum Finite Automata That Accept All The Strings Of a & b Such That String Contains At least 2 a's & ηb|w|≅0mod3

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

FINAL STATE Is q2,q3  Because Two Conditions Should Satisfy. CROSS CHECK babab // aa,bbb Read the full article

[smartcherryposts]

Construct The Minimal Finite Automata That Accept All The String Of a & b Such That There Is Even No. Of a And Even Number Of b

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

Check Previous Problem For Better Understanding     Read the full article

[smartcherryposts]

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

  Read the full article

[smartcherryposts]

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)   Read the full article