Self Doubt-TOC

Question

Number of $3$ state DFA drawn over alphabet $\left \{ a,b \right \}$ which doesnot accept empty language?

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I got ans as $4^{3}\times 3\times 2^{3}$

is it correct?

Comments

mam, i made a mistake on DFA with 2 final states ===> after seeing that link, i updated !

But i am not getting, how DFA with one final state is not matching with me ?

Moreover why you duplicated this question ?

 

DFA with one final state :-

let take A,B,C and A as the initial state, then

let B is the Final State, then

case i:- A ----> a will go A, b will go A ====> C have ----> a will go any of three, b will go any of three ==> 3*3 = 9

 

case ii:- A ----> a will go A, b will go C

     for accepting empty language, we can't reach any final state

    ====> C have ----> a can't go B==>2 possible, b can't go B==>2 possible ==> 2*2 = 4

 

case iii:- A ----> a will go C, b will go A

    ====> C have ----> a can't go B==>2 possible, b can't go B==>2 possible ==> 2*2 = 4

 

case iv:- A ----> a will go C, b will go C

    ====> C have ----> a can't go B==>2 possible, b can't go B==>2 possible ==> 2*2 = 4

 

for every case B have

 ----> a will go any of three, b will go any of three ==> 3*3 = 9

 

Total = ( 9 + 4 + 4 + 4 ) * 9 = 21 * 9 = 189

 

if we will make C as Final State, same procedure follow ===> 189

 

total for DFA with one final state = 2*189 = 378

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no @Shaik

I havenot duplicated, just very confused about this

So, self made this question and after that I find this link

Anyway let me check first

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@Shaik Masthan

yes logic looking correct