Minimal DFA

Question

Given following NFA

find the minimal equivalent DFA

Comments

approach i used is same as yours.

but answer is different.

Minimal DFA i obtained is accepting a(a+b)*. and there are 3 stated in minimal DFA. one initial state. one final state and one dead state.

in the table where you converted epsilon NFA to NFA, row 2nd and 3rd is incomplete, as state 2 and 3 both will go to all the states 1,2,3 on input a and input b. please check it once.

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can u pls give the detailed solution? I'm not understanding how a(a+b)* is coming @aambazinga

Answer 1

ð(q,x)=epsilon-closure(ð(q,epsilon-closure(x)))

epsilon-NFA to NFA 

	a	b
1	123
2	123	123
3	123	123

 

NFA to DFA

	a	b
1	*123	Dead
*123	*123	*123
Dead	Dead	Dead

123 is the final state.

Comments

thanks a lot... i understood now where i was going wrong

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@aambazinga

 

epsilon-NFA to NFA 

	a	b
1	123
2	123	123
3	123	12

which steps u taken last step I mentioned?

Is it taking first transition on alphabet, then epsilon transition ??

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@srestha
first epsilon, then alphabet, then epsilon

Answer 2

the above nfa is equivalent to  a(a+b)* for further full solution  see the image on the link; https://ibb.co/HTBkRzc

edit: i have corrected my image link

Comments

@aditi19 i have edited , i do not why it was not changed

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I'm not understand the conversion of epsilon NFA to NFA part
where did the transition (2,b)=1 go?
@rballiwal

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@aditi19 look at the transition between (1) and (2)