∈-NFA TO NFA

Question

Construct an eqv. NFA for the given ∈-NFA

Is this eqv. Nfa correct??

Or this one

Why is thr transition between q0 to q1 of 0,1 ?

Comments

FA doesnt contain more than one start state??

And nfa also doesnt contain more than one start state or this is only because ∈ nfa doesnt contain more than one start state??

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In any

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Okay..

Answer 1

$\epsilon$- closure $(q_0)=(q_0,q_1,q_2)$

$\epsilon$-closure$(q_1)=(q_1,q_2)$

$\epsilon$-closure$(q_2)=(q_2)$

We can design DFA directly by taking $\epsilon$- closure $(q_0)$ , i,e, $(q_0,q_1,q_2)$ as start state

$Q$\ $\Sigma$	$0$	$1$	$2$
->$(q_0,q_1,q_2)^*$	$(q_0,q_1,q_2)$	$(q_1,q_2)$	$(q_2)$
$(q_1,q_2)^*$	-	$(q_1,q_2)$	$(q_2)$
$(q_2)^*$	-	-	$(q_2)$
-	-	-	-

or NFA with q0 as start state and having states $q_0, q_1$ and $q_2$ where $q_2$ is final state

$Q$\ $\Sigma$	$0$	$1$	$2$
->$q_0$	$q_0,q_1,q_2$	$q_1,q_2$	$q_2$
$q_1$	-	$q_1,q_2$	$q_2$
$q_2^*$	-	-	$q_2$

And from NFA we can convert to DFA also 

Comments

∈ closure of a state q is all possible transition from q to any states?? Is it right?

∈ Nfa to nfa no. Of initial states and final states remain unchanged.

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$\epsilon$ -closure of $q$ , means all states reachable from $q$ with $\epsilon$ moves, including $q$ itself.

in equivalent NFA , initial state and final states remains unchanged.

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@praveen sir

we can't accept epsilon in new NFA which is constructed you ..

which is possible in the epsilon NFA given in the question.

i think the NFA provided by second diagram in the question is perfectly right.

can you please check.