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What are the final states of the DFA generated from the following NFA?

1. $q_{0}, q_{1}, q_{2}$
2. $[q_{0}, q_{1}], [q_{0}, q_{2}], [ ]$
3. $q_{0}, [q_{1}, q_{2}]$
4. $[q_{0}, q_{1}], q_{2}$
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the ∈ closure of the states which contain final states in its closure is also Final state

So q0 ,q1,q2 all are final states
by Loyal (9.9k points)
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But the dfa of this lang 0*1*2* has 4 final state and 1 trap state.
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No, 3 final states + 1 trap.
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OK I get it
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i think it is..

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@dexter bt this concept is used when we convert epsilon nfa to nfa ......bt here we r converting eps NFA to DFA.......
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In case of Nfa final state q0,q1,q2

but given is Dfa so final state should be

q0,q2,[q1,q2],[q0,q1,q2]

please guide if i am wrong
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given FA is NFA with epsilon moves.
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The final states I'm getting are [q0] , [q0,q1,q2] , [q1,q2] , [q2]. My question is You have concluded that q0,q1,q2 are final states...how???
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I am also getitng the same answer. Please guide , my xam is coming soon