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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. $[q_{0}, q_{1}], q_{2}$
in Theory of Computation by Boss (30.8k points) | 2.4k views

1 Answer

+7 votes
Best answer
Option a is the answer

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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0
But the dfa of this lang 0*1*2* has 4 final state and 1 trap state.
0
No, 3 final states + 1 trap.
0
OK I get it
+5

i think it is..

0
@dexter bt this concept is used when we convert epsilon nfa to nfa ......bt here we r converting eps NFA to DFA.......
+2
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
+1
given FA is NFA with epsilon moves.
0
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???
0
I am also getitng the same answer. Please guide , my xam is coming soon
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