7 votes 7 votes What are the final states of the DFA generated from the following NFA? $q_{0}, q_{1}, q_{2}$ $[q_{0}, q_{1}], [q_{0}, q_{2}], [ ]$ $q_{0}, [q_{1}, q_{2}]$ $[q_{0}, q_{1}], q_{2}$ Theory of Computation isro2013 theory-of-computation finite-automata + – makhdoom ghaya asked May 13, 2016 makhdoom ghaya 5.4k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
Best answer 9 votes 9 votes 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 Dexter answered May 13, 2016 • selected May 13, 2016 by Praveen Saini Dexter comment Share Follow See all 9 Comments See all 9 9 Comments reply Sourabh Kumar commented Jun 16, 2016 reply Follow Share But the dfa of this lang 0*1*2* has 4 final state and 1 trap state. 0 votes 0 votes Praveen Saini commented Jun 16, 2016 reply Follow Share No, 3 final states + 1 trap. 0 votes 0 votes Sourabh Kumar commented Jun 16, 2016 reply Follow Share OK I get it 0 votes 0 votes Praveen Saini commented Jun 16, 2016 reply Follow Share i think it is.. 6 votes 6 votes shikha rathore commented Jul 1, 2016 reply Follow Share @dexter bt this concept is used when we convert epsilon nfa to nfa ......bt here we r converting eps NFA to DFA....... 0 votes 0 votes One commented Aug 1, 2016 reply Follow Share 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 3 votes 3 votes Praveen Saini commented Aug 1, 2016 reply Follow Share given FA is NFA with epsilon moves. 1 votes 1 votes pradeepchaudhary commented Jun 28, 2019 reply Follow Share 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??? 1 votes 1 votes Pgoldar commented Dec 12, 2019 reply Follow Share I am also getitng the same answer. Please guide , my xam is coming soon 0 votes 0 votes Please log in or register to add a comment.