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 5.3k 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.