9 9 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 + – Misbah Ghaya 7.7k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
Best answer 9 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 1 flag: ✌ Spam (Sonu123x) Dexter comment Share Follow See all 10 Comments 10 10 Comments reply Show 7 previous comments pradeepchaudhary commented Jun 28, 2019 reply Follow flag 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??? 2 2 replyShare Pgoldar commented Dec 12, 2019 reply Follow flag I am also getitng the same answer. Please guide , my xam is coming soon 0 0 replyShare Vamsi_Krishna_Thullu commented Feb 13, 2025 reply Follow flag I also had the same doubt. I got [q0], [q0, q1, q2], [q1, q2], [q2] as final states and one dead state. After minimizing this DFA, I got { [q0], [q0,q1,q2] }, [q1, q2], [q2] as final states and one dead state. So, when we draw a minimal DFA using these states, then we have total 3 final states and one dead state. If we assume each of these final states as q0, q1, q2. then this might be the answer 0 0 replyShare Please log in or register to add a comment.
