0 votes 0 votes I know this is trivial but verify ! Theory of Computation language + – Jason GATE asked Feb 5, 2017 Jason GATE 383 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Option C. As there are no transitions on input symbols, so input is being accepted except epsilon. // Set of input symbols cannot be empty. If the state would jave been non-final then , option B would have been correct. Lucky sunda answered Feb 5, 2017 Lucky sunda comment Share Follow See all 4 Comments See all 4 4 Comments reply Jason GATE commented Feb 6, 2017 reply Follow Share Won't it accept all the Strings? 0 votes 0 votes Lucky sunda commented Feb 6, 2017 reply Follow Share No.Because there are on transactions for any input. They has to be there. 0 votes 0 votes shraddha_gami commented Feb 6, 2017 reply Follow Share It don't accept all strings because their is no self loop of alphabets... 1 votes 1 votes Jason GATE commented Feb 6, 2017 reply Follow Share Exactly !!! 0 votes 0 votes Please log in or register to add a comment.