1 votes 1 votes For the nfa given below, find δ*(q0, 1010) and δ* (q1,00). Theory of Computation theory-of-computation finite-automata + – Vicky rix asked Apr 2, 2017 retagged Jun 4, 2017 by Arjun Vicky rix 1.9k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes Please correct me if am wrong .... δ*(q0, 1010) = { q0 , q2 } δ* (q1,00) = { } (empty set) Vicky rix answered Apr 2, 2017 Vicky rix comment Share Follow See all 2 Comments See all 2 2 Comments reply rude commented Apr 2, 2017 reply Follow Share This is a very very bad practice. You just can not post the question as well as answer, with yourself. Whatever answer you are getting, write that in the question only that this is whai I am thinking, what do you guys think? this is the correct way of asking question. 0 votes 0 votes Heisenberg commented Apr 3, 2017 reply Follow Share relax, he has not committed a crime. Its ok. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes δ*(q0, 1010) = { q0 } and not { q0 , q2 } because even though we can reach q2 by using lambda, we haven't consumed the input 1010, which we should. δ* (q1,00) = { } (empty set) shraddha priya answered Apr 2, 2017 shraddha priya comment Share Follow See all 7 Comments See all 7 7 Comments reply Show 4 previous comments Angkit commented Apr 3, 2017 reply Follow Share At Q1,if we are giving 0 then we can get to Q0 ? 0 votes 0 votes Mandeep Singh commented Apr 3, 2017 reply Follow Share No there is no transition on 0 means if we get a 0 we will be in trap state and string will be unaccepted. Note: for simplicity we do not show trap states in NFA whereas it's a must in DFA. 1 votes 1 votes Angkit commented Apr 3, 2017 reply Follow Share got it.. thank you.. 0 votes 0 votes Please log in or register to add a comment.