0 votes 0 votes I'm getting its equaltion {anbn | n > 0} U {a} U {b} But given is {anbn | n >= 0} U {a} U {b} Whether epsilon is accepted or not?? Theory of Computation pushdown-automata npda theory-of-computation + – Ashwin Kulkarni asked Dec 24, 2017 Ashwin Kulkarni 865 views answer comment Share Follow See all 8 Comments See all 8 8 Comments reply Show 5 previous comments AakS commented Dec 24, 2017 reply Follow Share @srestha, epsilon acceptance is not possible. {a} and {b} will be accepted only when n=0 which is signified here by n>=0 in {anbn n>=0}...Am i ryt? 1 votes 1 votes joshi_nitish commented Dec 24, 2017 reply Follow Share it will accept, L = a + b +{anbn |n>=1} your answer is correct @Ashwin. 1 votes 1 votes saxena0612 commented Dec 24, 2017 reply Follow Share Starting state? 0 votes 0 votes Please log in or register to add a comment.