0 votes 0 votes Please check if solution is correct. Won't the grammar accept aabbb? Vikas Verma asked Nov 7, 2018 edited Nov 8, 2018 by Vikas Verma Vikas Verma 478 views answer comment Share Follow See all 10 Comments See all 10 10 Comments reply Show 7 previous comments Vikas Verma commented Nov 8, 2018 reply Follow Share Yes dilip, I understand that, But option C itself says no of a's is greater than b's right? But the grammar in the question accepts a string which is not having this property. 0 votes 0 votes Swapnil Naik commented Nov 8, 2018 reply Follow Share I watched the video associated with that question, according to the video the last transition is not b,z0,$\epsilon$ but it is $,z0,$\epsilon$. i.e. it can accept empty string. So now the above grammar accepts equal number of a's and b's. L = {$\epsilon$, ab,aabb,aaabbb,abab,aabbab,...} Now you take every prefix of above language, you will find no. of a's are greater than or equal to b's. 1 votes 1 votes `JEET commented Nov 8, 2018 reply Follow Share ok 0 votes 0 votes Please log in or register to add a comment.