1 votes 1 votes The language generated by the below grammar is S --> aSS / b Theory of Computation theory-of-computation finite-automata + – Vicky rix asked Apr 5, 2017 • retagged Jun 4, 2017 by Arjun Vicky rix 1.6k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes a w* bb where wϵ (a+b)* and number of b's= number of a's+1 is the language produced by the grammer It means any string having prefix a and suffix bb , Deepthi_ts answered Apr 11, 2017 • edited Apr 11, 2017 by Deepthi_ts Deepthi_ts comment Share Follow See all 2 Comments See all 2 2 Comments reply rajsh3kar commented Apr 12, 2017 reply Follow Share what about single 'b' 0 votes 0 votes Shubham Sharma 2 commented Apr 12, 2017 reply Follow Share 'b' is also derived from given grammar which can't be derived from your regular expression. 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes S --> aSS / b let us suppose the string is "aaabbbb" S-->aSS S-->aaSSS S-->aaaSSSS S-->aaabbbb The language generated by this grammer is a^nb^m such that n<m Tushar Dhoot answered Apr 5, 2017 Tushar Dhoot comment Share Follow See all 8 Comments See all 8 8 Comments reply 2018 commented Apr 5, 2017 reply Follow Share abb aabbb ababb aaabbbb abaabbb also string in language but in yr case it will not be accepted. 0 votes 0 votes Tushar Dhoot commented Apr 5, 2017 reply Follow Share Yes right... so number of b is greater that a 0 votes 0 votes 2018 commented Apr 5, 2017 reply Follow Share then also wrong bcz string bbba will be accepted according to u 0 votes 0 votes Shubham Sharma 2 commented Apr 5, 2017 reply Follow Share how "bbba" will be accepted if "b" is coming after "a" in the language generated. 0 votes 0 votes 2018 commented Apr 5, 2017 reply Follow Share i'was giving counter example to @Tushar Dhoot 0 votes 0 votes Shubham Sharma 2 commented Apr 5, 2017 reply Follow Share ok..what will be the correct language here? any idea.. 0 votes 0 votes 2018 commented Apr 5, 2017 reply Follow Share @shubham not getting exact language 0 votes 0 votes RAJKUMAR commented Jul 7, 2017 reply Follow Share this answer is best answer among these answer 0 votes 0 votes Please log in or register to add a comment.
1 votes 1 votes S --> aSS / b The language generated by this grammer is $ a^{*}(ab)^{*}a^{n}b^{m}$ where n<m and n>=0 and m>=1. Shubham Sharma 2 answered Apr 5, 2017 • edited Apr 12, 2017 by Shubham Sharma 2 Shubham Sharma 2 comment Share Follow See all 4 Comments See all 4 4 Comments reply Deepthi_ts commented Apr 11, 2017 i edited by Deepthi_ts Apr 11, 2017 reply Follow Share abab is also accepted by your regular expression which can't be derived from given grammer. 0 votes 0 votes Shubham Sharma 2 commented Apr 11, 2017 reply Follow Share m is greater than n , an extra b will always come at end in comparison with second last a. 0 votes 0 votes Deepthi_ts commented Apr 12, 2017 reply Follow Share aaaababbbbb this string is also derived from given grammar which can't be derived from your regular expression. 0 votes 0 votes Shubham Sharma 2 commented Apr 12, 2017 reply Follow Share It can be derived now. if Any other please correct me. 0 votes 0 votes Please log in or register to add a comment.