32 votes 32 votes Which of the following definitions below generate the same language as $L$, where $L=\{x^ny^n \text{ such that } n\geq 1 \}$? $E \rightarrow xEy\mid xy$ $x y \mid (x^+xyy^+$) $x^+y^+$ I only I and II II and III II only Theory of Computation gate1995 theory-of-computation easy context-free-language + – Kathleen asked Oct 8, 2014 • recategorized Apr 25, 2021 by Lakshman Bhaiya Kathleen 10.7k views answer comment Share Follow See all 6 Comments See all 6 6 Comments reply Show 3 previous comments RavGopal commented Jun 30, 2020 reply Follow Share @teluguenglish, when n=1,string would be ''xy", and in (Ⅰ),you can get "xy" because"xy" given as stopper. 1 votes 1 votes kshitij86 commented Feb 2, 2021 reply Follow Share Nice approach 0 votes 0 votes Manu Shaurya commented May 20, 2021 reply Follow Share @vaibhav101 Just to add something to your logic, L is Context Free but not regular. 0 votes 0 votes Please log in or register to add a comment.
Best answer 37 votes 37 votes Correct Option: A In the other two you can have any number of $x$ and $y$. There is no such restriction over the number of both being equal. Gate Keeda answered Oct 9, 2014 • edited May 5, 2021 by soujanyareddy13 Gate Keeda comment Share Follow See all 0 reply Please log in or register to add a comment.
4 votes 4 votes A is the answer B is wrong because xxyyy is there same apply for C also Murali answered Jul 13, 2015 Murali comment Share Follow See all 0 reply Please log in or register to add a comment.
1 votes 1 votes given language L is CFL , so we can not write a regular expression for this. but in option ii,iii it is written in regular expression so it is clearlly false. so option a is true abhishekmehta4u answered Mar 28, 2019 abhishekmehta4u comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes from the very easy eye We can say that i and iii are correct as n>=1 but as there is no opt so 1 is correct A arkaprabha1012 answered Aug 22, 2020 arkaprabha1012 comment Share Follow See all 0 reply Please log in or register to add a comment.