3 votes 3 votes $L_1 =\{a^n b^m c^n \mid m,n \geq 0\}$ and $L_2=\{ a^n b^n\mid n\geq 0\}$. If $L=L_2-L_1$ then $L$ is finite language regular language DCFL not DCFL Theory of Computation context-free-language identify-class-language + – Prateek Raghuvanshi asked Nov 10, 2017 Prateek Raghuvanshi 488 views answer comment Share Follow See all 4 Comments See all 4 4 Comments reply joshi_nitish commented Nov 10, 2017 reply Follow Share L=L2-L1 = { a^n b^n|n>=1} and hence it is DCFL 2 votes 2 votes Hemant Parihar commented Nov 10, 2017 reply Follow Share @manu, no it is not an empty string. Whatever be the common element between L1 and L2 should be removed from L2. Here common element is epsilon. 0 votes 0 votes Anu007 commented Nov 10, 2017 reply Follow Share an bm cn here we need to see a and b can be equal but c will also be equal so an bn can never be exept for n=0 generated by L1 0 votes 0 votes joshi_nitish commented Nov 11, 2017 reply Follow Share @Manu see in, L1 only epsilon is common with L2, because whenever you try to take string containing 'a' in L1, 'c' will also come in that string...but there is no string containing 'c' in L2, so only epsilon in L1 is common with L2, therfore L=L2-L1 = { a^n b^n|n>=1} 0 votes 0 votes Please log in or register to add a comment.
Best answer 6 votes 6 votes Option C. strings in L1 ={epsilon, ac, abc, aabcc, ...........} string in L2 = {epsilon , ab, aabb, aaabbb.........} in both language only epsilon is common L2-L1 = { ab, aabb, aaabbb........} ={a^n b^n | n>0} which is DCFL Akash Mittal answered Nov 10, 2017 selected Nov 11, 2017 by Manu Thakur Akash Mittal comment Share Follow See all 4 Comments See all 4 4 Comments reply Manu Thakur commented Nov 11, 2017 reply Follow Share if m=0, then L1 can generate ab, aabb, aaabbb, and so on.. right? 0 votes 0 votes Akash Mittal commented Nov 11, 2017 reply Follow Share if m=0, then it can only generate a^n c^n 0 votes 0 votes Manu Thakur commented Nov 11, 2017 reply Follow Share yes, i didn't notice! 0 votes 0 votes Kaluti commented Nov 14, 2017 reply Follow Share L = L2 - L1 can be written as L2 INTERSECTION L1' COMPLEMENT OF L1 IS DCFL L2 is already DCFL NOW intersection of two dcfl is not closed it means it may or may not be dcfl but here it is dcfl am i right? 0 votes 0 votes Please log in or register to add a comment.