2 votes 2 votes shefali1 asked Nov 20, 2017 shefali1 535 views answer comment Share Follow See all 10 Comments See all 10 10 Comments reply Show 7 previous comments arch commented Nov 21, 2017 reply Follow Share l2 is not dcfl its cfl only 0 votes 0 votes hs_yadav commented Nov 21, 2017 reply Follow Share no l2 is dcfl...it can be simply implemented using single stack....and deterministic transition... l2 is just representing aibjck ,j=k ??? 0 votes 0 votes Chhotu commented Nov 24, 2017 reply Follow Share Good question. 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes $L_{2}$ and $L_{4}$ both are DCFL. So answer is 2. In $L_{2}$ we have to only check j==k or not. Chhotu answered Nov 24, 2017 Chhotu comment Share Follow See all 0 reply Please log in or register to add a comment.