2 votes 2 votes shefali1 asked Nov 20, 2017 shefali1 540 views answer comment Share Follow See all 10 Comments See all 10 10 Comments reply Shivam Chauhan commented Nov 20, 2017 reply Follow Share $L_2$ and $L_4$ are CFL. Also DCFL 0 votes 0 votes shefali1 commented Nov 20, 2017 reply Follow Share plz explain in detail 0 votes 0 votes Shivam Chauhan commented Nov 20, 2017 reply Follow Share Is it correct? 0 votes 0 votes shefali1 commented Nov 20, 2017 reply Follow Share yes,plz explain each one in detail 0 votes 0 votes Shivam Chauhan commented Nov 20, 2017 reply Follow Share $L_2$ has both conditions i<=j and j>=i so it does not matter just check j = k. $L_4$ match i with j and if you get empty stack after all b finished then check c for even by alternating push and pop 3 votes 3 votes abhishek tiwary commented Nov 21, 2017 reply Follow Share in L3 and L4 single comparison is required where as in L1,L2 two comparison is required 0 votes 0 votes Akash Mittal commented Nov 21, 2017 reply Follow Share explain L2? 0 votes 0 votes 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.