0 votes 0 votes $\large L=\left \{ a^m b^n c^k | k<=min(m,n) \right \}$ is this context free? alok27 asked Sep 13, 2016 alok27 726 views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
2 votes 2 votes am bn ck where k<=min(m,n) is same as ( am bn ck where k<=m & k<=n) Clearly two condition with AND so it should not be a CFL. Digvijay Pandey answered Sep 13, 2016 Digvijay Pandey comment Share Follow See all 9 Comments See all 9 9 Comments reply ManojK commented Sep 13, 2016 reply Follow Share What if L = { am bn ck | k ≥ max(m,n) } ? 0 votes 0 votes Kapil commented Sep 13, 2016 reply Follow Share CSL .. 1 votes 1 votes ManojK commented Sep 13, 2016 reply Follow Share And what about L = { am bn ck | k $\large \leq$ min(m,n) } ? 0 votes 0 votes Kapil commented Sep 13, 2016 reply Follow Share again CSL and if L = { am bn ck | k >= min(m,n) } Then , NCFL 0 votes 0 votes ManojK commented Sep 13, 2016 reply Follow Share Can you distinguish L1 = { am bn ck | k ≤ min(m,n) } L2 = { am bn ck | k $\geq$ min(m,n) } 0 votes 0 votes Kapil commented Sep 13, 2016 reply Follow Share L1 = { am bn ck where k<=m AND k<=n } L2 = { am bn ck where k>=m OR k>=n } 0 votes 0 votes ManojK commented Sep 13, 2016 reply Follow Share In L1 can,t we check like this k<=n if k<=m. 0 votes 0 votes Digvijay Pandey commented Sep 13, 2016 reply Follow Share ^^ if k>m then it wont check for n. 0 votes 0 votes ManojK commented Sep 13, 2016 reply Follow Share Sir.For L1 can,t we say that if number of c's is less than number of a's or if number of c's is less than number of b's.? 0 votes 0 votes Please log in or register to add a comment.