2 2 votes How to check whether following language are context free 1)L= {a^nb^m ; n<=m<=3n} 2)L={w:2na(w) <=nb(w)<=3na(w) making NPDA or DPDA is not easy/convenient perhaps Theory of Computation + – Meenakshi Sharma 1.4k views answer comment Share Follow Print 0 reply Please log in or register to add a comment.
0 0 votes Both are not context free languages. 1)L= {a^nb^m ; n<=m<=3n} In this language m depends on n. m and n are related to each other with two conditions i.e. m>=n and m<=3n so m can take any value between n and 3n. Here we have two conditions on m which cant be solved by using PDA or NPDA . 2)L={w:2na(w) <=nb(w)<=3na(w) In this language the number of b's in the strings depends on the number of a's i.e. 2na(w) <=nb(w)<=3na(w) which again cant be solved by PDA or NPDA. Gaurav Joshi answered Jul 11, 2017 Gaurav Joshi comment Share Follow 0 reply Please log in or register to add a comment.
0 0 votes Cfl resist upto two comparisons Deepak Raj 1 answered Jul 13, 2017 Deepak Raj 1 comment Share Follow 0 reply Please log in or register to add a comment.