0 votes 0 votes L={a^n b^k, n <= k <=2n} 1.CFL 2.Non CFL parthbkgadoya asked Jan 30, 2017 parthbkgadoya 386 views answer comment Share Follow See 1 comment See all 1 1 comment reply saurabh rai commented Jan 30, 2017 reply Follow Share https://gateoverflow.in/92299/context-free-languagev 0 votes 0 votes Please log in or register to add a comment.
0 votes 0 votes it is CFL because it can Non-deterministic PDA can solve this. k will either be n,n+1,n+2,.....,2n. Now non-deterministic PDA can run them to find out which one satisfies it. Thus it is CFL. ankyAS answered Jan 30, 2017 ankyAS comment Share Follow See all 3 Comments See all 3 3 Comments reply Surajit commented Jan 30, 2017 reply Follow Share are you sure about this? I think 2 stacks might be required here and might be CSL.How to compare n<k and k<2n at the same time using one stack? We can push n no of a's at first then if you apply non determinism also how will you remember if it is less greater than n or less than 2n, you dont have a fixed value for n. 0 votes 0 votes Ashish321 commented Jan 31, 2017 reply Follow Share NPDA WILL ACCEPT THIS,BCZ NPDA IS SUM UP ALL STACK REQUAIRED IN THIS LANGUAGE,FOR ALL STATE AUTOMATION.TOTAL NO OF n STACK REQUIRED AT MOST. 0 votes 0 votes Surajit commented Jan 31, 2017 reply Follow Share now I see it thanks.. 0 votes 0 votes Please log in or register to add a comment.