0 votes 0 votes L1 is regular, L2 and L3 are CFL L1 is regular, L2 is CFL and L3 is CSL L1 is CFL but not regular,L2 is CSL but not CFL,L3 is CFL L1, L2 and L3 are CFL Theory of Computation made-easy-test-series regular-language context-free-language closure-property + – Sambhrant Maurya asked Jan 9, 2019 • edited Mar 3, 2019 by Rishi yadav Sambhrant Maurya 521 views answer comment Share Follow See all 2 Comments See all 2 2 Comments reply Ayush Upadhyaya commented Jan 14, 2019 reply Follow Share $L_1$ is regular because any number of form $2^k$ would be even, and any number of form $2^k+1$ is an odd number.So, your FA will be designed to accept all odd numbers. in L3, let ab=k $L_3=\{k^na^nb^n\}$-> well known CSL. Though I can still tick option 2, But I am not able to get a proof why L2 is CFL. 0 votes 0 votes Verma Ashish commented Mar 6, 2019 reply Follow Share @Ayush Upadhyaya Sir $2^k+1$ will not generate all odd numbers.. Like 7,11,... 0 votes 0 votes Please log in or register to add a comment.
