December2015 cs Q28
- Option : A
- Explanation : Language L1 is context free language as L1 = {anban|n > 0}, we can compare number of a’s before b and number of a’s after b both are equal using only one stack using push and pop function but we can not compare language L2. L2 = {an b an bn + 1|n > 0} using only one stack here we can compare all number of a’s before b and number of a’s after b both are equal using only one stack using push and pop function but we cannot compare all bn + 1 at that time. Hence, L1 is context free language and L2 is not context free language.
