181 views
Consider the following languages

$L_1$ = $\{a^nb^n\mid n \ge 0\}$
$L_2$ = Complement($L_1$)

Chose the appropriate option regarding the languages $L_1$ and $L_2$

(A) $L_1$ and $L_2$ are context free
(B) $L_1$ is context free but $L_2$ is regular
(C) $L_1$ is context free and $L_2$ is context sensitive
(D) None of the above
retagged | 181 views

$L_1$ is clearly a DCFL and DCFL is closed under complement. Hence, $L_2$ is also DCFL.
We can make a PDA for $L_2$ , using the same PDA for {aⁿbⁿ} as follows:

Start by pushing each 'a' on to stack. When b comes start popping. If 'a' comes after a 'b' or 'b' comes when the stack is empty, go to a final state from where the PDA accepts any string. Otherwise, at the end of the string, if stack is non-empty, accept the string and if stack is empty, reject the string.