1 votes 1 votes The language $\{ W^a X^b Y^{a+b} \mid a, b, >1\}$ is Regular Context-free but not regular Context sensitive but not context free Type$=0$ but not context sensitive Compiler Design nielit-2018 compiler-design context-free-language + – Arjun asked Dec 7, 2018 recategorized Oct 24, 2020 by Krithiga2101 Arjun 1.2k views answer comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes Is answer B? Because we can push W and X into the stack and for each Y we can pop one element of the stack.That means we can design a PDA for this grammar Priyadrasta Raut answered Feb 18, 2019 Priyadrasta Raut comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes B is correct. Its CFL or Type 2 language. Push a for each W. Push a for each X. Pop a for each Y. If the stack is empty then it is accepted else not. smsubham answered Mar 3, 2020 smsubham comment Share Follow See all 0 reply Please log in or register to add a comment.
0 votes 0 votes It is context free grammar but not regular because it requires to solve one stack of memory. topper98 answered Mar 18, 2020 topper98 comment Share Follow See all 0 reply Please log in or register to add a comment.