whenever there is 1 comparison condition and we can do it with help of 1 or 0 stack then it is CFL.
a. when a comes we do push(a) and put them in stack and when b comes we do pop(a). at end of input string stack will be empty -> it is CFL
b. it is non deterministic CFL. u and w are of finite length so could be treated as constant so language could be reduced to vcv^r where c = {ab,ba,bb,ba}.
c. for this we can compare w1 with w2 which would require only 1 stack for comparison. At last if stack is not empty tthen we accept the input else reject it -> cfl
Correct me if i am wrong.
