But we have or operation.Cant we handle in non deterministic manner?We have choice here ,first we can see first condition.If that fails then we can see second condition.?NPDA can create multiple copies??
2 is CFL not DCFL.
we first push for a then do nothing for b and then pop for c. if stack empty then it is CFL else not.
OR
we do nothing for a then push for b then do nothing for c and then pop for d. if stack empty then it is CFL else not.
L ={ a^n b^m c^x d^y | n=m or x=y} CFL but not DCFL
instead of this COPY 1: push a's , pop b's ---> in end stack empty , c ->ignore , d->ignore
can we do this???
push a's , pop b's ---> in end stack empty , c ->push c's , d->pop d's ---->in end stack is empty accept it?