(16)C:->on,n>=2
(17)B:->{0}
Here, first we find small string for hang.
we have 0,0,R and 0,0,L means if we have 0 input then we put 0 in stack and go to right and second one means is if we have 0 input then we put 0 in stack and go to left and if we get B(end i/p) then push B and go to right and finished.
So, if we have 00 string the we go fisrt right and then left then right .........continue never end so small string is o^2 then we say 0^n where n>=2.
and acceptable string is only "0"