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it will accept only {0} .if any string other than {0} will goes in infinite loop(i.e 00,000,000,0000).correct me if i am wrong?
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(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"

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