Answer
U can never get 9 0's
Work Step by Step
let us consider that the circle contains 9 0's after iterating the procedure.
We obtain 9 0's if all in the previous step were the same thus all bits were 0's or all bits were 1's.
If the previous step contained all 0's, then we have the same the case as the current iteration step. Since initially the circle does not contain only zeros had to contain something else then only zeros at some point and thus there exist a point where the circle contained only 1's.
A circle Contain only 1's if every pair of consecutive 9 digits is different, However, this is impossible because there were 5 1's and 4 0's thus
1's and 0's alternate, then we obtain that two 1's that will be next to each other (which would result in a one in next step). thus we obtained a contradiction and thus the assumption that the circle contains nine 0's after iterating the procedure is false. this then means that you can never get nine 0's.
