But this can be optimized further, i.e. the error-correction term don't need to be evaluated for every step. Instead it can be handled once every n steps - in this case n = int(11/3) = 3 - since you know that after the extra step has been taken, the error term can't overflow for a couple of steps. In this case, the worst-case error can be 2/11 after the correction, guaranteeing that 5/11, 8/11 will not overflow, i.e. you can always do at least 3 steps after a correction.

Another optimization is to work with signed numbers, and adjust the error term to work symmetrically around zero, to allow the processor flags to be used to indicate a correction, instead of all the time having to compare with 11.