Note that the time taken is linear with respect to the original execution.
A cycle of n instructions starting at instruction n0 will be detected in between n0 and n0+n iterations (i.e. <= 3*(n0+n) underlying instructions) since n0 iterations gets both into the cycle and the offset between them will become 0 some time in the next n iterations.
n could be very large, of course (e.g, using all of memory as a giant counter so the cycle length is huge), but the cycle detection is not really making your problem worse.
A cycle of n instructions starting at instruction n0 will be detected in between n0 and n0+n iterations (i.e. <= 3*(n0+n) underlying instructions) since n0 iterations gets both into the cycle and the offset between them will become 0 some time in the next n iterations.
n could be very large, of course (e.g, using all of memory as a giant counter so the cycle length is huge), but the cycle detection is not really making your problem worse.