 Solution

The second coordinate after two iterations satisfieswhenever those two divisions are needed, by the same remainder bound, now applied to $b_i>b_{i+1}>0$. Let $m=\lfloor {\log }_{2}b\rfloor$. If the Euclidean algorithm has not already stopped after $2m$ iterations, then its positive second coordinate is at most $1$: for $m\ge 1$,For $m=0$ the original $b$ is already $1$. A second coordinate of $1$ requires just one more division. ConsequentlyThis removes dependence on the potentially much larger input $a$. It improves the scale ${\log }_2(ab)$ to one involving only the smaller input; the rounded numerical bound need not be smaller for every pair. One may always take the minimum of the bounds in (ii) and (iii).