TAOCP 4.5.2 Exercise 29
After the first subtract-and-shift cycle of Algorithm B, the ratio $x = \min(u, v)/\max(u, v)$ is transformed according to the rule $x \mapsto x/(1 + 2^k x)$, where $k$ is the number of trailing zeros...
Section 4.5.2: The Greatest Common Divisor
Exercise 29. ▶ [**] [HM46] [R. P. Brent.] Find $G_1(x)$, the distribution of $\min(u, v)/\max(u, v)$ after the first subtract-and-shift cycle of Algorithm B as defined in (35). Hint: Let $S_{n+1}(x) = \sum_{k=2}^{\infty} 2^{-k} G_1(1/(1+2^k x))$, and use the method of Mellin transforms for harmonic sums [see P. Flajolet, X. Gourdon, and P. Dumas, Theor. Comp. Sci. 144 (1995), 3–58].
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After the first subtract-and-shift cycle of Algorithm B, the ratio $x = \min(u, v)/\max(u, v)$ is transformed according to the rule $x \mapsto x/(1 + 2^k x)$, where $k$ is the number of trailing zeros removed from the difference $|u - v|$. Let $S_{n+1}(x) = \sum_{k=2}^{\infty} 2^{-k} G_1(1/(1 + 2^k x))$ represent the contribution of all possible shifts, weighted by their probabilities $2^{-k}$. Using the method of Mellin transforms for harmonic sums, one can express the distribution function $G_1(x)$ as the solution to the functional equation
$G_1(x) = \sum_{k=1}^{\infty} 2^{-k} G_1\Bigl(\frac{1}{1 + 2^k x}\Bigr).$
This captures the expected distribution of the ratio after a single cycle, since each branch corresponds to subtracting the smaller number from the larger and then shifting out $k$ trailing zeros, with probability $2^{-k}$. The Mellin-transform technique then converts this harmonic sum into an analytic expression suitable for asymptotic or exact evaluation, giving the precise law of $\min(u, v)/\max(u, v)$ after one cycle.
$\boxed{G_1(x) = \sum_{k=1}^{\infty} 2^{-k} G_1\Bigl(\frac{1}{1 + 2^k x}\Bigr)}$