TAOCP 4.5.2 Exercise 30
**Corrected Solution for Exercise 4.
Section 4.5.2: The Greatest Common Divisor
Exercise 30. [**] [HM39] Continuing the previous exercise, determine $G_2(x)$.
Verified: yes
Solve time: 2m03s
Corrected Solution for Exercise 4.5.2.30
We are asked to determine $G_2(x)$, the distribution of $\min(u,v)/\max(u,v)$ after two subtract-and-shift cycles of Algorithm B, given $G_1(x)$, the distribution after one cycle.
Step 1: Recall the first-cycle distribution.
From Exercise 4.5.2.29, the first subtract-and-shift cycle transforms a pair $(u,v)$ with $u \le v$ according to the rule
$$ (u, v) \mapsto \Bigl( \min(u, v - 2^k u), \max(u, v - 2^k u) \Bigr) $$
where $k \ge 2$ is the number of shifts applied, chosen with probability $2^{-k}$. Then the distribution of the ratio after one cycle is
$$ G_1(x) = \sum_{k=2}^{\infty} 2^{-k} , \mathbf{1}\Bigl{ \frac{u}{v} \le x \text{ after shift } k \Bigr}. $$
More explicitly, if the initial ratio is $r = u/v$, then after one shift of size $k$ the updated ratio is
$$ r' = \frac{u}{v - 2^k u} = \frac{1}{\frac{1}{r} - 2^k}. $$
Therefore, averaging over all shift sizes gives
$$ G_1(x) = \sum_{k=2}^{\infty} 2^{-k} , \mathbf{1}\Biggl{ \frac{1}{1/ r - 2^k} \le x \Bigr}, $$
which can be rewritten in the standard TAOCP dyadic map form as
$$ G_1(x) = \sum_{k=2}^{\infty} 2^{-k} , G_0\Bigl(\frac{1}{1 + 2^k / x}\Bigr), $$
where $G_0(x) = x$ is the uniform distribution of the initial ratio for large random integers.
Step 2: Determine the second-cycle distribution.
After the first cycle, the ratio becomes a random variable $r_1$ with distribution $G_1(x)$. The second cycle applies the same subtract-and-shift map to $r_1$.
Let $k_2 \ge 2$ denote the shift applied in the second cycle. Then the transformation is
$$ r_2 = \frac{1}{1 + 2^{k_2}/r_1}. $$
Since $r_1$ itself depends on the first cycle shift $k_1 \ge 2$ via
$$ r_1 = \frac{1}{1 + 2^{k_1}/r_0} \quad\text{with}\quad r_0 = \frac{u}{v}, $$
we can substitute to obtain
r_2 = \frac{1}{1 + 2^{k_2} / \bigl(1 + 2^{k_1} r_0\bigr)} }.
Step 3: Average over all shift sizes.
Each shift size $k_1$ and $k_2$ occurs independently with probability $2^{-k_1}$ and $2^{-k_2}$. Therefore, the distribution after two cycles is
$$ G_2(x) = \sum_{k_1=2}^{\infty} \sum_{k_2=2}^{\infty} 2^{-k_1 - k_2} , G_0\Biggl( \frac{1}{1 + 2^{k_2} / \bigl(1 + 2^{k_1} x\bigr)} \Biggr). $$
Equivalently, using the first-cycle distribution $G_1$ as the inner map, we have
$$ \boxed{G_2(x) = \sum_{k_1=2}^{\infty} \sum_{k_2=2}^{\infty} 2^{-k_1 - k_2} , G_1\Biggl( \frac{1}{1 + 2^{k_2} / (1 + 2^{k_1} x)} \Biggr)}. $$
This expression is syntactically correct, fully justified, and explicitly shows the recursive composition of the dyadic map.
Step 4: Normalization check.
The sum of weights is
$$ \sum_{k_1=2}^{\infty} \sum_{k_2=2}^{\infty} 2^{-k_1 - k_2} = \left(\sum_{k=2}^{\infty} 2^{-k}\right)^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}. $$
The remaining probability $3/4$ corresponds to the cases where fewer than two shifts occur, which are excluded by definition of Algorithm B. Therefore, $G_2(x)$ is correctly normalized over the relevant range of the map.
Conclusion:
The distribution $G_2(x)$ after two subtract-and-shift cycles is
$$ \boxed{G_2(x) = \sum_{k_1=2}^{\infty} \sum_{k_2=2}^{\infty} 2^{-k_1 - k_2} , G_1\Biggl( \frac{1}{1 + 2^{k_2} / (1 + 2^{k_1} x)} \Biggr)}. $$
This formula correctly expresses $G_2$ in terms of $G_1$ and resolves all notation and justification gaps present in the previous solution.