TAOCP 4.5.2 Exercise 15

We are asked to determine the terminal values of $v_1$ and $v_2$ in **Algorithm X** from Section 4.

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Section 4.5.2: The Greatest Common Divisor

Exercise 15. [**] [M21] What are the values of $v_1$ and $v_2$ when Algorithm X terminates?

Verified: yes
Solve time: 2m18s

Solution

We are asked to determine the terminal values of $v_1$ and $v_2$ in Algorithm X from Section 4.5.2. Algorithm X is Knuth's notation for the binary GCD algorithm (Algorithm B) in this section. We summarize its essential properties before deducing the final values.

Algorithm X maintains two nonnegative integers $v_1$ and $v_2$, initially set to the inputs $u$ and $v$ (possibly after factoring out powers of two). The algorithm repeatedly applies the following steps:

  1. If $v_1$ is even, divide it by 2; if $v_2$ is even, divide it by 2.
  2. If both are odd, subtract the smaller from the larger.
  3. Repeat until one of the numbers becomes zero.

At termination, let $v_1 = x$ and $v_2 = y$. By the properties of the binary GCD algorithm:

  • The algorithm preserves the greatest common divisor: $\gcd(v_1, v_2) = \gcd(u, v)$ throughout.
  • Termination occurs when one of the values becomes zero.

Therefore, exactly one of $v_1$ or $v_2$ is zero at the end. The other value is equal to $\gcd(u, v)$, possibly multiplied by some power of 2 that was factored out at the beginning (depending on the variant of Algorithm X used in §4.5.2). In Knuth's notation (Algorithm B, Step B5):

$$ v_1 = \gcd(u, v), \quad v_2 = 0 $$

or symmetrically, if the roles are reversed:

$$ v_1 = 0, \quad v_2 = \gcd(u, v) $$

No intermediate values remain, because all reductions preserve the gcd and eventually reduce the other number to zero.

Thus, the terminal values are:

$$ (v_1, v_2) = (\gcd(u, v), 0) \quad \text{or} \quad (0, \gcd(u, v)) $$

depending on which variable becomes zero first. This fully answers the exercise.

Justification:

  • The subtraction step reduces the larger odd number by the smaller odd number, preserving the gcd.
  • Dividing by 2 preserves the gcd modulo powers of 2, which are factored out at initialization.
  • The loop terminates when one of the numbers is zero, leaving the other equal to $\gcd(u, v)$.

This conclusion is independent of the specific initial values $u$ and $v$.

Answer:

$$ \boxed{(v_1, v_2) = (\gcd(u, v), 0) \text{ or } (0, \gcd(u, v))} $$