Prime Numbers

464 Chapter 9 FAST ALGORITHMS FOR LARGE-INTEGER ARITHMETIC
can be computed with a shift into oblivion of the low-order zeros; note also for
theoretical convenience we may as well take v2(0) = ∞.)
1. [2’s power in gcd]
β =min{v2(x),v2(y)}; // 2 β gcd(x, y)
x = x/2 v2(x) ;
y = y/2 v2(y) ;
2. [Binary gcd]
while(x = y) {
(x, y) =(min{x, y}, |y − x|/2 v2(|y−x|) );
}
return 2 β x;
In actual practice on most machinery, the binary algorithm is often faster
than the Euclid algorithm; and as we have said, Lehmer’s enhancements may
also be applied to this binary scheme.
But there are other, more modern, enhancements; in fact, gcd enhancements
seem to keep appearing in the literature. There is a “k-ary” method
due to Sorenson, in which reductions involving k>2 as a modulus are performed.
There is also a newer extension of the Sorenson method that is claimed
to be, on a typical modern machine that possesses hardware multiply, more
than 5 times faster than the binary gcd we just displayed [Weber 1995]. The
Weber method is rather intricate, involving several special functions for nonstandard
modular reduction, yet the method should be considered seriously
in any project for which the gcd happens to be a bottleneck. Most recently,
[Weber et al. 2005] introduced a new modular GCD algorithm that could be
an ideal choice for certain ranges of operands.
It is of interest that the Sorenson method has variants for which the
complexity of the gcd is O(n 2 / ln n) as opposed to the Euclidean O(n 2 )
[Sorenson 1994]. In addition, the Sorenson method has an extended form for
obtaining not just gcd but inverse as well.
One wonders whether this efficient binary technique can be extended in the
way that the classical Euclid algorithm can. Indeed, there is also an extended
binary gcd that provides inverses. [Knuth 1981] attributes the method to
M. Penk:
Algorithm 9.4.3 (Binary gcd, extended for inverses). For positive integers
x, y, this algorithm returns an integer triple (a, b, g) such that ax + by = g =
gcd(x, y). We assume the binary representations of x, y, and use the exponent β
as in Algorithm 9.4.2.
1. [Initialize]
x = x/2 β ; y = y/2 β ;
(a, b, h) =(1, 0,x);
(v1,v2,v3) =(y, 1 − x, y);
if(x even) (t1,t2,t3) =(1, 0,x);
else {
(t1,t2,t3) =(0, −1, −y);