如何使用__builtin_ctz加速二进制GCD算法?

问题描述

clang和GCC具有int __builtin_ctz(unsigned)函数。这将以整数形式计算尾随零。 Wikipedia article on this family of functions提到可以使用__builtin_ctz加速二进制GCD算法,但我不知道如何。

二进制GCD的sample implementation看起来像这样:

unsigned int gcd(unsigned int u,unsigned int v)
{
    // simple cases (termination)
    if (u == v)
        return u;

    if (u == 0)
        return v;

    if (v == 0)
        return u;

    // look for factors of 2
    if (~u & 1) // u is even
        if (v & 1) // v is odd
            return gcd(u >> 1,v);
        else // both u and v are even
            return gcd(u >> 1,v >> 1) << 1;

    if (~v & 1) // u is odd,v is even
        return gcd(u,v >> 1);

    // reduce larger argument
    if (u > v)
        return gcd(u - v,v);

    return gcd(v - u,u);
}

我怀疑我可以如下使用__builtin_ctz

constexpr unsigned int gcd(unsigned int u,unsigned int v)
{
    // simplified first three ifs
    if (u == v || u == 0 || v == 0)
        return u | v;

    unsigned ushift = __builtin_ctz(u);
    u >>= ushift;

    unsigned vshift = __builtin_ctz(v);
    v >>= vshift;

    // Note sure if max is the right approach here.
    // In the if-else block you can see both arguments being rshifted
    // and the result being leftshifted only once.
    // I expected to recreate this behavior using max.
    unsigned maxshift = std::max(ushift,vshift);

    // The only case which was not handled in the if-else block before was
    // the odd/odd case.
    // We can detect this case using the maximum shift.
    if (maxshift != 0) {
        return gcd(u,v) << maxshift;
    }

    return (u > v) ? gcd(u - v,v) : gcd(v - u,u);
}

int main() {
    constexpr unsigned result = gcd(5,3);
    return result;
}

不幸的是,这还行不通。程序结果为4,应为1。那么,我在做什么错呢?如何在这里正确使用__builtin_ctzSee my code so far on GodBolt

解决方法

这是我从comments开始的迭代实现:

尽管尾部递归算法通常很优雅,但是迭代实现在实践中几乎总是更快。 (现代编译器实际上可以在非常简单的情况下执行此转换。)

unsigned ugcd (unsigned u,unsigned v)
{
    unsigned t = u | v;

    if (u == 0 || v == 0)
        return t; /* return (v) or (u),resp. */

    int g = __builtin_ctz(t);

    while (u != 0)
    {
        u >>= __builtin_ctz(u);
        v >>= __builtin_ctz(v);

        if (u >= v)
            u = (u - v) / 2;
        else
            v = (v - u) / 2;
    }

    return (v << g); /* scale by common factor. */
}

如上所述,|u - v| / 2步骤通常实现为非常有效的无条件右移,例如 shr r32 ,以除以(2)- (u)(v)都是奇数,因此|u - v|必须是偶数。

这不是严格必要的,因为“整理”步骤:u >>= __builtin_clz(u);将在下一次迭代中有效地执行此操作。

假设(u)(v)具有“随机”比特分布,则(n)通过 tzcnt 尾随零的概率为〜(1/(2^n))。该说明是对 bsf (IIRC的Haswell之前的__builtin_clz的实现)的改进。

,

感谢有帮助的评论员,我发现了一个关键错误:我应该使用min而不是max

这是最终解决方案:

#include <algorithm>

constexpr unsigned gcd(unsigned u,unsigned v)
{
    if (u == v || u == 0 || v == 0)
        return u | v;

    // effectively compute min(ctz(u),ctz(v))
    unsigned shift = __builtin_ctz(u | v);
    u >>= __builtin_ctz(u);
    v >>= __builtin_ctz(v);

    const auto &[min,max] = std::minmax(u,v);

    return gcd(max - min,min) << shift;
}

int main() {
    constexpr unsigned g = gcd(25,15); // g = 5
    return g;
}

此解决方案也很不错,nearly branch-free compile output

以下是到目前为止所有答案中的some benchmark results(我们实际上击败了std::gcd): benchmark