扩展右按位算法 [英] Expand Right Bitwise Algorithm

问题描述

最初，该帖子要求进行逆山羊和山羊操作，但是我意识到这远远超出了我的实际需要，所以我编辑了标题，因为我只需要一个

Originally this post requested an inverse sheep-and-goats operation, but I realized that it was more than I really needed, so I edited the title, because I only need an expand-right algorithm, which is simpler. The example that I described below is still relevant.

原始帖子:

我正在尝试找出如何进行逆向山羊和山羊的操作，或者甚至更好地执行右向翻转的操作.

I'm trying to figure out how to do either an inverse sheep-and-goats operation or, even better, an expand-right-flip.

根据 Hacker's Delight ，绵羊和山羊的行动可以用以下方式表示:

According to Hacker's Delight, a sheeps-and-goats operation can be represented by:

SAG(x, m) = compress_left(x, m) | compress(x, ~m)

根据此站点，可以通过以下方式找到反函数:

According to this site, the inverse can be found by:

INV_SAG(x, m, sw) = expand_left(x, ~m, sw) | expand_right(x, m, sw)

但是，我找不到关于expand_left和expand_right函数的任何代码.当然，它们是compress的逆函数，但是compress本身很难理解.

However, I can't find any code for the expand_left and expand_right functions. They are, of course, the inverse functions for compress, but compress is kind of hard to understand in itself.

示例:

为了更好地解释我在寻找什么，请考虑一组8位，例如:

To better explain what I'm looking for, consider a set of 8 bits like:

0000abcd

变量a，b，c和d可以为1或0.此外，还有一个掩码，用于重新定位位.因此，例如，如果掩码为01100101，则结果位将按以下方式重新定位:

The variables a, b, c and d may be either ones or zeros. In addition, there is a mask which repositions the bits. So for example, if the mask were 01100101, the resulting bits would be repositioned as follows:

0ab00c0d

这可以通过逆山羊和山羊操作来完成.但是，根据上述网站的本节，有一种更有效的方法他称为右上翻.在他的网站上，我无法弄清楚该怎么做.

This can be done with an inverse sheeps-and-goats operation. However, according to this section of the site mentioned above, there is a more efficient way which he refers to as the expand-right-flip. Looking at his site, I was unable to figure out how that can be done.

推荐答案

这是Hacker's Delight的expand_right，只说了expand，但它是right版本.

Here's the expand_right from Hacker's Delight, it just says expand but it's the right version.

unsigned expand(unsigned x, unsigned m) {
    unsigned m0, mk, mp, mv, t;
    unsigned array[5];
    int i;
    m0 = m;        // Save original mask.
    mk = ~m << 1;  // We will count 0's to right.
    for (i = 0; i < 5; i++) {
        mp = mk ^ (mk << 1);             // Parallel suffix.
        mp = mp ^ (mp << 2);
        mp = mp ^ (mp << 4);
        mp = mp ^ (mp << 8);
        mp = mp ^ (mp << 16);
        mv = mp & m;                     // Bits to move.
        array[i] = mv;
        m = (m ^ mv) | (mv >> (1 << i)); // Compress m.
        mk = mk & ~mp;
    }
    for (i = 4; i >= 0; i--) {
        mv = array[i];
        t = x << (1 << i);
        x = (x & ~mv) | (t & mv);
    }
    return x & m0;     // Clear out extraneous bits.
}

您可以使用expand_left(x, m) == expand_right(x >> (32 - popcnt(m)), m)制作left版本，但这可能不是最好的方法.

You can use expand_left(x, m) == expand_right(x >> (32 - popcnt(m)), m) to make the left version, but that's probably not the best way.

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