crc32_combine() 矩阵技巧的逆运算是什么？

Question

zlib 的 crc32_combine() 采用 crcA、crcB 和 lengthB 来计算 crcAB。

# returns crcAB
crc32_combine(crcA, crcB, lenB)

使用 Mark Adler 的精彩帖子 here and here I was able to produce crc32_trim_trailing.pl 中的概念，它采用 crcAB、crcB 和 lengthB 来计算 crcA（我用它来剥离已知长度和值的填充）。

# prints crcA
perl crc32_trim_trailing.pl $crcAB $crcB $lenB

不幸的是，这使用了所描述的慢速方法的原理，其中必须一次剥离每个空字节。它很慢，但是是一个很好的概念证明。

我一直在努力制作 crc32_trim_trailing 的快速版本，它利用了 Mark 的帖子中描述的矩阵技巧，并为 zlib 的 crc32_combine() 中的组合用例实现。

这是我尝试 crc32_trim_trailing.c。

/* crc32_trim_trailing.c
  This code is borrows heavily from crc32.c from zlib version 1.2.8, but has
  been altered.
*/

#include <stdio.h>

#define GF2_DIM 32      /* dimension of GF(2) vectors (length of CRC) */

/* ========================================================================= */
unsigned long gf2_matrix_times(mat, vec)
    unsigned long *mat;
    unsigned long vec;
{
    unsigned long sum;

    sum = 0;
    while (vec) {
        if (vec & 1)
            sum ^= *mat;
        vec >>= 1;
        mat++;
    }
    return sum;
}

/* ========================================================================= */
void gf2_matrix_square(square, mat)
    unsigned long *square;
    unsigned long *mat;
{
    int n;

    for (n = 0; n < GF2_DIM; n++)
        square[n] = gf2_matrix_times(mat, mat[n]);
}

/* ========================================================================= */
int main(int argc, char *argv[])
{
    unsigned long crc1;
    unsigned long crc2;
    int len2;

    sscanf(argv[1], "%lx", &crc1);
    sscanf(argv[2], "%lx", &crc2);
    sscanf(argv[3],  "%d", &len2);

    int n;
    unsigned long row;
    unsigned long even[GF2_DIM];    /* even-power-of-two zeros operator */
    unsigned long odd[GF2_DIM];     /* odd-power-of-two zeros operator */

    /* degenerate case (also disallow negative lengths) */
    if (len2 <= 0)
        return crc1;

    /* get crcA0 */
    crc1 ^= crc2;

    /* put operator for one zero bit in odd */
    odd[0] = 0x82608edbUL;          /* used sage math to get inverse matrix polynomial */
    row = 1;
    for (n = 1; n < GF2_DIM; n++) {
        odd[n] = row;
        row <<= 1;
    }

    /* put operator for two zero bits in even */
    gf2_matrix_square(even, odd);

    /* put operator for four zero bits in odd */
    gf2_matrix_square(odd, even);

    /* apply len2 zeros to crc1 (first square will put the operator for one
       zero byte, eight zero bits, in even) */
    do {
        /* apply zeros operator for this bit of len2 */
        gf2_matrix_square(even, odd);
        if (len2 & 1)
            crc1 = gf2_matrix_times(even, crc1);
        len2 >>= 1;

        /* if no more bits set, then done */
        if (len2 == 0)
            break;

        /* another iteration of the loop with odd and even swapped */
        gf2_matrix_square(odd, even);
        if (len2 & 1)
            crc1 = gf2_matrix_times(odd, crc1);
        len2 >>= 1;

        /* if no more bits set, then done */
    } while (len2 != 0);

    printf("\nCRC: %lx\n", crc1);

    return 0;
}

我将 xor 移动到矩阵乘法之前。这似乎没有问题，并通过对 crcAB 和 crcB 进行异或运算为我们提供了 crcA0。

接下来，使用 sage math，我找到了 crc32_combine() 中使用的初始矩阵的逆矩阵。

运行这些矩阵中的每一个通过 3 个正方形得到矩阵 crc32_combine() 用于添加 1 个空字节 (matrixA) 和它的逆 (matrixB)。

使用 sage math 我确认了以下内容。

矩阵A * 矩阵B = 恒等式
crc * identity = crc
crc * matrixA * matrixB = crc

代码：

M = MatrixSpace(GF(2),32,32)
A = M([0,1,1,1,0,1,1,1,0,0,0,0,0,1,1,1,0,0,1,1,0,0,0,0,1,0,0,1,0,1,1,0,
1,1,1,0,1,1,1,0,0,0,0,0,1,1,1,0,0,1,1,0,0,0,0,1,0,0,1,0,1,1,0,0,
0,0,0,0,0,1,1,1,0,1,1,0,1,1,0,1,1,1,0,0,0,1,0,0,0,0,0,1,1,0,0,1,
0,0,0,0,1,1,1,0,1,1,0,1,1,0,1,1,1,0,0,0,1,0,0,0,0,0,1,1,0,0,1,0,
0,0,0,1,1,1,0,1,1,0,1,1,0,1,1,1,0,0,0,1,0,0,0,0,0,1,1,0,0,1,0,0,
0,0,1,1,1,0,1,1,0,1,1,0,1,1,1,0,0,0,1,0,0,0,0,0,1,1,0,0,1,0,0,0,
0,1,1,1,0,1,1,0,1,1,0,1,1,1,0,0,0,1,0,0,0,0,0,1,1,0,0,1,0,0,0,0,
1,1,1,0,1,1,0,1,1,0,1,1,1,0,0,0,1,0,0,0,0,0,1,1,0,0,1,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,
0,0,0,0,0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0])

B = A^-1

I = A*B

print "matrixA"
print A.str()
print "matrixB"
print B.str()
print "identity"
print I.str()

N = MatrixSpace(GF(2),1,32)
THIS=N([1,1,1,1,1,1,1,0,1,1,1,0,1,1,1,0,1,0,0,0,0,0,1,0,0,0,1,0,0,1,1,1])

print "'this' crc * identity"
print THIS * I
print "'this' crc * maxtrixA"
print THIS * A
print "'this' crc * maxtrixA * matrixB"
print THIS * A * B

输出：

matrixA
[0 1 1 1 0 1 1 1 0 0 0 0 0 1 1 1 0 0 1 1 0 0 0 0 1 0 0 1 0 1 1 0]
[1 1 1 0 1 1 1 0 0 0 0 0 1 1 1 0 0 1 1 0 0 0 0 1 0 0 1 0 1 1 0 0]
[0 0 0 0 0 1 1 1 0 1 1 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1]
[0 0 0 0 1 1 1 0 1 1 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 0]
[0 0 0 1 1 1 0 1 1 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 0 0]
[0 0 1 1 1 0 1 1 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 0 0 0]
[0 1 1 1 0 1 1 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 0 0 0 0]
[1 1 1 0 1 1 0 1 1 0 1 1 1 0 0 0 1 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
matrixB
[1 0 1 0 1 0 0 1 1 1 0 1 0 0 1 1 1 1 1 0 0 1 1 0 1 0 1 0 0 1 1 0]
[0 1 0 1 0 1 1 1 0 1 1 0 0 1 1 0 1 1 0 1 0 0 0 0 1 1 1 1 1 0 1 1]
[1 0 1 0 1 1 1 0 1 1 0 0 1 1 0 1 1 0 1 0 0 0 0 1 1 1 1 1 0 1 1 0]
[0 1 0 1 1 0 0 1 0 1 0 1 1 0 1 0 0 1 0 1 1 1 1 0 0 1 0 1 1 0 1 1]
[1 0 1 1 0 0 1 0 1 0 1 1 0 1 0 0 1 0 1 1 1 1 0 0 1 0 1 1 0 1 1 0]
[0 1 1 0 0 0 0 1 1 0 1 0 1 0 0 0 0 1 1 0 0 1 0 0 1 1 0 1 1 0 1 1]
[1 1 0 0 0 0 1 1 0 1 0 1 0 0 0 0 1 1 0 0 1 0 0 1 1 0 1 1 0 1 1 0]
[1 0 0 0 0 0 1 0 0 1 1 0 0 0 0 0 1 0 0 0 1 1 1 0 1 1 0 1 1 0 1 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
identity
[1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1]
'this' crc * identity
[1 1 1 1 1 1 1 0 1 1 1 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 1 1 1]
'this' crc * maxtrixA
[1 1 0 0 0 0 0 0 0 1 0 1 1 1 1 0 0 0 1 0 1 1 0 1 1 1 0 1 1 0 1 0]
'this' crc * maxtrixA * matrixB
[1 1 1 1 1 1 1 0 1 1 1 0 1 1 1 0 1 0 0 0 0 0 1 0 0 0 1 0 0 1 1 1]

我使用 crc 和单位矩阵测试了 gf2_matrix_times()，正如预期的那样，crc 没有变化。

既然gf2_matrix_times(crc, matrixA)可以用来给crc加1个空字节，我本来希望gf2_matrix_times(crc, matrixB)可以用来去掉1个空字节。但是，这似乎不是开箱即用的。

此外，当 lengthB 为 1 时，sage 数学中的 crc * matrixA 产生的结果 (0xc05e2dda) 与 crc32_combine() 中的 crcA0 (0xa5f45be9) 不同。

为什么 sage math 和 gf2_matrix_times() 之间的 GF(2) 矩阵乘法存在差异？为什么当matrixA和matrixB逆时gf2_matrix_times(crc, matrixB)不反转gf2_matrix_times(crc, matrixA)？

Answer 1

我们将从查看标准 CRC-32 的简单按位实现开始（作为 CRC 的独立定义，此例程 returns 初始 CRC，即 CRC空字符串，当 data 是 NULL):

#include <stddef.h>
#include <stdint.h>

#define POLY 0xedb88320

uint32_t crc32(uint32_t crc, void const *data, size_t len) {
    if (data == NULL)
        return 0;
    crc = ~crc;
    while (len--) {
        crc ^= *(unsigned char const *)data++;
        for (int k = 0; k < 8; k++)
            crc = crc & 1 ? (crc >> 1) ^ POLY : crc >> 1;
    }
    crc = ~crc;
    return crc;
}

我们可以简化将 n 个零应用于 CRC：

uint32_t crc32_zeros(uint32_t crc, size_t n) {
    crc = ~crc;
    while (n--)
        for (int k = 0; k < 8; k++)
            crc = crc & 1 ? (crc >> 1) ^ POLY : crc >> 1;
    crc = ~crc;
    return crc;
}

现在让我们仔细看看单零位在CRC中的应用：

crc = crc & 1 ? (crc >> 1) ^ POLY : crc >> 1;

应用位时可以采用两条路径。在最后一个操作中，要么多项式与 CRC 异或，要么不是。如果我们想扭转这一局面，我们想知道它的走向。

我们可以通过查看结果的高位来判断。我们可以看到，如果多项式不是异或，那么高位一定是0。但是如果是异或呢？那样的话，结果的高位就是POLY的高位。我们可以看到高位是1。所以我们只看结果的高位就可以知道。事实上，任何有效的 CRC 多项式都必须是这种情况，因为所有的 x⁰ 项的系数都是 1。（该项在该反射多项式的高位。）

通过检查，我们可以轻松地反转该操作，其中 crc 进入是应用 0 位后的最终 CRC，而 crc 出来是应用 0 之前的 CRC位：

crc = crc & 0x80000000 ? ((crc ^ POLY) << 1) + 1 : crc << 1;

这将采用最终 CRC 并反转计算单个 0 位的 CRC 的操作。请注意，对于这种情况，我们必须插入会导致异或的低 1 位。

我们可以分解出 POLY 得到：

crc = crc & 0x80000000 ? (crc << 1) ^ ((POLY << 1) + 1) : crc << 1;

这与附加一个0位到非反射CRC的操作完全相同(POLY << 1) + 1，这只是 POLY 向左旋转一位。

然后我们可以编写一个函数从标准 CRC-32 中删除 n 零字节：

#define UNPOLY ((POLY << 1) + 1)

uint32_t crc32_remove_zeros(uint32_t crc, size_t n) {
    crc = ~crc;
    while (n--)
        for (int k = 0; k < 8; k++)
            crc = crc & 0x80000000 ? (crc << 1) ^ UNPOLY : crc << 1;
    crc = ~crc;
    return crc;
}

现在我们可以使用与 zlib 中相同的方法，但使用非反射 CRC，编写一个函数来从 O 中的 CRC-32 中删除 n 个零(log(n)) 时间。我们不需要反转任何矩阵，因为我们已经反转了原始操作。

剩下的留作reader的练习。

crc32_combine() 矩阵技巧的逆运算是什么？

What is the inverse of crc32_combine()'s matrix trick?

checksum

zlib

crc

matrix