matrix - crc32_combine() 的矩阵技巧的逆是什么？

Question

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

# returns crcAB
crc32_combine(crcA, crcB, lenB)

使用 Mark Adler 精彩帖子中的概念 here和 here我能够生产 crc32_trim_trailing.pl它需要 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则，其中每个空字节必须一次剥离一个。它很慢，但很好地证明了概念。

我一直在努力制作 crc32_trim_trailing 的快速版本，它利用 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。

这是我在 crc32_trim_trailing.c 上的尝试。

/* 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;
}

我将异或移到矩阵乘法之前。这似乎可以正常工作，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 我能够找到 crc32_combine() 中使用的初始矩阵的逆矩阵。

通过 3 个正方形运行这些矩阵中的每一个会导致矩阵 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 = 身份

crc * 身份 = crc

crc * 矩阵A * 矩阵B = 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。

由于 gf2_matrix_times(crc, matrixA) 可用于向 crc 添加 1 个空字节，我曾希望 gf2_matrix_times(crc, matrixB) 可用于删除 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 math 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？
当 matrixA 和 matrixB 是逆时，为什么 gf2_matrix_times(crc, matrixB) 不反转 gf2_matrix_times(crc, matrixA) ？

Answer 1

我们将首先查看标准 CRC-32 的简单按位实现(作为 CRC 的自包含定义，此例程返回初始 CRC，即空字符串的 CRC，当 data 为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;
}

我们可以简化申请 n CRC 的零:

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个零位在CRC中的应用:

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

应用该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。如果我们想扭转这种情况，我们想知道它往哪个方向发展。

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？在这种情况下，结果的高位是 POLY 的高位。 .我们可以看到高位是 1。所以我们可以通过查看结果的高位来判断。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。 (该项位于此反射多项式的高位。)

通过检查，我们可以轻松地反转该操作，这里是 crc进入的是应用 0 位后的最终 CRC，和 crc出来的是应用 0 位之前的 CRC:

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

这将采用最终的 CRC 并反转在单个 0 位上计算 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要得到:

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

这与使用多项式 (POLY << 1) + 1 将 0 位附加到非反射 CRC 的操作完全相同。 ，这只是 POLY向左旋转一位。

然后我们可以编写一个函数来删除 n来自标准 CRC-32 的零字节:

#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 相同的方法，但使用非反射 CRC，编写一个函数，以在 O(log(n)) 时间内从 CRC-32 中删除 n 个零。我们不需要反转任何矩阵，因为我们已经反转了原始操作。

其余部分留给读者作为练习。