用 Bit Packing 优化 C 中的矩阵乘法

Optimizing Matrix multiplication in C with Bit Packing

我目前正在尝试编写一种算法,用于使用位打包优化 GF(2) 上的矩阵乘法。矩阵 AB 均按列主顺序提供,因此我首先将 A 复制为行主顺序,然后将值打包为 8 位整数并使用奇偶校验来加快速度上操作。我需要能够测试高达 2048x2048 的方阵,但是,我当前的实现提供了高达 24x24 的正确答案,然后无法计算出正确的结果。任何帮助,将不胜感激。

//Method which packs an array of integers into 8 bits
uint8_t pack(int *toPack) {
    int i;
    uint8_t A;
    A = 0;
    for (i = 0; i < 8; i++) {
        A = (A << 1) | (uint8_t)toPack[i];
    }
    return A;
}

//Method for doing matrix multiplication over GF(2)
void matmul_optimized(int n, int *A, int *B, int *C) {
    int i, j, k;
    //Copying values of A into a row major order matrix.
    int *A_COPY = malloc(n * n * sizeof(int));
    int copy_index = 0;
    for (i = 0; i < n; i++) {
        for (j = 0; j < n; j++) {
            A_COPY[copy_index] = A[i + j * n];
            copy_index++;
        }
    }
    //Size of the data data type integers will be packed into
    const int portion_size = 8;
    int portions = n / portion_size;

    //Pointer space reserved to store packed integers in row major order
    uint8_t *compressedA = malloc(n * portions * sizeof(uint8_t));
    uint8_t *compressedB = malloc(n * portions * sizeof(uint8_t));

    int a[portion_size];
    int b[portion_size];
    for (i = 0; i < n; i++) {
        for (j = 0; j < portions; j++) {
            for (k = 0; k < portion_size; k++) {
                a[k] = A_COPY[i * n + j * portion_size + k];
                b[k] = B[i * n + j * portion_size + k];
            }
            compressedA[i * n + j] = pack(a);
            compressedB[i * n + j] = pack(b);
        }
    }

    //Calculating final matrix using parity checking and XOR on A and B
    int cij;
    for (i = 0; i < n; ++i) {
        for (j = 0; j < n; ++j) {
            int cIndex = i + j * n;
            cij = C[cIndex];
            for (k = 0; k < portions; ++k) {
                uint8_t temp = compressedA[k + i * n] & compressedB[k + j * n];
                temp ^= temp >> 4;
                temp ^= temp >> 2;
                temp ^= temp >> 1;
                uint8_t parity = temp & (uint8_t)1;
                cij = cij ^ parity;
            }
            C[cIndex] = cij;
        }
    }
    free(compressedA);
    free(compressedB);
    free(A_COPY);
}

我有两点意见:

  • 您可能应该将 cij 初始化为 0 而不是 cij = C[cIndex];。更新目标矩阵而不是存储 A * B 的结果似乎是不正确的。您的代码可能巧合地适用于小矩阵,因为目标矩阵 C 恰好是这个大小的全零。

  • 将分配大小计算为 malloc(n * n * sizeof(int)); 是有风险的,因为如果 int 小于 [=20],n * n 可能会溢出 int n =].考虑到您使用的大小,这可能不是问题,但始终使用 sizeof 作为第一个操作数以强制转换为以下操作数的 size_t 是个好主意:

    int *A_COPY = malloc(sizeof(*A_COPY) * n * n);