利用比特打包优化C中的矩阵乘法

Question

我目前正在尝试编写一种算法，使用比特打包优化GF（2）上的矩阵乘法。两个矩阵A和B都按列主要顺序提供，所以我首先将A复制到行主要顺序，然后将值打包成8位整数并使用奇偶校验检查来加速操作。我需要能够测试高达2048x2048的平方矩阵，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);
}

Answer 1

我有两个评论：

• 你可能应该将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。您的代码可能因巧合而适用于小矩阵，因为目标矩阵C恰好是此大小的全零。
• 将分配大小计算为malloc(n * n * sizeof(int));是有风险的，因为如果n * n小于int n，int可能会溢出size_t。鉴于你使用的大小，这可能不是一个问题，但总是使用sizeof作为第一个操作数强制转换为以下的size_t是一个好主意： int *A_COPY = malloc(sizeof(*A_COPY) * n * n);