问3×3和N×N方阵的乘法
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Code Review用户

提问于 2019-01-17 02:08:56

回答 1查看 567关注 0票数 0

我最近用C++编写了一个矩阵模块。

在开发过程中，我引用了一些源代码，发现了一个问题。

例如，矩阵乘法：

这种方法适用于所有N×N矩阵：

void multiplyMatrix(const float32* a, const float32* b, float32* dst,
                    int32 aColumns, int32 bColumns, int32 dstColumns, int32 dstRows) {

    for (int32 i = 0; i < dstRows; i++) {

        for (int32 j = 0; j < dstColumns; j++) 
            dst[i * dstColumns + j] = dotMatrix(a, b, aColumns, bColumns, j, i);

    }
}

float32 dotMatrix(const float32* a, const float32* b, 
                         int32 aColumns, int32 bColumns, 
                         int32 column, int32 row) {

    float32 result = 0.0f;

    int32 index = aColumns * row;
    for (int32 i = 0; i < aColumns; i++) {
        result += a[index++] * b[column];
        column += bColumns;
    }

    return result;
}

接下来，我编写了一个3x3矩阵类。

class Matrix3x3
{
public:
float32 m11, m12, m13, 
        m21, m22, m23, 
        m31, m32, m33;

float32 element[9];

void multiply(float32 ma11, float32 ma12, float32 ma13,
              float32 ma21, float32 ma22, float32 ma23,
              float32 ma31, float32 ma32, float32 ma33) {

    float32 temp1 = m11 * ma11 + m21 * ma12 + m31 * ma13;
    float32 temp2 = m12 * ma11 + m22 * ma12 + m32 * ma13;

    m13 = m13 * ma11 + m23 * ma12 + m33 * ma13;
    m11 = temp1;
    m12 = temp2;

    temp1 = m11 * ma21 + m21 * ma22 + m31 * ma23;
    temp2 = m12 * ma21 + m22 * ma22 + m32 * ma23;
    m23   = m13 * ma21 + m23 * ma22 + m33 * ma23;
    m21   = temp1;
    m22   = temp2;

    temp1 = m11 * ma31 + m21 * ma32 + m31 * ma33;
    temp2 = m12 * ma31 + m22 * ma32 + m32 * ma33;
    m31   = m13 * ma31 + m23 * ma32 + m33 * ma33;
    m32   = temp1;
    m33   = temp2;
}
}

显然，第一个很方便。

接下来，我测试了计算所需的时间：

        float32 e1[9];
        e1[0] = 2.1018f;   e1[1] = -1.81754f; e1[2] = 1.2541f;
        e1[3] = 0.54194f;  e1[4] = 2.75391f;  e1[5] = -0.1167f;
        e1[6] = -5.81652f; e1[7] = -7.9381f;  e1[8] = 4.2816f;

        float32 e2[9];
        e2[0] = 2.1018f;   e2[1] = -1.81754f; e2[2] = 1.2541f;
        e2[3] = 0.54194f;  e2[4] = 2.75391f;  e2[5] = -0.1167f;
        e2[6] = -5.81652f; e2[7] = -7.9381f;  e2[8] = 4.2816f;

        Matrix3x3 a;
        a.m11 = 2.1018f;   a.m12 = -1.81754f; a.m13 = 1.2541f; 
        a.m21 = 0.54194f;  a.m22 = 2.75391f;  a.m23 = -0.1167f;
        a.m31 = -5.81652f; a.m32 = -7.9381f;  a.m33 = 4.2816f;

        Matrix3x3 b = a;

        float64 timeSpent = 0;
        LARGE_INTEGER nFreq;
        LARGE_INTEGER nBeginTime;
        LARGE_INTEGER nEndTime;

        QueryPerformanceFrequency(&nFreq); // statistical frequency
        QueryPerformanceCounter(&nBeginTime);// start timer

        for (int32 i = 0; i < 100000; i++) {
            multiplyMatrix(e1, e2, dst, 3, 3, 3, 3);
        }

    QueryPerformanceCounter(&nEndTime); //end timer
            timeSpent = (float64)(nEndTime.QuadPart - nBeginTime.QuadPart) / (nFreq.QuadPart);

   printf("timeSpent1:%f\n", timeSpent);

   QueryPerformanceCounter(&nBeginTime);
   for (int32 i = 0; i < 100000; i++) {
            b.multiply(a.m11, a.m12, a.m13, 
                       a.m21, a.m22, a.m23, 
                       a.m31, a.m32, a.m33);
        }
   QueryPerformanceCounter(&nEndTime);
        timeSpent = (float64)(nEndTime.QuadPart - nBeginTime.QuadPart) / (nFreq.QuadPart);

printf("timeSpent2:%f\n", timeSpent);

产出：

timeSpent1:0.014277
timeSpent2:0.004649
timeSpent1:0.012684
timeSpent2:0.004522
.......
.......
timeSpent1:0.003414
timeSpent2:0.001166
timeSpent1:0.003407
timeSpent2:0.001242