矩阵乘法逐步加速演示

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原创最新推荐文章于 2026-03-25 01:54:35 发布 2.6k 阅读 AI 写同款· GEO 优化 ›

浅谈矩阵加速算法前言（看此文章请先了解一下什么是矩阵）矩阵加速是个很神奇的算法，它可以在极端数据下、级短时间内解决复杂问题，而且还很简单。我们从例题入手：例题一：Fibonacci第n项这个数据很大，不能常规地用递推求解。那我们不拐弯路，直接开始公式推导吧：因为（关键递推式），所以可以把放入一个矩阵乘法（n*m和m*q的矩阵）中。因为是2项乘2项，所以m=2，为了达到有规律的累乘，得保... 阅读详情

矩阵乘法

#include<iostream>
#include<ctime>
using namespace std;
int main() {
    int n = 1000;
    double* a = new double[n*n];
    double* b = new double[n*n];
    double* c = new double[n*n];
    int i, j, k;
    for (i = 0; i < n; ++i)
        for (j = 0; j < n; ++j) {
            a[i*n + j] = i + j;
            b[i*n + j] = i + j;
        }
    auto tic = clock();
    double s{ 0 };
    for (i = 0; i < n; ++i)
        for (j = 0; j < n; ++j) {
            s = 0.0;
            for (k = 0; k < n; ++k)
                s += a[i*n + k] * a[k*n + j];
            c[i*n + j] = s;
        }
    auto toc = clock() - tic;
    cout << toc << endl;
}

耗时 1600ms

后面矩阵转置

#include<iostream>
#include<ctime>
using namespace std;
int main() {
    int n = 1000;
    double* a = new double[n*n];
    double* b = new double[n*n];
    double* c = new double[n*n];
    int i, j, k;
    for (i = 0; i < n; ++i)
        for (j = 0; j < n; ++j) {
            a[i*n + j] = i + j;
            b[i*n + j] = i + j;
        }
    auto tic = clock();
    double s{ 0 };
    for (i = 0; i < n; ++i)
        for (j = 0; j < n; ++j) {
            s = 0.0;
            for (k = 0; k < n; ++k)
                s += a[i*n + k] * b[j*n + k];
            c[i*n + j] = s;
        }
    auto toc = clock() - tic;
    cout << toc << endl;
}

耗时 960ms

循环展开2个乘积

#include<iostream>
#include<ctime>
using namespace std;
int main() {
    int n = 1000;
    double* a = new double[n*n];
    double* b = new double[n*n];
    double* c = new double[n*n];
    int i, j, k;
    for (i = 0; i < n; ++i)
        for (j = 0; j < n; ++j) {
            a[i*n + j] = i + j;
            b[i*n + j] = i + j;
        }
    auto tic = clock();
    double s{ 0 };
    for (i = 0; i < n; ++i)
        for (j = 0; j < n; ++j) {
            s = 0.0;
            for (k = 0; k < n; k += 2)
                s += a[i*n + k] * b[j*n + k] +
                a[i*n + k + 1] * b[j*n + k + 1];
            c[i*n + j] = s;
        }
    auto toc = clock() - tic;
    cout << toc << endl;
}

耗时 560 ms

循环展开4个乘积

#include<iostream>
#include<ctime>
using namespace std;
int main() {
    int n = 1000;
    double* a = new double[n*n];
    double* b = new double[n*n];
    double* c = new double[n*n];
    int i, j, k, i1, j1;
    for (i = 0; i < n; ++i)
        for (j = 0; j < n; ++j) {
            a[i*n + j] = i + j;
            b[i*n + j] = i + j;
        }
    auto tic = clock();
    double s{ 0 };
    for (i = 0; i < n; ++i) {
        i1 = i * n;
        for (j = 0; j < n; ++j) {
            j1 = j * n;
            s = 0.0;
            for (k = 0; k < n; k += 4)
                s += a[i1 + k] * b[j1 + k] +
                a[i1 + k + 1] * b[j1 + k + 1] +
                a[i1 + k + 2] * b[j1 + k + 2] +
                a[i1 + k + 3] * b[j1 + k + 3];
            c[i*n + j] = s;
        }
    }
    auto toc = clock() - tic;
    cout << toc << endl;
}

耗时 480 ms

蛇形顺序

#include<iostream>
#include<ctime>
using namespace std;
int main() {
    int n = 1000;
    double* a = new double[n*n];
    double* b = new double[n*n];
    double* c = new double[n*n];
    int i, j, k, i1, j1;
    for (i = 0; i < n; ++i)
        for (j = 0; j < n; ++j) {
            a[i*n + j] = i + j;
            b[i*n + j] = i + j;
        }
    auto tic = clock();
    double s{ 0 };
    for (i = 0; i < n; ++i) {
        i1 = i * n;
        if (i % 2 == 0) {
            for (j = 0; j < n; ++j) {
                j1 = j * n;
                s = 0.0;
                for (k = 0; k < n; k += 4)
                    s += a[i1 + k] * b[j1 + k] +
                    a[i1 + k + 1] * b[j1 + k + 1] +
                    a[i1 + k + 2] * b[j1 + k + 2] +
                    a[i1 + k + 3] * b[j1 + k + 3];
                c[i*n + j] = s;
            }
        }
        else {
            for (j = n - 1; j >= 0; --j) {
                j1 = j * n;
                s = 0.0;
                for (k = 0; k < n; k += 4)
                    s += a[i1 + k] * b[j1 + k] +
                    a[i1 + k + 1] * b[j1 + k + 1] +
                    a[i1 + k + 2] * b[j1 + k + 2] +
                    a[i1 + k + 3] * b[j1 + k + 3];
                c[i*n + j] = s;
            }
        }
    }
    auto toc = clock() - tic;
    cout << toc << endl;
}