c – 优化Mandelbrot分形

2023年2月2日 159次阅读

这是一个以.ppm文件输出mandelbrot分形的代码.我该如何优化呢？

#include<bits/stdc++.h>
using namespace std;

int findMandelbrot(double cr, double ci, int max_iterations)
{
    int i = 0;
    double zr = 0.0, zi = 0.0;
    while (i < max_iterations && zr * zr + zi * zi < 4.0)
    {
        double temp = zr * zr - zi * zi + cr;
        zi = 2.0 * zr * zi + ci;
        zr = temp;
        ++i;
    }
    return i;
}

double mapToReal(int x, int imageWidth, double minR, double maxR)
{
    double range = maxR - minR;
    return x * (range / imageWidth) + minR;
}

double mapToImaginary(int y, int imageHeight, double minI, double maxI)
{
    double range = maxI - minI;
    return y * (range / imageHeight) + minI;
}

int main()
{
    ifstream f("input.txt");
    int imageWidth, imageHeight, maxN;
    double minR, maxR, minI, maxI;

    if (!f)
    {
        cout << "Could not open file!" << endl;
        return 1;
    }

    f >> imageWidth >> imageHeight >> maxN;
    f >> minR >> maxR >> minI >> maxI;

    ofstream g("output_image.ppm");
    g << "P3" << endl;
    g << imageWidth << " " << imageHeight << endl;
    g << "255" << endl;


    double start = clock();

    for (int i = 0; i < imageHeight; i++)
    {
        for (int j = 0; j < imageWidth; j++)
        {
            double cr = mapToReal(j, imageWidth, minR, maxR);
            double ci = mapToImaginary(i, imageHeight, minI, maxI);

            int n = findMandelbrot(cr, ci, maxN);

            int r = ((int)sqrt(n) % 256);
            int gr = (2*n % 256);
            int b = (n % 256);

            g << r << " " << gr << " " << b << " ";
        }
        g << endl;

        if(i == imageHeight / 2) break;
    }

    cout << "Finished!" << endl;

    double stop = clock();

    cout << (stop-start)/CLOCKS_PER_SEC;
    return 0;
}

我一直到imageHeight / 2,因为在Photoshop中我可以复制另一半.
我在考虑loghartimic的力量,但尝试了一些东西,只适用于整数…

最佳答案 所以这是热循环：

int i = 0;
double zr = 0.0, zi = 0.0;
while (i < max_iterations && zr * zr + zi * zi < 4.0)
{
    double temp = zr * zr - zi * zi + cr;
    zi = 2.0 * zr * zi + ci;
    zr = temp;
    ++i;
}
return i;

我知道如何在快速CPU指令中实现非整数功率但它不会让你脱离绑定,因为它根本不适用于复数.也不会使用std :: complex帮助.您不会为非内联支付任何费用,当然也无法在您找到它们时应用优化.所以我能做的最好的就是：

int i = max_iterations;
double zr = 0.0, zi = 0.0;
do {
    double temp = zr * zr - zi * zi + cr;
    zi = 2.0 * zr * zi + ci;
    zr = temp;
} while (--i && zr * zr + zi * zi < 4.0)
return max_iterations - i;

是的,我知道从循环中取出一个整数测试并没有太大的收获.我只发现了另一个优化器,你必须检查它是否真的更快：

int i = max_iterations;
double zr = 0.0, zi = 0.0;
do {
    double tempr = zr * zr - zi * zi + cr;
    double tempi = zr * zi;
    zi = tempi + tempi + ci;
    zr = tempr;
} while (--i && zr * zr + zi * zi < 4.0);
return max_iterations - i;

这就是所有人.

点赞