改进了我的Mandelbrot设置代码

时间:2013-04-20 19:10:25

标签: c colors set fractals mandelbrot

我在C中有以下Mandelbrot设置代码。我正在进行计算并为最终的分形图像创建.ppm文件。关键是我的分形图像是颠倒的,这意味着它旋转了90度。您可以通过执行我的代码来检查它: ./mandel> test.ppm

另一方面,我也想改变颜色。我想实现这个分形图像:

enter image description here

我的最后一个问题是我的代码没有检查代码的运行时间。我也有这部分的代码,但是当代码执行完成时,它不会打印运行时间。如果有人可以对我的代码进行适当的更改并帮助我实现这个分形图像,并显示已用时间,我会很高兴。

#include <math.h>
#include <stdlib.h>
#include <time.h>
#include <stdio.h>

void color(int red, int green, int blue)
{
    fputc((char)red, stdout);
    fputc((char)green, stdout);
    fputc((char)blue, stdout);
}

int main(int argc, char *argv[])
{
    int w = 600, h = 400, x, y; 
    //each iteration, it calculates: newz = oldz*oldz + p, where p is the current pixel, and oldz stars at the origin
    double pr, pi;                   //real and imaginary part of the pixel p
    double newRe, newIm, oldRe, oldIm;   //real and imaginary parts of new and old z
    double zoom = 1, moveX = -0.5, moveY = 0; //you can change these to zoom and change position
    int maxIterations = 1000;//after how much iterations the function should stop

    clock_t begin, end;
    double time_spent;

    printf("P6\n# CREATOR: E.T / mandel program\n");
    printf("%d %d\n255\n",w,h);

    begin = clock();

    //loop through every pixel
    for(x = 0; x < w; x++) 
    for(y = 0; y < h; y++)
    {
        //calculate the initial real and imaginary part of z, based on the pixel location and zoom and position values
    pr = 1.5 * (x - w / 2) / (0.5 * zoom * w) + moveX;
        pi = (y - h / 2) / (0.5 * zoom * h) + moveY;
        newRe = newIm = oldRe = oldIm = 0; //these should start at 0,0
        //"i" will represent the number of iterations
        int i;
        //start the iteration process
        for(i = 0; i < maxIterations; i++)
        {
            //remember value of previous iteration
            oldRe = newRe;
            oldIm = newIm;
            //the actual iteration, the real and imaginary part are calculated
            newRe = oldRe * oldRe - oldIm * oldIm + pr;
            newIm = 2 * oldRe * oldIm + pi;
            //if the point is outside the circle with radius 2: stop
            if((newRe * newRe + newIm * newIm) > 4) break;
        }

        color(i % 256, 255, 255 * (i < maxIterations));

    }

    end = clock();

    time_spent = (double)(end - begin) / CLOCKS_PER_SEC;
    printf("Elapsed time: %.2lf seconds.\n", time_spent);

    return 0;
}

1 个答案:

答案 0 :(得分:9)

第1部分: 您需要将循环的顺序交换为:

for(y = 0; y < h; y++)
for(x = 0; x < w; x++)

这将为您提供正确定向的分形。

第2部分: 为了节省打印时间,您应该将它打印到stderr,因为您要将ppm输出打印到stdout:

fprintf(stderr, "Elapsed time: %.2lf seconds.\n", time_spent);

第3部分: 要获得连续的平滑着色,您需要使用Normalized Iteration Count方法或类似方法。这是您着色部分的替代品,可以为您提供类似于您想要的东西:

    if(i == maxIterations)
        color(0, 0, 0); // black
    else
    {
        double z = sqrt(newRe * newRe + newIm * newIm);
        int brightness = 256. * log2(1.75 + i - log2(log2(z))) / log2(double(maxIterations));
        color(brightness, brightness, 255);
    }

它并不完全存在,因为我对Normalized Iteration Count方法进行了简单的近似实现。

Mandelbrot using some semi-continuous coloring

它不是一种完全连续的着色,但它有点接近。