7

我在 C 中有以下 Mandelbrot 集代码。我正在计算并为最终的分形图像创建一个 .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:

在此处输入图像描述

我的最后一个问题是我的代码没有检查我的代码的运行时间。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,并显示经过的时间,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;
}
4

1 回答 1

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:为了有时间打印出来,您应该将其打印到 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:要获得连续平滑的着色,您需要使用 Normalized Iteration Count 方法或类似的方法。这是您的着色部分的替代品,可为您提供与您想要的类似的东西:

    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,因为我对标准化迭代计数方法做了一个简单的近似实现。

Mandelbrot 使用一些半连续着色

它不是完全连续的着色,但它有点接近。

于 2013-04-20T21:13:40.867 回答