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我正在尝试用 C 代码创建一个 Mandelbrot 集；我的输出将是一个数据文件，一列实部和一列虚部要绘制在 Argand 或复平面上。我在标题中定义了所有复杂的数学内容，complex.h并且正在使用具有complex实double R部和double I虚部的结构。我正在尝试循环 dz 的谷值并更新 z iter=80 次，或者直到该点位于定义的半径之外。dz 是虚数，基本上在 [(dzrmin + i*dzimin), (dzrmax +i*dzimax)] 范围内。z 是当前复数，更新为 z^2 = dz + z。我有一个csum()对两个复数求和的函数和一个csquare()对一个复数求平方的函数。这是我的整个代码

    #include <stdio.h>
    #include <stdlib.h>
    #include <math.h>
    #include "complex.h"

    int main(void)
    {
        double dzrmin, dzrmax, dzimin, dzimax, dzr,dzi, r0,i0, maxrad; 
        int i,j,k,iter, Ndr,Ndi; 
        complex z, dz; 
        dzrmin = -2.1; 
        dzrmax = 0.6; 
        dzimin = -1.2; 
        dzimax = 1.2; 
        Ndr = 200 ; 
        Ndi = 180; 
        dzr = (dzrmax - dzrmin)/Ndr; 
        dzi = (dzimax - dzimin)/Ndi; 
        r0 = 0.0; 
        i0 = 0.0;
        z.R = r0; 
        z.I = i0; 
        dz.R = dzrmin; 
        dz.I = dzimin; 
        maxrad = 2.0; 
        iter = 80 ; 
        for(i = 0; i < Ndr; i++ )
        {
                    for(j = 0; j < Ndi; j++ )
                    {

                        for(k = 0; k< iter; k++ )
                        {
                    printf("%.6lf %.6lf\n", z.R, z.I ); 
                    z = csum( csquare( z ), dz ) ; 

                    if( cmag(z) > maxrad) k = iter ; 

                }

                dz.I += dzi ; 
            }
            dz.R += dzr; 
            z.R = r0; 
            z.I = i0; 
        }
        return 0; 
    }

和我的头文件complex.h

#include <stdlib.h>
#include <math.h>
typedef struct complexnumber{ double R; double I ; } complex ;
double cmag( complex z)                                                                                                                                             
{
    return pow( z.R*z.R + z.I*z.I, 0.5 ) ; 
}

complex csquare( complex z )             //returns square of a complex
{
    complex product ; 
    product.R = z.R*z.R - z.I*z.I ; 
    product.I = 2*z.R*z.I ; 
    return product ; 

}

complex csum( complex z1, complex z2) // sums two complex numbers 
{
    complex sum ; 
    sum.R = z1.R + z2.R ; 
    sum.I = z1.I + z2.I; 
    return sum ; 
}

我得到了一些实际值，它们很快变得非常大，并且位于半径 2 之外，然后是很多实部-nan和虚部-nan。关于我缺少什么的任何建议？

4

1 回答 1

1

我不确定您期望什么结果，但也许这个检测输出将帮助您诊断问题：

i = 0
i = 0, j = 0
i = 0, j = 0, k = 0: 0.000000 0.000000
i = 0, j = 1
i = 0, j = 1, k = 0: -2.100000 -1.200000
i = 0, j = 2
i = 0, j = 2, k = 0: 0.870000 3.853333
i = 0, j = 3
i = 0, j = 3, k = 0: -16.191278 5.531467
i = 0, j = 4
i = 0, j = 4, k = 0: 229.460353 -180.283027
i = 0, j = 5
i = 0, j = 5, k = 0: 20147.983719 -82736.760384
i = 0, j = 6
i = 0, j = 6, k = 0: -6439430272.999375 -3333957803.416253
i = 0, j = 7
i = 0, j = 7, k = 0: 30350987605860679680.000000 42937577616442236928.000000
i = 0, j = 8
i = 0, j = 8, k = 0: -922453122916892839668491877362832506880.000000 2606395772124638489891541615388582215680.000000
i = 0, j = 9
i = 0, j = 9, k = 0: -5942379156970062175257857153425511563624711669037998408592102022831643293646848.000000 -4808555839107517668369914051798230293111512606121703008400866410836391727464448.000000
i = 0, j = 10
i = 0, j = 10, k = 0: 12189660787377226979177299907109395027773453611835899247334853368367297050334425366457359535808690377953524893091770076537471791519164311649924318416907272192.000000 57148523986878398200502132726020983696472380197135570914789139569106551459309671849901109020634047615447493780748138450365838320365732961897986301575347306496.000000
i = 0, j = 10, k = 1: nan inf
i = 0, j = 10, k = 2: nan nan

请注意内部循环如何在第一次迭代中一直中断。我认为问题在于dz设置为-2.1 - 1.2i开始。但是，如果没有您要绘制的曼德布罗曲线的定义，我不知道应该发生什么。

这是由您的代码的这个单文件变体产生的，它使用了与 C99 标准冲突Complex的类型。complex复杂函数是静态的，因为当我编译时会关闭编译器警告：

$ gcc -O3 -g -std=c99 -Wall -Wextra -Wmissing-prototypes -Wstrict-prototypes cx.c -o cx

资源：

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

typedef struct Complex{ double R; double I; } Complex;

static double cmag(Complex z)
{
    return pow(z.R*z.R + z.I*z.I, 0.5);
}

static Complex csquare(Complex z)             //returns square of a complex
{
    Complex product;
    product.R = z.R*z.R - z.I*z.I;
    product.I = 2*z.R*z.I;
    return product;
}

static Complex csum(Complex z1, Complex z2) // sums two complex numbers
{
    Complex sum;
    sum.R = z1.R + z2.R;
    sum.I = z1.I + z2.I;
    return sum;
}

int main(void)
{
    double dzrmin, dzrmax, dzimin, dzimax, dzr, dzi, r0, i0, maxrad;
    int i, j, k, iter, Ndr, Ndi;
    Complex z, dz;
    dzrmin = -2.1;
    dzrmax = 0.6;
    dzimin = -1.2;
    dzimax = 1.2;
    Ndr = 200;
    Ndi = 180;
    dzr = (dzrmax - dzrmin)/Ndr;
    dzi = (dzimax - dzimin)/Ndi;
    r0 = 0.0;
    i0 = 0.0;
    z.R = r0;
    z.I = i0;
    dz.R = dzrmin;
    dz.I = dzimin;
    maxrad = 2.0;
    iter = 80;
    for (i = 0; i < Ndr; i++)
    {
        printf("i = %d\n", i);
        for (j = 0; j < Ndi; j++)
        {
            printf("i = %d, j = %d\n", i, j);
            for (k = 0; k < iter; k++)
            {
                printf("i = %d, j = %d, k = %d: ", i, j, k);
                printf("%.6lf %.6lf\n", z.R, z.I);
                z = csum(csquare(z), dz);
                if (cmag(z) > maxrad)
                    break;
            }
            dz.I += dzi;
        }
        dz.R += dzr;
        z.R = r0;
        z.I = i0;
    }
    return 0;
}
于 2013-01-20T01:41:27.387 回答