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java - 生成 Mandelbrot 分形时如何设置 c 的值?

转载 作者:行者123 更新时间:2023-12-01 09:08:17 25 4
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生成曼德尔布罗分形的方程为Zn+1 = Zn^2+C。问题是,在计算机程序中,C 用于缩放/分辨率和屏幕上的位置。我的问题是,我怎样才能得到这样的分形:

Wolfram

(方程 f(z) = sin(z/c), z0 = c )

cassinioval

我的代码(来自 Rosetta Code ):

public class MandelbrotSet extends JFrame {
private static final long serialVersionUID = 5513426498262284949L;
private final int MAX_ITER = 570;
private final double ZOOM = 150;
private BufferedImage image;
private double zx, zy, cX, cY, tmp;

public MandelbrotSet() {
super("Mandelbrot Set");
setBounds(100, 100, 800, 600);
setResizable(false);
setDefaultCloseOperation(EXIT_ON_CLOSE);
image = new BufferedImage(getWidth(), getHeight(), BufferedImage.TYPE_INT_RGB);
for (int y = 0; y < getHeight(); y++) {
for (int x = 0; x < getWidth(); x++) {
zx = zy = 0;
cX = (x - 400) / ZOOM;
cY = (y - 300) / ZOOM;
int iter = MAX_ITER;

while (zx * zx + zy * zy < 4 && iter > 0) {
tmp = zx * zx - zy * zy + cX;
zy = 2.0 * zx * zy + cY;
zx = tmp;
iter--;
}

image.setRGB(x, y, iter | (iter << 8));
}
}
}

@Override
public void paint(Graphics g) {
g.drawImage(image, 0, 0, this);
}

public static void main(String[] args) {
new MandelbrotSet().setVisible(true);;
}
}

最佳答案

通过三角定理

sin(A+i*B)=sin(A)*cos(i*B)+  cos(A)*sin(i*B) 
=sin(A)*cosh(B )+i*cos(A)*sinh(B )

并使用 z=x+i*yc=a+i*b

求商
(x+i*y)/(a+i*b)=(x+i*y)*(a-i*b)/(a*a+b*b)

对于上面的正弦表达式

A = (a*x+b*y)/(a*a+b*b)
B = (a*y-b*x)/(a*a+b*b)
<小时/>

在 javascript 中,生成此分形的小脚本如下所示:

function cosh(x) { return 0.5*(Math.exp(x)+Math.exp(-x)); }
function sinh(x) { return 0.5*(Math.exp(x)-Math.exp(-x)); }

function rgb(r,g,b) { return "rgb("+r+","+g+","+b+")"; }
var colors = new Array(24);
for(var k=0; k<8; k++) {
colors[ k] = rgb(k*64,(7-k)*64,(7-k)*64);
colors[ 8+k] = rgb((7-k)*64,k*64,(7-k)*64);
colors[16+k] = rgb((7-k)*64,(7-k)*64,k*64);
}

var cvs = document.getElementById('sine-fractal');
var ctx = cvs.getContext('2d');
var cx = 0.0, cy = 0.0;
var dx = 1.0;

var tiles = 100;
var scale = Math.min(cvs.width, cvs.height) / tiles;
ctx.scale(scale, scale);

function localx(i) { return cx-dx + 2*i*dx/tiles; }
function localy(j) { return cy-dx + 2*j*dx/tiles; }

for (var i = 0; i < tiles; i++) {
var a = localx(i);
for (var j = 0; j < tiles; j++) {
var b = localy(j);
var r2 = a*a + b*b;
var x = a, y = b;
var rounds = 0;
var max = 500;
while (x * x + y * y < 4 && rounds < max) {
var u = (a*x + b*y) / r2, v = (a*y - b*x) / r2;
x = Math.sin(u) * cosh(v);
y = Math.cos(u) * sinh(v);
rounds++;
}
ctx.fillStyle = colors[rounds % 24];
ctx.fillRect(i, j, 1, 1);
}
}
<canvas id='sine-fractal' width=200 height=200></canvas>

关于java - 生成 Mandelbrot 分形时如何设置 c 的值?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/41078505/

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