Question

我与一些曼德勒布罗人环绕的迷惑不解,因为我认为,它产生的景象相对应。我认为,我可能会试图解决把一个人打上 j印,看我可以做些什么的问题。我研究的是两种算法,即:

rel=“nofollow” http://library.thinkquest.org/26242/full/progs/a2.html

我将:

drawGraph: function(canvas,resolution,iterations,colors,coefficent){

                var context = canvas.getContext( 2d );

                for(var m = 0; m < resolution.x; m++){
                    for(var n = 0; n < resolution.y; n++){
                        var x = m, 
                            x2 = x*x,
                            y = n, 
                            y2 = y*y;

                        var i;
                        for(i = 1; i < iterations; i++){
                            if(x2 + y2 > 4) break;

                            var new_x = x2 - y2 + coefficent.a;
                            var new_y = 2*x*y + coefficent.b;

                            x = new_x;
                            y = new_y;
                        }

                        var color = i % colors;

                        DrawUtils.drawPoint(context,m,n,color); 
                    }
                }
            }

基本上用一个色体。

Then I tried this one:

我将:

drawGraph: function(canvas,resolution,iterations,colors,coefficent){

                var context = canvas.getContext( 2d );

                for(var m = 0; m < resolution.x; m++){
                    for(var n = 0; n < resolution.y; n++){
                        var x = 0,
                            y = 0,
                            x0 = ((m/resolution.x) * 3.5) - 2.5,
                            y0 = ((n/resolution.y) * 2) - 1;

                        var i = 0;
                        while(x*x + y*y < 4 && i < iterations){
                            var x_temp = x*x - y*y + x0;
                            y = 2*x*y + y0;
                            x  = x_temp;
                            i++;
                        }

                        var color = 0;
                        if(x*x + y*y >= 4){
                            color = i % colors;
                        }

                        DrawUtils.drawPoint(context,m,n,color);
                    }
                }
            }

产生黑箱。算法中混为一谈的我的措辞虽然说X0 和0 按比例计算是六分之一的因素,但在算法之后却说,系数c = x0 + iy0;因此,这并不意味着我不会将预先确定的系数移入功能?

在大多数这些测试中,我使用了0.25+0i系数,但我尝试了产生相同结果的其他方法。

我在这里做了什么错误?

Answer 1

第一点:你需要清楚地了解Julia各套和Mandelbrot之间的区别。两者都从不同角度深入了解<代码>f(z) = z^2 + c的特性。

<>>Julia <>/strong>。

关于<<>Mandelbrot的一组,我们做了一个图表,说明相同的初始<代码>z = 0为不同的<代码>c。

With that addressed...

您的第一部法典(试图为<条码>c设计的Julia, 载于<条码>、条码)的译本,您在您的第一页链接上从《经济、社会、文化权利国际公约》的译文并不正确。如果有

‘ run through every point on the screen, setting 
‘ m and n to the coordinates
FOR m = x_minimum TO x_maximum STEP x_resolution
             FOR n = y_minimum TO y_maximum STEP y_resolution
                          ‘ the initial z value is the current pixel,  
                          ‘ so x and y have to be set to m and n
                          x = m: y = n

页: 1

        for(var m = 0; m < resolution.x; m++){
            for(var n = 0; n < resolution.y; n++){

which is close, except for the crucial point that you are not taking any steps to implement STEP x_resolution. Your m is an integer that runs from 0 to resolution.x - 1 in steps of 1; and your x is set to m.

因此,你不从<条码>-2i到<条码>2+2i(见Julia套的体面观察室),而是从<条码>0到<条码>,到<条码>,x + Resolution.y i,其中将包含在低轴左上方的几页。

第二部法典(试图起草曼德尔布罗套)does 我可以立即看到什么是错的——我会辩论并看到:<代码>m/ Resolution.x总是0,因为@user973572建议这个问题。

Answer 2

在你的第一个例子中,我认为你不愿更新x2和 y2,因此,它们总是具有同样的价值。如果金额超过4美元,在检查之前需要更新x2和 y2。类似于

for(i = 1; i < iterations; i++){
    x2 = x*x,
    y2 = y*y
    if(x2 + y2 > 4) break;

which is probably wrong because I know nothing about javascript.

友情链接