1

我正在尝试在 AS3 上编写 sierpinski 三角形，但看起来我缺少一些东西。我被困在这行代码中并试图解决问题所在。我进行了计算，但似乎缺少一些东西。可以你帮忙？

public class Fract extends Sprite 
{
    private var rand:Function = Math.random;
    private var num:Number = 5;
    private var w:Number = stage.stageWidth;
    private var h:Number = stage.stageHeight;
    private var spr:Sprite;
    private var small:Sprite;

    public function Fract()
    {
        init();
    }

    private function init():void
    {   
        var i:Number;
        spr = new Sprite();
        small = new Sprite();
        addChild(spr);
        addChild(small);

        for (i = 1; i < num; i++)
        {
            drawTriangle(i);
        }
    }

    private function drawTriangle(i:Number):void
    {
        //CREATE P1 & P2
        var p1:Point = new Point((w / 4), (h - h / 4));
        var p2:Point = new Point((w - w / 4), p1.y);
        //Calculate distance between first two points
        var dist:Number = p2.x - p1.x;
        var dist2:Number = dist / 2;
        //Calculate p3y-so the triangle is equilateral
        var p3y:Number = Math.sqrt((dist * dist) - (dist2 * dist2));
        var p3:Point = new Point(p1.x + dist2, p1.y - p3y);

        if (i == 1)
        {
            spr.graphics.lineStyle(1, 0, 1);
            spr.graphics.beginFill(0, 1);
            spr.graphics.moveTo(p1.x, p1.y);
            spr.graphics.lineTo(p2.x, p2.y);
            spr.graphics.lineTo(p3.x, p3.y);
            spr.graphics.lineTo(p1.x, p1.y);
            spr.graphics.endFill();
        }
        else
        {
            var p4:Point = new Point(p1.x + (dist2 / i), p1.y - (p3y / i));
            var p5:Point = new Point(p4.x + (dist2 / (i - 1)), p4.y);
            var smallDist:Number = p5.x - p4.x;
            var smallDist2:Number = smallDist / 2;
            var p6y:Number = Math.sqrt((smallDist * smallDist) - (smallDist2 * smallDist2));
            var p6:Point = new Point(p4.x + smallDist2, p4.y + p6y);

            small.graphics.lineStyle(1, 0, 1);
            small.graphics.beginFill(0xffffff, 1);
            small.graphics.moveTo(p4.x, p4.y);
            small.graphics.lineTo(p5.x, p5.y);
            small.graphics.lineTo(p6.x, p6.y);
            small.graphics.lineTo(p4.x, p4.y);
            small.graphics.endFill();
        }
    }
}
4

1 回答 1

3

这可以在没有高超数学技能的情况下完成，因为您只需在一开始就进行复杂的计算，然后简单地使用三角形的边。它是一个递归函数，每轮调用自己 3 次，直到达到最大级别。我在下面包含了我的示例。请注意最大值的值，因为如果设置得太高，它很容易让您的计算机冻结一段时间......

这是一个非常有趣的问题！

public class Fract extends Sprite 
{
    private var w : Number = stage.stageWidth;
    private var h : Number = stage.stageHeight;
    private var max : uint = 6;

    public function Fract ()
    {
        init( );
    }

    private function init () : void
    {   
        var p1 : Point = new Point( (w / 4), (h - h / 4) );
        var p2 : Point = new Point( (w - w / 4), p1.y );
        var dist : Number = (p2.x - p1.x);
        var dist2 : Number = dist * .5;
        var p3 : Point = new Point( p1.x + dist2, p1.y - Math.sqrt( (dist * dist) - (dist2 * dist2) ) );

        drawTriangles( p1, p2, p3, 0, max );
    }

    private function drawTriangles (p1 : Point, p2 : Point, p3 : Point, level : uint, max : uint) : void
    {
        var spr : Sprite = new Sprite( );
        addChild( spr );
        var g : Graphics = spr.graphics;
        g.lineStyle( 1, 0, 1 );
        g.moveTo( p1.x, p1.y );
        g.lineTo( p2.x, p2.y );
        g.lineTo( p3.x, p3.y );
        g.lineTo( p1.x, p1.y );

        if (level < max) 
        {
            var p12 : Point = new Point( p1.x + (p2.x - p1.x) * .5, p1.y + (p2.y - p1.y) * .5 );
            var p23 : Point = new Point( p2.x + (p3.x - p2.x) * .5, p2.y + (p3.y - p2.y) * .5 );
            var p31 : Point = new Point( p3.x + (p1.x - p3.x) * .5, p3.y + (p1.y - p3.y) * .5 );
            drawTriangles( p1, p12, p31, level + 1, max);
            drawTriangles( p12, p2, p23, level + 1, max);
            drawTriangles( p31, p23, p3, level + 1, max);
        }
    }
}
于 2011-01-06T02:05:37.187 回答