如何简化这个绘制雪花的 Blockly 程序？

Question

问题是这样说的：

The program shown below on the left draws a snowflake. It uses 33 code blocks. Can you re-write it so that it uses at least one function and less than 30 blocks.

如何使用 Blockly 绘制这个？

Answer 1

经过一些研究，我认为递归函数不是这种情况下的解决方案，但在 python 中，我会尝试这样的方法

def drawLeaf(direction):
    // Draw your leaf in this direction using your blocks code

for direction in [0, 72, 144, 216, 288]:
    drawLeaf(direction)

Answer 2

首先，请看一下Finding coordinates of Koch Curve

这是一个 Java 代码片段（针对 Android），用于绘制 koch snowflake.It 的简单递归函数

public void drawfractal1(Canvas canvas , float x1 , float y1 , float x2 , float y2, int level){
        Paint paint = new Paint();
        int r = ThreadLocalRandom.current().nextInt(0, 256);
        int g = ThreadLocalRandom.current().nextInt(0, 256);
        int b = ThreadLocalRandom.current().nextInt(0, 256);
        paint.setColor(Color.rgb(r,g,b));
        paint.setStrokeWidth(7);

        if(level==0) {
            canvas.drawLine(x1,y1,x2,y2,paint);
        }
        else{
            float ux = x2-x1;
            float uy = y2-y1;
            float vx = y1-y2;
            float vy = x2-x1;

            float xa = x1 + ux/3;
            float ya = y1 + uy/3;
            float xb = x1 + ux/2 + (float) sqrt(3)*vx/6;
            float yb = y1 +uy/2 + (float)sqrt(3)*vy/6;
            float xc = x1 + 2*ux/3;
            float yc = y1 + 2*uy/3;
            drawfractal1(canvas , x1,y1,xa,ya,level-1);
            drawfractal1(canvas , xa,ya,xb,yb,level-1);
            drawfractal1(canvas , xb,yb,xc,yc,level-1);
            drawfractal1(canvas , xc,yc,x2,y2,level-1);
        }
    }

如何简化这个绘制雪花的 Blockly 程序？

How to simplify this Blockly program drawing a snowflake?

blockly