Question

I have task to write program allowing users to draw stars, which can differ in size and amount of arms. When I was dealing with basic stars I was doing it with GeneralPath and tables of points :

     int xPoints[] = { 55, 67, 109, 73, 83, 55, 27, 37, 1, 43 };
     int yPoints[] = { 0, 36, 36, 54, 96, 72, 96, 54, 36, 36 };
     Graphics2D g2d = ( Graphics2D ) g;
     GeneralPath star = new GeneralPath();
     star.moveTo( xPoints[ 0 ], yPoints[ 0 ] );
     for ( int k = 1; k < xPoints.length; k++ )
     star.lineTo( xPoints[ k ], yPoints[ k ] );
     star.closePath();
     g2d.fill( star );

What method should I choose for drawing stars with variable inner and outer radius, as well as different amount of arms ? This is what I should obtain :

alt text http://img228.imageshack.us/img228/6427/lab6c.jpg

Answer 1

Having n arms means you end up with 2n vertices, the even ones are on the outer circle, and the odd ones on the inner circle. Viewed from the center, the vertices are at evenly spaced angles (the angle is 2*PI/2*n = Pi/n). On an unit circle (r=1), the x,y coordinates of the points i=0..n is cos(x),sin(x). Multiply those coordinates with the respective radius (rOuter or rInner, depending of whether i is odd or even), and add that vector to the center of the star to get the coordinates for each vertex in the star path.

Here’s the function to create a star shape with given number of arms, center coordinate and outer, inner radius:

public static Shape createStar(int arms, Point center, double rOuter, double rInner) {
    double angle = Math.PI / arms;

    GeneralPath path = new GeneralPath();

    for (int i = 0; i < 2 * arms; i++) {
        double r = (i & 1) == 0 ? rOuter : rInner;
        Point2D.Double p = new Point2D.Double(
            center.x + Math.cos(i * angle) * r, 
            center.y + Math.sin(i * angle) * r);
        if (i == 0) {
            path.moveTo(p.getX(), p.getY());
        }
        else {
            path.lineTo(p.getX(), p.getY());
        }
    }
    path.closePath();
    return path;
}