Java中正方形的递归二分法

Question

我基本上是试着将左下角的正方形一分为二，直到零，然后用递归将剩余的方块左上角、右上角和右下角平分，直到所有的方块都填满为止。然而，我的代码执行底部正方形平分，但当我添加新行试图使右下角平分时停止。你知道为什么抽签会停止吗？我似乎不能在左下角的第一个平分之后实现一个简单的"+“形状的绘制。

package recursivepatterns;

import java.awt.Color;
import sedgewick.*;

public class PersianRug {


    public static void subDiv(double S, Color C, double x, double y) {
        //
        StdDraw.line(x+S/2, y, x+S/2, y+S);
        StdDraw.line(x, y+S/2, x+S, y+S/2);
    }


    /**
     * 
     * @param palette an array of Colors to choose from
     * @param llx lower left x coordinate of this rug square
     * @param lly lower left y coordinate of this rug square
     * @param size width (and therefore also height) of this rug square
     * @param north color index of the north side of this rug square
     * @param east color index of the east side of this rug square
     * @param south color index of the south side of this rug square
     * @param west color index of the west side of this rug square
     */

    private static void persianRug( 
            Color[] palette, 
            double llx, double lly, 
            double size, 
            int north, int east, int south, int west){
        //
        int c = 0; 
        Color C = palette[c];   

        subDiv(size,C,llx,lly); 

        if (size>0){
            // ll
            persianRug(palette,llx,lly,size/2, c, c, south, west); // north and east are fixed
            // lr
            StdDraw.setPenColor(StdDraw.RED);
            StdDraw.circle(0.5,0.5,size/2); // this tests that circles do in fact recursively get bigger
            persianRug(palette,llx+size/2,lly,size/2, c, c, south, west); // this line does not work at all

        }
        else return;
    }


    public static void main(String args[]) {
        //
        // Leave the following line commented out, but once you
        //   have things working, uncomment it, and also uncomment
        //   the similar line at the end of this method.
        // Uncommenting those lines will run the graphics code
        //   in double-buffering mode, so that your image will appear
        //   almost instantaneously, instead of being drawn one line
        //   at a time.
        //
        //  Here is the line to uncomment:
        //
        //StdDraw.show(10);   // don't forget to uncomment the other line at the end
        //


        //
        // Generate a palette of colors
        //
        Color[] palette = { StdDraw.BLUE, StdDraw.CYAN, StdDraw.DARK_GRAY,
                StdDraw.GRAY, StdDraw.GREEN, StdDraw.LIGHT_GRAY,
                StdDraw.MAGENTA, StdDraw.ORANGE, StdDraw.PINK, StdDraw.RED,
                StdDraw.WHITE, StdDraw.YELLOW };
        //
        // Draw the outermost square as a special case
        // Use color 0 for that
        //
        StdDraw.setPenColor(palette[0]);
        StdDraw.line(0, 0, 1, 0);
        StdDraw.line(1, 0, 1, 1);
        StdDraw.line(1, 1, 0, 1);
        StdDraw.line(0, 1, 0, 0);


        //
        // Kick off the recursion
        // Lower left is point (0,0)
        // Size of the square is 1
        // The color index of each surrounding side is 0
        //
        persianRug(palette, 0, 0, 1, 0, 0, 0, 0);
        //
        // Also uncomment this line when you have things working
        //   to speed up the drawing:
        //
        //StdDraw.show(10);
        //
    }

}