Question

I’m trying to learn Java from this book The Art and Science of Java

exercise in book The Art and Science of Java page 129

my problem is the pyramid should be centered in the window

import acm.program.*;
import acm.graphics.*;

public class pyramid extends GraphicsProgram {

    public void run() {


        for (int i = 0; i < BRICK_IN_BASE; i++) {
            for (int j = 0; j < i; j++) {

                int x = BRICK_WIDTH * j ;
                int y = BRICK_HEIGHT * i;

                GRect brick = new GRect(x, y, BRICK_WIDTH, BRICK_HEIGHT);
                add(brick);

            }
        }

    }



    private static final int    BRICK_WIDTH =30 ;
    private static final int    BRICK_HEIGHT =20 ;
    private static final int    BRICK_IN_BASE = 12;
}

Answer 1

my problem is the pyramid should be centered in the window

If you mean that you have a fixed area to draw in and the pyramid should be centered in that area:

1. get the maximum width of the pyramid: BRICK_IN_BASE * BRICK_WIDTH
2. start at x = (area.width-pyramid_width)/2
3. same for the height: BRICK_IN_BASE * BRICK_HEIGHT
4. start at y = (area.height-pyramid_height)/2

Besides that, note that for each level above the base you would have to add BRICK_WIDTH/2 to the x offset.

Edit (clarification):

The above points apply to getting the x and y coordinates to start drawing at. Note that

1. with “start at x” I meant the initial x offset. Your bricks’ x coordinate would be x_offset + BRICK_WIDTH * j
2. the same for start at y
3. for each level you have to increase x_offset by BRICK_WIDTH/2, i.e. int x = x_offset + (BRICK_IN_BASE - i - 1) * BRICK_WIDTH/2 + BRICK_WIDTH * j; (you’re starting at the top, so the level would have to be calculated as (BRICK_IN_BASE – i – 1) in order to get 11 for the top row and 0 for the bottom row).