checking if a point is inside a specified Rectangle

Question

ok, so i\'m doing an assignment for a Java class and one part of the assignment is to find out if a point is within the dimensions of a rectangle. so I created this code:

Answer 1

AWT Rectangle already has contains method. ( link )

Task seems about if you understand how naming spaces conflict. For example, if you are lazy (it's one of most admired qualities of a programmer), then you can write:

public static class Rectangle {
    java.awt.Rectangle _r;

    public Rectangle(int x, int y) {
        this._r = new java.awt.Rectangle(x, y);
    }
    public boolean contains(Point p) {
        return this._r.contains(p);
    }
}

You generally do not want to reimplementing features nor extend classes.

Answer 2

Though its a naive method, I tried the following concept:

If the Point (px,py) is inside the given rectangle, the sum of areas of triangles formed by joining 2 rectangle points and the given point (say in anti-clockwise or clockwise direction) would be equal to the sum of rectangle.

I have a picture for the same, but due to low reputation (as I am a newbie), cannot post it.

When I was formulating this into actual Java code, I had to handle a situation where the area value with decimal part having 15 9s was rounded to its nearest integer.

Refer this code:

import static java.lang.Math.sqrt;

public class PointInsideRect
{
    private static double square(double n)
    {
        return n*n;
    }
    private static double areaOfTriangle(
                int xa, int ya,
                int xb, int yb,
                int px, int py )
    {
        double side1 = sqrt(square(ya-yb) + square(xa-xb));
        double side2 = sqrt(square(ya-py) + square(xa-px));
        double side3 = sqrt(square(yb-py) + square(xb-px));

        double semi_perimeter = (side1 + side2 + side3) / 2;

        return sqrt(semi_perimeter
                    * ( semi_perimeter - side1 )
                    * ( semi_perimeter - side2 )
                    * ( semi_perimeter - side3 ));
    }

    private static double areaOfRect(
                int x1, int y1,
                int x2, int y2,
                int x3, int y3,
                int x4, int y4 )
    {
        double side1 = sqrt(square(y1-y2) + square(x1-x2));
        double side2 = sqrt(square(y2-y3) + square(x2-x3));
        return side1 * side2;
    }

    public boolean check(
                int x1, int y1,
                int x2, int y2,
                int x3, int y3,
                int x4, int y4, 
                int pointX, int pointY)
    {
        double tri1Area = areaOfTriangle(x1,y1, x2,y2, pointX,pointY);
        double tri2Area = areaOfTriangle(x2,y2, x3,y3, pointX,pointY);
        double tri3Area = areaOfTriangle(x3,y3, x4,y4, pointX,pointY);
        double tri4Area = areaOfTriangle(x4,y4, x1,y1, pointX,pointY);

        double rectArea = areaOfRect(x1,y1, x2,y2, x3,y3, x4,y4);

        double triAreaSum = tri1Area + tri2Area + tri3Area+ tri4Area;


        if(triAreaSum % Math.pow(10, 14) >= 0.999999999999999)
        {
            triAreaSum = Math.ceil(triAreaSum);
            System.out.println(triangleAreaSum);
        }
        return triAreaSum == rectArea;
    }


    public static void main(String[] args)
    {
        PointInsideRect inRect = new PointInsideRect();

        System.out.println(inRect.check(1,1, 1,3, 3,3, 3,1, 2,2));
    }
}

Answer 3

It looks ok to me. I would check that your test case actually has the numbers you think it does; I would also check that your accessors are all returning the right values (I can't tell you the number of times I've implemented getX() as {return this.y;}). Other than that it's anyone's guess.

Answer 4

Usually when dealing with computer graphics, the top left point is (0,0) and the bottom right corner is (width, height).

This means that you should reverse your conditions