Determine if two rectangles overlap each other?

Question

I am trying to write a C++ program that takes the following inputs from the user to construct rectangles (between 2 and 5): height, width, x-pos, y-pos. All of these rectangles will exist parallel to the x and the y axis, that is all of their edges will have slopes of 0 or infinity.

I've tried to implement what is mentioned in this question but I am not having very much luck.

My current implementation does the following:

// Gets all the vertices for Rectangle 1 and stores them in an array -> arrRect1
// point 1 x: arrRect1[0], point 1 y: arrRect1[1] and so on...
// Gets all the vertices for Rectangle 2 and stores them in an array -> arrRect2

// rotated edge of point a, rect 1
int rot_x, rot_y;
rot_x = -arrRect1[3];
rot_y = arrRect1[2];
// point on rotated edge
int pnt_x, pnt_y;
pnt_x = arrRect1[2]; 
pnt_y = arrRect1[3];
// test point, a from rect 2
int tst_x, tst_y;
tst_x = arrRect2[0];
tst_y = arrRect2[1];

int value;
value = (rot_x * (tst_x - pnt_x)) + (rot_y * (tst_y - pnt_y));
cout << "Value: " << value;  

However I'm not quite sure if (a) I've implemented the algorithm I linked to correctly, or if I did exactly how to interpret this?

Any suggestions?

Answer 1

This is from exercise 3.28 from the book Introduction to Java Programming- Comprehensive Edition. The code tests whether the two rectangles are indenticle, whether one is inside the other and whether one is outside the other. If none of these condition are met then the two overlap.

**3.28 (Geometry: two rectangles) Write a program that prompts the user to enter the center x-, y-coordinates, width, and height of two rectangles and determines whether the second rectangle is inside the first or overlaps with the first, as shown in Figure 3.9. Test your program to cover all cases. Here are the sample runs:

Enter r1's center x-, y-coordinates, width, and height: 2.5 4 2.5 43 Enter r2's center x-, y-coordinates, width, and height: 1.5 5 0.5 3 r2 is inside r1

Enter r1's center x-, y-coordinates, width, and height: 1 2 3 5.5 Enter r2's center x-, y-coordinates, width, and height: 3 4 4.5 5 r2 overlaps r1

Enter r1's center x-, y-coordinates, width, and height: 1 2 3 3 Enter r2's center x-, y-coordinates, width, and height: 40 45 3 2 r2 does not overlap r1

import java.util.Scanner;

public class ProgrammingEx3_28 {
public static void main(String[] args) {
    Scanner input = new Scanner(System.in);

    System.out
            .print("Enter r1's center x-, y-coordinates, width, and height:");
    double x1 = input.nextDouble();
    double y1 = input.nextDouble();
    double w1 = input.nextDouble();
    double h1 = input.nextDouble();
    w1 = w1 / 2;
    h1 = h1 / 2;
    System.out
            .print("Enter r2's center x-, y-coordinates, width, and height:");
    double x2 = input.nextDouble();
    double y2 = input.nextDouble();
    double w2 = input.nextDouble();
    double h2 = input.nextDouble();
    w2 = w2 / 2;
    h2 = h2 / 2;

    // Calculating range of r1 and r2
    double x1max = x1 + w1;
    double y1max = y1 + h1;
    double x1min = x1 - w1;
    double y1min = y1 - h1;
    double x2max = x2 + w2;
    double y2max = y2 + h2;
    double x2min = x2 - w2;
    double y2min = y2 - h2;

    if (x1max == x2max && x1min == x2min && y1max == y2max
            && y1min == y2min) {
        // Check if the two are identicle
        System.out.print("r1 and r2 are indentical");

    } else if (x1max <= x2max && x1min >= x2min && y1max <= y2max
            && y1min >= y2min) {
        // Check if r1 is in r2
        System.out.print("r1 is inside r2");
    } else if (x2max <= x1max && x2min >= x1min && y2max <= y1max
            && y2min >= y1min) {
        // Check if r2 is in r1
        System.out.print("r2 is inside r1");
    } else if (x1max < x2min || x1min > x2max || y1max < y2min
            || y2min > y1max) {
        // Check if the two overlap
        System.out.print("r2 does not overlaps r1");
    } else {
        System.out.print("r2 overlaps r1");
    }

}
}

Answer 2

Don't think of coordinates as indicating where pixels are. Think of them as being between the pixels. That way, the area of a 2x2 rectangle should be 4, not 9.

bool bOverlap = !((A.Left >= B.Right || B.Left >= A.Right)
               && (A.Bottom >= B.Top || B.Bottom >= A.Top));

Answer 3

struct rect
{
    int x;
    int y;
    int width;
    int height;
};

bool valueInRange(int value, int min, int max)
{ return (value >= min) && (value <= max); }

bool rectOverlap(rect A, rect B)
{
    bool xOverlap = valueInRange(A.x, B.x, B.x + B.width) ||
                    valueInRange(B.x, A.x, A.x + A.width);

    bool yOverlap = valueInRange(A.y, B.y, B.y + B.height) ||
                    valueInRange(B.y, A.y, A.y + A.height);

    return xOverlap && yOverlap;
}

Answer 4

This answer should be the top answer:

if the rectangles overlap then the overlap area will be greater than zero. Now let us find the overlap area:

if they overlap then the left edge of overlap-rect will be the max(r1.x1,r2.x1) and right edge will be min(r1.x2,r2.x2). so the length of the overlap will be min(r1.x2,r2.x2)-max(r1.x1,r2.x1)

so the area will be: area = (max(r1.x1, r2.x1) - min(r1.x2, r2.x2)) * (max(r1.y1, r2.y1) - min(r1.y2, r2.y2))

if area = 0 then they dont overlap. Simple isn't it?

Answer 5

Suppose that you have defined the positions and sizes of the rectangles like this:

My C++ implementation is like this:

class Vector2D
{
    public:
        Vector2D(int x, int y) : x(x), y(y) {}
        ~Vector2D(){}
        int x, y;
};

bool DoRectanglesOverlap(   const Vector2D & Pos1,
                            const Vector2D & Size1,
                            const Vector2D & Pos2,
                            const Vector2D & Size2)
{
    if ((Pos1.x < Pos2.x + Size2.x) &&
        (Pos1.y < Pos2.y + Size2.y) &&
        (Pos2.x < Pos1.x + Size1.x) &&
        (Pos2.y < Pos1.y + Size1.y))
    {
        return true;
    }
    return false;
}

An example function call according to the given figure above:

DoRectanglesOverlap(Vector2D(3, 7),
                    Vector2D(8, 5),
                    Vector2D(6, 4),
                    Vector2D(9, 4));

The comparisons inside the if block will look like below:

if ((Pos1.x < Pos2.x + Size2.x) &&
    (Pos1.y < Pos2.y + Size2.y) &&
    (Pos2.x < Pos1.x + Size1.x) &&
    (Pos2.y < Pos1.y + Size1.y))
                 ↓  
if ((   3   <    6   +   9    ) &&
    (   7   <    4   +   4    ) &&
    (   6   <    3   +   8    ) &&
    (   4   <    7   +   5    ))

Answer 6

Lets say the two rectangles are rectangle A and rectangle B. Let there centers be A1 and B1 (coordinates of A1 and B1 can be easily found out), let the heights be Ha and Hb, width be Wa and Wb, let dx be the width(x) distance between A1 and B1 and dy be the height(y) distance between A1 and B1.
Now we can say we can say A and B overlap: when

if(!(dx > Wa+Wb)||!(dy > Ha+Hb)) returns true