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

I have a very easy solution

let x1,y1 x2,y2 ,l1,b1,l2,be cordinates and lengths and breadths of them respectively

consider the condition ((x2

now the only way these rectangle will overlap is if the point diagonal to x1,y1 will lie inside the other rectangle or similarly the point diagonal to x2,y2 will lie inside the other rectangle. which is exactly the above condition implies.

Answer 2

Easiest way is

/**
 * Check if two rectangles collide
 * x_1, y_1, width_1, and height_1 define the boundaries of the first rectangle
 * x_2, y_2, width_2, and height_2 define the boundaries of the second rectangle
 */
boolean rectangle_collision(float x_1, float y_1, float width_1, float height_1, float x_2, float y_2, float width_2, float height_2)
{
  return !(x_1 > x_2+width_2 || x_1+width_1 < x_2 || y_1 > y_2+height_2 || y_1+height_1 < y_2);
}

first of all put it in to your mind that in computers the coordinates system is upside down. x-axis is same as in mathematics but y-axis increases downwards and decrease on going upward.. if rectangle are drawn from center. if x1 coordinates is greater than x2 plus its its half of widht. then it means going half they will touch each other. and in the same manner going downward + half of its height. it will collide..

Answer 3

For those of you who are using center points and half sizes for their rectangle data, instead of the typical x,y,w,h, or x0,y0,x1,x1, here's how you can do it:

#include <cmath> // for fabsf(float)

struct Rectangle
{
    float centerX, centerY, halfWidth, halfHeight;
};

bool isRectangleOverlapping(const Rectangle &a, const Rectangle &b)
{
    return (fabsf(a.centerX - b.centerX) <= (a.halfWidth + b.halfWidth)) &&
           (fabsf(a.centerY - b.centerY) <= (a.halfHeight + b.halfHeight)); 
}

Answer 4

Ask yourself the opposite question: How can I determine if two rectangles do not intersect at all? Obviously, a rectangle A completely to the left of rectangle B does not intersect. Also if A is completely to the right. And similarly if A is completely above B or completely below B. In any other case A and B intersect.

What follows may have bugs, but I am pretty confident about the algorithm:

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

bool is_left_of(Rectangle const & a, Rectangle const & b) {
   if (a.x + a.width <= b.x) return true;
   return false;
}
bool is_right_of(Rectangle const & a, Rectangle const & b) {
   return is_left_of(b, a);
}

bool not_intersect( Rectangle const & a, Rectangle const & b) {
   if (is_left_of(a, b)) return true;
   if (is_right_of(a, b)) return true;
   // Do the same for top/bottom...
 }

bool intersect(Rectangle const & a, Rectangle const & b) {
  return !not_intersect(a, b);
}

Answer 5

struct Rect
{
    Rect(int x1, int x2, int y1, int y2)
    : x1(x1), x2(x2), y1(y1), y2(y2)
    {
        assert(x1 < x2);
        assert(y1 < y2);
    }

    int x1, x2, y1, y2;
};

bool
overlap(const Rect &r1, const Rect &r2)
{
    // The rectangles don't overlap if
    // one rectangle's minimum in some dimension 
    // is greater than the other's maximum in
    // that dimension.

    bool noOverlap = r1.x1 > r2.x2 ||
                     r2.x1 > r1.x2 ||
                     r1.y1 > r2.y2 ||
                     r2.y1 > r1.y2;

    return !noOverlap;
}

Answer 6

It is easier to check if a rectangle is completly outside the other, so if it is either

on the left...

(r1.x + r1.width < r2.x)

or on the right...

(r1.x > r2.x + r2.width)

or on top...

(r1.y + r1.height < r2.y)

or on the bottom...

(r1.y > r2.y + r2.height)

of the second rectangle, it cannot possibly collide with it. So to have a function that returns a Boolean saying weather the rectangles collide, we simply combine the conditions by logical ORs and negate the result:

function checkOverlap(r1, r2) : Boolean
{ 
    return !(r1.x + r1.width < r2.x || r1.y + r1.height < r2.y || r1.x > r2.x + r2.width || r1.y > r2.y + r2.height);
}

To already receive a positive result when touching only, we can change the "<" and ">" by "<=" and ">=".