Two Rectangles intersection

问题

I have two rectangles characterized by 4 values each :

Left position X, top position Y, width W and height H:

X1, Y1, H1, W1
X2, Y2, H2, W2

Rectangles are not rotated, like so:

+--------------------> X axis
|
|    (X,Y)      (X+W, Y)
|    +--------------+
|    |              |
|    |              |
|    |              |
|    +--------------+
v    (X, Y+H)     (X+W,Y+H)

Y axis

What is the best solution to determine whether the intersection of the two rectangles is empty or not?

回答1:

if (X1+W1<X2 or X2+W2<X1 or Y1+H1<Y2 or Y2+H2<Y1):
    Intersection = Empty
else:
    Intersection = Not Empty

If you have four coordinates – ((X,Y),(A,B)) and ((X1,Y1),(A1,B1)) – rather than two plus width and height, it would look like this:

if (A<X1 or A1<X or B<Y1 or B1<Y):
    Intersection = Empty
else:
    Intersection = Not Empty

回答2:

Best example..

/**
 * 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);
}

and also one other way see this link ... and code it your self..

回答3:

Should the two rectangles have the same dimensions you can do:

if (abs (x1 - x2) < w && abs (y1 - y2) < h) {
    // overlaps
}

回答4:

I just tried with a c program and wrote below.

#include<stdio.h>

int check(int i,int j,int i1,int j1, int a, int b,int a1,int b1){
    return (\
    (((i>a) && (i<a1)) && ((j>b)&&(j<b1))) ||\ 
    (((a>i) && (a<i1)) && ((b>j)&&(b<j1))) ||\ 
    (((i1>a) && (i1<a1)) && ((j1>b)&&(j1<b1))) ||\ 
    (((a1>i) && (a1<i1)) && ((b1>j)&&(b1<j1)))\
    );  
}
int main(){
    printf("intersection test:(0,0,100,100),(10,0,1000,1000) :is %s\n",check(0,0,100,100,10,0,1000,1000)?"intersecting":"Not intersecting");
    printf("intersection test:(0,0,100,100),(101,101,1000,1000) :is %s\n",check(0,0,100,100,101,101,1000,1000)?"intersecting":"Not intersecting");
    return 0;
}

回答5:

Using a coordinate system where (0, 0) is the left, top corner.

I thought of it in terms of a vertical and horizontal sliding windows and come up with this:

(B.Bottom > A.Top && B.Top < A.Bottom) && (B.Right > A.Left && B.Left < A.Right)

Which is what you get if you apply DeMorgan’s Law to the following:

Not (B.Bottom < A.Top || B.Top > A.Bottom || B.Right < A.Left || B.Left > A.Right)

B is above A
B is below A
B is left of A
B is right of A

回答6:

If the rectangles' coordinates of the lower left corner and upper right corner are :
(r1x1, r1y1), (r1x2, r1y2) for rect1 and
(r2x1, r2y1), (r2x2, r2y2) for rect2
(Python like code below)

    intersect = False
    for x in [r1x1, r1x2]:
        if (r2x1<=x<=r2x2):
            for y in [r1y1, r1y2]:
                if (r2y1<=y<=r2y2):
                    intersect = True
                    return intersect
                else:
                    for Y in [r2y1, r2y2]:
                        if (r1y1<=Y<=r1y2):
                            intersect = True
                            return intersect
        else:  
            for X in [r2x1, r2x2]:
                if (r1x1<=X<=r1x2):
                    for y in [r2y1, r2y2]:
                        if (r1y1<=y<=r1y2):
                            intersect = True
                            return intersect
                        else:
                            for Y in [r1y1, r1y2]:
                                if (r2y1<=Y<=r2y2):
                                    intersect = True
                                    return intersect
    return intersect