Two Rectangles intersection - algorithm

Question

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?

Answer 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

Answer 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..

Answer 3

Should the two rectangles have the same dimensions you can do:
if (abs (x1 - x2) < w && abs (y1 - y2) < h) {
// overlaps
}

Answer 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;
}

Answer 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

Answer 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

Answer 7

Circle approach is more straightforward. I mean when you define a circle as a center point and radius. Same thing here except you have a horizontal radius (width / 2) and a vertical one (height /2) and 2 conditions for horizontal and for vertical distance.
abs(cx1 – cx2) <= hr1 + hr2 && abs(cy1 - cy2) <= vr1 + vr2
If you need to exclude the case with sides not intersecting filter out these with one rectangle smaller in both dimensions and not enough distance (between centers) from the bigger one to reach one of its edges.
abs(cx1 – cx2) <= hr1 + hr2 && abs(cy1 - cy2) <= vr1 + vr2 &&
!(abs(cx1 – cx2) < abs(hr1 - hr2) && abs(cy1 - cy2) < abs(vr1 - vr2) && sign(hr1 - hr2) == sign(vr1 – vr2))

Answer 8

Rectangle = namedtuple('Rectangle', 'x y w h')
def intersects(rect_a: Rectangle, rect_b: Rectangle):
if (rect_a.x + rect_a.w < rect_b.x) or (rect_a.x > rect_b.x + rect_b.w) or (rect_a.y + rect_a.h < rect_b.y) or (rect_a.y > rect_b.y + rect_b.h):
return False
else:
return True

Answer 9

if( X1<=X2+W2 && X2<=X1+W1 && Y1>=Y2-H2 && Y2>=Y1+H1 )
Intersect
In the question Y is the top position..
Note: This solution works only if rectangle is aligned with X / Y Axes.