Checking for intersection points between two rectangles?

Question

If I have two rectangles whose positions are deifned using two 2D vectors (i.e. top left, bottom right) how can I check for the points they are intersecting?

Answer 1

I assume you actually want the result of the intersection, not only the test if both rectangles intersect.

The intersection of rect1 = (l1, t1, r1, b1) and rect2 = (l2, t2, r2, b2) is again a rectangle:

rectIntersection = ( max(l1, l2), max(t1, t2), min(r1, r2), min(b1, b2) )

rectIntersection is of course empty if left >= right || top >= bottom assuming a rectangle is left/top-inclusive and right/bottom-exclusive.

The rectangles intersect if

l1 < r2 && l2<r1 && t1<b2 && t2<t1

Answer 2

Assuming the origin is on left-top of the screen.

If Check if the the top left of one rectangle (x3,y3) is lesser than the bottom right of another rectangle (x2,y2) then the two are intersecting.

The points are (x2,|y2-y3|) and (|x2-x3|,y2).

This is for a rectangle1 and reactangle2 to the right of rectangle1.

Apply inverse to the left translation.

Answer 3

Two rectangles overlap is there is at least one interior point X,Y common to both. Let the first rectable be {T1, L1, B1, R1} and the second {T2, L2, B2, R2} (top, left, bottom, right). Now it follows that (X>L1) and (X<R1) and (Y>T1) and (Y<B1), and similarly for rectangle 2. From (X>L1) and (X<R2) it follows that (L1<R2). Similarly, (L2<R1), (T1<B2) and (T2<B1).

These 4 conditions are thus necessary. It's not directly obvious that they are also sufficient, but that too is the case.

Answer 4

If you are interested more to a function to do the job, than to implement an algorithm,
check out IntersectRect Function on Windows.