Simplest way to calculate the intersect area of two rectangles

$\begingroup$

I have a problem where I have TWO NON-rotated rectangles (given as two point tuples {x1 x2 y1 y2}) and I like to calculate their intersect area. I have seen more general answers to this question, e.g. more rectangles or even rotated ones, and I was wondering whether there is a much simpler solution as I only have two non-rotated rectangles.

What I imagine should be achievable is an algorithm that only uses addition, subtraction and multiplication, possibly abs() as well. What certainly should not be used are min/max, equal, greater/smaller and so on, which would make the question obsolete.

Thank you!

EDIT 2: okay, it's become too easy using min/max or abs(). Can somebody show or disprove the case only using add/sub/mul?

EDIT: let's relax it a little bit, only conditional expressions (e.g. if, case) are prohibited!

PS: I have been thinking about it for a half hour, without success, maybe I am now too old for this :)

$\endgroup$ 2

1 Answer

$\begingroup$

Uses only max and min (drag the squares to see the calculation. Forget about most of the code, the calculation is those two lines with the min and max):

You can also reduce min to max here (or the opposite), i.e. $min\{a,b\} = -max\{-a,-b\}$.

---

First compute the bounding rectangles rect1 and rect2 with the following properties:

rect = { left: x1, right: x1 + x2, top: y1, bottom: y1 + y2,
}

The overlap area can be computed as follows:

x_overlap = Math.max(0, Math.min(rect1.right, rect2.right) - Math.max(rect1.left, rect2.left));
y_overlap = Math.max(0, Math.min(rect1.bottom, rect2.bottom) - Math.max(rect1.top, rect2.top));
overlapArea = x_overlap * y_overlap;

$\endgroup$ 4

Your Answer

Sign up or log in

Sign up using Google Sign up using Facebook Sign up using Email and Password Submit

Post as a guest

Post Your Answer Discard

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy