Simplest way to calculate the intersect area of two rectangles

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 🙂

Solutions Collecting From Web of "Simplest way to calculate the intersect area of two rectangles"

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):

http://jsfiddle.net/Lqh3mjr5/

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;