TITLE: Simplest way to calculate the intersect area of two rectangles
QUESTION [41 upvotes]: 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 :)

REPLY [74 votes]: 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;