This week's book giveaway is in the Mac OS forum. We're giving away four copies of a choice of "Take Control of Upgrading to Yosemite" or "Take Control of Automating Your Mac" and have Joe Kissell on-line! See this thread for details.

Here is a problem I ran into at work with some Web development. I came up with a solution, but I am worndering if you can come up with a slicker method.

Given X1,Y1,W1,H1 and X2,Y2,W2,H2 figure out if two rectangles overlap each other.

If X1+H1 < X2 or X2+H2 < X1 or Y1+W1 < Y2 or Y2+W2 < Y1, they don't overlap (assuming heights and widths are positive). If none of these holds, they overlap. I don't think you can simplify it any further than that.

are we assuming the edges are always parallel/perpendicular to the x and y axis?

There are only two hard things in computer science: cache invalidation, naming things, and off-by-one errors

Eric Pascarello
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Yes for this we are assuming they are parallel/perpendicular.

This is the basic idea I came up with, The first part forces the data to be in the correct orrientation to each other. This is done in JavaScript since that is what I was working with.

I replaced l and h with a and b respectively, because (1) l is a horrible name for a Java variable, and (2) to me names like l, h, w refer to the total height, length, or width. Since we're using a half-width and half height, I thought it better to remove a possible source of confusion. Anyway:

It might be possible to optimize this a bit more (deferring calculations of y and yMin, until they're really needed) but I think that would lose some clarity. [ December 31, 2004: Message edited by: Jim Yingst ]