A Point About Polygons
The test program (Listing 2) draws a random 40-sided polygon and then picks random points to throw at the inpoly() routine. Points the routine says are inside the polygon it draws red, points outside are blue.
Our first rendition of inpoly() had a subtle flaw which the test program made evident. The full story contains an embarrassing lesson. “It'll work,” we sneered, “We don't need to waste time on a full graphical test. Besides, it'd be too much fun.” After we found out our image maps had leaks, we wrote the test program. Figure 3 shows a close-up of the flaw.
Along a vertical line, all the colors are wrong. The flaw turned out to be that when our mindless mollusk crosses the bottom corner, the little hummer was counting the crossing of both edges! After that, he was always exactly wrong—he thought he was in when he was out, and he thought he was out when he was in. The solution must ensure that when our esteemed escargot crosses into the polygon corner, he counts exactly one crossing. Two is no good, and in fact, zero is just as bad—one is what we need. The reason the flaw in the close-up extends up from the corner is that the positive Y axis extends downward in screen coordinates.
I suspect this is a problem unique to the fixed-point world. I'm sure my fellow point-in-polygon smiths have either lucked out or dealt with it somehow. At least, I'd like to think so. (A lie-detector would peg me on that one. This article would be insufferably smug if I had found leaky corners in any of the other algorithms.) In my case, I realized I could not blindly count all crossings of the end point of each of the edges as a crossing. My first thought was to associate each end point with one—and only one—edge. This sounds fair and equitable, but like many things fitting that description, it just plain won't work. A problem turns up when Agent Snail just lightly nicks the corner of a polygon he's not inside at all. That's counted as one crossing, hence the snail report is bunk.
Since I abhor special cases, I sought something that would work in all cases.
The scheme for getting our faithful friend to count corner crossings correctly is to always count a crossing of the right end of each edge, but never the left end (right meaning positive X). In the figure, the black circles represent points our snail will count if he crosses; the white circles he won't count. When you put the polygon together, everything ends up the way we want. Nicking the corner means he counts either 0 crossings or two crossings. We don't care which; both are even and our snail knows he's outside. The circles with ones in them represent points counted once if the snail crosses them. This is fine, just like crossing the nearby sides.
It's time to analyze the guts of the inpoly() routine in Listing 3. This represents a slight modification of the snail's instructions. He plays a bit of a “she loves me, she loves me not” kind of game rather than counting up the crossings and then reporting whether the total is even or odd. He starts out assuming he's outside, and complements that assumption with each crossing. So much for the inside=!inside statement.
This if test happens inside a for loop that considers all of the edges of the polygon, one at a time. Each edge is a line segment that stretches between the corners (xold,yold) and (xnew,ynew). We've arranged it so (x1,y1) and (x2,y2) also represent the same edge, but the points are swapped, if necessary, to make it so x1 <= x2.
Now two things must be true for our ever-meticulous snail to count the crossing of this edge. First, the segment must straddle the Y axis (where the right end is counted but the left one is not). Second, straddling has to happen to the north of the snail's starting point. These are exactly the questions determined by the if statement's two pieces, on either side of the &&.
Now that first expression is a sneaky one, and I confess I might have preferred the less opaque code (x1 < xt && xt <= x2). You can see it does the same thing if you look carefully (very carefully—I was fooled for a while there). But I hate to fix something unless I've already broken it, if you know what I mean.
That north computation is the one I'm proud of because none of my esteemed fellow polygon smiths made one that doesn't need a divide. It does depend on the knowledge that (x2-x1) is positive. Other than that, it's just a transmogrification of that famous y=mx+b equation from high school algebra.
By the way, I've left out the case where an edge line segment stands straight up and down above the snail touchdown point. Such an edge would never be counted by Mr. Snail at all! That's because the == test would always be false, since xnew, xt and xold are all the same value. What's really wild is that's just what we want. In a sense, he's crossing three edges when we only want to count one. It turns out the adjacent line segment crossings are all we're interested in, and the rules already discussed work perfectly for them.
|Red Hat Enterprise Linux 7.1 beta available on IBM Power Platform||Jan 23, 2015|
|Designing with Linux||Jan 22, 2015|
|Wondershaper—QOS in a Pinch||Jan 21, 2015|
|Ideal Backups with zbackup||Jan 19, 2015|
|Non-Linux FOSS: Animation Made Easy||Jan 14, 2015|
|Internet of Things Blows Away CES, and it May Be Hunting for YOU Next||Jan 12, 2015|
- Designing with Linux
- Wondershaper—QOS in a Pinch
- Red Hat Enterprise Linux 7.1 beta available on IBM Power Platform
- Ideal Backups with zbackup
- Internet of Things Blows Away CES, and it May Be Hunting for YOU Next
- Slow System? iotop Is Your Friend
- New Products
- Purism Librem 15
- diff -u: What's New in Kernel Development
- Hats Off to Mozilla
Editorial Advisory Panel
Thank you to our 2014 Editorial Advisors!
- Jeff Parent
- Brad Baillio
- Nick Baronian
- Steve Case
- Chadalavada Kalyana
- Caleb Cullen
- Keir Davis
- Michael Eager
- Nick Faltys
- Dennis Frey
- Philip Jacob
- Jay Kruizenga
- Steve Marquez
- Dave McAllister
- Craig Oda
- Mike Roberts
- Chris Stark
- Patrick Swartz
- David Lynch
- Alicia Gibb
- Thomas Quinlan
- Carson McDonald
- Kristen Shoemaker
- Charnell Luchich
- James Walker
- Victor Gregorio
- Hari Boukis
- Brian Conner
- David Lane