PNPOLY

https://www.ecse.rpi.edu/Homepages/wrf/Research/Short_Notes/pnpoly.html

The C Code

Here is the code, for reference. Excluding lines with only braces, there are only 7 lines of code.

int pnpoly(int nvert, float *vertx, float *verty, float testx, float testy)
{
  int i, j, c = 0;
  for (i = 0, j = nvert-1; i < nvert; j = i++) {
    if ( ((verty[i]>testy) != (verty[j]>testy)) &&
	 (testx < (vertx[j]-vertx[i]) * (testy-verty[i]) / (verty[j]-verty[i]) + vertx[i]) )
       c = !c;
  }
  return c;
}

Argument	Meaning
nvert	Number of vertices in the polygon. Whether to repeat the first vertex at the end is discussed below.
vertx, verty	Arrays containing the x- and y-coordinates of the polygon's vertices.
testx, testy	X- and y-coordinate of the test point.

The Method

I run a semi-infinite ray horizontally (increasing x, fixed y) out from the test point, and count how many edges it crosses. At each crossing, the ray switches between inside and outside. This is called the Jordan curve theorem.

The case of the ray going thru a vertex is handled correctly via a careful selection of inequalities. Don't mess with this code unless you're familiar with the idea ofSimulation of Simplicity. This pretends to shift the ray infinitesimally down so that it either clearly intersects, or clearly doesn't touch. Since this is merely a conceptual, infinitesimal, shift, it never creates an intersection that didn't exist before, and never destroys an intersection that clearly existed before.

The ray is tested against each edge thus:

1. Is the point in the half-plane to the left of the extended edge? and
2. Is the point's Y coordinate within the edge's Y-range?

Handling endpoints here is tricky.