How do determine if a polygon is complex/convex/nonconvex?

Question

From the man page for XFillPolygon

· If shape is Complex, the path may self-intersect. Note that con‐ tiguous coincident points in the path are not treated as self- intersection.

· If shape is Convex, for every pair of points inside the polygon, the line segment connecting them does not intersect the path. If known by the client, specifying Convex can improve performance. If you specify Convex for a path that is not convex, the graphics results are undefined.

· If shape is Nonconvex, the path does not self-intersect, but the shape is not wholly convex. If known by the client, specifying Nonconvex instead of Complex may improve performance. If you specify Nonconvex for a self-intersecting path, the graphics results are undefined.

I am having performance problems with fill XFillPolygon and as the man page suggests the first step I want to take is to specify the correct shape of the Polygon ( I am currently using Complex to be on the safe side).

Is there an effecient algorithm to determine if a polygon (defined by a series of coordinates) is convex, non convex or complex.

Answer 1

Convex is fairly easy to test for: Consider each set of three points along the polygon. If every angle is 180 degrees or less you have a convex polygon. This test runs in O(n) time.

Note, also, that in most cases this calculation is something you can do once and save--most of the time you have a set of polygons to work with that don't go changing all the time.

Some more information on your data might help.

Edit: Peter Kirkham pointed out a polygon that this test won't catch. It's easy enough to catch his polygon, also: When you figure out each angle, also keep a running total of (180 - angle). For a convex polygon this will total 360.

Answer 2

You can make things a lot easier than the Gift-Wrapping Algorithm... that's a good answer when you have a set of points w/o any particular boundary and need to find the convex hull.

A polygon is a set of points in a list where the consecutive points form the boundary. It is much easier to figure out whether a polygon is convex or not (and you don't have to calculate any angles, either):

For each consecutive pair of edges of the polygon (each triplet of points), compute the dot product of the vectors defined by the edges pointing towards the points in increasing order. Take the dot product of these vectors:

 given p[k], p[k+1], p[k+2] each with coordinates x, y:
 dx1 = x[k+1]-x[k]
 dy1 = y[k+1]-y[k]
 dx2 = x[k+2]-x[k+1]
 dy2 = y[k+2]-y[k+1]
 dotproduct = dx1*dx2 + dy1*dy2

The polygon is convex if the dot products are either all positive or all negative. Otherwise the polygon is nonconvex.

If there are N points, make sure you calculate N dot products, e.g. be sure to use the triplets (p[N-2],p[N-1],p[0]) and (p[N-1],p[0],p[1]).