How to efficiently determine if a polygon is convex, non-convex or complex

algorithmcomputational-geometrygeometrypolygonxlib

From the man page for XFillPolygon:

  • If shape is Complex, the path may self-intersect. Note that contiguous 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 efficient algorithm to determine if a polygon (defined by a series of coordinates) is convex, non-convex or complex?

Best Solution

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.

In contrast, consider the case where the polygon is not self-intersecting, and it consists of a set of points in a list where the consecutive points form the boundary. In this case 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 z-component of the cross product of the vectors defined by the edges pointing towards the points in increasing order. Take the cross 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]
 zcrossproduct = dx1*dy2 - dy1*dx2

The polygon is convex if the z-components of the cross products are either all positive or all negative. Otherwise the polygon is nonconvex.

If there are N points, make sure you calculate N cross 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]).


If the polygon is self-intersecting, then it fails the technical definition of convexity even if its directed angles are all in the same direction, in which case the above approach would not produce the correct result.