No, there (still) is no solution in MATLAB for this. You could write it however. If you do, then post it on the file exchange. It should be doable. I might first start by computing the interior angle for each vertex. Any vertex in the polygon that has an interior angle that exceeds 180 degrees MUST be a vertex that will help divide the polygon into small pieces, since that vertex cannot be part of a convex polygon.
Next, count the number of such vertices.
- Consider if a polygon has NO interior angles that exceed 180 degrees. If so, then it is convex, and you are done.
- If there is only one such vertex, then connect that vertex to ANY other vertex that is not directly connected to the vertex in question. This forms a splitting of the polygon into exactly two shapes, each of which is convex, and again you are done.
- If there are exactly 2 such vertices, then you can connect the two. Again, this must form a splitting into two convex polygons.
- With 3 or more such vertices, you can reduce the problem recursively. You should find the number of convex polygons will be related to log2(n), where n is the number of vertices with interior angles that exceed 180 degrees.
For a simple example consider this example (probably invalid as a polygon, because it is a catholic one, i.e., it crosses itself):
px = rand(10,1);px(end+1) = px(1);
py = rand(10,1);py(end+1) = py(1);
intang = mod(atan2d(diff(py),diff(px)),360)
intang =
27.455
260.18
165.2
261.95
32.475
114.91
238.09
96.628
44.567
207.94
find(intang > 180)
ans =
2
4
7
10
I appended the first point to the end, to make it closed. The count of vertices with interior angles that exceed 180 degrees numbers 4.
