Up spake bulia byak:
(approximates (quotient (sum (n in 0 to (product 2 Pi)) ;; interval is 1 (radian) (sum (m in 0 to (product 2 Pi)) ;; interval is 1 (radian) Anm)) (product 2 Pi 2 Pi)) 0)
That's rather messy, slow and probably incorrect.
If you can code this in C/C++ using our livarot/Path.h class which...
(Yeah, I realize it's the user list, and I was recently chastised for proposing to make a patch here, but since you already wrote some code I though it would be OK in this case :)
It's OK to ask, but there's *NO* way I can do anything that involved (observe that I have mediocre C and ~ no C++ experience, and am not familiar with any of the code base). Sorry.
Also note that the above algorithm doesn't suggest any way of picking a point to test. Maybe you could start at the bboux center, then head in the `heaviest' direction, but I dont know if that would work reliably for concave shapes.
-trent PS: The above `code' is about as far from imperative code as the limit of a geometric series.