问题
I have a small set of N points in the plane, N < 50.
I want to enumerate all triples of points from the set that form a triangle containing no other point.
Even though the obvious brute force solution could be viable for my tiny N, it has complexity O(N^4).
Do you know a way to decrease the time complexity, say to O(N³) or O(N²) that would keep the code simple ? No library allowed.
Much to my surprise, the number of such triangles is pretty large. Take any point as a center and sort the other ones by increasing angle around it. This forms a star-shaped polygon, that gives N-1 empty triangles, hence a total of Ω(N²). It has been shown that this bound is tight [Planar Point Sets with a Small Number of Empty convex Polygons, I. Bárány and P. Valtr].
In the case of points forming a convex polygon, all triangles are empty, hence O(N³). Chances of a fast algorithm are getting low :(
回答1:
The paper "Searching for empty Convex polygons" by Dobkin, David P. / Edelsbrunner, Herbert / Overmars, Mark H. contains an algorithm linear in the number of possible output triangles for solving this problem.
A key problem in computational geometry is the identification of subsets of a point set having particular properties. We study this problem for the properties of convexity and emptiness. We show that finding empty triangles is related to the problem of determininng pairs of vertices that see each other in starshaped polygon. A linear time algorithm for this problem which is of independent interest yields an optimal algorithm for finding all empty triangles. This result is then extended to an algorithm for finding empty convex r-gons (r > 3) and for determining a largest empty convex subset. Finally, extensions to higher dimensions are mentioned.
回答2:
The sketch of the algorithm by Dobkin, Edelsbrunner and Overmars goes as follows for triangles:
for every point in turn, build the star-shaped polygon formed by sorting around it the points on its left. This takes N sorting operations (which can be lowered to total complexity O(N²) via an arrangement, anyway).
compute the visibility graph inside this star-shaped polygon and report all the triangles that are formed with the given point. This takes N visibility graph constructions, for a total of M operations, where M is the number of empty triangles.
Shortly, from every point you construct every empty triangle on the left of it, by triangulating the corresponding star-shaped polygon in all possible ways.
The construction of the visibility graph is a special version for the star-shaped polygon, which works in a traversal around the polygon, where every vertex has a visibility queue which gets updated.
The figure shows a star-shaped polygon in blue and the edges of its visibility graph in orange. The outline generates 6 triangles, and inner visibility edges 8 of them.
回答3:
for each pair of points (A, B):
for each of the two half-planes defined by (A, B):
initialize a priority queue Q to empty.
for each point M in the half plane,
with increasing angle(AB, AM):
if angle(BA, BM) is smaller than all angles in Q:
print A,B,M
put M in Q with priority angle(BA, BM)
Inserting and querying the minimum in a priority queue can both be done in O(log N) time, so the complexity is O(N^3 log N) this way.
回答4:
If I understand your questions, what you're looking for is https://en.wikipedia.org/wiki/Delaunay_triangulation
To quote from said Wikipedia article: "The most straightforward way of efficiently computing the Delaunay triangulation is to repeatedly add one vertex at a time, retriangulating the affected parts of the graph. When a vertex v is added, we split in three the triangle that contains v, then we apply the flip algorithm. Done naively, this will take O(n) time: we search through all the triangles to find the one that contains v, then we potentially flip away every triangle. Then the overall runtime is O(n2)."
来源：https://stackoverflow.com/questions/31249392/finding-all-empty-triangles