![]() Someone more mathematically gifted is surely able to find a better solution if so, please do comment / add your own answer. I've gone and implemented an algorithm that's a bit shaky but good enough for my purposes I've added it as an answer. You could also draw some triangles using this interactive. You can print off some square dotty paper, or some isometric dotty paper, and try drawing different triangles on it. (It's not even clear to me that the Delaunay triangulation is always a starting point with a path to that solution.) I'm gonna have to give it some more thought. We say that a shape tessellates if we can use lots of copies of it to cover a flat surface without leaving any gaps. It's not immediately obvious to me, though, if this is necessarily leading to the optimal solution. The idea would be to calculate the circumscribed circle for each polygon that results from the removal of a 'candidate' edge - and picking the one that only includes its own points in the resulting polygon. I was wondering about using circumscribed circles, like some Delaunay algorithms use, but haven't tried it yet. Therefor, I need some criterion to select which to keep and which to remove. However, there are many 'removable' edges that can only be removed if some other removable edge is not removed, and vice versa. ![]() Any final solution will have this edge removed. There are some which are obvious, as they are inside a 4-polygon of which all edges cannot be removed - such as the one indicated by the red arrow. Starting from a Delaunay triangulation, I identified all edges that could be removed in a first step. If someone can point me in the right direction, that would be appreciated. In that case the voronoi tesselation would be the sought tessellation. Voronoi tesselations are always convex I was wondering if there is a way to find a second set of points, so, that the original set of points forms the vertices of voronoi cells of the first. I have also seen many approaches to tessellating the plane with polygons of equal shape and without regarding any previously defined points. Can you make another one that looks different Sketch that one too. Use the blue tiles to create a tessellation by rhombuses. I have seen many approaches to optimal splitting up of a given set of points into triangles (in 2-D or simplices in n-D), but none for larger polygons. Use the green tiles to create a tessellation by equilateral triangles. Is there an algorithm to do this? (like Delaunay triangulation, but with polygons) I have a set of points in 2D, that I want to 'triangulate' with the lowest number of convex polygons.
0 Comments
Leave a Reply. |