Talk:Delaunay triangulation

message: I cannot add an algorithm to this page, because I'm the a inventor (discoverer?) of the algorithm.  
The algorithm is highly parallelizable and Self stabilizing.  
The 2-dimensional version is covered in this paper.
The N-dimensional version is proven correct in this document.
The algorithm relies on the property that if the graph is connected and is "locally delaunay", then it is the delaunay triangulation.  By "locally delaunay", I mean that a node's edges are the same as in a delaunay triangulation of a subgraph consisting of (1) the node, (2) its neighbors, and (3) it's neighbor's neighbors.
Mdnahas (talk) 00:41, 5 January 2022 (UTC)
thread_title:  Additional Algorithm 

message: I would like to rewrite
* This maximizes the minimum of all the angles of the triangles in the triangulation
as something like 
* This maximises the smallest angle found in the triangulation between two sides of the same triangle
so it can't mean
* For any of the triangles, this maximises the size of its smallest angle.
Have I got that right? Musiconeologist (talk) 00:01, 2 April 2024 (UTC)
replies: text: Suggestion: "between two sides of the same triangle" is equivalent to "at a vertex", since the angle between edges not of the same triangle must be bigger than within a triangle.  Which is more intuitive? —Tamfang (talk) 03:22, 2 April 2024 (UTC)
replies: text: @Tamfang Roughly equally intuitive, but switching attention to a new detail (the vertices)? "At a vertex" is a more comfortable phrase . . . But I realised the ambiguity still refuses to go away. Namely whether the maximum is for the set of angles at a given vertex, or for the union of all such sets. Musiconeologist (talk) 05:24, 2 April 2024 (UTC)

text: @Musiconeologist Have you tried searching for lay-accessible sources about Delaunay triangulation? This is a popular enough topic that I am guessing you can find some simple explanations that expert mathematicians and/or technical writers spent a lot of effort on clarifying already, and possibly even tested on an intended non-technical audience. –jacobolus (t) 03:51, 2 April 2024 (UTC)
replies: text: @Jacobolus I haven't! I simply got involved earlier this evening, because a random article linked to this one and I was dismayed to see a mass of symbols and unexplained terms instead of a straightforward description that would tell me what it was. Then I saw User:David Eppstein had already objected to the unnecessary notation, so I removed that and discovered the introduction was now comprehensible enough for me start simplifying the language. That's a good suggestion. Musiconeologist (talk) 04:25, 2 April 2024 (UTC)
thread_title:  Ambiguous sentence 

message: At first sight, the logic here needn't be true:
:For four or more points on the same circle . . . the Delaunay triangulation is not unique: each of the two possible triangulations that split the quadrangle into two triangles satisfies the "Delaunay condition" 
The logic assumes that the points are connected as a quadrilateral, and that the quadrilateral has no points inside it and is therefore inevitably divided into only two triangles. But we might have four widely-spaced points on a circle, with chains of multiple edges joining them. Is it known that there's no unique Delaunay triangulation in that situation? Not guaranteed to be unique seems more likely. Musiconeologist (talk) 20:39, 2 April 2024 (UTC)
replies: text: As long as you have a circle with more than three sites, empty of other sites, the Delaunay triangulation is non-unique. All triangulations of the sites on the circle can be part of a triangulation of the whole input that meets the usual definition of a DT (each triangle has an empty circumcenter). The number of triangulations of the sites on the circle is a Catalan number: 2 for four sites, 5 for five sites, 14 for six sites, etc. It is always more than one, for more than three sites. —David Eppstein (talk) 20:49, 2 April 2024 (UTC)
replies: text: We should take "two" out of "each of the two" if we're leading with "four or more" though. –jacobolus (t) 22:19, 2 April 2024 (UTC)
thread_title:  Four points on a circle 

message: The reference provided to show that "The Delaunay triangulation contains O\bigl(n^{\lceil d/2 \rceil}\bigr) simplices." does not give a proof of that. The author claims that another author proved it, but they just provide as reference a private communication, so there is no way to verify this. 150.241.212.145 (talk) 12:00, 15 April 2024 (UTC)
thread_title:  One of the claims contains a poor reference. 

message: One way to generate a D.t. is to map the points to a paraboloid, (x_0,...,x_n) → (x_0,...,x_n,\Sigma{x_i^2}), and take the convex hull; the triangulation is the reverse projection of the "underside" of this hull – but is there a name for, or anything interesting to say about, the upward-facing faces? —Tamfang (talk) 22:05, 15 April 2024 (UTC)
replies: text: They're the farthest-point Delaunay triangulation. —David Eppstein (talk) 22:34, 15 April 2024 (UTC)

text: One other thing probably worth mentioning, while on this topic, is that instead of projecting orthogonally onto a paraboloid we can also take the inverse sterographic projection onto a sphere, which lets us compute the delaunay triangulation of a sphere (which is also the convex hull) in terms of the delaunay triangulation of the plane, or vice versa. This sees use in computational geometry applications on the sphere. –jacobolus (t) 01:24, 16 April 2024 (UTC)
replies: text: That was the way K. Q. Brown originally did it. Nowadays you're more likely to see the orthogonal paraboloid version in textbooks. —David Eppstein (talk) 01:31, 16 April 2024 (UTC)
replies: text: As an aside: if you start with coordinates  in the plane, then the inverse stereographic projection is \left\langle \frac{\tfrac12(1 - x^2 - y^2)}{\tfrac12(1 + x^2 + y^2)}, \frac{x}{\tfrac12(1 + x^2 + y^2)}, \frac{y}{\tfrac12(1 + x^2 + y^2)}\right\rangle and the orthogonal projection to the paraboloid is \left\langle \tfrac12(1 - x^2 - y^2), x, y\right\rangle, so you can also get from the sphere to the paraboloid or vice versa by central projection through the origin. –jacobolus (t) 02:11, 16 April 2024 (UTC)
thread_title:  upside-down