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I need a 3D triangular sphere mesh with uniformly sampled vertices, say V, with a predefined adjacency. Is there a specific way to achieve that?

Zahir
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    What do you mean by predefined adjacency? So you only need the positions of the vertices? Is the adjacency guaranteed to resemble a sphere? – Nico Schertler Jun 19 '17 at 17:26
  • Yes, I need only positions of vertices. I have taken the adjacency from an existing mesh of genus 1, so that should resemble a sphere, I guess. – Zahir Jun 20 '17 at 04:41
  • Spheres have genus 0. So, no. – Nico Schertler Jun 20 '17 at 06:08
  • Possible duplicate of [Make a sphere with equidistant vertices](https://stackoverflow.com/questions/25012737/make-a-sphere-with-equidistant-vertices) – Spektre Jun 20 '17 at 07:34
  • uniformly sampled vertices mean that the vertices count is bound to specific numbers instead of arbitrary. See the duplicate QA for solutions. If you need just almost equidistant vertices use this [simple recursive sphere triangulation](https://stackoverflow.com/a/29139125/2521214) You can add the adjacency info to it really simply. – Spektre Jun 20 '17 at 07:37
  • @Spektre I'm pretty sure that adding the adjacency information is all but simple. What approach did you have in mind? – Nico Schertler Jun 20 '17 at 09:21
  • @NicoSchertler recursive subdivision of triangles (last link of mine)It is just a matter of including the neighbor info into the structure and pass it to the children ... – Spektre Jun 20 '17 at 12:40
  • @Spektre I still fail to see how you can force that subdivision scheme to produce a desired adjacency. That scheme turns a single triangle into four new triangles. And you need to do that to all triangles simultaneously in order to avoid t-junctions. So the only degree of freedom that you have is the number of subdivision steps. You cannot even produce a mesh with a desired number of vertices. – Nico Schertler Jun 20 '17 at 14:10
  • @NicoSchertler as I mentioned before the equidistant constraint make arbitrary number of vertices impossible (at least to my knowledge) – Spektre Jun 20 '17 at 14:18
  • Let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/147178/discussion-between-nico-schertler-and-spektre). – Nico Schertler Jun 20 '17 at 14:55
  • @NicoSchertler yes I agree but still arbitrary number of verticies is not achievable without serious density error. There are series of suitable verticies number for this ... see the [Wiki Geodesic_polyhedron](https://en.wikipedia.org/wiki/Geodesic_polyhedron). So you can still try to set number of verticies but it will round up to nearest achievable number ... – Spektre Jun 20 '17 at 15:17
  • @NicoSchertler The mesh I have is of genus 0. Apologies for the error. I am able to generate approximately equidistant points on the sphere for any arbitrary number of points. But still, the issue is how to impose the adjacency. – Zahir Jun 21 '17 at 11:05

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