# Regular tessellation of a closed surface

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Hi guys!

Is is possible to have a closed surface mesh being completely regular (all vertices having valence 6)?

Regards,
Dženan

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If your mesh has only triangular faces, the answer is no unless your mesh has the same topology as a torus. The reason is Euler's formula, which relates the number of vertices, edges and faces of a mesh to the "genus" - the number of handles of the mesh.

Here's a proof. Let v,e,f be the number of vertices, edges, and faces respectively.

If all vertices are regular, then each vertex is attached to 6 edges, and each edge is attached to 2 vertices. So 2e = 6v, or e = 3v.

For a triangular mesh, each edge belongs to two faces, and each face has three edges, so 2e = 3f. Using the above result, f = 2v.

Now using Euler's formula,
2 - 2g = v - e + f = v - 3v + 2v = 0.
So we must have g = 1, which means one handle (the same topology as a torus).

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Thanks, I didn't think of Euler's formula.

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There are two compact surfaces that have Euler characteristic 0: The torus and the Klein bottle. Of those, only the torus can be embedded in R^3, so Pragma's answer is probably what the OP was looking for. I just thought I would mention the Klein bottle for completeness.

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Well, I originally wanted to ask "can we triangulate surface of a sphere, but having all the vertices regular". The answer is clearly no. I sort of discovered this answer while trying to find such a triangulation - I just needed confirmation.
The reason I needed this - I want to generate refined mesh using subdivision, to see whether I can get away by only handling regular case.

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Quote:
 Original post by dzenanzWell, I originally wanted to ask "can we triangulate surface of a sphere, but having all the vertices regular". The answer is clearly no. I sort of discovered this answer while trying to find such a triangulation - I just needed confirmation. The reason I needed this - I want to generate refined mesh using subdivision, to see whether I can get away by only handling regular case.

The most you can get with regularity is an icosahedron (12 vertices of valence 5). You can then subdivide each triangular face into little triangles and get a mesh where most vertices have valence 6 and 12 of them have valence 5.

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Yes, I also found that as one of the approximate solutions.

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