Am I trying to do the impossible? (3d guru's apply)

[rachil0]

I think euler's formula for polyhedra indicates it can't be done, at least not with a texture that's a regular tiling of hexagons (ie each vertex having valence 3).

For any genus G polyhedron you must have V-E+F=2-2G, where V is the number of vertices, E is the number of edges and F is the number of facets. The sphere (and any polyhedron homeomorphic to a sphere) has genus 0.

But if you consider the regular tiling of hexagons in 2D, if you have F hexagons then you have 6F/2 edges (each hex has 6 edges but we double counted them because each edge is shared among two hexes). Also, the number of vertices is 6*F/3, each hex has 6 vertices and each vertex is shared 3 ways. Plug these numbers into eulers formula:

V-E+F = 2-2G
6F/3-6F/2+F = 2-2G
2F-3F+F = 2-2G
0 = 2-2G
G= 1

So when you tile regular hexagons like that you can only make genus 1 shapes - a torus. You can envision this tiling by taking the texture and wrapping it up into a cylinder, then rolling that cylinder back onto itself. You won't be able to texture or facetize a sphere seamlessly with hexagons unless you have some nodes with valence!=3. (Vertices like that are called extraordinary points in subdivision surfaces - in the infinitesimal limit they have weaker curvature guarantees that the other spots on the surface).

// brain.explode();

[deezbucks]

actually the euler characteristic of a sphere (or any surface homeomorphic to S^2) is 2. you are mistaking the euler characteristic of a torus of genus 1 (i.e. X(T) = 2 - 2g = 2-2(1) = 0).

for regular hexagons your counterexample argument still applies but you will arrive at the step 0 = 2. if the hexagons are not regular, the problem is slightly more complicated (find a relationship e and f, then use the given condition to put an upper bound on v in terms of f. This will
give you an upper bound on the euler characteristic, which will gie a contradiction), but it is still impossible to subdivide the surface of a sphere (or any surface homeomorphic to S^2) with 6-gons regular or not.

there are only 5 regular polyhedra, and the proof arises from a basic combinatorial argument. this can be found on google.

I think euler's formula for polyhedra indicates it can't be done, at least not with a texture that's a regular tiling of hexagons (ie each vertex having valence 3).

For any genus G polyhedron you must have V-E+F=2-2G, where V is the number of vertices, E is the number of edges and F is the number of facets. The sphere (and any polyhedron homeomorphic to a sphere) has genus 0.

But if you consider the regular tiling of hexagons in 2D, if you have F hexagons then you have 6F/2 edges (each hex has 6 edges but we double counted them because each edge is shared among two hexes). Also, the number of vertices is 6*F/3, each hex has 6 vertices and each vertex is shared 3 ways. Plug these numbers into eulers formula:

V-E+F = 2-2G
6F/3-6F/2+F = 2-2G
2F-3F+F = 2-2G
0 = 2-2G
G= 1

So when you tile regular hexagons like that you can only make genus 1 shapes - a torus. You can envision this tiling by taking the texture and wrapping it up into a cylinder, then rolling that cylinder back onto itself. You won't be able to texture or facetize a sphere seamlessly with hexagons unless you have some nodes with valence!=3. (Vertices like that are called extraordinary points in subdivision surfaces - in the infinitesimal limit they have weaker curvature guarantees that the other spots on the surface).

// brain.explode();