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Consider sequences of tesselations of the sphere. For instance, one such sequence might start with an icosahedron and proceed by subdividing each triangle face into 4 triangles and projecting the new vertices onto the sphere.

Some sequences of tesselations converge to the sphere in the sense that the maximum distance between the polyhedra and the sphere converges to 0. Some sequences possibly converge more strongly if the density of vertices converges to the uniform distribution over the sphere (this is not the case with the sequence offered as an example above).

  1. Are there convergent sequences of tesselations where the number of distinct triangles involved in each tesselation is bounded for the whole sequence, and if so, what is the lowest bound?

Rephrasing: is there an integer $k$ such that $\forall \epsilon > 0$ there is a polyhedron with triangular faces which approximate a unit sphere within $\epsilon$ and has at most $k$ distinct faces.

  1. If so, do some of these sequences also let the density of vertices approach the uniform distribution?

Clarification: we are looking at the number of distinct triangles. For example, an isocahedron has exactly 1 distinct triangle.

Further clarification: the rephrasing above is not quite accurate. As commenters point out, a simple rasterization of the sphere yields arbitrarily good approximations with just one triangle.

We are looking for polyhedra where the vertices lie on the sphere itself or, at least, convex polyhedra.

Arthur B
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  • It seems like you're asking several questions here. Am I right to think your main one is along the lines of "for a fixed distance tolerance $\epsilon$, what subdivision strategy approximates the sphere to within tolerance with the fewest triangles?" (possibly with the addendum "if there are several, which one is most uniform?"). – DCM Nov 14 '20 at 13:07
  • I think the "icosahedron followed by recursive subdivision" strategy isn't very efficient in this respect - I think it's better to start with the icosahedron then do the subdivision 'in one shot' rather than iteratively. One thing I'm curious about: is your ultimate aim to sample from a uniform distribution on the sphere, or to produce a triangulation? – DCM Nov 14 '20 at 13:17
  • Not with the fewest triangles, with the fewest distinct triangles. An isocahedron for instance has exactly 1 distinct triangle. – Arthur B Nov 14 '20 at 15:14
  • You could also phrase the first question as: Is there a finite set of triangles from which you can construct polyhedra that approximate spheres arbitrarily well? –  Nov 15 '20 at 01:28
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    That would be a slightly stronger claim. I'm open to the finite set changing as the degree of approximation gets better, so long as the cardinal stays the same. I can imagine situations where a triangle in the set has to become infinitely thin in the limit for instance. – Arthur B Nov 15 '20 at 06:32
  • Got it. I like the questions, and seeing no obvious constructions I’d expect negative answers, but I have no ideas on how to prove that. –  Nov 15 '20 at 08:24
  • @MattF. The stronger question you ask (“Is there a finite set of triangles from which you can […] approximate spheres arbitrarily well?”) has a negative answer, because a given finite set of spherical triangles can only be arranged in finitely many ways to tile the (finite-area!) sphere. – Gro-Tsen Nov 24 '20 at 10:12
  • @Gro-Tsen I don't get that counter argument. The set of triangles from which you build may be finite but we can use as many copies of each kind as you want. – Arthur B Nov 24 '20 at 10:50
  • These are flat triangles, not spherical triangles, so the set of triangles used doesn't necessarily determine the radius of the sphere they can approximate. – Arthur B Nov 24 '20 at 10:56
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    Here’s a negative answer to my question: If there are only finitely many triangles, then each vertex of the polyhedron has only finitely many possible arrangements. (Eg, we can’t fit more than five equilateral triangles around a single vertex.) For each of those arrangements, there’s a greatest distance from the faces at the vertex to theor circumscribing sphere (or greatest ratio of that distance to the radius of the circumscribing sphere). Thus the finite set of triangles can’t be used to approximate the sphere any better than the least of those greatest distances –  Nov 24 '20 at 12:14
  • Do you demand that the polyhedron be convex? Otherwise, there's a really boring $k = 1$ solution (for the weaker notion of convergence): rasterize the sphere by approximating it as the union of many tiny cubes, and then erect a shallow square-based pyramid on every square face that results. – Adam P. Goucher Nov 24 '20 at 15:16
  • Ah, Henry Segerman just made a similar point on Twitter. https://twitter.com/henryseg/status/1331254177952133126?s=19. What I had in mind was in fact stronger than convexity (I think) the vertices should lie on the sphere. – Arthur B Nov 24 '20 at 15:25
  • Somewhat analogous question to your $k=1$ case, with some partial answers. – Gro-Tsen Jan 20 '24 at 09:08

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