3 ms·

From the broken first attempt: Each face has 3 vertices, shared between 6 faces, so the total number of vertices is 65536, and they become faces of my die. Thi

by sp332 4y ago

From the broken first attempt: Each face has 3 vertices, shared between 6 faces, so the total number of vertices is 65536, and they become faces of my die. This should have tipped you off. If you have six triangular faces around every vertex, you have a flat Euclidean plane, not a sphere with positive curvature. For another example of this: https://m.youtube.com/watch?v=jfSTwqmrQDc https://m.youtube.com/watch?v=jfSTwqmrQDc

avmich 4y ago
That's why "don't do math after midnight". But I don't get what's wrong with repeated throws of tetrahedron - 2 bits of result at a time - or even icosahedron, with re-throws if one of 4 "wrong" sides is up - 4 bits at a "good" throw...
- thaumasiotes 4y ago
  You can also make repeated throws of an octahedron for 3 bits per throw. 16 is not a multiple of 3, but there's no difficulty with just discarding the 17th and 18th bits. So you'd need to roll 6d8 instead of 8d4.
tgb 4y ago
Pedantry: it doesn’t have to be flat - for example a triangulated parabola could also have that configuration. You only get a topological result from just knowing edge and vertex counts. Now if the triangles are identical equilateral then you’re in business. What if the triangles are all congruent but not equilateral? Can that even happen? That’s a fun one, so I won’t spoil it.
- sp332 4y ago
  Oh, do you think it could make a sphere then? If the angle at each triangle corner is a bit less than 60 degrees?
  - tgb 4y ago
    Not a sphere, we know that can’t happen due to the topological constraint you brought up. Instead picture a float plane tiled by equilateral triangles. Now mark each vertex as either low, middle, or high such that every triangle has one of each. Push the low ones below the plane and the high vertices above the plane by some amount x. Now it’s a bumpy plane full of triangles, each one is isosceles and all identical. I think that’s the only way to do this, but maybe there are more. Could we get a hyperbolic plane this way? Normally you squeeze extra triangles around each vertex to do that so I doubt it but maybe.

by sp332 4y ago

avmich 4y ago

That's why "don't do math after midnight". But I don't get what's wrong with repeated throws of tetrahedron - 2 bits of result at a time - or even icosahedron, with re-throws if one of 4 "wrong" sides is up - 4 bits at a "good" throw...

thaumasiotes 4y ago
You can also make repeated throws of an octahedron for 3 bits per throw. 16 is not a multiple of 3, but there's no difficulty with just discarding the 17th and 18th bits. So you'd need to roll 6d8 instead of 8d4.

tgb 4y ago

Pedantry: it doesn’t have to be flat - for example a triangulated parabola could also have that configuration. You only get a topological result from just knowing edge and vertex counts. Now if the triangles are identical equilateral then you’re in business. What if the triangles are all congruent but not equilateral? Can that even happen? That’s a fun one, so I won’t spoil it.

sp332 4y ago
Oh, do you think it could make a sphere then? If the angle at each triangle corner is a bit less than 60 degrees?
- tgb 4y ago
  Not a sphere, we know that can’t happen due to the topological constraint you brought up. Instead picture a float plane tiled by equilateral triangles. Now mark each vertex as either low, middle, or high such that every triangle has one of each. Push the low ones below the plane and the high vertices above the plane by some amount x. Now it’s a bumpy plane full of triangles, each one is isosceles and all identical. I think that’s the only way to do this, but maybe there are more. Could we get a hyperbolic plane this way? Normally you squeeze extra triangles around each vertex to do that so I doubt it but maybe.