Posted: Dec 21, 2012 11:21 am
Related to these polyhedra is List of convex uniform tilings, covering regular and some semiregular ones.
The regular ones are triangular, square, and hexagonal tilings; they have all three transitivities. That article also lists semiregular ones that are face-transitive or vertex-transitive; those two categories also have some edge-transitive ones.
As with their polyhedral counterparts, one can make the semiregular tilings from the regular ones by making pyramids on faces, edge rhombi, kites, and irregular pentagons.
The face-transitive ones I think can be called fair dartboards.
The mathematics behind them is interesting.
One starts with a theorem that mathematician Leonhard Euler had stated around 1750's: Euler's Formula For number of vertices V, edges E, and faces V, a polyhedron has this constraint on them:
V - E + F = 2
More generally, V - E + F = X
where X is a "topological invariant" of the surface, the "Euler characteristic". It is equal to 2(1-g), where g is the "genus" or number of "holes in the surface, or the number of cuts of the surface needed to make it sphere-like. A torus or doughnut shape needs one cut, thus, g = 1 and X = 0 for it. Two spheres need to be joined to make a single sphere-like shape, and that's a negative cut. g = -1 for them and X = 4, the sum of X for the individual spheres.
For a surface where V, E, and F may be infinite, that formula may not be very useful, but if the surface was formed by the tiling of some rectangle, then one can use that rectangle's properties. It has periodic boundary conditions, top edge mapping onto bottom edge and left edge mapping onto right edge. That gives this rectangle a torus-like topology, or g = 1 and X = 0. Thus, for it,
V - E + F = 0
The completely regular case is fairly easy. Let each face have n vertices/edges, and let each vertex have "rank" r faces/edges. Notice that duality interchanges r and n. One can easily show:
{V, E, F} = {1/r, 1/2, 1/n} * X / (1/r + 1/n - 1/2)
Solutions, in form (k0,k1) = (smaller, larger) of (r,n):
Polyhedron: X = 2 > 0
(2,k1) for any k1
(3,3), (3,4), (3,5)
Plane: X = 0
(3,6), (4,4)
Hyperbolic surface (infinite): X < 0
(3,k1), k1 >= 7
(4,k1), k1 >= 5
(k0,k1), k0 >= 5, k1 >= k0
The regular ones are triangular, square, and hexagonal tilings; they have all three transitivities. That article also lists semiregular ones that are face-transitive or vertex-transitive; those two categories also have some edge-transitive ones.
As with their polyhedral counterparts, one can make the semiregular tilings from the regular ones by making pyramids on faces, edge rhombi, kites, and irregular pentagons.
The face-transitive ones I think can be called fair dartboards.
The mathematics behind them is interesting.
One starts with a theorem that mathematician Leonhard Euler had stated around 1750's: Euler's Formula For number of vertices V, edges E, and faces V, a polyhedron has this constraint on them:
V - E + F = 2
More generally, V - E + F = X
where X is a "topological invariant" of the surface, the "Euler characteristic". It is equal to 2(1-g), where g is the "genus" or number of "holes in the surface, or the number of cuts of the surface needed to make it sphere-like. A torus or doughnut shape needs one cut, thus, g = 1 and X = 0 for it. Two spheres need to be joined to make a single sphere-like shape, and that's a negative cut. g = -1 for them and X = 4, the sum of X for the individual spheres.
For a surface where V, E, and F may be infinite, that formula may not be very useful, but if the surface was formed by the tiling of some rectangle, then one can use that rectangle's properties. It has periodic boundary conditions, top edge mapping onto bottom edge and left edge mapping onto right edge. That gives this rectangle a torus-like topology, or g = 1 and X = 0. Thus, for it,
V - E + F = 0
The completely regular case is fairly easy. Let each face have n vertices/edges, and let each vertex have "rank" r faces/edges. Notice that duality interchanges r and n. One can easily show:
{V, E, F} = {1/r, 1/2, 1/n} * X / (1/r + 1/n - 1/2)
Solutions, in form (k0,k1) = (smaller, larger) of (r,n):
Polyhedron: X = 2 > 0
(2,k1) for any k1
(3,3), (3,4), (3,5)
Plane: X = 0
(3,6), (4,4)
Hyperbolic surface (infinite): X < 0
(3,k1), k1 >= 7
(4,k1), k1 >= 5
(k0,k1), k0 >= 5, k1 >= k0