General honeycombs?

[Marek14]

I was wondering whether my approach to 2D tesselations could be generalized into 3D.

For 2D, I can take a given polygon sequence and compute the number and forms of uniform tesselations with that particular vertex figure. But when this goes to 3D, there is a problem: a polygon with given edge lengths can be always hammered to fit into a circle, and thus serve as a vertex figure, but a polyhedron with given edge lengths doesn't necessarily fit into a sphere.

Once I have a vertex figure that DOES fit into a sphere, plugging results from my 2D research into it should form a valid 3D honeycomb (although for more interesting results, it would be usually a honeycomb with mostly infinite hyperbolic cells).

Tetrahedron will obviously fit into a sphere, no matter what. But what about next simplest polyhedra -- triangular dipyramids and quadrangular pyramids with arbitrary edge lengths (though restricted to regular polygon chords)? Is there a simple way to determined which combinations fit and which don't? And do all formations that don't fit a sphere exist in some non-uniform spaces where curvature depends on direction?

[Klitzing]

At least for the quadrangular pyramid the answer is easy:
you just have the restriction that the (planar) quadrangle has to have a circum circle.
Then any tip atop - what so ever - would define an according circum sphere.

--- rk

[Marek14]

At least for the quadrangular pyramid the answer is easy:
you just have the restriction that the (planar) quadrangle has to have a circum circle.
Then any tip atop - what so ever - would define an according circum sphere.

--- rk

Hm, that opens some possibilities...

For uniform honeycombs, all four of the side edges must correspond to even polygons, but otherwise there seems to be a lot of options.

EDIT: Except that some pyramids might still be excluded because of impossible combination of side edges? Or is any combination of chords permissible?

The quadrangles that work for uniform tesselations are:

AAAA - a arbitrary - all four side edges must be the same. All four triangles are ASS.

AAAB - antiprismatic form - a divisible by 3 - a/a and a/b side edges could be same or different.
same: side triangles ASS, ASS, ASS, BSS
different: side triangles AST, ASS, AST, BTT - a must be even for this type to fit.

AAAB - hex form - a divisible by 6 - thinking about it, a hexagon in (6,6,6,3) has three different kinds of edges, so there are three possible side lengths here:
ASU, AST, ATU, BUU
Then we have three options with two side lengths:
AST, ASS, AST, BTT
ASS, AST, AST, BSS
AST, AST, ATT, BTT
And standard ASS, ASS, ASS, BSS with one side length.

Both these forms of AAAB have a special case with a=b, further options for AAAA when a is divisible by 6.

AABB - Three possible side lengths: a/a, a/b and b/b
AST, AST, BTU, BTU
With two-length options:
ASS, ASS, BST, BST
AST, AST, BST, BST
AST, AST, BTT, BTT (same as first option, just with A and B switched).
And one-length:
ASS, ASS, BSS, BSS

if a=b, additional options for AAAA with even a appear.

ABAB - only one type of edges, so only one possible side length: ASS, BSS, ASS, BSS

AABC - mixed form with a divisible by 4 and b and c even. a/a, a/b, b/c and a/c side edges could be all different. Leads to AAAB mixed type (for a = b or a = c), additional AABB options (for b = c) and additional AAAA options for (a = b = c).

ABAC - symmetrical, so only two possible side edges, a/b and a/c. Leads to AAAB semiregular type (for a = b or a = c), additional ABAB options (for b = c) and additional AAAA options for (a = b = c).

ABCD - all four types of edges can be different. Leads to AABC semiregular type (for two adjacent values equal), additional ABAC options (for two opposite values equal), additional AAAB semiregular, additional AABB, additional ABAB and additional AAAA.

So, all in all, we have:
AAAA - fully symmetrical with one side length, AAAB antiprismatic subtype with two, AAAB hex subtype with two, AABB subtype with up to three, AABC mixed subtype with up to four, ABAC subtype with two, ABCD subtype with up to four.
AAAB - semiregular - ABAC type with up to two side lengths, ABCD type with up to four.
AAAB - mixed - AABC type with up to four side lengths
AAAB - antiprismatic - fully symmetrical with up to two side lengths.
AAAB - hex - fully symmetrical with up to two side lengths.
AABB - fully symmetrical with up to three side lengths, AABC mixed type with up to four, ABCD type with up to four.
ABAB - fully symmetrical with one side length
AABC - semiregular - ABCD type with up to four side lengths
AABC - mixed - fully symmetrical with up to four side lengths
ABAC - fully symmetrical with up to two side lengths, ABCD type with up to four.
ABCD - fully symmetrical with up to four side lengths

[wendy]

The general case is much more complex.

In part, this is because 'aspect' is strictly 2d. So we can take a right-angle pentagon, and convert it into something like a half-hexagon (think mid-edge to mide-edge of hexagon), and actually render it as such. Likewise, we can make the group 2 2 2 2 into any size rectangle in euclidean space. According to John Conway, once you set the angles of a fundemental region, it also sets the size.

This means the only reliable method for general honeycombs is the simplex and Wythoff's construction.

There are of course, other methods. Similar to Johnson figures, one could search for chordal figures, that is, those figures that have as edges, the shortchords of polygons. This is how i discovered the octagonal ball, for example. That is, construct a vertex-figure oxqxo8ooooo&#tq, and then check that all of the faces belong to various uniform solids in 3d. Here, the polar figure is ox&#tq, that is a triangle prism tip, the other is xq&#tq which is a rhombo-Cuboctahedron.

If you can demonstrate a convex symmetry (that is, all angles are pi/integer), in rotation or reflection, the result works. The pt{3,5,3} belongs to this kind of approach. Here again i rely on some thing that JHC has demonstrated.

The lesson with the laminates is that the layer structure is more complex than the colour of the other vertex at the end of the line. In John Conway's orbifold, we find (1,2), which is a rotation, and [1,2] a reflection. The laminate edge is still 1,2, but might be governed by a transform of one symmetry to another. In say LC3, the base layer is a cubic to square layer xo2xx2xx&tx, but the subsequence layers, for example moves through the x, y, z axis, ie it is cyclic over three layers. This is governed by a symmetry cell.

[Marek14]

It seems more complicated.

For example: if I take a 3,3,3,4 quadrangle as base and put side edges equal to 4 (all lengths are to be understood as chords, of course), I'll get square antiprismatic prism. But if I put side edges equal to 6, I get something that probably can't be uniform. Truncated tetrahedra in such figure would have to have their hexagons adjacent to other truncated tetrahedra and/or to truncated octahedra, and it's not clear to me whether it's possible or not.

Next step would be putting side edges equal to 8 -- would THAT work? Then we'd have truncated cubes with octagons adjacent to either other truncated cubes or to (8,8,4) -- truncated quadlats. And, skipping even further, with side edges 12, we'd have truncated hexlats where dodecagons are joined to either other truncated hexlats or to truncated {6,4} pseudohedra. Here, the truncated {6,3} are already infinite, so whatever complicated pattern of these dodecagons works, it can definitely fit in there.

More generally, for which antiprism and which side edge there can be uniform tesselations?

[Mercurial, the Spectre]

Basically, the antiprism must have a circumradius of 1 or below assuming unit edge length. Then for other edge lengths, the total circumradius must also satisfy the requirement i.e. 2cos(pi/n)*circumradius is 1 or less for a chord length of n.

We have in 2D only one antiprism, which is the equilateral triangle. By scaling the triangle as a vertex figure we get the {n,3} family of polyhedra/tessellations. We can get up to 6 where it turns into a hexagonal tiling.

In 3D, we have an infinite family of antiprisms. Options include the 2-ap (tet), 3-ap (oct), 4-ap (squap), and 5-ap (pap).
For the tet, we can get the {n,3,3} family that works up to 5 if not considering hyperbolic tessellations.
For the oct, we have the {n,3,4} family with two members, {3,3,4} and {4,3,4}; the former is hex, and the latter is the cubic honeycomb.
For the squap, there is no uniform isogonal realization for unit edges, although the closest is the cube || oct.
For the pap, there is also no isogonal realization, although the {3,3,5} fits in with an icosahedral vertex figure. Ike is derived from pap by augmenting its pentagonal faces with pyramids. For other edge lengths it doesn't work..

In 4D, the only uniform antiprisms include the hex and gudap. Gudap is interesting because it has a unit circumradius, but I don't think a 4D tessellation is possible using it as a vertex figure. For the hex, we have the {3,3,3,4} and {4,3,3,4}.

More generally, since the only antiprism is the orthoplex, which is essentially a simplex antiprism, we have {3^n,4} and {4,3^n,4} as the only possible polytopes/tessellations that are derived from antiprisms. They are the n+2 orthoplex and the n+2 hypercubic honeycomb.

[Marek14]

Not really what I was asking for, though Plus, there are other options in 3D -- you don't have to use triangles in antiprisms, you can use hexagons, enneagons, dodecagons... any type of 3n-gons. As long as you repeat the pattern of attaching another type of polygon on every third side, they will fit together as a hyperbolic (3x,3x,3x,n) tiling with the same general features as antiprisms. (It's the only way to create an uniform (9,9,9,3) tiling.)