Facetings of expanded polytopes

Process of expanding the cube/octahedron into the small rhombicuboctahedron.
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The small cubicuboctahedron, an edge faceting of the small rhombicuboctahedron. It has triangles, squares and octagons for faces.
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The small rhombicubihedron, the inverse faceting of the small cubicuboctahedron. It has squares and octagons for faces.
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A few weeks ago I got mildly interested in facetings of expanded polytopes (with a focus on the expanded hypercubes), so I decided to make a short^{[dubious–discuss]} article on it here. I have no idea whether or not someone has figured this out before, although given that no one could (efficiently) solve a small challenge involving which facetings of the small exiocteractidiacosipentecontahexazetton are uniform, I somewhat doubt it despite its simplicity.

Explanation

Before beginning the article, let's get some specific terms out of the way:

In very simple geometric terms, expansion is an operation that expands the facets of the polytope outwards and "connects" them. A more formal definition for regular polytopes is that the first and last nodes of its Coxeter diagram are ringed.

A faceting is a (usually) non-convex variant of a polytope sharing the same vertex arrangement, although there are more specific variations focusing on sharing the same edge, face, cell, etc arrangement. A set of polytopes sharing the same vertex configuration is called an army whose convex hull is its general.

With that done, let's get into the actual article.

Expansion facetings

The facetings of expanded polytopes turns out to be a very broad category. Even with something as simple as the cuboctahedron, the expanded tetrahedron, there are already 56 facetings preserving tetrahedral symmetry. Due to this, we are restricting this category to a much smaller set of facetings, specifically those whose external facets are a subset of the general's facets with an additional inner facet in the non-simplicial orbit (unless the polytope in question is a simplex), as well as preserving symmetry.

This way, there are only four expansion facetings of the tetrahedron: the antitruncated tetrahedron, keeping the tetrahedral orbit faces, its inverse faceting, keeping its tetrahedral and edge orbit faces, the octahemioctahedron, keeping two tetrahedral orbit faces, and the cubihemioctahedron, keeping the edge orbit faces. As just shown, expansion facetings can be described by the facets they keep. In this manner, there's a natural ordering to them by describing which facets they keep and interpreting it as binary. This can be well seen in the expansion facetings of the small rhombicuboctahedron (or the small rhombicosidodecahedron or small quasirhombicosidodecahedron for that matter), whose components are not self dual. As there are three facet orbits to keep, there are 2³−2=6 expansion facetings, excluding the small rhombicuboctahedron itself (111) and the null faceting (000). They are:

Out of these, only the small rhombicubihedron (010) and small cubicuboctahedron (101) are uniform, while all others are semi-uniform. This is due to the fact that the two are the only ones with a uniform inner face, that being the octagon. All others have multiple edge lengths, but all elements are isogonal so they're still semi-uniform.

Expansion facetings behave similarly in higher dimensions, a list of facets are provided, being similarly indicable with binary. In general, there are 2ⁿ−2 expansion facetings of the expanded n-cube, with 2ⁿ⁻¹−2 being uniform. For example, the small disprismatotesseractihexadecachoron (expanded tesseract) has 14 expansion facetings in total but only 6 are uniform. Here's a list of them:

The inner facet problem

This is where things get interesting. Given an expansion faceting as its index, is it possible to determine if its uniform?

Looking at the external facets is useless as the faceting's external facets are all uniform given that they are a subset of an expanded hypercube, an uniform polytope. Solving this problem necessarily requires solving for its inner facet, something that seemingly no one has done before, or at least efficiently. In fact, the only person that tried to solve the challenge I mentioned at the start of this page said it took "at minimum two minutes" to solve one faceting.

Anyways, from dyadicity (every ridge must join two facets) it can be inferred that if a specific facet doesn't meet another because it is missing, there must be a facet from the inner facet to join there. This does not happen if this meeting happens between two external facets or doesn't happen at all. Therefore, the inner facet is the XOR of consecutive facets. Or, in more mathematical notation, I_j = F_j ⊕ F_j+1 for 0 ≤ j < n−1, with F being the faceting index, I being the inner facet index and n being the dimensionality of the polytope. Let's see a practical example...

Suppose we have a small exiocteractidiacosipentecontahexazetton faceting with 112 hepteracts, 448 triangular-penteractic duoprisms, 1120 tetrahedral-tesseractic duoprisms, 1792 cubic-pentachoral duoprisms, 1024 heptapetal prisms, 256 octaexa and 16 mystery small petihepteractihecatonicosoctaexon facetings as inner zetta. Is this polyzetton uniform? Well, it can be represented as the index 01111011. Let's use our method...

We apply the first XOR to the initial faceting.

0 1 1 1 1 0 1 1
1 0 0 0 1 1 0

This can be interpreted as, if the two numbers above are different then the number is 1, otherwise 0. Let's continue the chain.

0 1 1 1 1 0 1 1
1 0 0 0 1 1 0
1 0 0 1 0 1
1 0 1 1 1
1 1 0 0
0 1 0
1 1
0

This is the pyramidal representation of the faceting. With this, it is possible to see every inner elements of the polytope. Its last inner elements are the small cellibiprismatodispenteractitriacontaditeron, the small spinoprismatotesseractitesseractichoron, the small rhombicubihedron and the octagon. Since it ends with a uniform polygon, the polytope is uniform. In general, any polytope with a pyramidal representation ending with 0 is uniform. This is because an ending with 1 would imply it has two types of edges with length 1+√2 and 1. This representation also makes it clear this algorithm has a computational complexity of O(n²).

How about we try another one? This time with a non-uniform faceting. This one has 16+112 hepteracts, 1120 tetrahedral-tesseractic duoprisms, 1792 cubic-pentachoral duoprisms, 256 octaexa and 16 mystery small petihepteractihecatonicosoctaexon facetings, index 11011001. Let's get straight to the pyramidal represenation:

1 1 0 1 1 0 0 1
0 1 1 0 1 0 1
1 0 1 1 1 1
1 1 0 0 0
0 1 0 0
1 1 0
0 1
1

As you can see, it ends with a 1, making it not uniform. Its inner face is a tetrapod, and its inner cell is a polyhedron with that along with squares, index 110.

Now that we have seen how to get an inner facet from a faceting, is it possible to do the inverse, getting a faceting from an inverse faceting? The algorithm is simple; start with either 0 or 1 (these create an inverse faceting pair) and repeat, switching if the inner facet's index has 1. In more mathematical notation, F_j = F_j−1 ⊕ I_j−1 for 1 ≤ j < n, with F₀ being either 0 or 1.

Well... now that you have this knowledge, go try solving the original challenge! It's at #polytopes-general in the Polytope Discord. If you can't access that for whatever reason, here's the message:

I am also pretty sure this is the first use of emoji in this site's entire history.