Algorithms for finding uniform/scaliform polytopes

[mr_e_man]

I'm sure Polyhedron Dude, and the crowd on Discord, have attacked this problem.

Can anyone think of a systematic way to find all uniform polytopes, especially non-convex and non-Wythoffian ones?

Also, has it been proven that they're rigid, and form a discrete set (countably infinite)?

[Klitzing]

Coxeter et al. last century (1954) addressed the search for "uniform polyhedra", cf. their according paper, then in fact by means of using various facetings of the verfs of the already known convex cases. They then proclaimed that their listing is complete. It only was later (1975) by Skilling that this claim has been proven to be true. (He just added a single figure then, which however wasn't dyadic.)

The same way of exploiting the set of uniform polychora by means of faceting known vertex figures was then commenced by Bowers (after Dinogeorge was trying to do so by elaborating all Wythoffians). However this search project is still ongoing, as every now and then further uniform polychora pop up. - Within higher dimensions even less is known.

They search for scaliform polychora (they only occur as such from 4D on - within 3D they would coincide with the uniforms) was inaugurated by my own find of tutcup, then the first such at all. In those days I still called that one "weakely uniform". It was again Bowers, who proposed a positive name for those (since then they are called "scaliforms"), instead of that more diminuing working title. In those days soon all 4 convex scaliform polychora had been found, which up to now still seem to be all such. For a systematic way to exhaust scaliforms, esp. the non-convex ones, nothing is known to me. However, all alterprisms, gyroprisms, as well as all their higherdimensional analogues clearly will provide examples.

--- rk

[mr_e_man]

On second thought, maybe "scaliform" is too general... or not general enough. A scaliform polytope doesn't always have scaliform faces. A uniform polytope has uniform faces. An orbiform polytope has orbiform faces.

Does every orbiform polytope correspond to a spherical tiling (possibly overlapping itself)?
The problem I see is that a 2D face of a 4D polytope may pass through the origin, in which case its projection onto the sphere is 1D rather than 2D.
(This isn't a problem for a 2D face of a 3D polytope, because there is only one 2D subspace: the whole sphere. No choice is necessary.
Nor is it a problem for a 1D edge, because if the edge length equals the diameter of the sphere, then all edges coincide, and the polytope is degenerate.)

Can an orbiform polyhedron be flexible?
A vertex configuration like 3.4.(3/2).(4/3) is flexible. It appears in the internal blend (at the 2n-gon) of two n-gon cupolas, though such a polyhedron is degenerate (having two coplanar n-gons), and not flexible as a whole.

[mr_e_man]

I took a look at Skilling's paper.

A polyhedron vertex needs to have at least three edges. By uniformity, the vertex must be sent to the three adjacent vertices by some symmetry transformations. Essentially, Skilling took a symmetry group (the 3D ones are well known), enumerated all possible combinations of 3 transformations, applied these to an undetermined point in space, equated the resulting distances (edge lengths) to get a system of equations, and solved this to determine the point in space. Then the full symmetry group applied to that point generates the polyhedron's vertices, and there's a finite set of possible edges and faces connecting those vertices.

But the system of equations may be redundant, and have many solutions. He dealt with this possibility by hand, in the paper's appendix.