It's outstanding that all 4D dice with 5+ cells is possible, "gyrochora" being the general method.
Stella4D tells gyrochora are dual of step prisms, which can be created by Stella4D.
But what is a step prism? I can't find any reference to it in Google or DBpia.
What is a step prism?
 robertw
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Re: What is a step prism?
It was so long ago I worked on this that I really don't remember how they're defined, and possibly didn't understand it fully then! If you understand what gyrochora are, then these are their duals, so should have the dual properties.
I asked around though, and got a reply from Dr. Richard Klitzing (although I can't say I really followed it, but also haven't spent too long trying).
I asked around though, and got a reply from Dr. Richard Klitzing (although I can't say I really followed it, but also haven't spent too long trying).
When considering step prisms, first think about a purely ngonal multiprism {n} x {n} x {n} x .... x {n}, say k times. That one clearly would be a polytope living within 2k dimensions and having n^k vertices. In fact, its vertices thus can be coded by all the ktuples of (independently chosen) numbers from 1 to n.
The next thing is to select a specific subset of just n vertices therefrom, i.e. some specific ktuples only, and then consider the convex hull of those n points. For a typical selection this would still result in a 2k dimensional polytope.
The way of selection then is where the name step derives from. Moreover it assures the full dimensionality. You just will have to specify k1 bijective maps f_j from the set of numbers 1,..,n onto itself (i.e. one such permutation for each j=2,...,k)  for instance say f_j(i) = m*i mod n, for some (j dependent) number m  which step the vertices of the first ngon in a predefined distinct way through the other ones. That is, you just select those n specific ktuples, which start by 1,…,n respectively and follow in the remaining k1 entries each the globally chosen pattern: (a,b,c,...,k) = (id, f_2, f_3, ..., f_k)(a).

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Re: What is a step prism?
(Shouldn't the vertices be encoded by those of from 0 to n1?)robertw wrote: ↑Fri Aug 10, 2018 2:30 amWhen considering step prisms, first think about a purely ngonal multiprism {n} x {n} x {n} x .... x {n}, say k times. That one clearly would be a polytope living within 2k dimensions and having n^k vertices. In fact, its vertices thus can be coded by all the ktuples of (independently chosen) numbers from 1 to n.
The next thing is to select a specific subset of just n vertices therefrom, i.e. some specific ktuples only, and then consider the convex hull of those n points. For a typical selection this would still result in a 2k dimensional polytope.
The way of selection then is where the name step derives from. Moreover it assures the full dimensionality. You just will have to specify k1 bijective maps f_j from the set of numbers 1,..,n onto itself (i.e. one such permutation for each j=2,...,k)  for instance say f_j(i) = m*i mod n, for some (j dependent) number m  which step the vertices of the first ngon in a predefined distinct way through the other ones. That is, you just select those n specific ktuples, which start by 1,…,n respectively and follow in the remaining k1 entries each the globally chosen pattern: (a,b,c,...,k) = (id, f_2, f_3, ..., f_k)(a).
I see. So nm step prism is made by selecting n vertices out of nn duoprism, one each of each ngonal layers. f_j selects the vertices at even distance, so the resulting step prism is vertextransitive. (Hence its dual, the gyrochoron, is celltransitive.)
What is the symmetry of the nm step prism? I guess it's [n,2]⁺ in Coxeter notation, is it?
 robertw
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Re: What is a step prism?
Here's some further discussion from Dr. Richard Klitzing:
And an additional reply from Jonathan Bowers:Let's reduce to 4D. Then you'd start with an (n,n)duoprism. That one could be represented by its set of square faces, which altogether form the surface of a torus. When cut open you'll get an n x n grid of squares.
You could imagine that the cells of that duoprism, the individual nprisms would be represented behind the columns respectively before the rows. Esp. the ngons would be represented here as the vertical respectively horizontal edge sequences.Code: Select all
++++     ++++     ++++     ++++    
As I told yesterday you can number the vertices of those ngons. Thus any vertex here is represented by a pair of numbers, the position number wrt. the vertical and wrt. the horizontal edge sequence. This in fact is just the same as for the indices of matrix elements M=(m_ij)_ij.
The issue for step prisms now is to select a subset of n vertices from this set of n^2 vertices. In order to be still full dimensional, it is required to use one vertex from every horizontal and every vertical edge sequence each (and, for sure, n being large enough). As such this is like applying the numbers 1 to n repeatedly on that grid in a sudoku like manner.
Code: Select all
4321     1432     2143     3214    
And then selecting the subset of vertices with any single specific number. The step prism of consideration would be the convex hull of those n vertices. In fact, the given rule tells you how the starting vertex, chosen on the leftmost vertical line, steps through the grid while moving on towards the right.
In order to be full dimensional in our setup of 4D, n surely has to be 5 at least (because the smallest 4D figure is the 4simplex, having 5 vertices). Moreover this setup of step prisms shows that there are polytopes with any individual number of vertices within any (even) dimensional space, starting from the number of vertices of the according dimensional simplex, onwards.
Step prisms as they are on Stella which produce congruent vertices have vertices of the form {X,nX} which are a subset of k duoprism vertices where X are vertices of a kgon from 0k and n is an integer less than k/2 (note  we can use n>k/2, but kn and n lead to the same step prism but in a different orientation). If n=1, then it generates a diagonal polygon. The number n (which is called the 'step') is also in modk.
Sometimes two different steps lead to the same step prism  if m and n are steps and mnk =1 or 1, then they generate the same step prism but in a different orientation. Example 132 and 136 generate the same step prisms since 2*6 = 131
The dual of the step prisms are what Wendy calls step tegums and what I like to call gyrochora. These make some cool fair dice. Multiprisms can also generate step prisms where the vertices have the form {X,j1X,j2X,j3X,....,jnX}  jn's (all less than k/2) need to be different and neither equal to 1 or 0 to generate a fully dimensional polytope. I call their duals 'gyrotopes'