So I am thinking about vertex figures for Archimedean Honeycombs...
So the vertex figure is a polyhedron. If we put one of the Honeycomb's vertices to the Origin, with each neighbour of said vertex is a vertex of the polyhedron. If two neighbours share a face, they are connected by an edge in the vertex Figure. (I do not care about the vertex figure's faces)
Since all edges of the Honeycomb must have same length, the vertex figure must have a circumference.
So I want to build a certain honeycomb.
If my vertex figure has v vertices, I have 2v+1 degrees of freedom when placing them. (One for the circumradius r and two for polar coordinates) But since rotation shouldn't matter I can subtract 3.
That makes 2v-2 degrees of freedom.
Now each of the edges corresponds to a face of the hc. In Hyperbolic Space the inner angle of that face depends not only of the facetype (trigon, pentagon...) but also on its edge length.
So if the vertex figure has e edges,
We have e relations between r and coordinates.
So for eg pyramids we have 2v-2=e, meaning we expect a unique solution. (Which my code gives, as long as the hc is actually Hyperbolic)
But for prisms we have 2v-2>e. So my System is underdecided...
Does that mean I can wiggle my Honeycomb??
And for doublepyramids we have 2v-2<e. So my System is overdecided.
So I suppose only special symmetric cases have a solution??
Finding the right sidelength so that the six angles match up perfectly at the vertex was quite a math mess. It doesn't seem to work for all confugurations yet, but still, yay! ^^
My Patrons voted, my next Theme is Archimedean Honeycombs!
An Archimedean Honeycomb is vertex-transitive with regular faces.
So my vertex-figure-based approach stays feasible.
But the vertex figures are no longer platonic!
In Hyperbolic space the angle of a regular n-gon depends on its sidelength. So I need to find the sidelength for which the angles match up.
For trigonal vertex-figures I could solve this analytically.