Mike Stay asks
I haven't heard of polypolyhedra before. Is this what you're talking about? http://www.langorigami.com/science/polypolyhedra/polypolyhedra.php4
No. AFAIK, I made up the word independently, but the site above is using a different meaning. My meaning is that you have a bunch of (same-sized) regular solids, and you glue them together, matching up faces. Much like polyominos (squares) and polycubes. I have some counts for polypolygons (polypentagons &c) at http://crd.lbl.gov/~dhbailey/expmath/workshop2004/schroeppel-talk.pdf starting on page 23. The polygons don't tile the plane (exc. 3,4,6), so the polypolygons don't make good packing puzzles. Dan suggests using spherical & hyperbolic spaces. The 2D (surface of a sphere) case is finite, so I'd expect the same in 3D by analogy. I've always been curious about the hyperbolic cases, but never worked out the arithmetic. It's necessary to do exact symbolic arithmetic for this stuff: There are messy cases when things just touch exactly, and floating point isn't able to resolve if they are slightly separated, abut, or overlap. The planar case can have polygons that meet at bare vertices, or meet vertex-touches-edge, or edges meet that are offset. I imagine the 3D case is worse, with the only compensation being that there are only five solids to consider (and cubes are easy). Polypolygons (except for 3,4,6) have a dense set of possible center locations, but only one or two possible orientations. The 3D situation looks as if both center positions and polyhedron orientations are dense (except cubes). The flat-space 2D and 3D cases only require algebraic numbers. The spherical spaces can be tackled by embedding in the next larger Rn. Dan (or anyone): Does hyperbolic geometry require doing logs & exponential nasties, or is there a model of the space where the coordinates for polygons etc. can be done with algebraic numbers? I spoke too fast when I said "no four icosahedra vertices can fit around a point". My arithmetic shows a face-face angle of 138.2 degrees, and the vertex occludes a solid angle of (5 x 138.2 - 540) = 151 degrees excess, so 720/151 = 4.7... could fit around a point, based solely on angular-area considerations. I think that the icosahedron case, which requires big spherical regular pentagons, still cannot fit 4 onto a sphere because of wasted space when the pentagons are close together, but area considerations alone don't prove it. Incidentally, I get the following numbers (based only on angular area) for the maximum number of regular polyhedra around a point: tetrahedron 22 cube 8 octahedron 9 dodecahedron 4 icosahedron 4 (really 3 ?) The cube and dodecahedron cases are easily achievable, but I wonder about the others. Surely the recreational math people have looked at this? --------- Speaking of nasty floating point problems: The ill-condition of the matrix { 1/Fib(i+j) } makes it difficult to compute the inverse using limited-precision floating point. If we change the problem to {1/sqrt(Fib(i+j))} or even {sqrt(Fib(i+j)}, we wreck the rational arithmetic approach, and get a hard problem. Maybe for the next Oxford bloatnum problem list. Rich