Message #1238

From: Roice Nelson <>
Subject: {4,4,4}
Date: Tue, 02 Nov 2010 09:04:21 -0500

I was also thinking some yesterday about this thread between Andrey and
Nan. Puzzles with infinite-sided cells seemed understandable (now that
Andrey has shown the way!), but I had been having trouble wrapping my head
around puzzles with infinite-sided vertex figures. So although I could see
the {infinity,3} puzzle Andrey suggested would be a nice addition to
MagicTile (even if the programming might be difficult), thinking about
{3,infinity} was confounding. How would *that* behave?

I went back to Don’s pictures that Melinda emailed about, and looked at the
two views he has of the {3,infinity} tiling (the dual to the {infinity,3}).
What is interesting is that the cells are "ideal triangles" - all the
cell vertices lie at infinity, on the boundary of the disk. They are
infinite in extent *in multiple directions* (though not infinite in area).
So then I thought about how one would twist one of these cells, and
concluded that it would not be possible to (isometrically) twist a cell
without it overlapping other parts of the puzzle, since the vertices will
remain on the disk boundary through the motion. So if you want to have a
puzzle with an infinite vertex figure, you’ll be forced to have it behave in
an impossiball-like fashion (at least for 2D puzzles).

Furthermore, the slicing becomes degenerate for a {3,infinity} puzzle unless
you want make some modifications to the previous MagicTile generalization.
If you stick to circle slices (or sphere slices in H3), the only slicing
circle encompassing the entire cell is the disk boundary itself, and a twist
using that would be a rotation of the entire space. It would not permute
anything. One option would be to have slicing circles which did not contain
the entire cell though, and this could lead to some interesting puzzles.
That seems like the only good way forward to me actually.

I think the {4,4,4} situation will have some similarities to these
thoughts. Each cell in the {6,3,3} tiling only has one ideal point at
infinity, but the cells in the {4,4,4} tiling will have an infinite number
of ideal points. It seems the result is that any twist (which moves all the
material in a cell) will lead to overlapping material in the puzzle (?, I’m
not 100%). The slicing certainly suffers from the same issues described
above. It is the normal problems of puzzles without simplex vertex figures
multiplied by a gazillion!

Well, I could probably spew more, but I’ll stop in the hope that this
doesn’t all just sound like gibberish. I realize that if one
didn’t have some familiarity with the disk/ball models of hyperbolic spaces,
it probably would. (The first six chapters of Visual Complex
awesome for learning about these models btw.)


On Thu, Oct 28, 2010 at 1:28 PM, Andrey <> wrote:

> I’m not sure. If you take section of {4,4,4} with the plane going through
> the middle of the edge (and ortogonal to it), you’ll get {infinity,4}
> tiling. It’s difficult but not impossible to draw. And after some work with
> it (enumeration of areas, periodic colorings, etc) we can try to expand it
> to H3 space.
> {infinity, infinity} is very easy - in half-plane model. You draw series of
> half-circles (n,1/2), then take each of them and build the inversion of the
> drawing with respect to this circle. Result will look like fractal object
> based on continued fractions (with positive and negative quotients)…
> something like that. I never tried to draw it.
> Andrey

On Thu, Oct 28, 2010 at 1:01 PM, schuma <> wrote:

> I believe the first step to understand {4,4,4} is to understand {infinity,
> infinity} in the hyperbolic plane. What does {infinity, infinity} look like
> and how to draw it? It seems like no matter how you project it, you need to
> truncate not only each polygon and each vertex. After truncation, the
> picture would always look like an incomplete construction site.
> Nan