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Hi Anthony,

I maybe should walk down the hall... but this is easier.

Dual to the $d$-cubical tiling of $\mathbb R^d$ would be the "cross-polytope" tiling. This is the tiling made up of the duals -- the vertices are the centres of the $d$-cubes, the edges of the cross-polytope are the faces of the $d$-cubes, and so on.

To me it looks like you get your tiling from the cross-polytope tiling, by simply scaling up each tile appropriately -- scaling each tile at one of its vertices, and doing the scaling symmetrically with respect to the translation symmetry of the tiling. So as you scale, part of the tile vanishes (from a growing tile eating it up) and part gets created (via scaling).

Your picture appears to be consistent with something like that. Or rather than the cross-polytope tiling, it could the the same idea but with the cubical tiling.

edit:

Take this procedure for generating a tiling of $\mathbb R^n$. Let $M$ be an $n \times n$ invertible matrix with real entries. Let $\vec v \in \mathbb Z^n$. In the lexicographical order on $\mathbb Z^n$ we can lay down a "tile" being $[0,1]^n + M\vec v$, where whenever we place a tile, it overwrites any old tile that it may be placed on top of. Provided the norm of the matrix is small enough, this procedure writes over the entire plane. It produces a tiling a fair bit more general than what you're talking about. I'd call it perhaps a linear overlapping translation of the cubical tiling. But I don't know if there's standard names for such thing. Perhaps "lizard scales" ?

The tiling in your picture looks like something like this, with a 2x2 matrix with entries (left to right then top to bottom) $2/3, -1/3, 1/3, 3/4$. I seem to have forgotten how to typeset a $2\times 2$ matrix in mathjax or whatever is powering this website nowadays...

Does that make sense?

2 added 99 characters in body

Hi Anthony,

I maybe should walk down the hall... but this is easier.

Dual to the $d$-cubical tiling of $\mathbb R^d$ would be the "cross-polytope" tiling. This is the tiling made up of the duals -- the vertices are the centres of the $d$-cubes, the edges of the cross-polytope are the faces of the $d$-cubes, and so on.

To me it looks like you get your tiling from the cross-polytope tiling, by simply scaling up each tile appropriately -- scaling each tile at one of its vertices, and doing the scaling symmetrically with respect to the translation symmetry of the tiling. So as you scale, part of the tile vanishes (from a growing tile eating it up) and part gets created (via scaling).

Your picture appears to be consistent with something like that. Or rather than the cross-polytope tiling, it could the the same idea but with the cubical tiling.

1

Hi Anthony,

I maybe should walk down the hall... but this is easier.

Dual to the $d$-cubical tiling of $\mathbb R^d$ would be the "cross-polytope" tiling. This is the tiling made up of the duals -- the vertices are the centres of the $d$-cubes, the edges of the cross-polytope are the faces of the $d$-cubes, and so on.

To me it looks like you get your tiling from the cross-polytope tiling, by simply scaling up each tile appropriately -- scaling each tile at one of its vertices, and doing the scaling symmetrically with respect to the translation symmetry of the tiling. So as you scale, part of the tile vanishes (from a growing tile eating it up) and part gets created (via scaling).

Your picture appears to be consistent with something like that.