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4 votes
Accepted

For what $n$ do there exist non-periodic tilings with rotational symmetry of order $n$?

Jarkko Kari and Markus Rissanen construct such (even a substitutive one), called Sub Rosa, for even $n$ in [1]. ArXiv is https://arxiv.org/abs/1512.01402 The second-named author is Markus Rissanen, ...
Ville Salo's user avatar
  • 6,652
1 vote

Tiling with ten-fold symmetry and (unoriented) Penrose tiles?

The division of a pentagon into triangular pieces described here can be used to generate a tenfold quasilattice. At each node generated by any iteration additional edges are rendered in subsequent ...
Oscar Lanzi's user avatar
  • 2,370
22 votes
Accepted

Can you see through a cannonball packing?

Yes. View the FCC packing as a series of stacked square packings, with spheres of unit radii centered at the points $(2a,2b,2\sqrt{2}c)$ and $(2a+1,2b+1,(2c+1)\sqrt{2})$ for all $a,b,c,\in\mathbb Z$:  ...
RavenclawPrefect's user avatar
6 votes

Tiling with ten-fold symmetry and (unoriented) Penrose tiles?

We can start with a floret that looks like this: then follow the following substitution rules (repeatedly) : this yields a tiling, part of which is shown below. Some things to note about this ...
Andrew Bayly's user avatar
10 votes
Accepted

Tiling with one of each shape

Here is a picture (and extra characters to make it 30).
Martin Tancer's user avatar

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