2023-03-29 Penrose tiling for a floor

In a discussion on Slashdot, I got into a conversation with "dargaud" (who gives his home page as http://www.gdargaud.net/). He's interested in making a floor with a "Penrose" tiling - an aperiodic tiling with (if you get it right) pentagonal symmetry. So am I. The Slashdot discussion is here (search for users "dargaud" and "RockDoctor", which is me ; but this may disapear, so I've archived the relevant bit here - it's a Box link. Seems to work though).

Both of us have been looking at doing a floor with a Penrose tiling. The Slashdot report is of a single shape that can achieve this. (I missed that from the summary, when it should have been the most prominent point. But "Meh". Bad summary writing/ editing ; or the editor thought "aperiodic tiling" was something novel, which it hasn't been since the early 1970s. Whatever, hardly news on SLashdot. Bitching about the editors is something I try to avoid, but sometimes your patience bends) From comments, this singular shape includes convex and concave sections of the perimiter that enforce Penrose's "matching rules" to enforce against getting either periodic patterns, or patterns that can't be continued to infinity.

This image

(lifted from Wiki) shows the components of a two-rhomb tiling system with "matching rules" expressed as both edge decorations and surface paintings.

The way I'd decided to do it is to use these rhombs, but without the decorations. I'd have to plan carefully to get the tiling right, before removing my marks on the prefabricated tiles. Or putting the marks on the floor, I haven't decided. The rhombs have angles of 36, 144, 36, and 144° for the "thin", and 72, 108, 72, and 108° for the "fat".

My next step was to calculate the sizes for making such tiles ("rhombs", rhombuses") with an edge of 10 cm. I've lost that calculation.

… and I've just re-done the calculation, then

… plotted them up. Note - the dimensions for the laminate tile are just a guess - and I might not use laminate in any case - maybe glue lino tiles onto sheet lino, so I can include some "bodge" allowance. But the main purpose of this picture is to show the effect of tile size on wastage, and how I'd "fit" the tile size to the actual dimensions of laminate sheet (or lino tiles) to optimise usage. Even if I decide to do it with lino, the same thinking would apply. Whatever materials I use, they're not going to be free, so I don't want to buy more than I have to.

The paper that triggered the Slashdot story was from ArXiv : An aperiodic monotile. The preprint doesn't indicate submission of the paper to a journal, but that may be normal practice in maths these decades. They have a pun in their terminology - they call their shape an "einstein" - nothing to do with a cetain theoretical physicist, but from the German "ein stein", meaning "one shape", referring to the "monotile" nature of their discovery - that this shape (actually a large, if not infinite, range of shapes) can tile the plane infinitely and hierarchically (which they claim to prove means aperiodically).

The "einstein" they present however has a seeming implicit triangular grid - which is quite distinct from the pentagonal symmetry in the Penrose multi-tile examples. So already I'm less than interested in it. My inner mineralogist wants to see pentagons! Time for an illustration, I think.

You see what I mean about the underlying triangular grid? That, I would definitely not go to the effort of building - far too pedestrian.

Their dark shaded tile looks, to me, to be composed of 4 congruent sub units, irregularly arranged ; and each of those is composed of four convex (so, relatively easy to cut) "kites" which also look to be congruent. I'm going to have to read further to find out what their geometry is, but they look like Penrose's "kites" from the "Kite and Dart" constructions from Penrose on Wikipedia :

Are this paper's "kites" (they use that term) the same shape as Penrose's? The authors specifically reference Penrose's kites in their description of the history of the problem.

Their next figure also shows the underlying trigonal (or is it hexagonal) pseudo-symmetry from their tilings. The note this themselves "Finally, we have noticed that these chains seem to impart a rough hexagonal arrangement to the hats, which is particularly clear in the triangular and parallelogram-shaped structures that are surrounded by chains."

20 pages into an 89 page paper, and I've seen nothing that looks like a definition of the "hat", or indeed, it's component "kites". I strongly suspect they're "Penrose Kites", as above, but if they say so, then I've missed the statement apart from the hint noted above. They mention that their "polykite" (also, "hat? Or have I missed something?) has sides of either 1 or $\sqrt{3}$ - which is slightly worrying because I'd expect those nuimbers to come out of 30° and 60° triangles, not those with 36° or 72° angles (where I'd expect to see lengths of 1 and [$\sqrt{5}-1$]. Distinguishing those by eye ... I'd need a definition. If it is a 30-60 structure though, that would go a long way to explaining the (pseudo-)trigonal symmetry. (Hey, I had to learn some MAthML!)

Working further, I get to Lemma A.1, where their kite is defined as having sides of 1 and $\text{exception 25:}$ (and since it's a kite, not a parallelogram, the sequence must be 1, 1, $\text{exception 25:}$, $\text{exception 25:}$. That can't be the "Penrose kite", but mut be one composed of 30°-60°-90° triangles. Which would at least make manufacturing them in large numbers relatively easy.

Figures A.8 and A.9 make the case again.

The final 60-odd pages of the paper are enumrations of possible (and impossible) neighbouring sides of the tiles, from which, I assume they can derfive their aperiodic claims.

I've no need to go further into this rabbit hole. I'm sure they've got their maths right, but aesthetically, the resultant pseudo-trigonal symmetries are not what I'm looking for. It'll be the fat-thin Penrose rhombs for me (or possibly "kites and darts", assembling those from their component triangles, which I can make by cutting strips, then cutting strips into triangles, then re-assembling.

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