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Pieter Mostert
@pieter@mathstodon.xyz
mastodon 4.7.2I'm a former potter with a maths background, and currently work with data at a rural education NGO in South Africa.
113 Followers
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15 Posts
Joined May 27, 2023
Peter Selinger now has an app
https://www.mathstat.dal.ca/~selinger/hat-partition/ that lets you create hat tilings using the Markov partition described in his paper with Sébastien Labbé (https://arxiv.org/abs/2604.20964). For a printable version of this, see Sébastien's blog post: http://www.slabbe.org/blogue/2026/03/a-construction-of-the-hat-tilings-by-a-markov-partition/
As a follow up to last week's post on Markov partitions for Hat (hat and turtle) and Spectre (hats in turtles and turtles in hats) tilings (@pieter@mathstodon.xyz), here is a way of colouring such tilings. For each control point, we colour the tile according to its distance from the boundary of the fractal window it lies in.
The first image shows a patch of a turtle tiling using the colour map in the second. Control points that lie near the boundary of the fractal window are close to flipping between the two states of a Conway worm, so this is a nice way of highlighting subsets of tiles that are 'close' to being Conway worms.
One of the things on my to-do list is to create an animation of a patch as the triangular grid moves slowly relative to the underlying pattern, but if anyone wants to take a stab at this, please do.
(1/2)
Replying to
We can now change the projection again, going from the \((1,0)\) projection to the \((1, -1)\) projection, dragging the regions along as before. The result is that the triangles spread out further, but the grid points in adjacent triangles no longer lie on a common rectangular grid (at least, not one with the same distance between points). This is why my construction here https://mathstodon.xyz/@pieter/110520611405361464
required two steps.
But if we shift the control point to the position Peter chose, the grids become compatible, yet the regions (which must move along with the control points) fortunately don't overlap.
So far I've ignored the tiles with less frequent handedness. From the second image, we can deduce regions that must correspond to their control points by considering their 'H8' neighbourhoods and looking at the intersections of the regions associated to these tiles (If this isn't clear, this post on empires of turtle tilings may help explain the general principle: https://mathstodon.xyz/@pieter/111696562690124891)
(4/n)
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As Arnaud Cheritat explains here https://www.math.univ-toulouse.fr/~cheritat/2023-monotile/4D-lift/page-4.html given a tiling in the Hat or Spectre* family, you can construct a MLD tiling of hexagons and rhombs (see also @mathBlock@mathstodon.xyz (https://mathblock8128.wordpress.com/2023/09/16/spar-lines-hexagons-and-tiling-a-finite-boundary/ Fig 22) and @jsmith@mathstodon.xyz (https://arxiv.org/pdf/2403.01911 Fig 3.1) who point out that there is a aperiodic set of prototiles consisting of a rhomb and two types of comets with directed, coloured edges). Given a base vertex, you can assign to each vertex in a comet-rhomb tiling a pair of complex numbers in \(\Lambda^2\), where \(\Lambda = \mathbb{Z}[\xi]\) (\(\xi = e^{2\pi i/6}\)) is the set of Eisenstein integers. By including edges, we obtain a directed graph with vertices in \(\Lambda^2\). This is the 'lift' of the tiling. A tiling using a particular tile set in the Hat or Spectre family is then a projection of this graph, where the hat tile corresponds to projecting onto \((1,-1)\) and the turtle with projecting onto \((1,\xi^2)\).
[You could also do this using the vertices of the hat tile, but I find the comet-rhomb version simpler since there aren't as many points, and projections to tiles in the hat family are all of the form \((1, \sigma) \) for \(\sigma \in \mathbb{C}\) of modulus 1.]
*By the Spectre family, I mean tilings using Tile(\(a,b\)) and Tile(\(b,a\)) (in the notation of https://arxiv.org/pdf/2303.10798) with appropriately marked edges
(2/n)
Replying to
To get the turtle Markov partition I started off with, you can follow the same steps, but using the projection along \((1, \xi^2)\) instead of \((1,-1)\), and them moving the control point to the underside of the turtle's shell. By good fortune, the shifted regions don't overlap in this case either.
(5/n)
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Lastly, we turn our attention to the Spectre tiling. Things are much the same as before, although since the deflation involves a reflection, we need to use a vector \(v\) such that \(𝑣𝑀=λ𝑣̅,\)
instead of a left eigenvector.
The two images show Markov partitions (I'm assuming) for hats-in-turtles and turtles-in-hats respectively.
(6/6)
Replying to
The deflation operation can be abelianised (in some sense) to give a 2 x 2 complex matrix with eigenvalues \(\phi\) and \(\phi^{-1}\), where \(\phi\) is the square of the golden mean. If we project the lifted centres of tiles with more frequent handedness onto the left eigenspace of \(\phi^{-1}\) (spanned by \((1, -\xi\phi)\)) and colour the projected points according to the orientation of the tile, we get a pattern which is the superposition of the two triangles shown, translated so their centres coincide.
On the other hand, consider the projection onto the first coordinate. For a given tile centre, we can track its position as we continuously deform the projection from that eigenspace projection to the \((1,0)\) projection, and we can translate the region that the eigenspace projection of the point landed in as we do so, keeping the relative position of the projected point within the region fixed. This spreads out the regions, and we obtain the periodic pattern shown in the second image.
I've overlain this with hexagons, which are the images of the tiles with more frequent handedness under this projection. A short line indicates the orientation of the lifted tile. In the notation of the previous post, these lines extend from \((0,0)\) towards \((\xi^2,0)\) (for a tile in that particular orientation)
(3/n)
For any fellow fans of Angine de Poitrine, they just released their second album: https://anginedepoitrine.bandcamp.com/album/vol-ii
If you aren't (yet) a fan, take a listen to their KEXP performance: https://www.youtube.com/watch?v=0Ssi-9wS1so
Angine de Poitrine - Full Performance (Live on KEXP)
Replying to
@simontatham@hachyderm.io @csk@mathstodon.xyz There's definitely more work needed for tilings by spectres, or versions of the Hat family other than the hat or turtle, but since the hats-in-turtles tiling is based on a discrete grid, and because there exists a transformation of a tiling by spectres to a hats-in-turtles tiling that translates tiles by a finite distance, given a point, you just need to calculate a finite patch of hats-in-turtles which is guaranteed to cover the point once the patch is transformed to spectres.
The relation between representations by points and sequences of supertiles is fairly simple if you ignore the edge cases. Any point inside the fractal region determines a nested sequence of fractal regions, each of which contains the point, and is one of the elements of the partition of the previous region, as shown in the diagram. So the list of supertile types and relations can be used to write the coordinates of the point as a \(\beta\) expansion of the form \(\sum_{n=0}^\infty c_n \phi^{-n}\) where (\phi\) is the square of the golden mean, and the \(c_n\)'s come from a finite set of complex numbers
Replying to
@csk@mathstodon.xyz Nice to see this implemented, Craig.
I'm curious how you determine which fractal shape, if any, a control point lies in. Is it anything like the iterative process I describe here: https://mathstodon.xyz/@pieter/115951754157537975
I wish I'd thought to install an AI-slop blocker for youtube sooner. This seems to be working great so far, although unfortunately there isn't a version for Android: https://addons.mozilla.org/en-US/firefox/addon/ai-channel-blocker-for-youtube/
