Post #2923563
2026-04-29 01:14 UTC
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)
Replies (1)
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@pieter@mathstodon.xyz 2026-04-29 01:28
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)