## Tilings in hyperbolic tessellations

Higher-dimensional geometry (previously "Polyshapes").

### Tilings in hyperbolic tessellations

Here's a question: I'm still following the hyperbolic game HyperRogue. It keeps growing and exploring new concepts.

One important concept is: which shapes can tile the hyperbolic plane if they have to be comprised from the basic hexagons and heptagons of the underlying (6,6,7) tessellation? Zeno has found one such configuration made from 40 cells. Clearly, all configurations that tile must have number of cells that's a multiple of 10 (with hexagons and heptagons in 7:3 ratio). Any such configuration with 10k cells has area k*pi and the resulting tiling can be considered as a variant of {3k+6,3} tessellation as the underlying (6,6,7) only permits 3-valent vertices.

So: any configuration with 10 cells (smallest possible) will have 9 neighbours around every tile (assuming uniformity). I've found a 10-cell configuration that tiles (hexagon, 3 surrounding hexagons + 3 surrounding heptagons, 3 hexagons at distance 2, looks like a fat triangle), and yesterday I discovered that it has 4 distinct ways to cover the hyperbolic plane uniformly as long as its 3-fold symmetry stays unbroken. There might be more ways to use it if you allow it to be surrounded in an asymmetric way.
Marek14
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### Re: Tilings in hyperbolic tessellations

The smallest toric cell has integer numbers of all elements.

So for x7o3o, you have per heptagon, 7 * 1/2 = 3:60, 7 * 1/3 = 2:40 , so you need six heptagons to make these integer. This has 6h, 21e, 14v in an area of 1 (21-6-14)

So if you are working with regular figures, note that 7,3, and 7,4 and 7,14 share symmetries.

For 7,4 we have per heptagon, 7 * 1/2 = 3:60 edges and 1:90 vertices, giving 4h, 14e, 7c in an area of 3. The same area is covered by 18 heptagons of 7,3.

For 14,7 we have per 14-gon, one cell, 7 edges, and two vert, giving an area of 7-1-2 = 4. These cells are covered by 24 cells of 7,3 which is a known regular tiling. (You have every petrie polygon at 8 )

Coxeter + Moser say that there there is a 7,3 with a petrie polgyon of 9, which leads to 36 heptagons, which is double the minimum for 7,4.

One can find the corresponding x3x7o by dividing the heptagon count by 0;36, ie *10/3.
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wendy
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### Re: Tilings in hyperbolic tessellations

Not sure how that is actually relevant, though. I already know the numbers necessary to get a tiling configuration, what I need to know is how to find the specific configurations that work.
Marek14
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### Re: Tilings in hyperbolic tessellations

The first question was how to even describe a tile without showing it. In the end, I described the edge of the pattern by detailing how you have to move to go around it. For example, one 10-cell tile in (6,6,7) is described like this:

(6/5, 7/1, 6/3, 7/2)^3

This says that there's a hexagon with five adjacent cells outside the pattern (that means it's only connected to the rest of the pattern by 1 cell), a heptagon with 1 outer cell, a hexagon with 3 outer cells and heptagon with two outer cells, the whole thing repeated 3 times. The 7/1 and 7/2 actually describe the same heptagon which is traversed twice.

Then I partition the cells adjacent to the tile into consecutive blocks that belong to the adjacent tiles. I automatically eliminate overly long blocks that can't fit.

Finally, I try to complete the tiling based on these partitions.

For this tile, I found 5 ways to tile the hyperbolic plane with it:

http://steamcommunity.com/sharedfiles/f ... =503384852
http://steamcommunity.com/sharedfiles/f ... =503402779
http://steamcommunity.com/sharedfiles/f ... =503402809
http://steamcommunity.com/sharedfiles/f ... =503402834
http://steamcommunity.com/sharedfiles/f ... =503402858
Marek14
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