What regular or semiregular hyperbolic tiling has the smallest average tile area?

200 Views Asked by At

I have noticed that hyperbolic tilings tend to be rather "sparse" in that each tile takes up a lot of space. If I remember correctly, for a given curvature the area of any tile in a given hyperbolic tiling is fixed, unlike in the Euclidean plane, so, what regular or semiregular hyperbolic tiling is "densest" - has the smallest average tile area for a set curvature - out of all possible ones? Is this known? If drawn in the Poincare disk, it would be possible to clearly see more tiles in that tiling than in any other, I imagine.

1

There are 1 best solutions below

4
On

Here is the list of Archimedean tilings: https://pastebin.com/4AvJ527V

The number "arcm" is how many tiles fit in area $4\pi$; it is usually an integer.

The number "dual" is for the dual tiling.

The symbol can be entered in http://www.plunk.org/~hatch/HyperbolicApplet/ or in HyperRogue ( http://www.roguetemple.com/z/hyper/online2.php?c=-d:arcm ) to draw the tiling.

Then you see the areas of each individual tile, and its percentage (e.g. 7: 0.390412x0.3 means that heptagons have area 0.390412 and 30% tiles are heptagons in this tiling).

Increasing faces decreases the area, so (3,3,3,3,3,4) is clearly the best solution with valence 6, and (3,3,3,3,3,3,3) is clearly the best solution with valence 7 or more. Then we have an exhaustive list of Archimedean tessellations with valence up to 5, and with dual >= 28.3688 (it is a bit easier to compute dual than arcm; but all tiles with arcm >= 40 are found).

The "list" lists some nice tilings for comparison (the list is taken from HyperRogue).

enter image description here

The greatest number is 124 for (3,3,3,3,7). The heptagon is almost 10x larger than the triangle, though, so maybe not very good for your application? (HyperRogue uses (6,6,7) because it has small tiles which are still of roughly similar sizes)

If you want small tiles, the binary tiling can have congruent tiles which are as small you want (they are "long" but very narrow). The Goldberg-Coxeter construction can be used to obtain nice tessellations with small, almost regular tiles.