3 ms·

> Oh, and that goes for every vertex of every tile in the infinite tiling, not just for a single tile. Could it possibly not? i.e. is there any way to tile n-g

> Oh, and that goes for every vertex of every tile in the infinite tiling, not just for a single tile. Could it possibly not? i.e. is there any way to tile n-gons in the (A + B*sqrt(3))[] kernel that has vertices not in that kernel? What comes to mind is the counterexample of plain unit squares tiled with strips at irrational offsets from one another. But what about for n != 4? Does every tiling have some vertex-vertex co-location?

jameshart 3y ago
These aren't general n-gons on vertices in that set - the Einstein tiles have additional constraints: The angles of the Einstein tile are all ±90 or ±120° - so the edges are all parallel to one of the 12 30° meridians. And the edge lengths are all integer multiples of a common unit. And as you place these tiles edge-to-edge, and match vertices, the same applies to the edges that make up each subsequent tile. So each line takes integer steps along lines that are either parallel to the axes or are at a 30° angle from an axis. If we take 2 as the shortest length of an edge (so all the edges of the shape are either 2 or 4 units long), then: - if we move from a point 2 units along an edge parallel to an axis, we add or subtract 2 from one of our cartesian coordinates. - if we move 2 units along an edge at 30° from an axis, we must add or subtract 1 from one of our coordinates, and √3 from the other (it's the hypotenuse of a 30°/60° right triangle) Those are the only moves you can make, so in an infinite tiling of Einstein tiles, the vertices will all fall on those points.
- basil-rash 3y ago
  Right, but I’m trying to generalize the claim to beyond these tiles. The claim is that for every vertex-vertex tiling of n-gons with n not all equal to 4 and vertices in some kernel (A + B*sqrt(C)), for integer ABC, all verticies in the tiling will also be in that same kernel. Which is to say there’s nothing interesting about these Einstein tilings maintaining their NMsqet(3) kernel.

jameshart 3y ago

These aren't general n-gons on vertices in that set - the Einstein tiles have additional constraints: The angles of the Einstein tile are all ±90 or ±120° - so the edges are all parallel to one of the 12 30° meridians. And the edge lengths are all integer multiples of a common unit. And as you place these tiles edge-to-edge, and match vertices, the same applies to the edges that make up each subsequent tile. So each line takes integer steps along lines that are either parallel to the axes or are at a 30° angle from an axis. If we take 2 as the shortest length of an edge (so all the edges of the shape are either 2 or 4 units long), then: - if we move from a point 2 units along an edge parallel to an axis, we add or subtract 2 from one of our cartesian coordinates. - if we move 2 units along an edge at 30° from an axis, we must add or subtract 1 from one of our coordinates, and √3 from the other (it's the hypotenuse of a 30°/60° right triangle) Those are the only moves you can make, so in an infinite tiling of Einstein tiles, the vertices will all fall on those points.

basil-rash 3y ago
Right, but I’m trying to generalize the claim to beyond these tiles. The claim is that for every vertex-vertex tiling of n-gons with n not all equal to 4 and vertices in some kernel (A + B*sqrt(C)), for integer ABC, all verticies in the tiling will also be in that same kernel. Which is to say there’s nothing interesting about these Einstein tilings maintaining their NMsqet(3) kernel.