The hat needs its mirror image. Months later, a tile that does not
The hat answered the einstein problem in March 2023, and almost immediately somebody pointed at the reflections. A hat tiling uses both the tile and its mirror image. If you are cutting shapes out of paper that is one shape, since you can turn a piece over. If you are cutting them out of something with a front and a back — or thinking of the tile as a physical object — it is arguably two.
Nobody thought this was a serious flaw. Everybody wanted it gone anyway. In May 2023, two months later, the same four authors published a tile that never needs its reflection at all.
First, the objection is about something real. H is the only one of the four clusters containing a reflected hat, so “can a hat tiling avoid reflections” becomes “can a patch avoid H”, and that is a count anyone can run.
The black tick is the limiting share, one third. Note the T seed: its first generation is a single H and three P halves, so it begins at 25% and climbs. φ⁴ = 6.854102 is irrational, so this density is a second, independent obstruction to any period — and it is also the objection the spectre was built to answer, since a shape needing its own reflection struck some people as two shapes.
Try it: Try all four seeds. Past the first generation, every patch contains H tiles, and the share settles on exactly one third. Reflected hats are forced at a density of one in φ⁴ + 1 — about 12.7% — which is also, incidentally, a second proof of aperiodicity, since that density is irrational.
The hat is not an isolated shape but one point on a continuous path of thirteen-sided polygons, all with the same angles, differing only in the lengths of two classes of edge. Walk to the point where the two classes have equal length and you reach Tile(1, 1): twelve edges of length 1 and one of length 2, which is fourteen unit lengths in all. Split that double edge at its midpoint and you have the equilateral fourteen-gon the shape is usually drawn as, with a straight angle at the new vertex. This is what the second paper is built on.
At a = b the two edge classes have the same length, so as a thirteen-gon the tile has twelve edges of length 1 and one of length 2 — fourteen unit lengths in all. Split that double edge at its midpoint and you get the equilateral fourteen-gon the shape is usually drawn as, with 1 straight angle where the hat’s double edge used to be. The area comes to 3√3 + 3 = 8.196152, between the hat’s eight kites and the turtle’s ten. This is not the spectre. The tile in “A chiral aperiodic monotile” has these edges replaced by curves, and with straight edges the shape admits periodic tilings.
Try it: Move between the three named members. The vertex count never changes, the angles never change, and the area runs 8√3 for the hat, 3√3 + 3 for Tile(1, 1), 10√3 for the turtle. Note the honest warning under the panel: with straight edges this shape is not the spectre and does not tile aperiodically.
Tile(1, 1) with straight edges admits periodic tilings. The spectre is that shape with its edges replaced by curves, chosen so that a copy and its mirror image can never both appear. Equal edge lengths are what make this possible: all fourteen edges can carry the same curve, because they are all the same length.
The paper’s particular curves are not derived anywhere in this module, so the panel below uses an S-curve of our own to show the mechanism instead. Everything it claims about that S-curve is computed.
An edge has to serve two tiles at once: one walks it from P to Q and the other from Q to P, so the curve must be the same set of points either way round. A curve with half-turn symmetry about the edge’s midpoint — an S — has exactly that. And the same S is what rules the mirror out: reflecting the tile bulges every S the other way, so a reflected copy fits happily against other reflected copies and never against an unreflected one. The blue outline is that reflected decoration, separated from the amber one by up to 0.161 at a bulge of 14% of each edge’s length.
Try it: An edge is walked in opposite directions by the two tiles that share it, so its curve must look the same either way round — an S does, by half-turn symmetry about the midpoint. Turn on the reflected decoration and see why the same S excludes the mirror: reflecting bulges every curve the other way, so a reflected copy fits other reflected copies and nothing else.
It would be natural to guess that the spectre works by being its own mirror image, so that the question of reflections stops arising. It does not. Every member of the family is chiral, Tile(1, 1) exactly as much as the hat, and equalising the edge lengths buys no symmetry whatever.
| member | sides | distinct edge lengths | interior angles | same as its mirror? |
|---|---|---|---|---|
| The hatTile(1, √3) | 13 | 3 | identical | no |
| Tile(1, 1)the spectre skeleton | 13 | 2 | identical | no |
| The turtleTile(√3, 1) | 13 | 3 | identical | no |
| Tile(2, 1)an unnamed member | 13 | 3 | identical | no |
| Tile(1, 2)another | 13 | 2 | identical | no |
The comparison is on the turning sequence — the edge lengths in order with the angles between them — which pins a polygon down up to a rigid motion. The mirror’s sequence is the same walk taken backwards with every turn negated, and no cyclic shift makes the two agree for any member of the family. The angles themselves never change along the family at all; only the lengths move.
Try it: Read the last column. The spectre’s trick runs the other way from the natural guess — instead of making reflections harmless, the curves make them unusable. That is what “chiral aperiodic monotile” means in the title of the paper: not a tile that lacks handedness, but a tiling that uses only one of them.
Worth saying plainly. The curved spectre is not implemented here. The function called spectreSkeleton() returns Tile(1, 1) with straight edges, its docstring says so, and its tests say so. Nothing in this module claims that any member of the family tiles the plane, aperiodically or otherwise — the tilings you have been looking at throughout are of the H, T, P and F clusters, which is a different and fully verified construction. What the last page does have is the family itself, and the reason it exists.