Four clusters, one substitution, and an inflation factor that cannot be rational
Proving that a tile can tile the plane takes one construction. Proving that it can never do so periodically is a claim about every tiling that shape admits — infinitely many of them, most of which nobody will ever draw. The way out is to show that every one of them has a rigid internal structure, and then that the structure and a period cannot coexist.
The hats themselves are the wrong unit for this. They do not obey a substitution rule. What the proof does first is group them into four clusters — H, T, P and F — show that every hat tiling decomposes into those clusters and in only one way, and then show that the clusters themselves group into larger clusters of the same four kinds, forever.
Each cluster type, when it becomes a larger cluster, is built from a fixed recipe of the four kinds. Write those four recipes as a matrix and the entire growth of the tiling is one linear map applied over and over. Its largest eigenvalue is φ⁴ = (7 + 3√5)/2 = 6.854101966…, and that is the factor the tile count multiplies by at every step.
| gen | H | T | P | F | matrix total | tiles in the patch | growth |
|---|---|---|---|---|---|---|---|
| 0 | 1 | 0 | 0 | 0 | 1 | 1 | — |
| 1 | 3 | 1 | 1.5 | 3 | 8.5 | 13 | 13.0000 |
| 2 | 19 | 3 | 13.5 | 21 | 56.5 | 100 | 7.6923 |
| 3 | 129 | 19 | 91.5 | 147 | 386.5 | 727 | 7.2700 |
| 4 | 883 | 129 | 625.5 | 1,011 | 2,648.5 | 5,104 | 7.0206 |
| 5 | 6,051 | 883 | 4,285.5 | 6,933 | 18,152.5 | 35,317 | 6.9195 |
The matrix is H → 3H + 1T + 1.5P + 3F; T → 1H + 1.5P; P → 2H + 2P + 2F; F → 2H + 1.5P + 3F. Its characteristic polynomial factors as λ(λ − 1)(λ² − 7λ + 1), whose largest root is (7 + 3√5)/2 — which is φ⁴ exactly.
The half-tiles are the paper’s own accounting, not a convenience: a P or an F straddles two supertiles and each owns half of it. That is why the matrix total is fractional and why the patch, which must emit whole tiles, holds roughly twice as many. Both grow by the same factor, and it is the factor that matters.
Try it: Change the seed and watch the growth column. After the first step it comes down towards 6.854 from above, whichever tile you start from — the boundary of a small patch is a large fraction of it, and that fraction shrinks as the patch grows. The eigenvalue is what is left. The first step is the exception: a T supertile has only four children, so it starts below and jumps up.
Suppose a hat tiling had a period: a translation v, not zero, carrying the tiling exactly onto itself. Because the decomposition into clusters is unique, that translation has to carry the clusters onto clusters too — and then the level-2 clusters onto level-2 clusters, and so on at every level.
But the clusters grow. Areas multiply by φ⁴ each level, so lengths multiply by φ² ≈ 2.618, without limit. Fix any v you like and go far enough up the hierarchy: eventually a single supertile is enormous compared with v, and a shift by v lands inside the same supertile rather than on a different one — which contradicts it being a symmetry of a decomposition into distinct supertiles. The only vector that survives every level is zero.
| level | area of a H supertile | × previous | linear size, relative to level 0 |
|---|---|---|---|
| 0 | 33 | — | 1.000 |
| 1 | 177 | 5.363636 | 2.316 |
| 2 | 1,185 | 6.694915 | 5.992 |
| 3 | 8,097 | 6.832911 | 15.664 |
| 4 | 55,473 | 6.851056 | 41.000 |
| 5 | 380,193 | 6.853659 | 107.336 |
| 6 | 2,605,857 | 6.854037 | 281.008 |
| 7 | 17,860,785 | 6.854093 | 735.687 |
| 8 | 122,419,617 | 6.854101 | 1926.054 |
| 9 | 839,076,513 | 6.854102 | 5042.476 |
| 10 | 5,751,115,953 | 6.854102 | 13201.373 |
The area column is exact: level-0 areas of 33, 9, 16 and 15 unit lattice triangles, multiplied by the substitution matrix. The ratio settles on φ⁴ = 6.854102, so lengths grow by φ² = 2.618034 and the supertiles run away from any fixed vector you care to name. Levels above 4 are not drawn — tracing a level-5 outline means walking 35,317 tiles — but the arithmetic continues without them.
Try it: Set the candidate period as long as the slider allows, then walk the level up. The blue arrow is drawn at the same scale as the supertile every time, and by level 4 it has all but vanished. The table carries the same arithmetic out to level 10, where an H supertile covers 174 million times the area of the H it stands for.
The argument above is only as good as the word unique. What makes it work is that a tile’s position in the hierarchy is not a choice: given the tiling, each tile belongs to one level-1 supertile, which belongs to one level-2 supertile, and the chain is forced all the way up.
Click any tile below and the instrument prints its chain. Turn on the supertile outlines and move the level slider to watch the same tiles regrouped at each scale — the regrouping is nested, never crossing, which is the visual form of the same fact.
Generations stop at 6, which is 242,962 tiles. The substitution keeps going — generation 7 from an H is 1,667,659 — but building that takes about 0.7 s and 25 MB, so the instrument does not offer it.
Try it: Set the level to 1 and then to 4 with the outlines on. No outline ever crosses another: the level-4 supertiles are unions of level-3 ones, exactly. Then click a tile and read the chain — a nesting like P ⊂ F₁ ⊂ H₂ ⊂ H₃, one line that fixes the tile’s place at every scale at once.
There is a second argument, shorter and entirely numerical. A periodic tiling has a fundamental domain: one finite patch, repeated. Every proportion you could measure in it — how many H per T, what fraction of hats are reflected — is a count of things in that one patch divided by another count in the same patch. It is a ratio of integers.
In the hat tiling those proportions are the matrix’s dominant eigenvector, and they are irrational. The cleanest one: φ⁴ unreflected hats for every reflected hat. A tiling in which one hat in 7.854 is a mirror image cannot be built by repeating any finite patch, because 7.854 is not a fraction.
The black tick on each bar is the limit. The bar is where the proportion stands after the chosen number of substitutions, starting from one tile of each type — and where it starts does not matter, because the matrix has a single dominant eigenvalue.
The reflected hats are counted through H, the only metatile that contains one. The paper draws 4 hats in an H, 1 in a T, and 2 each in a P and an F; exactly one hat per H is reflected. On the limiting proportions that gives φ⁴ unreflected hats for every reflected one, and φ⁴ is irrational — so no finite patch, repeated, could produce this tiling.
Try it: Drag the substitution count up from zero. The proportions start at a quarter each and walk to their limits within a dozen steps. H settles on exactly one third — the only rational one of the four — while P goes to √5 − 2 and F to 2 − φ.
Both arguments start from the hierarchy, and the hierarchy is the part this page assumes rather than proves. Showing that every hat tiling decomposes into H, T, P and F clusters — and in exactly one way — is the long half of the 2023 paper, done by an exhaustive analysis of how hats can meet. What the machinery above adds is that once you have it, aperiodicity follows twice over. The tile that needs no reflections at all came two months later.