Cut every tile into smaller copies, inflate, and repeat forever
A substitution is a recipe saying what each kind of tile turns into: a fixed arrangement of tiles of the same kinds. Apply it to a single tile and you get a patch. Apply it to that patch and you get a bigger one. Keep going and the patch grows past any bound you name, and every stage contains every stage before it.
Hats do not obey such a rule directly, which is why the proof works with four clusters of hats instead — H, T, P and F. This page is about the rule those four obey, and about one way of picturing it that is wrong.
Each of the four cluster types becomes a larger cluster — its supertile — assembled from a fixed list of the four kinds. H′ takes thirteen pieces, T′ takes four, and P′ and F′ take ten and eleven.
Solid pieces are whole tiles the supertile owns outright; the faded ones are tiles it owns half of, the other half belonging to a neighbouring supertile. The dashed outline through the white dots is the supertile’s marked corners, and there are exactly as many of them as the H metatile has vertices — which is the correspondence that lets a supertile stand in for a tile at the next level up.
Try it: Step through all four rules. T′ is the odd one out — four pieces where the others take ten or more, which is why a T supertile grows more slowly at first and why T ends up the rarest tile in the finished tiling.
Some of the pieces in those rules are faded, and the reason is one of the places this construction goes wrong if you are not careful. A P or an F is bisected: cut corner to corner, with one half going to each of two neighbouring supertiles. Neither owns the whole tile, and the substitution matrix has to count it as a half in each.
The two halves share the cut and nothing else, and their areas add to the whole tile’s exactly — 8 plus 8 is 16. Each half’s corner list is a subsequence of the whole tile’s, so a supertile that owns a half still knows where the whole tile sits. That is what lets two neighbouring supertiles agree that they are talking about the same P rather than each drawing their own.
Try it: Compare P and F. The cut runs corner to corner in both, so each half’s corners are a subsequence of the whole tile’s — a half is not a new shape, it is a whole tile with a note attached. That is what lets two supertiles agree they are talking about the same tile instead of each producing their own copy.
The usual way to describe a substitution is: blow the tile up by a fixed factor, then cut the enlarged copy into tiles of the original size. For the Penrose rhombs that is exactly right. Here it is not right at all.
An H supertile is combinatorially equivalent to an H: it has the same nine corners in the same order, with the same sequence of edge labels round the outside. It is not similar to an H — its shape is different, and different again at the next level. The area of an H supertile is not φ⁴ times the area of an H; it is 5.36 times, and the ratio only creeps up to φ⁴ = 6.854 as the levels go by.
| level | corners | area, unit triangles | area ÷ previous | T : H |
|---|---|---|---|---|
| 0 | 9 | 33 | — | 0.272727 |
| 1 | 9 | 177 | 5.3636 | 0.322034 |
| 2 | 9 | 1,185 | 6.6949 | 0.311392 |
| 3 | 9 | 8,097 | 6.8329 | 0.309374 |
| 4 | 9 | 55,473 | 6.8511 | 0.309069 |
| 5 | 9 | 380,193 | 6.8537 | 0.309025 |
The corner count never changes — 9 at every level — and the edge labels round the supertile read AXXBXXBXX, the same cyclic sequence as round the tile itself. That is what “combinatorially equivalent” means. The shape changes: the outlines above do not lie on top of one another. And the area ratio starts nowhere near φ⁴ and only settles on 6.8541 as the levels go up. So “inflate by φ² and cut it up” is the asymptotic behaviour, not a description of any single step.
Try it: The outlines are each scaled to the same size, so similar shapes would land on top of one another. They do not. Watch the area ratio column climb towards 6.854 without ever being it, and the T-to-H column settle on an irrational limit — the number that rules out a period.
The instrument now has the control this page earns: a generations slider that applies the substitution, and a seed choice for which tile to start from. Colouring by generation shades the tiles by which supertile they belong to at a chosen level, and the outlines draw those supertiles directly.
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 generations to 1 and compare with the rule pictures above — it is the same thirteen pieces. Then go to 5 and turn on the supertile outlines with the level at 4: the patch is thirty-five thousand tiles, and the outlines show it is still just the same thirteen-piece rule, applied five times.