The count of reflected hats is never one out

The tile that needs no reflection measured how often a hat tiling uses its minority handedness, on patches of a few dozen tiles, and got a number near one in eight. It then quoted the exact ratio, the fourth power of the golden ratio, from the paper that announced the tile, and it was careful about why: no measurement on a finite patch can distinguish an irrational number from a rational one nearby, and it is the irrationality that matters. That essay gave one line to where the number comes from, in passing. The reflected tiles sit in one kind of cluster, it said, and cluster counts at successive levels are the entries of an integer matrix’s powers.

This essay computes it. The table of how clusters make clusters takes four rows, and solving it needs nothing beyond the arithmetic of $\mathbb{Q}(\sqrt{5})$. What it gives is more than the one number. It gives the exact frequency of each cluster, one of which turns out to be a third on the nose. And the rest of the table’s arithmetic gives something that the ratio could not have said and that a measurement could never have shown: the reflected count in a cluster of any size is never more than half a tile from where the exact share puts it. A tile that never repeats nevertheless keeps its books to within one tile at every scale, and that is the same fact as its having sharp diffraction, seen from an unexpected side.

The table, and what is taken on trust #

The convention here is the published one. In the paper by Smith, Myers, Kaplan and Goodman-Strauss from 2023, any tiling by hats breaks up in exactly one way into clusters of four kinds, called H, T, P and F. They hold four, one, two and two hats. Each H contains one reflected hat, and no other cluster contains any. Those clusters then group into larger clusters of the same four kinds, and so on upward without end, so each hat tiling is a hierarchy of clusters with no top.

The table is quoted, and so is the theorem behind it. That every hat tiling has this hierarchy is a hard result about forced local configurations, proved in that paper by a long case analysis, and it is not reproved here. One tile, and no period found the hat by a search and could show only that it covers large patches, which is the reason for caution about what any calculation on this page establishes. Everything below is conditional on the table and the theorem, and everything below is exact.

The frequencies, solved rather than estimated #

Write $M$ for the table as a four-by-four matrix of non-negative integers. If a patch holds a vector $c$ of clusters at one level, it holds $cM$ at the next. So a level-$k$ H cluster holds the top row of $M^k$, and after a few levels the rows of $M^k$ all point in the same direction: that is the Perron–Frobenius theorem, since $M$ has a positive power. The common direction is the left eigenvector for the largest eigenvalue, and normalised to sum to one, it is the frequency of each kind of cluster in every hat tiling at every level.

The largest eigenvalue is $\varphi^4 = (7 + 3\sqrt{5})/2 \approx 6.854$, so one level multiplies the number of clusters by about seven. The eigenvector is solved here by row reduction with exact pairs of rationals standing for $a + b\sqrt{5}$, so no rounding enters.

The answer is tidy enough to read aloud. $H$ has frequency exactly $1/3$, and there is no golden ratio in it. $T$ has $1/(3\varphi^4)$, and the form of that is immediate from the table: a T is only ever made from an H, one apiece, so $T$ at each level is $H$ at the level below, shrunk by the growth factor. $P$ has $\sqrt{5} - 2$ and $F$ has $(3 - \sqrt{5})/2$.

The hats follow by weighting. A cluster mix of those frequencies holds $4/3 + 1/(3\varphi^4) + 2(\sqrt{5} - 2) + (3 - \sqrt{5})$ hats per cluster, of which $1/3$ are reflected, and that simplifies to

$$\frac{\text{unreflected}}{\text{reflected}} = \frac{7 + 3\sqrt{5}}{2} = \varphi^4.$$

So the number that the earlier essay quoted comes out of a table of sixteen small integers. It is also exact, and the arithmetic that gives it says so without approximation: the program’s output is the pair of rationals $7/2$ and $3/2$, not a decimal.

That is the proof of non-periodicity promised in that essay, stated in two sentences. A periodic tiling has a rational share of reflected tiles, since a period cell holds a whole number of each. Every hat tiling has the hierarchy, hence this share, and the share is irrational, so no hat tiling has a period. The irrationality does the work, and it has been computed rather than cited, although the hierarchy it rests on has not.

The same style of argument, a frequency that cannot be rational, is what the Penrose tilings use with two tiles and the golden ratio. How often each patch occurs did it for a chain. There is one difference worth noticing. For Penrose tiles the irrational ratio is between two shapes, so it is possible in advance that a tiling could have any proportion, whereas the hat has one shape and two orientations of it. What the ratio constrains is a choice no shape forces, the handedness of a tile at a position, and that was the whole question that made a reflection-free variant important.

A cluster, level by level #

Exact counts need only integer multiplication, so they can be taken as far as anyone likes.

The ratio settles quickly, gaining more than a decimal place at each level and landing on alternate sides of the limit. Twenty-seven over four, the rational number the earlier essay used as an example that a small patch could not rule out, is passed for good at the third level. So far this is what one expected.

The right-hand column is the surprise. The error is the actual reflected count minus the count the irrational share predicts for that many hats, and it does not grow. A level-eight cluster has seventeen million hats and two million reflected ones, and the two numbers are related by an irrational ratio, so the exact count can never be an integer. The prediction is out, but by less than half a tile. At level ten, 821 million hats, it is out by exactly the same amount as at level eight, to the fourth decimal. It alternates between $0.4824$ and $-0.1843$ and it keeps doing so.

This is not what an arbitrary tiling would do, and there is a direct comparison.

The straight line is the square-root growth of independent chance. At $1.8 \times 10^{12}$ hats its expected error is about 449,000 tiles. The hat’s error at that size is 0.48. That is a ratio near a million, and the gap keeps widening. So the reflected tiles are spread with a regularity that no random arrangement has at any size, in a tiling that has no period. This is the difference that order is not periodicity is about, and it can now be stated as a number: how far a count strays from its average.

Why the error is bounded #

The explanation is the part of the matrix that the frequencies did not use. $M$ has four eigenvalues, and the frequency used only the largest.

The characteristic polynomial is $x^4 - 7x^3 + 7x - 1 = (x^2 - 1)(x^2 - 7x + 1)$, which gives the roots at once: $\varphi^4$ and $\varphi^{-4}$ from the second factor, and $1$ and $-1$ from the first.

Any count in a level-$k$ cluster is a fixed combination of the $k$-th powers of those four numbers. The part from $\varphi^{4k}$ grows, and is exactly what the frequency predicts. Subtract that, which is what the error does, and what remains is a combination of $\varphi^{-4k}$, $1^k$ and $(-1)^k$. The first dies away. The second is constant. The third flips sign every level. So the error must settle to a constant plus or minus another constant, and from the table its two values are $0.4824$ and $-0.1843$, the sum and difference of those constants.

Nothing has been said about geometry in that argument. It is a statement about the table. It is also about as strong as such a statement can be. A number greater than one whose conjugates all lie within the unit circle is called a Pisot number, and $\varphi^4$ is one, with the single conjugate $\varphi^{-4}$. Inflation tilings whose growth factor is of this kind are the ones known to have sharp diffraction, and a growth factor of the other kind generally gives none. The table’s two extra eigenvalues of size exactly one are the reason the error settles to two values rather than to zero. In the Fibonacci chain, by contrast, the matrix has only the eigenvalues $\varphi$ and $-1/\varphi$, and the same error shrinks towards zero with the level.

Two numbers the hierarchy can never change #

An eigenvalue of size one is more than a bound. It means something is conserved, and here it can be written out, because the eigenvalues are integers and so their eigenvectors are too. Solve the table with the eigenvalue $1$ and the answer is a linear combination of the cluster counts that no level of the hierarchy alters:

$$\#H + \#T - \#F.$$

A single H has the value one. Its next level, three H, one T, three P and three F, has $3 + 1 - 3 = 1$. Go up as many levels as anyone cares to compute. At level ten, with thousands of millions of clusters of each kind, the three counts differ from one another by enormous amounts, and the combination of them is still exactly one. A T has the value one as well, a P has the value zero and an F has minus one, and those values also last for ever.

The eigenvalue $-1$ gives the second, $-3\#H + 3\#T + 2\#P + \#F$, which changes sign at every level and never changes size. For an H it reads $-3$, then $3$, then $-3$, without end.

So a cluster of any size carries two small integers that were fixed when its ancestor at the bottom was chosen. They do not grow, they cannot drift, and they are the reason that the error takes exactly two values: the reflected count is a fixed combination of the growing part and those two invariants, and once the growing part has been taken out they are all that is left. It is hard to think of a reason to expect this in a tiling known above all for lacking any regularity of the ordinary kind. No geometric meaning for either invariant is offered here. Each is a fact about the table, tested at every level by exact integers, and whether $\#H + \#T - \#F$ counts something visible in a tiling, a boundary index of some sort for instance, is a question the arithmetic raises and does not answer.

There is a connection in the other direction too, and it was not obvious in advance that one existed. How much of the hat is a crystal put scatterers on the corners of a patch and found sharp reflections at positions no lattice accounts for, and it noted that the hat’s tilings are described by a cut-and-project scheme: a slice through a higher-dimensional lattice. That result is due to Socolar, and independently to Baake, Gähler and Sadun, both in 2023. A cut-and-project set has bounded discrepancy in exactly the sense here, because whether a point appears is decided by whether its internal coordinate falls in a bounded window. So a count’s bounded error and a pattern’s sharp diffraction are two views of one structure. The bound that appears in a table of sixteen integers is what the diffraction pattern’s sharp peaks require. The table cannot prove those peaks exist. It is merely the arithmetic that any such explanation would have to agree with, and it does.

A check that knows nothing of clusters #

All of this has been arithmetic on a quoted table, and it should be tried against something independent. The search that found the hat grows patches by exact covering, with no notion of a cluster. It can report how many tiles of each handedness a patch has.

The comparison is honest but weak. The patches are small, and a patch has edge tiles, whose handedness nothing forces. The result lies around the computed share, on either side of it, and that is all it can do. In particular the bounded error of the clusters does not transfer to these patches, for a reason that matters. A level-$k$ cluster is a very special region, one whose boundary is shaped to fit the hierarchy. An arbitrary disc cuts through clusters at every level along its edge, and each of them contributes its own imbalance. For an arbitrary region of a hat tiling, nothing above bounds the error by a constant. The boundary argument gives something growing with the length of the edge, which in the plane is the same order as the square root of the area, and so on a disc this page cannot claim to beat chance at all. A disc’s true error may well be far smaller than that bound, since the window of a cut-and-project scheme gives good control of many shapes, but that has not been measured here. Saying “never one out” of an arbitrary shape would go beyond the evidence, and the title of this essay means the clusters.

A stronger independent check would need clusters identified in a patch by a procedure that does not assume the table, and that has not been done. The agreement of the exact count with a measured patch is therefore a check that nothing has gone badly wrong, not a confirmation of the table.

What the numbers do not show #

Several things stand outside this calculation, and they should be listed plainly.

The hierarchy is quoted. If the theorem were wrong, the arithmetic here would be a true statement about a table with no bearing on hats. The theorem has been checked independently by others, including by the coordinate and computer proofs that followed the paper, but no such check is carried out on this page.

The eigenvalue argument shows a bound on the count of one thing, reflected tiles, in clusters. It shows nothing about the positions of those tiles. A row of clusters could have the right count while being arranged in any way the table allows. The table does not distinguish orders of clusters inside a parent; only the geometry does that, and that geometry is exactly what makes the tiling force itself.

And the sign of the error has no physical meaning at any one level. It alternates because the $-1$ eigenvalue alternates, and the $-1$ eigenvalue exists because of how the table is written: taking two levels at a time gives a matrix with no negative eigenvalue and a single constant error. Whether the alternating pair reflects something about the tiling, such as a two-coloured structure on the levels, or is only an artefact of how a level was defined, is not settled by counting.

What was required of the arithmetic #

Every count on this page uses arbitrary-precision integers, so a level-fourteen cluster has its exact hat count rather than a rounded one, and the frequency is computed in pairs of exact rationals. A refusal is included for the obvious wrong answers: a reflected count that does not grow with the level, a ratio equal to $27/4$. Those tests are what let the bound be a statement about the arithmetic rather than about the precision of a floating-point number, whose rounding, at $10^{12}$ tiles, would be larger than the half a tile being claimed.

Still open: the corners #

How much of the hat is a crystal asked for the plane’s share rather than one patch’s: how much of what a hat tiling scatters lies in the reflections of an ordinary crystal with partly filled sites. That is a question about the frequencies of corners, which are determined by these same clusters: each cluster kind holds a known number of corners of each kind, so the corner frequencies in the plane are a linear function of the exact eigenvector computed above. Doing it requires the clusters’ actual geometry, placed on the kite grid, and not just their counts. The result would be an exact number in $\mathbb{Q}(\sqrt{5})$ against the sixty-three per cent that patch measured, and it may or may not be a simple one. The geometry has not been built here. The first thing it would show is whether the clusters’ corners sit on the same lattice as the kite grid at every level, which is what the earlier essay’s patch suggested and did not prove.