Tiles That Never Repeat

There are two rhombi. Both have all four sides the same length. One is fat, with angles of 72 and 108 degrees; the other is thin, with angles of 36 and 144. They are the two Penrose rhombi, and the matching rules specify which tile edges may meet. Infinite tilings obeying those rules have no non-zero translation that leaves the whole pattern unchanged, although local arrangements do repeat. This is the aperiodicity theorem cited from Penrose and de Bruijn. A finite patch alone cannot establish it, and this page does not reproduce its proof.

What you can watch here is the rule that builds them, which is called deflation. Cut each rhombus in half along a diagonal, so that everything is a triangle. Then replace each triangle by smaller triangles of the same two shapes, arranged so that they exactly fill the one they replaced. A thin half becomes one thin and one fat. A fat half becomes one thin and two fat.

Starting with one thin half, one replacement gives two triangles. Eight replacements give 1,597, all inside the original outline and arranged by the same rule. The laboratory counts these pieces exactly rather than estimating their number from the picture.

The patch never grows. What changes is that the pieces are shorter by a factor of the golden ratio each round, so there are more of them, and after a few rounds the figure stops looking like a subdivided triangle and starts looking like the tiling in the photographs.

Start with the smallest thing there is: one thin half-rhombus. Two of its sides are the rhombus's edges and are equal; call that length 1. The third side is the diagonal it was cut along, and it is shorter, by exactly a factor of $\phi =1.618\dots$. So the very first figure on this page, before any rule is applied, already contains the only proportion the whole tiling is made of.

Now apply the rule once. The long side is cut at the point that divides it in the ratio $1:1/\phi$, that point is joined across to the far corner, and there are two triangles where there was one: a thin one and a fat one. Both of their rhombus edges are $1/\phi$, which is why the next round can do the same thing again. The rule is self-similar because $\phi$ is the number for which cutting this way leaves you the same shape smaller.

The counting comes next and it is the part that pays. Write $t$ for the number of thin halves and $f$ for the number of fat ones. The two rules say

a thin half makes one thin and one fat; a fat half makes one thin and two fat.

so after one more round there are $t+f$ thin halves and $t+2f$ fat ones. That is the whole of the arithmetic on this page. Start at $t=1$, $f=0$ and turn the handle:

Fat pulls ahead, because a fat half makes two more of itself and a thin one makes only one. And the ratio $f/t$ goes 0, 1, 1.5, 1.6, 1.615, 1.6176, 1.61798, which is not wandering: it is closing on $\phi =1.61803\dots$ from below, faster each round, and never reaching it.

Why $\phi$? Because if the ratio settles anywhere at all, it settles somewhere that the rule leaves alone. Suppose $f/t\to r$. Then the new ratio is $(t+2f)/(t+f)$, which in terms of $r$ is $(1+2r)/(1+r)$, and if that is also $r$ then

which is the defining equation of the golden ratio. Layer 5 does this properly, without assuming the limit exists.

This page has two things on it and they are not the same kind of thing at all.

A patch is finite. "It never repeats" is about the plane. However deep you deflate, you have a bounded piece of paper with a few thousand triangles on it. Repetition is the statement that some translation carries the whole tiling onto itself, and no finite piece can rule that out or in.

So the beautiful five-fold patch on this page is not evidence for the famous claim. It is an illustration of it. That distinction is the site's whole subject and this is the page where it is sharpest, because the picture is so convincing and so completely silent.

What the page can settle for itself is the counting, and the counting is not a lesser subject. The ratio going to $\phi$ is proved here, in layer 5, in four lines, from a matrix with four entries in it. The counts at every depth are exact whole numbers rather than measurements off a drawing. Those are small results and they are genuinely this page's own.

And then the failure preset, which is where the two halves meet.

The shapes are not what forbid repetition. The markings are. Take the same two rhombi, ignore the arrows, and lay them out in a lattice. They tile the plane with no gaps and no overlaps in a pattern that repeats every few tiles, and you can watch it do so.

That is worth doing because it is the mistake the picture invites. A reader who has been shown a gorgeous non-repeating patch and told "these two shapes never repeat" has been told something false about the shapes. The tiles alone are perfectly happy to repeat. It takes the matching rules, which are marks on the edges rather than properties of the outlines, to stop them, and that is the difficult and beautiful thing Penrose actually found.

Let $T$ denote the thin half-rhombus, with angles 36°, 72°, 72°, and $F$ the fat one, with angles 108°, 36°, 36°, both scaled so their two equal sides have length 1. Let $\sigma$ be the substitution that replaces each triangle by the smaller triangles named in layer 1, positioned as the kernel places them.

1. $\sigma$ is a partition: the triangles $\sigma$ puts inside a triangle $X$ cover $X$ exactly, with disjoint interiors. Proved, in layer 5, by writing the pieces down.
2. Writing $(t_n,f_n)$ for the counts after $n$ rounds from a patch $(t_0,f_0)$, $$\left(\begin{array}{c}{t}_{n+1}\\ f_{n+1}\end{array}\right)=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)\left(\begin{array}{c}{t}_n\\ f_n\end{array}\right).$$ Proved, immediately from 1.
3. For any starting patch with $t_0+f_0\ge 1$ and $t_0,f_0\ge 0$, $f_n/t_n\to \phi$, and the error is $O({\phi }^{-4n})$. Proved, in layer 5.
4. From $(t_0,f_0)=(1,0)$, the counts at depths 0 to 12 are $t$ = 1, 1, 2, 5, 13, 34, 89, 233, 610, 1597, 4181, 10946, 28657 and $f$ = 0, 1, 3, 8, 21, 55, 144, 377, 987, 2584, 6765, 17711, 46368. Exhaustively checked over those thirteen depths, in exact whole numbers, and cross-checked against the drawn patch to depth 8.
5. The two rhombi, as unmarked shapes, admit a tiling of the plane invariant under a lattice of translations. Proved, by the exhibition drawn in the failure preset.
6. With Penrose's matching rules, the two rhombi tile the plane, and no such tiling is invariant under any translation. Proved by Penrose and by de Bruijn, cited, and not reproduced here.

Statement 6 is the reason anybody has heard of these tiles, and it is the one thing on this page that this page does not do. Statement 5 is what stops 6 from being misread as a fact about the outlines.

1. The substitution is a partition. Take the thin triangle with apex $a$ and base $bc$, so $|ab|=|ac|=1$ and $|bc|=1/\phi$. Let $p$ be the point on $ab$ with $|ap|=1/\phi$. The segment $cp$ cuts the triangle into $△cpb$ and $△pca$, which is a partition because $p$ lies strictly between $a$ and $b$. It remains to identify the two pieces. In $△apc$ the angle at $a$ is 36°, so

so $|pc|=1/\phi$. Then $△cpb$ has $|cp|=|cb|=1/\phi$ and $|pb|=1-1/\phi =1/{\phi }^2$, which is a thin triangle at scale $1/\phi$; and $△pca$ has $|pc|=|pa|=1/\phi$ and $|ca|=1=\phi \cdot (1/\phi )$, which is a fat triangle at the same scale. The fat case is the same computation twice: cut $ba$ at $q$ with $|bq|=1/{\phi }^2$ and $bc$ at $r$ with $|br|=1/\phi$, and the segments $qr$ and $ra$ cut the triangle into three pieces which the cosine rule identifies as one thin and two fat, all at scale $1/\phi$. ∎

Two consequences follow at once and both are used everywhere on this page. The total area is unchanged, since the children fill the parent exactly. And if no two tiles of a patch overlap, then no two tiles of its deflation overlap either, because each child lies inside its own parent. That second one is why the patch at depth 8 is a tiling: not because anything was checked, but because depth 0 was a tiling and the property is inherited.

2. The matrix. A thin triangle is replaced by one thin and one fat; a fat triangle by one thin and two fat. Counting over the whole patch gives $t_{n+1}=t_n+f_n$ and $f_{n+1}=t_n+2f_n$, which is the matrix in statement 2. ∎

3. The ratio. Let $M=\left(\begin{array}{cc}1& 1\\ 1& 2\end{array}\right)$. Its characteristic polynomial is ${\lambda }^2-3\lambda +1$, whose roots are

using ${\phi }^2=\phi +1=(3+\sqrt{5})/2$. Since ${\lambda }_{+}\ne {\lambda }_{-}$ the matrix is diagonalisable. An eigenvector for ${\lambda }_{+}={\phi }^2$ satisfies $t+f={\phi }^{2}t$, that is $f=({\phi }^2-1)t=\phi t$, so $v_{+}=(1,\phi )$; similarly $v_{-}=(1,-1/\phi )$. Write the starting patch as $(t_0,f_0)=\alpha v_{+}+\beta v_{-}$. Both entries of $M$'s first column are positive and $M^2$ has all entries positive, so $M$ is a primitive non-negative matrix and $\alpha >0$ for any non-zero non-negative start. Then

since ${\lambda }_{-}/{\lambda }_{+}={\phi }^{-4}$. So the ratio converges to $\phi$, and the error falls by a factor of ${\phi }^4\approx 6.85$ each round, which is what the last column of the counts view shows. ∎

Notice what this does not say. It says nothing about where any tile is. The whole argument is about two numbers, and it would be exactly the same argument for any substitution with these multiplicities and any geometry at all.

5. The unmarked shapes tile the plane periodically. Take unit vectors $e_0,e_1,e_2$ at 0°, 36° and 72°. The parallelogram spanned by $e_0$ and $e_2$ is a fat rhombus, since the angle between them is 72°; translating it by every multiple of $e_0$ gives a band of the plane, bounded below by the line through the origin in direction $e_0$ and above by the parallel line through $e_2$. The parallelogram spanned by $e_0$ and $e_1$ is a thin rhombus, and its band sits on top of the first one with the corners meeting, because both bands have their corners spaced one edge apart in the direction $e_0$ and the upper band is started at $e_2$. Repeating the pair of bands, translated each time by $e_1+e_2$, covers the plane. The result is carried onto itself by $e_0$ and by $e_1+e_2$, so it is periodic, and it uses both rhombi and nothing else. ∎

What is not proved here. Statement 6, which is the theorem. Also not proved here: that the substitution patches, continued for ever, give a tiling of the whole plane. Both are real mathematics and both are cited rather than reproduced. And the "matching rule" this laboratory checks is a smaller and more local thing than Penrose's arrows: it is the list of ways two triangles are found meeting along an edge in the patches this page actually deflates, which is a description of those patches and not a theorem about all tilings. It is enough to point at the exact edges where the periodic lattice does something the substitution never does, and the ledger says that is all it is.

What this laboratory computes. The counts are exact whole numbers at any depth, computed as bigint iterations of a two-by-two matrix with no floating-point step in the path, so the ratio view is exact arithmetic printed as arithmetic. The drawing is not: coordinates are ordinary double-precision numbers, and they are used for drawing and for nothing else. The two are compared rather than conflated. At every depth up to 8 the number of triangles the drawing produces is checked against the number the matrix predicts, and the drawn patches are checked pairwise for overlaps and vertex by vertex for gaps at the depths the ledger names.

Why it stops at eight rounds. Depth 8 from the sun is 15,970 triangles. That is already past what an SVG stays legible at and it is a browser layout rather than a computation, so the cap is a drawing decision. Nothing about the mathematics stops there, and the counts view goes on printing exact figures well past the point where the picture would be a grey rectangle.

What the drawn patch is not. It is not evidence that no Penrose tiling repeats. It is not even evidence that the substitution continued for ever tiles the plane, which is a separate theorem requiring an argument about the limit rather than about any stage of it. If you want the first, the citations are Penrose and de Bruijn and this page has not reproduced either.

Where the mathematics is still moving. Not on this particular pair of tiles, which have been understood since the seventies. The live questions in the area are about how few tiles can force aperiodicity and what "force" should mean: the single aperiodic tile found in 2023, and the long-running work on which substitution rules produce tilings with which spectral and diffraction properties. That last one is the reason physicists care, because a diffraction pattern with five-fold symmetry is what quasicrystals produce and what the classical crystallography said could not exist.

Read this next to The Sequence That Reads Itself Aloud. Same argument, one dimension down. There a substitution acts on strings, the growth rate is again the dominant eigenvalue of a matrix built from the rule, and the answer is Conway's constant instead of $\phi$. Seeing it twice makes clear that the eigenvalue is doing the work and the geometry is along for the ride.