Two algorithms for randomly generating aperiodic tilings

[Simon Tatham, 2023-04-10]

[Part of a series: Penrose and hats | Spectres]


In the 1970s, Roger Penrose discovered several sets of polygons which will tile the plane, but only aperiodically, without the tiling repeating in a fixed pattern.

Samples of his two best-known tiling types, called P2 and P3, are shown below. P2 is composed of quadrilaterals usually referred to as ‘kite’ and ‘dart’; P3 is composed of two sizes of rhombus, known as ‘rhombs’ in this context for some reason:

P2 tiling: kites and darts
P3 tiling: thin and thick rhombs

(Strictly speaking, any one of these polygons can tile the plane periodically, because so can any quadrilateral at all. The tiling is only forced to be aperiodic if you restrict the ways the tiles can connect. You could do this by putting jigsaw-piece protrusions and holes on the edges, but that’s ugly, so people normally prefer to show the plain quadrilaterals and let the matching rules be implicit. In this article I’ll ignore this detail just like everyone else.)

In 2023, a long-open question was answered: can you force aperiodicity in the same way using only one type of tile, instead of Penrose’s two? You can! The shape in question is described as a ‘hat’ by its discoverers Smith, Myers, Kaplan and Goodman-Strauss, and a sample of its tiling is shown below:

Hat monotiling

(There’s a quibble with this one too: the hat shape is asymmetric, and in order to tile the whole plane, you need to use it in both handednesses. For some purposes this still means you need two types of tile! But from a mathematical perspective it’s more natural to regard reflections as not a fundamentally different shape.)

One of the games in my puzzle collection, namely “Loopy” (also known as Slitherlink), can be played on a lot of different tilings of the plane. Most of them are periodic, but all three of these aperiodic tilings are also perfectly feasible to play Loopy on. So the question arises: how do you best get a computer to generate pieces of these tilings?

One really obvious option would be to pre-generate a fixed patch of each tiling, and set Loopy puzzles on that by just varying the puzzle solution (the closed loop you’re trying to reconstruct from clues). But that’s boring – surely the whole point of a tiling that does not repeat is that it shouldn’t be the same every time! So when you request a new game, you should get a different patch of the tiling itself, as well as a different loop to find in it.

So our mission goal is: given an output region and a source of random numbers, generate a randomly selected patch of each of these three aperiodic tilings.

In this article I’ll discuss two very different algorithms for doing this, and say which one I prefer. (Spoiler: there will be a very definite one that I prefer.)

Substitution systems

All three of the tilings we’ll discuss here can be generated recursively by the same basic method.

For each of these tilings, there’s a system for starting with an existing tiling, and then substituting each tile for smaller tiles according to a fixed rule. This generates a new tiling, using the same set of tiles, but more of them at a smaller scale.

Because the output tiling is of the same type, it follows that you can substitute that in the same way, and so on. So if you repeat this many times, you can start from a single initial tile, and expand it into as large a region of the plane as you like.

One slight complication, in every case, is that the most convenient substitution system is not based on the actual tiles you want to end up with. In each case, you have to substitute a different system of tiles, and after you’ve done that enough, convert the result into the final tiles you want as output.

Penrose tilings: half-tile triangles

The substitution systems for the Penrose P2 and P3 tilings work in very similar ways, so I’ll describe both of them together.

In both cases, the most convenient way to think of the substitution system is to imagine each of the starting tiles being divided in half, to make two isosceles triangles. That way, the substitution can replace each of those triangles with either two or three smaller triangles, exactly filling the same area as the original triangle. (If you didn’t do this, then most smaller tiles would half-overlap two larger tiles, and it would be more awkward to keep track of which tiles you’d expanded from where.)

The divisions for each tile type look like this. For the P2 tiling, the kite and dart are each divided down their line of symmetry. For the P3 tiling, the thin rhomb is cut along its short diagonal, and the thick one along its long diagonal:

[p2-tile-acute.svg] [p2-tile-obtuse.svg]
P2 tile divisions
[p3-tile-acute.svg] [p3-tile-obtuse.svg]
P3 tile divisions

(Therefore, in the P2 tiling, the acute isosceles triangles are larger than the obtuse ones, whereas in the P3 tiling, vice versa.)

The two halves of each original tile are shown in different colours, because they will be treated differently by the substitution rules. Specifically, they will be mirror images of each other.

Here are the substitution rules for each of those types of triangle:

[p2-acute-undivided.svg] [arrow.svg] [p2-acute-divided.svg]
[p2-acute-reflected-undivided.svg] [arrow.svg] [p2-acute-reflected-divided.svg]
[p2-obtuse-undivided.svg] [arrow.svg] [p2-obtuse-divided.svg]
[p2-obtuse-reflected-undivided.svg] [arrow.svg] [p2-obtuse-reflected-divided.svg]
P2 substitution rules
[p3-acute-undivided.svg] [arrow.svg] [p3-acute-divided.svg]
[p3-acute-reflected-undivided.svg] [arrow.svg] [p3-acute-reflected-divided.svg]
[p3-obtuse-undivided.svg] [arrow.svg] [p3-obtuse-divided.svg]
[p3-obtuse-reflected-undivided.svg] [arrow.svg] [p3-obtuse-reflected-divided.svg]
P3 substitution rules

To see the complete system in action, select the radio buttons below to see the effect of running a different number of iterations, starting from a single obtuse triangle of each tiling. In each step, you can see that every large triangle transforms into smaller ones in accordance with the rules above.

P2: [p2-r0.svg] [p2-r1.svg] [p2-r2.svg] [p2-r3.svg] [p2-r4.svg] [p2-r5.svg] [p2-r6.svg] [p2-r7.svg] [p2-r8.svg]
P3: [p3-r0.svg] [p3-r1.svg] [p3-r2.svg] [p3-r3.svg] [p3-r4.svg] [p3-r5.svg] [p3-r6.svg] [p3-r7.svg] [p3-r8.svg]
Interactive demo of iterated P2 and P3 substitution

So one way to generate Penrose tilings is to expand a starting triangle like this, and when you’ve expanded it enough times, glue each pair of adjacent half-tiles back together into a single kite, dart or rhomb.

Hat tiling: a system of four metatiles

The hat tiling can also be generated by a substitution system. In this case, however, the system is based on a set of four tiles that have no obvious relation to the actual hat shape!

The paper refers to these as “metatiles”, to avoid confusion with the hat itself (the single tile). The four metatiles are shown below, with their substitution rules:

[hat-exp-H-before.svg] [arrow.svg] [hat-exp-H-after.svg]
[hat-exp-T-before.svg] [arrow.svg] [hat-exp-T-after.svg]
[hat-exp-P-before.svg] [arrow.svg] [hat-exp-P-after.svg]
[hat-exp-F-before.svg] [arrow.svg] [hat-exp-F-after.svg]
Hat metatile substitution rules

The arrows on three of the metatile types indicate orientation: those three metatiles are symmetric, but in the final conversion to hats, each one will generate an entirely asymmetric pattern of hats. So the substitution process has to remember what every metatile’s orientation should be, in order not to get the last step wrong.

(The remaining metatile is already asymmetric, so it doesn’t need an arrow.)

In these substitution diagrams, the area of the original large metatile is completely covered by the smaller ones, and the smaller ones also overlap the edges. This means that when two metatiles are placed next to each other, their expansions will overlap too. In fact, they will always overlap in a way that causes some of the resulting metatiles to coincide.

Here’s an illustration of the expansion in action, expanding a single starting metatile for a few levels of recursion:

[hat-r0.svg] [hat-r1.svg] [hat-r2.svg] [hat-r3.svg]
Interactive demo of iterated hat metatile substitution

So, similarly to the previous section, if you want to generate a large patch of hat tiling, you can start with a single metatile and expand it repeatedly until you think it’s big enough. Finally, each metatile is converted into either 1, 2 or 4 hats according to the following rules:

[hat-exp-H-before.svg] [arrow.svg] [hat-final-H.svg]
[hat-exp-T-before.svg] [arrow.svg] [hat-final-T.svg]
[hat-exp-P-before.svg] [arrow.svg] [hat-final-P.svg]
[hat-exp-F-before.svg] [arrow.svg] [hat-final-F.svg]
Rules for converting hat metatiles to hats

(Incidentally, the hat in the centre of the cluster of four expanded from the hexagonal metatile, shown here in the darkest red colour, is the only one of these hats that is reflected compared to the rest. So you can readily identify the rare reflected hats in this tiling, if you’re generating it using this metatile system.)

However, beware! The final expansion step isn’t quite as easy as it looks, because the expansion into hats is distorting.

As I’ve drawn the metatiles here, the substitution of one set of metatiles for a smaller set is geometrically precise: each large metatile from one iteration of the algorithm can be overlaid on the next iteration and the vertices will align exactly as shown in the diagrams above. But when each metatile expands into a cluster of hats, the clusters aren’t quite exactly the same sizes as the metatiles they replace. So the precise positions in the plane will vary in a nonlinear way, and there’s no simple formula that matches a metatile’s coordinates to the coordinates of its hats.

The paper offers an alternative version of this scheme in which the smallest level of metatiles do match up geometrically to the hats they turn into – but in that version, the downside is that each size of metatiles is not precisely the same shape as the next size. Instead, the metatiles themselves gradually distort as they expand.

Either way, this makes it more difficult to perform the overall procedure. With the fixed-shape metatiles as shown here, in the final conversion to hats, you can’t just iterate over each metatile and independently write out the coordinates of its hats; with the alternative version, you have the same problem during the expansion of metatiles into smaller ones. In each case, you have to do some kind of a graph search (breadth-first or depth-first, whichever you find easiest), choosing one metatile to start with and processing the rest in an order that guarantees that each metatile you process is adjacent to one you’ve already placed. That way you can figure out the location of each output hat or metatile, by knowing of an existing adjacent thing it has to fit together with.

Here’s an example, which also illustrates the distortion. This shows the same set of metatiles as in step 2 of the iteration above, and its expansion into hats. The top left corners of the two images are aligned – but you can see that at the other end, the corresponding pieces of tiling are quite offset from each other.

[hat-r2.svg] [hat-f2.svg]
Interactive demo of converting hat metatiles to hats

Choosing a random patch from a fixed expansion

In the previous section, I’ve shown how all three of our aperiodic tiling types (P2, P3, hats) can be generated by repeated subdivision of a starting tile, followed by a final postprocessing step. (In the two Penrose cases, gluing half-tiles back together into whole ones; in the hat case, substituting the four metatiles for their hat clusters.)

This is all very well if you want to generate a fixed piece of tiling. But what if you want a different piece every time?

One obvious answer is: make a fixed patch of tiles much larger than you need, and select a small region from it at random.

This works basically OK in principle. But if you implement it naïvely, it’s very slow, because you spend a lot of effort on generating all the fine detail in huge regions of the tiling that you’re just going to throw away. And the more uniform a probability distribution you want for your output region, the larger the fixed region you have to generate, and the more pointless effort you waste.

We can save a lot of that effort by not being quite so naïve about the procedure. Suppose that, instead of generating a patch of tiling and then deciding which part of it to return, we instead decide in advance where our output region is going to be within the starting tile. Then, in each iteration, we can identify a tile that’s completely outside the output region and won’t contribute any tiles to it, and discard it early, before we waste any more time expanding it into lots of subtiles that will all be thrown away.

For example, here’s an illustration of how much work this might save when selecting a small rectangle from the same P3 example pictured above:

[p3-subrect-r0.svg] [p3-subrect-r1.svg] [p3-subrect-r2.svg] [p3-subrect-r3.svg] [p3-subrect-r4.svg] [p3-subrect-r5.svg] [p3-subrect-r6.svg] [p3-subrect-r7.svg] [p3-subrect-r8.svg]
Interactive demo of P3 substitution pruned for the target area

At every stage, we identify triangles – large or small – that are out of our region, and stop bothering to expand them any further. This bounds the wasted effort at a much more acceptable level. Compare the number of triangles in the final pruned output with the number in the unpruned version!

This technique is easy to apply for Penrose tilings, because the expansion process is geometrically exact. At every stage of the process, the outline of each triangle is exactly the boundary of the region that will contain all its eventual output tiles. So it’s easy to identify whether a triangle intersects the target region.

But in the hats tiling, where the expansion process is geometrically distorting, it’s not so easy. In order to work out whether a given metatile is going to contribute to your target region, first you have to work out what coordinates in the current layer of metatile expansion correspond to the target region in the final hat tiling. This is computationally tricky: even if you could find an exact formula that translates each corner of your target region into its coordinates at the current iteration, then you’re not done, because it’s also not guaranteed that the image of a straight line between two of those corners will still be straight after the distortion is applied. In other words, if you’re trying to construct a patch of hats to fill a specific rectangle, you might find the region of metatiling you need to keep isn’t even rectangular!

It might be possible to get this technique to work regardless, by computing a conservative approximation to the target region. That way maybe you do a little more work than you need, computing a few metatiles that never contribute hats to the output but your approximation couldn’t quite prove it; but the effort saving would still be large compared to the naïve approach. But this is likely to be tricky, and has plenty of room for subtle bugs. Err in one direction, and you do more work than you need; err in the other direction and you generate wrong output.

This entire approach has another downside. As I mentioned above, if you want your random patch of tiling to be selected uniformly from the limiting probability distribution of finite pieces of tiling over the whole plane, then you can’t achieve it this way, in an exact manner. You can only approximate, by running a large enough number of iterations that the huge patch you’re selecting a rectangle from is close enough to that limiting distribution. And the closer you want the distribution, the more you have to expand, and the more work you have to do.

More subtly, this means you have to figure out in advance how many iterations you want to run, based on the size of your target region and the error threshold you’re prepared to tolerate in probability distribution. That involves a fiddly formula which is easy to get wrong, introducing subtle bugs you’ll probably never spot.

In the next section, I’ll describe a completely different approach that avoids all these problems, and works for hats as easily as it does for Penrose tiles.

Combinatorial coordinates

The alternative approach begins with the following observation.

Suppose that, in any of these recursive substitution systems, whenever we substitute one larger tile for a number of smaller ones, we assign a different label to each smaller tile.

For example, in the P2 tiling, an acute triangle expands into three smaller triangles; so we could number them 0, 1, 2 in some arbitrary but consistent way. And an obtuse triangle expands into just two, which we could call 0 and 1. Or, perhaps more elegantly, we can observe that no two triangles expanded from the same larger triangle are the same type (counting reflections as different), so we could simply label each triangle by one of the four types (two types of acute, two types of obtuse).

For the metatiles in the hat system, each metatile has between 7 and 13 children, and some of them are the same type as each other. So the problem is larger – but not more difficult. We can still just assign each child a numeric label in an arbitrary way, so that the hexagonal metatile has children labelled from 0 to 12 inclusive, the triangle from 0 to 6, etc.

Then, after you’ve run the expansion process some particular number of times, each tile of the output can be uniquely identified by the sequence of labels it generated as it was expanded. You could identify some particular tile by saying something like: “From the starting tile, take the child with this label; from there, take the child with that label, then the child with the other label, …” and after you’ve listed the label at every step of the expansion, your listener knows exactly how to find the same tile you were thinking of.

Now, supposing I tell you the coordinates of one particular output tile, specified in this way. Can you figure out the coordinates of one of its neighbours, sharing an edge with it?

Assuming the tile you want is not on the very edge of the giant starting tile, yes, you can. And you can do it entirely by manipulating the string of labels, without ever having to think about geometry.

Penrose tilings: indexing triangle edges

We’ll tackle the Penrose tilings first. In this system, the smallest unit we ever deal with is a single half-tile triangle, so we’ll assign coordinates to each of those.

I’ll label the tile types with letters. For the P2 tiling, the two handednesses of acute triangle are called A and B; for the P3 they’re C and D. The obtuse triangles in P2 will be U and V; the ones for P3 will be X and Y.

Here’s an illustration that shows the coordinates of each output triangle for the first few layers of expansion, in both tiling types. The coordinates are written with the label of the smallest triangle first, followed by its parents in increasing order of size. (This matches the natural way to store them in an array, because that way, the fixed array index 0 corresponds to the real output triangles.) So whenever a triangle is divided into two or three smaller triangles, the coordinates for each smaller triangle are derived from the original triangle by appending a new type label to the front.

P2: [p2-labelled-r0.svg] [p2-labelled-r1.svg] [p2-labelled-r2.svg] [p2-labelled-r3.svg]
P3: [p3-labelled-r0.svg] [p3-labelled-r1.svg] [p3-labelled-r2.svg] [p3-labelled-r3.svg]
Interactive demo of iterated P2 and P3 substitution with combinatorial coordinates

We’ll also need to index the edges of each triangle in a unique way, so as to describe unambiguously which neighbour of a triangle we’re going to ask for. For example, we could say that we always assign index 0 to the base of the isosceles triangle, and the two equal sides are assigned indexes 1 and 2, in anticlockwise order from the base. So when you want to know about a particular neighbour of your starting tile, you’ll identify which neighbour by giving the index of an edge: “What triangle is on the other side of edge {0/1/2} of my current one?”

Then, for each substitution rule, we can make a map of which smaller triangles border on which others, and which of their edges correspond:

[p2-acute-edgenum-undivided.svg] [arrow.svg] [p2-acute-edgenum-divided.svg]
[p2-acute-edgenum-reflected-undivided.svg] [arrow.svg] [p2-acute-edgenum-reflected-divided.svg]
[p2-obtuse-edgenum-undivided.svg] [arrow.svg] [p2-obtuse-edgenum-divided.svg]
[p2-obtuse-edgenum-reflected-undivided.svg] [arrow.svg] [p2-obtuse-edgenum-reflected-divided.svg]
P2 substitution rules, with edge and type labels
[p3-acute-edgenum-undivided.svg] [arrow.svg] [p3-acute-edgenum-divided.svg]
[p3-acute-edgenum-reflected-undivided.svg] [arrow.svg] [p3-acute-edgenum-reflected-divided.svg]
[p3-obtuse-edgenum-undivided.svg] [arrow.svg] [p3-obtuse-edgenum-divided.svg]
[p3-obtuse-edgenum-reflected-undivided.svg] [arrow.svg] [p3-obtuse-edgenum-reflected-divided.svg]
P3 substitution rules, with edge and type labels

From these maps, with a little recursion, you can work out everything you need.

For example, suppose you’re in a type B triangle, and you want to see what’s on the other side of its edge #0. Look at the type of its parent triangle. If the parent is type A or type U, then the maps above for those triangle types show that edge #0 of the type-B sub-triangle is also edge #1 of a type-U triangle. So you can adjust the lowest-order label in your coordinate list to specify a different sub-triangle, and return success.

More symbolically: to move across edge #0 of any triangle whose coordinate string begins with BA or BU, just replace the initial B with a U.

On the other hand, suppose your type-B triangle is expanded from another type-B triangle. In that case, the map for the parent triangle says it doesn’t know what’s on the far side of your edge #0 – that’s off the edge of the map.

But that’s all right: recurse further up to find a map on a larger scale! We know that going out of edge #0 of the smaller type-B triangle means going out of edge #1 of the larger type-B triangle. So ask the same type of question one layer further up (perhaps recursing again, if necessary).

When the recursive call returns, it will give you an entire updated set of coordinates identifying the second smallest triangle you’re going to end up in, and which edge of it is on the other side of this one’s edge #1. So, finally, you can append the lowest-order coordinate to that, by figuring out which of the smaller triangles of the new large triangle is the one you need to end up at.

In our example case, there’s one final step. Going out of edge #0 of our original B meant going out of edge #1 of the larger B. But that edge is divided into two segments, each belonging to a different sub-triangle. So we need to remember which segment of the larger edge we were crossing: were we to the left or the right of the division?

We’ll always expect to find that the incoming edge of the new large triangle is subdivided into segments in the same way, and the map for that triangle will let us find which sub-triangle corresponds to each segment of the outer edge. So you can still figure out which sub-triangle you end up in.

For example, suppose we had coordinates ending BB, and our recursive call (“we’re going out of edge #1 of a B, what happens next?”) rewrote the second B to an A (perhaps modifying further labels above that) and told us we were coming in along edge #2 of the A. Then we consult the A map, and we see that its edge #2 is indeed divided in two, and coming in via the right-hand one of those segments leads us in the A child. So we end up with AA as our new lowest-order coordinates (plus whatever rewrites were made at higher orders by the recursion).

(You can check these examples in the expansion shown above. In the 3-iteration P2 diagram, there are triangles labelled BABU and BUUU, and as described above, the edge #0 of each one – that is, its short base edge – connects to UABU or UUUU respectively, with only the lowest-order label differing. But the triangle labelled BBBU is a more difficult case, requiring a rewrite at the second layer: its edge #0 connects to a triangle labelled AABU.)

Hats: indexing kite edges, and dealing with non-uniqueness

For the hats tiling, indexing the edges of the tiles isn’t quite so convenient. The hat polygon itself has 13 edges, and the maps for which ones border on each neighbouring hat are pretty complicated. Not only that, but because the metatile expansions overlap each other, every tile potentially has multiple different, equally legal, lists of coordinates. And the metatiles don’t necessarily meet edge-to-edge, in the sense that one edge of this metatile might join up with the edges of two other metatiles.

To solve the first of those problems, I found the easiest thing is to stop considering a whole hat at a time, and use a smaller and more regular unit. The hat tiling can be made to align neatly with a periodic tiling of the plane with kites, shaped so that six of them joined at the pointy ends make a regular hexagon and three joined at the blunt ends make an equilateral triangle:

A hat aligned to its underlying grid

If you align the hats to this tiling, then every hat occupies exactly eight whole kites, as shown above. And the kites are simply shaped, with just four edges. So I found it’s easier to consider each individual kite to have a set of combinatorial coordinates, and to devise an algorithm that will tell you the coordinates of a neighbouring kite, given the coordinates of the starting kite and which of the four edges you want to head out of.

The coordinates of an individual kite look something like this:

and so on. At the highest-order end, this terminates with you knowing the outermost metatile type, but not anything about its parent, or which child of that parent it might be.

To navigate this coordinate system, we’ll make a set of giant lookup tables. But first, we’ll have to assign numeric indexes to all three of the expansion processes involved: metatiles to smaller metatiles, then to hats, then to kites. It doesn’t matter how we do this, as long as we do it consistently.

[hat-exp-H-before.svg] [arrow.svg] [hat-exp-H-after-coords.svg]
[hat-exp-T-before.svg] [arrow.svg] [hat-exp-T-after-coords.svg]
[hat-exp-P-before.svg] [arrow.svg] [hat-exp-P-after-coords.svg]
[hat-exp-F-before.svg] [arrow.svg] [hat-exp-F-after-coords.svg]
Metatiles with child metatile indexes
[hat-exp-H-before.svg] [arrow.svg] [hat-final-H-coords.svg]
[hat-exp-T-before.svg] [arrow.svg] [hat-final-T-coords.svg]
[hat-exp-P-before.svg] [arrow.svg] [hat-final-P-coords.svg]
[hat-exp-F-before.svg] [arrow.svg] [hat-final-F-coords.svg]
Metatiles with hat indexes
Single hats with kite indexes

The first type of lookup table we’ll need, I call a kitemap. For each type of second-order metatile (i.e. each possible value of m2), you expand it into its set of first-order metatiles, and then into hats, and then into kites, and you keep the coordinate labels you generated at each stage. To illustrate this, here’s the kitemap for the triangular metatile. (It make a manageable example, because it’s the smallest. The others are similar but larger.)

[hat-metamap0.svg] [hat-metamap1.svg] [hat-kitemap-hats.svg] [hat-kitemap.svg]
Interactive demo of building one of the four kitemaps

From this visual map, you can read off the coordinates of each kite adjacent to a starting kite. For example, consider the kite labelled 7.3.0, at the top right of the central hat in this diagram: you can see that the four kites bordering it are 0.0.0, 6,3,0, 3.1.1 and 4.1.1, and you can identify which edge of the kite takes you to each of those neighbours. So if the coordinates of your current kite started with (khc) = (7, 3, 0), then you would know how to generate the coordinates of each neighbouring kite, simply by rewriting those three low-order labels.

The version of the kitemap used by the algorithm is not a visual map like this: it’s a lookup table, or rather one for each of the four metatile types. Each table is indexed by the triple (khc) and a particular kite edge, and it tells you the new values of (khc) corresponding to the kite on the far side of that edge.

So, to compute the coordinates of a neighbouring kite, you start by choosing the kitemap lookup table corresponding to the metatile type m2, and look up (khc, kite edge) in it. If it has an answer, you’re done – just replace the three low-order indices in your input coordinates with the ones you got out of the kitemap lookup table, and you’ve got a set of coordinates for the new kite.

(Of course, when you rewrite c in the coordinates, you must also rewrite m to match the type of the new first-order metatile. To do this, you just need a much simpler lookup table of which type of metatile each child is.)

But sometimes this doesn’t work. For example, suppose you’re in the kite labelled 0.0.4 in the above map, right at the bottom. Then traversing one of the four kite edges will take you inwards to 1.0.4, but for the other three, this map has no answer. In cases like this, the entry in the kitemap lookup table for “what happens if I head in this direction from here?” will be a special value meaning “don’t know, that’s off the edge of the map.”

By analogy with the Penrose case, an obvious thing to try here might be to specify (in some way) which edge of the kitemap you’re stepping off, and then recurse to a higher layer to find out where you’ve come back in to some other kitemap. If we had made smaller kitemaps based only on a single first-order metatile (i.e. just containing 1, 2 or 4 hats), then we would have no choice but to do exactly this, because all first-order metatiles’ hat expansions are disjoint. But hat expansions of metatiles have extremely twiddly and complicated edges, and that sounded like a very tricky thing to get right.

And we don’t have to! There’s a nicer way, based on the fact that the metatile expansions overlap each other. If we’re about to head off the outermost border of the kitemap, that must mean that we’re near the edge of the expansion of our second-order metatile m2. So the first-order metatile we’re in must be one of the outermost metatiles in that expansion – which means it overlaps with the expansion of another second-order metatile, or maybe even two of them. And among the second-order metatiles whose expansions contain this particular first-order one, there must be at least one in which we’re about to step inwards towards the centre, not outwards towards the edge.

In other words, the fact that there are multiple valid coordinates for the same kite is not a problem after all. It’s an opportunity! If we can rewrite two layers of metatile child index in our coordinate system, then we can find an equivalent representation of our current kite, which places it in a kitemap that we’re not about to step off the edge of. And now we’ve completely avoided having to deal with those long crinkly complicated edges of all the maps.

To implement this step, we make a second type of map, which I call a metamap. This time, take each third-order metatile type m3, and expand it twice, into second- and then into first-order metatiles. As you do that, keep track of all the different ways that each first-order metatile is generated (there will be more than one if it’s in the overlap between the expansions of two or three second-order metatiles). So we end up with a map in which some metatiles have multiple coordinates:

[hat-metamap0.svg] [hat-metamap1.svg] [hat-metamap2.svg]
Interactive demo of building one of the four metamaps

Finally, we translate that map into a lookup table indexed by any of the coordinate pairs in this diagram, giving all the other coordinate pairs equivalent to it. This lookup table allows you to rewrite the part of the coordinate system that says

“… metatile type m, which is child c of type m2, which is child c2 of m3 …”

by selecting the metamap for tile type m3, and looking up the pair (cc2) in it. The metamap might return you one or two alternative (cc2) pairs, and for each one, you can substitute those in your coordinates (which doesn’t change which kite you’re referring to). This will also give you a new value of m2, so you’re representing your existing location relative to a different kitemap which overlaps the previous one. Now you can try looking up your new (khc) triple in it (using the rewritten value of c), and this time, perhaps you’re not on the edge of the map any more, and can find the coordinates of the neighbouring kite successfully.

If even that doesn’t work, it’s because you’re on the outer edge of the metamap, meaning you’re near the edge of the expansion of the third-order metatile m3. And now every layer above this looks the same – so you can try applying the same kind of metamap rewrite one layer up, by looking up (c2c3) in the metamap for m4 and seeing if that permits a more useful rewrite one layer down. If even that doesn’t help, try further up still, and so on.

Making it up as you go along

In both of the previous sections, I’ve described a technique for computing the coordinates of each smallest-size element of the tiling (a Penrose half-tile, or a single kite), using a recursive algorithm which looks at higher- and higher-order labels in the coordinate system as necessary.

Of course, there’s one obvious way this can fail. You can only store a finite number of coordinates. What if the recursion tries to go all the way off the top, and look at a higher-order coordinate than you have at all?

One obvious option is: report failure. If for some reason you had decided in advance that you were only interested in some specific fixed patch of tiling, and that nobody should ever go beyond the edges of that at all, then a failure of this type should be interpreted as “Clonk! You have run into the wall, and can’t go that way.”

(Perhaps if you had decided to set a text adventure game in the 7-level expansion of a type-A Penrose half-tile, or something along those lines, then this might be what you’d actually want! Each individual triangle would correspond to a room, and not all rooms would have exits in all three directions.)

But in the main situation I’m discussing here, we don’t ever want to report failure. Our aim is to generate a randomly selected patch of tiling to cover a particular target rectangle. Just because that rectangle turns out to protrude over the edge of the highest-order tile we currently know about doesn’t mean there’s no possible way to extend the tiling in that direction; it just means we haven’t yet made a decision about which way to extend it.

If you imagine an infinite tiling of the entire plane in any of our tiling systems, then in principle, any given element has a set of coordinates that go on forever to higher- and higher-order, larger and larger, supertiles. If we had started by inventing an infinite sequence of coordinates for our starting element, then there would be no problem with running out of coordinates during the recursion – we could recurse as high as necessary.

Of course, you can’t invent infinitely many things up front in a computer. But what you can do is to invent them lazily, by inventing the first few, and being prepared to generate a new one as and when it’s requested.

So this is what we do. In our algorithm, we invent some random coordinates of our original starting element, up to a certain point. And then, if the recursion ever tries to go beyond the level we know about, we simply invent another layer, on demand, append it to the coordinates of the starting element, and pretend it was there all along, and that our starting element has always had that extra coordinate. So if any other coordinate list lying around elsewhere in the software is shorter than the current coordinates of the starting position, then we can extend it by appending the extra elements that have been appended to the starting position since it was generated. (Because it has the semantics “When we last got here, we didn’t know what the next few levels looked like, but now we do know”.)

Of course, when you make up a parent coordinate at random, you have to make it consistent with the rules about which tiles can be parents of which other tiles. In the Penrose system, for example, you can see from the maps in a previous section that the type-B triangle can be a child of A or U or another B, but it can’t be a child of a V. So if your current topmost tile type is B, then you must select the next one up by randomly choosing from the set {A, U, B}, and not from all four tile types. Similarly in the hats system, not all metatiles can appear as children of other metatiles: the triangular metatile only appears in the expansion of the hexagonal one, so if your current topmost metatile is a triangle, then you only have one choice for what the next larger one will be. And even if your current metatile occurs as a child of all the metatile types, you also have to decide what its child index will be relative to its parent, and that index must be chosen in a way that matches the tile types.

Better still, once you’ve finished generating your output patch of tiling, you can retrieve the full set of coordinates you ended up having to invent for the starting position – and those can act as a short and convenient identifier that allows another person running the same algorithm to generate the identical piece of tiling, because they should never find themselves needing a coordinate you hadn’t already generated. That would only happen if they tried to extend your tiling into a larger region than you had tried yourself.

Another thing you can do here is to choose the next parent using a deliberately biased probability distribution, to match the limiting distribution over the whole plane.

In each of these tiling systems, there’s a system of linear equations that tells you how many tiles of each type you have after an expansion, if you know how many tiles of each type you started with. By eigenvalue/eigenvector analysis, you can process these equations to determine the proportions of the various tile types occurring overall – or, more precisely, the limit of the proportions in a very large finite patch, as the patch size tends to ∞.

This overall distribution of tile types, in turn, can be processed into a set of conditional probabilities, each answering a question of the form “Of all the tiles of type t in the plane, what proportion are child #n of tile type u?” So if you work out all those probabilities, then you can select the next coordinate pair (nu) in a way that matches that limiting distribution.

(Also, when you randomly select your initial tile, you should do it according to the original, unconditional, version of the limiting distribution.)

The effect of this refinement is that our piece of tiling will be generated as if the coordinates of the starting element had been chosen uniformly at random from the whole plane! That’s normally a meaningless concept, but in this case it (just about) makes sense, because the number of output patches of tiling you could possibly return is finite, and the distribution of those converges to a limit if you choose uniformly from larger and larger regions of the plane.

Compare this to the recursive-expansion technique for generating tilings. In that technique, your output distribution is – by construction – the one you’d get by choosing from an actual fixed patch of tiling of a finite (if large) size, because that’s exactly what you are doing. If you start from a larger and larger patch, your output distribution approaches the idealised limit, but in order to get it closer and closer, you have to do more and more work. Here, we can directly generate a tiling from precisely the limiting distribution, and we didn’t have to do any extra work at all to get there!

Generating the output tiling

So, now we have a method for moving around a tiling and determining a string of coordinates describing each smallest-sized element we come to. What do we do with that?

Our ultimate aim is to actually generate a set of tiles that cover the specified target area. For this, we don’t really need the full set of higher- and higher-order coordinates at all. We only need to know the lowest-order tile type labels: enough to know which kind of thing appears next in the tiling. Those low-order labels are the output of the algorithm. The higher-order parts of the coordinate system are just an internal detail of how the algorithm keeps track of where it had got to. (Also, as I mention in the previous section, they can also act as an identifier to make a particular run repeatable.)

So, to use this technique to actually generate a Penrose tiling, a simple approach is to use a graph search (say, breadth-first). Pick an initial starting triangle, and place it somewhere in your target area (the centre is an obvious choice, but anywhere will do). Then, repeatedly, find some triangle edge that you don’t know what’s on the other side of yet, and compute the combinatorial coordinates of the triangle on the other side, by starting from the coordinates of the triangle you already have. The lowest-order label in the new triangle’s coordinates will tell you which shape of triangle it is (acute or obtuse), and which of its edges corresponds to the edge you just traversed. That’s enough information to calculate the coordinates of the new triangle’s third vertex, starting from the two endpoints of that edge. Now add the two new edges of that triangle to your queue of edges to try later. Then go back and pick another edge to explore, and so on.

At every stage, if you generate a triangle that’s completely outside the target area, there’s no need to store it or to queue up its other edges. If you discard out-of-bounds triangles as you get to them, then this search process will eventually terminate, because the queue of edges that still need exploring has run out, and you’ve covered the whole target region in half-tile triangles.

Every time you generate the coordinates of a triangle, you find out which of the four triangle types it is: not just whether it’s acute or obtuse, but also which handedness it is. So you know which half of an output tile each triangle is. Therefore, as you go along, you can also generate the full output Penrose tile corresponding to every triangle. This will generate most of them twice (any tile with both halves inside the target area), but as long as you notice that and de-duplicate them, that’s fine.

For generating the hat tiling, you could do the same kind of thing – but there’s an even easier approach that doesn’t need a graph search at all, using the fact that the whole tiling lives on a fixed grid of kites.

For the hat tiling, the easiest approach is to specify your target region as a connected set of kites, and simply iterate over that set in some fixed way. For example, to fill a rectangular region, you might process the kites in the region in raster order, iterating along each row from left to right and then going on to the next row. Or you might find it easier to process alternate rows in opposite orders. As long as you process the kites in an order that means every kite (after the first one) is adjacent to a kite you’ve already found the coordinates of, anything will work. At the end of the iteration, you know the combinatorial coordinates of every kite in your region.

Once you’ve done that, you can generate the output in whichever way is easiest. For every kite in the region, you know which kite it is out of the 8 making up its containing hat, and that (together with knowing whether the hat is a reflected one) is enough to generate its complete outline. So you could do that for every kite in the region, and discard duplicates.

But an even easier way than that is to only generate an output hat when the iteration finds a kite which was #0 in its hat, and even then, discard it if that whole hat doesn’t fit in the output area. This guarantees to generate every valid output hat exactly once, so no de-duplication is even needed.

(If you need to generate hats that cover your output area, instead of the subset of hats that fit entirely inside it, then that algorithm can still be used – just expand the output region by the maximum width of a hat on each side, so that any hat that intersects the true region must fall entirely within the expanded one.)

As long as you’re only looking at the three coordinates (khm) of each kite (which kite it is, in which hat, of which type of first-order metatile), it doesn’t matter that multiple different coordinates can refer to the same location, because they’ll all agree on those three points. The lowest-order coordinate index that can differ is c, the question of which child of which second-order metatile the first one is. So the information you really need (kite index, and whether the hat is reflected) is easy to read off reliably.


I’ve mentioned in passing a few ways that this combinatorial coordinate technique is superior to recursive expansion. But let’s bring them all together:

Uniform distribution. By generating each coordinate with the correct probability distribution, you can arrange that your output patch of tiling is chosen from precisely the overall limiting distribution, and not just some approximation to it based on a large finite region.

No need to engage with geometry of higher-order tiles. The question of what points in the plane might be the vertices of any higher-order tile just never comes up, in this system. The only thing we ever track about the higher-order tile layers is their combinatorial nature: what tiles exist at all, which ones are children of which, which ones are adjacent to which, which ones overlap which. This isn’t such a huge advantage in the Penrose tilings, where the higher-order geometry was still fairly easy. But in the hats system, the distortion of the metatiles is made completely irrelevant by treating them purely combinatorially, so we never have to worry about it at all.

No risk of overflow. In the recursive expansion technique, we need to generate actual coordinates for all the tiles at every layer. If we’re choosing a small output region from a really big patch, this might mean we have to worry about floating-point precision loss in the largest scales – or alternatively integer overflow, if we use exact integer-based coordinates (see the appendix below). But in this system, the only output points we ever generate are the ones in the actual output region, so we only need to ensure that our number representation is accurate enough for that. The much larger tiles surrounding the output region are still described in this algorithm, but instead of being described by very large single numbers (representing vertices in the plane), they’re described by a long string of small numbers (the labels in the combinatorial coordinate list). It’s as if we had transferred the working state from machine integers or floats into a bignum representation – just a slightly weird two-dimensional mixed-base one.

No manual tuning needed. In the recursive expansion technique, you have to think about the size of output region you want, and choose up front how many recursion levels you’re going to perform (also based on what error you’re prepared to tolerate from the ideal probability distribution). But in this system, you find out after the algorithm finishes how many levels of coordinates it turned out to need to invent. You don’t have to make a decision up front at all, which means you don’t need to debug the code that does it.

(In particular, you might not end up with the same number of coordinate labels, in two runs on the same target area with different random numbers. It will depend on whether the starting element’s randomly chosen coordinates happened to put it close to a high-order tile boundary.)

Close to linear time. The number of coordinate labels we end up dealing with is variable. But on average it will be proportional to the log of the number of output tiles. For each output tile we generate, we perform a bounded number of operations that step from the coordinates of one element to an adjacent one. Each of those steps will take O(log n) time, or maybe O(log2 n) for the hats system which might need to recurse back and forth rewriting metatile labels. But that’s only an upper bound: most coordinate rewrites will be able to succeed by adjusting only the lowest-order labels, and the occasions of recursing higher up will become progressively rarer as you ascend the layers.

I don’t have a formal proof that the average overall running time is any better than O(n log2 n). But it seems intuitively clear that it will in practice be much closer to linear time than that expression makes it look!

Only needs to use logarithmic memory. If you’re generating a hats tiling to cover a rectangular region by iterating over the kites in that region, you can arrange to keep a very small number of coordinate lists live at one time – just two or three is enough to guarantee that every kite you process is adjacent to one of a small number of past kites that you’re still keeping the coordinates for. There’s no need to store the full coordinates of every kite for the duration of the algorithm: you can emit each hat as you come to it in a streaming manner, and forget nearly everything about the ones you’ve already generated.

So this system is suitable for generating absolutely enormous patches of hat tiling, with only a modest cost in memory usage. The running time is close to linear in the number of output hats, and the average memory usage is just O(log n). Probably whatever is printing out the hats after you generate them will have higher cost than that!

I think it should be possible to work this way in the Penrose case too, even without a convenient underlying grid to iterate over. But it would be fiddly. I think you could do it by iterating along a sequence of horizontal lines of the output region, at a spacing small enough to guarantee any triangle must intersect at least one line. Then, instead of adding both of each new triangle’s outgoing edges to a queue, instead choose just one of them to traverse next, by picking the one that intersects the horizontal line you’re currently following. Meanwhile, a similar iteration in the perpendicular direction is finding all the triangles on the left edge of the region, to begin each horizontal iteration at. To avoid emitting any triangle twice, check to find how far above the lowest point of the triangle the current line hits it; if that’s greater than the line spacing then you know a previous sweep must have already caught it.

I’m fairly sure this would work, but I haven’t tried generating Penrose tilings in this low-memory mode. I’ve only tried the more obvious breadth-first search. Also, this would be slower by a constant factor in payment for the memory saving – you’d visit most triangles more than once.


I’ve presented two algorithms for generating aperiodic tilings at random. The first algorithm, picking a random region from a large fixed area, is efficient enough for Penrose tilings, but works rather badly for hats because of the distortion problem. The second, combinatorial coordinates, is efficient for both types of tiling, and especially so for hats, where the other algorithm performed worse. Also, it has many other assorted advantages, listed in the previous section.

The combinatorial coordinates system described here is in use in the live version of Loopy for generating hat tilings, and is working very nicely.

The Penrose tilings in Loopy are generated by the other algorithm, picking a random region from a fixed patch. At the time Penrose tilings were implemented in Loopy, neither I nor the implementor had thought of the combinatorial coordinates system.

Now that I have thought of it, it’s tempting to rewrite Loopy’s Penrose tiling code using the shiny new algorithm. But on balance I’m not sure that I want to, because its textual game descriptions are based on specifying the coordinates of the random sub-region. So if I threw out the existing code, then existing game descriptions and save files would stop working.

(I could keep the old code, solely for parsing old game descriptions, and switch to combinatorial coordinates for new ones. But that seems wasteful in another sense – I’d end up with two pieces of complicated code where only one is really needed!)

However, I have implemented a combinatorial-coordinate Penrose generator as prototype code, and it’s pretty convenient. If I were starting from scratch in Loopy with no backwards compatibility considerations, I’d certainly choose that algorithm now!

Appendix: exact coordinate systems

In both the Penrose and hat tilings, it’s useful to have a representation for storing the coordinates of vertices of the tiling which allows you to compute without floating-point rounding errors. That way, you can easily check whether a vertex of the Penrose tiling is the same one you’ve seen before, or keep track of the coordinates of the current kite in the underlying grid of the hats tiling, without having to deal with the awkwardness that when you get back to something that logically should be the same tile or vertex, its coordinates have changed a tiny bit.

Complex roots of unity

For the Penrose tiling, one neat approach is as follows. Consider the coordinates to be complex numbers, rather than just 2-element vectors. Then let t be the complex number cos(π/5) + i sin(π/5), which has modulus 1 and argument π/5. Then, multiplying any other number by t has the effect of rotating it by exactly a tenth of a turn around the origin.

Therefore, we know that t5 = −1, because rotating by a tenth of a turn five times must rotate by half a turn, exactly negating the number you started with. So t is a zero of the polynomial z5 + 1. This polynomial factorises as (z + 1)(z4 − z3 + z2 − z + 1), and t is not a zero of the first factor (or else it would just be −1). So it must be a zero of the second factor, meaning that t4 = t3 − t2 + t − 1.

So, suppose you have two complex numbers, each expressed as a linear combination of the first four powers of t, say a + bt + ct2 + dt3 and w + xt + yt2 + zt3. Then you can multiply those polynomials together, multiply out the product, and reduce every term containing t4 or higher by applying the identity t4 = t3 − t2 + t − 1. This gives you the product of the two numbers, also in the form of a linear combination of the first four powers of t. Better still, if the the two values you’re multiplying have all four coefficients integers, then so does the product.

Also, t has further useful properties. It turns out that t + 1/t = φ, the golden ratio. This is also the ratio of the two side lengths of the triangles making up a Penrose tiling. And 1/t is just t9 = −t4 = 1 − t + t2 − t3. So φ itself has a representation in this system, with all four coordinates integers: φ = 1 + t2 − t3. So does its inverse, because 1/φ = φ − 1 = t2 − t3. So you can scale a vector length up or down by a factor of φ, by simply multiplying by an appropriate tuple of four integers, and still leave all four of its coordinates as integers in this system.

This gives you everything you need to calculate the vertices of Penrose tilings without rounding error, using either of the methods described in this article (recursive subdivision of a large starting tile, or breadth-first search via combinatorial coordinates). The coordinates of a single triangle can all be computed from each other by a combination of rotating by t and scaling up or down by φ; once you have one triangle, you can similarly compute the coordinates of the triangle next to it; and in the subdivision process the different sizes of triangle are scaled by φ each time. And every coordinate you need will be described by a tuple (abcd) of four integers, so they’re easy to check for equality!

The hat tiling is simpler, because it lives entirely on the underlying kite tiling. There are lots of ways to represent the coordinates of vertices of that tiling, and they need not be mathematically clever; any old ad hoc approach will be good enough. However, a similar trick to the above is a particularly nice approach: let s = cos(π/3) + i sin(π/3) represent a sixth of a turn around the origin, in which case s2 = s − 1 for similar reasons to above. Then you can represent all your vertices as integer combinations of 1 and s, and compute products in the same way as before, so you can rotate a sixth of a turn by multiplying by s or by 1/s = s5 = −s2 = −s + 1.

Separate x and y coordinates using √5

One thing that isn’t convenient, in the above Penrose representation using t, is checking whether a vertex is inside or outside your intended target rectangle. The four coordinates in the tuple (abcd) are pretty abstract, and have no obvious relation to x and y coordinates in the plane.

Eventually, of course, you’ll have to convert them back to actual complex numbers, to plot your tiling on the screen or the page. This is done by these formulae, which give you separate real and imaginary parts (i.e. x and y coordinates) for all the numbers involved:

t = cos(π/5) + i sin(π/5)
t2 = cos(2π/5) + i sin(2π/5)
t3 = cos(3π/5) + i sin(3π/5)

So you could just use those formulae to convert all your coordinates into floating point as you generate the tiling, and use the floating-point coordinates to decide whether any part of a tile is inside your output rectangle.

But if you’d rather do even that check without rounding error, there’s a more exact method available. It happens that cos(π/5) = φ/2 = (√5 + 1)/4, and cos(2π/5) = (√5 − 1)/4. The sines appearing in the imaginary parts don’t have nearly such nice algebraic representations (in fact they’re roots of a horrible quartic), but their ratio does: the ratio sin(2π/5)/sin(π/5) turns out to be just φ again.

(The t3 term can be reduced to the existing numbers by observing that sin(3π/5) = sin(2π/5), and cos(3π/5) = −cos(2π/5).)

So if we define X = 1/4 and Y = sin(π/5)/2 to be our basic units of distance in the x and y directions respectively, then the formulae above reduce to

t = (1 + √5)X + 2iY
t2 = (−1 + √5)X + (1 + √5)iY
t3 = (1 − √5)X + (1 + √5)iY

in which each of the x and y coordinates is expressed as an integer combination of 1 and √5 (times some fixed scale unit which we mostly ignore).

In this representation, you can exactly compare two numbers of that form to determine which is greater. Start by subtracting the two, giving a single number of the form a + b√5; then the problem reduces to testing its sign. If a and b have the same sign as each other, then the answer is obvious; if they have opposite sign, then you just need to know which of a and b√5 has greater magnitude, which is the same as asking which of a2 and 5b2 is bigger. And you can do that calculation in integers!

So if you scale your output rectangle just once to write its width as a multiple of X and its height as a multiple of Y, then you can do the entire process of tiling generation, including bounds checking, in exact integer arithmetic, and convert back to floating point only for the final job of actually plotting the points.

(You could also choose to use these √5-based coordinates for the whole job, instead of converting from the t representation. It’s slightly more inconvenient to do rotation and scaling by φ in this system, because some of the operations you need will involve integer division, whereas in the t representation, it’s all just multiplication and addition. But the divisions will be by small constant integers, and if you generate correct coordinates then they’ll never leave a remainder. So you might decide that this is less hassle overall than having two complicated coordinate systems and a conversion in between. It’s up to you.)

Appendix: interleaving the Penrose tiling types

Here’s an interesting thing about the P2 and P3 tilings.

Both tiling types have substitution systems that involve the same two types of isosceles triangle, each in two mirror-image forms: one with an acute apex angle, and one obtuse.

In P2, the acute triangle is the larger one, and the substitution step divides it into three smaller triangles – but two of them are the same two triangles that the obtuse triangle divides into. So you could imagine dividing the acute triangle first into a smaller acute triangle and an obtuse triangle of the current size, and then completing the procedure by dividing all the obtuse triangles.

In P3, the situation is exactly reversed. The obtuse triangle is larger, and its three substituted tiles include two that could also have been obtained by first making an acute triangle and then subdividing that.

It turns out that, by separating each tiling’s subdivision into these two phases, they can be interleaved! Here’s a single set of substitution rules that divide both acute and obtuse triangles into just two pieces:

[p2-obtuse-undivided-uniform.svg] [arrow.svg] [p2-obtuse-divided-uniform.svg]
[p2-obtuse-reflected-undivided-uniform.svg] [arrow.svg] [p2-obtuse-reflected-divided-uniform.svg]
[p3-acute-undivided-uniform.svg] [arrow.svg] [p3-acute-divided-uniform.svg]
[p3-acute-reflected-undivided-uniform.svg] [arrow.svg] [p3-acute-reflected-divided-uniform.svg]
Interleaved P2/P3 substitution rules

The rule with this substitution system is that, in each iteration, you only use the rules for one type of triangle – whichever one is currently larger. If you have larger obtuse triangles than acute, then you only divide up the obtuse triangles; this makes the obtuse triangles smaller and leaves the acute ones the same size (including the extra ones you just made). That means, in the next iteration, the acute triangles are larger, so this time you only subdivide those ones. So in each iteration, you use whichever rule you didn’t use the previous time.

Here’s an illustration of the resulting interleaved substitution system. In each iteration, the appropriate boundaries between mirror-image triangles are drawn fainter, so you can see that in alternate iterations they combine into kites and darts, or into rhombs, depending on which tile type is currently the larger one:

[p3-r0-uniform.svg] [p2-r1-uniform.svg] [p3-r1-uniform.svg] [p2-r2-uniform.svg] [p3-r2-uniform.svg] [p2-r3-uniform.svg] [p3-r3-uniform.svg] [p2-r4-uniform.svg] [p3-r4-uniform.svg] [p2-r5-uniform.svg] [p3-r5-uniform.svg] [p2-r6-uniform.svg] [p3-r6-uniform.svg] [p2-r7-uniform.svg] [p3-r7-uniform.svg] [p2-r8-uniform.svg] [p3-r8-uniform.svg]
Interactive demo of interleaved P2/P3 substitution

This isn’t of practical use while constructing the tilings, but isn’t it pretty?


Some useful links if you’d like to know more: