# Beyond the wall

Working with aperiodic tilings using finite-state transducers

[Simon Tatham, 2024-06-10]

[Part of a series: Penrose and hats | Spectres | finite-state transducers | more transducers]

## Introduction

In two articles last year, I described a very convenient recursive algorithm for generating random patches of aperiodic tilings in a computer program, using a system I called ‘combinatorial coordinates’. The first article introduced the idea, and applied it to the well-known Penrose tilings and the ‘hat’ tiling (newly discovered at the time). A few months later, the ‘Spectre’ tiling was discovered, and I wrote a second article extending the technique to cover that one too.

In this article, I’ll present a variation of that algorithm, derived by applying regular-language theory and finite state machines to the combinatorial coordinate strings.

The resulting algorithm does the same job as the previous one, but faster and more simply. (Or rather, it’s faster and simpler at run time – but at the cost of a more complicated step building the lookup tables for the tiling you want to use. But you only have to do that part once, and I’ve provided software below that knows how!)

But it doesn’t just do the same thing better. It also does more. It can handle cases that would have made the previous algorithm fail and crash the program – in fact that’s why I started looking for it. Those cases include some particularly pretty instances of the tilings, so I’ll show lots of fun pictures along the way!

### A couple of things before we start

This article is structured as a journey of discovery: I’m telling the story of how this happened to me, starting with a puzzle to solve, some techniques I invented to solve it, and other things I found out by applying those techniques once I had them. But I’m not claiming to have been the first discoverer or inventor of all of this: this isn’t an announcement of novel mathematical results. Some of the ideas here certainly had been thought of before: see the references at the end. So I only ‘discovered’ those parts in the sense that I hadn’t known them before, and there was no obvious place I was able to learn them before having to find them out the hard way.

A lot of the diagrams in this article are interactive, in the same style as the first of the two previous articles. When you see a diagram with radio buttons above it, then it’s really multiple diagrams: select different radio buttons to flip between images.

### Previously on ‘Combinatorial Coordinates’…

I’m going to refer back to details of the two previous articles a lot, but I can at least give a very quick recap of the general ideas to begin with. If you’ve got the previous articles fresh in your mind already, you can safely skip this section. If this recap leaves you completely in the dark, the two previous articles are linked above.

All of these tilings are generated from substitution systems, in which one tiling is expanded into a smaller-scale one by replacing every tile with some particular combination of smaller tiles. The ‘combinatorial coordinates’ of a tile describe its position within this hierarchy, without reference to geometry: you say which tile type it is, and which type its larger parent is, and the parent of that, and so on.

Also, for each link in that chain, you have to specify which child of the parent each tile is. For Penrose tiles (at least, the way I’ve handled them by cutting each tile into two triangles), that doesn’t need any extra information, because no tile has two children of the same type, so the sequence of triangle types itself is enough. But for the hat and Spectre tilings, you have to assign a numeric index to each child tile in each expansion, and include those indices in the coordinate strings.

In a tiling of the whole infinite plane, each tile has an infinite sequence of combinatorial coordinates, because the parent tiles keep getting bigger without bound; alternatively, if you’re considering only a finite patch of tiling contained within some top-level supertile, then you have a finite string of coordinates describing each small tile’s location within that patch.

In the previous two articles, I developed a recursive algorithm that takes the combinatorial coordinates of a tile as input, plus a specification of one edge of that tile, and it calculates what’s on the other side of that edge, in the same form: it tells you the coordinates of the neighbouring tile, and says which edge of that tile connects to the specified edge of the input one. This is done by pure string processing, considering nothing but the coordinate strings themselves (plus a set of lookup tables describing the substitution system). The general idea (in the simplest case) is that at each step from a smaller tile to its supertile, you find your smaller tile in a map of the supertile’s expansion; if the edge you’re trying to traverse is inside the expansion then you know what tile (and which edge of it) is on the far side and you’re done; if it’s on the edge of the whole expansion then you work out which edge of the supertile it corresponds to, and recurse one level higher to ask the same kind of question again.

This made a very convenient system for generating random patches of tiling to play puzzle games on, because you don’t have to worry about the geometric coordinates of the higher-order tiles. In the hat and Spectre systems, those coordinates are distorted in a complicated way, so it’s a big advantage not to have to deal with them at all.

## My code crashed, and the reason turned out to be interesting

I stumbled on the first clue that started this article purely by mistake. When I was writing the code to generate the Spectre tiling for the game ‘Loopy’ in my puzzle collection, one of my half-written test programs crashed unexpectedly.

Well, not very unexpectedly. Half-written test programs crash all the time – that’s what makes them half-written. Normally you smite your forehead, fill in some immediately obvious missing bit, and carry on. And in one sense, this case was no exception: I fixed it quickly, and it didn’t slow me down for long in the process of getting Spectre tilings into Loopy.

But after I’d fixed it, I started thinking harder about the implications of what had happened.

### How the crash had happened

The program in question was my first attempt to compute a full coordinate transition in the Spectre tiling. I’d written a first draft of the recursive transition algorithm, and it was time to test it. So I wrote a test program which contained a sample list of coordinates of a single Spectre tile, and told it to find the coordinates of one of the neighbouring tiles.

In the finished version of the code, the initial coordinate list is invented lazily, and randomly. That is, each higher-order coordinate is not generated until the first time the algorithm needs to know it, and then a random-number generator chooses from the possible options according to the right probability distribution. That way, you get your patch of tiling chosen uniformly from all the possibilities, and the output coordinate string describing the patch is exactly long enough to communicate everything you need, with no wasted space.

But this was my very first test of the transition code, and I hadn’t bothered to connect up the random number generator yet. So my ‘generate an extra coordinate’ function did something more trivial: given a list of legal outputs, it just picked the first one every time.

I thought that would be fine for a first attempt, and I could sort out the random numbers later. But in fact that turned out to be the reason my code had crashed. I needed the random numbers, for even the first test run to work!

At every step in the recursive transition algorithm, the recursion might terminate, or it might decide it needs to recurse again, depending on the state of the algorithm and the value of the next coordinate. When you generate coordinates randomly, this is reliable, because sooner or later, the random number generator will pick a coordinate that causes the recursion to terminate. But by making a fixed choice instead of a random choice, I’d caused the algorithm to always decide it needed to recurse again. So it went into a loop calling itself recursively, overflowed the stack, and crashed.

This didn’t take me long to figure out. As soon as I recognised the type of crash as a stack overflow, it was easy to spot that the algorithm was making the same decisions in every recursion, and therefore always recursing. I connected up the random number generator and tried again. No more problem.

### Why that was interesting

That particular crash stuck in my memory, even after I’d fixed it, and I kept on thinking about it.

I’ve forgotten exactly what the coordinates were in the original crashing code, but it’s not too hard to reconstruct some input that causes an endless recursion. In fact, you can find cases where the code makes the same choice at every recursion level, in an unchanging endless loop.

I’ll show a specific example of how that could happen, starting from one of the diagrams from the previous article. This is showing one of the nine hexagonal metatile types (type Y), and how it expands into smaller hexagons.

In this figure, the four edges labelled “0.4”, “1.4”, “2.4”, “3.4” in the diagram on the right correspond to the edge labelled “4” in the single hex on the left. And one of those edges is also edge #4 of a Y-type hex: edge 2.4 of the expansion diagram is also edge 4 of the hex labelled “7 (Y)”.

So, suppose the coordinate transition algorithm is given a Y-type hex, and told to find out what’s on the other side of its edge #4. When it looks at the next coordinate up, it discovers that that Y hex is child #7 of another Y hex. In other words, it realises it’s not just in any Y hex: it’s in the one labelled “7 (Y)” on this diagram.

Edge #4 of that hex is on the outside of this map, so it can’t terminate the recursion. Instead, it must recurse to the next level to find out what larger hexagon lies next to this one.

But that puts it back in the position of trying to traverse edge #4 of a Y-type hex, which is exactly where it started!

It’s fine for this to happen a finite number of times during a run of the algorithm. In fact, it’s inevitable, on any large enough input. Within any full Spectre tiling of the plane, cases of this problem must occur infinitely often at every finite depth. There will always be nth-order Y hexes, for any n; each of those has a chain of Y-type children somewhere inside it, with each one being child #7 of the previous one. So, for any n, there will be cases where the recursive transition algorithm performs n successive steps leading it from “edge #4 of a Y” to “edge #4 of the next higher Y” – but after that, something different happens, the recursion terminates, and the algorithm completes the whole transition successfully.

The problem arises when the function that invents the next higher-order coordinate always returns the same value. If it decides that this Y is child #7 of another Y, every single time, then every recursive call to the transition function will recurse again into an identical one, and nothing different will ever happen to terminate the recursion.

So that explains the crash, in terms of the mechanism of the code. But what does this say about the geometry of the tiling?

Suppose that we started with a single Y hex, and expanded it repeatedly, according to the rules in the previous article. We’d get a sequence of larger and larger patches of tiling made from the nine hex types. Each one would be the nth-order expansion of the original Y hex; each one would contain, within it, many Y hexes of the smallest size; and in each of these patches, one particular Y hex would be the one that was child #7 of child #7 of child #7 of … of child #7 of the starting Y hex.

Now suppose that we overlay all of these larger and larger patches on top of each other, so that the special ‘child #7 n times’ hexes all occupy the same position, with the same orientation. When any two patches overlap, they must agree with each other, so you never get any tiles partially overlapping. Then we do the final expansion from hexes to Spectres, and the same thing happens. You can see the result being built up gradually here, by selecting the radio buttons to advance through the images:

(The expansion from hexes to Spectres has flipped the orientation of everything; sorry about that!)

The union of all those finite tiling patches covers an infinite region of the plane. Every tile in that region is required to exist, as a consequence of the original premise that there existed a tile whose infinitely long coordinate string repeated “child #7 of a Y” forever.

But it’s not the whole of the plane. There’s a complicated crinkly boundary, and none of our finite tiling patches specifies anything on the far side of it!

We already know that if you try to cross edge #4 of that original, central, lowest-order Y hex, the coordinate transition algorithm will recurse for ever. This isn’t an isolated phenomenon. It’s also true of every other edge on the same boundary. Every tile in this whole infinite region has coordinate string that is eventually of the form “Y that is child #7 of a Y” repeating out to infinity, and every edge on the boundary is part of edge #4 of a sufficiently high-order one of those Y hexes. As you move further and further around the region, more and more of the low-order coordinates vary, but only finitely many. Every tile’s coordinates eventually settle down to the same ultimate repetition, in the limit. So any attempt to get the recursive algorithm to cross that boundary will end up, sooner or later, trying to cross edge #4 of one of the higher-order expanded Y hexes, and get into the same infinite loop.

In my imagination, I like to think of the Spectre tiling as being like a landscape of fields, with shallow walls between them. Around the edge of the expansion of each lowest-order metatile, imagine a wall of height 1; around the edge of each second-order metatile, some of those walls grow to height 2; third-order metatile boundaries have height 3, and so on. If you imagine walking around that landscape using the recursive coordinate transition algorithm, the height of each wall indicates how many recursive steps the algorithm needs to make, before it can climb all the way over the wall to get to the other side.

It’s easy to imagine that all the walls might have finite height. And in many of the possible tilings of the whole infinite plane with Spectres, that’s true. But in this particular tiling, containing our endless “child #7 of a Y” tile, it isn’t true. In this tiling, there’s an infinitely high wall, extending forever in both directions (although not in a straight line).

Now I hope you see why I was fascinated by this crash. A routine kind of question that I’ve been dealing with all my life – “why isn’t my program working?” – had turned into a much more interesting question. What’s on the other side of that wall?

## What’s on the other side of the wall?

If you’re investigating a particular tiling of the plane with Spectres (or Penrose tiles, or hats) and you find one of these infinite supertile boundaries barring your way, what does it mean? Can the partial tiling on one side of the boundary be extended to a tiling of the whole plane, filling in the far side? If so, is there a unique way to do it?

### Existence?

A natural first thought is: who says there’s anything on the far side of the wall?

Perhaps, when you find an infinitely high boundary in your tiling, it’s an indication that your attempt to tile the whole plane has simply failed. Perhaps it will turn out that there’s no possible way to tile the whole plane by extending the partial tiling you’ve got so far. Perhaps the only thing you can do is to back up and try again, this time not choosing such a silly sequence of starting coordinates.

That’s certainly a possibility I considered. But I wasn’t convinced I believed it, for two reasons.

One reason has to do with the self-similarity of these tilings. For any finitely long section of the infinite wall in the previous section, you can find a finite patch of tiling containing a section looking just like it, by taking only a finite number of coordinates to be “child #7 of a Y”, and after that, choosing some other coordinate that makes the recursion terminate. And there will be a perfectly sensible answer to the question of what’s on the far side of that finite piece of wall.

In other words, you can find valid pieces of Spectre tiling that fit perfectly to the far side of the wall. Each of those pieces is finitely large, but they’re unbounded in size: you can find one with any finite size you like, no matter how big.

That doesn’t guarantee that all those finite pieces agree with each other. Or even that some subset of them agree with each other well enough to form a consistent tiling of the whole plane. But there’s no guarantee that they don’t, either. So we shouldn’t automatically assume that there’s no possible answer.

The other reason I had for optimism is that I thought of a case I could answer. If we swap from the Spectre tiling to one of the Penrose tilings, there are analogous cases in which there definitely is something you can put on the far side of the wall. I’ll show one.

In my previous article I handled Penrose tilings by dividing each one into two isosceles triangles, because that way the substitution system is particularly simple: each triangle expands into either 2 or 3 smaller triangles exactly covering the same area. Here’s a reminder of the substitution rules for the P2 tiling (the one with kites and darts).

Suppose we start with the type-A triangle, and expand it twice. Part way along the expansion of edge #1 of the original triangle, we see a smaller type-A triangle, with edge #1 on the exterior. Its coordinates within the larger triangle are AB: that is, it’s the A child of the B child of the larger A.

Now we do the same as we did with Spectres in the previous section. We iterate that double-expansion step repeatedly, obtaining larger and larger triangular patches of tiling; in each of those patches, we focus on the particular type-A triangle which has the above relationship to its parent in every double-expansion, so that its coordinates within the starting triangle are of the form ABAB…ABAB. Then we overlay all our finite patches in such a way that our designated triangle is always in the same place.

Just as in the previous section, this construction covers part of the plane with P2 tiles. That central triangle has an infinite coordinate string consisting of endless repetitions of ABABAB… forever. All the other triangles in this half-plane have coordinate strings differing in the low-order components, but they all eventually settle down to repeating ABABAB forever. And the boundary of our half-plane – drawn in a thick line at the bottom of the image sequence above – is an infinitely high wall, which the recursive algorithm can’t step across, because as soon as it got to the trailing ABABAB… section of the coordinate string, it would be trapped into an endless recursion, just as in the Spectre case.

But the difference between this and the Spectre case is that for this partial tiling, we can easily construct something from first principles that fits on the other side of the line. Just take the mirror image!

You can see from the expansion diagrams above that the four triangle types come in two mirror-image pairs: A and B are mirror images of each other, and so are U and V. And the edge constraints permit edge 1 of an A triangle to connect to edge 2 of a B triangle, i.e. to its mirror image.

The expansions of the triangles are symmetric in the same way: the expansion of the mirror image of a triangle is the same as the mirror image of its expansion. Therefore, the nth-order expansion of an A triangle must be able to connect to the corresponding expansion of a B, in the same way: everything along that edge must fit correctly.

But everything along our boundary edge is part of an iterated expansion of edge 1 of an A triangle. That’s how we constructed it in the first place. So the whole boundary must be able to fit alongside its mirror image.

So if we simply reflect our half-plane of tiling in the boundary, transforming triangle types A↔B and U↔V in the process, we obtain a second half-plane that is guaranteed to fit alongside the original one:

So we’ve successfully constructed a Penrose tiling of the whole plane, extending the half-plane. If you try to move around this tiling using the recursive coordinate-transition algorithm, there’s still an infinitely high wall down the middle which the algorithm can’t climb over – but now the wall is transparent, and we can see through it to the mirror-image world on the far side. So at least we know there’s something there.

Moreover, we know what combinatorial coordinates must be of the triangles on the far side of the wall. They’re the mirror images of the ones on the near side.

What’s on the other side of the wall from that starting type-A triangle whose coordinates go ABABAB… forever? It’s a type-B triangle whose coordinates go BABABA… forever. The recursive algorithm can’t compute that, but it begins to look as if it’s true, none the less.

Perhaps we’re looking at a limitation of the recursive algorithm in particular, and not a fundamental uncomputability. Perhaps there’s a more sophisticated algorithm that we can use instead of the recursive one, which can step through these walls?

### Uniqueness?

In that example, I haven’t really proved that the neighbour of ABABAB… is BABABA…. All I’ve proved is that one possible neighbour of ABABAB… is BABABA…. So the next question is: might there be other possibilities? Put another way: is it possible that more than one half-plane of P2 tiling would fit perfectly next to the half-plane we constructed in the previous section?

To begin answering that question, let’s go back to something I said at the start of the previous section. I said: given any finitely long segment of one of these infinitely high (and infinitely long) boundaries, you can find finite patches of the tiling containing a finitely high boundary of the same shape as that segment. And it’s perfectly possible to use the recursive algorithm to cross that wall, because it only needs to recurse for finitely many steps. So perhaps we could generate some examples of those finite patches, and see if they all look consistent with each other.

This is an easy experiment to try, because all it needs is the code I already had for randomised coordinate transitions. We just make a set of coordinates that repeat ABABAB for some finite number of iterations, and then ask the randomised transition algorithm to tell us what’s on the other side of edge #1 of the lowest-order A triangle – under the usual rules that if (or rather, when) it recurses beyond the end of the finite coordinate string we’ve provided, it’s allowed to invent further parent tile types at random until it finds one that terminates the recursion. So the input coordinate string is extended by a few levels, and we get an output string that matches it.

If you try that, you get answers looking like this:

In this list, every output string starts with BABABA…. So did the rest of the 1000 example strings I generated. And if you increase the number of copies of ABABAB at the start of the input string, then the number of BABABA at the start of the output string increases by the same amount.

If you interpret each of these random transitions as describing a finite patch of tiling with a line through it mimicking our infinite boundary, then this is saying that all those finite patches have something on the far side of that line matching our mirror-image half-plane.

This certainly makes it look as if the far side of the boundary is unique. If there were another equally valid half-plane of tiling we could put there, then surely we’d expect this random generation experiment to generate some patches consistent with the mirror-image half-plane, and some consistent with whatever the alternative was.

We can do the same experiment with the original Spectre example. I won’t list the example coordinate strings, because they’re longer and uglier in the Spectre system (you have to specify the child index and the parent type at every step, because just saying the tile type leaves some ambiguities). But the upshot is that if you generate random transitions across edge #4 of a Y hex that is child #7 of another Y, and so on for a large finite number of iterations, the answers all start off by saying that you come in to edge #3 of an S hex that is child #3 of another S hex, and so on. So, in that situation too, it looks as if there might be a unique “other half” of the plane, that fits perfectly to the opposite side of our original wall.

But this is a long way from proving it! All we have here is some probabilistic experimental evidence.

## Computing coordinate transitions on line

Let’s forget for a moment that we haven’t proved any of these observations, and think about what it might mean if they turn out to be true.

The recursive algorithm for computing coordinate transitions generates the very first symbol of the output string last. It recurses along the input coordinate string to the point where it finds a transition within a tile expansion (rather than off the edge of one), and then it turns round and unwinds its stack, generating the symbols of the output coordinate string in reverse order.

But in the previous section, it appeared that a transition across edge #1 of a Penrose half-tile with coordinates starting ABABAB… always delivers output starting with BABABA…. Assuming that’s true, it shouldn’t be necessary to go all the way to the end of the recursion to find out how the output string starts. As soon as we’ve seen the first few symbols of the input string, we already know the output string is going to start with a B, no matter what happens later on in the string. At some point very early in the run, our algorithm has seen enough information that it “ought” to be able to know that, even though it won’t actually get round to working it out until the end of its run.

So it might be possible to rewrite our coordinate transition system as an on-line algorithm: one which can begin producing output even before it’s seen all of its input. You feed in the coordinates of your input string one by one, and at some point, it knows the first coordinate of the output string, and can print it. Keep feeding it input, and it keeps producing output.

The algorithm’s output certainly won’t be able to keep pace precisely with the input. Sometimes it will need to look a few symbols ahead in the input, before it can be sure of the next output symbol. So the output will lag behind the input.

But if things always work the way they seem to in this ABABAB… → BABABA… case, then we can hope the lag will be bounded. That is, there will be some number k such that we can always know the first n symbols of the output if we’ve seen n + k symbols of input.

If the lag is bounded, then that suggests that the algorithm should only need to remember a finite amount of information from what it’s already seen. In other words, maybe it can be modelled as a finite state machine.

Of course, this is all speculation: we haven’t proved that this will work. We’ve just seen some evidence suggesting that it might. But techniques for building finite state machines are self-checking: if they don’t work, they report failure or run for ever, and don’t output an incorrect result. So we don’t need to prove in advance that this idea will work. We can just try to construct a finite state machine for doing this job, and see if it turns out to work!

### Building a finite-state transducer

The most common type of finite state machine, seen in regular language theory, is a recogniser: you feed it an input string, and the only output it generates is “yes” or “no”, at the end, if the string matched the language it recognises. But here, we want to convert our input string into an entire output string, so we need a different type of finite state machine, called a transducer. In a transducer, each state transition is annotated with a string of output (maybe one symbol, or more than one, or sometimes the empty string): when the machine’s input causes it to take that transition, it writes the corresponding output.

To build one of these, we use similar techniques to the standard NFA→DFA conversion used in regular language theory. But first we need some preparation.

We begin by making an ordinary (recogniser-style) finite state machine, whose job is to detect whether two input coordinate strings are adjacent to each other. That is, if you feed it both the input and the output of any successful run of the recursive algorithm, it should report “yes”, and if you feed it any other pair of coordinate strings, it should report “no”.

The recogniser will receive the two strings in an interleaved form, so that each ‘symbol’ consists of a pair of coordinates, one from each string. Also, the initial input symbol must include the edge indices of the input and output tile, because those are a vital part of the input and output of the transition algorithm, along with the tile coordinates.

We’ll continue to use P2 as our example tiling. Here’s a randomly generated input and output pair from the recursive algorithm: edge #1 of a triangle with coordinates ABBU should connect to edge #2 of a triangle with coordinates BUUU. If we fed that to our adjacency recogniser, it would receive a sequence of symbols looking something like this:

1. (A, 1, B, 2) – the type of the two lowest-order triangles, and which edges of the former and the latter are claimed to be adjacent
2. (B, U) – the first-order supertile type in each string
3. (B, U) – the second-order supertile type in each string
4. (U, U) – the third-order supertile type in each string

and, since that’s where the recursive algorithm finished having to recurse, we’d expect the adjacency recogniser to accept at this point, and report “yes, these two strings match correctly.”

This adjacency-recognising state machine is reasonably simple to generate directly, starting from the transition maps of the Penrose triangles, which I show again here for convenience:

The basic idea is that the state machine is divided into two similar parts, each one corresponding to one of the two input strings. Each sub-machine consumes the half of the input symbol that relates to its own string, and keeps track of what triangle type it’s in, and where in that triangle it is. Then they ‘compare notes’ at each step to make sure they remain consistent.

The best way to show this is to illustrate how the machine thinks, as it works through an example. Here’s the way it processes the string above:

What’s going on in those diagrams:

1. Receive the input symbol (A, 1, B, 2). So the left sub-machine is trying to go out of edge #1 of an A, and the right one is trying to go out of edge #2 of a B. So far, so good: those are the same type of edge – the tiling rules permit them to connect.
2. Receive the input symbol (B, U).
1. On the left, the A is now known to be part of a larger B; on the right, the B is part of a U. Both target edges are on an edge of the larger triangle; they’re part of the same edge type of the larger triangle; that edge is divided into two pieces in the expansion, and the two sub-machines agree on which part of the edge they’re going out of (namely the top half, when the two are shown facing each other as here). In other words, this looks good: everything matches.
2. So now we forget about the smaller triangles, and focus on the next layer up. All we remember is that the left sub-machine is trying to go out of edge #2 of a B, and the right out of edge #0 of a U.
3. Receive the input symbol (B, U).
1. On the left, the previous B is part of another larger B; similarly on the right, the U is part of a larger U. Again, we’re on the edge of the larger triangle in both cases (this time, the whole edge, not part of it), and the edge types match.
2. Again, forget about the smaller triangles and move up. Now the left sub-machine is trying to go out of edge #0 of a B, and the right out of edge #1 of a U.
4. Receive the input symbol (U, U). Now both sides are trying to make a transition in the interior of a U triangle. On the left, we’re trying to cross from the B sub-triangle into the U; on the right we’re going from the U into the B. In other words, the two sides are crossing the same edge inside the expansion, in opposite directions. Success! The machine accepts.

(Compare this with how the simpler recursive algorithm would handle the same coordinate transition. As it recurses deeper and deeper, it imagines the left-hand diagram in each of the above steps; when it reaches a transition within the top-level U triangle, it terminates the recursion, and as it unwinds its stack, it goes through the right-hand diagrams in reverse order, calculating the output coordinates from largest to smallest.)

In other words, each state of the adjacency recogniser consists of two (tile, edge) pairs, one for each of the input strings. To compute each state transition, you use each half of the input symbol to locate one of those (tile, edge) pairs in a supertile.

The first thing you find out is whether the edge of the previous tile is on the outside of the new expansion, or on the inside. If it’s on the inside (as in step 4 above), we’re making a transition entirely within a supertile – or, at least, that half of the machine thinks so. So you check that the other half agrees, and that the positions within the supertile match, and then accept.

Otherwise, you’re making a transition out of the supertile. So again you check that the other half agrees, and that the details of the new edges make sense, and then you make a state transition to two (tile, edge) pairs representing the supertiles, ready for the next input symbol.

If the states aren’t consistent, for any reason, then the adjacency recogniser rejects the input string. This can happen in lots of ways:

• One half of the machine believes it’s trying to make a transition within its supertile, while the other half thinks it’s trying to step off the edge of its supertile.
• Both halves are trying to make a transition within a supertile, but the two supertiles aren’t the same type.
• The two supertiles are the same type, but the two positions within it don’t match – they’re not opposite sides of the same edge within the supertile’s expansion.
• Both halves are trying to step off an edge of the supertile, but the types of the supertile edges don’t match – they aren’t edges that the tiling rules would ever allow to connect.
• The supertile edge types do match, but that supertile edge is divided into multiple sub-edges in the expansion, and the two halves aren’t trying to cross the same sub-edge. For example, in step 2a above, if one side had been trying to cross the longer top section of the edge and the other the shorter bottom section.

We’ll need to add complications later, when we tackle more difficult tilings. But for Penrose tilings, which are the simplest, this is enough.

One last thing: the plan is for all this machinery to end up handling infinitely long sequences of coordinates. So after the machine ‘accepts’ after finitely many steps, as it did in step 4 in our example above, its job isn’t done! If you’ve found a pair of finite coordinate strings that map to each other, then a pair of infinite coordinate sequences starting with those prefixes are also legitimately adjacent, as long as the remainder of the two sequences are identical. So even the accepting states of the adjacency recogniser still need to do work: their remaining job is to receive an input symbol, check that the two halves of it are identical, and reject if not. (Also, check that it even makes sense – that is, the new supertile is consistent with the previous tile.)

#### Reinterpret the recogniser as a non-deterministic transducer

Now we’ve got a state machine that can take two input strings s and t, and tell us whether t is the correct output of a coordinate transition starting from s. But what we want is a machine that can take s by itself as input, and generate t as output.

In a sense, we’ve already got one!

We take our adjacency-recognising machine, and simply reinterpret one of the strings as its output. That is, in each state transition triggered by a particular pair of symbols (xy) from the two input strings, we pretend it’s a state transition triggered by just the input x, generating y as output.

The problem with this is that now our machine is non-deterministic. As an adjacency recogniser, it was deterministic: given any input symbol pair, there was only one transition the machine could possibly take. (Or none, if the input was illegal.) But there will be plenty of cases where inputs (xy) and (xz) both trigger different valid transitions from the same state i, say to states j and k respectively. So if we reinterpret the machine description so that only x is the input, then we have two possible transitions: we could go to state j and output y, or to state k and output z, and we don’t have enough information to know which is right.

Of course, this is what we expect. We already know that there are cases where the nth output coordinate can’t be determined from just the first n input coordinates; the algorithm will sometimes have to look further ahead to find out which of multiple possibilities is right. The nondeterminism in this machine exactly reflects that: we expect there to be exactly one correct output in the end, but at the moment we take a transition on each input symbol, we don’t always know what it is.

#### Then make it deterministic

How do you turn a non-deterministic finite state transducer into a deterministic one? Pretty much the same way you do for recognisers.

In regular language theory, there’s a standard construction that turns a non-deterministic recogniser (in this context often called ‘NFA’, for Nondeterministic Finite Automaton) into a deterministic one (‘DFA’). The trick is that each DFA state represents a set of possible states of the NFA, with the semantics “My NFA must be in one of these states, but I don’t know which.” You compute a transition between DFA states by going through every NFA state in the source set, and following all possible transitions from any of those states on the input character, to construct the set of NFA states you could possibly be in after seeing that character. Then you look for a DFA state corresponding to that set, or make a new one if it doesn’t already exist, and that’s the destination of this state transition in the DFA.

Essentially the same technique works for transducers, except that we have to deal with the extra complication of the output. If we collect a set of possible NFA transitions, and they don’t all produce output, or don’t all produce the same output, what do we do about it?

Answer: instead of each DFA state representing a set of NFA states, it represents a set of pairs (ip), where i is a state of the NFA, and p is a string of pending output. The pending output was generated by whatever path through the NFA we followed to get here, but it hasn’t yet been emitted from the DFA, because we’re not yet sure whether that path was the right one.

Then, if our source set contains a pair (ip), and NFA state i contains a transition on the input symbol which goes to NFA state j and outputs some data x, we enter the pair (jp + x) into our destination set. So we haven’t actually emitted the output for that NFA transition: we’ve just appended it to the end of our “pending” string.

For example, here’s a tiny made-up fragment of nondeterministic transducer:

On each edge, the input symbol is written in black, and the output in red. So this says that from the start state 0, on receiving the input symbol ‘a’, we can either go to state 1 and output ‘x’, or to state 2 and output ‘y’.

When we make a deterministic machine out of this, we’ll have an initial DFA state {(0, “”)} (simply being in state 0 with an empty output string). On the input symbol ‘a’, it will transition to {(1, “x”), (2, “y”)}, which says that we followed one of those two transitions, but don’t know which yet. This also means we don’t yet know what the NFA’s output will turn out to have been – so our DFA can’t emit any output at all on this state transition. It has to wait to see what comes next before it knows which of ‘x’ and ‘y’ starts the output string.

Once we’ve constructed the complete set of pairs that correspond to our destination DFA state, we do one extra step. We collect together the pending output strings from all the pairs we have, and see if they all start with a common prefix. If they do, then that prefix is guaranteed to be the right thing to output next, no matter which of these NFA states we’re in. So we remove that common substring from the pending output of every pair, and write it into our DFA state transition as actual output. Then we find, or make, a DFA state corresponding to the set of pairs we have left. So the DFA we construct will produce each symbol of output as soon as it’s sure what it is – as soon as it’s ruled out every case where some other symbol might have been correct.

Suppose the example fragment above continued like this:

From the DFA state {(1, “x”), (2, “y”)}, what should we do on the input symbol ‘b’? From state 1 we can take the transition to state 3, outputting ‘z’, or to 4 outputting ‘w’; from state 2 we have no options at all. So we make the set {(3, “xz”), (4, “xw”)}. But in that set, all the possibilities have ‘x’ as the first character of the output. So we remove it from all of them, and emit it as output from the DFA, because we’re now sure of it. So instead of making a DFA state for {(3, “xz”), (4, “xw”)}, we make one for {(3, “z”), (4, “w”)}, and on the transition to it the DFA outputs ‘x’.

On the other hand, if we’re in {(1, “x”), (2, “y”)} and we see the symbol ‘c’, then there are no possible transitions from NFA state 1, and the only thing we can do is to travel from 2 to 4, outputting ‘y’. So we construct the set {(4, “yy”)}. But now everything in our set agrees that the first two characters of the output should be “yy”, because there’s only one thing in the set. So on this DFA transition we generate the two output characters “yy”, and move to a DFA state representing {(4, “”)}.

So the DFA fragment constructed from the NFA fragment above would look like this:

If we feed “ab” to this machine, then it outputs ‘x’ on the second transition, but still isn’t sure which of ‘z’ and ‘w’ comes next. But if we feed it “ac” then it can produce the full output string “yy”, and end up with no output still buffered.

Then, just like the more usual subset algorithm for making recogniser DFAs, we keep on going: for every DFA state we’ve generated, we compute all the outgoing transitions, and whenever one goes to a DFA state we haven’t explored yet, we queue it up to investigate later. Repeat until we’ve explored all reachable states, and the algorithm terminates.

There’s only one problem with this algorithm: it’s not guaranteed to terminate!

When you do this for recognisers, it is guaranteed to terminate, because if the number of NFA states is finite, then so is the number of possible sets of NFA states. The DFA might be exponentially large compared to the NFA (and in occasional nasty cases that actually happens), but sooner or later, the construction algorithm will run out of ways to generate a new DFA state that doesn’t match any of the existing ones, so the algorithm will terminate, and the output machine will be finite.

But here, our DFA states aren’t sets of NFA states; they’re sets of pairs each containing an NFA state and a string. And strings can be as long as you like, so the set of possible pairs is infinite. So there’s no guarantee that the procedure terminates after constructing a finite number of states. It might continue forever, constructing DFA states containing longer and longer strings of unemitted pending output.

But it’s worth a try! We can run the algorithm and see if it works.

### An example transducer

And it does! Running this algorithm on the triangle-based P2 substitution system, it terminates and reports success.

The resulting state machine has 31 states. I show it as a Graphviz diagram here just to prove that it exists, but I don’t expect it to be very informative, because Graphviz makes an impenetrable tangle of it, and I haven’t found any nicer way to lay it out:

The starting state is state 0, at the top. As in the previous section’s examples, edges are labelled with their input symbol in black, and output symbols (if any) in red. The first input symbol consists of a tile-type letter and an edge number, like A2 or V1; after that, all further input symbols are just single letters. The same is true for the output symbols.

The four nodes at the bottom with double-circle outlines are accepting states. If you reach any one of these, then the transducer has reported success: it’s consumed enough of the input string to complete a coordinate transition, and generated the corresponding output, just the same as the recursive algorithm would.

Of course, your actual tile coordinate string might contain more coordinates than a given transition needed to modify. In that situation, you’d want to copy the remaining unchanged coordinates from the input into the output. For that purpose, the accepting states have those state transitions written out explicitly, so that if you prefer, you can keep running the state machine, ignoring the fact that it’s now in an accepting state.

(There are multiple accepting states so that each one can reject input coordinates that don’t make sense at all, such as claiming a U tile has parent B. There’s one accepting state for each triangle type, and each one has transitions for only the triangle types which can sensibly be a parent of its own type. So those four states between them form a little secondary state machine which recognises legal coordinate strings.)

## Using the transducer

OK, now we’ve proved the concept, by successfully making a transducer for at least one tiling. What’s it good for?

### A better alternative to the recursive algorithm

Firstly, and most obviously, you can use this in place of the recursive algorithm I described in my previous articles, if you’re trying to generate random patches of tiling.

That is: given a finitely long coordinate string of a tile and an edge of that tile to traverse, you feed it to the transducer, symbol by symbol, and concatenate the symbols that come out. The result will give you the coordinates of the neighbour tile, and which edge of that one is adjacent to the edge you originally specified.

If all goes well, the transducer reaches an accepting state. This should be a state of the DFA in which every (state, pending output) pair describes an accepting state of the adjacency recogniser, with the pending output string empty. In other words, the input and output strings so far are precisely a pair you could have got from the recursive algorithm – if you’d given it that input, it would have returned you that output, without having to look at any further symbols than the ones you’ve already seen. So you can proceed just as if the recursive algorithm had returned success: if there are any further coordinates in your input string, copy them to the output unchanged, and you’re done.

But, just as in the previous articles, there’s a risk that you reach the end of the finite coordinate string before that happens. So you respond in the same way as before: choose an extra higher-order supertile at random, pretend it was there all along, and feed it to the transducer as an extra input. If it still hasn’t accepted, do the same thing again. Sooner or later you’ll get lucky – just as, sooner or later, generating these same extra coordinates in the middle of the recursive algorithm would manage to find one that terminated the recursion.

What are the advantages of doing this job with a transducer instead of recursion?

Simpler lookup tables. In the recursive algorithm you need multiple complicated tables, one way or another. At least one table per tile; maybe separate tables for transitions within a tile and transitions into it from outside; for the hat system’s overlapping metatiles you need those ‘metamaps’ that show how to compute the redundant representations of a coordinate string. Here, there’s just one lookup table, indexed by your current state and the next coordinate symbol.

Guaranteed linear time in the string length. The recursive algorithms I’ve described sometimes need to recurse multiple times. In the hat tiling, you might need to recurse back and forth around the string rewriting coordinate pairs using the metamap, until you find one where you can make the lowest-level transition without falling off the edge of the kitemap. In the Spectre tiling, you might need a second recursive call due to the zero-thickness spur in one of the expansion diagrams. Only the Penrose tilings are simple enough not to have either of these problems.

But processing a string with a finite state machine never has to backtrack. If you feed it N input symbols, it gives you N output symbols in O(N) worst-case time, no matter what.

(However, those awkward properties of the tiling still make life difficult! But the difficulty happens at the point when you construct the state machine. Once you’ve managed that, using it at run time is the easy part. I’ll discuss the difficulties in later sections.)

Constant memory usage. Of course, a recursive algorithm that recurses N levels deep requires Ω(N) stack to store the state of where it had got to. A finite state machine doesn’t need to remember where it was in all the previous characters of the string – it has already produced output for most of them and forgotten about them completely.

Unified algorithm for all systems. There’s only one way to feed a string to a transducer. So you can write a single engine for doing it, and use it for any tiling system that this method can handle. No special-case code per tiling, like the metamap business in my original hat algorithm.

None of these is a huge advantage. The lower space and time costs are a nice improvement in theoretical terms, but in practice, your coordinate strings are all short enough that it doesn’t matter a great deal (because they grow like the log of the size of patch you’re generating). I think the software engineering advantages are more important than the performance: only one piece of code to handle all tilings, and it’s very simple.

But even if all these advantages are small, they all point in the same direction – all of them are in favour of using a transducer instead of recursion. So this is my new recommended approach for generating random tiling patches, if you’re in a position to use it.

### Infinitary transition algorithm for eventually periodic coordinates

But I didn’t go to all this effort just to slightly optimise something I already had a way to do. I wanted to do the impossible! We wanted this transducer so that it can bypass those infinitely high walls.

So, what happens if we ask this state machine to try to cross the wall in our symmetric Penrose tiling, by asking it to cross edge #1 of a tile whose coordinates are an infinitely long string endlessly repeating ABABABAB…?

We can go through the state machine diagram above, and figure it out:

• Start in state 0.
• Feed in the symbol A1 (starting tile type and edge to cross), which takes us to state 8, emitting no output yet.
• Feed in B, which goes to state 23, outputting the symbol B2.
• Feed in A, which goes to state 20, outputting A.
• Feed in B, which goes to state 23, outputting B.
• Now we’re back in state 23, which is the same state we were in two symbols ago; and we’re about to feed in another instance of AB, just as we did last time. So it’s clear that we’re going to repeat the previous two transitions and their output, and then keep doing that. So, from here on, we expect output A,B,A,B,… forever.

In other words, we’ve managed to describe the infinite output string corresponding to the infinite input we had in mind. If our input string goes A1 B and then repeats AB forever, then the output string starts B2 A, and then repeats BA forever. In other words, if you cross the infinite wall at edge #1 of the tile with coordinates ABABAB… then you end up coming in through edge #2 of a tile with coordinates BABABA….

And that’s exactly the string we thought it should be, from the symmetry argument. But this time, we computed it, without having to depend on symmetry. So this technique can still generate an answer in cases that aren’t symmetric.

To use this ‘infinitary’ version of the coordinate transition algorithm, you have to be able to represent an infinite input string in a finite amount of space, and also work out a representation for the output string. We did that in the example above by knowing that the string was going to end up repeating itself.

In other words, suppose you start with an eventually periodic coordinate string, represented by two finite strings (sr), describing an infinite string that starts with a copy of s and then repeats r forever. Then you can feed any string of that form to this transducer, and generate an output string in the same form, by keeping track of both your position in r (once you reach it at all) and the current state of the transducer. When both of those variables are in a state they’ve been in before, you know that the machine will spend the rest of eternity generating more copies of whatever output it’s produced since the last time you were here. So you can express the output string in the same (sr) form.

So now we’ve removed the randomness from the coordinate transition algorithm, and also eliminated the risk of infinite recursion. We can cross those infinitely high walls just as easily as finite ones!

### Some pretty pictures: pentagonally symmetric Penrose tilings

I’ve made you suffer through a lot of maths and computer science, so let’s take a break for some pictures.

Using this technique, we can generate specific instances of the Penrose tilings, carefully chosen to have symmetry. With the recursive algorithm, these would have been difficult, because the symmetry tends to give rise to infinite supertile boundaries, causing the recursion to fail. But now we have this on-line algorithm which can handle infinite boundaries, that’s not a problem any more.

We’ve already seen an example of a P2 tiling chosen to have reflective symmetry about a single straight line. But we can do better than that. There are also two P2 tilings of the plane with pentagonal symmetry about a centre point – five-way rotational symmetry, and reflection in five axes.

To construct one of those using the new algorithm, all we have to do is to work out the combinatorial coordinates of one of the tiles at the centre. In this case, the trick is to start with a single triangle type, say A, and expand it four times. After that, one of the corners of the starting triangle will be occupied by the same corner of a smaller version of the same triangle. So then we iterate that quadruple expansion and overlay the larger and larger pieces, just as we did before.

The reason we have to expand four times is because the triangle type in that corner cycles through all four triangle types before coming back to its starting point. So the coordinate string for that triangle will also repeat with period 4. In fact it will be AVBUAVBUAVBUAVBU… forever.

In the previous two cases where we deliberately constructed an infinite boundary, it looked as if we’d made a partial tiling that filled “half” the plane, in the sense that we only expected to need one other partial tiling to fit to it. But in this case, we’ve fillled only “one tenth” of the plane – we’re going to need nine more rotated and reflected images of this infinite sector to fill the rest.

That’s all we need to generate a symmetric tiling: we just feed that repeating coordinate into the code, place the corresponding triangle somewhere in the middle of the picture, and hit ‘go’!

I said there were two P2 tilings with pentagonal symmetry. Given the cyclic coordinate string we have here, it’s easy to guess what the other one will be: surely we’ll obtain it by starting the same cyclic sequence of four tile types at a different point. And indeed, with coordinates VBUAVBUAVBUAVBUA…, we get this symmetric tiling, which has five darts around the centre point instead of five kites:

You might ask: the sequence of coordinates AVBUAVBUAVBU… repeats with period 4. So maybe there are two more possibilities available by shifting twice more, to get coordinates starting with BUAV or with UAVB? Do those give further symmetric tilings?

No, they don’t: they give the same two again, rotated by a tenth of a turn. Each of the straight-line boundaries in these pictures is a line of reflection symmetry, like the one in the ABABAB… example in an earlier section. And just as in that case, the coordinates of each mirror-image tile is obtained by swapping A↔B and U↔V. So we’ve already seen a triangle with coordinates BUAVBUAVBUAV…: it’s the mirror image of AVBUAVBUAVBU…, which we used to generate the first of the two diagrams above. The ten sectors of that diagram alternate AVBU… and BUAV… as you go round the origin. Similarly, VBUA… is the mirror image of UAVB…, so those two tile types alternate around the origin of the second image.

That’s enough kites and darts. What about the other Penrose tiling, the P3 one made of thin and thick rhombs?

The transducer-constructing algorithm works just as well for the P3 tiling as it does for P2. (Not surprisingly, since as I mentioned in a previous article, the P2 and P3 tilings can be obtained from each other by performing a sort of ‘half-expansion’ at each step. So if one works, you’d expect the other one to work too.)

And just like the P2 tiling, P3 has two particular instances with pentagonal symmetry, and you can generate them using exactly the same technique. This time, the coordinates of a central triangle don’t cycle through the four triangle types: instead, they repeat XYYX XYYX XYYX…, or the cyclic variation YYXX YYXX YYXX….

As a result, both symmetric P3 tilings have five thick rhombs around the centre of symmetry, so they don’t look as obviously different from each other as the P2 ones! But if you look at the pattern near the centre, you can see that they’re not the same: in one of them, a ring of thin rhombs encircles the middle five thick ones, whereas in the other, ten thick rhombs point outwards in a sort of star shape.

## Building a transducer for the Spectre tiling

I began this article with an example of an infinite wall in the Spectre tiling, because that’s the one I noticed first, because it happened when I was writing a Spectre generator. But ever since then, I’ve been talking about Penrose tilings, because those are easier to work with and reason about.

So, now we’ve solved the easy case, it’s time to get back to the original more complicated one! Let’s build a finite state transducer for the Spectre substitution system.

### Spurious construction failure

If you try that in the most obvious way, constructing an adjacency-recogniser state machine exactly the way I described in a previous section, something goes wrong. The construction runs successfully, and outputs a transducer … but it’s subtly incomplete. There are valid coordinate strings you can use as input, for which the recursive transition algorithm successfully generates a result, but which the transducer will reject, because at some point it reaches a state where it has no legal transition on the next symbol.

The problem arises because of this expansion rule in the Spectre system, which turns the S hex type into a Spectre, but has a zero-thickness ‘spur’ sticking out of one corner:

As I described in the previous article, the recursive algorithm has to compensate for this spur by sometimes calling its recursive subroutine twice. You try to enter an S hex along edge #1, or the wrong part of edge #0, and when you consult the map above, you discover you haven’t landed in the actual Spectre, you’ve landed on one side of the spur – and so you have to call the next-level recursion again, to step straight back off the other side of the spur and see where you end up after that. Crossing the spur costs an extra function call.

But our adjacency recogniser doesn’t know anything about spurs, and isn’t set up to account for those extra function calls. So if you feed it the input and output of a run of the recursive algorithm that had to cross a spur, it will reject that pair of strings, when it should have accepted them.

This particular spur isn’t too hard to handle, because in the Spectre tiling system, the S hex always appears in that three-hex cluster with a G and D hex, always in the same relative orientation:

This means that all the edges affected by the spur are on the interior of that G,D,S cluster of lowest-order hexes, and all three of those hexes will be expanded from the same larger hex. So any coordinate transition involving a spur is also going to finish without having to recurse higher.

Because of that convenient property, there are two different ways we could modify the substitution system itself to make this spur go away. We could replace the G, D and S hex types with a single combined ‘trihex’ tile, fusing together their expansion maps, so that the spur just became a pair of ordinary edges between two of the Spectres in the combined GDS expansion. Alternatively, we could merge the lowest two levels of the substitution system, so that instead of each second-order hex expanding into 7 or 8 first-order hexes each of which becomes either 1 or 2 Spectres, we’d have a set of transition maps that turn second-order hexes directly into 8 or 9 Spectres each – bypassing the intermediate stage where the spur lives.

But instead, I’m going to show an improvement to the method of constructing the adjacency recogniser, which allows it to handle this spur completely automatically, with no manual hacking needed. This will be useful later, because there will be more substitution systems with spurs!

To begin with, let’s see how the existing adjacency recogniser would think, if it saw a transition of this kind:

What’s going on in the above diagrams:

1. We receive the starting symbol identifying the potentially matching edges: edge #6 of the left Spectre and edge #1 of the right.
2. We receive a symbol telling us that the left Spectre is expanded from a D hex, and the right Spectre is #1 of the two from a G hex. Edge #6 of the left Spectre is part of edge #0 of the D hex, and edge #1 of the right Spectre is part of edge #4 of the G hex.
3. So now we’re expecting edge #0 of a D hex to match edge #4 of a G. But the hex matching rules say they don’t. Reject the input string!

The recursive algorithm would also run into this difficulty, but it would know what to do. It would know what edge #0 of the D hex does meet: edge #1 of the S hex. So it would go down a layer to find out where in the S Spectre it had landed. It would find it had landed on the spur – and it would recurse back up again to see what was on the far side of the spur, and end up finding the G hex, after a detour.

But in this context, we don’t have the luxury of stepping down a layer and back up again. An adjacency recogniser has to run in a single pass. We have to do something at this layer that will stop us from being confused about the spur!

The answer is to consider the geometry of the tiles. If edge #6 of one Spectre coincides with edge #1 of another Spectre, then several other pairs of edges are required to coincide, by the shapes of the tiles. And some of those pairs of edges correspond to edges of the hexes that do officially meet.

The revised algorithm’s thought process goes like this:

1. As before, we start by trying to match edge #6 of the left Spectre to edge #1 of the right.
2. We find out the parent hex types, and discover that edges #6 and #1 of the Spectres don’t correspond to matching edges of the hexagon. But this time, we don’t give up…
3. We imagine how these tiles would fit together geometrically. If the specified pair of edges meet, then three other pairs of edges meet too – and vice versa. All four edge pairs meet, or none.
4. So the question “do edges #6 and #1 of these Spectres meet?” is equivalent to “do edges #4 and #3 meet?”, or “#5 and #2”, or “#7 and #0”. If we can get a positive answer to any of these questions, we’re good.
5. The promising edges are #4,#5 on the left and #3,#2 on the right, because those are expanded from a different pair of edges of the parent D and G hexes: edge #1 of the D and edge #3 of the G. It also looks promising that these are the whole expansion of that pair of edges.
6. So we pretend those were the edges we’d originally been asked about, and proceed on that basis. Now we’re trying to match edge #1 of the D to edge #3 of the G – and those do potentially match!

The adjacency recogniser has kind of ‘slid sideways’ along the edges of the two Spectres, to a different pair of edges that must meet if and only if the original edges met, and don’t have a spur in the way.

With that problem solved, I was able to build a working finite-state transducer for the Spectre tiling. The resulting state machine has 151 states – nearly five times as many as the state machine for Penrose P2 that I exhibited in full above! That’s some kind of a measure of how much more complicated the Spectre tiling is than the Penrose ones.

### Testing that the transducer is complete

When I found out that my first Spectre transducer didn’t handle all possible inputs, it took me by surprise: an attempt to generate an actual tiling crashed. You don’t really want to wait until run time to find out that kind of problem. You’d rather be able to know in advance whether your transducer is complete.

So here’s a technique for testing exhaustively that it’s capable of handling all valid input coordinate strings.

You do this by first making a second state machine that recognises valid coordinate strings. Then you analyse the two state machines jointly, by doing a graph search over their Cartesian product – that is, the space of pairs of states, one from each machine. So you’re generating an exhaustive list of states the transducer can be in, for each state of the coordinate-recogniser. And then you expect that the transducer state should never fail to have a valid outgoing transition on a symbol, if the coordinate-recogniser state that goes with it has one.

And making a recogniser for valid coordinate strings is no trouble at all: you just need one state for every tile type, and a transition from each tile type to any tile type that can legitimately be its parent. It’s the simplest state machine in this whole article.

So now you can automatically detect whether your transducer is complete. And once I’d added the system above for handling spurs, this test passed, where it had previously failed. So I was confident that I had a working Spectre transducer.

It’s also possible to do this same check directly on the adjacency recogniser, to see if there’s any input for which it doesn’t know of any possible output. Simply discard one half of the input from every transition of the adjacency recogniser, and you’ve turned it into an ordinary recogniser NFA for the language of inputs it will accept. Then you can turn that into a DFA, or directly check it against the DFA for legal coordinate strings, in exactly the same way.

### Crossing infinite walls in the Spectre tiling

Now we have a Spectre transducer, we can go straight to generating Spectre tilings using the infinitary algorithm I described in a previous section.

As before, we represent an infinitely long string of Spectre tile coordinates in a form consisting of an initial segment and a repeating segment, so that we can represent any infinite string as long as it’s eventually periodic. Then we use the finite-state transducer to convert an eventually periodic input string into an eventually periodic output string, by detecting when it starts repeating.

We’ll start with the case I originally described, where at every stage of the recursion you’re trying to exit edge #4 of a Y hex, and because each Y turns out to be child #7 of another Y, this continues forever. When I was experimenting with the recursive algorithm to try to guess what might fit on the other side of the resulting wall, it looked as if this boundary might match up to one where at every layer you’re trying to exit edge #3 of an S hex.

And indeed this is what happens. If we make a coordinate description in which each Y hex is child #7 of another Y, and ask the algorithm to cross a Spectre edge corresponding to edge #4 of the lowest-order Y, then it delivers an output string full of S hexes, in which each one is child #3 of another S, just as we’d guessed. The resulting full tiling of the plane is shown here: the Y side of the wall is the smaller top left region, and the rest of the picture is the previously unreachable S side.

Success! We’ve finally answered the original question of what was on the far side of the first infinite wall we found.

It’s almost a disappointment that it just looks like more of the same kind of Spectre tiling, isn’t it? There’s no obvious symmetry here; no noticeable change of quality between the two sides of the boundary. The boundary itself is the most interesting thing, with its self-similarly crinkly shape – but if I hadn’t drawn it in a thick line, you’d never have been able to pick it out, or even detect that this wasn’t a perfectly ordinary patch of Spectre tiling. There’s nothing special to be seen at all.

(And that’s not so surprising, given that any finite patch of this tiling, even one crossing the boundary, must exactly match a finite patch that occurs infinitely often in every infinite Spectre tiling of the plane. That must be true, because that’s how we constructed the transducer that computes transitions across the wall.)

So here’s a different example, which I didn’t discover until I was actually writing up this article. Let’s go back to the Y hex expansion and look at it again:

Previously, I pointed out that edge #4 of the “Y (7)” subtile is part of edge #4 of the larger Y supertile. But there’s a second example of this phenomenon in the same diagram. Edge #5 of the “Y (4)” subtile is part of edge #5 of the supertile. So if we make a set of coordinates that say that each Y is child #4 of another Y, instead of child #7, then we should find another tiling with an infinite wall, different from the previous one.

And we do. But this one is more interesting, because if you ask the transducer to compute the coordinates of the neighbouring hex, you find … that the output is exactly the same as the input! The result of crossing edge #5 of a Y whose coordinates repeat (Y, 4) forever is that you find you’ve come back in to edge #5 of another Y hex, whose coordinates also repeat (Y, 4) forever.

So we’ve got two tiles in the plane with the same combinatorial coordinates: the same sequence of supertile types, and the same position within each of those supertiles. What are the consequences of that?

One consequence is that you have to avoid an embarrassing bug in your software. If you’re generating pictures of tilings and your code uses a data structure that maps the combinatorial coordinates of a tile to its position in the plane (perhaps so you can recognise a tile you’ve seen before by finding its coordinates in this map), then you’ll find it fails in this case, because two different tile positions need to occupy the same slot in your data structure. So instead you have to write the code the other way round, using positions in the plane as your keys, and combinatorial coordinates as values. Then it’s no problem to have multiple tiles with the same coordinate sequence.

(In fact, the same issue arises in the symmetric Penrose tilings I showed earlier. Each of those divides the plane into ten sectors with infinitely high walls between them, but there are only two types of sector, alternating around the origin. So for every tile, there are four others elsewhere in the plane with the same coordinate sequence.)

But another consequence is that the two sides of this wall must be exactly the same shape, because the coordinate string of our starting tile determines the entire shape of the boundary it’s on. So we expect that this boundary should have order-2 (180°) rotational symmetry, and not only that, so does the entire tiling that it bisects.

And it does! So, from first principles, we’ve constructed a Spectre tiling of the plane with order-2 rotational symmetry:

If I hadn’t happened to spot that from just this expansion diagram, here’s another way I could have found it. In the previous Spectre article I showed a randomly generated patch of tiling using the hexagons themselves, rather than the underlying Spectres. Here’s another one, without the distracting jigsaw edges I used in the previous article:

Looking at that patch, you can see cases where two Y hexes are adjacent to each other, with the orientation arrows on them pointing in directions 180° apart. I’ve marked one of them with a red circle.

So I could have started by spotting that in the diagram, and asking myself: “What if we take two Y hexes connected in that way, expand them repeatedly, and overlay the expansions with their points of symmetry aligned?” And you’d see that the two Y hexes connected to each other along edge #5, so you’d look at the Y expansion map to see what happened when you put edge #5 of the whole map next to another copy of itself, and that would tell you what sub-tile was at the centre of the edge. From there you could figure out how to do the expansion.

The infinitary transition algorithm for eventually periodic coordinates shortcuts the work of doing the actual expansion. With this algorithm ready to hand, we only need to make a starting coordinate string describing one of those hexes, and then the software does the rest automatically.

So let’s try another feature that I’ve marked with a circle on that same diagram: we can see three F hexes adjacent, with their arrows in 120° rotational symmetry: each F hex’s edge #4 connects to the next one’s edge #3. Looking at the F expansion map, that looks as if you’d need the F hex to be child #6 of another F. So we can immediately generate a Spectre tiling with three-way symmetry, by making a coordinate string repeating (F, 6) forever.

Similarly, there are triangles of three Y hexes with the arrows all pointing outwards, i.e. with each Y’s edge #2 connecting to the next one’s edge #1, which suggests that a different three-way symmetric Spectre tiling is obtained by repeating (Y, 1).

And here they are: two different three-way symmetric Spectre tilings.

## The hats tiling

We’ve successfully built transducers for the Penrose tilings (both P2 and P3), and for the Spectre tiling. But we haven’t tackled the hat tiling yet. Obviously, that’s the next thing to do!

### A non-overlapping substitution system for hats

But we can’t do this using the standard substitution system for the hats tiling, as described in the original paper, and as I used in my previous article for randomly generating patches of hat tiling. Why not? Because, in that system, the expansions of adjacent metatiles overlap each other, so that the same hat can have many different possible coordinates, all valid at once.

In the previous article, I found that overlap quite useful. It allows a kind of ‘atlas’ of the tiling: for any direction you might try to step off a hat, you can find some metatile expansion in which that step doesn’t take you off the edge of the expansion, and if that’s not the metatile expansion your current coordinates describe, there’s a simple way to translate coordinates between all the patches that overlap at any given point. So you can convert one coordinate string into an equivalent one when necessary. That way I completely avoided having to work out how all the complicated crinkly edges of the metatile expansions matched up to each other.

But here, our test of a successful transducer construction is that it should deliver a unique output for any input – so if there fundamentally isn’t a unique coordinate for the neighbour of any input hat, then that test would report failure, even if the construction had really succeeded.

So the first thing we need is a different substitution system for the hat tiling, in which the tile expansions don’t overlap, so that every hat has a unique coordinate string.

The original hat paper mentions the existence of one. In the diagrams of the metatiles’ expansions, some of the sub-metatiles are shown with dotted lines, meaning that if those are omitted, then the overlaps go away. But it’s not described in full: the details of how the edges fit together are left to the reader to work out. That was exactly the part I didn’t feel like doing when I first looked at the paper, which is why I used the overlapping system instead and found a way to deal with the overlaps…

… but happily, since then, someone else has figured it out and written down all the details, so I still didn’t have to work it out myself! The paper “Dynamics and topology of the Hat family of tilings” contains just the missing details I wanted: in their Figure 4 they’ve divided the boundary of each metatile into edges of five different types, and their Figure 5 shows exactly how each of those edge types expands to multiple edges when the metatiles are expanded. (And the caption for Figure 5 suggests that the authors of that paper agreed with me that it wasn’t trivial!)

Given those details, we can calculate all the edge mappings in a way that’s very similar to the Spectre system. These diagrams assign an index to each part of every metatile’s boundary (often regarding a single straight edge of the polygon as divided into more than one segment). The expansion diagrams number all the internal edges of the tiles, so that you can see which pairs of edge segments of neighbouring tiles meet; and each edge segment on the perimeter is assigned a two-part number indicating which edge of the higher-order tile it counts as part of, and which smaller segment of that edge’s expansion it is. These diagrams contain enough information to make a set of lookup tables that would let you run the recursive algorithm for the hats tiling using this non-overlapping form of the substition system.

The substitution rules for turning each metatile into hats are the same as before, except that now we also have labels on all the edges showing how they match up to the next metatile:

Looking at those diagrams, we can see a few of those awkward zero-thickness spurs again, like the one we saw above in the Spectre system. They appear when two adjacent edges of a higher-order metatile expand to paths of lower-order edges in such a way that one path starts by retracing the last step of the other one. This happened only once in the Spectre tiling, but it happens multiple times here: once in the expansion of the P metatile into hats, twice in the expansion of T to hats, and three times in the expansion of H to other metatiles.

This is the point where it’s a good thing I automated the spur-handling technique I described earlier. With that technique done automatically rather than manually, these spurs are no more trouble than in the Spectre tiling, even though they occur at an infinite number of levels of the system (every time metatiles are expanded from more metatiles) instead of only a single level (hexagons to Spectres). With the same spur-handling code I described above, the adjacency matcher passes the exhaustive completeness test: it knows of at least one legal neighbour string for every valid input string.

Great! Now we try to build a deterministic transducer from that.

### Ambiguity

When I described the technique for converting an adjacency recogniser into a deterministic transducer, I mentioned that the algorithm wasn’t guaranteed to terminate. If it turns out that there’s no limit to how far the algorithm might need to look ahead in order to decide the next output, then the attempt to construct a finite-state transducer will fail, because the number of states just isn’t finite: the algorithm just keeps making more and more states containing longer and longer strings of pending output with different first symbols, and doesn’t reliably reach a point where it can decide that one of them is wrong and start emitting the other one.

For the Penrose tilings and the Spectre tiling, this possible failure didn’t happen. But for the hats tiling, it does! There are cases where the coordinate transition algorithm needs to look ahead by an unbounded distance to determine the next output coordinate.

When I saw this result, I wasn’t sure that I was looking at a real phenomenon and not a bug. So my next job was to find ways to cross-check it.

First, I had to find out what input string can trigger this problem, and what the two different possible outputs might be. To find that out, I ran the transducer construction with diagnostic printouts, so that it would display the full details of each state it discovered and what transition reached it. Then, when it reached a state with lots of buffered output, I could see what input would lead to that state, and what possible outputs it was considering.

The diagnostics showed that the problem input (or at least one of them) was a coordinate string starting like this:

`(hat, 11), (F, 0), (F, 5), (F, 5), (F, 5), (F, 5), (F, 5), (F, 5),`

That is, we’re trying to cross edge #11 of a hat, which is hat #0 expanded from an F metatile, which is child #5 of a bigger F, which is child #5 of a still bigger F, and so on, with each subsequent F being child #5 of a bigger one.

Cross-referencing those numbers to the expansion diagrams above which list all the child and edge indices, we find that edge #11 of hat #0 of an F metatile is part of the expansion of edge #0 of the F itself. In the next level of diagrams, we look at edge #0 of an F which is child #5 of a larger F: that’s part of edge #1 of the larger F. And in the next level, edge #1 of an F which is child #5 of another F is still part of edge #1 of the larger F.

So we’ve reached another situation like the one in the Spectre diagram I showed right at the start, in which you’re trying to cross edge #1 of an F, and you recurse one level up and find that now you’re trying to cross edge #1 of the next larger F, so you’re back in the same case again. In other words, this coordinate string surely identifies a point on an infinite supertile boundary: if we tried to process it with the recursive algorithm, it would recurse forever. And that’s just the kind of place where we might expect a problem in constructing a transducer – so that’s a good sign, because this seems to be making sense so far!

So, if the input identifies an infinite supertile boundary, and the machine doesn’t seem to be able to decide what to output … maybe that indicates that we’ve found an infinite boundary with two things that could fit to the far side?

My diagnostics also showed me the two output strings that the machine couldn’t decide between. They were:

`(hat, 4), (H, 0), (H, 7), (H, 7), (H, 7), (H, 7), (H, 7), (H, 7),`
`(hat, 0), (H, 2), (H, 8), (H, 8), (H, 8), (H, 8), (H, 8), (H, 8),`

in which not only does every higher-order metatile coordinate disagree, but we aren’t even sure of which edge of the neighbouring hat we’re going to find. The two possible outputs describe hats that aren’t in the same orientation, with edge #11 of the input hat meeting edge #4 or #0 respectively of the output one.

Of course, the diagnostics didn’t show these strings continuing in the same fashion all the way to infinity, because I had to interrupt the program after it had only buffered a finite amount of output. So I only had evidence that a large finite number of (F, 5) pairs in the input appeared to be able to match output in either of these two forms, with a similar number of (H, 7) or (H, 8) pairs.

But when I printed out the state machine for the adjacency recogniser itself and followed the transitions it would make on each of those (input, output) pairs, I found that it did recognise both of them as legitimate, and for each one, ended up in a state where a further pair of coordinates from that input would cause it to remain in the same state. So it really does look as if the infinite string that extends the above input can go with either of the infinite strings extending those two possible outputs.

We can also check this using the recursive algorithm, which is older and simpler technology, and doesn’t have any difficulty with spurs, so it doesn’t depend on any of the horribly complicated code I wrote to try to work around them. Just as I did for Penrose tilings during my earlier investigation, we can give the recursive algorithm a finitely long string of (F, 5) pairs as input, and let it produce some possible outputs by extending that string with random extra coordinates.

So I did that, and the results confirmed my previous observations:

No matter how long a string of (F, 5) coordinates I put in the input, the recursive algorithm was still able to generate output strings full of (H, 7) or full of (H, 8), depending on what random coordinates it made up to put on the end of my input string. This is completely different from the results in the earlier table, where I did this for Penrose tilings, and found that the output always started with the same coordinates, no matter what random things were appended to the end.

So I believe it. The hat substitution system has a fundamentally different character from the Penrose and Spectre ones in the following respect: when you encounter an infinitely high wall in this system, there isn’t always a unique answer to the question of what’s on the other side!

### A wall with two possible other sides

Of course, the next question is: what do the two possible far sides of this wall look like?

The first question is how to find out the answer! After all, the whole point is that I didn’t manage to generate a useful transducer this time, which is usually the tool that makes this job easy.

You could do it by generating a finite approximation to each version, whose coordinates are a finitely long prefix of the full infinite string, and just use a long enough prefix that the wrong parts round the outside are too far away to appear in the picture.

But there’s a nicer way. Even if you don’t manage to build a transducer, you can use the adjacency recogniser by itself to try to find out the neighbour of a specific infinite coordinate string C. You do this by another automaton-manipulation exercise: start from the adjacency recogniser, and construct a subset of it by constraining one of the two input strings to be C. This gives you a much smaller DFA that matches every possible infinite string that could be a neighbour of C. Then you can analyse that to find out whether the output is unique, by removing dead-end states until there are none left, and then just tracing forward from the start state to generate output, stopping if you still have multiple choices.

This is pretty slow. For every coordinate transition you have to build a new state machine! But it works for normal coordinate transitions which only cross a finitely high boundary. It can even cross infinite boundaries if there’s only one choice of what can go on the far side. The only time it fails is when crossing an infinite boundary where the far side is ambiguous. And even then, it will notice, and report failure, instead of recursing forever.

So you can use that technique to draw any tiling involving an ambiguous boundary, by providing it with a starting tile from each side of the boundary. Then it wouldn’t matter if the algorithm wasn’t able to cross the border, because every hat in the plane would be reachable from one of the starting hats.

In fact I didn’t even have to do that, because of another curiosity. The (F, 5) coordinate string I started with has two possible neighbours, containing infinitely many (H, 7) or (H, 8). But it turns out that each of those strings only has one possible neighbour – it can only go with the (F, 5) string!

So I was able to generate both of the diagrams below from just one starting hat – the one on the variable side of the boundary. This NFA-pruning transition algorithm can cross the boundary in one direction, to get to the invariant (F, 5) side. It just can’t cross back again to the side it started from – but that’s all right, because it doesn’t need to.

Anyway, here are the actual pictures. When I saw them, they took me completely by surprise. Most of the two alternative far sides match exactly! There’s just one path of hats that differ between the two, and that path meanders through the image on a self-similar course – rather like the boundary itself, but touching it only once and then wandering off in a different direction. Here I show the boundary in the usual thick black line, and I’ve also shaded the hats that aren’t the same between the two versions:

I wasn’t 100% sure what I’d expected to see, but it wasn’t that! I’d definitely imagined that, somehow, everything on one side of the wall would manage to be completely different between the (H, 7) and (H, 8) layouts, and only the (F, 5) layout on the other side would be constant. But in fact the path on which the two versions differ seems to have nothing much to do with the boundary. In fact, just as in the other tiling types, you wouldn’t be able to spot where the infinitely high boundary was, if I hadn’t marked it with a thick line. It isn’t the interesting part of this picture!

However, now that we’ve seen what this looks like, it’s possible to derive it again from first principles. Let’s go back to the original, overlapping, form of the hat metatile system, and look at the expansion of the H metatile:

The expansion diagram is very close to symmetric under a 120° rotation. All the tile types remain the same, and only the directions of the orientation arrows change. And not even all of those: if you rotate 120° anticlockwise, then the new topmost H tile is still oriented with its arrow downwards, and the P in the upper left is still oriented with its arrow inwards, because the H and P pair in the bottom right rotate into those same orientations when you move them up to the top.

So, let’s imagine that we made two tilings of the whole plane by starting with an H metatile in each of those two orientations, and repeatedly expanding it, at every stage identifying the previous patch of tiling with the expansion of the H in its bottom left (since in both cases that H has the same orientation as the larger starting one). A lot of the first-order expansion would be the same as before, and that similarity would be repeated on larger and larger scales, leading to very large sections of the whole plane matching perfectly.

Another thing we can see from this analysis is that we expect every part of the tiling to be covered by the same type of metatile in both variants. The only thing that should differ is the orientation of metatiles. And not even all the metatiles: the F metatiles don’t change position at all between these rotated expansions – and a good job too, because they’re completely asymmetric.

And, indeed, that matches what we see in the tilings above. The path of varying tiles is composed of hats expanded only from H, T and P metatiles; all the blue hats (from F metatiles) are invariant. Each pair of green hats (from a P) reverses into a P pointing the other way; yellow hats from a T rotate by 120°; groups of four red hats from an H rotate by 120°, which only alters three of the actual hats, because – just like the metatiles – one of the hats rotates into exactly the position that another has rotated out of.

Someone who was paying proper attention to the expansion diagrams could probably have spotted this possibility from first principles in March 2023, without needing any help. (Indeed, probably someone else has found this pair of tilings by now, although I haven’t heard of it.) But I didn’t. I had to discover it the hard way, by trying to make a transducer, having it fail, looking up the coordinates that went wrong, and trying to relate those back to the original diagrams!

### An alternative substitution system

It’s very interesting that we’ve found this pair of extremely similar instances of the hat tiling. It’s pretty cool that we were led to it by regular language theory (even if, with hindsight, there were easier ways). But practically speaking, it’s an inconvenience! As I said in a previous section, now that I’ve developed this transducer technique, I think it’s a really nice way to handle tiling generation – even better than the recursive algorithm I’ve been enthusing about in the previous two articles. So my first emotion, when I found I couldn’t make one for the hat tiling, was disappointment rather than fascination. I wanted to be able to recommend this state-machine technique for general use across all of these tiling types!

But it’s too early to give up hope. There’s an important question we haven’t answered: is the ambiguity in the hat tiling a property of the tiling itself, or just a property of the particular substitution system I’m using? In other words: is there a different substitution system for the hat tiling that avoids the ambiguity?

I spent a while trying to find modifications of the HTPF tiling system that encoded just enough extra information to disambiguate that ambiguous case. I wasn’t successful.

But in May 2024, somebody else was! Bowen Ping and Brad Klee have derived a completely different substitution system for the hat tiling, which looks a lot more like the Spectre one than the original HTPF system. The metatiles are all hexagonal; each one expands to just one hat, except for a single metatile that expands to two hats, one of which is the rare reflected one. And it turns out – although as far as I know this wasn’t anything to do with the problem the authors were trying to solve – that this system for generating hat tilings avoids the ambiguity, and allows the successful generation of a transducer!

I’ll show the 10-hex expansion rules in full. First, here are the rules for expanding each hex to more hexes. Every input hex delivers 7 output hexes in the same pattern, except for the M hex, which has one missing:

(The letters I’ve used for these hex types are my own choices. I’ve avoided I for the usual reason that it can be confused with a digit, and avoided F and H because those are metatile names in the other hat system.)

And here are the rules for expanding each hex into an output hat – again, except for the M hex, which expands to two hats, with the one on top being the reflected ‘antihat’. They all look very similar, but the important point isn’t the single hat in each case (except M): it’s the specification of which edges of the hat correspond to which edges of the hexagon.

Again, there are some zero-width spurs. Quite a lot of them, in fact: seven of the hex types have a spur in their expansion to hexes, and the D hex has a spur in its expansion to hats as well. But the same technique I described for the Spectre tiling is enough to deal with them, and construct an adjacency matcher that passes the completeness test: it knows at least one legal output for any input. And when I tried to convert it to a deterministic transducer, that was successful too: this system has exactly one legal output for any input. Success!

For Penrose tiles and Spectre tiles, the first thing I did at this point was to figure out input coordinate strings that would generate patterns with rotational or reflective symmetry, because they’re pretty. But in this case, that wasn’t the first thing on my mind. The first thing was: how does this substitution system deal with the ambiguous pair of hat tilings in the previous section?

By eyeball comparison, I managed to work out a few things about how the HTPF system matches up to this one. The result was enough to find a pair of infinite coordinate strings that both correspond to the (F, 5) infinite supertile. One is to take the A hex type and make it child #2 of another A hex forever; the other is to do exactly the same thing with the E hex type. These two supertiles correspond to versions of (F, 5) that go with the (H, 7) and (H, 8) variants respectively.

Here are the two resulting images. As before, I’ve shaded the path of hats that are different in shape and layout between the two tilings:

There’s lots of interesting stuff here!

My first question was: how has this substitution system avoided the ambiguity?

In the infinite supertile at the bottom of the picture, corresponding to the invariant (F, 5) section, there’s just one visible difference: the topmost hat on the left is a different colour, indicating that it’s expanded from a different first-order hexagon type. (Its higher-order hexagon types are all different too, of course; it’s just that I only used the lowest-order one to decide the hat colour.)

In fact, looking more closely, quite a lot of hats near the varying path change colour. It’s as if the path of changing actual hats has a sort of ‘aura’ around it, composed of hats that change colour but not shape. And that aura extends just far enough to cross the infinite boundary – so you can disambiguate which version of the path you want to see, by specifying what colour that one corner hat should be, because it will be different in the two cases.

But also: look what the infinite boundaries are doing!

In the HTPF version of this picture, there was just one infinite supertile boundary, separating the invariant (F, 5) supertile from the two alternative ones (H, 7) and (H, 8) that fitted to it. But in this system, each of those two alternatives is itself divided into two infinite supertiles, so that the diagram as a whole contains three. Not only that, but the boundary between them isn’t in the same place in both versions. In fact the connectivity between the regions isn’t the same either: in the (A, 2) picture one region separates the other two, but in the (E, 2) picture there’s a point where all three regions connect. Even more interestingly, it looks as if half of that extra boundary is the same, and only the part on the left changes!

I have no idea what the mathematical significance of any of that is (if any), but it’s fascinating, and again not what I’d have expected!

You might wonder what the other supertiles are, in these two diagrams. In the (A, 2) diagram – corresponding to (H, 7) in the HTPF version – the other two supertiles are (D, 0) on the left, and (K, 2) in the upper right. In the (E, 2) diagram, corresponding to (H, 8), they’re (M, 0) in the middle and (D, 2) at the top. (My notation for an infinite supertile with eventually periodic coordinates is to simply list the eventually-repeating section. If it has length >1 – which we’ve seen in the Penrose examples, and there will be a couple more below – then it makes sense to list it starting at a multiple of its own length, to distinguish supertiles whose repeating sections are rotations of each other.)

Finally, I’ll show some symmetric instances of the hat tiling, just as I’ve done for the other two tilings earlier. Looking at the HTPF system, it’s always seemed clear to me that there are two instances of the hat tiling with order-3 rotational symmetry: if you start by putting three F metatiles with their points together, and expand that figure once, you find the expansion still has three F around the centre point but this time with their other 120° corners meeting. Expanding again, the Fs turn round again and you’re back to having the points together. So you expect two symmetric hat tilings, by choosing which of those orientations to take as the lowest order of metatiles and expand to hats.

In the 10-hex system, the corresponding coordinates also involve a two-way alternation. You need a G hex, which is child #6 of an E, which is child #2 of a G, and so on. By taking the lowest-order hex to be one or the other of those, we can generate the pair of symmetric tilings using our nice efficient 10-hat transducer.

There’s one more symmetric hat tiling, which occurs when two F metatiles meet back to back. Expanding that figure repeatedly about its centre gives a tiling with 2-way rotational symmetry. In the 10-hex system, it corresponds to a G hex being child #4 of another hex forever.

So here are the three symmetric instances of the hat tiling:

## What does all of this mean, in mathematical terms?

In this article so far, I’ve talked a lot about the mechanism of constructing and using these transducers, but not a lot about what it means that we can build one, or can’t.

To begin with: do we even have any guarantee that a tiling constructed by using a transducer to cross an infinite supertile boundary is consistent? It looks plausible in the examples so far, but can we be sure that if you walk a certain number of steps along the boundary from your starting tile, and then cross it using an infinitary step, then you get the same coordinate string that you’d have got if you crossed at the original location and then walked along the far side of the boundary to the same place?

Also, in tilings where you can construct a transducer, does that really prove that there’s only one possible way to extend a tiling of the plane to the other side of an infinite supertile boundary? Or might the transducer be restricting itself to a subset of the possibilities, so that there might be a completely different answer that the transducer could never generate?

The second of those questions is the easier one to start answering, because we’ve already seen an example! Consider the case of the two very similar instances of the hat tiling. Even in the 10-hex substitution system, there’s an infinite supertile – the one shown at the bottom of the image – in which all the hats are exactly the same in both versions. The only thing that differs is the sequence of types of hexagonal supertile.

But that doesn’t stop the actual hats from fitting together on the two sides of the boundary!

Suppose you’d never heard of the HTPF substitution system, and you only knew of the 10-hex system for the hat tiling. Suppose you’d written a hat tiling generator, using the recursive algorithm, with coordinates based on the 10-hex system. Suppose it crashed the first time you tested it, for the same reason my Spectre one did – you hadn’t connected up the random numbers yet, and you’d accidentally fed in a fixed sequence of coordinates that defined an infinite supertile boundary. And suppose, by really bad luck, your fixed coordinate sequence described that infinite supertile boundary, from our HTPF ambiguous pair.

You might wonder the same thing as I did: whether the far side of the boundary existed, and whether it was unique. You might do exactly what I did – figure out how to build a transducer. And since you’re using the 10-hex hat system, you’d find the transducer was successfully constructed (as I did, in the Spectre case). You’d use it to generate what you thought was the unique far side of your boundary. And then somebody would come along and exhibit a different pattern of hats that also fits to the far side of your original boundary, and you’d be completely startled. Perhaps you’d even be angry. The transducer has somehow let you down – it promised uniqueness, and didn’t deliver!

What’s the loophole? The answer is that the transducer promises a unique sequence of coordinates that’s the neighbour of a given input sequence. But maybe what we really wanted to know is whether there’s a unique layout of tiles that’s the neighbour of some existing layout. And in this case, there are two sequences of coordinates that give rise to exactly the same layout of hats on the near side of the boundary. The transducer gives you a unique neighbour for your sequence, and for the other sequence, but really, each one also fits to the other.

But wait – how do I know that’s not true for Spectres? For all I know, the example I originally found could be in exactly this situation, and I might just not have found the other sequence yet!

So the existence of a finite-state transducer doesn’t guarantee everything we’d like it to. What does it guarantee? What can we say about tilings for which a transducer exists?

We derived our transducer from a state machine that recognises the language of pairs of finite coordinate strings that the original recursive algorithm can transform into each other: that is, the relative coordinates of two neighbouring sub-tiles within the (iterated) expansion of some individual high-order (but finite-order) supertile. So any pair of strings that appear as the input and output of the transducer must have the property that each finite prefix of the pair of strings must also be a prefix of some valid finitely long coordinate transition. Otherwise the adjacency recogniser would have rejected the pair before that prefix was complete, and the transducer wouldn’t have generated it as output.

There’s probably a better name, but I think of this property as “local plausibility”: the pair of neighbouring tiles is surrounded by larger and larger neighbourhoods in which the context – including the supertile types – looks like things you’d also see in ordinary finite situations.

The good news is that I think this “local plausibility” property does mean that the resulting tiling is consistent. If you walk any finite number of steps along one side of the boundary, then there’s some finite supertile that contains all the steps you’ve taken. And the counterpart of that finite supertile on the far side of the boundary must fit to it, because of local plausibility: there’s some ordinary finite situation in which the same two supertiles are adjacent in the same way. So if you walk in a loop across the boundary and back, you should never find an inconsistency.

What about uniqueness? The transducer can only tell us that there’s a unique locally plausible extension of the half-tiling we started with, because locally plausible tilings are all it knows about. So it can’t tell us what other possibilities might exist that aren’t locally plausible.

And that’s what went wrong in this case of the ambiguous hat tiling. There are two ways to extend the original infinite supertile to a full hat tiling of the plane; but in the 10-hex system, you have to make a choice about which coordinate string you use to represent the supertile, and whichever one you pick, only one of the extensions is locally plausible.

So there’s a unique locally plausible extension, but that doesn’t mean there’s a unique extension of any kind.

So, does a case like this also exist in the Spectre tiling? I have no idea! I’d be interested in ways to find out.

## Other possible applications for state machines

I came up with this transducer idea as a means of investigating those infinitely high walls. I think it’s also a good system for generating finite patches of tiling – I recommend it over the previous recursive algorithm.

But I don’t think that’s the only thing a transducer – or even the simpler adjacency recogniser – can possibly be useful for. Here are some other thoughts.

### Generating coordinates from a tiling patch

My main use of transducers is to generate a patch of tiling, starting from a string of coordinates. I think it should also be possible to use them to go in the other direction.

Suppose somebody shows me a patch of Penrose or Spectre or hat tiling, without any coordinate labels attached. I’d like to be able to find a string of coordinates in my preferred substitution system which give rise to precisely that patch, containing the same tiles connected in the same ways. (For example, that would let me recreate the same patch in my own software, and also find legal ways to extend it.) But I wasn’t given any coordinates at all: all I have is a complicated drawing that shows the lowest-level tiles, and how they connect to each other.

If I have a transducer for my substitution system, I think I could use it to answer this question, in the following way.

I maintain a DFA that matches a regular language of coordinate strings. I choose a starting tile of the tiling patch, and initialise my DFA to the one that matches any legal coordinate string for the type of that starting tile.

Now I walk around the tiling patch I’m given, stepping from one tile to the next, until I’ve visited every tile. Every time I step from a tile t to another tile u, I combine my DFA of possible coordinate strings for t with my transducer, to produce a DFA of possible coordinates for u, under the constraint that the first symbol of the output (giving the lowest-level tile type, and saying which edge of it corresponds to the edge of t I just stepped off) must match what’s in the input tiling patch. This should be a simple exercise in DFA manipulation.

So, as I walk around the patch, my DFA is updated in two ways. Firstly, it updates with my current position. But it also becomes more and more refined, because it has to match only the coordinates that could have given rise to every tile I’ve visited so far.

So when I’ve covered the whole patch, I don’t just have one possible coordinate for the tile I end up at. I have a DFA that recognises the language of all the possible coordinates! And it’s easy to pick an arbitrary one to return.

(It would be nice if we could combine this coordinate-finder idea with coordinate generation, to produce a system that can translate a string of coordinates between two substitution systems for the same tiling. I’m pretty sure you could do that slowly for a finite patch, by generating the whole patch of interest in one system and walking around all of it with the coordinate finder I describe above. But whether you can do it fast, or for infinite coordinate sequences, is another matter. The translations I’ve done in this article, between the HTPF and 10-hex systems for hats, were done by hand, based on the coordinates in question being particularly simple.)

### Choosing coordinates with interesting properties

Another use of a finite state machine that understands your coordinate system is that you can explore it to find interesting corners.

I’ve shown several examples of the Spectre and hat tilings with rotational symmetry. These all have the property that, at the centre of symmetry, two or three tiles meet which have identical combinatorial coordinates – two copies of the same infinite supertile share a boundary, and their apex tiles share an edge.

If I hadn’t already known or guessed that some symmetric instances of these tilings existed, I could have found them in an automated way, by searching the adjacency recogniser state machine. Like several previous examples in this article, you do this by constraining its input – but instead of constraining one side of the input to a fixed string and seeing what comes out as the other, you constrain the two sides of the input to be the same as each other (apart from the edge indices at the start). This search will deliver a DFA that matches the full set of coordinate strings of tiles that can possibly meet another copy of themselves.

In fact, one of the tilings in this article was found that way. I spotted the three symmetric Spectre tilings by eye, and I’d already guessed that the two 3-way symmetric hat tilings must exist, but I overlooked the 2-way symmetric hat tiling, until this search pointed it out to me.

Another search of this kind lets you find strings of coordinates that give rise to infinite supertile boundaries in the first place. If I hadn’t started this whole investigation by stumbling on one, could I have found it by a systematic search?

Yes, I could. If you delete all the accepting states from an adjacency recogniser, then any infinite walk in the remaining directed graph spells out a pair of infinite coordinate strings that have an infinitely high wall between them, because precisely the criterion for a wall being finitely high is that the adjacency recogniser enters an accepting state. So it would be easy to find any of the infinite walls I’ve mentioned already: for the ones in which a single coordinate is repeated forever, you just look for any vertex with an edge back to itself, and for the ones with longer period (like ABAB… or AVBUAVBU… in the P2 tilings) you look for a longer circuit.

This analysis also shows that there must exist an uncountable number of other infinite supertile boundaries, whose coordinates aren’t periodic: as soon as you can find more than one different circuit from the same vertex back to itself, you can construct a coordinate string interleaving those circuits in any pattern you like. I’ve only been showing boundaries derived from strings with a short period here because those tend to be the pretty ones. There are lots of others.

## Future possibilities

What’s left to do with all of this?

I’d like to have a try at writing the coordinate-finder I mentioned just now. I expect the DFA manipulation to be reasonably easy, but the hard part would be finding a practical way for a user to input an existing patch of tiling that they wanted to determine coordinates for. (You wouldn’t want to type in edge indices all the time!)

Another thing I’d like to improve is the techniques for handling spurs and other awkwardnesses in tiling descriptions. The technique I described in detail above was enough to cope with all the spurs in the Spectre 9-hex system, the hat HTPF system, and the new hat 10-hex system – but it doesn’t cope with everything. In particular, here’s a substitution system that my software still can’t build a working adjacency recogniser for:

This substitution system generates the same Penrose P2 tiling we’ve already got working, but this time I haven’t done my usual trick of cutting each kite and dart into two triangles. Instead I’ve left them whole, which makes life easier for an end user trying to generate the tiling – they wouldn’t have to do the postprocessing step of expanding each triangle into its full tile and discarding duplicates.

This system works fine if you’re using the basic recursive algorithm: it might have to recurse back and forth crossing spurs, but it gets there in the end. But I haven’t been able to build a working adjacency recogniser for it, and I’ve tried fairly hard – I added lots of extra sophistication to my code to handle the awkward cases that come up, but never got the result to pass my completeness check. There always seemed to be another awkward case I hadn’t handled yet.

(Why’s it hard? I’m not sure I even have a complete list of the problems. But one example is that, in this system, two tiles sharing an edge can be expanded from higher-order tiles that only share a vertex, because the darts in the expansions stick out by such a long distance from the original tile outlines.)

And this system, doing the same for the P3 tiling, is perhaps even worse, because in a sense, one of the expansions isn’t even connected:

Maybe one day I’ll have another try.

## Software

If you’d like to play with all of this stuff yourself, I’ve uploaded my software for general substitution-system handling into my git repository collection. It’s written in Sage, an extension of Python to be (among other things) a symbolic algebra system.

You can clone my git repository with this command:

`git clone https://git.tartarus.org/simon/tilings.git`

Or you can browse the repository on the web.

I warn you that this isn’t quite up to my usual standards of solid production code! This is research code: when I wrote it I was trying to find out whether things would work at all, so I didn’t optimise or refactor or polish very much, for fear that I’d waste my time. I’ve written some hasty documentation in `README.txt`, but for more detail, you might have to examine the example tiling files yourself, or the code. Or ask me specific questions, and I’ll try to extend the documentation to include answers to them.

Perhaps the most immediately useful thing you could do with that software is to generate the actual working transducers for Penrose P2 and P3, Spectre 9-hex, and hat 10-hex. If you want to actually use the transducer algorithm to generate tilings, calculating those state machines is the hard part of the job, and my software will do it for you. The rest of the algorithm is just a matter of processing strings using the transducer, making up random extra coordinates if you need to, and plotting the output in whatever graphics system you want to use.

This software will do all of that too – you can use it to directly generate pretty pictures – but it’s quite slow. To make a significantly large piece of tiling I’d probably want to write the output engine in a faster language!

## Acknowledgments and references

I’m indebted to Robin Houston for helpful discussions while I was investigating this, and for reading the draft article.

I’m also grateful to Bowen Ping and Brad Klee, firstly for inventing their 10-hex substitution system for the hat tiling, and secondly for kindly permitting me to publish it here. Without that system, this article would have been very different, and would have ended in a shamefaced apology that this really nice algorithm only worked properly for two of the three aperiodic tilings I’ve been discussing.

As I said in the introduction, I’m not claiming to have discovered all the mathematical ideas here for the first time. I thought of them by myself, but at least some of them had been thought of independently before. The first paper I link to below describes the concept of ‘addresses’ in a substitution tiling, which are equivalent to the thing I’ve been calling ‘combinatorial coordinates’; it mentions the possibility of infinite-order supertiles that aren’t the whole plane; and it describes something very like my adjacency-recogniser state machine, specifically mentioning the possible application of using it to match up infinite supertiles whose boundaries fit together. So I definitely don’t get credit for any of those ideas – they predate my investigation here by 25 years.

Some other useful resources:

Aperiodic hierarchical tilings
A 1999 paper by Chaim Goodman-Strauss (24 years before co-discovering the hat tiling), discussing substitution tilings in general and the notion of addressing a tile by its place in the hierarchy.
An aperiodic monotile
The paper introducing the hat tiling, and showing the original (overlapping) substitution system of H,T,P,F metatiles.
Dynamics and topology of the Hat family of tilings
I haven’t used the main subject of this paper, but in passing, it provides full details of the non-overlapping version of the H,T,P,F substitution system.
A chiral aperiodic monotile
The paper introducing the Spectre tiling, with full details of its substitution system.
HatHexagons and HatHexagonalTiling in the Wolfram Function Repository
Bowen Ping and Brad Klee’s own code exhibiting the 10-hex substitution system for the hat tiling.

## Appendix: Batman turns up where you least expect him

The previous two articles both ended with an irrelevant but pretty appendix. Let’s do that again!

In a previous section I derived the combinatorial coordinate representation of the two pentagonally symmetric instances of the P2 tiling. Those aren’t new, of course: they’re well known in the Penrose tiling literature, usually under the names “infinite sun pattern” (the one with a decagon of five kites at the centre) and “infinite star pattern” (with a star of five darts).

That wasn’t too hard, because the coordinates system makes it easy to work out the representation of a pattern that’s completely symmetric. But there’s another quite famous P2 tiling instance (for example, Martin Gardner chose it to be the cover of a book) which is very nearly symmetric, but not quite. It’s known as the “cartwheel pattern”:

From a distance it looks as if this pattern has the symmetries of a decagon: 10-way rotational symmetry, and reflection. But it’s not quite true. The whole pattern really does have reflective symmetry about the vertical axis, but the rotational symmetry isn’t quite complete. If you rotate the whole picture a tenth of a turn at a time, the blue figure inside the central decagon – known as ‘Batman’ – will gradually turn upside down. And the ten radiating ‘spokes’, highlighted here in red, aren’t quite symmetric either: they come in two different handednesses, arranged in an irregular pattern around the circle. So when you rotate the whole picture, some of the spokes rotate into the place of one with the opposite handedness.

I thought it would be fun to try to figure out the combinatorial coordinates of a tile in this pattern, so that it would be easy to draw it at large scale. When I did it, I got a surprise.

Since this picture is mostly symmetric, it makes sense to start by wondering: what happens if we expand each half-tile triangle via the usual expansion rules? We’d surely get another picture with the same ‘mostly symmetric’ property, and maybe after some number of iterations it would look like the same one again?

The easiest way to try that is to start from the central ‘Batman in a decagon’ figure, divide its kites and darts into the triangles I use for my coordinate system, and expand it a few times to see what happens:

Just what we were hoping for! Expanding this decagon once, we get a pattern that has a smaller copy of the same decagon figure again at its centre – but upside down. Expanding a second time, it’s the right way up again. And repeating the expansion twice more, we see that this procedure by itself is producing the spokes and symmetric sectors on the outside.

Of course, I haven’t proved that that will continue. But we can at least see that the essential symmetry is likely to continue: since expanding the original decagon 4 times produces a lot of mostly-symmetric exterior with a decagon at the centre, it follows that expanding another 4 times will introduce the same chunk of tiles in place of the smaller central decagon, and expand the mostly-symmetric exterior we already have into more mostly-symmetric stuff. So it’s pretty convincing.

What about actually finding the coordinates? This is another exercise of the same type we’ve done several times in this article. Choose a triangle in the middle of the largest Batman; expand twice to get a smaller Batman the same way up; determine the coordinates of the corresponding triangle of the smaller Batman inside the larger original triangle; then we’ll have a finite string of coordinates that we can repeat forever, to obtain the infinite coordinate sequence for that triangle in the whole plane. Since the two triangles will share an edge, we also expect that there will be an infinite supertile boundary down the vertical axis, but that’s fine, our transducer can cope with that.

Great! So let’s try it. We’ll choose the type-A triangle in the middle of Batman, just to the right of the centre line.

The smaller instance of the same triangle is an A, which is a child of a B, which is a child of the larger A we started with. Iterating this forever, that must mean the type-A triangle in the middle of the Batman figure has the infinite coordinate sequence ABABAB…

Wait … that looks very familiar, doesn’t it? It’s exactly the same coordinate sequence I chose as my arbitrary starting example earlier, when I was trying to convince myself that it was even possible for an infinite supertile boundary to have something on the far side at all. It was the first example I found, because it’s a particularly simple way to expand a triangle and choose a similarly oriented sub-triangle part way along one of its edges – you only have to expand by two steps, which is the fewest possible.

In the course of doing this investigation and writing this article, I’d already found the coordinates describing the cartwheel pattern, completely by accident – and I’d never noticed! If you took that ‘simplest possible’ reflection-symmetric P2 tiling I showed in the previous section, turned it 90° so that the axis was vertical, and coloured it correctly, it would be the cartwheel pattern.

Batman was hiding behind me all the time!