A NOTE ON APERIODIC AMMANN TILESSHIGEKI AKIYAMAAbstract. We present a variant of Ammann tiles consisting of twosimilar rectilinear hexagons with edge subdivision, which can tile theplane but only in non periodic ways. A special matching rule, ghost
marking, plays a key role in the proof.

We shall show that the set of tiles in Figure 1 is aperiodic, that is, it tiles
the plane but only in non periodic way.
c1c2
111c1c11c2

c1c2c1

1c1c

1c1

1c2Figure 1. Similar Ammann tiles A and B√
Two shapes are √similar to each other by the ratio c = τ where τ is thegolden mean (1 + 5)/2. These shapes belong to the ‘Ammann A2’ family
having length parameters. On this choice of length, Grünbaum-Shephard[9] remarked‘However it does not seem possible to devise a simple matching condition
for these tiles involving markings which are such that the similarity between
the tiles extends to markings as well.’and suggested a complicated matching rule including directions and a certain
gluing rule of edges to force aperiodicity. Figure 2 shows their markings withedge direction in [9]. Their two tiles have exactly similar markings.
Our marking is a simpliﬁcation of it, but they are diﬀerent between twotiles. Markings of edges in Figure 1 indicate that they are cut at thoseplaces, and thus the tile A is treated as a decagon and B an octagon. Our
matching rule is simple, we assume that our tiling is edge-to-edge, that is,
only the edges of equal lengths can meet in the tiling and they must exactly
12SHIGEKI AKIYAMAFigure 2. Markings by Grünbaum-Shephard in [9]agree from one end to the other. We prove that this simple matching ruleprovides aperiodic tiles, regardless of the above quoted remark.
Theorem 1. The decagon A and the octagon B form a set of aperiodic tiles.
A tile is a closed polygon in R2 . Denote by A a ﬁnite set of tiles. A tiling
T of R2 is a collection of tiles which covers R2 without interior intersections,and each tile is congruent to an element of A under rigid motion (action ofisometry of R2 : composition of rotation, translation, and reﬂection). A tilingT has a period p ∈ R2 when the tiling exactly matches with its translation
by p. A patch of a tiling T is a ﬁnite set of tiles in T whose union ishomeomorphic to a closed ball. A tiling T is non periodic if the only periodof T is 0. A set of tiles A is called aperiodic if A generates a tiling, but
all the tilings generated by A are non periodic. To enforce aperiodicity, weoften introduce certain matching rules (conditions) on the surfaces of tiles.
A well known example of aperiodic tiles is due to Penrose which consistsof two kinds of tiles: kites and darts with matching rules. Ammann gave
several sets of aperiodic tiles, one of which is a target of this paper. See[9, 13] for tilings by Penrose, Ammann and more with historical accounts,
and [3] for the proof of aperiodicity of Ammann tiles. It had been an open
question to give a single tile which forms a set of aperiodic tiles. Recently
Socolar-Taylor [15] gave such a hexagonal tile with matching rules. Theirmatching rules aﬀect non adjacent tiles but their inﬂuence is local. Readers
can ﬁnd further discussion to clarify ambiguities in the deﬁnition of numberof tiles, periods and matching rules in [16].
The shape in Figure 1 is rediscovered by Scherer who considered a dissection of a polygon X into several pieces, all of which is similar to theoriginal X. Seemingly unaware of the result of Ammann, he found a solution ‘Golden Bee’, which is a polygon composed of two similar copies ofitself (see Figure 4). The term ‘Bee’ comes from the shape of the alphabet
character ‘b’. Completing the study of Scherer, Schmerl [12] showedthat√
the only polygons that have this property are a right triangle, a 2 : 1 sizedparallelogram, or the Golden Bee. See also [2] for a similar question ondissection, assuming that all pieces are similar to each other, but they are
not necessarily similar to the original shape.

A NOTE ON APERIODIC AMMANN TILES

3

Our set of tiles consists of two similar rectilinear hexagons in Figure 1. A
counter-clockwise simple closed curve consisting of line segments of lengths
c3 , c2 , c, 1/c2 , 1/c, 1 in this order forms a boundary of B and we make turnsof right angle at the end points of segments. We turn right at the junction of edges of length 1/c2 and 1/c which forms a dent of B, and turnleft at all other corners. The tile B carries two markings within. Themarkings depicted in B indicate that we deal with this tile as an octagon
rather than a hexagon. Thus we consider the length of the edges of B asc, 1/c, 1, 1/c2 , c, 1/c2 , 1/c, 1 in counter-clockwise order where c3 = c + 1/c,c2 = 1 + 1/c2 . The shape of A is c-times larger than B and A has fourmarkings which cut three edges. So we consider A a decagon of edge lengths
1/c2 , 1, 1, c, 1/c, 1, 1/c2 , 1/c, 1, c in counter-clockwise order where c4 = 1 +1 + 1/c2 , c3 = c + 1/c and c2 = 1 + 1/c2 .
It turned out that there are 8 tiles up to translation in this tiling because
the right angle rotation does not appear (see Figure 11).
Proof of the Theorem. Hereafter, to avoid description of rigid motion,congruent tiles are not distinguished and designated by the same symbol
when there is no room of confusion. An edge of length k is called k-edge.
A standard way to prove non periodicity of a tiling is to show uniquecomposition property (c.f. [9, 3, 18]). Assume, for a given tiling by A andB, there is a unique way to compose them as a super-tiling by tiles cA andcB and also we can repeat this discussion. Then a period p of a tiling must
be a period of the super-tiling. Iterating this discussion, the length of p
must be smaller than cn : the diameter of a ball which can be inscribed bothin cn A and in cn B, which shows p = 0. To prove aperiodicity of A andB using unique composition property, we show that all tiling by A and B
can be regarded as a tiling by super-tiles cA and cB having exact enlarged
matching rules. As a result, we shall show that by this simple marking, theset belongs to Ammann A2 aperiodic family in the sense of [3], and that all
tiling by A and B are non periodic.
Let us start with the fact that cB = A discarding markings. We observethat it is impossible to tile the plane only by A. Indeed, the only possibility
to ﬁll the dent of A by A is depicted in Figure 3 by edge-to-edge condition
and the northern remaining part can not be ﬁlled only by A. It is also plain
to see that the dent of B is ﬁlled only by a suitable reﬂection of A as inFigure 4. Therefore a tiling by A and B must contain at least one A and B.
For any given tiling by A and B, one can think of the union of B and A inFigure 4 as a super-tile cA, because its shape is a c times similar extension ofA and inherits markings scaled by c with a new marking from the boundary
broken line. One marking of A becoming a dent of cA should be erased andwe further remove two markings to obtain cA. The remaining A tiles which
do not ﬁll the dent of B, can be seen as cB by erasing two markings. See
Figure 5 for the super-tiles cA and cB. We have to show that this removal
does no harm. To this matter, let us call these four markings to be removed,ghost markings. We observe that all ghost markings appear at the junction

4SHIGEKI AKIYAMA

Figure 3. Filling the dent of A by A

Figure 4. Golden Bee: Filling the dent of B

Figure 5. Super-tiles cA and cB: broken lines indicate 4
ghost markings and the former boundaryof 1-edge and 1/c2 -edge. Our aim is to show that the set of tilings by cA
and cB with ghost markings and the one without them are the same.
Now we come to a technical point of the proof. Note that inﬂuence ofremoval of a usual marking is intricate, it does not simply mean to relax or tostrengthen the matching rule: on one hand we can no longer cut this c2 -edge
into 1-edge and 1/c2 -edge at this point, but on the other hand this c2 -edge

A NOTE ON APERIODIC AMMANN TILES

5

becomes orientation insensitive and thus c2 -edges may match in two ways.
To cope with this logical diﬃculty, we treat ghost markings in a special way:
we admit the freedom to ignore it or to use it to cut the edge. In other words,
two edges meeting at a ghost marking can be used as a single c2 -edge or thecombination of 1-edge and 1/c2 -edge, as we wish wherever they are andindependently. In short, we may think that ‘ghosts exist or not exist’ as welike. Under this treatise, we have an inclusion T1 ⊃ T2 where T1 is the set oftilings by cA and cB with ghost markings and T2 is the one without them.1
We then prove in the sequel that in any tiling in T1 , every ghost marking
must meet another ghost marking across the edges, which shows that ghostmarkings are dispensable, i.e., T1 = T2 . Once we know this equality, thesuper-tiling by cA and cB appeared in the previous paragraph surely falls
into T1 . Thus the proof of unique composition property will be ﬁnished.
Pick a tiling of T1 . First, we show that ghost markings can not be used
to cut c2 -edges. To see this, it suﬃces to examine the cases where two 1/c2 edges meet but their ghost markings do not. There are 10 cases to check,
up to symmetry as in Figure 6. We prove that they do not appear in T1 , byFigure 6. Illegal patches of the 1-st kindshowing that they can not extend to large patches. Indeed, as indicated byblack spots, we can either immediately ﬁnd a corner or dent which can not
be ﬁlled, or two corners that we can not simultaneously ﬁll. Second, we showthat there is no tiling in T1 that two c2 -edges meet but their ghost markings
do not match. We check such 10 cases up to symmetry as in Figure 7 inthe similar manner. As black spots indicate, for 9 cases, we can show in at
most two steps. The remaining case without a black spot requires further
classiﬁcation. It is reduced to four conﬁgurations in Figure 8 and in each
1Ignore markings when we compare two tilings to have the inclusion.

6

SHIGEKI AKIYAMA

Figure 7. Illegal patches of the 2nd kindcase we ﬁnd a dent which can not be ﬁlled. Thus we showed that a tiling inFigure 8. The remaining caseby cA and cB with ghost markings is possible only when ghost markings are
not essentially in use, i.e., T1 ⊂ T2 . We have shown the unique composition
property.
It remains to show that there exists a tiling by A and B. We have seenthat the super-tile cA is subdivided into A and B and cB is made fromA in Figure 5. Let us discard all markings for the moment and consider
this Figure 5 as a substitution rule: giving the images of A and B by thecomposition of two maps: c-multiplication and subdivision into A and B.
We can iterate this c-multiplication subdivision and create patches which
spirally grow like Figure 9. This gives a tiling of the plane by A and B,
which is called a ﬁxed point of the substitution rule. We may view thisﬁxed point as a super-tiling by c3 A and c3 B with markings indicating thesubdivision structure by A and B. Then c3 A is the composition of three A’s
and two B’s, and c3 B is a composition of two A’s and one B. Boundaries ofA and B give natural markings of super-tiles c3 A and c3 B as black points inFigure 10. Conﬁrm that they are exactly located at the c3 scaled markingsof A and B. As the tiling by c3 A and c3 B surely exists, we know that there

A NOTE ON APERIODIC AMMANN TILES7Figure 9. Inﬂation subdivision

Figure 10. Markings from c3 A and c3 B
exists a tiling by A and B with the markings scaled down by ratio 1/c3 .
Figure 11 gives a patch of tiling by A and B.
Figure 11. A patch of tiling by A and B
Aperiodic tiles give a good model of quasi-crystals. The diﬀraction pattern of Penrose tilings shows 10-fold symmetry, which simulates the discovery of quasi-crystals by Shechtman [14]. The reason of 10-fold symmetryis explained from the work of de Bruijn [4], who proved that Penrose tiling

8SHIGEKI AKIYAMA

is a cut and projection tiling, i.e., its reference points set is the projectionof 5 dimensional lattice points in some band. Frettlöh [5] showed that thisAmmann tiling is a cut and project tiling as well, and it is an ‘algebraic
dual’ of golden triangle tiling (c.f. [8, 6, 7]). An algorithm for pure discrete tiling dynamics in [1] directly tells the existence of an underlying cutand projection scheme. A nice feature of such aperiodic tiles related to cutand projection scheme is that local interactions (matching rules) alone force
pure point diﬀraction in diﬀraction. Therefore it is not necessary to assume
higher dimensional lattice model within real substances. See [17, 10, 11].
As in [9], Ammann found several markings on A2 family. One of themis a slight modiﬁcation of A and B which we call A′ and B ′ : tile A is
shrunk by ratio 1/c to x-axis and B to y-axis. He put several diﬀerent kindsof markings to have aperiodicity. Figure 12 shows his impressive markings
called ‘Ammann bars’ where the segments on A′ and B ′ must extend tostraight lines across the edges. The tiling by A′ and B ′ is a 1/c-shrunk of111Τ

1

Τ21Τ
1ΤΤ

Τ
1

ΤΤ

Figure 12. Ammann Bars
the one by A and B in one direction. However for A′ and B ′ , we do notﬁnd edge-to-edge matching rules like A and B which enforce aperiodicity.
Our construction is sensitive to the aﬃne transform which changes lengthsof the edges.
Our method of ghost markings might apply to situations where all tiles
are similar. Especially it is challenging to ﬁnd a rep-tile that our method
works. Since it is composed of same-sized similar copies of itself, it would
produce a single aperiodic tile with edge-to-edge matching rules. However
the author did not ﬁnd such an example so far.
This work is done while my stay in Kochi University in September 2011.
The author wishes to express his gratitude to K. Komatsu, H. Hayashi andH. Ei for invitation, stimulating discussion and pointing a missing case (!)
in Figure 6, and J.-Y. Lee and D. Frettlöh for discussion and supplying relevant references. He also thanks anonymous referees who gave him essential
advices on the readability of the original manuscript.

A NOTE ON APERIODIC AMMANN TILES

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