On the Structure of Ammann A2 Tilings

Abstract

We establish a structure theorem for the family of Ammann A2 tilings of the plane. Using that theorem we show that every Ammann A2 tiling is self-similar in the sense of Solomyak (Discret Comput Geom 20:265–279, 1998). By the same techniques we show that Ammann A2 tilings are not robust in the sense of Durand et al. (J Comput Syst Sci 78(3):731–764, 2012).

Appendix A

1.1 A.1 Proof of Lemma 4.1

• (a) Every small tile F from S produces the same tile $(F\cup \text {the brother of }F)$ in $\sigma ^{-1}S$ and $\sigma ^{-1}T$. Every large tile G from S produces the same tile $(F\cup \text { the sister of }G)$ in $\sigma ^{-1}S$ and $\sigma ^{-1}T$, if the sister of G is in S (and hence in T). Otherwise S (and hence T) contains a tile H that covers the cavity of G. This tile is different from the sister of G. Thus in this case the tile G produces itself in both $\sigma ^{-1}S$ and $\sigma ^{-1}T$.

• (b) By item (a) both tilings $\sigma ^{-1}S$ and $\sigma ^{-1}T$ are subsets of $\sigma ^{-1}(S\cup T)$. Hence $\sigma ^{-1}S\cup \sigma ^{-1}T\subset \sigma ^{-1}(S\cup T)$. This inclusion cannot be proper, as $\sigma ^{-1}S\cup \sigma ^{-1}T$ and $\sigma ^{-1}(S\cup T)$ tile the same set.

1.2 A.2 Proof of Proposition 3.1

(a) By induction: we will show that if a tiling T is A2 then so is $\sigma T$. Assume that T is an A2 d-tiling. Color in the tiling T all large and small tiles as shown in Fig. 4(b,c). Such coloring will be called canonical. Assume that in the canonical coloring of T the colors and orientations in all pairs of adjacent tiles match. We have to show that the same holds for $\sigma T$.

To verify this, cut all large d-tiles from T, as shown in Fig. 4(a). Then change the colored segments in the original canonical coloring using the following substitution:

$$\begin{aligned} \overrightarrow{6}&\rightarrow \overrightarrow{5}\\ \overrightarrow{5}&\rightarrow \overrightarrow{4}\\ \overrightarrow{4}&\rightarrow \overrightarrow{3}\\ \overrightarrow{3}&\rightarrow \overleftarrow{4}\overleftarrow{6}. \end{aligned}$$

The reverse arrows in $\overleftarrow{4}$ and $\overleftarrow{6}$ mean that we reverse orientation. After that color the cut by colors $\overrightarrow{4}$, $\overrightarrow{5}$, $\overrightarrow{6}$, as shown in Fig. 4(a).

As the transformation of colors of every segment does not depend on the tile it belongs to, it does not destroy the matching requirement. Therefore it remains to verify that after transformation we get the canonical coloring of $\sigma T$. This can be verified just by comparing Fig. 4(a), 4(b) and 4(c): the transformation of colors applied to Fig. 4(c) produces the coloring as in Fig. 4(b) and the transformation of colors applied to Fig. 4(b) produces the coloring as in Fig. 4(a).

(b) Let T be an A2 tiling of a convex set. We have to show that for every small tile $H\in T$ there is a large G located as shown in Fig. 4(a). Indeed, the cavity in H formed by arrows $\overrightarrow{5},\overrightarrow{6}$ is somehow filled by another tile G in T. Notice that only large tiles have a right angle with arrows $\overrightarrow{5},\overrightarrow{6}$ thus G is a large tile. There is only one such angle in every large tile and only one way to properly attach a large tile to a small tile to fill the gap, namely the way shown in Fig. 4(a).

(c) Let T be an A2 tiling of a convex set. We have to show that $\sigma ^{-1}T$ is again an A2 tiling. This is done in a way similar to that in the proof of item (a).

The tiling $\sigma ^{-1}T$ and its canonical coloring can be obtained from the canonical coloring of T in two steps: erase sides $\overleftarrow{4},\overleftarrow{5},\overleftarrow{6}$ shared by each pair (sister, brother), to get the tiling $\sigma ^{-1}T$ and then replace the colors using the map:

$$\begin{aligned} \overrightarrow{6}\, \overrightarrow{4}&\rightarrow \overleftarrow{3}\\ \overrightarrow{5}&\rightarrow \overrightarrow{6}\\ \overrightarrow{4}&\rightarrow \overrightarrow{5}\\ \overrightarrow{3}&\rightarrow \overrightarrow{4}. \end{aligned}$$

Note that, after erasing sides shared by each pair of small and large tile, every occurrence of arrow $\overrightarrow{6}$ in a tile from T is followed by an arrow $\overrightarrow{4}$ belonging to the same side of the same tile. This is easy to verify looking at Fig. 4(b,c) (the arrow $\overrightarrow{6}$ has two occurrences on sides of the large tile and two occurrences on sides of the small tile; all they are followed by $\overrightarrow{4}$ except for the side of a small tile that belongs to the cavity—but such sides have been erased). An arrow $\overrightarrow{4}$ is replaced using the first rule, if it follows $\overrightarrow{6}$, and using the third rule otherwise. In the obtained coloring the colors and orientations match, as the transformation of every arrow does not depend on the tile whose side it belongs to.

It remains to verify that the resulting coloring of $\sigma ^{-1}T$ is indeed canonical. This can be verified by comparing Fig. 4(a–c), as in the proof of item (a).

1.3 A.3 Proof of Proposition 3.2

Items (a) and (c) follow from the following

Claim

Assume that T is a finite d-supertile or an infinitely composable d-tiling. Assume that H is any of its tiles (small or large). Then there is a unique sequence $H_0\subset H_1\subset H_2\subset \cdots$ that starts with H and ends with [T], if T is a finite supertile, and is infinite, if T is infinitely composable, and such that $H_i$ is either the son, or the daughter of $H_{i+1}$ for all i, and $S_d(H_i)\subset T$ for all i.

Proof

For finite supertiles the statement can be proven by induction on the level of T. If T is a supertile of level $-1$ or 0, then the statement is obvious. Otherwise T is a disjoint union of supertiles $T'$ and $T''$ of smaller levels, which tile the son and the daughter of T, respectively. Since $T'$ and $T''$ are disjoint, we have either $H\in T'$, or $H\in T''$. In the first case the last but one tile in the sought sequence $H_0,H_1,H_2,\ldots$ must be $[T']$ and the statement for T follows from the induction hypothesis for $T'$. Similarly, in the second case the statement for T follows from the induction hypothesis for $T''$.

For infinitely composable tilings we are unable to use similar arguments, since the sequence $H_0, H_1, H_2,\ldots$ must be infinite. Let us first prove that such sequence exists. For every l consider the tiling $\sigma ^{-l}T$ obtained from T by l compositions. The tiling T is a disjoint union of supertiles $S_d(G)$ where $G\in \sigma ^{-l}T$. Let $H_l$ denote the (unique) tile from $\sigma ^{-l}T$ such that the supertile $S_d(G)$ contains H.

We claim that for all l the tile $H_l$ is either the son, or the daughter of $H_{l+1}$, or $H_l$ coincides with $H_{l+1}$. Indeed, the tiling $\sigma ^{-l}T$ is the decomposition of the tiling $\sigma ^{-l-1}T$. If $H_{l+1}$ is a small tile in the tiling $\sigma ^{-l-1}T$, then its decomposition coincides with it, thus $H_{l+1}$ is in $\sigma ^{-l}T$ and hence $H_l$ and $H_{l+1}$ coincide. Otherwise the tiling $\sigma ^{-l}T$ contains the result of decomposition of $H_{l+1}$, that is, the son F and the daughter G of $H_{l+1}$. Since

$$\begin{aligned} H\in S_d(H_{l+1})=S_d(F)\sqcup S_d(G), \end{aligned}$$

the tile H belongs either to $S_d(F)$, or to $S_d(G)$. In the first case $H_l$ must be equal to F and otherwise $H_l=G$.

Removing repetitions from the sequence $H_0, H_1,H_2, \ldots$ we obtain the sought sequence of tiles.

Let us prove now that that such chain is unique. Assume that there are two such chains $H_0\subset H_1\subset H_2\subset \cdots$ and $G_0\subset G_1\subset G_2\subset \cdots$ Let us show by induction on n that $H_n=G_n$. By assumption we have $H_0=G_0=H$.

Induction step: assume that $H_n=G_n$. By way of contradiction, assume that $H_{n+1}\ne G_{n+1}$. Then the tile $H_n=G_n$ is the son of $H_{n+1}$ and the daughter of $G_{n+1}$ (or the other way around, but the other case in entirely similar). Let l stand for the level of the supertile $S_{d}(H_n)=S_{d}(G_n)$. The levels of supertiles $S_{d}(H_{n+1})$ and $S_{d}(G_{n+1})$ are $l+1$ and $l+2$ respectively. By Lemma 4.1(a) both tilings $\sigma ^{-l}S_{d}(H_{n+1})$, $\sigma ^{-l}S_{d}(G_{n+1})$ are included into $\sigma ^{-l}T$. The first tiling consists of $H_n$ and its sister. The second one consists of $H_n$ and the son and the daughter of the brother of $H_n$. Hence the tiling $\sigma ^{-l}T$ contains the sister of $H_n$ and the son of the brother of $H_n$, which overlap (see Fig. 20).

Fig. 20

The sister S (marked grey) of the large tile $H_n$ overlaps with the son B of the brother of $H_n$

The obtained contradiction proves that $H_{n+1}=G_{n+1}$. $\square$

(a) Let T be an infinite supertile with representation $H_0\subset H_1\subset H_2\subset \cdots$ and H any of its tiles. Consider any n such that $H\in S_d(H_n)$. By the claim there is a chain of tiles $G_0\subset G_1\subset \cdots \subset G_l$ such that $G_0=H$, $G_l=H_n$ and $G_i$ is either the son, or the daughter of $G_{i+1}$ for all i. Then the sequence

$$\begin{aligned} G_0, G_1,\ldots , G_l, H_{n+1}, H_{n+2},\ldots \end{aligned}$$

is the sought representation of T. The uniqueness part of the claim implies that such representation is unique.

(b) Let F be any tile from T. Then there are k, l such that $F\in S_d(H_k)$ and $F\in S_d(G_l)$. Let $I_0,I_1,\ldots ,I_a$ denote the representation of $S_d(H_k)$ that starts with F and $J_0,J_1,\ldots ,J_b$ the representation of $S_d(G_l)$ that starts with F. Then both sequences of tiles

$$\begin{aligned} I_0,I_1,\ldots ,I_a,H_{k+1},H_{k+2},\ldots \qquad J_0,J_1,\ldots ,J_b,G_{l+1},G_{l+2},\ldots \end{aligned}$$

are representations of T that both start with F. By item (a) these sequences coincide. Note that $H_{k+i}$ is the $(a+i)$th term in the first representation and $G_{l+j}$ is the $(b+j)$th term in the second representation (for all i, j). Hence $H_{b+k+i}=G_{a+l+i}$ for all i.

(c) Consider the chain $H_0\subset H_1\subset H_2\subset \cdots$ existing by the claim. Then the union $\bigcup _{n=1}^{\infty }S_d(H_n)$ is the sought supertile. To prove uniqueness, notice that the representation of any supertile S satisfying the statement and starting with H must satisfy the claim.

1.4 A.4 Proof of Proposition 3.5

‘If’ part. Assume that $\alpha =u\gamma$ and $\beta =v\gamma$ where $w(u)=w(v)$. Then the supertiles with succinct representations (H, u) and (G, v) have the same level (equal to $w(u)=w(v)$) and hence are congruent. This implies that the infinite supertiles with succinct representations $(H,u\gamma )$ and $(G,v\gamma )$ are congruent as well.

‘Only if’ part. We are given congruent infinite supertiles T, S with succinct representations $(H,\alpha )$ and $(G,\beta )$, respectively. W.l.o.g. we may assume that the tilings T, S coincide (otherwise we apply to H the isometry f that maps T to S and obtain another succinct representation $(f(H),\alpha )$ of the supertile S).

Thus we are given two different succinct representations $(H,\alpha )$ and $(G,\beta )$ of the same infinite supertile S. By Proposition 3.2(b) the corresponding representations $(H=H_0), H_1,H_2,\ldots$ and $(G=G_0), G_1,G_2,\ldots$ have the same tail, that is, $H_{n+i}=G_{m+i}$ for some m, n and all $i\geqslant 0$. Thus $\alpha =u\gamma$ and $\beta =v\gamma$ where (H, u) and (G, v) are succinct representations of the d-supertile $S_d(H_n)=S_d(G_m)$, and $\gamma$ is the infinite s-l-sequence with

$$\begin{aligned} \gamma _i={\left\{ \begin{array}{ll} s,&{} \text {if }H_{n+i}=G_{m+i}\hbox { is the daughter of }H_{n+i+1}=G_{m+i+1},\\ l&{} \text {otherwise}. \end{array}\right. } \end{aligned}$$

We have $w(u)=w(v)$, since (H, u) and (G, v) are succinct representations of the same supertile.