Distortion Reversal in Aperiodic Tilings

DISTORTION REVERSAL IN APERIODIC TILINGS arXiv:2111.07523v1 [math.DS] 15 Nov 2021 LOUISA BARNSLEY, MICHAEL BARNSLEY, AND ANDREW VINCE Abstract. It is proved that homeomorphic images of certain two-dimensional aperiodic tilings, such as Ammann-A2 tilings, are recognizable, in both mathematical and practical senses. One implication of the results is that it is possible to search for distorted aperiodic structures in nature, where they may be hiding in plain sight. 1. Introduction Herein we prove that homeomorphic images of certain two-dimensional selfsimilar aperiodic tilings are recognizable, in both mathematical and practical senses. In the practical sense our results say it is possible to search for and recognize distorted aperiodic structures, for example in natural and physical settings. We recall that the mathematical discovery of aperiodic tilings by Penrose in 1972 and Ammann in 1975 preceded the discovery by Shechtman in 1982 of quasicrystals, and the subsequent examples of quasicrystals in meteorites in 2009 [13]. In this paper we focus on Ammann-A2 tilings [2], but the ideas are more generally applicable. Our results illustrate that aperiodic tilings can be retrieved from their images distorted by unknown homeomorphisms. For example, we show how distorted Ammann-A2 tilings, such as the ones in Figure 1, are recognizable both practically and mathematically. In Figure 2 an algorithm (without knowledge of the distorting homeomorphism) is applied to the distorted tiling to successively retrieve a patch of the original Ammann-A2 tiling. The situation is a bit strange: the unknown distortion may be such as to change the Hausdorff dimension of the boundaries of the tiles in a very rough way, or to transform an Ammann-A2 tiling to a tiling by triangles. Nevertheless, the key properties of the tiling may still be discernable. There is a substantial literature on the occurrence of two-dimensional tilings in physics [14]. Many of these are distortions of tilings by hexagons. Many naturally occurring tilings have an average of six faces per tile, and typical tiles are hexagons. Ammann-A2 tiles are hexagonal and comprise among the simplest of self-similar tilings. Despite this, they do not seem to show up in nature. Are they hiding in plain sight? We provide some tools which may be applied to this question. 2. Organization and Main Results Let P be a finite aperiodic prototile set. This means that there exist tilings of the plane by copies of the tiles in P , but every such tiling is non-periodic, i.e., has 2020 Mathematics Subject Classification. 52C20, 05B45. Key words and phrases. Ammann tiling, aperiodic tiling, distortion. This work was partially supported by a grant from the Simons Foundation (#322515 to Andrew Vince). 1 2 L. BARNSLEY, M. BARNSLEY, AND A. VINCE Figure 1. This illustrates a patch of an Ammann-A2 tiling, top left, and three different homeomorphisms of it. Properties of the Ammann-A2 tiling may be determined from a distorted image. no translational symmetry. Such tilings have been objects of fascination, both in research and recreational mathematics, since their introduction in the 1960’s. The most well-known such prototile set consists of the two Penrose tiles. To develop the ideas in this paper we concentrate, because of their simplicity, on tilings based on the set of two Ammann-A2 tiles. There are uncountably many corresponding Ammann-A2 tilings, obtained either by matching rules or by substitution rules. Basic properties of such Ammann-A2 tilings are provided in Section 3. Many non-periodic tilings have a hierarchical structure, meaning that if T is such a tiling, then there exist a sequence of tilings T = T 0 , T 1 , T 2 , . . . , such that, for all n ≥ 1, each tiling in T n is the non-overlapping union of tiles in T n−1 . For the Ammann-A2 tilings, the tiling T n , appropriately scaled, is again an Ammann-A2 tiling. Specifically, there is a constant s such that Tn := sn T n is an Ammann-A2 tiling for all n. There is a well-defined procedure, which we call amalgamation, taking Tn to Tn+1 , and denoted Tn+1 = a(Tn ). Amalgamation and hierarchy for the Ammann-A2 tilings is explored in Section 4. Here is the question that is the subject of this paper. Suppose that we are given a distorted version T ′ of an Ammann-A2 tiling T . Specifically, there is an unknown (not revealed to us) homeomorphism h of the plane such that T ′ = h(T ). Is it possible to retrieve the undistorted Ammann-A2 tiling T from its distorted image T ′ - without knowledge of h? As long as the distance between any point x in the plane and its image h(x) is bounded, we answer this question in the affirmative. DISTORTION REVERSAL IN APERIODIC TILINGS 3 Figure 2. Successive images, obtained by amalgamation, converge towards patches of an Ammann-A2 tiling. Suppose that amalgamation can be carried over to the distorted tiling T ′ . Then Section 5 contains the following somewhat surprising result: if Tn′ = an (T ′ ), n ≥ 0, are the tilings obtain by successive application of the amalgamation operator a applied to the distorted image T ′ , then the sequence {Tn } of tilings converges, in the sense of Theorem 1 in Section 5 and Remark 1, to the original tiling T . This is illustrated in Figure 2. The above distortion reversal result is predicated on being able to amalgamate tiles of the distorted tiling T ′ , a tiling in which metric and geometric properties of the original tiling are lost. Only the combinatorial properties of T are carried over to T ′ . In Section 6 we obtain, completely combinatorially, the needed amalgamation operator a applied to T ′ . This is done in Algorithm A in Section 6; see Theorem 2. Of course, the plane and tilings of it, are unbounded. We would like to perform the distortion reversal efficiently on a large patch of the tiling T ′ , say on a patch that contains the disk DR of radius R centered at the origin. Algorithm B, a variation of Algorithm A, accomplishes this. Theorem 3 in Section 6 states that each application of the combinatorial amalgamation applied to such a patch can be performed in running time O(R2 ). An alternative method for reversing the distortion in an Ammann-A2 tiling is provided in Section 7. It uses the combinatorial amalgamation operator applied to the distorted tiling T ′ to generate an infinite binary string c, called the code of the tiling. The code is the input for an iterated function system based method to retrieve the original tiling T . This method, encapsulated in Theorem 4 of Section 7, has the advantage that there is no restriction on the homeomorphism h. It has the disadvantage that there is no sequence of increasingly accurate images converging 4 L. BARNSLEY, M. BARNSLEY, AND A. VINCE to the original tiling T ; it simply produces T or, in finite time, an arbitrarily large patch of T . 3. Ammann-A2 Tilings In this paper a tile is defined as a set in the plane homeomorphic to a closed disk. A tiling of the plane is a set of non-overlapping tiles whose union is R2 . By non-overlapping is meant that the intersection of any two distinct tiles has empty interior. For all tilings in this paper, the intersection of any two distinct tiles is connected (possibly empty). A patch of a tiling is a finite number of tiles whose union is a topological disk. The classical Ammann-A2 hexagon G, sometimes referred as the golden bee, is depicted in Figure 3. It is the only polygon, other √ than any right triangle and any parallelogram with side lengths in the ratio 2 : 1, that is the non-overlapping union of two smaller similar copies of itself [12]. These two smaller copies are isometric to sG and s2 G, where s = 1/τ and τ is the square root of the golden ratio, i.e., τ is the positive real root of x4 − x2 − 1 = 0. Figure 3. Golden bee G An Ammann-A2 tiling is a tiling of the plane by non-overlapping isometric copies of sG and s2 G, which we will refer to as the large and small tiles, respectively. The tiling must obey matching rules dictated by the elliptical markings on the upper tiles of the left panel in Figure 4. Other alternative but equivalent sets of markings are also possible, for example the markings shown in the right panel of Figure 4. Although the decorations on small and large tiles at the top of the left panel and the decorations on the small and large tiles of the right panel differ, matching according to either set of decorations define the Ammann-A2 tilings, as observed in [7, p. 551]; see also [1, 2, 4, 10]. Hereafter we will call an Ammann-A2 tiling simply an Ammann tiling. A portion of an Amman tiling is shown in Figure 5. Two tilings T and T ′ are said to be congruent if there is an isometry U such that T ′ = U (T ). Much has been written on Ammann tilings both in mathematics journals, for example [4, 7] and in recreational sources, notably [8, 11]. There are uncountable many Ammann tilings, each being non-periodic. Every Ammann tiling is repetitive and every pair of Ammann tilings are locally isomorphic. A tiling T is repetitive, also called quasiperiodic, if, for every finite patch P of T , there is a real number R such that every ball of radius R contains a patch congruent to P . Two tilings are locally isomorphic if any patch in either tiling also appears in the other tiling. DISTORTION REVERSAL IN APERIODIC TILINGS 5 Figure 4. Two equivalent matching rules for the Amman tilings. As explained in the paragraph above, the matching rules dictated by the decorated pair of small and large tiles at the upper left and the matching rules dictated by the decorated pair of small and large tiles at the right both define the Ammann tilings. As explained in Section 4, the figure at the lower left shows how a small Ammann tile meets its large partner. The Ammann tilings may be defined, as an alternative to matching rules, by substitution rules, see [4, 6]. Another method of construction appears in [3] and is used in Section 7 of this paper. Figure 5. Left: a circular patch of an Ammann-A2 tiling, with markings as in Figure 4. Right: the same patch, with tiles in various colors, and part of a possible continuation using faded tiles. 4. Amalgamation and the Hierarchy Let T be an Ammann tiling. For each small tile s in T , there is a partner, defined as the unique large tile b in T such that the union of s and b, called the amalgamation of s and b, is a hexagon congruent to G. Any two distinct small tiles 6 L. BARNSLEY, M. BARNSLEY, AND A. VINCE have distinct partners. The tiling obtained from T by amalgamating each small tile with its partner is denoted A(T ). The scaled tiling a(T ) := sA(T ) is an Ammann tiling called the amalgamation of T . Proposition 4.1. If T is an Ammann tiling, then so is a(T ). Proof. The upper left and right panels of Figure 4 show two different decorations on the small and large tiles, call them the D1 and the D2 decorations, respectively. Each small tile in a(T ) (left tile in the right panel of Figure 4) is a (scaled by s) large tile in T (right tile in the left panel); and each large tile in a(T ) (right tile in the right panel) is the scaled amalgamation of a small and large tile in T (bottom of the left panel). As previously noted, the matching rules dictated by the D1 and D2 decorations are equivalent in that they both define the Ammann tilings. Therefore if T is an Ammann tiling, then so is a(T ).  Related comments on matching rules for the Ammann tilings are given in [1, 2, 4]. In particular see [1] for a discussion of replacement of markings and the notion of ghost markings. Equivalence of different sets of markings is discussed in [10]. The sequence of tilings H(T ) := {T, A(T ), A2 (T ), . . . }, consisting of Ammann tilings at larger and larger scales, will be referred to as the hierarchy of T . In general, if t is any tile at any level of the hierarchy, then t is the non-overlapping union of tiles in T . Denote this tiling of t by T (t). The tiling T (t) is sometimes referred to as a supertile in the tilings literature; see for example [5, 7, 9]. Any large tile t in An (T ) is congruent to τ n−1 (G). For such a large tile T in An (T ), denote T (t) by Tn which, by Proposition 4.2 below, is well-defined, independent of t. The tiling T6 appear in Figure 6. Figure 6. The supertile T6 . Proposition 4.2. If p and q are congruent tiles at any level of the hierarchy, then T (p) and T (q) are congruent. DISTORTION REVERSAL IN APERIODIC TILINGS 7 Proof. It is true at level 1; A(T ) is the non-overlapping union of a small and a large Ammann tile in a unique way. Proceeding by induction on the level, assume that the statement of the proposition is true at level n − 1, and consider congruent tiles p and q of An (T ). Then either (1) p and q are small tiles of An (T ), in which case they are large tiles of An−1 (T ), or (2) p and q are large tiles of An (T ), in which case they are the amalgamation of a large and a small tile of An−1 (T ). In case (1) we have that T (p) and T (q) are congruent by the induction hypothesis. In case (2) the decomposition of p into a non-overlapping union of a large and small tile of An−1 (T ) is the same as for q. Moreover, by the induction hypothesis, the decompositions of large and small tiles of An−1 (T ) is unique. Therefore we again have that T (p) and T (q) are congruent.  Lemma 1 below is needed to prove a main result in Section 6. We first introduce a useful combinatorial notion. Define a planar map M to be a 2-cell embedding of a locally finite simple graph Γ in the plane. By locally finite is meant that the degree of each vertex is finite, and by simple is meant no loops or multiple edges. The faces of M are the closures of the connected components of R2 \ Γ. By 2-cell embedding is meant that each face of M is homeomorphic to a closed disk. Two planar maps M and M ′ with underlying graphs Γ and Γ′ , respectively, are isomorphic if there is a graph isomorphism taking Γ to Γ′ that preserves faces. A tiling T can be considered a planar map. A vertex of T is the intersection point of three or more distinct tiles of T , assuming that the intersection is not empty, and an edge of T is the intersection, if not empty or a single point, of two distinct tiles of T . The faces of T are the tiles of T . From an Ammann tiling T , a tiling Tb is constructed as follows. Note that the degree (number of incident edges) of any vertex of an Ammann tiling T is either 3 or 4. Color red each edge of T (and its two incident vertices) that joins two vertices of degree 4. For any red vertex lying on only one red edge, remove the color red from that vertex and remove the incident edge. Let Tb be the tiling induced by the red edges and vertices, i.e., by removing all edges and vertices not colored red. For any face f of Tb, it’s red boundary ∂f together with all enclosed tiles of T is a finite tiling which is denoted T (f ). Lemma 1. With notation as above, if T is an Ammann tiling, then Tb has the following properties. (1) If f is a tile of Tb, then T (f ) = T4 or T (f ) = T5 . (2) Tb = A5 (T ). Proof. In Tn , any edge adjacent to exactly one tile is called a boundary edge. A boundary vertex is a vertex that lies on a boundary edge. In Tn , color edges and vertices red by the same rule as used to color T , but, in addition, color red any edge (and incident vertex) joining a vertex of degree 4 to a boundary vertex and color all boundary edges (and incident vertices) red. As previously, for any red vertex lying on only one red edge, remove the color red from that vertex and remove the incident edge. Denote by Tcn the tiling induced by only the red edges and vertices. Figure 6 shows the tiling T6 with red edges designated and the corresponding tiling Tb6 . Note that, for any red edge e in T6 not on the boundary, one incident tile is the reflection of the other incident tile in the line that extends e. Also note that Tb6 has two faces, say f and f ′ such that T (f ) = T4 and T (f ′ ) = T5 . We claim that the 8 L. BARNSLEY, M. BARNSLEY, AND A. VINCE same is true for all Tn , n ≥ 6, i.e., (1) for any red edge e of Tn not on the boundary, one incident tile is the reflection of the other in the line that extends e (and this implies that any red vertex of Tn , not on the boundary, has degree 4), and (2) for any tile f of Tcn it is the case that T (f ) = T4 or T (f ) = T5 . The claim will be proved by induction on n. Assume that assertions (1) and (2) above are true for Tn and Tcn . Obtain Tn+1 by first subdividing each large tile in Tn into one large and one small tile, then enlarging the resulting tiling by a factor of τ . Now color the edges of Tn+1 as previously prescribed. If e is a red edge in Tn and f and f ′ are the adjacent tiles, then either they are both small tiles in Tn , and hence not subdivided, or they are both large tiles in Tn , and hence subdivided in exactly the same way, so that the reflection property remains true in Tn+1 . Therefore, each red edge in Tn induces either one or two red edges in Tn+1 . Let E be this set of ′ induced red edges and let Tn+1 be Tn+1 , but with only E and the boundary edges (and incident vertices) colored red. Since each tile of Tcn is either T4 or T5 , each ′ ′ tile in Tn+1 is either T5 or T6 . But when additional red faces are added to Tn+1 to obtain Tn+1 , no additional red edges are added to each T5 because T5 has no degree 4 internal (not on the boundary) vertices, and the additional red edges added to each T6 , as shown in Figure 6, divides each T6 into one T4 and one T5 . Our claim has now been proved. To extend the result from Tcn to Tb, let f be any tile of Tb. If f lies in the interior (no edge of f on the boundary) of a tile in An (T ) for some n, then by the paragraphs above, T (f ) is either T4 or T5 . So assume that there is an edge e of f that lies the on the boundary of some tile tn of An (T ) for all n sufficiently large, say n ≥ n0 . Denote by Tn the subset ofSthe tiling T that lies in tn . Then Tn0 ⊂ Tn0 +1 ⊂ Tn0 +2 ⊂ · · · . The nested union n≥n0 Tn is an Ammann tiling of a proper subset of the plane. But it is known [4] that this occurs only if it is a tiling of a half-space bounded by the line L that extends e or a quadrant of the plane bounded by two perpendicular rays L1 and L2 . Moreover, if T is an Ammann tiling of the half plane, then the only extension to an Ammann tiling of the entire plane is obtained by reflecting T in the line L of T . Similarly for a tiling obtained from a tiling T of a quadrant by reflecting in the two perpendicular border lines L1 and L2 . In the half plane case, the edges and vertices of T on L will be colored red, and in the quadrant case, the edges and vertices of T on L1 and L2 will be colored red. So again, T (f ) is either T4 or T5 , and statement (1) of Lemma 1 is proved. Concerning statement (2), the tiles in both A5 (T ) and Tb are congruent copies of 5 τ (G) and τ 4 (G). Let t ∈ Tb. If s is an small tile in Aj (T ), j < 5, that is contained in t, and b is its partner, then the shapes dictate that b is also contained in t. Therefore Tb = A5 (T ).  5. Bounded Distortion Homeomorphism Definition 1. A homeomorphism h of the plane has bounded distortion if there is a constant C such that |x − h(x)| ≤ C for all x ∈ R2 . In this case, h is said to have bounded distortion C. DISTORTION REVERSAL IN APERIODIC TILINGS 9 Lemma 2. Let h be a homeomorphism of the plane with bounded distortion C. If φr is a similarity transformation with scaling ratio r, then 1 |x − φ−1 ◦ h ◦ φ(x)| ≤ C r for all x ∈ R2 . Proof. Let x ∈ R2 , y = φr (x), z = h(y), w = φ−1 r (z). Then −1 −1 |x − φ−1 r ◦ h ◦ φr (x)| = |x − w| = |φr (y) − φr (z)| = = 1 |y − z| r 1 1 |y − h(y)| ≤ C r r for all x ∈ R2 .  Let X denote the set of tilings as defined in Section 3. For a tiling T ∈ X, let ∂T denote the union of the boundaries of the tiles of T . Define a distance function on X as follows. d(T, T ′ ) = dH (∂T, ∂T ′ ). where dH is the Hausdorff distance. Intuitively, the d-distance between tilings is small if the tilings are almost the same over the entire plane. The next corollary follows immediately from Lemma 2. Corollary 1. If h a homeomorphism of bounded distortion C and φr is a similarity transformation with scaling ratio r, then
1 C d φ−1 r ◦ h ◦ φr )(T ), T ≤ r for all tilings T ∈ X. If T is an Ammann tiling and h is a homeomorphism of the plane, then h acts on T . Let T ′ = h(T ) be the image of an Ammann tiling T under a homeomorphism h. The small and large tiles in T ′ are defined to be the images of the small and large tiles in T . If s′ is a small tile in T ′ corresponding to a small tile s in T , then the partner of s′ is the image under h of the partner of s. Therefore the amalgamation a(T ′ ) and the hierarchy H(T ′ ) can be defined for T ′ exactly as they are for T . Theorem 1. Let T be an Ammann tiling, h a bounded distortion homeomorphism of the plane, and T ′ = h(T ). Then
lim d ak (T ′ ), ak (T ) = 0. k→∞ Proof. Let H(T ) = {T = T0 , T1 , T2 , . . . } be the hierarchy of T and H(T ′ ) = {T ′ = T0′ , T1′ , T2′ , . . . } the hierarchy of T ′ . Then ak (T ) = sk Tk ak (T ′ ) = sk Tk′ = sk h(Tk ) = sk h(s−k ak (T )) = (sk ◦ h ◦ s−k ) ak (T )
= (τ −k ◦ h ◦ τ k ) ak (T ) where we recall that τ = s−1 . By Corollary 1 we have

C d ak (T ′ ), ak (T ) = d (τ −k ◦ h ◦ τ k ) ak (T )), ak (T ) ≤ k . τ
10 L. BARNSLEY, M. BARNSLEY, AND A. VINCE From this it follows that
lim d ak (T ′ ), ak (T ) = 0. k→∞  Remark 1. Theorem 1 does not state that ak (T ′ ) becomes arbitrary close to T as k → ∞. It does imply, however, that an arbitrarily large finite patch of the tiling ak (T ′ ) becomes arbitrarily close to a patch Pk of T as k → ∞. This follows from Proposition 4.1 and the fact that any two Ammann tilings are locally isomorphic. 6. Combinatorial Amalgamation Let T be the set of all images of Ammann tilings of the plane under homeomorphisms of the plane. Essential to Theorem 1 is being able to apply the amalgamation operator a : T → T. Let T be an Ammann tiling and T ′ = h(T ), where h is a homeomorphism of the plane. In this section it is proved that a(T ′ ) can be determined combinatorially, without knowledge of the homeomorphism h. Theorem 2. Let T be an Ammann tiling, h a homeomorphism of the plane and T ′ = h(T ). Then Algorithm A below computes a(T ′ ), without knowledge of h. The proof of Theorem 2 appears after a few simple lemmas. With terminology from Section 4, the following lemma is clear. Lemma 3. If T is a tiling of the plane and h is a homeomorphism of the plane, then as planar maps T and h(T ) are isomorphic. Two planar maps M1 (left) and M2 (right) are shown in Figure 7 (ignoring the dots and arrows). Recalling the definition of Tn from Section 4, the following two lemmas are apparent by inspection. Lemma 4. The two planar maps M1 and M2 in Figure 7 are isomorphic to T4 and T5 , respectively, considered as planar maps. Lemma 5. Each of the two planar maps in Figure 7, ignoring the arrows, has trivial automorphism group. Figure 7. Rule for locating the partner of a small tile. DISTORTION REVERSAL IN APERIODIC TILINGS 11 Each arrow in Figure 7 points from a face bounded by a 4-cycle to an adjacent face. A consequence of Lemma 5 is that, if M is any planar map isomorphic to M1 or M2 in Figure 7, then arrows can be uniquely assigned to M such that there is an arrow from face f1 to f2 in M if and only if there is an arrow from φ(f1 ) to φ(f2 ) in M1 or M2 , where φ is the isomorphism. Algorithm A Input: A tiling T ′ that is the image of an Amman tiling, distorted by an unknown homeomorphism. Output: The amalgamated tiling a(T ′ ). Perform the following steps to obtain the tiling a(T ′ ) from T . (1) Color red each edge (and its two incident vertices) of T ′ that joins two vertices of degree 4. But, for any red vertex lying on only one red edge, remove the color red from that vertex and remove the incident edge. c′ be the tiling induced by just the red edges and vertices of T ′ , i.e., (2) Let T by removing all edges and vertices not colored red. c′ , let T ′ (f ) be the set of tiles of T ′ contained in t. (3) For every tile f ∈ T Viewed as a planar map, each T ′ (f ) is isomorphic to one of the planar maps in Figure 7. (This is shown to be the case in the proof below of c′ do: Theorem 2.) For every f ∈ T • For every tile s of T ′ (f ) that is bounded by a 4-cycle, do: – Locate the partner b of s using the relevant arrow in Figure 7. – Replace the two tiles s and b in T ′ by their union to form a single tile in A(T ′ ), and hence in a(T ′ ) after scaling by s = 1/τ . Proof of Theorem 2. The notation in this proof follows the notation in Section 4 just prior to Lemma 1. Let s′ be any small tile in T ′ . It must be shown that Algorithm A pairs s′ with its large partner tile in T ′ . The tile s is contained in a unique tile f ∈ A5 (T ) = Tb, the last equality by statement (2) of Lemma 1. Therefore s′ := h(s) is contained in f ′ := h(f ) ∈ A5 (T ′ ). Since f ∈ Tb, we know that f ′ ∈ h(Tb). By Lemma 3, T and h(T ), as well as Tb and h(Tb), are isomorphic as planar maps. Therefore, by statement (1) of Lemma 1, if T (f ′ ) denotes the set of all tiles in T ′ contained in f ′ , then T (f ′ ) is isomorphic to T4 or T5 . By Lemma 4, we know that T (f ′ ) is isomorphic to one of the planar maps in Figure 7. Note that the boundary of any face in T ′ is a 4-cycle if the corresponding tile is small and a 5 or 6-cycle if the corresponding tile is large. Therefore, by Lemma 5 and the remarks following it, the partner of s′ ∈ f ′ is uniquely determined by the arrows in Figure 7. This completes the proof of Theorem 2.  Algorithm A acts on the tiling T ′ of the entire plane and hence does not run in finite time. From a practical point of view, it may be asked whether it is possible to efficiently compute the amalgamation of the distorted tiling T ′ on a finite subset of the plane, for instance on a patch containing the disk DR of radius R centered at the origin. We tweak Algorithm A to obtain a combinatorial Algorithm B that 12 L. BARNSLEY, M. BARNSLEY, AND A. VINCE applies to a finite patch of a tiling. Algorithm B computes a(T ′ ) on DR and runs in time quadratic in R. This is the content of Theorem 3 below. Figure 8 illustrates Algorithm A above and Algorithm B below, showing the red edges of a distorted Ammann tiling before and after the red edges that do not belong to cycles are removed. The notation T ′ |R and a(T ′ |R) in Theorem 3 are defined as follows. If T ′ is a tiling, T ′ |R the set of tiles of T ′ with non-empty intersection with DR . If T ′ is an Ammann tiling or a homeomorphic image of an Ammann tiling, the amalgamation a(T ′ |R) is obtained by replacing every small tile t in T |R by the union of t and its partner - even if the partner of t has empty intersection with DR . Lemma 6. Let R be a positive real number. If T is an Ammann tiling and h is a homeomorphism of bounded distortion C, then (1) h(A5 (T ))|R ⊂ DR+2C+τ 6 , (2) the number √ of tiles2 of h(T ) contained in the disk DR is less than 5)(R + C) . − τ π( 15 4 Proof. Concerning statement (1), if x is any point in a tile t′ ∈ h(A5 (T ))|R, and y is any point in t′ ∩ DR , then by the triangle inequality and the fact that h is of bounded distortion C, we have |x| ≤ |y| + |x − y| ≤ R + diam(t′ ) = R + diam(h(t)) ≤ R + diam(t) + 2C ≤ R + diam(τ 5 G) + 2C = R + τ 6 + 2C, where diam denotes the diameter of the set. Concening statement (2), because h is of bounded distortion C, we have DR ⊆ h(DR+C ). Each small tile in an Ammann tiling is congruent to s2 G. Therefore √ 15 2 there are less than area(DR+C )/area(s G) = τ π( 4 − 5)(R + C)2 Ammann tiles in DR+C .  Algorithm B Input: A tiling T ′ that is the image of an Amman tiling, distorted by an unknown homeomorphism of bounded distortion C, and a positive real number R. Output: A subset T ′′ of the amalgamated tiling a(T ′ ) that contains a(T ′ |R). b = R + 2C + τ 6 (1) Set R b Perform the following steps to obtain the tiling T ′′ from the tiling T ′ |R: b red according to the following rules: (2) Color some edges and vertices of T ′ |R b that joins • Color red each edge (and its two incident vertices) of T ′ |R two vertices of degree 4. • Remove the color red from any red vertex or edge not lying on a red b cycle in T ′ |R. c′ be the tiling induced by the red edges and vertices of T ′ |R. b (3) Let T DISTORTION REVERSAL IN APERIODIC TILINGS 13 Figure 8. Left: a distorted Amman tiling with edges that connect vertices of degree 4 marked in red. Right: the same tiling after removal of red edges that do not lie on red cycles. c′ , let T ′ (t) be the set of tiles of T ′ |R b contained in t. (4) For every tile t ∈ T ′ Viewed as planar map, each T (t) is isomorphic to one of the planar maps c′ do: in Figure 7. For every t ∈ T • For every tile s of T ′ (t) that is bounded by a 4-cycle, do: – Locate the partner b of s using the the relevant arrow in Figure 7. b by their union to form a – Replace the two tiles s and b in T ′ |R ′′ single tile in T . Theorem 3. Let T be an Ammann tiling, h a homeomorphism of the plane of bounded distortion, and T ′ = h(T ). Then Algorithm B above computes a(T ′ |R) in time quadratic in R, without knowledge of h. Proof. In addition to what was shown in the proof of Theorem 2, it must be verified c′ constructed in Algorithm B is contained in D b , which follows that the finite tiling T R from statement (2) of Lemma 1 and startement (1) of Lemma 6. 14 L. BARNSLEY, M. BARNSLEY, AND A. VINCE Concerning the running time of Algorithm B, it is clear that the number of steps b By statement (2) of is O(n), where n is the sum of the number of tiles in T ′ |R. ′ b Lemma 6, the number of tiles in T |R is less than 15 √ 15 √ b + C)2 = τ π( − 5)(R + 3C + τ 6 )2 = O(R2 ).  τ π( − 5)(R 4 4 7. Direct Retrieval of an Ammann tiling from a Distorted Image Let T be an Ammann tiling and T ′ = h(T ), where h is a homeomorphism of the plane. In this section we extract from T ′ , without using h, an infinite binary string θ = θ1 θ2 θ3 · · · . A tiling T (θ) will be combinatorially constructed from θ such that T = T (θ). Definition 2. Let S be the set of all pairs (T ′ , t′ ), where T ′ = h(T ) and t′ = h(t) for some Ammann tiling T , large tile t ∈ T , and homeomorphism h of the plane. Let {1, 2}∞ = {c1 c2 c3 · · · : ci ∈ {1, 2} for all i}, i.e., the set of infinite binary stings. Also let {1, 2}∗ denote the set of finite binary strings. To define a map g : S → S and a function c : S → {1, 2}∞ , called the code of (T ′ , t′ ), let (T ′ , t′ ) ∈ T. Define (Te, e t) = g(T ′ , t′ ) as follows. There are two cases. Case 1. If t′ has a partner in T ′ , then let Te = A(T ′ ), the amalgamation of T ′ , and let e t be union of t′ and its partner, a large tile in Te. Define g(T ′ , t′ ) = (Te, e t) and e c(T ′ , t′ ) = 1. Case 2. If t′ has no partner in T ′ , then let Te = A2 (T ′ ), the second amalgamation of T ′ . Then t′ is a small tile in A(T ′ ) that has a partner. In Te let e t be the union ′ ′ ′ ′ e e e of t and its partner in A(T ). Let g(T , t ) := (T , t). Note that t is a large tile in Te. Define e c(T ′ , t′ ) = 2. Define a sequence of tilings T0′ = T ′ , T1′ , T2′ , . . . and tiles t′0 = t′ , t′1 , t′2 , . . . , recur′ sively by (Tn+1 , t′n+1 ) = g(Tn′ , t′n ). Then define the code of (T ′ , t′ ) to be the binary string c(T ′ , t′ ) = e c(T0′ , t′0 ) e c(T1′ , t′1 ) e c(T3′ , t′3 ) · · · , and note that, as done in Section 6, the amalgamation used to determine the code c(T ′ , t′ ) of T ′ can be done combinatorially without knowing the homeomorphism. Theorem 4 below states that c(T ′ , t′ ) completely determines the original Ammann tiling T , and can be used to generate T . An intuitive way to see this is to start with a patch P0 consisting of a single large Ammann tile t. Let c(T, t) = c1 c2 · · · . If c1 = 1, embed P0 in a patch P1 as shown at the start of the bottom row in Figure 9; if c1 = 2, embed P0 in a patch P1 as at the start of the top two rows in Figure 9. Continue in this way to get a nested sequence of patches as in Figure 9. The nested union is T . Remark 2. It is well known that there are a couple of special cases where the nested union ∪{Pn ; n ≥ 0} is a tiling that fills a half plane or a quadrant of the plane. Therefore, in this section such tilings will also be classified as Ammann tilings. If T is an Ammann tiling of the half plane, then a tiling of the entire plane satisfying the matching rules can be obtained by reflecting T in the border line of T . These tilings will not be classified as Amman tilings in this section. Similarly for a tiling obtained from a tiling T of a quadrant by reflecting in the two border rays. DISTORTION REVERSAL IN APERIODIC TILINGS 15 A more rigorous formulation of the above method of retrieving the original tiling from the code uses the following combinatorial construction. Recall that s = 1/τ , where τ is the square root of the golden ratio. The fact that the golden bee G is the non-overlapping union of two small similar copies of itself can be stated more precisely as follows. G = f1 (G) ∪ f2 (G), where f1 (G) and f2 (G) are non-overlapping and (7.1) f1        x 0 −s x s = + , y s 0 y 0 f2    2 x s = y 0 0 −s2     x 0 + . y 1 Figure 9. The starts of three possible sequences of embeddings, from left to right, as described in the text. The initial large tile, the patch P0 , is shown shaded and has a fixed location in the plane. For σ = σ1 σ2 . . . σk ∈ {1, 2}∗ let e(σ) := k X σi i=1 and let σ∗ := σk denote the last term in the string σ. We introduce the notation fσ := fσ1 ◦ fσ2 ◦ · · · ◦ fσk f−σ := fσ−1 ◦ fσ−1 ◦ · · · ◦ fσ−1 . 1 2 k 16 L. BARNSLEY, M. BARNSLEY, AND A. VINCE and θ|k := θ1 θ2 . . . θk ∞ for θ = θ1 θ2 , · · · ∈ {1, 2} . By convention, θ|0 is the empty string ∅; e(∅) = 0, and f∅ is the identity. Let W (θ, k) := {σ ∈ {1, 2}∗ : e(σ∗ ) > e(σ) − e(θ|k) ≥ 0}. Intuitively, the finite strings σ ∈ W (θ, k) are those whose sum differs from the sum of the elements of θ|k by zero or one. This forces each tile t(θ, k, σ) in Definition 3 below to be congruent to either a large or small Ammann tile. Definition 3. For each θ ∈ {1, 2}∞ , a tiling T (θ) is constructed in three steps. (1) A single tile. For each integer k ≥ 1 and each σ ∈ [N ]∗ , construct a single tile t(θ, k, σ) that is similar to G: t(θ, k, σ) := (f−(θ | k) ◦ fσ )(sG). (2) A patch of tiles. Form a patch T (θ, k) of tiles given by: T (θ, k) := {t(θ, k, σ) : σ ∈ W (θ, k)} . (3) A tiling. The tiling T (θ), depending only on θ ∈ {1, 2}∞ , is the union of the patches T (θ, k), which is known [3] to be a nested union: [ T (θ) := T (θ, k). k≥1 ∞ The set {1, 2} is called the parameter set. For each parameter θ, the tiling T (θ) is called the θ-tiling. Modulo Remark 2, a θ-tiling is an Ammann tiling. Theorem 4. Let T ′ = h(T ) be the image of Ammann tiling T under a homeomorphism h of the plane. Let t′ be any large tile in T ′ , and let θ := c(T ′ , t′ ) be the code of (T ′ , t′ ). Then T = T (θ), independent of which large tile t′ ∈ T ′ is chosen. Proof. Let t = h−1 (t′ ) ∈ T , and note that c(T ′ , t′ ) = c(T, t). Let θ := c(T, t). It must be shown that T (θ) = T for any large tile t ∈ T . Since an Ammann tiling is determined by its code, it suffices to show that, for any large tile t ∈ T , there is a large tile p ∈ T (θ) such that that c(T (θ), p) = θ. A tile t(θ, k, σ) ∈ T (θ) is a large tile congruent to sG in T (θ) if e(σ)−e(θ | K) = 0 and a small tile congruent to s2 G if e(σ) − e(θ | K) = 1. From the definitions, this implies that t(θ, k, σ) ∈ T (θ) is a small tile if and only if σ∗ = 2, i.e., σ = σ1 σ2 · · · σj−1 2, and the tile t(θ, k, ω), where ω = σ1 σ2 · · · σj−1 1, is the large partner of t(θ, k, σ). Let p = t(θ, k, σ), where σ = σ1 σ2 · · · σj , be an arbitrary tile in T (θ). Then winding through Definition 3, the code of c(T (θ), p) is given by c(T (θ), p) = σj σj−1 · · · σ1 θk+1 θk+2 · · · . Consider two cases. If the original large tile t ∈ T has a partner, then θ1 = 1, i.e., c(T, t) = 1θ2 θ3 · · · . In T (θ), let p = t(θ, 1, 1). Then the code of c(T (θ), p) is given by c(T (θ), p) = 1θ2 θ3 · · · . If the original large tile t ∈ T has no partner, then θ1 = 2, i.e., c(T, t) = 2θ2 θ3 · · · . In T (θ), let p = t(θ, 1, 2). Then the code of c(T (θ), p) is given by c(T (θ), p) = 2θ2 θ3 · · · . In either case c(T (θ), p) = θ.  Remark 3. An arbitrarily large patch of the original tiling T can be obtained in finite time and efficiently from the distorted tiling T ′ by first using the combinatorial DISTORTION REVERSAL IN APERIODIC TILINGS 17 amalgamation Algorithm B to obtain a finite concatenation θ|k of a code θ for T ′ and then by using the procedure in Definition 3 to construct the patch T (θ, k). aknowlegements We thank Vanessa Robbins for interesting conversations in relation to this work. References [1] S. Akiyama, A note on aperiodic Ammann tiles, Discrete and Computational Geometry,48 (2012), 702-710. [2] R. Ammann, B. Grunbaum, G. C. Shephard, Aperiodic tiles, Discrete and Computational Geometry 8(1992), 1-25. [3] M. F. Barnsley, A. Vince, Self-similar polygonal tilings, Amer. Math. Monthly 124 (2017), 905-921. [4] B. Durand, A. Shen, N. Vereshchagin, On the structure of Ammann-A2 tilings, Discrete and Computational Geometry 63 (2020), 577-606. [5] N. P. Frank, L. Sadun, Fusion tilings with infinite local complexity, Topology Proceedings 43 (2012), [6] D. Frettlöh, E. Harriss, F. Gähler: Tilings encyclopedia, https://tilings.math.uni-bielefeld.de/ [7] B. Grünbaum and G. S. Shephard, Tilings and Patterns, W. H. Freeman and Company, New York, 1987. [8] M. Gardner, Extraordinary nonperiodic tilings that enriches the theory of tiles, Scientific American 236 (1977), 110-121. [9] C. Goodman-Strauss, Matching rules and substitution tilings, Annals of Mathematics 147 (1998), 181-223. [10] A. Korotin, Local rules for self-similar tilings of the plane by polygons, Honours thesis (in Russian), 2016. [11] M. Senechal, The mysterious Mr. Ammann, Math. Intelligencer 26 (2004), 10-21. [12] J. H. Schmerl, Dividing a polygon into two similar polygons, Discrete Math., 311 (2011), 220-231. [13] P. J. Steinhardt, The Second Kind of Impossible: The Extraordinary Quest for a New Form of Matter, Simon & Shuster, 2018. [14] D. Weaire, N. River, Soap, cells and statistics - random patterns in two dimensions, Contemporary Physics, 50 (2009), 199-239. Mathematical Sciences Institute, Australian National University, Canberra, ACT, Australia Email address: [email protected] Mathematical Sciences Institute, Australian National University, Canberra, ACT, Australia Email address: [email protected] Department of Mathematics, University of Florida, Gainesville, FL, USA Email address: [email protected]