1 Introduction
One of the most well-known aperiodic tilings was discovered by Penrose. In its original version, four shapes derived from the regular pentagon can be used to tile the plane, and none of the allowed tilings are periodic [Reference Penrose48]. Penrose tilings were soon given an equivalent description in terms of multigrids or cut and project schemes [Reference Govert de Bruijn12]; see also [Reference Grünbaum and Shephard24, §10] and [Reference Baake and Grimm5, §6.2]. The aperiodic structure of Penrose tilings is explained by the properties of a specific irrational number: the positive root
$\varphi $
of the polynomial
$x^2-x-1$
, also known as the golden ratio or golden mean. For example, in the kite-and-dart version of the Penrose tilings, the ratio of kites to darts is equal to the golden ratio [Reference Penrose49].
Recently, the discovery of an aperiodic monotile [Reference Smith, Myers, Kaplan and Goodman-Strauss58] attracted a lot of attention [Reference Socolar60, Reference Baake, Gähler and Sadun6, Reference Akiyama and Araki2]. Smith and coauthors presented a single shape, a 13-edge polygon called the hat, whose isometric copies tile the plane but never periodically. Again, the golden ratio appears in tilings by the hat. In a tiling by isometric copies of the hat, both the hat and its mirror image appear (up to orientation preserving isometries – that is, translations and rotations). The frequency of the hat and its mirror image in a tiling are not equal. The ratio of the most frequent orientation of the hat to the least frequent one is equal to the fourth power of the golden ratio.Footnote 1 Two months later, the same authors discovered another aperiodic tile called Spectre, which does not need its mirror image to tile the plane [Reference Smith, Myers, Kaplan and Goodman-Strauss59]. Tilings by the Spectre are not all combinatorially equivalent to tilings by the hat: some are periodic (if the reflected tile is allowed). But every tiling by the hat tile is combinatorially equivalent to some Spectre tiling.
Other examples of aperiodic tilings are related to the golden mean, including Ammann A2 L-shaped tiles [Reference Ammann, Grünbaum and Shephard4] (also studied in [Reference Akiyama1, Reference Durand, Shen and Vereshchagin15]); see Figure 1. The golden mean also appears in the description of tilings generated by the Jeandel–Rao aperiodic set of 11 Wang tiles [Reference Jeandel and Rao26]: the frequency of the tiles [Reference Labbé34], the inflation factor of its self-similarity [Reference Labbé33, Reference Labbé35], and the slopes of its nonexpansive directions [Reference Labbé, Mann and McLoud-Mann36] are all expressed in
$\mathbb {Q}(\phi )$
.
Figure 1 Two shapes belonging to the Ammann A2 family. The matching conditions are given by what are called Ammann bars appearing as dashed and solid lines in the interior of the tiles and which must continue straight across the edges of the tiling. This is a reproduction of Figure 10.4.1 from [Reference Grünbaum and Shephard24]. See also Figure 12 from [Reference Akiyama1].
It is then natural to ask whether there are aperiodic tilings out there such that the ratios of tile frequencies are not in
$\mathbb {Q}(\varphi )$
. It turns out that there are many. Recall that the first examples of aperiodic tilings provided by Berger [Reference Berger8], simplified by Knuth [Reference Knuth30] and Robinson [Reference Robinson53], are described by substitutions whose inflation factor is an integer (2 in this case). Many other substitutive and aperiodic planar tilings have an integer inflation factor and are listed in [Reference Baake and Grimm5, §6.4]. It includes the chair tiling [Reference Robinson52], the sphinx tiling [Reference Solomyak63], the
$(1+\varepsilon +\varepsilon ^2)$
-tiling [Reference Penrose50] and the Taylor and Socolar-Taylor tilings [Reference Socolar and Taylor61].
Many substitution tilings with non-integer inflation factor are known. Various types of planar aperiodic substitution tilings with n-fold rotational symmetry involving cyclotomic numbers were described in recent years [Reference Gähler, Kwan and Maloney22, Reference Kari and Rissanen29, Reference Frettlöh, Say-awen and De Las Peñas17, Reference Pautze47, Reference Kari and Lutfalla28]; see the sections [Reference Frettlöh18, §1.7] and [Reference Baake and Grimm5, §7.3]. Examples of algebraic non-Pisot aperiodic tilings were portrayed in [Reference Baake and Grimm5, §6.5]. Moreover, substitution tilings with transcendental inflation factor were recently proposed in [Reference Frettlöh, Garber and Mañibo19] using compact alphabets.
Closer to golden mean are other algebraic integers, starting with those of degree two, for which aperiodic tilings exist. In Ammann A4 and A5 aperiodic tilings [Reference Grünbaum and Shephard24], the ratio of frequency of the two involved tiles is
$\sqrt {2}$
[Reference Ammann, Grünbaum and Shephard4, p. 22]. Nowadays these tilings are known as Ammann–Beenker tilings [Reference Baake and Grimm5, §6.1] since their algebraic properties were independently described in [Reference Beenker7]. In [Reference Ammann, Grünbaum and Shephard4], the question whether there exist sets of aperiodic prototiles associated with irrational numbers other than
$\sqrt {2}$
and the golden ratio was mentioned. But they had ‘no conjecture concerning the characterization of all numbers that are possible for such ratios’ of frequencies of tiles.
The inflation factor of Ammann–Beenker substitution tilings is
$1+\sqrt {2}$
[Reference Baake and Grimm5, Prop. 6.2]. This number is sometimes called the silver mean because its continued fraction expansion is
$[2;2,2,\dots ]$
, where that of the golden mean is
$[1;1,1,\dots ]$
. The golden mean and the silver mean belong to a larger family made of the positive root of the polynomial
$x^2-nx-1$
, where n is a positive integer:
$$\begin{align*}\beta_n= \frac{n+\sqrt{n^2+4}}{2} = n + \frac{\displaystyle 1}{\displaystyle n + \frac{\displaystyle 1}{\displaystyle n + \frac{\displaystyle 1}{\displaystyle \dots}}}. \end{align*}$$
We refer to this root as the
$n^{th}$
metallic mean [46]. These numbers were called silver means [Reference Schroeder, Fractals and Laws56] and noble means in [Reference Baake and Grimm5, § 4.4] (note that noble numbers was already defined in [Reference Schroeder, Fractals and Laws56, Appendix B, p. 392–394] for a different meaning). Observe also that the definition of metallic means from [Reference de Spinadel13] is larger, as it contains all positive roots of polynomial
$x^2-px-q$
, where p and q are positive integers. In this contribution, we consider only the metallic means, in the sense of de Spinadel, which are algebraic units; that is,
$p\geq 1$
and
$q=1$
.
When a tiling space is preserved by a substitution, it is also preserved by powers of this substitution. Since odd-powers of metallic means are metallic means, we know substitution tilings for infinitely many other metallic means. In particular, the inflation factor of the third power of the substitution for Penrose tilings is the
$4^{th}$
metallic mean
$\beta _1^3 = \beta _4$
. Also, the inflation factor of the third power of the substitution for Ammann–Beenker tilings is the cube of the silver ratio, which is the
$14^{th}$
metallic mean
$\beta _2^3 = \beta _{14}$
, etc. For more information, we refer the reader to the OEIS [45] where indices of metallic means that are powers of other metallic means are listed as sequence A352403.
In recent years, new discoveries were made in the theory of quasicrystals related to metallic mean numbers. A self-similar hexagonal quasicrystal whose inflation factor is the
$3^{rd}$
metallic mean (also called bronze-mean) was described in [Reference Dotera, Bekku and Ziherl14]. It is given by a substitution rule involving a small and a large equilateral triangles and a rectangle; see [Reference Frettlöh, Harriss and Gähler20]. Their construction was further extended to every
$(3n)^{th}$
metallic mean in [Reference Nakakura, Ziherl, Matsuzawa and Dotera44] where
$n\geq 1$
is a positive integer.
Our contribution
In this contribution, we introduce a new family of aperiodic tiles using the oldest known shape for aperiodic tiles: the unit square. Unit squares with labeled edges and tilings of the plane by infinitely many translated copies of them were considered by Wang [Reference Wang66] with the condition that adjacent tiles must share the same label on the common edge. Such tiles are nowadays called Wang tiles. A set of Wang tiles is aperiodic if it admits at least one valid tiling, and none of them is periodic. The first known aperiodic set of tiles was discovered by Berger [Reference Berger8]: a set of 20426 Wang tiles. Many smaller examples were discovered thereafter, and we refer the reader to [Reference Jeandel and Rao26] for an overview of these developments leading to the discovery of the smallest possible size (
$=11$
) for an aperiodic set of Wang tiles.
For every positive integer n, we construct a set
${\mathcal {T}}_n$
made of
$(n+3)^2$
Wang tiles, and we consider the subshift
${\Omega }_n$
defined as the set of valid configurations
$\mathbb {Z}^2\to {\mathcal {T}}_n$
over these tiles. We also say that
${\Omega }_n$
is a Wang shift because it is a subshift defined from a set of Wang tiles. The set
${\mathcal {T}}_n$
is the disjoint union of 5 sets of tiles:
-
•
$n^2$
white tiles, -
• n yellow horizontal stripe tiles and n yellow vertical stripe tiles,
-
• n blue horizontal stripe tiles and n blue vertical stripe tiles,
-
•
$n+1$
green horizontal overlap tiles and
$n+1$
green vertical overlap tiles, -
•
$7$
junction tiles.
We observe that the sum of cardinalities of the five subsets is
$n^2+2n+2n+2(n+1)+7=(n+3)^2$
. The sets
${\mathcal {T}}_n$
of Wang tiles for
$n=1,2,3,4,5$
are shown in Figure 2, and rectangular valid tilings over the sets
${\mathcal {T}}_n$
for
$n=1,2,3,4$
are shown in Figure 3, Figure 4, Figure 5 and Figure 6.
Figure 2 Metallic mean Wang tile sets
${\mathcal {T}}_n$
for
$n=1,2,3,4,5$
.
Figure 3 A valid
$17\times 23$
pattern with Wang tile set
${\mathcal {T}}_1$
.
Figure 4 A valid
$17\times 23$
pattern with Wang tile set
${\mathcal {T}}_2$
.
Figure 5 A valid
$17\times 23$
pattern with Wang tile set
${\mathcal {T}}_3$
.
Figure 6 A valid
$17\times 23$
pattern with Wang tile set
${\mathcal {T}}_4$
.
The family of Wang shift
$({\Omega }_n)_{n\geq 1}$
has too many nice properties to hold in one article. In this first article dedicated to its study, we focus on its substitutive properties. Its dynamical properties and the consideration of
${\mathcal {T}}_n$
as the set of instances of a computer chip will be considered separately in a follow-up contribution.
The main result of the current contribution is to prove that the Wang shift
${\Omega }_n$
is self-similar for every integer
$n\geq 1$
. The self-similarity is given by a 2-dimensional substitution over an alphabet of size
$(n+3)^2$
. The self-similarity is not a bijection, but informally it is essentially one. This is formalized with the terminology of recognizability (one-to-one up to a shift) and surjectivity up to a shift. See Section 2 for the definition of Wang shifts and Section 3 for the definition of 2-dimensional substitutions, self-similarity and recognizability.
Theorem A. For every integer
$n\geq 1$
, the set
${\mathcal {T}}_n$
containing
$(n+3)^2$
Wang tiles defines a Wang shift
${\Omega }_n$
which is self-similar. More precisely, there exists an expansive and recognizable 2-dimensional substitution
${\omega }_n:{\Omega }_n\to {\Omega }_n$
which is onto up to a shift – that is, such that
${\Omega }_n=\overline {{\omega }_n({\Omega }_n)}^\sigma $
.
The proof of Theorem A is the same for every integer
$n\geq 1$
. Indeed, we show that every configuration in
${\Omega }_n$
can be decomposed uniquely into rectangular blocks that we call return blocks. These return blocks and their right, top, left and bottom labels are in bijection with an extended set
${\mathcal {T}}_n^{\prime }\supset {\mathcal {T}}_n$
of Wang tiles. Then we show that in the extended Wang shift
${\Omega }_n^{\prime }\supseteq {\Omega }_n$
defined from the extended set
${\mathcal {T}}_n^{\prime }$
of Wang tiles, only the tiles in
${\mathcal {T}}_n$
appear. Thus,
${\Omega }_n^{\prime }\subseteq {\Omega }_n$
. This shows that
${\Omega }_n={\Omega }_n^{\prime }$
and that
${\Omega }_n$
is self-similar.
As a corollary, we deduce that the Wang shift
${\Omega }_n$
is aperiodic.
Corollary B. For every integer
$n\geq 1$
, the Wang shift
${\Omega }_n$
is aperiodic.
Our second result is that the self-similarity is primitive. As in the 1-dimensional case, we say that a 2-dimensional substitution
$\omega $
is primitive if there exists
$m\in \mathbb {N}$
such that, for every
$a,b\in \mathcal {A}$
, the letter b occurs in
$\omega ^m(a)$
.
Theorem C. For every integer
$n\geq 1$
, the 2-dimensional substitution
${\omega }_n:{\Omega }_n\to {\Omega }_n$
is primitive. The Perron–Frobenius dominant eigenvalue of the incidence matrix of
${\omega }_n$
is
$\beta _n^2$
, the square of the
$n^{th}$
metallic mean number, and the inflation factor of
${\omega }_n$
is
$\beta _n$
.
Our third result is that the Wang shift
${\Omega }_n$
is minimal; that is, if
$X\subseteq {\Omega }_n$
is a nonempty closed shift-invariant subset, then
$X={\Omega }_n$
. Equivalently, every shift orbit is dense, which implies that every configuration in
${\omega }_n$
is uniformly recurrent. Every small set of aperiodic Wang tiles does not satisfy this property. For instance, the Robinson Wang shift is not minimal [Reference Gähler, Julien and Savinien21], and neither is the Jeandel–Rao Wang shift [Reference Labbé35]. The proof of minimality is based on a criterion involving the patterns of shapes
$1\times 2$
,
$2\times 1$
and
$2\times 2$
and their images under the substitution; see Lemma 10.4.
Theorem D. For every integer
$n\geq 1$
, the Wang shift
${\Omega }_n$
is minimal and is equal to the substitutive subshift
${\Omega }_n=\mathcal {X}_{{\omega }_n}$
.
In a tiling of the plane by the two shapes shown in Figure 1 respecting the matching condition, there appear what are called Ammann bars. In this case, the slopes of the Ammann bars take four different values: two slope values for the dashed Ammann bars and two slope values for the solid Ammann bars. As explained in [Reference Grünbaum and Shephard24, p.594–598], the solid bars can be regarded as the edges of a new tiling by rhombs and parallelograms, for which the dashed bars can be regarded as markings on the tiles specifying the matching conditions. Sixteen parallelogram tiles arise from this construction which can be recoded as 16 Wang tiles. As we show in Theorem E, the Ammann 16 Wang tiles are equivalent to
${\mathcal {T}}_1$
, the first member of the family
${\mathcal {T}}_n$
when
$n=1$
.
Theorem E. When
$n=1$
, the set
${\mathcal {T}}_n$
is equal, up to symbol relabeling, to the Ammann set of 16 Wang tiles.
Thus, the family
$({\mathcal {T}}_n)_{n\geq 1}$
can be considered as an extension of the Ammann set of Wang tiles to the metallic mean numbers.
Structure of the article
In Section 2, we present preliminaries on dynamical systems, subshifts and Wang shifts. In Section 3, we recall definitions of 2-dimensional substitutions. In Section 4, we introduce two Wang shifts
${\Omega }_n\subseteq {\Omega }_n^{\prime }$
defined by the sets
${\mathcal {T}}_n\subseteq {\mathcal {T}}_n^{\prime }$
of Wang tiles. In Section 5, we define two substitutions
${\omega }_n^{\prime }:{\Omega }_n^{\prime }\to {\Omega }_n^{\prime }$
and
${\omega }_n:{\Omega }_n\to {\Omega }_n$
. In Section 6, we describe the return blocks in the Wang shifts
${\Omega }_n$
and
${\Omega }_n^{\prime }$
, and we prove that every configuration in the Wang shift
${\Omega }_n$
can be desubstituted into a configuration from
${\Omega }_n^{\prime }$
. In Section 7, we prove that tiles in
${\mathcal {T}}_n^{\prime }\setminus {\mathcal {T}}_n$
do not appear in configurations of
${\Omega }_n^{\prime }$
. Thus,
${\Omega }_n^{\prime }\subseteq {\Omega }_n$
. Observe that Section 7 depends on the results from Section 5 and Section 6. In Section 8, we prove that
${\Omega }_n$
is self-similar and aperiodic. In Section 9, we prove that the self-similarity is primitive. In Section 10, we prove that
${\Omega }_n$
is minimal. In Section 11, we state some questions raised by the current work. The article finishes with two appendices. Section A (Appendix A) gathers pictures of the substitutions
${\omega }_n$
for
$1\leq n\leq 5$
. In Section B (Appendix B), we prove the self-similarity of
${\Omega }_n$
when
$n=2$
using computer explorations.
2 Preliminaries on Wang shifts
This section follows the preliminary section of the chapter [Reference Labbé37].
2.1 Topological dynamical systems
Most of the notions introduced here can be found in [Reference Walters65]. A dynamical system is a triple
$(X,G,T)$
, where X is a topological space, G is a topological group and T is a continuous function
$G\times X\to X$
defining a left action of G on X: if
$x\in X$
, e is the identity element of G and
$g,h\in G$
, then using additive notation for the operation in G, we have
$T(e,x)=x$
and
$T(g+h,x)=T(g,T(h,x))$
. In other words, if one denotes the transformation
$x\mapsto T(g,x)$
by
$T^g$
, then
$T^{g+h}=T^g T^h$
. In this work, we consider the Abelian group
$G=\mathbb {Z}\times \mathbb {Z}$
.
If
$Y\subset X$
, let
$\overline {Y}$
denote the topological closure of Y and let
$\overline {Y}^T:=\cup _{g\in G}T^g(Y)$
denote the T-closure of Y. A subset
$Y\subset X$
is T
-invariant if
$\overline {Y}^T=Y$
. A dynamical system
$(X,G,T)$
is called minimal if X does not contain any nonempty, proper, closed T-invariant subset. The left action of G on X is free if
$g=e$
whenever there exists
$x\in X$
such that
$T^g(x)=x$
.
Let
$(X,G,T)$
and
$(Y,G,S)$
be two dynamical systems with the same topological group G. A homomorphism
$\theta :(X,G,T)\to (Y,G,S)$
is a continuous function
$\theta :X\to Y$
satisfying the commuting property that
$S^g\circ \theta =\theta \circ T^g$
for every
$g\in G$
. A homomorphism
$\theta :(X,G,T)\to (Y,G,S)$
is called an embedding if it is one-to-one, a factor map if it is onto, and a topological conjugacy if it is both one-to-one and onto and its inverse map is continuous. If
$\theta :(X,G,T)\to (Y,G,S)$
is a factor map, then
$(Y,G,S)$
is called a factor of
$(X,G,T)$
and
$(X,G,T)$
is called an extension of
$(Y,G,S)$
. Two dynamical systems are topologically conjugate if there is a topological conjugacy between them.
2.2 Subshifts and shifts of finite type
In this section, we introduce multidimensional subshifts, a particular type of dynamical systems [Reference Lind and Marcus40, §13.10], [Reference Schmidt55, Reference Lind39, Reference Hochman25]. Let
$\mathcal {A}$
be a finite set,
$d\geq 1$
, and let
$\mathcal {A}^{\mathbb {Z}^d}$
be the set of all maps
$x:\mathbb {Z}^d\to \mathcal {A}$
, equipped with the compact product topology. An element
$x\in \mathcal {A}^{\mathbb {Z}^d}$
is called configuration, and we write it as
$x=(x_{\boldsymbol {m}})=(x_{\boldsymbol {m}}:{\boldsymbol {m}}\in \mathbb {Z}^d)$
, where
$x_{\boldsymbol {m}}\in \mathcal {A}$
denotes the value of x at
${\boldsymbol {m}}$
. The topology on
$\mathcal {A}^{\mathbb {Z}^d}$
is compatible with the metric defined for all configurations
$x,x'\in \mathcal {A}^{\mathbb {Z}^d}$
by
$\mathrm {dist}(x,x')=2^{-\min \left \{\Vert {\boldsymbol {n}}\Vert \,:\, x_{\boldsymbol {n}}\neq x^{\prime }_{\boldsymbol {n}}\right \}}$
where
$\Vert {\boldsymbol {n}}\Vert = |n_1| + \dots + |n_d|$
. The shift action
$\sigma :{\boldsymbol {n}}\mapsto \sigma ^{\boldsymbol {n}}$
of the additive group
$\mathbb {Z}^d$
on
$\mathcal {A}^{\mathbb {Z}^d}$
is defined by
$$ \begin{align} (\sigma^{\boldsymbol{n}}(x))_{\boldsymbol{m}} = x_{{\boldsymbol{m}}+{\boldsymbol{n}}} \end{align} $$
for every
$x=(x_{\boldsymbol {m}})\in \mathcal {A}^{\mathbb {Z}^d}$
and
${\boldsymbol {n}}\in \mathbb {Z}^d$
. If
$X\subset \mathcal {A}^{\mathbb {Z}^d}$
, let
$\overline {X}$
denote the topological closure of X and let
${\overline {X}^{\sigma }}:=\{\sigma ^{\boldsymbol {n}}(x)\mid x\in X, {\boldsymbol {n}}\in \mathbb {Z}^d\}$
denote the shift-closure of X. A subset
$X\subset \mathcal {A}^{\mathbb {Z}^d}$
is shift-invariant if
${\overline {X}^{\sigma }}=X$
. A closed, shift-invariant subset
$X\subset \mathcal {A}^{\mathbb {Z}^d}$
is a subshift. If
$X\subset \mathcal {A}^{\mathbb {Z}^d}$
is a subshift, we write
$\sigma =\sigma ^X$
for the restriction of the shift action (2.1) to X. When X is a subshift, the triple
$(X,\mathbb {Z}^d,\sigma )$
is a dynamical system and the notions presented in the previous section hold.
A configuration
$x\in X$
is periodic if there is a nonzero vector
${\boldsymbol {n}}\in \mathbb {Z}^d\setminus \{{\boldsymbol {0}}\}$
such that
$x=\sigma ^{\boldsymbol {n}}(x)$
, and otherwise it is nonperiodic. We say that a nonempty subshift X is aperiodic if the shift action
$\sigma $
on X is free.
For any subset
$S\subset \mathbb {Z}^d$
let
$\pi _S:\mathcal {A}^{\mathbb {Z}^d}\to \mathcal {A}^S$
, denote the projection map which restricts every
$x\in \mathcal {A}^{\mathbb {Z}^d}$
to S. A pattern is a function
$p\in \mathcal {A}^S$
for some finite subset
$S\subset \mathbb {Z}^d$
. To every pattern
$p\in \mathcal {A}^S$
corresponds a subset
$\pi _S^{-1}(p)\subset \mathcal {A}^{\mathbb {Z}^d}$
called cylinder. A nonempty set
$X\subset \mathcal {A}^{\mathbb {Z}^d}$
is a subshift if and only if there exists a set
$\mathcal {F}$
of forbidden patterns such that
$$ \begin{align} X = \{x\in\mathcal{A}^{\mathbb{Z}^d} \mid \pi_S\circ\sigma^{\boldsymbol{n}}(x)\notin\mathcal{F} \text{ for every } {\boldsymbol{n}}\in\mathbb{Z}^d \text{ and } S\subset\mathbb{Z}^d\}; \end{align} $$
see [Reference Hochman25, Prop. 9.2.4]. A subshift
$X\subset \mathcal {A}^{\mathbb {Z}^d}$
is a subshift of finite type (SFT) if there exists a finite set
$\mathcal {F}$
such that (2.2) holds. In this article, we consider shifts of finite type on
$\mathbb {Z}\times \mathbb {Z}$
– that is, the case
$d=2$
.
2.3 Wang shifts
A Wang tile is a tuple of four colors
$(a,b,c,d)\in I\times J\times I\times J$
where I is a finite set of vertical colors and J is a finite set of horizontal colors; see [Reference Wang66, Reference Robinson53]. A Wang tile is represented as a unit square with colored edges:
For each Wang tile
$\tau =(a,b,c,d)$
, let 
, 
, 
, 
denote respectively the colors or labels of the right, top, left and bottom edges of
$\tau $
.
Let
$\mathcal {T}=\{t_0,\dots ,t_{m-1}\}$
be a set of Wang tiles as the one shown in Figure 7. A configuration
$x:\mathbb {Z}^2\to \{0,\dots ,m-1\}$
is valid with respect to
$\mathcal {T}$
if it assigns a tile in
$\mathcal {T}$
to each position of
$\mathbb {Z}^2$
, so that contiguous edges of adjacent tiles have the same color; that is,
for every
${\boldsymbol {n}}\in \mathbb {Z}^2$
where
${\boldsymbol {e}}_1=(1,0)$
and
${\boldsymbol {e}}_2=(0,1)$
. A finite pattern which is valid with respect to
$\mathcal {U}$
is shown in Figure 8.
Figure 7 The set of 3 Wang tiles introduced in [Reference Wang66] using letters
$\{A,B,C,D,E\}$
instead of numbers from the set
$\{1,2,3,4,5\}$
for labeling the edges. Each tile is identified uniquely by an index from the set
$\{0,1,2\}$
written at the center each tile.
Let
$\Omega _{\mathcal {T}}\subset \{0,\dots ,m-1\}^{\mathbb {Z}^2}$
denote the set of all valid configurations with respect to
$\mathcal {T}$
, called the Wang shift of
$\mathcal {T}$
. To a configuration
$x\in \Omega _{\mathcal {T}}$
corresponds a tiling of the plane
$\mathbb {R}^2$
by the tiles
$\mathcal {T}$
where the unit square Wang tile
$t_{x({\boldsymbol {n}})}$
is placed at position
${\boldsymbol {n}}$
for every
${\boldsymbol {n}}\in \mathbb {Z}^2$
, as in Figure 8. Together with the shift action
$\sigma $
of
$\mathbb {Z}^2$
,
$\Omega _{\mathcal {T}}$
is a SFT of the form (2.2) since there exists a finite set of forbidden patterns made of all horizontal and vertical dominoes of two tiles that do not share an edge of the same color.
A configuration
$x\in \Omega _{\mathcal {T}}$
is periodic if there exists
${\boldsymbol {n}}\in \mathbb {Z}^2\setminus \{0\}$
such that
$x=\sigma ^{\boldsymbol {n}}(x)$
. A set
$\mathcal {T}$
of Wang tiles is periodic if there exists a periodic configuration
$x\in \Omega _{\mathcal {T}}$
. Originally, Wang thought that every set
$\mathcal {T}$
of Wang tiles is periodic as soon as
$\Omega _{\mathcal {T}}$
is nonempty [Reference Wang66]. Wang noticed that if this statement were true, it would imply the existence of an algorithm solving the domino problem – that is, taking as input a set of Wang tiles and returning yes or no whether there exists a valid configuration with these tiles. Berger, a student of Wang, later proved that the domino problem is undecidable and he also provided a first example of an aperiodic set of Wang tiles [Reference Berger8]. A set
$\mathcal {T}$
of Wang tiles is aperiodic if the Wang shift
$\Omega _{\mathcal {T}}$
is a nonempty aperiodic subshift.
2.4 Directional determinism
A set
$\mathcal {T}$
of Wang tiles is called SW-deterministic if there do not exist two different tiles in
$\mathcal {T}$
that would have the same colors on their bottom and left edges, respectively [Reference Kari and Papasoglu27]. In other words, for all colors
$C_1$
and
$C_2$
, there exists at most one tile in
$\mathcal {T}$
whose bottom and left edges have colors
$C_1$
and
$C_2$
, respectively.
Let
$S=\{a_1,a_1+1,\dots ,b_1\}\times \{a_2,a_2+1,\dots ,b_2\}$
be a rectangular support where
$a_1,b_1,a_2,b_2$
are integers such that
$a_1\leq b_1$
and
$a_2\leq b_2$
. Let
$p:S\to \mathcal {T}$
be a valid rectangular pattern over the tiles
$\mathcal {T}$
. We say that the bottom labels of p and top labels of p are, respectively, the sequences
read on the pattern from left to right. Also, we say that the left labels of p and right labels of p are, respectively, the sequences
read on the pattern from bottom to top.
As shown in the next lemma, the local definition of SW-deterministic sets of Wang tiles extends into a wider property on rectangular patterns.
Lemma 2.1. Let
$\mathcal {T}$
be a SW-deterministic set of Wang tiles. If p and q are two rectangular valid patterns with the same shape, the same sequence of bottom labels and the same sequence of left labels, then
$p=q$
.
Proof. By contradiction, suppose that there are two distinct rectangular patterns p and q whose sequence of bottom labels is X and sequence of left labels is Y. Since p and q are distinct, there exists a position
$k\in \mathbb {N}^2$
such that
$p_k\neq q_k$
. Consider such a position in the support of p and q which minimizes the norm
$\Vert k\Vert _1$
. Since the position is minimal, every tile at position smaller in norm is the same in both patterns. In particular, it implies that 
and 
. The set of Wang tile
${\mathcal {T}}_n$
is SW-deterministic. This implies that 
and 
. Since the four labels of the Wang tiles are the same, we must have
$p_k=q_k$
, a contradiction. We conclude the uniqueness of the rectangular pattern.
NW-, NE- and SE-deterministic sets of Wang tiles are defined analogously. Recall that it was shown in [Reference Kari and Papasoglu27] that there exist aperiodic tile sets that are deterministic in all four directions simultaneously.
3 Preliminaries on 2-dimensional substitutions
Rectangular 2-dimensional substitutions and their symbolic dynamical systems were considered in [Reference Mozes43]. For a certain class of 2-dimensional substitution systems, it was shown how to construct a set of Wang tiles such that the associated Wang shift is an almost everywhere one-to-one extension of the substitution system [Reference Mozes43, Theorem 4.5]. This result was generalized later for geometrical substitutions over polygonal tiles [Reference Goodman-Strauss23].
In this section, we introduce
$2$
-dimensional substitutions. Our definition and the one presented in [Reference Mozes43] are incomparable. On the one hand, we restrict to the deterministic case (every letter has a unique image). On the other hand, we extend to different alphabets
$\mathcal {A}$
and
$\mathcal {B}$
for the domain and codomain. The section follows the preliminary section of the chapter [Reference Labbé37].
3.1 d-dimensional word
We denote by
$\{{\boldsymbol {e}}_k|1\leq k\leq d\}$
the canonical basis of
$\mathbb {Z}^d$
where
$d\geq 1$
is an integer. If
$i\leq j$
are integers, then 
denotes the interval of integers
$\{i, i+1, \dots , j\}$
. Let
${\boldsymbol {n}}=(n_1,\dots ,n_d)\in \mathbb {N}^d$
and
$\mathcal {A}$
be an alphabet. We denote by
$\mathcal {A}^{{\boldsymbol {n}}}$
the set of functions
An element
$u\in \mathcal {A}^{\boldsymbol {n}}$
is called a d
-dimensional word of size
${\boldsymbol {n}}=(n_1,\dots ,n_d)\in \mathbb {N}^d$
on the alphabet
$\mathcal {A}$
. We use the notation 
when necessary. The set of all finite d-dimensional words is
$\mathcal {A}^{*^d}=\bigcup _{{\boldsymbol {n}}\in \mathbb {N}^d} \mathcal {A}^{\boldsymbol {n}}$
. A d-dimensional word of size
${\boldsymbol {e}}_k+\sum _{i=1}^d{\boldsymbol {e}}_i$
is called a domino in the direction
${\boldsymbol {e}}_k$
. When the context is clear, we write
$\mathcal {A}$
instead of
$\mathcal {A}^{(1,\dots ,1)}$
. When
$d=2$
, we represent a d-dimensional word u of size
$(n_1,n_2)$
as a matrix with Cartesian coordinates:
$$ \begin{align*} u= \left(\begin{array}{ccc} u_{0,n_2-1} &\dots & u_{n_1-1,n_2-1} \\ \dots &\dots & \dots \\ u_{0,0} &\dots & u_{n_1-1,0} \end{array}\right). \end{align*} $$
Let
${\boldsymbol {n}},{\boldsymbol {m}}\in \mathbb {N}^d$
and
$u\in \mathcal {A}^{\boldsymbol {n}}$
and
$v\in \mathcal {A}^{\boldsymbol {m}}$
. If there exists an index i such that
$n_j=m_j$
for all
$j\in \{1,\dots ,d\}\setminus \{i\}$
, then the concatenation of u and v in the direction
${\boldsymbol {e}}_i$
is defined: it is the d-dimensional word
$u\odot ^i v$
of size
$(n_1,\dots ,n_{i-1},n_i+m_i,n_{i+1},\dots ,n_d)\in \mathbb {N}^d$
given as
$$ \begin{align*} (u\odot^i v) ({\boldsymbol{a}}) = \begin{cases} u({\boldsymbol{a}}) & \text{if}\quad 0 \leq a_i < n_i,\\ v({\boldsymbol{a}}-n_i{\boldsymbol{e}}_i) & \text{if}\quad n_i \leq a_i < n_i+m_i. \end{cases} \end{align*} $$
The notation
$u\odot ^i v$
was used in [Reference Charlier, Kärki and Rigo11].
The following equation illustrates the concatenation of
$2$
-dimensional words in the direction
${\boldsymbol {e}}_2$
:
$$\begin{align*}\left(\begin{array}{ccccc} 4 & 5 \\ 10 & 5 \end{array}\right) \odot^2 \left(\begin{array}{ccccc} 3 & 10 \\ 9 & 9 \\ 0 & 0 \\ \end{array}\right) = \left(\begin{array}{ccccc} 3 & 10 \\ 9 & 9 \\ 0 & 0 \\ 4 & 5 \\ 10 & 5 \end{array}\right) \end{align*}$$
and in the direction
${\boldsymbol {e}}_1$
:
$$\begin{align*}\left(\begin{array}{ccccc} 2 & 8 & 7 \\ 7 & 3 & 9 \\ 1 & 1 & 0 \\ 6 & 6 & 7 \\ 7 & 4 & 3 \end{array}\right) \odot^1 \left(\begin{array}{ccccc} 3 & 10 \\ 9 & 9 \\ 0 & 0 \\ 4 & 5 \\ 10 & 5 \end{array}\right) = \left(\begin{array}{ccccc} 2 & 8 & 7 & 3 & 10 \\ 7 & 3 & 9 & 9 & 9 \\ 1 & 1 & 0 & 0 & 0 \\ 6 & 6 & 7 & 4 & 5 \\ 7 & 4 & 3 & 10 & 5 \end{array}\right). \end{align*}$$
Let
${\boldsymbol {n}},{\boldsymbol {m}}\in \mathbb {N}^d$
and
$u\in \mathcal {A}^{\boldsymbol {n}}$
and
$v\in \mathcal {A}^{\boldsymbol {m}}$
. We say that u occurs in v at position
${\boldsymbol {p}}\in \mathbb {N}^d$
if v is large enough; that is,
${\boldsymbol {m}}-{\boldsymbol {p}}-{\boldsymbol {n}}\in \mathbb {N}^d$
and
$$\begin{align*}v({\boldsymbol{a}}+{\boldsymbol{p}}) = u({\boldsymbol{a}}) \end{align*}$$
for all
${\boldsymbol {a}}=(a_1,\dots ,a_d)\in \mathbb {N}^d$
such that
$0\leq a_i<n_i$
with
$1\leq i\leq d$
. If u occurs in v at some position, then we say that u is a d-dimensional subword or factor of v.
3.2 d-dimensional language
A subset
$L\subseteq \mathcal {A}^{*^d}$
is called a d-dimensional language. A language
$L\subseteq \mathcal {A}^{*^d}$
is called factorial if for every
$v\in L$
and every d-dimensional subword u of v, we have
$u\in L$
. All languages considered in this contribution are factorial. Given a configuration
$x\in \mathcal {A}^{\mathbb {Z}^d}$
, the language
$\mathcal {L}(x)$
defined by x is
$$ \begin{align*} \mathcal{L}(x) = \{u\in\mathcal{A}^{*^d} \mid u\text{ is a } d\text{-dimensional subword of } x\}. \end{align*} $$
The language of a subshift
$X\subseteq \mathcal {A}^{\mathbb {Z}^d}$
is
$\mathcal {L}_X = \cup _{x\in X} \mathcal {L}(x)$
. Conversely, given a factorial language
$L\subseteq \mathcal {A}^{*^d}$
, we define the subshift
$$ \begin{align*} \mathcal{X}_L = \{x\in\mathcal{A}^{\mathbb{Z}^d}\mid \mathcal{L}(x)\subseteq L\}. \end{align*} $$
A d-dimensional subword
$u\in \mathcal {A}^{*^d}$
is legal (or allowed) in a subshift
$X\subset \mathcal {A}^{\mathbb {Z}^d}$
if
$u\in \mathcal {L}_X$
, and it is illegal in X if
$u\notin \mathcal {L}_X$
[Reference Baake and Grimm5]. A language
$L\subseteq \mathcal {A}^{*^d}$
is illegal in a subshift
$X\subset \mathcal {A}^{\mathbb {Z}^d}$
if
$L\cap \mathcal {L}_X=\varnothing $
.
3.3 d-dimensional morphisms
Let
$\mathcal {A}$
and
$\mathcal {B}$
be two alphabets. Let
$L\subseteq \mathcal {A}^{*^d}$
be a factorial language. A function
$\omega :L\to \mathcal {B}^{*^d}$
is a d
-dimensional morphism if for every i with
$1\leq i\leq d$
, and every
$u,v\in L$
such that
$u\odot ^i v$
is defined and
$u\odot ^i v\in L$
, we have that the concatenation
$\omega (u)\odot ^i \omega (v)$
in direction
${\boldsymbol {e}}_i$
is defined and
$$ \begin{align*} \omega(u\odot^i v) = \omega(u)\odot^i \omega(v). \end{align*} $$
Note that the left-hand side of the equation is defined since
$u\odot ^i v$
belongs to the domain of
$\omega $
. A d-dimensional morphism
$L\to \mathcal {B}^{*^d}$
is thus completely defined from the image of the letters in
$\mathcal {A}$
, so we sometimes denote a d-dimensional morphism as a rule
$\mathcal {A}\to \mathcal {B}^{*^d}$
when the language L is unspecified.
As noticed by [Reference Mozes43, p.144], the images under the morphism of any two letters appearing in the same row of a word from L have the same height. Symmetrically, the images under the morphism of any two letters appearing in the same column of a word from L have the same width.
Let
$L\subseteq \mathcal {A}^{*^d}$
be a factorial language and
$\mathcal {X}_L\subseteq \mathcal {A}^{\mathbb {Z}^d}$
be the subshift generated by L. A d-dimensional morphism
$\omega :L \to \mathcal {B}^{*^d}$
can be extended to a continuous map
$\omega :\mathcal {X}_L\to \mathcal {B}^{\mathbb {Z}^d}$
(over the topology of subshifts, as defined in Section 2.2) in such a way that the origin of
$\omega (x)$
is at position
${\boldsymbol {0}}$
in the word
$\omega (x_{\boldsymbol {0}})$
for all
$x\in \mathcal {X}_L$
. More precisely, the image under
$\omega $
of the configuration
$x\in \mathcal {X}_L$
is
where
${\boldsymbol {1}}=(1,\dots ,1)\in \mathbb {Z}^d$
, 
for all
$n\in \mathbb {N}$
and 
. We say that the map
$\omega :\mathcal {X}_L\to \mathcal {B}^{\mathbb {Z}^d}$
is a d
-dimensional substitution.
In general, the image of a subshift under a d-dimensional substitution might not be closed under the shift. But the closure under the shift of the image of a subshift
$X\subseteq \mathcal {A}^{\mathbb {Z}^d}$
under
$\omega $
is the subshift
$$ \begin{align*} \overline{\omega(X)}^{\sigma} = \{\sigma^{\boldsymbol{k}}\omega(x)\in\mathcal{B}^{\mathbb{Z}^d} \mid {\boldsymbol{k}}\in\mathbb{Z}^d, x\in X\} \subseteq \mathcal{B}^{\mathbb{Z}^d}. \end{align*} $$
This motivates the following definition.
Definition 3.1. Let X, Y be two subshifts and
$\omega :X\to Y$
be a d-dimensional substitution. If
$Y=\overline {\omega (X)}^\sigma $
, then we say that
$\omega $
is onto up to a shift.
3.4 Self-similar subshifts
In this section, we consider languages and subshifts defined from morphisms leading to self-similar structures. In this situation, the domain and codomain of morphisms are defined over the same alphabet. Formally, we consider the case of d-dimensional morphisms
$\mathcal {A}\to \mathcal {B}^{*^d}$
where
$\mathcal {A}=\mathcal {B}$
.
The definition of self-similarity depends on the notion of expansiveness. It avoids the presence of lower-dimensional self-similar structure by having expansion in all directions.
Definition 3.2. We say that a d-dimensional morphism
$\omega :\mathcal {A}\to \mathcal {A}^{*^d}$
is expansive if for every
$a\in \mathcal {A}$
and
$K\in \mathbb {N}$
, there exists
$m\in \mathbb {N}$
such that 
.
Definition 3.3. A subshift
$X\subseteq \mathcal {A}^{\mathbb {Z}^d}$
is self-similar if there exists an expansive d-dimensional morphism
$\omega :\mathcal {A}\to \mathcal {A}^{*^d}$
such that
$X=\overline {\omega (X)}^\sigma $
.
Self-similar subshifts can be constructed by iterative application of a morphism
$\omega $
starting with the letters. The language
$\mathcal {L}_\omega $
defined by an expansive d-dimensional morphism
$\omega :\mathcal {A}\to \mathcal {A}^{*^d}$
is
$$ \begin{align*} \mathcal{L}_\omega = \{u\in\mathcal{A}^{*^d} \mid u\text{ is a }d\text{-dimensional subword of } \omega^n(a) \text{ for some } a\in\mathcal{A}\text{ and } n\in\mathbb{N} \}. \end{align*} $$
The substitutive shift
$\mathcal {X}_\omega =\mathcal {X}_{\mathcal {L}_\omega }$
defined from the language of
$\omega $
is a self-similar subshift since
$\mathcal {X}_\omega =\overline {\omega (\mathcal {X}_\omega )}^{\sigma }$
holds.
A d-dimensional morphism
$\omega :\mathcal {A}\to \mathcal {A}^{*^d}$
is primitive if there exists
$m\in \mathbb {N}$
such that for every
$a,b\in \mathcal {A}$
, the letter b occurs in
$\omega ^m(a)$
. Note that if
$\omega $
is primitive, then the Perron–Frobenius theorem applies for its incidence matrix
$M_\omega =(|\omega (a)|_b)_{(b,a)\in \mathcal {A}\times \mathcal {A}}$
; see [Reference Queffélec51].
3.5 d-dimensional recognizability
The definition of recognizability dates back to the work of Host, Quéffelec and Mossé [Reference Mossé42]. The definition introduced below is based on some work of Berthé et al. [Reference Berthé, Steiner, Thuswaldner and Yassawi9] on the recognizability in the case of S-adic systems where more than one substitution is involved.
Definition 3.4 (recognizable).
Let
$X\subseteq \mathcal {A}^{\mathbb {Z}^d}$
and
$\omega :X\to \mathcal {B}^{\mathbb {Z}^d}$
be a d-dimensional substitution. If
$y\in \overline {\omega (X)}^{\sigma }$
(i.e.,
$y=\sigma ^{\boldsymbol {k}}\omega (x)$
for some
$x\in X$
and
${\boldsymbol {k}}\in \mathbb {Z}^d$
, where
$\sigma $
is the d-dimensional shift map), we say that
$({\boldsymbol {k}},x)$
is an
$\omega $
-representation of y. We say that it is centered if
$y_{\boldsymbol {0}}$
lies inside of the image of
$x_{\boldsymbol {0}}$
(i.e., if 
coordinate-wise). We say that
$\omega $
is recognizable in
$X\subseteq \mathcal {A}^{\mathbb {Z}^d}$
if each
$y\in \mathcal {B}^{\mathbb {Z}^d}$
has at most one centered
$\omega $
-representation
$({\boldsymbol {k}},x)$
with
$x\in X$
.
The self-similarity of
${\Omega }_n$
allows us to conclude aperiodicity of the Wang shift using well-known arguments (see [Reference Solomyak62, Reference Mossé42], who showed that recognizability and aperiodicity are equivalent for primitive substitutive sequences).
The following statement corresponds to only one of the directions (the easy one) of the equivalence which does not need the notion of primitivity. It was proved for 2-dimensional substitutions in [Reference Labbé33]; see also [Reference Labbé37, Proposition 3.6].
Proposition 3.5 [Reference Labbé33, Proposition 6].
Let
$\omega :\mathcal {A}\to \mathcal {A}^{*^d}$
be an expansive d-dimensional morphism. Let
$X\subseteq \mathcal {A}^{\mathbb {Z}^d}$
be a self-similar subshift such that
$\overline {\omega (X)}^\sigma =X$
. If
$\omega $
is recognizable in X, then X is aperiodic.
4 The family of metallic mean Wang tiles
For every integer
$n\in \mathbb {Z}$
, we write
$\overline {n}$
to denote
$n+1$
and
$\underline {n}$
to denote
$n-1$
:
$$ \begin{align*} \overline{n} &:= n+1,\\ \underline{n} &:= n-1. \end{align*} $$
For every Wang tile
$\tau =(a,b,c,d)$
, we define its symmetric image under the positive diagonal as
$\widehat {\tau }=(b,a,d,c)$
:
4.1 The tiles
For every integer
$n\geq 1$
, let
$$\begin{align*}V_n = \{(v_0,v_1,v_2)\in\mathbb{Z}^3\colon 0\leq v_0\leq v_1\leq 1\text{ and } v_1\leq v_2\leq n+1\} \end{align*}$$
be a set of vectors having non-decreasing entries with an upper bound of 1 on the middle entry and an upper bound of
$n+1$
on the last entry. The label of the edges of the Wang tiles considered in this article are vectors in
$V_n$
. To lighten the figures and the presentation of the Wang tiles, it is convenient to denote the vector
$(v_0,v_1,v_2)\in V_n$
more compactly as a word
$v_0v_1v_2$
. For instance, the vector
$(1,1,1)$
is represented as
$111$
.
For every integer
$n\geq 1$
, we define the following set of Wang tiles whose labels belong to the set
$V_n$
. We have
$n^2$
white tiles whose labels all start with
$11$
:
We have horizontal stripe tiles whose top and bottom labels all start with
$11$
and whose left and right labels start with
$0$
. These are divided into four sets according to the first two digits of the left and right labels which can be
$00$
(associated with color blue) or
$01$
(associated with color yellow).
The set
$B_n^{\prime }$
of horizontal blue tiles are those such that both left and right labels start with
$00$
and are identified with a horizontal blue stripe. The set
$Y_n$
of horizontal yellow tiles are those such that both left and right labels start with
$01$
and are identified with a horizontal yellow stripe. The set
$G_n$
of horizontal green tiles are those such that the left label starts with
$00$
and right label starts with
$01$
and are identified with a green region intersecting blue and yellow horizontal stripes. The set
$A_n$
of horizontal antigreen tiles contains the tiles whose left label starts with
$01$
and whose right label starts with
$00$
. They are identified with non-intersecting blue and yellow horizontal stripes and no green intersecting region.
The tiles in
$A_n$
are called ‘antigreen’ because they are ‘against the system’ as shown later in Lemma 7.2. Antigreen tiles do not appear in any valid configuration, but they are needed as they play an important role in the description of the substitutive structure of the valid configurations allowed by these tiles; see Proposition 5.9 and Proposition 6.7.
We also have vertical stripe tiles which are the symmetric images of the horizontal stripe tiles under a reflection over the positive diagonal:
Finally, we have 9 junction tiles (the gray region is drawn in blue or yellow color depending on the specific values of
$k,\ell ,r,s$
):
We may observe that white tiles and junction tiles are closed under the reflection over the positive diagonal:
$$\begin{align*}\widehat{W_n}=W_n\ \text{and } \widehat{J_n^{\prime}}=J_n^{\prime}. \end{align*}$$
Junction tiles play a similar role to junction tiles in [Reference Mozes43]; thus, we reuse the same vocabulary.
4.2 The extended set
${\mathcal {T}}_n^{\prime }$
of metallic mean Wang tiles
For every integer
$n\geq 1$
, the extended set of metallic mean Wang tiles is the union of all of the tiles defined above:
$$\begin{align*}\mathcal{T}^{\prime}_n = W_n \cup B^{\prime}_n \cup G_n \cup Y_n \cup A_n \cup \widehat{B^{\prime}_n} \cup \widehat{G_n} \cup \widehat{Y_n} \cup \widehat{A_n} \cup J^{\prime}_n. \end{align*}$$
The set
${\mathcal {T}}_n^{\prime }$
of tiles defines the extended metallic mean Wang shift
$$\begin{align*}\Omega^{\prime}_n=\Omega_{\mathcal{T}^{\prime}_n}. \end{align*}$$
The set
$\mathcal {T}^{\prime }_n$
contains
$n^2+2(n+1+n+1+n+n)+9=n^2+8n+13$
Wang tiles. The set
$\mathcal {T}^{\prime }_n$
of Wang tiles for
$n=4$
is shown in Figure 9.
Figure 9 Extended metallic mean Wang tile sets
$\mathcal {T}^{\prime }_n$
for
$n=4$
. The junction tiles in
$\mathcal {D}$
are shown with a
$\times $
-mark in their center.
4.3 The subset
${\mathcal {T}}_n$
of metallic mean Wang tiles
We need to define an important subset of the extended metallic mean Wang tiles
${\mathcal {T}}_n^{\prime }$
because some of the tiles are not necessary as they do not appear in any valid configurations of
${\Omega }_n^{\prime }$
. For example, we can observe that no tile of
${\mathcal {T}}_n^{\prime }$
has label
$00\overline {n}$
on the left or bottom. Therefore, the last horizontal blue tile and last vertical blue tile which use label
$00\overline {n}$
on their top or right edge admit no immediate surroundings with tiles in
${\mathcal {T}}_n^{\prime }$
. As shown in Section 7 using results proved in Section 5 and Section 6, other tiles from
${\mathcal {T}}_n^{\prime }$
do not admit arbitrarily large surroundings. Therefore, it is convenient to remove them.
Let
be the set containing the last blue horizontal and vertical tiles as well as two of the junction tiles. For every positive integer n, we delete the four tiles of
$\mathcal {D}$
from
${\mathcal {T}}_n^{\prime }$
as well as all of the antigreen tiles. We obtain the following subset of metallic mean Wang tiles
$$ \begin{align*} {\mathcal{T}}_n&= {\mathcal{T}}_n^{\prime} \setminus \left(A_n\cup\widehat{A_n} \cup \mathcal{D} \right)\\ &= W_n \cup B_n \cup G_n \cup Y_n \cup \widehat{B_n} \cup \widehat{G_n} \cup \widehat{Y_n} \cup J_n, \end{align*} $$
where
$B_n=B_n^{\prime }\setminus \mathcal {D}$
is the remaining set of n horizontal blue stripe tiles and
$J_n=J_n^{\prime }\setminus \mathcal {D} =\left \{ j_n^{0,0,0,0}, j_n^{0,1,0,0}, j_n^{0,0,0,1}, j_n^{0,1,0,1}, j_n^{1,1,0,1}, j_n^{0,1,1,1}, j_n^{1,1,1,1} \right \}$
is the remaining set of
$7$
junction tiles. The set
${\mathcal {T}}_n$
contains
$n^2+2(n+n+1+n)+7=(n+3)^2$
Wang tiles. It is shown in Figure 2 for
$n=1,2,3,4,5$
.
The set
${\mathcal {T}}_n$
of tiles defines the Metallic mean Wang shift
$$\begin{align*}{\Omega}_n=\Omega_{{\mathcal{T}}_n}, \end{align*}$$
which is a subshift of
${\Omega }_n^{\prime }$
because
${\mathcal {T}}_n\subset {\mathcal {T}}_n^{\prime }$
.
Remark 4.1. The reader may wonder why we need to introduce the extended set
${\mathcal {T}}_n^{\prime }$
if only the tiles in the subset
${\mathcal {T}}_n$
appear in configurations of
${\Omega }_n^{\prime }$
. This is because the extended set is needed to describe and prove the self-similarity of
${\mathcal {T}}_n$
in Theorem A. In the proof (using the vocabulary of supertiles from the only article published by Ammann [Reference Ammann, Grünbaum and Shephard4]), we show that if the markings of the supertiles at level k are in bijection with the tiles in
${\mathcal {T}}_n$
, then the markings of the supertiles at level
$k+1$
are in bijection with tiles in
${\mathcal {T}}_n^{\prime }$
(not
${\mathcal {T}}_n$
!). In other words, we cannot get rid of the ghost tiles in
${\mathcal {T}}_n^{\prime }\setminus {\mathcal {T}}_n$
because they keep reappearing at the next level of the hierarchy in bigger sizes.
4.4 The Ammann aperiodic set of 16 Wang tiles
A reproduction of the Ammann aperiodic set of 16 Wang tiles [Reference Grünbaum and Shephard24, p.595, Figure 11.1.13] is shown in Figure 10. The Ammann set of 16 Wang tiles corresponds to
${\mathcal {T}}_1$
.
Figure 10 Left: a reproduction of the Ammann aperiodic set of 16 Wang tiles [Reference Grünbaum and Shephard24, p.595, Figure 11.1.13]. Middle: the Ammann aperiodic set of 16 Wang tiles in the same order but with coloring corresponding to the white, yellow, green, blue and junction tiles of the set
${\mathcal {T}}_1$
. Right: The set
${\mathcal {T}}_1$
of Wang tiles whose edge labels are vectors in
$\mathbb {N}^3$
. The sets are equivalent up to a bijection of the edge labels.
Theorem E. When
$n=1$
, the set
${\mathcal {T}}_n$
is equal, up to symbol relabeling, to the Ammann set of 16 Wang tiles.
Proof. The following is a bijection from the labels of the Ammann set of 16 Wang tiles and the labels of the tiles in
${\mathcal {T}}_1$
:
$$ \begin{align*} 1 \mapsto 112,\quad 2 \mapsto 111,\quad 3 \mapsto 001, \quad 4 \mapsto 011, \quad 5 \mapsto 012, \quad 6 \mapsto 000. \end{align*} $$
See Figure 10 (note that the order of the tiles is not the same).
Thus, the family
$({\mathcal {T}}_n)_{n\geq 1}$
can be considered as a generalization of the Ammann aperiodic set of 16 Wang tiles.
4.5 Symmetric properties
The set
${\mathcal {T}}_n$
has nice symmetric properties. The first being that it is closed under the mirror image through the positive diagonal; that is,
$\widehat {{\mathcal {T}}_n}={\mathcal {T}}_n$
. Another less evident observation is that the set
${\mathcal {T}}_n$
is equivalent to its image under a half-turn rotation up to the application of an involution of
$V_n\setminus \{(0,0,\overline {n})\}$
applied on the edge labels of the Wang tiles.
Lemma 4.2. Let
$\sigma :(i,j,k) \mapsto (i,1+i-j,n+1+i-k)$
(an involution on
$V_n\setminus \{(0,0,\overline {n})\}$
). Then,
Proof. When rotating the tiles of
${\mathcal {T}}_n$
by half a turn and applying the map
$\sigma $
on the resulting labels, we may observe that yellow tiles become blue tiles and vice versa, white tiles are mapped to white tiles, junction tiles are mapped to junction tiles and green tiles are mapped to green tiles.
This translates into the existence of nontrivial reflection symmetry and rotational symmetry for the Wang shift
${\Omega }_n$
. As we show in this article, it has no translational symmetries.
4.6 Directional determinism
We show in this section that the sets
${\mathcal {T}}_n$
and
${\mathcal {T}}_n^{\prime }$
are SW- and NE-deterministic.
Lemma 4.3. The sets
${\mathcal {T}}_n$
and
${\mathcal {T}}_n^{\prime }$
are SW- and NE-deterministic. However, the sets
${\mathcal {T}}_n$
and
${\mathcal {T}}_n^{\prime }$
are neither NW- nor SE-deterministic.
Proof. Let us show that
${\mathcal {T}}_n^{\prime }$
is SW-deterministic. Let
$s,t\in {\mathcal {T}}_n^{\prime }$
be such that 
and 
for some vectors
$u=(u_0,u_1,u_2),v=(v_0,v_1,v_2)\in V_n$
.
-
• If
$u_0=0$
,
$v_0=0$
, then
$s,t\in J^{\prime }_n$
. -
• If
$u_0=1$
,
$v_0=1$
, then
$s,t\in W_n$
. -
• If
$u_0=0$
,
$v_0=1$
,
$u_1=0$
and
$v_2=n$
, then
$s,t\in B^{\prime }_n$
. -
• If
$u_0=0$
,
$v_0=1$
,
$u_1=0$
and
$v_2=\overline {n}$
, then
$s,t\in G_n$
. -
• If
$u_0=0$
,
$v_0=1$
,
$u_1=1$
and
$v_2=n$
, then
$s,t\in A_n$
. -
• If
$u_0=0$
,
$v_0=1$
,
$u_1=1$
and
$v_2=\overline {n}$
, then
$s,t\in Y_n$
. -
• If
$u_0=1$
,
$v_0=0$
,
$v_1=0$
and
$u_2=n$
, then
$s,t\in \widehat {B^{\prime }_n}$
. -
• If
$u_0=1$
,
$v_0=0$
,
$v_1=0$
and
$u_2=\overline {n}$
, then
$s,t\in \widehat {G_n}$
. -
• If
$u_0=1$
,
$v_0=0$
,
$v_1=1$
and
$u_2=n$
, then
$s,t\in \widehat {A_n}$
. -
• If
$u_0=1$
,
$v_0=0$
,
$v_1=1$
and
$u_2=\overline {n}$
, then
$s,t\in \widehat {Y_n}$
.
One can observe that each of the subsets
$W_n$
,
$B^{\prime }_n$
,
$G_n$
,
$Y_n$
,
$A_n$
,
$J^{\prime }_n$
is SW-deterministic. By symmetry,
$\widehat {B^{\prime }_n}$
,
$\widehat {G_n}$
,
$\widehat {Y_n}$
and
$\widehat {A_n}$
are SW-deterministic. We conclude that
$s=t$
. Thus,
${\mathcal {T}}_n^{\prime }$
is SW-deterministic. Using a similar argument, one can observe that
${\mathcal {T}}_n^{\prime }$
is NE-deterministic. By restriction, the subset
${\mathcal {T}}_n\subset {\mathcal {T}}_n^{\prime }$
is SW- and NE-deterministic.
However,
${\mathcal {T}}_n$
is neither NW- nor SE-deterministic because the subset
$J_n$
is neither NW- nor SE-deterministic. By extension, the extended set
${\mathcal {T}}_n^{\prime }$
is neither NW- nor SE-deterministic.
5 A substitution
${\Omega }_n\to {\Omega }_n$
The goal of this section is twofold. First, we introduce a 2-dimensional substitution
${\Omega }_n\to {\Omega }_n$
deduced from a substitution
$\tau _n:V_n\to (V_n)^*$
defined on the boundary labels of the Wang tiles. Then, we characterize the possible valid rectangular tilings with external labels in the image of
$\tau _n$
; see Proposition 5.9.
5.1 A 1-dimensional substitution for the boundary
It is convenient to define, for every integer
$n\geq 1$
, the following map:
$$\begin{align*}\begin{array}{rccl} \tau_n:&V_n & \to & (V_n)^*\\ &xyz & \mapsto & \begin{cases} 0(x{-}y{+}1)n \cdot (11n)^{z-x-1} \cdot (11\overline{n})^{n+1-z} & \text{ if } x\neq z,\\ 0(x{-}y{+}1)\overline{n}\cdot (11\overline{n})^{n-z} & \text{ if } x=z. \end{cases} \end{array} \end{align*}$$
The above formula declines into the following five cases:
$$ \begin{align} \begin{aligned} \tau_n(000) &=01\overline{n}\cdot(11\overline{n})^{n},\\ \tau_n(111) &=01\overline{n}\cdot(11\overline{n})^{n-1},\\ \tau_n(00\overline{i})&=01n\cdot(11n)^{i}\cdot(11\overline{n})^{n-i} &&\qquad\text{ if }0\leq i\leq n,\\ \tau_n(01\overline{i})&=00n\cdot(11n)^{i}\cdot(11\overline{n})^{n-i} &&\qquad\text{ if }0\leq i\leq n,\\ \tau_n(11\overline{i})&=01n\cdot(11n)^{i-1}\cdot(11\overline{n})^{n-i} &&\qquad\text{ if }1\leq i\leq n. \end{aligned} \end{align} $$
For example, when
$n=1$
,
$n=2$
or
$n=4$
, we have
$$\begin{align*}\tau_1 \left\{\begin{array}{l} 000\mapsto 012,112\\ 001\mapsto 011,112\\ 002\mapsto 011,111\\ 011\mapsto 001,112\\ 012\mapsto 001,111\\ 111\mapsto 012\\ 112\mapsto 011\\ \end{array}\right., \qquad \tau_2 \left\{\begin{array}{l} 000\mapsto 013,113,113\\ 001\mapsto 012,113,113\\ 002\mapsto 012,112,113\\ 003\mapsto 012,112,112\\ 011\mapsto 002,113,113\\ 012\mapsto 002,112,113\\ 013\mapsto 002,112,112\\ 111\mapsto 013,113\\ 112\mapsto 012,113\\ 113\mapsto 012,112\\ \end{array}\right., \qquad \tau_4 \left\{\begin{array}{l} 000\mapsto 015,115,115,115,115\\ 001\mapsto 014,115,115,115,115\\ 002\mapsto 014,114,115,115,115\\ 003\mapsto 014,114,114,115,115\\ 004\mapsto 014,114,114,114,115\\ 005\mapsto 014,114,114,114,114\\ 011\mapsto 004,115,115,115,115\\ 012\mapsto 004,114,115,115,115\\ 013\mapsto 004,114,114,115,115\\ 014\mapsto 004,114,114,114,115\\ 015\mapsto 004,114,114,114,114\\ 111\mapsto 015,115,115,115\\ 112\mapsto 014,115,115,115\\ 113\mapsto 014,114,115,115\\ 114\mapsto 014,114,114,115\\ 115\mapsto 014,114,114,114\\ \end{array}\right.. \end{align*}$$
The map
$\tau _n$
was discovered during computer explorations. It appears naturally when searching for a self-similarity for the tilings in
${\Omega }_n$
; see Appendix B in Section B and in particular the output of the computation performed at line 84.
Lemma 5.1. For every
$(x,y,z)\in V_n$
, the map
$\tau _n$
satisfies the following:
-
• the length of
$\tau _n(xyz)\in (V_n)^*$
is
$n+1-x$
; -
• the first item of
$\tau _n(xyz)$
is
$0(x{-}y{+}1)n$
or
$0(x{-}y{+}1)\overline {n}$
; -
• there are
$z-x$
occurrences of
${*}{*}n$
in the image of
$\tau _n(xyz)$
.
In particular,
$\tau _n$
is injective.
Proof. The three items follow from the definition. We prove that
$\tau _n$
is injective. Assume that
$xyz\neq x'y'z'$
. We want to show that
$\tau _n(xyz)\neq \tau _n(x'y'z')$
. If
$x\neq x'$
, then the images are distinct because their lengths are different. If
$x=x'$
and
$y\neq y'$
, then the images are distinct because the second digit of their first item satisfies
$x{-}y{+}1\neq x{-}y'{+}1$
. If
$x=x'$
,
$y=y'$
and
$z\neq z'$
, then the images are distinct because there are
$z-x$
occurrences of
${*}{*}n$
in the image of
$\tau _n(xyz)$
.
5.2 A substitution
${\omega }_n^{\prime }$
for the tiles in
${\mathcal {T}}_n^{\prime }$
Let
be the set of possible values for the bottom right part of a junction tile in
$J_n^{\prime }$
.
Lemma 5.2. Let
$n\geq 1$
be an integer. For every
$v\in V_n$
, there exist a unique bottom right part
$q\in Q_n$
and a unique sequence
$t_1t_2\dots t_{k-1}$
of tiles in
${\mathcal {T}}_n^{\prime }$
such that
$q\,t_1t_2\dots t_{k-1}$
is a valid horizontal strip of tiles from left to right whose sequence of bottom labels is
$\tau _n(v)$
where
$k=|\tau _n(v)|$
.
Moreover, if
$\gamma $
is the sequence of top labels of
$t_1t_2\dots t_{k-1}$
and
$\delta $
is its right-most right label – that is, the right label of
$t_{k-1}$
(see Figure 11) – then the following statements hold.
-
• If
$v=00i$
with
$0\leq i\leq \overline {n}$
, then-
– if
$0\leq i\leq n$
, then
$\gamma =(111)^i\cdot (112)^{n-i}$
and
$\delta =01\overline {n}$
, -
– if
$i=n+1$
, then
$\gamma =(111)^{n}$
and
$\delta =00\overline {n}$
.
-
-
• If
$v=01i$
with
$1\leq i\leq \overline {n}$
, then-
– if
$1\leq i\leq n$
, then
$\gamma =(111)^i\cdot (112)^{n-i}$
and
$\delta =01n$
, -
– if
$i=n+1$
, then
$\gamma =(111)^{n}$
and
$\delta =00n$
.
-
-
• If
$v=11i$
with
$1\leq i\leq \overline {n}$
, then-
– if
$1\leq i\leq n$
, then
$\gamma =(111)^{i-1}\cdot (112)^{n-i}$
and
$\delta =01n$
, -
– if
$i=n+1$
, then
$\gamma =(111)^{n-1}$
and
$\delta =00n$
.
-
In particular, no antigreen tiles appear in the horizontal strip. Also, if
$v\in V_n\setminus \{00\overline {n}\}$
, then the last blue tile does not appear in the strip.
Figure 11 A horizontal strip of tiles from
${\mathcal {T}}_n^{\prime }$
made of a bottom right part q of a junction tile and a sequence
$t_1t_2\dots t_{k-1}$
of horizontal stripe tiles. The bottom labels of the strip is
$\tau _n(v)$
for some
$v\in V_n$
. The top labels of the horizontal stripe tiles is
$\gamma \in (V_n)^*$
and its right-most right label is
$\delta \in V_n$
.
Proof. Assume
$v=00i$
with
$0\leq i\leq \overline {n}$
. The following three cases occur.
-
• If
$i=0$
, then the sequence of bottom labels is
$\tau _n(000) = 01\overline {n}\cdot (11\overline {n})^n$
, the sequence of top labels is
$(112)^n$
and the right-most right label is
$01\overline {n}$
. -
• If
$1\leq i\leq n$
, then the sequence of bottom labels is
$\tau _n(00i) = 01n\cdot (11n)^{i-1}\cdot (11\overline {n})^{n+1-i}$
, the sequence of top labels is
$(111)^i\cdot (112)^{n-i}$
and the right-most right label is
$01\overline {n}$
. -
• If
$i=n+1$
, then the sequence of bottom labels is
$\tau _n(00i) = 01n\cdot (11n)^{n}$
, the sequence of top labels is
$(111)^{n}$
and the right-most right label is
$00\overline {n}$
.
The
$n+1$
tiles of the strip for the three cases are illustrated in Figure 12. We observe that the last blue tile (the blue horizontal stripe tile with left label
$00n$
) is used in the strip only when
$i=n+1$
.
Figure 12 Horizontal strip with bottom word
$\tau _n(00i)$
with
$0\leq i\leq n$
.
Assume
$v=01i$
with
$1\leq i\leq \overline {n}$
. The following two cases occur.
-
• If
$1\leq i\leq n$
, then the sequence of bottom labels is
$\tau _n(01i) = 00n\cdot (11n)^{i-1}\cdot (11\overline {n})^{n+1-i}$
, the sequence of top labels is
$(111)^i\cdot (112)^{n-i}$
and the right-most right label is
$01n$
. -
• If
$i=n+1$
, then the sequence of bottom labels is
$\tau _n(00i) = 00n\cdot (11n)^{n}$
, the sequence of top labels is
$(111)^{n}$
and the right-most right label is
$00n$
.
The
$n+1$
tiles of the strip for the two cases are illustrated in Figure 13.
Figure 13 Horizontal strip with bottom word
$\tau _n(01i)$
with
$1\leq i\leq n+1$
.
Assume
$v=11i$
with
$1\leq i\leq \overline {n}$
. The following three cases occur.
-
• If
$i=1$
, then the sequence of bottom labels is
$\tau _n(111) = 01\overline {n}\cdot (11\overline {n})^{n-1}$
, the sequence of top labels is
$(112)^{n-1}$
and the right-most right label is
$01n$
. -
• If
$2\leq i\leq n$
, then the sequence of bottom labels is
$\tau _n(11i) = 01n\cdot (11n)^{i-2}\cdot (11\overline {n})^{n+1-i}$
, the sequence of top labels is
$(111)^{i-1}\cdot (112)^{n-i}$
and the right-most right label is
$01n$
. -
• If
$i=n+1$
, then the sequence of bottom labels is
$\tau _n(11i) = 01n\cdot (11n)^{n-1}$
, the sequence of top labels is
$(111)^{n-1}$
and the right-most right label is
$00n$
.
The n tiles of the strip for the three cases are illustrated in Figure 14.
Figure 14 Horizontal strip with bottom word
$\tau _n(11i)$
with
$1\leq i\leq n+1$
.
Since
${\mathcal {T}}_n=\widehat {{\mathcal {T}}_n}$
, Lemma 5.2 has a symmetric version describing the vertical strip of tiles from
${\mathcal {T}}_n^{\prime }$
with left labels equal to
$\tau _n(u)$
for some
$u\in V_n$
. Lemma 5.2 and its symmetric version can be used together to construct valid rectangular patterns with external boundaries given by the images under the map
$\tau _n$
; see Figure 16.
Figure 15 The global shape of a rectangular pattern whose sequence of bottom labels is
$\tau _n(v)$
and sequence of left labels is
$\tau _n(u)$
. The pattern is split into four disjoint parts: the junction tile, the left column, the bottom row and the white tiles.
Figure 16 Left: some antigren tile in
${\mathcal {T}}_4^{\prime }$
. Middle: the images under
$\tau _4$
of the labels of the tile form the boundary labels of a rectangle. Right: there is a unique rectangular pattern with such boundary words and tiles in
${\mathcal {T}}_4^{\prime }$
. As shown in Lemma 5.3, this holds for every
$n\geq 1$
and for every tile in
${\mathcal {T}}_n^{\prime }$
allowing to define the map
${\omega }_n^{\prime }$
.
Lemma 5.3. Let
$\alpha ,\beta ,u,v\in V_n$
. If 
, then there exists a unique valid rectangular pattern with tiles in
${\mathcal {T}}_n^{\prime }$
whose right, top, left and bottom labels are respectively
$\tau _n(\alpha )$
,
$\tau _n(\beta )$
,
$\tau _n(u)$
and
$\tau _n(v)$
.
Proof. Let
$u,v\in V_n$
. For every tile in
${\mathcal {T}}_n^{\prime }$
, the left label starts with
$0$
if and only if the right label starts with 0, and, symmetrically, the bottom label starts with 0 if and only if the top label starts with 0. Since we have
$\tau _n(V_n)\subset \{00n,01n,01\overline {n}\}\cdot \{11n,11\overline {n}\}^*$
, any valid rectangular pattern with tiles in
${\mathcal {T}}_n^{\prime }$
whose sequence of bottom labels is
$\tau _n(v)$
and sequence of left labels is
$\tau _n(u)$
can be split into four disjoint regions: a junction tile at the bottom left corner, a row of horizontal stripe tiles at the bottom, a column of vertical stripe tiles on the left and a rectangular pattern of white tiles for the remaining rectangle; see Figure 15.
(Existence) Let
$u,v,\alpha ,\beta \in V_n$
be such that 
. First, we show that the junction tile at the bottom left corner of the rectangular pattern with bottom labels
$\tau _n(v)$
and left labels
$\tau _n(u)$
is one of the 9 junction tile in
${\mathcal {T}}_n^{\prime }$
. For every
$u,v\in V_n$
, we have
$\tau _n(u),\tau _n(v)\in \{00n,01n,01\overline {n}\}\cdot (V_n)^*$
. For every
$x,y\in \{00n,01n,01\overline {n}\}$
, there exists a unique junction tile in
${\mathcal {T}}_n^{\prime }$
with bottom label x and left label y.
It remains to show the existence of tiles from
${\mathcal {T}}_n^{\prime }$
to cover the bottom row, the left column and the region of white tiles while respecting the label constraints; see Figure 15. Again, we proceed case by case.
Suppose that t is a junction tile in
${\mathcal {T}}_n^{\prime }$
; that is,
$u,v\in \{00n,01n,01\overline {n}\}$
. We have
$|\tau _n(u)|=|\tau _n(v)|=n+1$
. In order to formalize the argument that follows, it is practical to define the following two maps on the subset
$\{00n,01n,01\overline {n}\}\subset V_n$
:
$$\begin{align*}\begin{array}{rccl} \sigma:&\{00n,01n,01\overline{n}\}&\to& V_n\\ &00n &\mapsto& 01\overline{n},\\ &01n &\mapsto& 01n,\\ &01\overline{n}&\mapsto& 00n, \end{array} \quad \text{ and } \quad \begin{array}{rccl} \mu:&\{00n,01n,01\overline{n}\}&\to& V_n\\ &00n &\mapsto& 000,\\ &01n &\mapsto& 001,\\ &01\overline{n}&\mapsto& 011. \end{array} \end{align*}$$
Notice that
$\alpha =\mu (v)$
and
$\beta =\mu (u)$
and
$\sigma $
is an involution. Also, if
$v\in \{00n,01n,01\overline {n}\}$
, then
$\tau _n\circ \mu (v)=\sigma (v)\cdot (11\overline {n})^n$
. From Lemma 5.2, there exists a unique choice of tiles for the bottom row whose sequence of top labels is
$(111)^n$
and right-most right label is
$01\overline {n}$
if
$v=00n$
,
$01n$
if
$v=01n$
,
$00n$
if
$v=01\overline {n}$
. In other words, the right-most right label of the bottom row is
$\sigma (v)$
. Symmetrically, there exists a unique choice of tiles for the left column whose sequence of right labels is
$(111)^n$
and top-most top label is
$\sigma (u)$
. Since the bottom row is of length n, and white tiles increase the last digit by one, the remaining region can be uniquely filled with white tiles such that the sequence of right labels of the rectangular pattern is
$\sigma (v)\cdot (11\overline {n})^n=\tau _n\circ \mu (v)=\tau _n(\alpha )$
. Symmetrically, the sequence of top labels of the rectangular pattern is
$\sigma (u)\cdot (11\overline {n})^n=\tau _n\circ \mu (u)=\tau _n(\beta )$
.
Suppose that 
is a white tile in
${\mathcal {T}}_n^{\prime }$
; that is,
$u=11j$
with
$1\leq j\leq n$
and
$v=11i$
with
$1\leq i\leq n$
. Also,
$\alpha =11\overline {j}$
and
$\beta =11\overline {i}$
. We have
$|\tau _n(u)|=|\tau _n(v)|=n$
. From Lemma 5.2, there exists a unique choice of tiles for the bottom row whose sequence of top labels is
$(111)^{i-1}\cdot (112)^{n-i}$
and right-most right label is
$01n$
. From a symmetric version of Lemma 5.2, there exists a unique choice of tiles for the left column whose sequence of right labels is
$(111)^{j-1}\cdot (112)^{n-j}$
and top-most top label is
$01n$
. The remaining region can be uniquely filled with white tiles. In this case, the sequence of right labels of the rectangular pattern is
$01n\cdot (11n)^{j-1}\cdot (11\overline {n})^{n-j}=\tau _n(11\overline {j})=\tau _n(\alpha )$
. Symmetrically, the sequence of top labels of the rectangular pattern is
$01n\cdot (11n)^{i-1}\cdot (11\overline {n})^{n-i}=\tau _n(11\overline {i})=\tau _n(\beta )$
.
Suppose that t is a horizontal stripe tile in
${\mathcal {T}}_n^{\prime }$
. We have
$u=00j\text { with }0\leq j\leq n$
or
$u=01j\text { with }1\leq j\leq n$
. Also
$v\in \{11n,11\overline {n}\}$
. Let
$$\begin{align*}\beta= \begin{cases} 111 & \text{ if } u=00j\text{ with }0\leq j\leq n,\\ 112 & \text{ if } u=01j\text{ with }1\leq j\leq n, \end{cases} \quad \text{ and } \quad \alpha= \begin{cases} 00\overline{j} & \text{ if } v=11n,\\ 01\overline{j} & \text{ if } v=11\overline{n}. \end{cases} \end{align*}$$
Also,
$|\tau _n(u)|=n+1$
and
$|\tau _n(v)|=n$
. There are two cases for v:
-
• If
$v=11n$
, then from Lemma 5.2, there exists a unique choice of tiles for the bottom row whose sequence of top labels is
$(111)^{n-1}$
and right-most right label is
$01n$
. -
• If
$v=11\overline {n}$
, then from Lemma 5.2, there exists a unique choice of tiles for the bottom row whose sequence of top labels is
$(111)^{n-1}$
and right-most right label is
$00n$
.
There are two cases for u:
-
• If
$u=00j$
with
$0\leq j\leq n$
, then from the symmetric version of Lemma 5.2, there exists a unique choice of tiles for the left column whose sequence of right labels is
$(111)^{j}\cdot (112)^{n-j}$
and top-most top label is
$01\overline {n}$
. -
• If
$u=01j$
with
$1\leq j\leq n$
, then from the symmetric version of Lemma 5.2, there exists a unique choice of tiles for the left column whose sequence of right labels is
$(111)^{j}\cdot (112)^{n-j}$
and top-most top label is
$01n$
.
Thus, the remaining region can be uniquely filled with white tiles, and the sequence of right labels of the rectangular pattern is
$$\begin{align*}\tau_n(\alpha)= \begin{cases} 01n\cdot(11n)^{j}\cdot(11\overline{n})^{n-j}=\tau_n(00\overline{j}) & \text{ if } v=11n,\\ 00n\cdot(11n)^{j}\cdot(11\overline{n})^{n-j}=\tau_n(01\overline{j}) & \text{ if } v=11\overline{n}. \end{cases} \end{align*}$$
Symmetrically, the sequence of top labels of the rectangular pattern is
$$\begin{align*}\tau_n(\beta)= \begin{cases} 01\overline{n}\cdot(11\overline{n})^{n-1}=\tau_n(111) & \text{ if } u=00j \text{ with } 0\leq j\leq n,\\ 01n\cdot(11\overline{n})^{n-1}=\tau_n(112) & \text{ if } u=01j\text{ with }1\leq j\leq n. \end{cases} \end{align*}$$
Suppose that t is a vertical stripe tile in
${\mathcal {T}}_n^{\prime }$
. A rectangular pattern respecting the constraints can be obtained by taking the image under reflection of the rectangular pattern constructed above for when t is a horizontal stripe tile in
${\mathcal {T}}_n^{\prime }$
.
(Uniqueness) Uniqueness follows from Lemma 2.1 and Lemma 4.3.
Following Lemma 5.3, we define the following map:
For example, the map
$\omega _1^{\prime }$
is illustrated in Figure 17.
Figure 17 The substitution
$\omega _1^{\prime }$
. An
$\times $
-mark indicates the tiles in
$J_1^{\prime }\setminus J_1$
.
Lemma 5.4. The map
${\omega }_n^{\prime }$
defines a
$2$
-dimensional substitution
${\omega }_n^{\prime }:{\Omega }_n^{\prime }\to {\Omega }_n^{\prime }$
.
Proof. From Lemma 5.3, for every tile
$t\in {\mathcal {T}}_n^{\prime }$
, the image
${\omega }_n^{\prime }(t)$
is a valid rectangular pattern over the Wang tiles
${\mathcal {T}}_n^{\prime }$
. Moreover, if
$s\odot ^1t\in ({\mathcal {T}}_n^{\prime })^{*^2}$
is a valid horizontal domino, then
${\omega }_n^{\prime }(s\odot ^1t)$
is a valid rectangular pattern over the Wang tiles
${\mathcal {T}}_n^{\prime }$
. Similarly, if
$s\odot ^2t\in ({\mathcal {T}}_n^{\prime })^{*^2}$
is a valid vertical domino, then
${\omega }_n^{\prime }(s\odot ^2t)$
is a valid rectangular pattern over the Wang tiles
${\mathcal {T}}_n^{\prime }$
. Thus, if
$y\in {\Omega }_n^{\prime }$
is a valid configuration, then
${\omega }_n^{\prime }(y)$
is also a valid configuration. Therefore,
${\omega }_n^{\prime }(y)\in {\Omega }_n^{\prime }$
.
5.3 A substitution
${\omega }_n$
for the tiles in
${\mathcal {T}}_n$
Not all tiles of
${\mathcal {T}}_n^{\prime }$
appear in the image of a tile under the substitution
${\omega }_n^{\prime }$
. For example, it follows from Lemma 5.2 that antigreen stripe tiles do not appear in the images of tiles under
${\omega }_n^{\prime }$
. Therefore, the substitution
${\omega }_n^{\prime }$
is not primitive.
As it can be observed in Figure 17, some tiles in
${\mathcal {T}}_1^{\prime }\setminus {\mathcal {T}}_1$
appear in the images of
$\omega _1^{\prime }$
. Namely, the images of the antigreen tiles under
$\omega _1^{\prime }$
contain junction tiles in
$J_1^{\prime }\setminus J_1=\{j_1^{0,0,1,1},j_1^{1,1,0,0}\}$
. As shown in the next lemma, this is the only exception.
Lemma 5.5. Let
$n\geq 1$
be an integer and
$t\in {\mathcal {T}}_n^{\prime }$
. The pattern
${\omega }_n^{\prime }(t)$
contains a tile in
${\mathcal {T}}_n^{\prime }\setminus {\mathcal {T}}_n$
if and only if
$n=1$
and t is an antigreen tile.
Proof. Let
$n\geq 1$
be an integer. (
$\impliedby $
) If
$n=1$
, the set of antigreen tiles in
${\mathcal {T}}_1^{\prime }$
is
$A_1\cup \widehat {A_1}=\{a_1^1,\widehat {a_1^1}\}$
. In Figure 17, we observe that
$\omega _1^{\prime }(a_1^1)$
contains the junction tile
$j_1^{1,1,0,0}\in {\mathcal {T}}_1^{\prime }\setminus {\mathcal {T}}_1$
and
$\omega _1^{\prime }(\widehat {a_1^1})$
contains the junction tile
$j_1^{0,0,1,1}\in {\mathcal {T}}_1^{\prime }\setminus {\mathcal {T}}_1$
.
(
$\implies $
) Let 
. The labels of the boundary of
${\omega }_n^{\prime }(t)$
are 
. Suppose that the pattern
${\omega }_n^{\prime }(t)$
contains a tile in
${\mathcal {T}}_n^{\prime }\setminus {\mathcal {T}}_n$
. We have
$v\in V_n\setminus \{00\overline {n}\}$
. From Lemma 5.2, the bottom row of the pattern
${\omega }_n^{\prime }(t)$
does not contain the last blue tile. Also,
$u\in V_n\setminus \{00\overline {n}\}$
. From the symmetric version of Lemma 5.2, the left column of the pattern
${\omega }_n^{\prime }(t)$
does not contain the last blue tile. From Lemma 5.2, the pattern
${\omega }_n^{\prime }(t)$
does not contain any antigreen (vertical or horizontal) stripe tile. Therefore, the pattern
${\omega }_n^{\prime }(t)$
must contain a junction tile in
$J_n^{\prime }\setminus J_n=\{j_n^{1,1,0,0},j_n^{0,0,1,1}\}$
.
Suppose that
${\omega }_n^{\prime }(t)$
contains the junction tile
$j_n^{1,1,0,0}$
. Therefore, we must have
$\tau _n(v)\in 01\overline {n}\cdot (V_n)^*$
and
$\tau _n(u)\in 00n\cdot (V_n)^*$
. Thus,
$v\in \{000,111\}$
and
$u=01j$
with
$1\leq j\leq n$
. We proceed case by case.
-
• Assume
$v=000$
. The only tile
$t\in {\mathcal {T}}_n^{\prime }$
with bottom label
$v=000$
is a blue or green vertical stripe tile whose left label is
$u=11n$
or
$u=11\overline {n}$
, a contradiction. -
• Assume
$v=111$
and
$n>1$
. The only tile
$t\in {\mathcal {T}}_n^{\prime }$
with bottom label
$v=111$
is a white tile whose left label is
$u=11i$
with
$1\leq i\leq n$
, a contradiction. -
• Assume
$v=111$
and
$n=1$
. The only tile
$t\in {\mathcal {T}}_1^{\prime }$
with bottom label
$v=111$
is a white tile whose left label is
$u=111$
, a blue horizontal stripe tile whose left label is
$000$
or
$001$
, or an antigreen tile
$a_1^1$
whose left label is
$011$
. Only the antigreen tile does not yield a contradiction with the value of u given above. Thus,
$t=a_1^1$
.
Symmetrically, if
${\omega }_n^{\prime }(t)$
contains the junction tile
$j_n^{0,0,1,1}$
, we conclude that
$n=1$
and
$t=\widehat {a_1^1}$
.
A consequence of Lemma 5.5 is that if
$n\geq 2$
and
$t\in {\mathcal {T}}_n^{\prime }$
, then the pattern
${\omega }_n^{\prime }(t)$
contains only tiles from
${\mathcal {T}}_n$
. Also for every
$n\geq 1$
and
$t\in {\mathcal {T}}_n$
, the pattern
${\omega }_n^{\prime }(t)$
contains only tiles from
${\mathcal {T}}_n$
. Thus, it becomes natural to restrict the substitution
${\omega }_n^{\prime }$
to the set
${\mathcal {T}}_n$
. We obtain the following map
${\omega }_n={\omega }_n^{\prime }|_{{\mathcal {T}}_n}$
:
The substitutions
${\omega }_n$
for
$n=1,\dots ,5$
are illustrated in Figure 31, Figure 32, Figure 33, Figure 34 and Figure 35 (in the appendix).
Lemma 5.6. The map
${\omega }_n$
defines a
$2$
-dimensional substitution
${\omega }_n:{\Omega }_n\to {\Omega }_n$
such that
$\overline {{\omega }_n({\Omega }_n)}^\sigma \subset {\Omega }_n$
.
Proof. From Lemma 5.4, the map
${\omega }_n^{\prime }$
defines a
$2$
-dimensional substitution
${\Omega }_n^{\prime }\to {\Omega }_n^{\prime }$
. From Lemma 5.5,
${\omega }_n^{\prime }(x)\in {\Omega }_n$
for every configuration
$x\in {\Omega }_n$
. Thus,
${\omega }_n^{\prime }({\Omega }_n)\subseteq {\Omega }_n$
. The restriction of
${\omega }_n^{\prime }$
to
${\Omega }_n$
is
${\omega }_n$
, so that
${\omega }_n({\Omega }_n)\subseteq {\Omega }_n$
. Since
${\Omega }_n$
is a subshift, it is closed under the shift. Therefore,
$\overline {{\omega }_n({\Omega }_n)}^\sigma \subset {\Omega }_n$
.
The goal of the next section is to show that
${\Omega }_n=\overline {{\omega }_n({\Omega }_n)}^\sigma $
– namely, that every configuration in
${\Omega }_n$
can be desubstituted using
${\omega }_n$
. The proof of this is completed in Section 8. Following the above discussion, the
$2$
-dimensional substitution
${\omega }_n^{\prime }$
is not primitive, but we show in Section 9 that the substitution
${\omega }_n$
is primitive.
5.4 A sufficient and necessary condition
The goal of this section is to show that the sufficiency in the statement of Lemma 5.3 is also a necessity – namely, that every rectangular pattern, with external boundary labeled by images under
$\tau _n$
, is obtained from a tile in
${\mathcal {T}}_n^{\prime }$
. The precise statement is given in Proposition 5.9.
For every integer
$n\geq 1$
, let
$$\begin{align*}Z_n=\{v_0v_1v_2\in V_n \mid v_0 = 0\} \end{align*}$$
be the set of vectors of
$V_n$
such that the first entry is zero and let
$$\begin{align*}M_n=\{v_0v_1v_2\in V_n \mid v_2\geq n\} \end{align*}$$
be the set of vectors of
$V_n$
such that the last entry is n or
$\overline {n}$
.
Lemma 5.7. If
$$ \begin{align} \begin{aligned} (u,v) &\in (\{11\overline{n}\}\times V_n\setminus Z_n) \cup(V_n\setminus Z_n\times \{11\overline{n}\})\\ &\qquad\cup(\{00\overline{n}\}\times M_n\cap Z_n) \cup(M_n\cap Z_n\times \{00\overline{n}\})\\ &\qquad\cup(\{00\overline{n},01\overline{n}\}\times M_n\setminus Z_n) \cup(M_n\setminus Z_n\times \{00\overline{n},01\overline{n}\}), \end{aligned} \end{align} $$
then there exists a unique valid rectangular pattern with tiles in
${\mathcal {T}}_n^{\prime }$
whose right, top, left and bottom labels are respectively R, T,
$\tau _n(u)$
and
$\tau _n(v)$
for some
$R,T\in (V_n)^*$
, and there is no
$(\alpha ,\beta )\in V_n\times V_n$
such that
$R=\tau _n(\alpha )$
and
$T=\tau _n(\beta )$
.
Proof. Suppose that
$u\in V_n\setminus Z_n$
and
$v=11\overline {n}$
. We have
$|\tau _n(u)|=|\tau _n(v)|=n$
. Since
$v=11\overline {n}$
, then from Lemma 5.2, there exists a unique choice of tiles for the bottom row whose sequence of top labels is
$(111)^{n-1}$
and right-most right label is
$00n$
. There are two cases to consider for u:
-
• If
$u=11j$
with
$1\leq j\leq n$
, then from a symmetric version of Lemma 5.2, there exists a unique choice of tiles for the left column whose sequence of right labels is
$(111)^{j-1}\cdot (112)^{n-j}$
. -
• If
$u=11\overline {n}$
, then from a symmetric version of Lemma 5.2, there exists a unique choice of tiles for the left column whose sequence of right labels is
$(111)^{n-1}$
.
In both cases, the remaining region of the rectangular pattern can be uniquely filled with white tiles. The sequence of right labels of the rectangular pattern starts with
$00n$
. Such a sequence cannot be written as an image under the map
$\tau _n$
because there is no
$\alpha \in V_n$
such that
$\tau _n(\alpha )$
starts with
$00n$
and is of length n.
Suppose that
$u\in M_n\cap Z_n=\{00n,00\overline {n},01n,01\overline {n}\}$
and
$v=00\overline {n}$
. We have
$|\tau _n(u)|=|\tau _n(v)|=n+1$
. From Lemma 5.2, there exists a unique choice of tiles for the bottom row whose sequence of top labels is
$(111)^n$
. Since
$v=00\overline {n}$
, from a symmetric version of Lemma 5.2, there exists a unique choice of tiles for the left column whose sequence of right labels is
$(111)^n$
and top-most top label is
$00\overline {n}$
. Since the bottom row is of length n, and white tiles increase the last digit by one, the remaining region can be uniquely filled with white tiles. Since the sequence of top labels of the rectangular pattern starts with
$00\overline {n}$
, it cannot be written as an image under the map
$\tau _n$
.
Suppose that
$u\in \{00\overline {n},01\overline {n}\}$
and
$v\in M_n\setminus Z_n=\{11n,11\overline {n}\}$
. We have
$|\tau _n(u)|=n+1$
and
$|\tau _n(v)|=n$
. From Lemma 5.2, there exists a unique choice of tiles for the bottom row whose sequence of top labels is
$(111)^n$
. Symmetrically, there exists a unique choice of tiles for the left column whose sequence of right labels is
$(111)^n$
and top-most top label is in
$\{00n,00\overline {n}\}$
. The remaining region can be uniquely filled with white tiles. The sequence of top labels is in
$\{00n,00\overline {n}\}\cdot (11\overline {n})^{n-1}$
. Such a sequence cannot be written as an image under the map
$\tau _n$
because there is no
$\alpha \in V_n$
such that
$\tau _n(\alpha )$
starts with
$00n$
or
$00\overline {n}$
and is of length n.
Suppose that
$u=11\overline {n}$
and
$v\in V_n\setminus Z_n$
, or
$u=00\overline {n}$
and
$v\in M_n\cap Z_n$
, or
$u\in M_n\setminus Z_n$
and
$v\in \{00\overline {n},01\overline {n}\}$
. A rectangular pattern respecting the constraints can be obtained by taking the image under reflection of the rectangular pattern constructed above.
Proposition 5.8. Let
$u,v\in V_n$
. There exists a valid rectangular pattern of tiles in
${\mathcal {T}}_n^{\prime }$
whose sequence of bottom labels is
$\tau _n(v)$
and sequence of left labels is
$\tau _n(u)$
if and only if
$$ \begin{align} (u,v) \in (V_n\setminus Z_n\times V_n\setminus Z_n) \cup(M_n\cap Z_n\times M_n\cap Z_n) \cup(M_n\setminus Z_n\times Z_n) \cup(Z_n\times M_n\setminus Z_n). \end{align} $$
Proof. Let
$u,v\in V_n$
. (
$\implies $
) We show the contrapositive – namely, that if (5.5) does not hold, then there is no rectangular pattern with
$\tau _n(u)$
as the sequence of labels on the left and
$\tau _n(v)$
as the sequence of labels at the bottom. If (5.5) does not hold, then
$$ \begin{align*} (u,v)\in &\big((Z_n \times V_n\setminus Z_n)\cup (V_n\setminus Z_n \times Z_n)\cup (Z_n \times Z_n)\big)\\ &\qquad\qquad\setminus \big( (M_n\cap Z_n\times M_n\cap Z_n) \cup(M_n\setminus Z_n\times Z_n) \cup(Z_n\times M_n\setminus Z_n) \big) \\ &=(Z_n \times V_n\setminus (M_n\cup Z_n))\cup (V_n\setminus (M_n\cup Z_n) \times Z_n) \cup(Z_n \times Z_n\setminus M_n) \cup(Z_n\setminus M_n \times Z_n). \end{align*} $$
There are four cases to consider:
-
• Assume
$u\in Z_n$
and
$v\in V_n\setminus (M_n\cup Z_n)$
. We have
$v=11j$
with
$1\leq j<n$
. From Lemma 5.2, the bottom row of the rectangular pattern has at least one label
$112$
on its top. Since the difference between the last digit of the top label and the last digit of the bottom label of a white tile is 1 and the maximal last digit of a white tile in
${\mathcal {T}}_n$
is
$\overline {n}$
, the height of the white tile region is at most
$n-1$
; see Figure 18. Thus,
$|\tau _n(u)|\leq n$
. This is incompatible with
$u\in Z_n$
because
$u\in Z_n$
implies that
$|\tau _n(u)|=n+1$
.Figure 18 The height of a valid vertical column made entirely of white tiles from
${\mathcal {T}}_n$
is at most
$n-1$
if the bottom label of the bottom-most tile is
$112$
or if the top label of the top-most tile is
$11n$
. -
• Assume
$u\in V_n\setminus (M_n\cup Z_n)$
and
$v\in Z_n$
. This case also leads to a contradiction following an argument symmetric to the previous one. -
• Assume
$u\in Z_n$
and
$v\in Z_n\setminus M_n$
. We have
$v=00j$
with
$0\leq j<n$
or
$v=01j$
with
$1\leq j<n$
. In both cases, we have from Lemma 5.2 that the bottom row of the rectangular pattern has at least one label
$112$
on its top. For the same reason as in the first item, the height of the rectangular pattern is
$|\tau _n(u)|\leq n$
. This is incompatible with
$u\in Z_n$
because
$u\in Z_n$
implies that
$|\tau _n(u)|=n+1$
. -
• Assume
$u\in Z_n\setminus M_n$
and
$v\in Z_n$
. This case also leads to a contradiction following an argument symmetric to the previous one.
(
$\impliedby $
) Let
Notice that
$Q\subset P$
and
$$ \begin{align} \begin{aligned} P\setminus Q &= (\{11\overline{n}\}\times V_n\setminus Z_n) \cup(V_n\setminus Z_n\times \{11\overline{n}\})\\ &\qquad\cup(\{00\overline{n}\}\times M_n\cap Z_n) \cup(M_n\cap Z_n\times \{00\overline{n}\})\\ &\qquad\cup(\{00\overline{n},01\overline{n}\}\times M_n\setminus Z_n) \cup(M_n\setminus Z_n\times \{00\overline{n},01\overline{n}\}). \end{aligned} \end{align} $$
We assume that (5.5) holds; that is,
$(u,v)\in P$
. There are two cases to consider.
-
• If
$(u,v)\in Q$
, then, from Lemma 5.3, there exists a valid rectangular pattern with tiles in
${\mathcal {T}}_n^{\prime }$
whose left and bottom labels are respectively
$\tau _n(u)$
and
$\tau _n(v)$
. -
• If
$(u,v)\in P\setminus Q$
, then using (5.6) and Lemma 5.7, there exists a valid rectangular pattern with tiles in
${\mathcal {T}}_n^{\prime }$
whose left and bottom labels are respectively
$\tau _n(u)$
and
$\tau _n(v)$
.
Proposition 5.9. Let
$\alpha ,\beta ,u,v\in V_n$
. There exists a valid rectangular pattern with tiles in
${\mathcal {T}}_n^{\prime }$
whose right, top, left and bottom labels are respectively
$\tau _n(\alpha )$
,
$\tau _n(\beta )$
,
$\tau _n(u)$
and
$\tau _n(v)$
if and only if 
.
Proof. Let
$\alpha ,\beta ,u,v\in V_n$
. (
$\impliedby $
) The existence of the rectangular pattern was proved in Lemma 5.3.
(
$\implies $
) Suppose that there exists a valid rectangular pattern with tiles in
${\mathcal {T}}_n^{\prime }$
whose right, top, left and bottom labels are respectively
$\tau _n(\alpha )$
,
$\tau _n(\beta )$
,
$\tau _n(u)$
and
$\tau _n(v)$
. From Proposition 5.8,
$(u,v)$
satisfies (5.5); that is
$(u,v)\in P$
. From Lemma 5.7,
$(u,v)$
does not satisfy (5.4) because all boundary words can be written as the image of
$\tau _n$
. Thus,
$(u,v)\notin P\setminus Q$
using (5.6). We conclude that
$(u,v)\in Q$
. Thus, there exists
$\alpha ',\beta '\in V_n$
such that 
. From Lemma 5.3, there exists a valid rectangular pattern with tiles in
${\mathcal {T}}_n^{\prime }$
whose right, top, left and bottom labels are respectively
$\tau _n(\alpha ')$
,
$\tau _n(\beta ')$
,
$\tau _n(u)$
and
$\tau _n(v)$
. From Lemma 2.1, we must have
$\tau _n(\alpha ')=\tau _n(\alpha )$
and
$\tau _n(\beta ')=\tau _n(\beta )$
because
${\mathcal {T}}_n^{\prime }$
is SW-deterministic from Lemma 4.3. Since
$\tau _n$
is injective over the set
$V_n$
, we have
$\alpha =\alpha '$
and
$\beta =\beta '$
.
Proposition 5.9 is used in Lemma 6.6 in order to desubstitute configurations in
${\Omega }_n$
over tiles in
${\mathcal {T}}_n$
. Nevertheless, considering tiles in
${\mathcal {T}}_n^{\prime }$
is necessary for Proposition 5.9 to hold for every integer
$n\geq 1$
. Following Lemma 5.5, Proposition 5.9 can be restated as follows when
$n\geq 2$
.
Corollary 5.10. Suppose that
$n\geq 2$
is an integer and let
$\alpha ,\beta ,u,v\in V_n$
. There exists a valid rectangular pattern with tiles in
${\mathcal {T}}_n$
whose right, top, left and bottom labels are respectively
$\tau _n(\alpha )$
,
$\tau _n(\beta )$
,
$\tau _n(u)$
and
$\tau _n(v)$
if and only if 
.
Proof. Let
$\alpha ,\beta ,u,v\in V_n$
. (
$\implies $
) follows from Proposition 5.9 since
${\mathcal {T}}_n\subset {\mathcal {T}}_n^{\prime }$
.
(
$\impliedby $
) From Lemma 5.5, for every tile
$t\in {\mathcal {T}}_n^{\prime }$
, the rectangular pattern
${\omega }_n^{\prime }(t)$
satisfies the boundary conditions, and it contains only the tiles from the set
${\mathcal {T}}_n$
.
6 A desubstitution
${\Omega }_n\leftarrow \Omega ^{\prime }_n$
In this section, we decompose configurations in
${\Omega }_n^{\prime }$
and in
${\Omega }_n$
into rectangular blocks called return blocks. The external boundary labels of the return blocks within a configuration in
${\Omega }_n$
behave like a new set
${\mathcal {T}}_n^{\prime }$
of Wang tiles which contains
${\mathcal {T}}_n$
as a subset.
6.1 Return blocks in the Wang shift
${\Omega }_n^{\prime }$
In this section, we study some properties of the Wang shift
${\Omega }_n^{\prime }$
defined by the Wang tiles
${\mathcal {T}}_n^{\prime }$
. Since
${\mathcal {T}}_n \subset {\mathcal {T}}_n^{\prime }$
for every
$n\geq 1$
, we have
${\Omega }_n\subset {\Omega }_n^{\prime }$
. Thus, the properties shown for
${\Omega }_n^{\prime }$
also hold for
${\Omega }_n$
.
A tiling with the set
$\mathcal {T}^{\prime }_4$
is shown in Figure 19. We observe the presence of rows and columns of colored tiles. At the intersection of these colored rows and columns are junction tiles. In other words, the set of positions of junction tiles in the figure is the Cartesian product of two sets. Also, the distance between two consecutive junction tiles in the same row or column is 4 or 5. In the following lemmas, we prove that these observations hold in general.
Figure 19 A valid
$15\times 15$
pattern using the extended set
${\mathcal {T}}_4^{\prime }$
of Wang tiles. Note that it contains some antigreen tiles.
Lemma 6.1. Let
$n\geq 1$
be an integer. For every valid configuration
$c\in {\Omega }_n^{\prime }$
, the distance between two consecutive occurrences of junction tiles in the same row is n,
$n+1$
or
$n+2$
.
Also, the sequence of bottom labels of the tiles between two consecutive junction tiles (including the left junction tile but not the right one) belongs to
$\{00n,01n,01\overline {n}\} \cdot \{11n,11\overline {n}\}^*$
.
Proof. The horizontal Rauzy graph restricted to tiles whose vertical edge labels are starting with zero is shown in Figure 20. An arc in the horizontal Rauzy graph links two tiles
$s\to t$
if and only if the right label of tile s is equal to the left label of tile t. The graph allows to visualize the combinatorial structure between two consecutive junction tiles on the same horizontal row within a configuration of
${\Omega }_n^{\prime }$
.
Figure 20 Combinatorial structure between two consecutive junction tiles on the same horizontal row within a configuration of
${\Omega }_n^{\prime }$
. The nodes of the graph are placed such that any two tiles appearing in the same column have the same last digit for its left or right labels. The length of a path from a junction tile to a junction tile is n,
$n+1$
or
$n+2$
.
The right label of a junction tile is
$000$
,
$001$
or
$011$
, which implies that the last digit of the right label of a junction tile is
$0$
or
$1$
. The left label of a junction tile is
$00n$
,
$01n$
or
$01\overline {n}$
, which implies that the last digit of the left label of a junction tile is n or
$\overline {n}$
. Since the last digit increases by 1 from the left label to the right label of every intermediate tile (a tile appearing in between two consecutive junction tiles in the same row), the number of tiles in between two consecutive junction tiles on the same row is at least
$n-1$
and at most
$\overline {n}-0=n+1$
. We conclude that the distance (number of edges in the Rauzy graph) between two consecutive junction tiles in the same row is n,
$n+1$
or
$n+2$
. In particular, it is at least n.
The bottom label of a junction tile is in the set
$\{00n,01n,01\overline {n}\}$
. The bottom label of every intermediate tile is
$11n$
or
$11\overline {n}$
. Therefore, the sequence of bottom labels of the tiles between two consecutive junction tiles (including the left junction tile but not the right one) belongs to
$\{00n,01n,01\overline {n}\} \cdot \{11n,11\overline {n}\}^*$
; see Figure 20.
Lemma 6.2. Let
$n\geq 1$
be an integer. For every valid configuration
$c\in {\Omega }_n^{\prime }$
, the distance between two consecutive occurrences of a vertical stripe tile (blue, green, yellow or antigreen) in the same row is
$n-1$
, n or
$n+1$
.
Proof. The horizontal Rauzy graph restricted to vertical edge labels starting with
$1$
is shown in Figure 21. An arc in the horizontal Rauzy graph links two tiles
$s\to t$
if and only if the right label of tile s is equal to the left label of tile t. The graph allows to visualize the combinatorial structure between two consecutive vertical stripe tiles on the same horizontal row within a configuration of
${\Omega }_n^{\prime }$
.
Figure 21 Combinatorial structure between two consecutive vertical stripe tile on the same horizontal row within a configuration of
${\Omega }_n^{\prime }$
. The length of a path from a vertical stripe tile to a vertical stripe tile is
$n-1$
, n or
$n+1$
.
The right label of a vertical stripe tile is
$111$
or
$112$
, which implies that the last digit of the right label of a vertical stripe tile is
$1$
or
$2$
. The left label of a vertical stripe tile is
$11n$
or
$11\overline {n}$
, which implies that the last digit of the left label of a vertical stripe tile is n or
$\overline {n}$
. Since the last digit increases by 1 from the left label to the right label of every intermediate tile (a tile appearing in between two consecutive vertical stripe tiles in the same row), the number of tiles in between two consecutive vertical stripe tiles on the same row is at least
$n-2$
and at most
$\overline {n}-1=n$
. We conclude that the distance (number of edges in the Rauzy graph) between two consecutive vertical stripe tiles in the same row is
$n-1$
, n or
$n+1$
. In particular, it is at most
$n+1$
.
Lemma 6.3. Let
$n\geq 1$
be an integer. For every valid configuration
$c\in {\Omega }_n^{\prime }$
, there exist two strictly increasing sequences
$A,B:\mathbb {Z}\to \mathbb {Z}$
such that the following hold.
-
1. The set of positions of junction tiles in the configuration c is the Cartesian product
$c^{-1}(J_n^{\prime }) = A(\mathbb {Z})\times B(\mathbb {Z})$
. -
2. The distance between two consecutive occurrences of junction tiles in the same row is n or
$n+1$
; that is,
$A(k+1)-A(k)\in \{n,n+1\}$
for every
$k\in \mathbb {Z}$
. -
3. The distance between two consecutive occurrences of junction tiles in the same column is n or
$n+1$
; that is,
$B(k+1)-B(k)\in \{n,n+1\}$
for every
$k\in \mathbb {Z}$
.
Proof. (1) Let
$$ \begin{align*} E &= \{(\alpha_1\alpha_2\alpha_3, \beta_1 \beta_2 \beta_3, \gamma_1\gamma_2\gamma_3, \delta_1\delta_2\delta_3) \in{\mathcal{T}}_n^{\prime}\mid\alpha_1=0\}\subset {\mathcal{T}}_n^{\prime},\\ F &= \{(\alpha_1\alpha_2\alpha_3, \beta_1 \beta_2 \beta_3, \gamma_1\gamma_2\gamma_3, \delta_1\delta_2\delta_3) \in{\mathcal{T}}_n^{\prime}\mid\beta_1=0\}\subset {\mathcal{T}}_n^{\prime}. \end{align*} $$
Tiles in E have zero as the first coordinate of their right and left edge labels since
$\alpha _1=\gamma _1$
. Tiles in F have zero as the first coordinate of their top and bottom edge labels since
$\beta _1=\delta _1$
. Notice that we have
$E\cup F \subset Y_n\cup \widehat {Y_n} \cup G_n\cup \widehat {G_n} \cup B_n^{\prime }\cup \widehat {B_n^{\prime }}\cup J_n^{\prime } \cup A_n\cup \widehat {A_n}$
and
$E\cap F = J_n^{\prime }$
. Let
$c\in {\Omega }_n^{\prime }$
be a valid configuration. The positions of tiles from E in c are contiguous rows; that is, there exists
$B\subset \mathbb {Z}$
such that
$c^{-1}(E)=\mathbb {Z}\times B$
. The positions of tiles from F in c are contiguous columns; that is, there exists
$A\subset \mathbb {Z}$
such that
$c^{-1}(F)=A\times \mathbb {Z}$
. Therefore, the set of positions of junction tiles in c is given by the Cartesian product of A and B:
$$\begin{align*}c^{-1}(J_n^{\prime}) = c^{-1}(E \cap F) = c^{-1}(E) \cap c^{-1}(F) = (\mathbb{Z}\times B) \cap (A\times\mathbb{Z}) = A\times B. \end{align*}$$
The fact that the sets A and B are the images of increasing maps
$\mathbb {Z}\to \mathbb {Z}$
follows from observations (2) and (3) proved below.
(2) From Lemma 6.1, the distance between two consecutive occurrences of junction tiles in the same row is n,
$n+1$
or
$n+2$
. From Lemma 6.2, the distance between two consecutive occurrences of a vertical stripe tile (blue, green, yellow or antigreen) in the same row is
$n-1$
, n or
$n+1$
. Since vertical strips and junction tiles are vertically aligned, the difference between two consecutive elements of
$A\subset \mathbb {Z}$
is n or
$n+1$
. Also, if
$a\in A$
, then
$a+n\in A$
or
$a+n+1\in A$
. Also
$a-n\in A$
or
$a-n-1\in A$
. Thus, A is the image of an increasing map
$A:\mathbb {Z}\to \mathbb {Z}$
such that
$A(k+1)-A(k)\in \{n,n+1\}$
for every
$k\in \mathbb {Z}$
.
(3) From the symmetry of the set
${\mathcal {T}}_n^{\prime }$
of tiles, the same observation holds for the distance between consecutive junction tiles in the same column.
Lemma 6.3 means that we can subdivide valid configurations in
${\Omega }_n^{\prime }$
by rectangular patterns containing a unique junction tile at their bottom left corners; see Figure 22.
Figure 22 Return blocks appearing in Figure 19. Each return block contains a unique junction tile at its bottom left corner.
Proposition 6.4. Every configuration in
${\Omega }_n^{\prime }$
can be divided uniquely into rectangular blocks of sizes
$n\times n$
,
$n\times \overline {n}$
,
$\overline {n}\times n$
and
$\overline {n}\times \overline {n}$
with a unique junction tile at their bottom left corners.
Proof. Let
$c\in {\Omega }_n^{\prime }$
be a configuration. Let
$A,B:\mathbb {Z}\to \mathbb {Z}$
be the two increasing maps from Lemma 6.3 such that
$c^{-1}(J_n^{\prime })=A(\mathbb {Z})\times B(\mathbb {Z})$
. For every
${\boldsymbol {\ell }}=(\ell _1,\ell _2)\in \mathbb {Z}^2$
, the pattern appearing in c at support
$[A(\ell _1),A(\ell _1+1)-1]\times [B(\ell _2),B(\ell _2+1)-1]$
is a rectangular pattern containing a unique junction tile at its bottom left corner.
We call such a rectangular pattern described in Proposition 6.4 a return block (to a junction tile) (see Figure 23), following the terminology of return words in combinatorics on words [Reference Durand16, Reference Vuillon64]. While the classical notion of return word is to a single pattern, here the notion of return block is to a subset of tiles – namely, the junction tiles. From Proposition 6.4, the width (and height) of these blocks is n or
$n+1$
. On the right of the junction tile within a return block is the bottom row where horizontal blue, green, yellow or antigreen tiles appear. Similarly, above the junction tile within a return block is the left column where vertical blue, green, yellow or antigreen tiles appear.
Figure 23 A return block is split into four disjoint parts: the junction tile, the left column, the bottom row and the white tiles. Both its width W and its height H take values in the set
$\{n,n+1\}$
.
We may observe that the sequences of bottom labels of a return block made of tiles in
${\mathcal {T}}_4^{\prime }$
appearing completely in Figure 19 and in Figure 22 are in the set
$$ \begin{align} \left\{ \begin{array}{l} 004\cdot 114\cdot 115\cdot 114\cdot 115,\\ 004\cdot 115\cdot 114\cdot 115\cdot 115,\\ 015\cdot 114\cdot 115\cdot 115\cdot 115,\\ 014\cdot 114\cdot 115\cdot 115,\\ 014\cdot 115\cdot 115\cdot 115,\\ 015\cdot 115\cdot 115\cdot 115 \end{array} \right\} \not\subset \tau_n(V_n). \end{align} $$
In particular,
$004\cdot 114\cdot 115\cdot 114\cdot 115$
does not belong to the image of
$\tau _n$
when
$n=4$
. But the sequence of bottom labels of a return block has a particular structure for configurations in
${\Omega }_n$
. This is the subject of the next section.
6.2 Return blocks in the Wang shift
${\Omega }_n$
When considering configurations in
${\Omega }_n$
instead of
${\Omega }_n^{\prime }$
, there are no antigreen tiles in the row between two consecutive junction tiles. Thus, Figure 20 simplifies to Figure 24. In particular, in the bottom row of a return block within a configuration in
${\Omega }_n$
, the horizontal blue, green and yellow stripes appear in this order (when they appear). The same observation holds for the left column of a return block ordered from bottom to top.
Figure 24 Combinatorial structure between two consecutive junction tiles on the same horizontal row within a configuration of
${\Omega }_n$
. The nodes of the graph are placed such that any two tiles appearing in the same column have the same last digit for its left or right labels.
Surprisingly, when the tiles are restricted to the set
${\mathcal {T}}_n$
, the boundary of the return blocks can be decoded using the map
$\tau _n$
defined in Section 5.
Lemma 6.5. Let r be a return block appearing in a configuration
$c\in {\Omega }_n$
. The sequences of bottom labels of tiles in the bottom row of the return block r (from left to right) belong to the set
$$ \begin{align*} \tau_n(V_n)=\;& \{01\overline{n}\cdot (11\overline{n})^{i} \mid n-1\leq i\leq n\}\\ &\cup\{01n \cdot (11n)^{i}(11\overline{n})^{j} \mid i,j\geq0, n-1\leq i+j\leq n\}\\ &\cup\{00n \cdot (11n)^{i}(11\overline{n})^{j} \mid i,j\geq0, i+j=n \}. \end{align*} $$
Proof. From Lemma 6.1, the sequence of bottom labels of the tiles between two consecutive junction tiles (including the left junction tile but not the right one) belongs to
$\{00n,01n,01\overline {n}\} \cdot \{11n,11\overline {n}\}^*$
.
In the bottom row of every return block within a configuration in
${\Omega }_n$
, there is no antigreen stripe tile, and the horizontal blue, green and yellow stripe tiles appear in this order: blue
$\to $
green
$\to $
yellow. Since the bottom label of a blue horizontal stripe tile is
$11n$
and the bottom label of a green or yellow horizontal stripe tile is
$11\overline {n}$
, the sequence of bottom labels of tiles in a horizontal row starting from a junction tile and ending before the next occurrence of a junction tile is in the set
$$\begin{align*}\{00n,01n,01\overline{n}\} \cdot (11n)^*(11\overline{n})^*. \end{align*}$$
Some more restrictions are imposed:
-
• If it starts with
$00n$
, the length of the sequence is
$n+1$
. Indeed, if the bottom label of a junction tile is
$00n$
, then its right label is
$000$
with last digit
$0$
. From Figure 24, the width of the return block containing this junction tile must be
$n+1$
or
$n+2$
. A return block of width
$W=n+2$
is impossible from Proposition 6.4. Thus, the width of the return bock is
$W=n+1$
. -
• Also, if it starts with
$01\overline {n}$
, the next label is not
$11n$
and has to be
$11\overline {n}$
. Indeed
$01\overline {n}$
is the bottom label of a junction tile with right label
$011$
, and
$011$
must be the left label of a yellow horizontal stripe tile with bottom label
$11\overline {n}$
; see Figure 24.
Restricting the sequences to those of lengths n or
$\overline {n}$
, we have that the sequence of bottom labels of tiles in the bottom row of the return block r (from left to right) belongs to the set
$$ \begin{align*} & \{01\overline{n}\cdot (11\overline{n})^{i} \mid n-1\leq i\leq n\}\\ &\cup\{01n \cdot (11n)^{i}(11\overline{n})^{j} \mid i,j\geq0, n-1\leq i+j\leq n\}\\ &\cup\{00n \cdot (11n)^{i}(11\overline{n})^{j} \mid i,j\geq0, i+j=n \}\\ =\;&\{\tau_n(111), \tau_n(000)\}\\ &\cup\{\tau_n(00i)\mid 1\leq i \leq n+1\} \cup\{\tau_n(11i)\mid 2\leq i \leq n+1\}\\ &\cup\{\tau_n(01i)\mid 1\leq i \leq n+1\}\\ =\;&\tau(V_n).\\[-38pt] \end{align*} $$
6.3 Desubstitution
${\Omega }_n\leftarrow \Omega ^{\prime }_n$
In this section, we prove that every valid configuration with the tiles
${\mathcal {T}}_n$
can be desubstituted into a valid configuration over
$\mathcal {T}^{\prime }_n$
using the substitution
${\omega }_n^{\prime }$
. It is based on the following lemma which relates return blocks in
${\Omega }_n$
to tiles of
${\mathcal {T}}_n^{\prime }$
.
Lemma 6.6. Let
$y\in {\Omega }_n$
be a configuration. For every return block r appearing in y, there exists a unique tile 
such that
$r={\omega }_n^{\prime }(t)$
with external labels 
.
Proof. Let
$y\in {\Omega }_n$
be a configuration. From Proposition 6.4, the configuration y can be divided into return blocks, – that is, rectangular blocks of sizes
$n\times n$
,
$n\times \overline {n}$
,
$\overline {n}\times n$
or
$\overline {n}\times \overline {n}$
with a unique junction tiles at the bottom left corner; see Figure 23.
Let r be a return block appearing in y. From Lemma 6.5, the sequences of bottom labels of tiles in the bottom row of the return block r (from left to right) belong to the set
$\tau _n(V_n)$
. By symmetry and since r is surrounded by returns blocks, this also holds for the right, top and left labels of r. Therefore, let
$\alpha ,\beta ,\gamma ,\delta \in V_n$
such that the right, top, left and bottom labels of the return block r are respectively
$\tau _n(\alpha )$
,
$\tau _n(\beta )$
,
$\tau _n(\gamma )$
and
$\tau _n(\delta )$
. From Proposition 5.9, 
. From Lemma 5.3, there exists a unique rectangular pattern with these external labels. Thus,
$r={\omega }_n^{\prime }(t)$
.
Proposition 6.7. Let
$n\geq 1$
be an integer. For every configuration
$y\in {\Omega }_n$
, there exist a unique configuration
$x\in \Omega ^{\prime }_n$
and a unique vector
${\boldsymbol {k}}\in \{0,1,\dots ,n\}^2$
such that
$y=\sigma ^{\boldsymbol {k}}({\omega }_n^{\prime }(x))$
.
Proof. Let
${\omega }_n^{\prime }:\Omega ^{\prime }_n\to {\Omega }_n^{\prime }$
be the 2-dimensional substitution defined in (5.2).
Let
$y\in {\Omega }_n$
be a configuration. From Lemma 6.3, there exist two strictly increasing sequences
$A,B:\mathbb {Z}\to \mathbb {Z}$
such that the set of positions of junction tiles in the configuration y is the Cartesian product
$A(\mathbb {Z})\times B(\mathbb {Z})$
. Also, the distance between two consecutive occurrences of junction tiles in the same row or the same column is n or
$n+1$
; that is,
$A(\ell +1)-A(\ell )\in \{n,n+1\}$
and
$B(\ell +1)-B(\ell )\in \{n,n+1\}$
for every
$\ell \in \mathbb {Z}$
. We may suppose without loss of generality that the sequences A and B are defined in such a way that the sequences take nonnegative values for nonnegative indices exclusively. In other words,
$A(\ell )\geq 0$
if and only if
$\ell \geq 0$
and
$B(\ell )\geq 0$
if and only if
$\ell \geq 0$
.
For every
${\boldsymbol {\ell }}=(\ell _1,\ell _2)\in \mathbb {Z}^2$
, consider the return block
$y|_{S_{\boldsymbol {\ell }}}$
of support
$S_{\boldsymbol {\ell }}=[A(\ell _1),A(\ell _1+1)-1]\times [B(\ell _2),B(\ell _2+1)-1]$
. From Lemma 6.6, there exists a unique tile
$x_{\boldsymbol {\ell }}\in {\mathcal {T}}_n^{\prime }$
such that
$y|_{S_{\boldsymbol {\ell }}}={\omega }_n^{\prime }(x_{\boldsymbol {\ell }})$
. Let
${\boldsymbol {k}}=(-A(-1),-B(-1))$
. The configuration
$\sigma ^{-{\boldsymbol {k}}}(y)$
has a junction tile at the origin
$(0,0)$
. The configuration
$x=(x_{\boldsymbol {\ell }})_{{\boldsymbol {\ell }}\in \mathbb {Z}^2}$
belongs to
${\Omega }_n^{\prime }$
and satisfies that
${\omega }_n^{\prime }(x)=\sigma ^{-{\boldsymbol {k}}}(y)$
. Thus,
$y=\sigma ^{{\boldsymbol {k}}}{\omega }_n^{\prime }(x)$
.
Proposition 6.8. For every integer
$n\geq 1$
, the 2-dimensional substitution
${\omega }_n^{\prime }:\Omega ^{\prime }_n\to {\Omega }_n^{\prime }$
satisfies
${\Omega }_n\subseteq \overline {{\omega }_n^{\prime }(\Omega ^{\prime }_n)}^\sigma $
.
Proof. From Proposition 6.7, for every configuration
$y\in {\Omega }_n$
, there exist a unique configuration
$x\in \Omega ^{\prime }_n$
and a unique vector
${\boldsymbol {k}}\in \{0,1,\dots ,n\}^2$
such that
$y=\sigma ^{\boldsymbol {k}}({\omega }_n^{\prime }(x))$
. Therefore,
${\Omega }_n\subseteq \overline {{\omega }_n^{\prime }(\Omega ^{\prime }_n)}^\sigma $
.
7 Tiles in
${\mathcal {T}}_n^{\prime }\setminus {\mathcal {T}}_n$
are illegal so that
${\Omega }_n^{\prime }={\Omega }_n$
By definition
${\mathcal {T}}_n\subset {\mathcal {T}}_n^{\prime }$
, so that
${\Omega }_n\subseteq {\Omega }_n^{\prime }$
. In this section, we prove that in every configuration of the Wang shift
$\Omega ^{\prime }_n$
defined from the set
$\mathcal {T}^{\prime }_n$
, only the tiles from
${\mathcal {T}}_n$
appear; that is,
$\Omega ^{\prime }_n\subseteq {\Omega }_n$
.
7.1 Illegal tiles
Recall that the additional tiles are
$$\begin{align*}\mathcal{T}^{\prime}_n\setminus {\mathcal{T}}_n = A_n\cup\widehat{A_n} \cup \{j_n^{0,0,1,1}, j_n^{1,1,0,0}\} \cup \{b_n^n,\widehat{b_n^n}\}. \end{align*}$$
The proof that these tiles do not appear in any configuration in
${\Omega }_n^{\prime }$
follows from the following lemmas. The easiest is to show that no configuration contains the last blue tile because the argument is very local.
Lemma 7.1. A valid configuration in
$\Omega ^{\prime }_n$
contains no blue tile in
$\{b_n^n,\widehat {b_n^n}\}$
.
Proof. Let
$c\in \Omega ^{\prime }_n$
be a valid configuration. The configuration c does not contain the tile
$b_n^n = $
because no tile from
$\mathcal {T}^{\prime }_n$
has left label
$00\overline {n}$
. Similarly, the configuration c does not contain the tile
$\widehat {b_n^n}$
because no tile from
$\mathcal {T}^{\prime }_n$
has bottom label
$00\overline {n}$
.
Then, we show that no configuration of
${\Omega }_n^{\prime }$
contains any antigreen tile. The argument is more difficult because antigreen tiles admit large surroundings; see Figure 19. As seen in the figure and proved in the next lemma, the presence of an antigreen tile forces the presence of another antigreen tile a few rows below that is closer to the left to a junction tile.
Lemma 7.2. A valid configuration in
$\Omega ^{\prime }_n$
contains no antigreen tile from the set
$A_n\cup \widehat {A_n}$
.
Proof. Let
$c\in \Omega ^{\prime }_n$
be a valid configuration. Recall that 
. The configuration c does not contain the tile
$a_n^n$
because
$a_n^n$
has left label
$00\overline {n}$
, but no tile from
$\mathcal {T}^{\prime }_n$
has left label
$00\overline {n}$
. Similarly, the configuration c does not contain the tile
$\widehat {a_n^n}$
because
$\widehat {a_n^n}$
has top label
$00\overline {n}$
, but no tile from
$\mathcal {T}^{\prime }_n$
has top label
$00\overline {n}$
.
Suppose by contradiction that
$a_n^i$
appears in the configuration c for some integer i such that
$1\leq i\leq n-1$
. Let
$A,B:\mathbb {Z}\to \mathbb {Z}$
be the two increasing maps from Lemma 6.3 such that
$c^{-1}(J_n^{\prime })=A(\mathbb {Z})\times B(\mathbb {Z})$
. Suppose that
$a_n^i$
appears at position
${\boldsymbol {\ell }}=(\ell _1,\ell _2)\in \mathbb {Z}^2$
. Let
${\boldsymbol {k}}=(k_1,k_2)\in \mathbb {Z}^2$
be such that
$A(k_1)\leq \ell _1<A(k_1+1)$
and
$B(k_2)\leq \ell _2<B(k_2+1)$
. Note that we must have
$B(k_2)=\ell _2$
. Suppose that the occurrence
${\boldsymbol {\ell }}$
is chosen such that
$\ell _1-A(k_1)$
is the minimum among all occurrences of the tile
$a_n^i$
in c – in other words, such that the distance to the nearest junction tile to its left on the same row is minimal. Since the bottom and top labels of
$a_n^i$
start with
$1$
, the column
$\ell _1$
in the configuration c contains no junction tile; thus,
$A(k_1)\neq \ell _1$
and
$\ell _1-A(k_1)\geq 1$
. There are two cases to consider.
Case
$\ell _1-A(k_1)=1$
. In this case, the tile at position
$(A(k_1),B(k_2))$
is a junction tile with right label
$011$
and bottom label
$01\overline {n}$
. Also, the antigreen tile at position
$(\ell _1,\ell _2)$
is
$a_n^1$
. Below the antigreen tile are white tiles, and below the junction tile is a yellow or green tile that we show in gray in Figure 25.
Figure 25 The presence of the antigreen
$a_n^1$
leads to a contradiction.
So the unit parts of horizontal edge labels decrease by one at each level from top to bottom until we reach the white tile at position
$(\ell _1,B(k_2-1)+1)$
with bottom label 111 and a tile at position
$(A(k_1),B(k_2-1)+1)$
with bottom label
$0{*}2$
. The tile at position
$(\ell _1,B(k_2-1))$
must be a green or blue tile with left label
$00{*}$
. The tile at position
$(A(k_1),B(k_2-1)$
must be a junction tile, but there are no junction tiles with top label
$0{*}2$
. So this case leads to a contradiction.
Case
$\ell _1-A(k_1)>1$
. This means that tiles in the column to the left of
$a_n^i$
do not contain junction tiles. On Figure 20, we observe that only the yellow tile
$y_n^{i-1}$
has right label
$01i$
. Thus, the tile to the left of
$a_n^i$
at position
$(\ell _1-1,\ell _2)$
needs to be the yellow tile
$y_n^{i-1}$
. For every integer j such that
$B(k_2-1)<j<B(k_2)$
, the tiles at positions
$(\ell _1-1,j)$
and
$(\ell _1,j)$
are white tiles. So the unit parts of the horizontal edge labels decrease by one at each level from top to bottom. Thus, the tile at position
$(\ell _1-1,B(k_2-1))$
has top label
$112$
, and the tile at position
$(\ell _1,B(k_2-1))$
has top label
$111$
. The situation is illustrated in Figure 26.
Figure 26 The presence of the antigreen
$a_n^i$
leads to a contradiction.
Since
$112$
and
$111$
are the labels of consecutive horizontal edges, we deduce from Figure 20 that the tile at position
$(\ell _1-1,B(k_2-1))$
must be an antigreen tile as well. We observe that this antigreen tile is closer in distance to a junction tile to its left on the same row. This is a contradiction with the minimality of
$\ell _1-A(k_1)$
. Thus, the configuration c does not contain the antigreen tile
$a_n^i$
.
Finally, by contradiction, suppose that the tile
$\widehat {a_n^i}$
appears in the configuration c. Since
${\mathcal {T}}_n^{\prime }$
is symmetric – that is,
$\widehat {{\mathcal {T}}_n^{\prime }}={\mathcal {T}}_n^{\prime }$
– the symmetric configuration
$\widehat {c}$
is also a valid configuration in
${\Omega }_n^{\prime }$
. Thus, the configuration c contains the tile
$a_n^i$
which contradicts the conclusion of the previous paragraph.
The previous lemma implies that the pattern shown in Figure 19 cannot be extended to a valid configuration in
${\Omega }_n^{\prime }$
.
Lemma 7.3. A valid configuration in
$\Omega ^{\prime }_n$
contains no junction tile from the set
$\{j_n^{0,0,1,1},j_n^{1,1,0,0}\}$
.
Proof. Recall that
Let
$x\in \Omega ^{\prime }_n$
be a valid configuration. We first prove that x does not contain the tile
$j_n^{0,0,1,1}$
. By contradiction, suppose that the tile
$j_n^{0,0,1,1}$
appears in the configuration x at some position
${\boldsymbol {\ell }}\in \mathbb {Z}^2$
. Consider the return block containing this junction tile and let W be its width and H be its height.
The bottom label of the junction tile
$j_n^{0,0,1,1}$
is
$00n$
, and its right label is
$000$
with last digit
$0$
. From Figure 20, the width of the return block containing this junction tile must be
$n+1$
or
$n+2$
. A return block of width
$W=n+2$
is impossible from Proposition 6.4. Thus, the width of the return bock is
$W=n+1$
.
If
$n>1$
, then we have
$W=n$
, which is a contradiction. Indeed, the tile appearing above the junction tile
$j_n^{0,0,1,1}$
must be a vertical stripe tile with right label
$112$
, either yellow or antigreen. From the observation made in Figure 18, the width of this return block is
$W=n$
.
If
$n=1$
, three different junction tiles can appear on top of
$j_n^{0,0,1,1}$
. All of them have right label
$001$
. On the right of
$j_n^{0,0,1,1}$
, there may be a green or a blue tile, both of them having top label
$111$
. We get the following picture where we illustrate the blue or green tile in gray.
But no tiles from
${\mathcal {T}}_1^{\prime }$
have left label
$001$
and bottom label
$111$
; see Figure 27. Thus, no tile can be placed at position
${\boldsymbol {\ell }}+(1,1)$
. This is a contradiction.
Figure 27 Extended metallic mean Wang tile sets
$\mathcal {T}^{\prime }_n$
for
$n=1$
.
Finally, by contradiction, suppose that the tile
$j_n^{1,1,0,0}=\widehat {j_n^{0,0,1,1}}$
appears in the configuration x. Since
${\mathcal {T}}_n^{\prime }$
is symmetric – that is
$\widehat {{\mathcal {T}}_n^{\prime }}={\mathcal {T}}_n^{\prime }$
– the symmetric configuration
$\widehat {x}$
is also a valid configuration in
${\Omega }_n^{\prime }$
. Thus, the configuration x contains the tile
$j_n^{0,0,1,1}$
, which contradicts the first part of the proof.
We may now prove the following result.
Proposition 7.4. For every integer
$n\geq 1$
,
${\Omega }_n^{\prime }={\Omega }_n$
.
Proof. Since
${\mathcal {T}}_n\subseteq {\mathcal {T}}_n^{\prime }$
, we have
${\Omega }_n\subseteq {\Omega }_n^{\prime }$
.
Let
$c\in \Omega ^{\prime }_n$
be a valid configuration. From Lemma 7.1, the configuration c contains no blue tile in
$\{b_n^n,\widehat {b_n^n}\}$
. From Lemma 7.2, the configuration c contains no antigreen tile from
$A_n\cup \widehat {A_n}$
. From Lemma 7.3, the configuration c contains no junction tile from the set
$\{j_n^{0,0,1,1},j_n^{1,1,0,0}\}$
. Thus, the range of c is
$c(\mathbb {Z}^2)\subset {\mathcal {T}}_n$
. Thus,
$c\in {\Omega }_n$
, from which we conclude that
${\Omega }_n^{\prime }\subseteq {\Omega }_n$
.
8
${\Omega }_n$
is self-similar and aperiodic
In this section, we show that
${\Omega }_n$
is self-similar and aperiodic. We prove Theorem A below after recalling its statement.
Theorem A. For every integer
$n\geq 1$
, the set
${\mathcal {T}}_n$
containing
$(n+3)^2$
Wang tiles defines a Wang shift
${\Omega }_n$
which is self-similar. More precisely, there exists an expansive and recognizable 2-dimensional substitution
${\omega }_n:{\Omega }_n\to {\Omega }_n$
which is onto up to a shift – that is, such that
${\Omega }_n=\overline {{\omega }_n({\Omega }_n)}^\sigma $
.
Proof. Let
$n\geq 1$
be an integer. From Proposition 6.8, the 2-dimensional substitution
${\omega }_n^{\prime }:{\Omega }_n^{\prime }\to {\Omega }_n^{\prime }$
defined in (5.2) satisfies
${\Omega }_n\subseteq \overline {{\omega }_n^{\prime }(\Omega ^{\prime }_n)}^\sigma $
. From Proposition 7.4, we have
${\Omega }_n^{\prime }={\Omega }_n$
. The restriction of
${\omega }_n^{\prime }$
to
${\Omega }_n$
is the 2-dimensional substitution
${\omega }_n:{\Omega }_n\to {\Omega }_n$
defined in (5.3). From Lemma 5.6,
${\omega }_n({\Omega }_n)\subset {\Omega }_n$
. Therefore, we have
$$\begin{align*}{\Omega}_n \subseteq\overline{{\omega}_n^{\prime}(\Omega^{\prime}_n)}^\sigma =\overline{{\omega}_n^{\prime}({\Omega}_n)}^\sigma =\overline{{\Omega}_n({\omega}_n)}^\sigma \subseteq{\Omega}_n. \end{align*}$$
Therefore,
${\omega }_n$
is in fact a 2-dimensional substitution
${\Omega }_n\to {\Omega }_n$
satisfying
${\Omega }_n=\overline {{\omega }_n({\Omega }_n)}^\sigma $
. The 2-dimensional substitution
${\omega }_n$
is recognizable following Proposition 6.4 since every configuration in
${\Omega }_n$
can be uniquely divided into return blocks. The 2-dimensional substitution
${\omega }_n$
is expansive (the image of every tile contains a junction tile and the image of every junction tile has a height and width at least 2). Hence, the Wang shift
${\Omega }_n$
is self-similar with respect to the substitution
${\omega }_n$
.
Proof of Corollary B.
From Theorem A, we have that the Wang shift
${\Omega }_n$
is self-similar, satisfying
${\Omega }_n=\overline {{\omega }_n({\Omega }_n)}^\sigma $
. Since the substitution
${\omega }_n$
is expansive and recognizable, it follows from Proposition 3.5 that
${\Omega }_n$
is aperiodic.
9 The self-similarity is primitive
Substitutive shifts obtained from expansive and primitive morphisms are interesting for their properties. As in the 1-dimensional case, we say that
$\omega $
is primitive if there exists
$m\in \mathbb {N}$
such that for every
$a,b\in \mathcal {A}$
, the letter b occurs in
$\omega ^m(a)$
. In this section, we show that the 2-dimensional substitution
${\omega }_n$
is primitive.
Lemma 9.1. For every integer
$n\geq 1$
, the 2-dimensional substitution
${\omega }_n:{\Omega }_n\to {\Omega }_n$
is primitive.
Proof. The proof follows from the following observations about the substitution
${\omega }_n$
:
-
• in the image of every tile in
${\mathcal {T}}_n$
under
${\omega }_n$
, there is some junction tile; -
• in the image of every junction tile, there is a white tile
$w_{n}^{1,1}$
; -
• in the image of the white tile
$w_{n}^{1,1}$
, there is the junction tile
$j_n^{1,1,1,1}$
; -
• in the image of the junction tile
$j_n^{1,1,1,1}$
, there are the junction tile
$j_n^{0,0,0,0}$
, all white tiles
$W_n$
including the white tile
$w_{n}^{1,1}$
, and all blue tiles
$B_n\cup \widehat {B_n}$
including the blue tiles
$\{b_{n}^0,\widehat {b_{n}^0}\}$
(all blue tiles appear in the image because the left and bottom label of
$j_n^{1,1,1,1}$
is
$01\overline {n}$
; see Lemma 5.2 and Figure 13); -
• in the image of the blue tiles
$\{b_{n}^0,\widehat {b_{n}^0}\}$
, there are all yellow tiles
$Y_n\cup \widehat {Y_n}$
including the yellow tiles
$\{y_{n}^1,\widehat {y_{n}^1}\}$
(all yellow tiles appear in the images because the left label of
$b_n^0$
is
$000$
and the bottom label of
$\widehat {b_n^0}$
is
$000$
; see Lemma 5.2 and Figure 12); -
• in the image of yellow tiles
$Y_n\cup \widehat {Y_n}$
, there are the junction tiles
$\{j_n^{0,1,0,0},j_n^{0,0,0,1}\}$
; -
• in the image of
$Y_n\cup \widehat {Y_n}\cup \{j_n^{0,0,0,0}\}$
, there are all green tiles
$G_n\cup \widehat {G_n}$
:-
– green tiles
$g_n^n$
and
$\widehat {g_n^n}$
appear in the image of
$j_n^{0,0,0,0}$
because the left and bottom label of
$j_n^{0,0,0,0}$
is
$00n$
; see Lemma 5.2 and Figure 12; -
– green tiles
$g_n^i$
and
$\widehat {g_n^i}$
for
$0\leq i < n$
appear in the images of the yellow tiles because the bottom label of
$\widehat {y_n^{\overline {i}}}$
is
$01\overline {i}$
; see Lemma 5.2 and Figure 13);
-
-
• in the image of
$j_n^{0,0,0,0}$
, there is the junction tile
$j_n^{0,1,0,1}$
; -
• in the image of the blue tiles
$\{y_{n}^1,\widehat {y_{n}^1}\}$
, there are the green tiles
$\{g_{n}^0,\widehat {g_{n}^0}\}$
; -
• in the image of the green tiles
$\{g_{n}^0,\widehat {g_{n}^0}\}$
, there are the junction tiles
$\{j_n^{0,1,1,1},j_n^{1,1,0,1}\}$
.
The tiles that can be obtained from the successive application of the substitution
${\omega }_n$
are shown in Figure 28. The graph in the figure shows that every tile appears at distance 7 of every tile in
${\mathcal {T}}_n$
. Thus, for every tile
$t\in {\mathcal {T}}_n$
, the pattern
$({\omega }_n)^7(t)$
contains all tiles of
${\mathcal {T}}_n$
. Therefore, we conclude that
${\omega }_n$
is primitive.
The exponent
$7$
deduced in the previous proof is not sharp, as computations illustrate that for every integer
$n\geq 2$
, the incidence matrix of
$({\omega }_n)^4$
is already positive, while the incidence matrix of
$(\omega _1)^5$
is positive.
Figure 28 When an arrow appears linking sets of tiles
$S\to T$
and vertex T has in-degree one, it means that
$T\subseteq { \bigcup }_{s\in S}\{t\in {\mathcal {T}}_n\mid t\text { occurs in }{\omega }_n(s)\}$
; that is, every tile
$t\in T$
appears in the image of some tile
$s\in S$
under the substitution
${\omega }_n$
. When two arrows
$S\to T$
and
$S'\to T$
appear, it means that every tile
$t\in T$
appears in the image of some tile
$s\in S\cup S'$
under the substitution
${\omega }_n$
. The figure illustrates that for every tile
$t\in {\mathcal {T}}_n$
, the pattern
$({\omega }_n)^7(t)$
contains every tile of
${\mathcal {T}}_n$
. This shows the primitivity of the substitution
${\omega }_n$
.
Lemma 9.2. The Perron–Frobenius dominant eigenvalue of the incidence matrix of
${\omega }_n$
is
$\beta _n^2$
, the square of the
$n^{th}$
metallic mean number, and the inflation factor of
${\omega }_n$
is
$\beta _n$
.
Proof. We may deduce the dominant eigenvalue of the incidence matrix of
${\omega }_n$
from that of a simpler substitution. For every integer
$n\geq 1$
, let
$\rho _n$
be the following 1-dimensional substitution:
$$\begin{align*}\begin{array}{rccl} \rho_n:&\{\texttt{a},\texttt{b}\}^* & \to & \{\texttt{a},\texttt{b}\}^*\\ & \texttt{a}&\mapsto&\texttt{a}\texttt{b}^n\\ & \texttt{b}&\mapsto&\texttt{a}\texttt{b}^{n-1} \end{array}. \end{align*}$$
The incidence matrix of
$\rho _n$
is
$$\begin{align*}\left( \begin{array}{cc} 1 & 1\\ n & n-1 \end{array} \right) \end{align*}$$
whose characteristic polynomial is
$x^2-nx-1$
. The Perron–Frobenius dominant eigenvalue of the incidence matrix of
$\rho _n$
is the positive root
$\beta _n$
of the polynomial
$x^2-nx-1$
. Since
$\rho _n$
is primitive, the growth rate of
$|\rho _n^k(u)|$
is independent of
$u\in \{\texttt {a},\texttt {b}\}$
and is equal to
$\beta _n$
[Reference Queffélec51, Corollary 5.2]. In other words, for every
$u\in \{\texttt {a},\texttt {b}\}$
, we have
$$ \begin{align} \lim_{k\to\infty} |\rho_n^k(u)|^{\frac{1}{k}} = \beta_n. \end{align} $$
We observe that the
$2$
-dimensional substitution
${\omega }_n$
is an extension of the direct product
$\rho _n\times \rho _n$
of the 1-dimensional substitution
$\rho _n$
with itself. By extension, we mean the existence of a map
$$\begin{align*}\begin{array}{rccl} \zeta:&{\mathcal{T}}_n&\to&\{\texttt{a},\texttt{b}\}\times\{\texttt{a},\texttt{b}\}\\ &t&\mapsto& \begin{cases} (\texttt{a}, \texttt{a}) & \text{ if } t \in J_n,\\ (\texttt{b}, \texttt{a}) & \text{ if } t \in B_n\cup Y_n\cup G_n,\\ (\texttt{a}, \texttt{b}) & \text{ if } t \in \widehat{B_n}\cup \widehat{Y_n}\cup \widehat{G_n},\\ (\texttt{b}, \texttt{b}) & \text{ if } t \in W_n,\\ \end{cases} \end{array} \end{align*}$$
such that
$(\rho _n\times \rho _n)\circ \zeta =\zeta \circ {\omega }_n$
.
Since
${\omega }_n$
is primitive, the dominant eigenvalue
$\lambda $
of the incidence matrix of the substitution
${\omega }_n$
is equal to the growth rate of 
as
$k\to \infty $
, where
$t\in {\mathcal {T}}_n$
is any tile and 
denotes the cardinality of the support of a rectangular pattern
$p\in ({\mathcal {T}}_n)^{*^2}$
. Let
$t\in {\mathcal {T}}_n$
such that
$\zeta (t)=(t_1,t_2)$
for some
$t_1,t_2\in \{\texttt {a},\texttt {b}\}$
. Since
$\zeta $
is a tile to tile map, it preserves the area. Thus, we have
Therefore, the incidence matrices of the substitutions
${\omega }_n$
and
$\rho _n\times \rho _n$
have the same Perron-Frobenius dominant eigenvalue, and it is equal to
$\beta _n^2$
.
The inflation factor is the factor of the homogeneous dilation associated with the stone inflation constructed from the direct product
$\rho _n\times \rho _n$
[Reference Baake and Grimm5, § 5.6] (for example, a stone inflation for
$\rho _4\times \rho _4$
is shown in Figure 30 when
$n=4$
). The inflation factor of the stone inflation of
$\rho _n\times \rho _n$
is
$\beta _n$
as it multiplies distances between points by
$\beta _n$
and the areas by
$\beta _n^2$
.
Theorem C. For every integer
$n\geq 1$
, the 2-dimensional substitution
${\omega }_n:{\Omega }_n\to {\Omega }_n$
is primitive. The Perron–Frobenius dominant eigenvalue of the incidence matrix of
${\omega }_n$
is
$\beta _n^2$
, the square of the
$n^{th}$
metallic mean number, and the inflation factor of
${\omega }_n$
is
$\beta _n$
.
Proof. From Lemma 9.1,
${\omega }_n$
is primitive. The Perron–Frobenius dominant eigenvalue of the incidence matrix of
${\omega }_n$
and its inflation factor is computed in Lemma 9.2.
From Perron–Frobenius theorem, the primitivity of the substitution
${\omega }_n$
implies that every Wang tile in
${\mathcal {T}}_n$
appears with positive frequency in a configuration in the substitutive subshift
$\mathcal {X}_{{\omega }_n}$
generated by the substitution
${\omega }_n$
. The frequencies of the tiles are given by the entries of the right-eigenvector of the incidence matrix of
${\omega }_n$
normalized so that the sum of its entries is 1.
10
${\Omega }_n$
is minimal
The goal of this section is to prove that
${\Omega }_n$
is minimal. To prove minimality, we need more notions. We use the method proposed in [Reference Labbé37, §3.3].
10.1 A criterion for minimality of a self-similar subshift
Recall that a subshift X is self-similar if
$X=\overline {\omega (X)}^\sigma $
for some expansive d-dimensional substitution; see Definition 3.2 and Definition 3.3. First we recall Lemma 3.8 from [Reference Labbé37].
Lemma 10.1. Let
$\omega :\mathcal {A}\to \mathcal {A}^{*^d}$
be an expansive and primitive d-dimensional morphism. Let
$X\subseteq \mathcal {A}^{\mathbb {Z}^d}$
be a nonempty subshift such that
$X=\overline {\omega (X)}^{\sigma }$
. Then
$\mathcal {X}_\omega \subseteq X$
.
Proof. The language of X is also self-similar satisfying
$\mathcal {L}(X)=\mathcal {L}(\omega (\mathcal {L}(X)))$
. Recursively,
$\mathcal {L}(X)=\mathcal {L}(\omega ^m(\mathcal {L}(X)))$
for every
$m\geq 1$
. Since X is nonempty, there exists a letter
$a\in \mathcal {A}$
such that for all
$m\geq 1$
, the d-dimensional word
$\omega ^m(a)$
is in the language
$\mathcal {L}(X)$
. From the primitivity of
$\omega $
, there exists
$m\geq 1$
such that
$\omega ^m(a)$
contains an occurrence of every letter of the alphabet
$\mathcal {A}$
. Therefore, every letter is in
$\mathcal {L}(X)$
, and the d-dimensional word
$\omega ^m(a)$
is in the language
$\mathcal {L}(X)$
for all letters
$a\in \mathcal {A}$
and all
$m\geq 1$
. So we conclude that
$\mathcal {L}(\mathcal {X}_\omega )\subseteq \mathcal {L}(X)$
and
$\mathcal {X}_\omega \subseteq X$
.
Proving that a self-similar d-dimensional subshift X satisfying
$X=\overline {\omega (X)}^{\sigma }$
is equal to
$\mathcal {X}_\omega $
can be tricky. As illustrated in the following example, it depends on the combinatorics of the substitution.
Example 10.2. Consider the following 2-dimensional substitution
$\nu $
over alphabet
$\{a,b,c\}$
:
$$\begin{align*}\nu: a\mapsto \left(\begin{array}{ccccc} c&c&c&c&c\\ c&c&c&c&c\\ c&c&a&c&c\\ \end{array}\right),\quad b\mapsto \left(\begin{array}{ccccc} c&c&b&c&a\\ c&c&c&c&c\\ c&c&c&c&c\\ \end{array}\right),\quad c\mapsto \left(\begin{array}{ccccc} c&c&a&c&c\\ c&c&c&b&c\\ c&c&c&c&c\\ \end{array}\right). \end{align*}$$
We may observe that the vertical domino
$\left (\begin {smallmatrix}a\\b\end {smallmatrix}\right )$
does not belong to the language of the substitutive subshift
$\mathcal {X}_\nu $
since it does not appear in any of the k-th image of any letter under the substitution. But one can see that the vertical domino
$\left (\begin {smallmatrix}a\\b\end {smallmatrix}\right )$
is preserved by the substitution. Therefore, there exists a configuration x containing a single vertical domino
$\left (\begin {smallmatrix}a\\b\end {smallmatrix}\right )$
which is fixed by the substitution. Thus, we have
$$\begin{align*}\varnothing\neq \mathcal{X}_\nu\subsetneq \mathcal{X}_\nu\cup\{\sigma^n(x)\,|\,n\in\mathbb{Z}^2\}. \end{align*}$$
The subshift
$\mathcal {X}_\nu \cup \{\sigma ^n(x)\,|\,n\in \mathbb {Z}^2\}$
is self-similar, but it is not minimal because it contains a proper nonempty subshift.
Therefore, to conclude that we have the equality
$\mathcal {X}_\omega =X$
for a self-similar subshift X, it is convenient to consider the domino patterns of size
$1\times 2$
and
$2\times 1$
straddling the images of the two letters of a domino as well as the
$2\times 2$
patterns straddling the images of the four letters of
$2\times 2$
pattern. More precisely, we need to consider the following directed graphs:
-
• Let
$G_\omega ^{2\times 2}=(V_\omega ^{2\times 2},E_\omega ^{2\times 2})$
be the directed graph whose vertices and edges are
$$ \begin{align*} V_\omega^{2\times2} &= \left\{ \left(\begin{smallmatrix} a&b\\ c&d \end{smallmatrix}\right) \in\mathcal{A}^{2\times 2} \mid a\equiv_1 b, c\equiv_1 d, a\equiv_2 c, b\equiv_2 d \right\},\\ E_\omega^{2\times2} &= \left\{ \left(\begin{smallmatrix} e&f\\ g&h \end{smallmatrix}\right) \to \left(\begin{smallmatrix} a&b\\ c&d \end{smallmatrix}\right) \middle| \begin{array}{l} a \text{ is the bottom right letter of } \omega(e),\\ b \text{ is the bottom left letter of } \omega(f),\\ c \text{ is the top right letter of } \omega(g),\\ d \text{ is the top left letter of } \omega(h)\\ \end{array} \right\}. \end{align*} $$
-
• Let
$G_\omega ^{2\times 1}=(V_\omega ^{2\times 1},E_\omega ^{2\times 1})$
be the directed graph whose vertices and edges are -
• Let
$G_\omega ^{1\times 2}=(V_\omega ^{1\times 2},E_\omega ^{1\times 2})$
be the directed graph whose vertices and edges are
Finally, for every directed graph
$G=(V,E)$
, we define the set of recurrent vertices – that is, those belonging to a cycle of the graph:
Example 10.3. The graphs
$G_\nu ^{2\times 2}$
,
$G_\nu ^{2\times 1}$
and
$G_\nu ^{1\times 2}$
for the 2-dimensional substitution
$\nu $
defined in Example 10.2 are shown in Figure 29. The recurrent vertices of the graphs are as follows:
In particular, we observe that the vertical domino
$\left (\begin {smallmatrix}a\\b \end {smallmatrix}\right )$
belongs to a cycle of
$G_\nu ^{1\times 2}$
, even though it is not in the language
$\mathcal {L}(\mathcal {X}_\nu )$
.
Figure 29 The graphs
$G_\nu ^{2\times 2}$
,
$G_\nu ^{2\times 1}$
and
$G_\nu ^{1\times 2}$
for the substitution
$\nu $
.
The recurrent vertices of the three graphs
$G_\omega ^{2\times 2}$
,
$G_\omega ^{2\times 1}$
and
$G_\omega ^{1\times 2}$
provide a criteria for the minimality of a self-similar subshift
$X={\overline {\omega (X)}^{\sigma }}$
. Lemma 3.7 and Lemma 3.9 from [Reference Labbé37] gave hypothesis under which an expansive and primitive 2-dimensional substitution has a unique nonempty self-similar subshift. The following lemma is a relaxed version which allows to conclude that a self-similar subshift is minimal even when the 2-dimensional substitution admits more than one self-similar subshift (some made of configurations which are not uniformly recurrent).
Lemma 10.4. Let
$X={\overline {\omega (X)}^{\sigma }}$
be a nonempty self-similar subshift where
$\omega :\mathcal {A}\to \mathcal {A}^{*^d}$
is an expansive and primitive
$2$
-dimensional morphism. The following are equivalent:
-
(i)
for every size
$s\in \{2\times 2, 2\times 1, 1\times 2\}$
, -
(ii)
$X=\mathcal {X}_\omega $
, -
(iii) X is minimal.
for every size 
An element
$u\in \mathcal {A}^{\boldsymbol {n}}$
is called a d
-dimensional word of size
${\boldsymbol {n}}=(n_1,\dots ,n_d)\in \mathbb {N}^d$
on the alphabet
$\mathcal {A}$
. We use the notation 
when necessary.
Proof. Assume that
$X=\overline {\omega (X)}^{\sigma }$
for some
$\varnothing \neq X\subseteq \mathcal {A}^{\mathbb {Z}^d}$
.
(i)
$\implies $
(ii) From Lemma 10.1, we have
$\mathcal {X}_\omega \subseteq X$
. Let
$z\in \mathcal {L}(X)$
. We want to show that
$z\in \mathcal {L}(\mathcal {X}_\omega )$
. Since
$\omega $
is expansive, let
$m\in \mathbb {N}$
such that the image of every letter
$a\in \mathcal {A}$
by
$\omega ^m$
is larger than z; that is, 
for all
$a\in \mathcal {A}$
. We have
$z\in \mathcal {L}(X)=\mathcal {L}\left (\omega ^m(\mathcal {L}(X))\right )$
. By the choice of m, z cannot overlap more than two blocks
$\omega ^m(a)$
in the same direction. Thus, there exists a word
$u\in \mathcal {L}(X)$
of size
$1\times 1$
,
$2\times 1$
,
$1\times 2$
or
$2\times 2$
such that z is a subword of
$\omega ^m(u)$
. If u is of size
$1\times 1$
, then
$z\in \mathcal {L}(\mathcal {X}_\omega )$
. We may assume that the word u has the smallest possible rectangular size
$s\in \{2\times 1, 1\times 2, 2\times 2\}$
.
We have
$u\in V_\omega ^{s}$
. Since
$u\in \mathcal {L}(X)$
and X is self-similar, there exists a sequence
$(u_k)_{k\in \mathbb {N}}$
with
$u_k\in V_\omega ^{s}\cap \mathcal {L}(X)$
for all
$k\in \mathbb {N}$
such that
$$\begin{align*}\cdots\rightarrow u_{k+1} \rightarrow u_k \rightarrow \cdots\rightarrow u_1 \rightarrow u_0 = u \end{align*}$$
is a left-infinite path in the graph
$G_\omega ^{s}$
. Since
$V_\omega ^{s}$
is finite, there exist some
$k,k'\in \mathbb {N}$
with
$k<k'$
such that
$u_k=u_{k'}$
. Thus, 
and u is a subword of
$\omega ^k(u_k)$
. From the hypothesis, we have
$u_k\in \mathcal {L}(\mathcal {X}_\omega )$
. Since
$\omega $
is primitive, there exists
$\ell $
such that
$u_k$
is a subword of
$\omega ^\ell (a)$
for every
$a\in \mathcal {A}$
. Therefore, z is a subword of
$\omega ^{m+k+\ell }(a)$
for every
$a\in \mathcal {A}$
. Then
$z\in \mathcal {L}(\mathcal {X}_\omega )$
and
$\mathcal {L}(X)\subseteq \mathcal {L}(\mathcal {X}_\omega )$
. Thus,
$X\subseteq \mathcal {X}_\omega $
and
$X=\mathcal {X}_\omega $
.
(ii)
$\implies $
(i) If
$X=\mathcal {X}_\omega $
, then
$\mathcal {L}(X)=\mathcal {L}(\mathcal {X}_\omega )$
. Thus, 
for every size
$s\in \{2\times 2, 2\times 1, 1\times 2\}$
.
(ii)
$\implies $
(iii) The substitutive shift of
$\omega $
is well defined since
$\omega $
is expansive, and it is minimal since
$\omega $
is primitive, using standard arguments [Reference Queffélec51, §5.2].
(iii)
$\implies $
(ii) From Lemma 10.1, we have
$\mathcal {X}_\omega \subseteq X$
. Since X is minimal, we conclude that
$\mathcal {X}_\omega =X$
.
10.2 The Wang shift
${\Omega }_n$
is minimal when
$n\geq 2$
The proof that the Wang shift
${\Omega }_n$
is minimal needs to be split into two cases. When
$n=1$
, configurations in
$\Omega _1$
have consecutive rows containing junction tiles, whereas this does not happen when
$n\geq 2$
. This affects the language of patterns of vertical domino support. In particular, a vertical domino made of two junction tiles may appear in the language of
${\Omega }_n$
when
$n=1$
. In this section, we consider the case
$n\geq 2$
.
Lemma 10.5. Let
$n\geq 2$
be an integer. The following vertical dominoes appear in the language of the substitutive subshift
$\mathcal {X}_{{\omega }_n}$
:
$$ \begin{align*} \mathcal{L}_{1\times 2}(\mathcal{X}_{{\omega}_n}) &\supseteq \left\{ \left(\begin{array}{c} j_n^{0,1,0,0}\\ \widehat{g_n^{n-1}} \end{array}\right), \left(\begin{array}{c} j_n^{0,1,0,1}\\ \widehat{g_n^{n-1}} \end{array}\right), \left(\begin{array}{c} j_n^{0,1,0,0}\\ \widehat{y_n^{n-1}} \end{array}\right), \left(\begin{array}{c} j_n^{0,1,0,1}\\ \widehat{y_n^{n-1}} \end{array}\right), \left(\begin{array}{c} j_n^{0,1,1,1}\\ \widehat{y_n^{n-1}} \end{array}\right) \right\}\\ &\quad\cup \left\{ \left(\begin{array}{c} j_n^{1,1,0,1}\\ \widehat{g_n^{n}} \end{array}\right), \left(\begin{array}{c} j_n^{1,1,0,1}\\ \widehat{y_n^{n}} \end{array}\right), \left(\begin{array}{c} j_n^{1,1,1,1}\\ \widehat{y_n^{n}} \end{array}\right), \left(\begin{array}{c} j_n^{0,0,0,0}\\ \widehat{b_n^{n-1}} \end{array}\right), \left(\begin{array}{c} j_n^{0,0,0,1}\\ \widehat{b_n^{n-1}} \end{array}\right) \right\}\\&\quad\cup \left\{ \left(\begin{array}{c} g_n^{i-1}\\ w_n^{i,n} \end{array}\right), \left(\begin{array}{c} g_n^{i}\\ w_n^{i,n} \end{array}\right) \middle| \; 1\leq i\leq n \right\}\\ &\quad\cup \left\{ \left(\begin{array}{c} b_n^{i-1}\\ w_n^{i,n-1} \end{array}\right) \middle| \; 1\leq i\leq n \right\} \cup \left\{ \left(\begin{array}{c} b_n^{i}\\ w_n^{i,n-1} \end{array}\right) \middle| \; 1\leq i\leq n-1 \right\}\\ &\quad\cup \left\{ \left(\begin{array}{c} y_n^{i-1}\\ w_n^{i,n} \end{array}\right) \middle| \; 2\leq i\leq n \right\} \cup \left\{ \left(\begin{array}{c} y_n^{i}\\ w_n^{i,n} \end{array}\right) \middle| \; 1\leq i\leq n \right\}. \end{align*} $$
Proof. We show that every vertical domino listed above appears in the image of some tile under the application of the 2-dimensional substitution
${\omega }_n$
. Below, we use the notation
$p \xrightarrow {{\omega }_n} q$
to denote that q is a pattern appearing in the image
${\omega }_n(p)$
. We have
$$ \begin{align*} &j_n^{0,1,0,1} \xrightarrow{{\omega}_n} \left(\begin{array}{c}\widehat{b_n^{0}}\\ j_n^{0,0,0,0} \end{array}\right) \xrightarrow{{\omega}_n} \left(\begin{array}{ccccc} \mathbf{j_n^{1,1,0,1}} & y_n^1 & y_n^2 & \dots & y_n^n\\ \mathbf{\widehat{g_n^{n}}} & w_n^{1,n} & w_n^{2,n} & \dots & w_n^{n,n} \end{array}\right), \\ &j_n^{0,1,0,1} \xrightarrow{{\omega}_n} \left(\begin{array}{c}\widehat{w_n^{1,1}}\\ b_n^0 \end{array}\right) \xrightarrow{{\omega}_n} \left(\begin{array}{ccccc} \mathbf{j_n^{1,1,1,1}} & y_n^1 & y_n^2 & \dots & y_n^{n-1}\\ \mathbf{\widehat{y_n^{n}}} & w_n^{2,n} & w_n^{3,n} & \dots & w_n^{n,n} \end{array}\right), \\ &j_n^{0,1,0,1} \xrightarrow{{\omega}_n} \left(\begin{array}{c}\widehat{w_n^{1,1}}\\ b_n^0 \end{array}\right) \xrightarrow{{\omega}_n} \left(\begin{array}{c}j_n^{1,1,1,1}\\ \widehat{y_n^{n}} \end{array}\right) \xrightarrow{{\omega}_n} \left(\begin{array}{ccccc} \mathbf{j_n^{0,0,0,0}} & b_n^0 & b_n^1 & \dots & b_n^{n-1}\\ \mathbf{\widehat{b_n^{n-1}}} & w_n^{1,n-1} & w_n^{2,n-1} & \dots & w_n^{n,n-1} \end{array}\right), \\ &g_n^0 \xrightarrow{{\omega}_n} \left(\begin{array}{c}\widehat{y_n^1}\\ j_n^{0,1,1,1} \end{array}\right) \xrightarrow{{\omega}_n} \left(\begin{array}{c}\mathbf{j_n^{0,0,0,1}}\\ \widehat{\mathbf{b_n^{n-1}}} \end{array}\right) \xrightarrow{{\omega}_n} \left(\begin{array}{ccccc} \mathbf{j_n^{0,1,0,0}} & b_n^1 & b_n^2 & \dots & b_n^{n-1}\\ \mathbf{\widehat{g_n^{n-1}}} & w_n^{1,n-1} & w_n^{2,n-1} & \dots & w_n^{n-1,n-1} \end{array}\right). \end{align*} $$
Also,
$$ \begin{align*} &g_n^1 \xrightarrow{{\omega}_n} \left(\begin{array}{c}\widehat{g_n^1}\\ j_n^{0,1,0,1}\\ \end{array}\right) \xrightarrow{{\omega}_n} \left(\begin{array}{c}\mathbf{j_n^{0,1,0,1}}\\ \mathbf{\widehat{g_n^{n-1}}} \end{array}\right). \end{align*} $$
Since
$n\geq 2$
, we have
$$ \begin{align*} &j_n^{1,1,1,1} \xrightarrow{{\omega}_n} \left(\begin{array}{c}\widehat{b_n^0}\\ j_n^{0,0,0,0} \end{array}\right) \xrightarrow{{\omega}_n} \left(\begin{array}{c}y_n^1\\ w_n^{1,n} \end{array}\right) \xrightarrow{{\omega}_n} \left(\begin{array}{c}\mathbf{j_n^{0,1,0,0}}\\ \mathbf{\widehat{y_n^{n-1}}} \end{array}\right), \\ &w_n^{1,1} \xrightarrow{{\omega}_n} \left(\begin{array}{c}w_n^{2,2}\\ y_n^1 \end{array}\right) \xrightarrow{{\omega}_n} \left(\begin{array}{c}\mathbf{j_n^{0,1,0,1}}\\ \mathbf{\widehat{y_n^{n-1}}} \end{array}\right), \\ &g_n^0 \xrightarrow{{\omega}_n} \left(\begin{array}{c}w_n^{2,1}\\ \widehat{b_n^1} \end{array}\right) \xrightarrow{{\omega}_n} \left(\begin{array}{c}\mathbf{j_n^{1,1,0,1}}\\ \mathbf{\widehat{y_n^{n}}} \end{array}\right), \\ &w_n^{1,1} \xrightarrow{{\omega}_n} \left(\begin{array}{c}\widehat{y_n^1}\\ j_n^{1,1,1,1} \end{array}\right) \xrightarrow{{\omega}_n} \left(\begin{array}{c}g_n^0\\ w_n^{1,n} \end{array}\right) \xrightarrow{{\omega}_n} \left(\begin{array}{c}\mathbf{j_n^{0,1,1,1}}\\ \mathbf{\widehat{y_n^{n-1}}} \end{array}\right).\\[-44pt] \end{align*} $$
Lemma 10.6. The following four
$2\times 2$
patterns belong to the language of the substitutive subshift
$\mathcal {L}(\mathcal {X}_{{\omega }_n})$
:
Proof. We show that every pattern listed above appears in the image of some tile under some repeated application of the 2-dimensional substitution
${\omega }_n$
. Below, we use the notation
$p \xrightarrow {{\omega }_n} q$
to denote that q is a pattern appearing in the image
${\omega }_n(p)$
. The four patterns can be obtained in a few steps when applying the substitution
${\omega }_n$
on the tiles
$j_n^{1,1,1,1}$
and
$y_n^1$
. We have the following:
Lemma 10.7. Let
$n\geq 2$
be an integer. The following two vertical dominoes are illegal in
${\Omega }_n$
:
Proof. Let
$c\in {\Omega }_n$
be a valid configuration. Let
$A,B:\mathbb {Z}\to \mathbb {Z}$
be the two increasing maps from Lemma 6.3 such that
$c^{-1}(J_n)=A(\mathbb {Z})\times B(\mathbb {Z})$
.
Suppose that
$j_n^{1,1,1,1}$
appears at position
${\boldsymbol {\ell }}=(\ell _1,\ell _2)\in \mathbb {Z}^2$
and that
$\widehat {g_n^n}$
appears at position
$(\ell _1,\ell _2-1)$
in c. Let
${\boldsymbol {k}}=(k_1,k_2)\in \mathbb {Z}^2$
be such that
$A(k_1)\leq \ell _1<A(k_1+1)$
and
$B(k_2)\leq \ell _2<B(k_2+1)$
. Since
$j_n^{1,1,1,1}$
is a junction tile, we must have
$A(k_1)=\ell _1$
and
$B(k_2)=\ell _2$
. At position
$(\ell _1,\ell _2-2)$
, there must be a blue tile
$\widehat {b_n^{n-1}}$
since only this tile has top label
$00n$
when
$n\geq 2$
. The current situation is illustrated below.
Consider the return block with support
$[A(k_1-1),A(k_1))\times [B(k_2),B(k_2+1))$
. It has label
$01\overline {n}$
at the far right of its bottom row. From Lemma 5.2, the width of this return block cannot be n, so it has to be
$$\begin{align*}A(k_1)-A(k_1-1)=n+1. \end{align*}$$
Now consider the return block with support
$[A(k_1-1),A(k_1))\times [B(k_2-1),B(k_2))$
. The white tile at position
$(A(k_1)-1,\ell _2-2)$
has right label
$11n$
. From the observation made in Figure 18, the width of this return block is
$$\begin{align*}A(k_1)-A(k_1-1)=n. \end{align*}$$
This is a contradiction. Thus,
$\left (\begin {array}{c} j_n^{1,1,1,1}\\ \widehat {g_n^{n}} \end {array}\right ) \notin \mathcal {L}({\Omega }_n)$
.
The same contradiction is obtained if we suppose that
$j_n^{0,1,1,1}$
appears at position
${\boldsymbol {\ell }}=(\ell _1,\ell _2)\in \mathbb {Z}^2$
and that
$\widehat {g_n^{n-1}}$
appears at position
$(\ell _1,\ell _2-1)$
in c. Indeed, a blue tile with left label
$11n$
is also forced to appear at position
$(\ell _1,\ell _2-2)$
.
Note that Lemma 10.7 cannot be extended to the case
$n=1$
.
Proposition 10.8. For every integer
$n\geq 2$
, the Wang shift
${\Omega }_n$
is minimal and is equal to the substitutive subshift
${\Omega }_n=\mathcal {X}_{{\omega }_n}$
.
Proof. Let
$n\geq 2$
be an integer. From Theorem C, the 2-dimensional substitution
${\omega }_n$
is primitive. Also,
${\omega }_n$
is expansive. From Theorem A, the Wang shift
${\Omega }_n$
is self-similar, satisfying
${\Omega }_n={\overline {{\omega }_n({\Omega }_n)}^{\sigma }}$
. Therefore, we may use Lemma 10.4 to show that the Wang shift
${\Omega }_n$
is minimal and
$\mathcal {X}_{{\omega }_n}={\Omega }_n$
. From Lemma 10.4, our goal is show that
for every size
$s\in \{2\times 2, 2\times 1, 1\times 2\}$
.
Case
$s=1\times 2$
. We have
However, we can estimate the set of vertical dominoes in
$\mathcal {L}({\Omega }_n)$
by the pair of tiles sharing the same label on the common horizontal edge excluding the two illegal dominoes from Lemma 10.7:
$$ \begin{align*} &\mathcal{L}_{1\times 2}({\Omega}_n) \cap \left\{ \left(\begin{array}{c} a\\ c \end{array}\right) \middle| \; a \text{ is a junction tile or a horizontal stripe tile } \right\}\\ &\subseteq \left\{ \left(\begin{array}{c} a\\ c \end{array}\right) \middle| \begin{array}{l} a\in\{ j_n^{0,1,0,0}, j_n^{0,1,0,1}, j_n^{0,1,1,1} \}\\ c\in\{\widehat{g_n^{n-1}},\widehat{y_n^{n-1}}\} \end{array} \right\} \setminus \left\{ \left(\begin{array}{c} j_n^{0,1,1,1}\\ \widehat{g_n^{n-1}} \end{array}\right) \right\} \hspace{10mm} \text{(tiles sharing edge label } 01n) \\ &\quad\cup \left\{ \left(\begin{array}{c} a\\ c \end{array}\right) \middle| \begin{array}{l} a\in\{ j_n^{1,1,0,1}, j_n^{1,1,1,1} \}\\ c\in\{\widehat{g_n^{n}},\widehat{y_n^{n}}\} \end{array} \right\} \setminus \left\{ \left(\begin{array}{c} j_n^{1,1,1,1}\\ \widehat{g_n^{n}} \end{array}\right) \right\} \hspace{10mm} \text{(tiles sharing edge label } 01\overline{n}) \\ &\quad\cup \left\{ \left(\begin{array}{c} a\\ c \end{array}\right) \middle| \begin{array}{l} a\in\{ j_n^{0,0,0,0}, j_n^{0,0,0,1} \}\\ c\in\{\widehat{b_n^{n-1}}\} \end{array} \right\} \hspace{15mm} (\text{tiles sharing edge label } 00n) \\ &\quad\cup \left\{ \left(\begin{array}{c} b_n^i\\ w_n^{k,n-1} \end{array}\right) \middle| \begin{array}{l} 0\leq i\leq n-1,\\ 1\leq k\leq n \end{array} \right\} \hspace{15mm} (\text{tiles sharing edge label } 11n) \\ &\quad\cup \left\{ \left(\begin{array}{c} y_n^i\\ w_n^{k,n} \end{array}\right) \middle| \begin{array}{l} 1\leq i\leq n,\\ 1\leq k\leq n \end{array} \right\} \hspace{15mm} (\text{tiles sharing edge label } 11\overline{n}) \\ &\quad\cup \left\{ \left(\begin{array}{c} g_n^i\\ w_n^{k,n} \end{array}\right) \middle| \begin{array}{l} 0\leq i\leq n,\\ 1\leq k\leq n \end{array} \right\} \hspace{15mm} (\text{tiles sharing edge label } 11\overline{n}). \end{align*} $$
Note that
Thus, we can compute the intersection of the two sets, and using Lemma 10.5, we obtain
Case
$s=2\times 1$
. The condition is satisfied because this case is symmetric to the case
$s=1\times 2$
.
Case
$s=2\times 2$
. The tiles appearing on the corners of images of letters under
${\omega }_n$
are quite restricted. Therefore, we have the following inclusion:
The above set has size
$2\times 4\times 4\times 2=64$
. Of those, only four belong to
$\mathcal {L}({\Omega }_n)$
because the choice made for the tile b imposes a unique choice for the tiles a, d and c. Thus, using Lemma 10.6, we obtain
From Lemma 10.4, we conclude that the Wang shift
${\Omega }_n$
is minimal and
${\Omega }_n=\mathcal {X}_{{\omega }_n}$
.
10.3 The Wang shift
${\Omega }_n$
is minimal when
$n=1$
From Theorem E,
${\mathcal {T}}_n$
is equivalent to the 16 Ammann Wang tiles when
$n=1$
. We know from [Reference Grünbaum and Shephard24] that the 16 Ammann Wang tiles are self-similar and that the self-similarity is recognizable (the decomposition of every configuration into the 16 supertiles shown in [Reference Grünbaum and Shephard24, Figure 11.1.6] is unique). This corresponds to the case
$n=1$
of Theorem A proved here. Therefore, from Lemma 10.1, we have
$\mathcal {X}_{\omega _1}\subseteq \Omega _1$
. The goal of this section is to prove that the equality holds and therefore that
$\Omega _1$
is minimal. Note that minimality of
$\Omega _1$
was not proved in [Reference Grünbaum and Shephard24], neither in the more recent works about Ammann A2 tilings [Reference Akiyama1, Reference Durand, Shen and Vereshchagin15].
The proof made in the previous section for
$n\geq 2$
does not directly work for
$n=1$
because it is not true anymore that next to a junction tile is never a junction tile. Indeed, when
$n=1$
, two junction tiles can be adjacent horizontally or vertically. This observation changes the description of vertical and horizontal dominoes that appear in the language.
Adapting the proof made above for
$n\geq 2$
to the case
$n=1$
is possible. But, instead of doing this, we have chosen to provide a proof based on computer experiments in order to check that the criterion provided in Lemma 10.4 is satisfied. We hope that it may be useful to study other examples.
Lemma 10.9. The Wang shift
$\Omega _1$
is minimal and
$\Omega _1=\mathcal {X}_{\omega _1}$
.
Proof. From Theorem C, the 2-dimensional substitution
$\omega _1$
is primitive. Also,
$\omega _1$
is expansive. From Theorem A, the Wang shift
$\Omega _1$
is self-similar, satisfying
$\Omega _1={\overline {\omega _1(\Omega _1)}^{\sigma }}$
. Therefore, we may use Lemma 10.4 to show the minimality of
$\Omega _1$
.
We compute below the patterns in
$\mathcal {L}_s(\Omega _1)$
and
$\mathcal {L}_s(\mathcal {X}_{\omega _1})$
for every size
$s\in \{2\times 2, 2\times 1, 1\times 2\}$
. As we observe below, these sets are equal. Therefore, it is not necessary to compute 
. We define
$\omega _1$
as a
$2$
-dimensional substitution over the alphabet
$\{0,1,2,\dots ,15\}$
according to the labeling of the tiles shown in Figure 31. We compute the patterns of size
$s\in \{2\times 2, 2\times 1, 1\times 2\}$
in the substitutive subshift
$\mathcal {X}_{\omega _1}$
:
We choose a solver to compute the dominoes and
$2\times 2$
patterns below. Three reductions are available: to a mixed-integer linear program, to a SAT instance or to an exact cover problem solved with Knuth’s dancing links algorithm [Reference Knuth31]. We use Knuth’s algorithm because it performs well and it is in SageMath by default.
We define the set
${\mathcal {T}}_1$
of Wang tiles in an order consistent with the labeling of the tiles with the indices in the set
$\{0,1,2,\dots ,15\}$
as shown in Figure 31. We compute the patterns of size
$s\in \{2\times 2, 2\times 1, 1\times 2\}$
in the Wang shift
$\Omega _1$
:
We compare the sets of horizontal dominoes, vertical dominoes and
$2\times 2$
patterns computed above within the language of the substitutive subshift
$\mathcal {X}_{\omega _1}$
and within the language of the Wang shift
$\Omega _1$
. We observe their equality:
Therefore, the above computations prove that we have the following equality:
$$\begin{align*}\mathcal{L}_s(\Omega_1) = \mathcal{L}_s(\mathcal{X}_{\omega_1}) \end{align*}$$
for every size
$s\in \{2\times 2, 2\times 1, 1\times 2\}$
. Thus, for every size
$s\in \{2\times 2, 2\times 1, 1\times 2\}$
, we have
From Lemma 10.4, we conclude that
$\Omega _1$
is minimal and
$\Omega _1=\mathcal {X}_{\omega _1}$
.
10.4 Proof of Theorem D
Theorem D. For every integer
$n\geq 1$
, the Wang shift
${\Omega }_n$
is minimal and is equal to the substitutive subshift
${\Omega }_n=\mathcal {X}_{{\omega }_n}$
.
11 Open questions
Note that the
$n^{th}$
metallic mean is a quadratic Pisot unit; that is, it is an algebraic unit of degree two, and all its algebraic conjugates have modulus strictly less than one. The other quadratic Pisot units are the positive roots of
$x^2-nx+1$
for
$n\geq 3$
. The family of quadratic Pisot units has nice properties [Reference Borwein and Hare10, Reference Komatsu32, Reference Masáková, Pastirčáková and Pelantová41]; see also [Reference Akiyama and Komornik3]. The continued fraction expansion of the positive root of
$x^2-nx+1$
is
$[n-1; (1, n-2)^\infty ]$
. In particular, it is not purely periodic.
Question 1. Let
$\beta $
be a positive quadratic Pisot unit which is not a metallic mean. Can we construct a self-similar set of Wang tiles whose inflation factor is
$\beta $
?
An alternative question is about those quadratic integers whose continued fraction expansion is purely periodic.
Question 2. Let
$\beta $
be a positive quadratic integer whose continued fraction expansion is purely periodic. Does there exist a set of Wang tiles such that the shift is self-similar with inflation factor equal to
$\beta $
?
The procedure explained in [Reference Grünbaum and Shephard24, p.594–598] starts from the Ammann A2 shapes shown in Figure 1 and constructs a set of 16 Wang tiles which we show in Theorem E to be equivalent to the set
${\mathcal {T}}_1$
. A question we can ask is whether this construction can be inverted. More precisely, starting from the Ammann set of 16 Wang tiles, can we recover the two Ammann shapes shown in Figure 1 with their Ammann bars? In general, we ask the following question.
Question 3. For every integer
$n\geq 1$
, can we find geometrical shapes with Ammann bars on them such that encoding their tilings by rhombi along a pair of Ammann bars is equivalent to the tiles
${\mathcal {T}}_n$
?
Theorem E together with the discussion [Reference Grünbaum and Shephard24, p.594–598] is an answer to Question 3 when
$n=1$
. An answer to Question 3 would shed light on Mr. Ammann’s remarkable insights [Reference Senechal57].
Relation to the work of Mozes
Let
$n\geq 1$
be an integer and recall the 1-dimensional substitution
$$\begin{align*}\rho_n = \begin{cases} \texttt{a} \mapsto \texttt{ab}^{n}\\ \texttt{b} \mapsto \texttt{ab}^{n-1} \end{cases} \end{align*}$$
over alphabet
$\{\texttt {a},\texttt {b}\}$
defined in the proof of Lemma 9.2. The incidence matrix of
$\rho _n$
is
$\left (\begin {smallmatrix} 1&1\\ n&n-1 \end {smallmatrix}\right )$
whose characteristic polynomial is
$x^2-nx-1$
, and whose Perron–Frobenius dominant eigenvalue is the
$n^{th}$
metallic mean. A right dominant eigenvector is
$\left (\begin {smallmatrix} 1\\ \beta _n-1 \end {smallmatrix}\right )$
and a left dominant eigenvector is
$\left (\begin {smallmatrix} n & \beta _n-1 \end {smallmatrix}\right )$
. Following the theory on inflation tilings [Reference Baake and Grimm5, § 6], a stone inflation associated with the substitution
$\rho _n$
gives a volume of n to the letter
$\texttt {a}$
and a volume of
$\beta _n-1$
to the letter
$\texttt {b}$
. The stone inflation induced by the direct product
$\rho _n\times \rho _n$
of the substitution
$\rho _n$
with itself in the sense of [Reference Mozes43, § 6] is shown in Figure 30; see also [Reference Baake and Grimm5, Example 5.9]. Note that another substitution with same inflation factor and often used in examples illustrating metallic means is
$\texttt {a}\mapsto \texttt {a}^n\texttt {b}, \texttt {b}\mapsto \texttt {a}$
[Reference Baake and Grimm5, Remark 4.7].
Figure 30 Stone inflation associated with the direct product of the substitution
$\rho _n$
with itself with inflation factor equal to
$\beta _n$
, the
$n^{th}$
metallic mean. The size of the rectangles are given by the entries of a Perron–Frobenius dominant left-eigenvector of the incidence matrix of
$\rho _n$
. The figure is drawn with parameter
$n=4$
. Color is added to the tiles to differentiate them and visually link them to the tiles in
${\mathcal {T}}_n$
.
From the work of Mozes [Reference Mozes43], we know that there exists a tiling system given by a finite set of tiles and a finite set of matching rules such that the tiling system is a symbolic extension of the substitutive dynamical system generated by the 2-dimensional substitution
$\rho _n\times \rho _n$
over a four-letter alphabet. Since the substitution
$\rho _n\times \rho _n$
is recognizable (or has ‘unique derivation’, using the vocabulary of Mozes), the tiling system constructed by Mozes is even measure-theoretically isomorphic to the substitutive dynamical system. Note that the construction of an equivalent tiling system out of a substitution was extended to geometric substitutions [Reference Goodman-Strauss23].
In this contribution, we provide an explicit construction of a tiling system
${\Omega }_n$
which is a symbolic extension of the 2-dimensional substitutive subshift defined by
$\rho _n\times \rho _n$
. The set of Wang tiles deduced from [Reference Mozes43] when applied on
$\rho _n\times \rho _n$
would be much larger than
$(n+3)^2$
. This raises a question about the optimality of a tiling system for 2-dimensional substitutions.
Question 4. Is the size of
${\mathcal {T}}_n$
optimal? In other words, does there exist a set
$\mathcal {T}$
of Wang tiles of cardinality
$\#\mathcal {T}<(n+3)^2$
such that the Wang shift
$\Omega _{\mathcal {T}}$
is isomorphic to the 2-dimensional substitutive subshift
$\mathcal {X}_{\rho _n\times \rho _n}$
?
A Appendix A: The substitutions
${\omega }_n$
for
$1\leq n\leq 5$
Figure 31 Substitution
$\omega _1$
.
Figure 32 Substitution
$\omega _2$
.
Figure 33 Substitution
$\omega _3$
(rotated 90 degrees counterclockwise).
Figure 34 Substitution
$\omega _4$
(rotated 90 degrees counterclockwise).
Figure 35 Substitution
$\omega _5$
(rotated 90 degrees counterclockwise).
B Appendix B: Proving the self-similarity of
$\Omega _2$
in SageMath
In this section, we illustrate how Theorem A can be proved in SageMath for a specific but not too big integer
$n\geq 1$
. Since the proof of Theorem A given in this article was deduced from such computer experiments performed for small values of n, we hope that the approach shown below can be used to study and show the self-similarity of other aperiodic set of Wang tiles.
We use here a method proposed in [Reference Labbé35] to study the substitutive structure of the Jeandel–Rao Wang shift [Reference Jeandel and Rao26]. The method is based on the notion of marker tiles (not to be confused with the notion of marker used in Lemma 10.1.8 from [Reference Lind and Marcus40]). A nonempty subset
$M\subset \mathcal {A}$
is called markers for the direction
${\boldsymbol {e}}_2$
within a subshift
$X\subset \mathcal {A}^{\mathbb {Z}^2}$
if for every configuration
$x\in X$
, the positions of the markers are nonadjacent rows; that is,
$x^{-1}(M)=\mathbb {Z}\times P$
for some set
$P\subset \mathbb {Z}$
such that
$1\notin P-P$
. A symmetric definition holds for markers for the direction
${\boldsymbol {e}}_1$
. It was proved that the existence of marker tiles allows to decompose uniquely a Wang shift. Informally, marker tiles are merged with the tiles that appear just on top of (or just below) them. Remaining tiles are kept unchanged. The search for markers and the construction of the substitution is performed by two algorithms FindMarkers and FindSubstitution. Their pseudocode can be found in [Reference Labbé35]; see also the chapter [Reference Labbé37] where a simpler example is considered.
Below, we prove the self-similarity of
${\Omega }_n$
when
$n=2$
using SageMath [Reference Developers54] with optional package slabbe [Reference Labbé38]. The algorithms FindMarkers and FindSubstitution are used twice horizontally and then twice vertically. The computations show that every configuration in
$\Omega _2$
can be decomposed uniquely into 25 supertiles. The 25 supertiles are equivalent to the original set of 25 tiles. Thus, the Wang shift
$\Omega _2$
is self-similar and we compute the self-similarity.
We choose a solver to search for markers and desubstitutions below.
First, we define the set
${\mathcal {T}}_2$
of Wang tiles.
Then, we search for markers for the direction
${\boldsymbol {e}}_1$
(such markers appear on nonadjacent columns). We fusion the markers with the possible tiles appearing on their right (thus the marker appear on the left side of each pair).
The resulting set of Wang tiles (shown above at the source of the arrows) is obtained by concatenating the top and bottom labels of the merged pairs:
It turns out that tiles with indices 11, 14, 20, 27 are not needed within the above set of tiles as they do not have a surrounding of radius 2 as confirmed by the following computation. Thus, they cannot appear in any tiling. In fact, they correspond to antigreen tiles and other tiles proved to be illegal in Section 7. We compute the remaining twenty five tiles below.
We confirm that the set
$U_5$
is equivalent to the set
${\mathcal {T}}_n$
of Wang tiles we started with. We extract the bijection s6 between the indices of the tiles. Also, it gives a bijection for the horizontal edge labels and vertical edge labels. Both are equal. This bijection corresponds to the map
$\tau _n$
when
$n=2$
defined in Section 5.1.
One may compare the bijection computed above with the map
$\tau _n$
defined in Section 5. The only difference is that the image of the label
$003$
does not appear in the computed bijection above because it is does not appear as an edge label in the set
${\mathcal {T}}_2$
.
The self-similarity is:
The characteristic polynomial of the incidence matrix of the self-similarity is:
The self-similarity shown with the associated Wang tiles:
Acknowledgements
The author is thankful to the reviewers for their careful reading and remarks which led, in particular, to an improved and more formal proof of the self-similarity of
${\omega }_n$
. The author also wants to thank Dirk Frettlöh for making him aware of other existing aperiodic substitutive tilings involving metallic mean numbers, including the bronze mean [Reference Dotera, Bekku and Ziherl14].
Competing interest
The authors have no competing interests to declare.
Funding statement
This work was partly funded from France’s Agence Nationale de la Recherche (ANR) projects CODYS (ANR-18-CE40-0007) and IZES (ANR-22-CE40-0011). It was also supported by grants from the Symbolic Dynamics and Arithmetic Expansions (SymDynAr) Project, co-funded by ANR (ANR-23-CE40-0024) and FWF (https://doi.org/10.55776/I6750), the Austrian Science Fund.
Reproducibility statement
All results proved in this article are proved by hand except the proof of Lemma 10.9 and the computations performed in Appendix B, which are based on the open-source mathematical software SageMath [Reference Developers54] and the optional package slabbe [Reference Labbé38]. All SageMath input/output blocks in this article were created using the sageexample environment with SageTeX version 2021/10/16 v3.6 and with the following software versions:
The fact that these software are open-source means that anyone is free to use, reproduce, verify, adapt for their own needs all of the computations performed therein according to the GNU General Public License (version 2, 1991, http://www.gnu.org/licenses/gpl.html).
The contents of all of the sageexample environments from the tex source are gathered in the file demos/arXiv_2312_03652_doctest.sage autogenerated by SageTeX when running pdflatex. This file is included in the slabbe package and available at https://gitlab.com/seblabbe/slabbe/. It allows to make sure that future releases of the package do not break the code included in this article. It is possible to reproduce all computations present in this article and check that all outputs are correct, by doctesting this file – that is, by running the command sage -t demos/arXiv_2312_03652_doctest.sage. It should output All tests passed! and [67 tests, 31.02s wall] (most probably with a different timing).
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