Facet Connectedness of Arithmetic Discrete Hyperplanes with Non-Zero Shift

Abstract

We present a criterion for the arithmetic discrete hyperplane to be facet connected when $\theta$ is the connecting thickness . We encode the shift $\mu$ in a numeration system associated with the normal vector and we describe an incremental construction of the plane based on this encoding. We deduce a connectedness criterion and we show that when the Fully Subtractive algorithm applied to has a periodic behaviour, the encodings of shifts $\mu$ for which the plane is connected may be recognised by a finite state automaton.

1 Introduction

The arithmetic discrete hyperplane with normal vector , shift $\mu \in \mathbb {R}$ and thickness $\theta \in \mathbb {R}$, is the set of points in $\mathbb {Z}^d$ defined by

where $\langle \cdot , \cdot \rangle$ denotes the usual scalar product on $\mathbb {R}^d$ [1, 9].

For $\kappa$ in $\{0,\dots ,d-1\}$, two points and in $\mathbb {Z}^d$ are $\kappa$-neighbours if and only if and . In particular, and are $(d-1)$-neighbours if and only if for some i, where is the canonical basis of $\mathbb {R}^d$. A set $S \subset \mathbb {Z}^d$ is $\kappa$-connected if and only if for all and in S, there exists a path in S where and are $\kappa$-neighbours for all $i=1,\dots ,n-1$.

Many works have been devoted to determining the conditions under which the hyperplane is $\kappa$-connected, see [3,4,5, 7] among others. The set of thicknesses $\theta$ for which is non-empty and $\kappa$-connected is a right unbounded interval of $\mathbb {R}^+$ the lower bound of which, denoted by , is called the $\kappa$-connecting thickness of with shift $\mu$. In this work, we consider only the $(d-1)$-connectedness which is the only case where a general algorithm is known to compute the $\kappa$-connecting thickness. We shall therefore drop the subscript $\kappa$ in all denotations and speak of connectedness and connecting thickness.

By definition of , is empty or disconnected for all and is non-empty and connected for all . The question which arises naturally is whether is connected or not. In [5], we showed that is almost always disconnected, excepted when belongs to some specific set which was proven to be Lebesgue negligible by Kraaikamp and Meester [8]. In this case, which does not actually depend on $\mu$ and will therefore be denoted simply by . We showed that for such vectors, may be connected or not, depending on the value of $\mu$. In particular, is always connected while is always disconnected.

In the present work, we investigate the general case where $\mu$ is arbitrary. We first set our main definitions and recall how the connecting thickness may be computed by means of the Fully Subtractive algorithm. Then we introduce a numeration system associated with the normal vector , in which the shift $\mu$ may be encoded as an infinite sequence in $\{0,1\}^\omega$. Based on this encoding, we describe an incremental construction of and we show that, provided that the encoding of $\mu$ has some property called bi-admissibility, we actually construct the connected component of in . We first give a purely geometrical connectedness criterion based on a particular projection of a connected component of . Then we investigate specific patterns occurring in the encoding of $\mu$ and we introduce the notion of minimal interior patterns which allows us to deduce a more operational connectedness criterion. Finally, we study the case where the Fully Subtractive algorithm applied to the normal vector has a periodic behaviour. We show that in this case, the language of encodings of shifts $\mu$ for which is connected is a regular language which may therefore be recognised by a finite state automaton.

2 Preliminaries

The $(d-1)$-connecting thickness may be computed by means of the Fully Subtractive algorithm as follows. Let be the vector $(1,\dots ,1)$. For each $k \in \{1,\dots ,d\}$, we define $\gamma _k$ as

Given a vector , the algorithm computes a sequence of pairs where . For all $n\ge 1$, if has at least two components and no component is 0 then where $\delta _n$ is the index of a minimal coordinate of . If some component becomes 0 in the process, it is simply erased and we continue with . If has only one component left, the algorithm stops. This sequence has the property that for all $n\ge 1$, is non-empty and connected if and only if is non-empty and connected [4, 5].

If the process terminates with , which happens if and only if for some integral vector  [5], then . Otherwise, converges to and which does not depend on $\mu$ and will therefore be denoted simply by .

The set $\mathcal {K}_d$ is defined as the set of vectors such that the Fully Subtractive algorithm erases no component and for all $n\ge 1$. If then and . We proved in [5] that may be connected only if for some $n \ge 1$, belongs to some $\mathcal {K}_{d'}$, in which case, for all $m\ge n$. Kraaikamp and Meester proved in [8] that $\mathcal {K}_d$ is Lebesgue negligible when $d\ge 3$.

From now on, is a vector in $\mathcal {K}_d$ with $d \ge 3$ and .

Given the sequence of vectors computed by the Fully Subtractive algorithm, we define the directive sequence $\Delta = (\delta _n)_{n\ge 1}$ where $\delta _n$ is the index of the minimal component of . This minimal component is unique since otherwise would have a zero component, which is impossible by definition of $\mathcal {K}_d$. The sequence $\Delta$ is therefore well defined. For all $n\ge 1$ we set $\Delta ^n = \delta _n\delta _{n+1}\dots$ and .

To each vector in $\mathcal {K}_d$ is associated a unique sequence $\Delta$ which has the property that each $k \in \{1,\dots ,d\}$ occurs infinitely many times in $\Delta$. Conversely, each sequence $\Delta \in \{1,\dots ,d\}^\omega$ in which each $k\in \{1,\dots ,d\}$ occurs infinitely many times is associated with a unique direction in $\mathcal {K}_d$, which means that the vector is determined by $\Delta$ up to a multiplicative factor [5, 6].

For all $n\ge 1$, we set and where is the transpose of $\gamma _i$. We have and for all $n\ge 1$, $\Omega ^n=\sum _{i\ge n}\theta _i$. In particular, $\Omega = \Omega ^1 = \sum _{i\ge 1} \theta _i$.

We denote by $\mathbb {Z}^\omega$ the set of sequences of integers and by $\mathbb {Z}^\star 0^\omega$ the set of sequences with finitely many non-zero terms. The length of $\sigma \in \mathbb {Z}^\omega$, denoted by $|\sigma |$, is the smallest index $n\ge 0$ such that $\forall m>n,\ \sigma _m=0$. If no such index exists then $|\sigma |=+\infty$. If $\sigma$ is a finite sequence then $|\sigma |$ denotes its natural length. We define the mappings $\psi ^\Delta$ on $\mathbb {Z}^\star 0^\omega$ by $\psi ^\Delta (\sigma ) = \sum _n \sigma _n\,T^\Delta _n$ and on $\mathbb {Z}^\omega$ by when this sum converges. It is in particular the case when $\sigma$ is bounded. Given $\sigma \in \{0,1\}^\omega$ we denote by $\widetilde{\sigma }$ the sequence obtained by replacing zeros with ones and ones with zeros in $\sigma$. We have . When $\sigma \in \mathbb {Z}^\star 0^\omega$, we have . These definitions are naturally extended to the case where $\sigma \in \mathbb {Z}^\star$ by $\psi ^\Delta (\sigma ) = \psi ^\Delta (\sigma 0^\omega )$ and .

If $\xi ^{\Delta ,k}_n$ is the index of the $n^{\mathrm {th}}$ occurrence of k in $\Delta$, we recall [5, 6] that for all $k\in \{1,\dots ,d\}$ and for all $n\ge 1$, we have

$$\begin{aligned} \theta _1+\dots +\theta _{\xi ^{\Delta ,k}_1}= & {} v_k \end{aligned}$$

(1)

$$\begin{aligned} \theta _{\xi ^{\Delta ,k}_n+1}+\dots +\theta _{\xi ^{\Delta ,k}_{n+1}}= & {} \theta _{\xi ^{\Delta ,k}_n} \end{aligned}$$

(2)

In the sequel, we symmetrise our problem by studying the connectedness of so that . We have if and only if there exists such that . In this case, we have which is disconnected when $d\ge 3$ [5]. In all other cases, is connected if and only if is.

We may also reduce the study to the case where $\mu \in [0;\Omega ]$. Indeed, if $q=\left\lfloor {\mu /v_1}\right\rfloor$ and $\mu '=\mu -q\,v_1$, then , and . Thus is a translate of by an integral vector and is connected if and only if is.

3 $\Delta$-Numeration System

We showed in [5] that , meaning that . We now extend this result.

Theorem 1

To prove this theorem, we need a technical lemma.

Lemma 2

• 1. $2 \theta _1 < \Omega$.

• 2. If $d \ge 3$, $j < n$ and $\delta _j \ne \delta _n$, then $\theta _j + \sum _{i=1}^n \theta _i > \Omega$.

• 3. If $d\ge 3$ and then $v_i+v_j > \Omega$ for all $i \ne j$.

Proof of Theorem 1. The inclusion is obvious since for all $\sigma \in \{0,1\}^\omega$, we have .

Now let $x \in [0;\Omega ]$, $r_1 = x$ and for all $n \ge 1$, $r_{n+1} = r_n - \sigma _n\,\theta _n$ where $\sigma _n = 1$ if $r_n \ge \theta _n$ and $\sigma _n=0$ otherwise. Then we have $x = \sum _{k<n} \sigma _k\,\theta _k + r_n$ and $r_n \in [0;\Omega ^{n}]$ for all n. Indeed, we obviously have $r_n \ge 0$ for all n. By hypothesis, we have $r_1 = x \in [0;\Omega ^{1}] = [0;\Omega ]$. Now, for $n\ge 1$, assume that $r_n \le \Omega ^{n}$ and let us show that $r_{n+1} \le \Omega ^{n+1}$. If $r_n < \theta _n$ then $\sigma _n=0$ and $r_{n+1} = r_n < \theta _n$. By Lemma 2 applied to we get $2\,\theta _n < \Omega ^{n}$ and thus $\theta _n < \Omega ^{n}-\theta _n = \Omega ^{n+1}$. If $r_n \ge \theta _n$, then $\sigma _n=1$ and $r_{n+1} = r_n-\theta _n \le \Omega ^{n}-\theta _n = \Omega ^{n+1}$.

Since and , we have $\lim _{n\rightarrow \infty } r_n = 0$ hence $\lim _{n\rightarrow \infty } \sum _{k<n} \sigma _k\,\theta _k = \sum _{n\ge 1} \sigma _n\,\theta _n = x$. Then we have $\sigma = (\sigma _n)_{n\ge 1} \in \{0,1\}^\omega$ and .   $\square$

The sequence $(\theta _n)_{n\ge 1}$ may be seen as the basis of a numeration system in which we can encode any $\mu \in [0;\Omega ]$ as a sequence $\overline{\mu }\in \{0,1\}^\omega$. Note that by considering a bi-infinite sequence $\Delta$, we would define a full numeration system in which any non negative real number may be encoded as $\sigma .\sigma '$ with $\sigma \in \{0,1\}^\star$ and $\sigma ' \in \{0,1\}^\omega$.

Definition 3

($\Delta$-admissible sequence) 

• A sequence $\sigma \in \{0,1\}^\omega$ is $\Delta$-admissible iff there does not exist $k\in \{1,\dots ,d\}$ and an index $n_0$ such that $\sigma _n=0 \iff \delta _n = k$ for all $n\ge n_0$.

• A sequence $\sigma \in \{0,1\}^\omega$ is $\Delta$-bi-admissible iff $\sigma$ and $\widetilde{\sigma }$ are $\Delta$-admissible.

• If there exist k and $n_0$ such that $\sigma _n=0 \iff \delta _n = k$ for all $n\ge n_0$, then k is unique. In this case, we say that $\sigma$ is $\Delta$-non-admissible of type k.

A $\Delta$-non-admissible sequence is analogous to the writing $1=0.9999\dots$ in the classical numeration system in base 10. For instance, if $\Delta =(12213)^\omega$ then $\sigma = 11001(01101)^\omega$ is $\Delta$-non-admissible of type 1. We have actually . This principle is general: for all $\Delta$-non-admissible sequence $\sigma$, a finite sequence $\sigma '$ exists such that .

Proposition 4

Let $\sigma \in \{0,1\}^\omega$ be a $\Delta$-non-admissible sequence of type k, $n_0$ such that $\delta _{n_0}=k$ and $\sigma _n = 0 \iff \delta _n = k$ for all $n\ge n_0$. Then .

Proof

For all $n\ge 0$, let $\zeta _n = \xi ^{\Delta ,k}_n$. Let $r_0$ be such that $n_0 = \zeta _{r_0}$ and for all $r\ge r_0$, let $\sigma ^{r} = \sigma _1\dots \sigma _{\zeta _r-1}1$. We have $\sigma ^{r_0} = \sigma _1\dots \sigma _{n_0-1}1$ and by Eq. 2, for all $r\ge r_0$, . Then

and since and , we get the result at the limit.   $\square$

If a sequence $\sigma$ is not $\Delta$-bi-admissible then, by applying this transformation to either $\sigma$ or $\widetilde{\sigma }$, we get in all cases a $\Delta$-bi-admissible sequence. Hence the corollary:

Corollary 5

For all $\sigma \in \{0,1\}^\omega$, there exists a $\Delta$-bi-admissible sequence $\sigma '$ such that .

Definition 6

($\Delta$-normalised sequence). A sequence $\sigma \in \{0,1\}^\omega$ is $\Delta$-normalised if and only if it is $\Delta$-admissible and it does not contain a sub-sequence $01^{k}$ with $k\ge 1$ at a position i such that $\delta _i=\delta _{i+k}$.

Proposition 7

For all $\mu \in [0;\Omega ]$, there exists a unique $\Delta$-normalised sequence $\sigma \in \{0,1\}^\omega$ such that .

Definition 8

($\Delta$-normal form). Given a sequence $\sigma \in \mathbb {N}^\omega$ such that , the unique $\Delta$-normalised sequence $\sigma ' \in \{0,1\}^\omega$ such that is called the $\Delta$-normal form of $\sigma$ and is denoted by $\sigma \mathord \Downarrow ^\Delta$.

Note that given a sequence $\sigma$, there does not always exist a sequence $\sigma '$ which is both $\Delta$-bi-admissible and $\Delta$-normalised and such that . For instance, if $\Delta =(123)^\omega$ then $\sigma = (100)^\omega$ is $\Delta$-normalised but not $\Delta$-bi-admissible because $\widetilde{\sigma } = (011)^\omega$ is not $\Delta$-admissible. Since the $\Delta$-normal form is unique, there is no $\Delta$-normalised and $\Delta$-bi-admissible sequence $\sigma '$ such that .

4 Incremental Construction of

In [5] and [6], we presented an incremental construction of when $\mu =0$. We now extend this construction to the case where $\mu$ is arbitrary.

As said before, we may always reduce the study of the connectedness of to the case where $\mu \in [0;\Omega ]$. Then there exists a sequence $\overline{\mu }= (\mu _i)_{i \ge 1} \in \{0,1\}^\omega$ such that . We define the sequence $(\mathcal {S}^{\Delta ,\overline{\mu }}_n)_{n\ge 0}$ of subsets of $\mathbb {Z}^d$ by and for all $n\ge 1$,

For example, let where $\alpha \in [0;1]$ and $\alpha +\alpha ^2+\alpha ^3=1$. We have $\Delta =(123)^\omega$. The picture below shows $\mathcal {S}^{\Delta ,\overline{\mu }}_{12}$ for $\overline{\mu }=(01)^\omega$ and $\overline{\mu }=(011001)^\omega$. We have and . We shall see in Sect. 6 that is connected in the first case and disconnected in the second case.

We observe that the geometry of $\mathcal {S}^{\Delta ,\overline{\mu }}_{n}$ does not seem to depend on $\overline{\mu }$. Only the position with respect to the origin changes. Indeed, we have

Where $\mathcal {S}^\Delta _n$ stands for $\mathcal {S}^{\Delta ,0^\omega }_n$. For all $\overline{\mu }\in \{0,1\}^\omega$ and all $n \ge 0$, $\mathcal {S}^{\Delta ,\overline{\mu }}_n$ is a translate of $\mathcal {S}^\Delta _n$ by an integral vector, and therefore inherits topological properties of $\mathcal {S}^\Delta _n$, especially the fact that $\mathcal {S}^{\Delta ,\overline{\mu }}_n$ is symmetric, connected and circuit-free [5, 6].

Let $\mathcal {S}^{\Delta ,\overline{\mu }}_\infty$ be the limit of the sequence $(\mathcal {S}^{\Delta ,\overline{\mu }}_n)_{n\ge 0}$. We have

and as an immediate corollary, we get that for all $\overline{\mu }\in \{0,1\}^\omega$, $\mathcal {S}^{\Delta ,\overline{\mu }}_\infty$ is connected and circuit-free.

When $\mu =0$, we have  [5]. The next theorem establishes the relationship between $\mathcal {S}^{\Delta ,\overline{\mu }}_\infty$ and .

Theorem 9

If $d\ge 3$, $\overline{\mu }\in \{0,1\}^\omega$ is a $\Delta$-bi-admissible sequence and , then $\mathcal {S}^{\Delta ,\overline{\mu }}_\infty$ is the connected component of in .

Proof

(sketch). We observe that by hypothesis, we have $\mu \in \left[ 0;\Omega \right]$ and thus . Then we prove that any neighbour in of a point belongs to $\mathcal {S}^{\Delta ,\overline{\mu }}_\infty$. Let $\Sigma ^{\overline{\mu }} = (\{0,1\}^\omega - \overline{\mu }) \cap \{-1,0,1\}^\star 0^\omega$, i.e. the set of sequences $\sigma \in \{-1,0,1\}^\star 0^\omega$ such that $\sigma +\overline{\mu }\in \{0,1\}^\omega$. We have $\mathcal {S}^{\Delta ,\overline{\mu }}_\infty = \psi ^\Delta (\Sigma ^{\overline{\mu }})$. We consider a sequence $\sigma \in \Sigma ^{\overline{\mu }}$ such that and we assume . Then we exhibit a sequence $\tau \in \Sigma ^{\overline{\mu }}$ such that .   $\square$

The fact that $\overline{\mu }$ is $\Delta$-bi-admissible is crucial in the theorem above. Let us consider for instance $\Delta =(123)^\omega$ and $\overline{\mu }= (011)^\omega$. This sequence is $\Delta$-non-admissible of type 1 since for $n\ge 1$, we have $\mu _n=0 \iff \delta _n=1$. We have and hence . In this case, which is connected and is therefore equal to the connected component of . However, we may show that where is the connected component of in which is disconnected. Similarly, $\overline{\mu }= (100)^\omega$ is $\Delta$-admissible but not $\Delta$-bi-admissible since $\widetilde{\overline{\mu }} = (011)^\omega$ which is not $\Delta$-admissible. In this case, and we have which is connected. But .

Proposition 10

If $\overline{\mu }$ is a $\Delta$-non-admissible sequence of type k and $\overline{\mu }'$ is a $\Delta$-admissible sequence such that $|\overline{\mu }'| < +\infty$ and , then where is the connected component of in .

When is disconnected, the incremental construction builds only , the connected component of in . We may build the other connected components of as follows. Given a point , we have . Since , we have and there exists a $\Delta$-bi-admissible encoding $\overline{\mu }'$ of . Then $\mathcal {S}^{\Delta ,\overline{\mu }'}_\infty$ is the connected component of in which means that is the connected component of in . This way, we may build all the connected components provided that we know a starting point in each of them. may even be built directly by initialising the construction with instead of . Note that initialising the construction with without considering $\overline{\mu }'$ does not give the result. It would build which is different from .

For instance, let us consider a vector the directive sequence of which is $\Delta =(1213)^\omega$, $\overline{\mu }=(1000)^\omega$ which is $\Delta$-bi-admissible and . Then has two connected components. The point belongs to and a $\Delta$-bi-admissible encoding of is $\overline{\mu }'=(0011)^\omega$. Figure 1 shows and .

Fig. 1.

The two connected components of when $\Delta =(1213)^\omega$

5 Connectedness Criterion

Let $\mathcal {B}$ be the closed ball $[-\frac{1}{2},+\frac{1}{2}]^d$, the vector $(1,\dots ,1)$ and the orthogonal projection on the plane . By a slight abuse of notation, for any set $S\subset \mathbb {R}^d$, we denote by and the interior and the boundary of in .

Theorem 11

Let , $\mu \in \left[ 0,\Omega \right]$ and the connected component of in . Then is connected if and only if is empty.

The proof of this theorem requires two additional lemmas.

Lemma 12

For all , we have .

Lemma 13

Let such that and . Then .

Proof of Theorem 11. We first note that since $\mu \in \left[ 0,\Omega \right]$ and thus . Since , the discrete plane is $(d-1)$-separating [1] which implies and therefore . If is connected then hence . Its boundary in is therefore empty.

Assume now that is disconnected and let . Then contains no point of the form . Indeed, if belongs to then it is connected to in and cannot belong to . But these points are the only which could belong to such that . We deduce that is disjoint from which therefore has a non-empty boundary in .   $\square$

Definition 14

(Interior and border of $\mathcal {S}^{\Delta ,\overline{\mu }}_n$). Let :

• is interior in $\mathcal {S}^{\Delta ,\overline{\mu }}_n$ if and only if is contained in ;

• is on the border of $\mathcal {S}^{\Delta ,\overline{\mu }}_n$ if and only if it is not interior in $\mathcal {S}^{\Delta ,\overline{\mu }}_n$.

We saw earlier that $\mathcal {S}^{\Delta ,\overline{\mu }}_n = \mathcal {S}^\Delta _n - \psi ^\Delta (\mu _1\dots \mu _n)$. Thus a point is interior in $\mathcal {S}^{\Delta ,\overline{\mu }}_n$ if and only if is interior in $\mathcal {S}^\Delta _n$. Therefore we shall mainly study $\mathcal {S}^\Delta _n$. All results will apply immediately to $\mathcal {S}^{\Delta ,\overline{\mu }}_n$ by translation.

Proposition 15

Let $\sigma \in \{0,1\}^n$ such that $\psi ^\Delta (\sigma )$ is interior in $\mathcal {S}^\Delta _n$.

• For all $\sigma ' \in \{0,1\}^\star$, $\psi ^\Delta (\sigma \sigma ')$ is interior in $\mathcal {S}^\Delta _{n+|\sigma '|}$.

• For all $\Delta _0 \in \{0,\dots ,d\}^\star$ and all $\sigma _0 \in \{0,1\}^{|\Delta _0|}$, $\psi ^{\Delta _0\Delta }(\sigma _0\sigma )$ is interior in $\mathcal {S}^{\Delta _0\Delta }_{|\Delta _0|+n}$.

Again, we need two technical lemmas.

Lemma 16

Let be the 0-neighbourhood of in $\mathbb {Z}^d$. A point is interior in $\mathcal {S}^\Delta _n$ if and only if for all such that , and all i, j such that $i\ne j$, $y_i \le 0$ and $y_j \ge 0$, $\mathcal {S}^\Delta _n$ contains at least one point in .

Lemma 17

Let $\Delta =\delta _1\dots \delta _{n_0} \in \{1,\dots ,d\}^{n_0}$, $\Delta ' \in \{1,\dots ,d\}^\omega$ and where is the transpose of $\gamma$. For all $n\ge 0$, we have

$$ \mathcal {S}^{\Delta \Delta '}_{n_0+n} = \mathcal {S}^{\Delta }_{n_0} + \varGamma ^{\Delta }(\mathcal {S}^{\Delta '}_n) $$

The picture below shows an example of this composition.

Proof of Proposition 15 (sketch). The first point is obvious since $\mathcal {S}^\Delta _n \subset \mathcal {S}^\Delta _m$ for all $n\le m$.

We prove the second point for $|\Delta _0|=1$, i.e. $\Delta _0 = \delta \in \{1,\dots ,d\}$, and $\sigma _0 = \varepsilon \in \{0,1\}$. The general result follows by induction. We consider more generally a set $\mathcal {S}$ having the property of Lemma 12. Using Lemmas 16 and 17, we determine that we have to prove that if is interior in $\mathcal {S}$, then for all such that , and all i, j such that $i\ne j$, $y_i \le 0$ and $y_j \ge 0$, $\mathcal {S}$ contains at least one point in

We consider 18 cases according to the respective values of $\varepsilon$, i, j and $y_\delta$.   $\square$

Lemma 16 allows us to determine whether a point is interior in $\mathcal {S}^\Delta _n$ by looking only at its 0-neighbourhood. Using this lemma we may compute all possible configurations in which is interior. These are subsets of the 0-neighbourhood of which satisfy the conditions of the lemma. By Lemma 12, we may restrict this neighbourhood to the points such that . We may also keep only the configurations which are minimal with respect to inclusion. Then is interior in $\mathcal {S}^\Delta _n$ if and only if its 0-neighbourhood in $\mathcal {S}^\Delta _n$ contains one of these minimal configurations.

We find 64 minimal configurations in dimension 3 and 1498 in dimension 4. If we consider only configurations which appear in planes the coordinates of the normal vector of which are sorted in ascending order, we find, in dimension 3, the 21 minimal configurations shown below. Other minimal configurations are obtained by permuting the coordinates.

Lemma 18

If for all n, is at bounded distance of the border of $\mathcal {S}^{\Delta ,\overline{\mu }}_n$, then there exists $n_0$ such that is on the border of $\mathcal {S}^{\Delta ^{n_0},\overline{\mu }^{n_0}}_n$ for all n, where $\overline{\mu }^{n_0}$ denotes the sequence $(\mu _n)_{n\ge n0}$.

Proof

Since is at bounded distance of the border of $\mathcal {S}^{\Delta ,\overline{\mu }}_n$, $\mathcal {S}^{\Delta ,\overline{\mu }}_\infty$ has a non-empty border.

If is on the border of $\mathcal {S}^{\Delta ,\overline{\mu }}_\infty$ then we may take $n_0=1$. Otherwise, let m be such that is interior in $\mathcal {S}^{\Delta ,\overline{\mu }}_m$ or equivalently, $\psi ^\Delta (\mu _1\dots \mu _m)$ is interior in $\mathcal {S}^\Delta _m$. For all $k\ge 0$, we have $\mathcal {S}^\Delta _{m+k} = \mathcal {S}^{\delta _1\dots \delta _m}_m+ \varGamma ^{\delta _1\dots \delta _m}(\mathcal {S}^{\Delta ^{m+1}}_k)$. The distance from $\psi ^{\Delta ^{m+1}}(\mu _{m+1}\dots \mu _{m+k})$ to the border of $\mathcal {S}^{\Delta ^{m+1}}_k$ is strictly less than the distance from $\psi ^\Delta (\mu _1\dots \mu _{m+k})$ to the border of $\mathcal {S}^{\Delta }_{m+k}$. By repeating this operation, we find eventually $n_0$ such that the distance from $\psi ^{\Delta ^{n_0}}(\overline{\mu }^{n_0})$ to the border of $\mathcal {S}^{\Delta ^{n_0}}_{k}$ is zero for all $k \ge 0$. This means that $\psi ^{\Delta ^{n_0}}(\overline{\mu }^{n_0})$ is on the border of $\mathcal {S}^{\Delta ^{n_0}}_{k}$, or equivalently is on the border of $\mathcal {S}^{\Delta ^{n_0},\overline{\mu }^{n_0}}_{k}$.   $\square$

Proposition 19

Let $\mu \in [0;\Omega ]$ and $\overline{\mu }\in \{0,1\}^\omega$ a $\Delta$-bi-admissible sequence such that . Then is connected if and only if for all $m\ge 1$, there exists $n\ge m$ such that $\psi ^{\Delta ^{m}}(\mu _m\cdots \mu _n)$ is interior in $\mathcal {S}_{n-m+1}^{\Delta ^{m}}$.

Since each point in $\mathcal {S}^\Delta _n$ may be coded by a word in $\{0,1\}^n$, we consider the language of words which code interior points. We define the language $\mathcal {L}^{\Delta }$ as the set of words $\sigma \in \{0,1\}^\star$ such that $\psi ^\Delta (\sigma )$ is interior in $\mathcal {S}^\Delta _{|\sigma |}$ and for all , we define the language as

Intuitively, is the set of codes of points such that and belong to the same $\mathcal {S}^\Delta _n$, i.e. .

Lemma 20

For all :

• 1.  ;

• 2. is non-empty if and only if or .

The language $\mathcal {L}^\Delta$ may then be defined from the minimal interior configurations we defined earlier. If $\mathcal {C}$ is the set of these minimal configurations, then

We saw that if $\psi ^\Delta (\mu _1\dots \mu _n)$ is interior in $\mathcal {S}^\Delta _n$, then $\psi ^\Delta (\mu _1\dots \mu _m)$ is interior in $\mathcal {S}^\Delta _m$ for all $m\ge n$. Then there exists a minimal index $n_1$ such that $\psi ^\Delta (\mu _1\dots \mu _{n_1})$ is interior in $\mathcal {S}^\Delta _{n_1}$, meaning that $\psi ^\Delta (\mu _1\dots \mu _{n_1-1})$ is on the border of $\mathcal {S}^\Delta _{n_1-1}$. Similarly, there exists a maximal index $n_0$ such that $\psi ^{\Delta ^{n_0}}(\mu _{n_0}\dots \mu _{n_1})$ is interior in $\mathcal {S}^{\Delta ^{n_0}}_{n_1-n_0+1}$, which means that $\psi ^{\Delta ^{n_0+1}}(\mu _{n_0+1}\dots \mu _{n_1})$ is on the border of $\mathcal {S}^{\Delta ^{n_0+1}}_{n_1-n_0}$. Thus the pair $(\delta _{n_0}\dots \delta _{n_1},\mu _{n_0}\dots \mu _{n_1})$ is a minimal interior pattern. For all $\Delta '$ and all $\overline{\mu }'$, if there exists an index i such that $\delta '_i\dots \delta '_{i+n_1-n_0} = \delta _{n_0}\dots \delta _{n_1}$ and $\mu '_i\dots \mu '_{i+n_1-n_0}=\mu _{n_0}\dots \mu _{n_1}$, then $\psi ^{\Delta '}(\mu '_1\dots \mu '_m)$ is interior in $\mathcal {S}^{\Delta '}_m$ for all $m \ge i+n_1-n_0$.

Definition 21

(Interior Patterns)

• A pattern is an element of $\cup _{n\ge 0} (\{1,\dots ,d\}^n \times \{0,1\}^n)$.

• A pattern $(\delta _1\dots \delta _n,\sigma _1\dots \sigma _n)$ is interior if and only if $\psi ^{\delta _1\dots \delta _n}(\sigma _1\dots \sigma _n)$ is interior in $\mathcal {S}^{\delta _1\dots \delta _n}_n$.

• An interior pattern $(\delta _1\dots \delta _n,\sigma _1\dots \sigma _n)$ is minimal if and only if neither $(\delta _2\dots \delta _n,\sigma _2\dots \sigma _n)$ nor $(\delta _1\dots \delta _{n-1},\sigma _1\dots \sigma _{n-1})$ is interior.

We set $\mathcal {L}^\Delta _n = \mathcal {L}^{\Delta ^n}$ and

$$ \mathcal {M}^\Delta _n = \mathcal {L}^\Delta _n \cap \{0,1\}\overline{\mathcal {L}^\Delta _{n+1}} \cap \overline{\mathcal {L}^\Delta _n}\{0,1\} $$

where $\overline{\mathcal {L}}= \{0,1\}^\star \setminus \mathcal {L}$.

$\mathcal {M}^\Delta _n$ is the set of words $\sigma \in \{0,1\}^\star$ such that $(\delta _n\dots \delta _{n+|\sigma |-1},\sigma )$ is a minimal interior pattern. Thus the set of all minimal interior patterns which may appear in $(\Delta ,\overline{\mu })$ is

$$ \bigcup _{n\ge 1} \{(\delta _n\dots \delta _{n+|\sigma |-1},\sigma ) \mid \sigma \in \mathcal {M}^\Delta _n\} $$

and the set of all minimal interior patterns is

$$ \bigcup _{\Delta \in \{1,\dots ,d\}^\omega } \{(\delta _1\dots \delta _{|\sigma |},\sigma ) \mid \sigma \in \mathcal {M}^\Delta _1\} $$

Theorem 22

Let $\mu \in [0;\Omega ]$ and $\overline{\mu }$ a $\Delta$-bi-admissible encoding of $\mu$. Then is connected if and only if $(\Delta ,\overline{\mu })$ contains infinitely many minimal interior patterns.

6 The Periodic Case

We consider now the specific case where $\Delta$ is periodic, that is $\Delta =(\delta _1\dots \delta _p)^\omega$ for some $p\ge 1$ and $\{\delta _1,\dots ,\delta _p\}=\{1,\dots ,d\}$. This means that after p application of the Fully Subtractive algorithm to the normal vector , we obtain a vector which is proportional to . In other words, for some . The value $\beta$ is the inverse of the Pisot eigenvalue of $\gamma _{\delta _1}^{-1}\dots \gamma _{\delta _p}^{-1}$ which was proven to be Pisot in [2].

For each , there exists $\pi \in \{0,1\}^\star$ such that . The language is obviously recognised by the infinite automaton which contains for each word $\sigma \in \{0,1\}^\star$ a state labelled with $\sigma$ and for each $\alpha \in \{0,1\}$ the transition . A state is final if and only if and $|(\sigma +\pi )\mathord \Downarrow ^\Delta 0^\omega | \le |\sigma |$. When $\Delta$ is periodic, this infinite automaton is actually equivalent to a finite one, which means that is a regular language.

Theorem 23

If $\Delta$ is periodic then is a regular language for all .

Proof

(sketch). We saw that is non-empty if and only if or and . We have obviously . It is therefore sufficient to prove the result when . A word $\sigma \in \{0,1\}^\star$ belongs to if and only if $\psi ^\Delta (\sigma +\pi )$ belongs to $\mathcal {S}^\Delta _{|\sigma |}$, i.e. if and only if and $|(\sigma +\pi )\mathord \Downarrow ^\Delta 0^\omega | \le |\sigma |$.

We consider an incremental construction of the automaton in which states are labelled with triples $(\sigma ,k,n) \in \{0,1\}^\star \times (\mathbb {N}\setminus \{0\}) \times (\mathbb {N}\setminus \{0\})$. The initial state is $(\pi \mathord \Downarrow ^\Delta ,1,1)$ and we add two special states: a final state $\top$ and a trap state $\bot$ with the transitions and . For each state $(\sigma ,k,n)$ and each $\alpha \in \{0,1\}$ we have a transition where ‘’ is a function which simplifies states. If then . If and $|(\sigma +0^{k-1}1)\mathord \Downarrow ^{\Delta ^n}0^\omega | \le k$ then . In all other cases, the simplification of $(\sigma ,k,n)$ consists in removing from $\sigma \mathord \Downarrow ^{\Delta ^n}$ a useless prefix of size $\ell$: if $\sigma \mathord \Downarrow ^{\Delta ^n} = \tau _1\dots \tau _\ell \sigma '$, then the result of the simplification is $(\sigma ',k-\ell ,1+(n+\ell -1)\bmod p)$. Intuitively, a prefix is useless if it does not actually participate in the computation of the $\Delta ^n$-normal form of $\sigma +0^{k-1}\mu$, i.e. for all $\mu \in \{0,1\}^\star$,

which may be tested in finite time. We first show that k is bounded by $p+2$. Then, from Lemma 17 we have $\mathcal {S}^\Delta _{p+n} =\mathcal {S}^{\delta _1\dots \delta _p}_p+\varGamma ^{\delta _1\dots \delta _p}(\mathcal {S}^\Delta _n)$ where $\varGamma ^{\delta _1\dots \delta _p}$ is the inverse of a Pisot operator [2]. Using this decomposition, we are able to prove that $2\mathcal {S}^\Delta _m \cap \mathcal {S}^\Delta _\infty \subset \mathcal {S}^\Delta _{m+r}$ for some fixed r. We deduce that after m steps of the construction of the automaton, $|(\sigma +\pi )\mathord \Downarrow ^\Delta 0^\omega |$ has not grown more than r. Using the fact that k is bounded, we deduce that after simplification, the length of $\sigma$ is also bounded. Since $\Delta$ is periodic, we may replace n with $1+(n-1) \bmod p$ which is also bounded. Therefore, the automaton has only finitely many states.   $\square$

From this theorem, we deduce that $\mathcal {L}^\Delta _n$ and $\mathcal {M}^\Delta _n$ are regular languages for all n and may therefore be recognised by finite state automata. In particular, the words $\sigma$ such that $\psi ^\Delta (\sigma )$ is interior in $\mathcal {S}^\Delta _{|\sigma |}$ are recognised by a finite state automaton. Since $\Delta$ is periodic, the sequences $(\mathcal {L}^\Delta _n)_{n\ge 1}$ and $(\mathcal {M}^\Delta _n)_{n\ge 1}$ are also periodic and their periods divide the period of $\Delta$.

For example, if , where $\alpha \in [0;1]$ and $\alpha +\alpha ^2+\alpha ^3=1$, then $\Delta = (123)^\omega$ and we find for all n, $\mathcal {M}^\Delta _n = \mathcal {M}\cup \widetilde{\mathcal {M}}$ where

$$ \mathcal {M}= 1111 \,|\,0111 \,|\,1(100)^+11 \,|\,1(010)^+10 \,|\,0(001)^+010 $$

As announced in the example under Sect. 4, we deduce that is connected when and disconnected when .

If we take , where $\alpha \in [0;1]$ and $\alpha ^3-\alpha ^2-3\,\alpha +1=0$, then $\Delta =(1213)^\omega$. We find for all $n\ge 0$, $\mathcal {M}^\Delta _{2\,n+1} = \emptyset$ and $\mathcal {M}^\Delta _{2\,n+2} = \mathcal {M}\cup \widetilde{\mathcal {M}}$ where

$$ \mathcal {M}= 01001 \,|\,(00 \,|\,10 \,|\,11)000 \,|\,(00 \,|\,10 \,|\,11)(0101)^+1 \,|\,10(1000)^\star 10(11 \,|\,010 \,|\,0011) $$

In order for to be disconnected, $\overline{\mu }$ must have a suffix which avoids completely the minimal interior patterns. As stated by the next theorem, this happens almost never which means that is actually almost always connected.

Theorem 24

The set is Lebesgue negligible.