Computational aspects of relaxation complexity: possibilities and limitations

Abstract

The relaxation complexity ${{\,\mathrm{rc}\,}}(X)$ of the set of integer points X contained in a polyhedron is the smallest number of facets of any polyhedron P such that the integer points in P coincide with X. It is a useful tool to investigate the existence of compact linear descriptions of X. In this article, we derive tight and computable upper bounds on ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$, a variant of ${{\,\mathrm{rc}\,}}(X)$ in which the polyhedra P are required to be rational, and we show that ${{\,\mathrm{rc}\,}}(X)$ can be computed in polynomial time if X is 2-dimensional. Further, we investigate computable lower bounds on ${{\,\mathrm{rc}\,}}(X)$ with the particular focus on the existence of a finite set $Y \subseteq \mathbb {Z}^d$ such that separating X and $Y \setminus X$ allows us to deduce ${{\,\mathrm{rc}\,}}(X) \ge k$. In particular, we show for some choices of X that no such finite set Y exists to certify the value of ${{\,\mathrm{rc}\,}}(X)$, providing a negative answer to a question by Weltge (2015). We also obtain an explicit formula for ${{\,\mathrm{rc}\,}}(X)$ for specific classes of sets X and present the first practically applicable approach to compute ${{\,\mathrm{rc}\,}}(X)$ for sets X that admit a finite certificate.

1 Introduction

A successful approach for solving discrete optimization problems is based on integer programming techniques. To this end, (i) a suitable encoding $X \subseteq \mathbb {Z}^d$ of the discrete problem’s solutions together with an objective $c \in \mathbb {R}^d$ has to be selected, and (ii) a linear system $Ax \le b$, $Cx = f$ defining a polyhedron $P \subseteq \mathbb {R}^d$ has to be found such that $P \cap \mathbb {Z}^d = X$. In the following, we refer to such a polyhedron P as a relaxation of X. Then, the discrete problem can be tackled by solving the integer program $\max \{ {c}^\intercal {x} : Ax \le b,\; Cx = f,\; x \in \mathbb {Z}^d \}$, which can be solved, e.g., by branch-and-bound or branch-and-cut techniques (see, e.g., Conforti et al. [10] or Schrijver [31]).

To solve these integer programs efficiently, the focus in Step (ii) was mostly on identifying facet defining inequalities of the integer hull of P. Such inequality systems, however, are typically exponentially large and one may wonder about the minimum number of facets of any relaxation of X, which allows to compare different encodings X of the discrete problem. Kaibel and Weltge [24] called this quantity the relaxation complexity of X, denoted ${{\,\mathrm{rc}\,}}(X)$, and showed that certain encodings of, e.g., the traveling salesman problem or connected subgraphs, have exponentially large relaxation complexity. They also introduced the quantity ${{\,\mathrm{rc}\,}}_{\mathbb {Q}}(X)$, which is the smallest number of facets of a rational relaxation of X, and posed the question whether ${{\,\mathrm{rc}\,}}(X) = {{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$ holds in general. Recently, this was answered affirmatively for $d \le 4$, see [3]. The same authors also showed that ${{\,\mathrm{rc}\,}}(X)$ is computable if $d \le 3$ using an algorithm with potentially super-exponential worst case running time. Computability for $d \ge 4$ and explicit formulas for ${{\,\mathrm{rc}\,}}(X)$ for specific sets X, however, are still open problems.

In this article, we follow the line of research started in [3] and derive a tight upper bound on ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$ for arbitrary dimensions, which is based on a “robustification” of ${{\,\mathrm{rc}\,}}(X)$ (Sect. 2). We also point out when this upper bound can be used to compute ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$ exactly. Computable lower bounds on ${{\,\mathrm{rc}\,}}(X)$ are dealt with in Sect. 3, with a particular focus on the question about the existence of finite certificates for the size of a minimal relaxation. In Sect. 4 we focus on dimension $d=2$ and show that there is a (weakly) polynomial time algorithm to compute ${{\,\mathrm{rc}\,}}(X)$ in this case. Section 5 derives an explicit formula for the relaxation complexity for integer points in rectangular boxes and crosspolytopes. In Sect. 6 we conclude the article with a discussion of numerical experiments. We also discuss our first practically applicable implementation to compute ${{\,\mathrm{rc}\,}}(X)$ in cases where this quantity is known to have a finite certificate.

An extended abstract of this article appeared as [2] in the conference proceedings of IPCO 2021.

Notation and Terminology. For a set $X \subseteq \mathbb {R}^d$, we denote by ${{\,\mathrm{conv}\,}}(X)$ its convex hull and by ${{\,\mathrm{aff}\,}}(X)$ its affine hull. The boundary and interior of X are denoted by ${{\,\mathrm{bd}\,}}(X)$ and $\mathrm {int}(X)$, respectively. A set $X \subseteq \mathbb {Z}^d$ is called lattice-convex if ${{\,\mathrm{conv}\,}}(X) \cap \mathbb {Z}^d = X$. The dimension of a set X, denoted by $\dim (X)$, is the dimension of its affine hull. For two sets $X, Y \subseteq \mathbb {R}^d$, $X + Y = \{x + y : x \in X,\; y \in Y\}$ is their Minkowski sum. We define $[n] {:}{=}\{1, \dots , n\}$, $[n]_0 {:}{=}\{0,1,\dots ,n\}$, and $e_i$ to be the i-th standard unit vector in $\mathbb {R}^d$. Given two subsets $A, B \subseteq \mathbb {R}^d$, we say that a polyhedron $P \subseteq \mathbb {R}^d$ weakly separates A from B, if $A \subseteq P$ and $B \subseteq \mathbb {R}^d \setminus \mathrm {int}(P)$.

A crucial concept is that of observers [3]. For $X \subseteq \mathbb {Z}^d$ with ${{\,\mathrm{conv}\,}}(X) \cap \mathbb {Z}^d = X$, an observer is a point $z \in \mathbb {Z}^d \setminus X$ such that ${{\,\mathrm{conv}\,}}(X \cup \{z\}) \cap \mathbb {Z}^d = X \cup \{z\}$. The set of all observers of X is denoted by ${{\,\mathrm{Obs}\,}}(X)$. The relevance of the observers for finding ${{\,\mathrm{rc}\,}}(X)$ comes from the fact that any linear system $Ax \le b$ separating X and ${{\,\mathrm{Obs}\,}}(X)$ also separates X and $\mathbb {Z}^d \setminus X$, see [3]. This motivates, for $X, Y \subseteq \mathbb {Z}^d$, to introduce ${{\,\mathrm{rc}\,}}(X,Y)$ (resp. ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X,Y)$) as the smallest number of inequalities in a (rational) system $Ax \le b$ separating X and $Y \setminus X$.

Two reoccurring lattice-convex sets are the discrete standard simplex $\varDelta _d {:}{=}\{ \mathbf {0}, e_1,\ldots ,e_d\}$ and the discrete standard cross-polytope $\lozenge _d {:}{=}\left\{ \mathbf {0}, \pm e_1, \ldots , \pm e_d \right\}$.

Related Literature. Kaibel and Weltge [24] introduced the notion of relaxation complexity and derived a concept for deriving lower bounds on ${{\,\mathrm{rc}\,}}(X)$. Computability of ${{\,\mathrm{rc}\,}}(X)$, for $d = 2$, has been shown by Weltge [35] who also derived a lower bound on ${{\,\mathrm{rc}\,}}(X)$ only depending on the dimension of X. In [3], this lower bound has been improved and computability also for $d=3$ has been established. The interplay between the number of inequalities in a relaxation and the size of their coefficients has been investigated in [19]; see also [20] for a lower bound on the relative size of coefficients in a relaxation. For $X \subseteq \{0,1\}^d$, Jeroslow [21] derived an upper bound on ${{\,\mathrm{rc}\,}}(X, \{0,1\}^d)$, which is an important subject in the area of social choice, see, e.g., Hammer et al. [18] and Taylor and Zwicker [33]. The problem of separating two sets by a function from a given family is known as the binary classification problem in the machine learning community. The support vector machine is the binary classification by means of linear functions, see, e.g., Cristiani and Shawe-Taylor [12]. Our investigations can be interpreted as a variation on the support-vector-machine theme, as we deal with the binary classification using the maximum of (a possibly small) number of linear functions.

Besides the integer programming model $\max \{ {c}^\intercal {x} : Ax \le b,\; Cx = f,\; x \in \mathbb {Z}^d \}$ that we study in this article, an alternative way to optimize over X is to solve a linear optimization problem over $P = {{\,\mathrm{conv}\,}}(X)$. Since P might have exponentially many facets, many researchers have investigated extended formulations of P, which are polyhedra Q in a higher-dimensional space that linearly project onto P. Such extended formulations might have significantly less facets; see the surveys by Conforti et al. [9] and Kaibel [23] and the references therein. Alternatively, one can consider extended relaxations of X, i.e., mixed-integer programming models that allow to use auxiliary (integer) variables; see, e.g., Bader et al. [4], Cevallos et al. [7], or Weltge [35, Ch. 7.1].

2 Computable bounds on the relaxation complexity

Let $X \subseteq \mathbb {Z}^d$. Throughout the article, we mostly consider that X belongs to the subfamily

$$\begin{aligned} \mathcal {C}(\mathbb {Z}^d) {:}{=}\left\{ X \subseteq \mathbb {Z}^d : X \text { is finite, lattice-convex and full-dimensional} \right\} \end{aligned}$$

of lattice-convex sets. Finiteness is a crucial property for many arguments to work, but at the same time it is natural since the usual combinatorial optimization problem deals with integer points in a certain polytope. Unbounded lattice-convex sets occur in the context of covering problems and cutting plane generation from corner polyhedra (cf. [8, 10]). Some of our techniques do apply in this extended setting, but we stick within the finite scope of Weltge and Kaibel [24, 35] here. Assuming X to be full-dimensional is no loss of generality: The relaxation complexity as defined by Kaibel and Weltge [24] only concerns the inequalities of a linear description of X, while equations can be added at no cost.

Besides the theoretical lower bounds on ${{\,\mathrm{rc}\,}}(X)$ based on hiding sets (see Sect. 2.2 below), to the best of our knowledge no systematic way for deriving algorithmically computable lower and upper bounds on ${{\,\mathrm{rc}\,}}(X)$ and ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$ has been discussed so far. One of our goals in this work is to make progress on this problem. To this end, note that ${{\,\mathrm{rc}\,}}(X,Y) \le {{\,\mathrm{rc}\,}}(X)$ for any $Y \subseteq \mathbb {Z}^d$. This bound is best possible if ${{\,\mathrm{rc}\,}}(X,Y) = {{\,\mathrm{rc}\,}}(X)$, which leads us to the following concept.

Definition 1

Let $X \subseteq \mathbb {Z}^d$ be lattice-convex. Then, $Y \subseteq \mathbb {Z}^d$ is called a certificate for ${{\,\mathrm{rc}\,}}(X)$, if ${{\,\mathrm{rc}\,}}(X, Y) = {{\,\mathrm{rc}\,}}(X)$.

Analogously, we define certificates for ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$ and the related operators that we describe below. In view of [3, Prop. 4.3], a certificate for X is given by its set of observers ${{\,\mathrm{Obs}\,}}(X)$. Since ${{\,\mathrm{Obs}\,}}(X)$ can be infinite, we might not be able to systematically compute ${{\,\mathrm{rc}\,}}(X, {{\,\mathrm{Obs}\,}}(X))$. If Y is finite, however, ${{\,\mathrm{rc}\,}}(X,Y)$ can be computed using a mixed-integer program, see [3] and Sect. 6, which suggests the following lower bound on ${{\,\mathrm{rc}\,}}(X)$:

For $t \in \mathbb {Z}_{>0}$, let $B_t = [-t,t]^d \cap \mathbb {Z}^d$ be the set of integer points with coordinates bounded by t in absolute value. Now, it is clear that ${{\,\mathrm{rc}\,}}(X,B_t) \le {{\,\mathrm{rc}\,}}(X,B_{t'})$, for every $t \le t'$. Thus, the parameter

$$\begin{aligned} {{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X) {:}{=}\max _{t \in \mathbb {Z}_{>0}} \,{{\,\mathrm{rc}\,}}(X,B_t) \end{aligned}$$

is well-defined and satisfies ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X) \le {{\,\mathrm{rc}\,}}(X)$. Moreover, this is an identity if and only if ${{\,\mathrm{rc}\,}}(X)$ admits a finite certificate.

In the remainder of this section, we also introduce a computable (and eventually tight) upper bound on ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$ and we show that this always admits a finite certificate. Moreover, we also review the idea of so-called hiding sets and introduce hiding graphs which will be most useful for our computational approach discussed in Sect. 6.

2.1 A computable upper bound on ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$

To find an upper bound on ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$, we introduce a robustification of ${{\,\mathrm{rc}\,}}(X)$ by enforcing that a relaxation is not allowed to support ${{\,\mathrm{conv}\,}}(X)$. To make this precise, for $\varepsilon > 0$, let $X_\varepsilon {:}{=}X + \varepsilon \lozenge _d$, where $\varepsilon \lozenge _d = \left\{ \mathbf {0}, \pm \varepsilon e_1,\ldots ,\pm \varepsilon e_d\right\}$ is the discrete $\ell _1$-ball with radius $\varepsilon$. By construction, we clearly have $X \subseteq \mathrm {int}({{\,\mathrm{conv}\,}}(X_\varepsilon ))$.

Definition 2

Let $X \in \mathcal {C}(\mathbb {Z}^d)$ and let $\varepsilon > 0$. A polyhedron $Q \subseteq \mathbb {R}^d$ is called an $\varepsilon$-relaxation of X if $X_\varepsilon \subseteq Q$ and $\mathrm {int}(Q) \cap \mathbb {Z}^d = X$. The $\varepsilon$-relaxation complexity ${{\,\mathrm{rc}\,}}_{\varepsilon }(X)$ is the smallest number of facets of an $\varepsilon$-relaxation of X.

Alternatively speaking, an $\varepsilon$-relaxation of X is a polyhedron Q such that $\mathrm {int}(Q) \cap \mathbb {Z}^d = X$, and for which the minimum of $\Vert x-z\Vert _1$, over any $x \notin \mathrm {int}(Q)$ and $z \in X$, is at least $\varepsilon$. Further, in complete analogy to ${{\,\mathrm{rc}\,}}(X,Y)$, we define ${{\,\mathrm{rc}\,}}_{\varepsilon }(X,Y)$ to be the smallest number of inequalities necessary to separate $X_{\varepsilon }$ and $Y \setminus X_{\varepsilon }$.

Our first observation is that ${{\,\mathrm{rc}\,}}_\varepsilon (X)$ always admits a finite certificate $Y^\varepsilon \subseteq \mathbb {Z}^d$. The existence of such a finite certificate is noteworthy, because, in general, ${{\,\mathrm{rc}\,}}(X)$ might not admit a finite certificate, see Theorem 11 below.

Lemma 3

Let $X \in \mathcal {C}(\mathbb {Z}^d)$ and let $\varepsilon >0$ be such that X admits an $\varepsilon$-relaxation. Then, there exists an explicitly computable finite set $Y^\varepsilon \subseteq \mathbb {Z}^d$ such that ${{\,\mathrm{rc}\,}}_\varepsilon (X) = {{\,\mathrm{rc}\,}}_\varepsilon (X,Y^\varepsilon )$.

Proof

We show that there exists a sufficiently large and explicit constant $c_{d,\varepsilon ,X} > 0$ such that every $\varepsilon$-relaxation Q of X is contained in $P_{X,\varepsilon } {:}{=}{{\,\mathrm{conv}\,}}(X) + c_{d,\varepsilon ,X} \cdot [-1,1]^d$. Then, the finite set $Y^\varepsilon {:}{=}P_{X,\varepsilon } \cap \mathbb {Z}^d$ is explicitly computable such that ${{\,\mathrm{rc}\,}}_\varepsilon (X) = {{\,\mathrm{rc}\,}}_\varepsilon (X,Y^\varepsilon )$.

In order to find $P_{X,\varepsilon }$, we first note that by definition $X + \varepsilon \lozenge _d \subseteq {{\,\mathrm{conv}\,}}(X_\varepsilon )$. Now, let Q be an arbitrary $\varepsilon$-relaxation of X and let $q \in Q$ be some point therein. As Q weakly separates X from $\mathbb {Z}^d \setminus X$, the set ${{\,\mathrm{conv}\,}}(X \cup \{q\})$ contains at most $|X |+1$ integer points. For $v \in X$, let

$$\begin{aligned} K_{v,q} {:}{=}{{\,\mathrm{conv}\,}}\left( (v + \varepsilon \lozenge _d) \cup \{q,2v-q\}\right) \ \text {and} \ K'_{v,q} {:}{=}{{\,\mathrm{conv}\,}}\left( (v + \varepsilon \lozenge _d) \cup \{q\}\right) . \end{aligned}$$

The set $K_{v,q}$ is convex and centrally symmetric around the integer point v. Moreover, $K'_{v,q} \subseteq Q$, so that we would get a contradiction if $|K'_{v,q} \cap \mathbb {Z}^d | > |X | + 1$. A classical extension of Minkowski’s first fundamental theorem in the Geometry of Numbers is due to van der Corput [11] who proved that, for every convex body $K \subseteq \mathbb {R}^d$ that is centrally symmetric around an integer point, the inequality ${{\,\mathrm{vol}\,}}(K) \le 2^d \cdot |K \cap \mathbb {Z}^d |$ holds. Without loss of generality, assume that the $\ell _\infty$-norm $N {:}{=}\Vert v-q\Vert _\infty$ of $v-q$ is attained at the last coordinate. Then, $K_{v,q}$ contains the crosspolytope $C_{v,q} {:}{=}{{\,\mathrm{conv}\,}}(\{v \pm \varepsilon e_1,\ldots ,v \pm \varepsilon e_{d-1},q,2v-q\})$, and so by van der Corput’s result we have

$$\begin{aligned} |K'_{v,q} \cap \mathbb {Z}^d |&\ge \tfrac{1}{2} |K_{v,q} \cap \mathbb {Z}^d | \ge \tfrac{1}{2^{d+1}} {{\,\mathrm{vol}\,}}(K_{v,q}) \ge \tfrac{1}{2^{d+1}} {{\,\mathrm{vol}\,}}(C_{v,q}) = \tfrac{1}{2^{d+1}} \tfrac{2^d}{d!} \varepsilon ^{d-1} N. \end{aligned}$$

To avoid the discussed contradiction, the last term must be upper bounded by $|X |+1$, which translates to the inequality

$$\begin{aligned} \Vert v-q\Vert _\infty = N \le \frac{2 \, d! \, (|X |+1)}{\varepsilon ^{d-1}} {=}{:}c_{d,\varepsilon ,X}. \end{aligned}$$

Since $q \in Q$ and $v \in X$ were arbitrary, this indeed shows that the $\varepsilon$-relaxation Q of X is contained in ${{\,\mathrm{conv}\,}}(X) + c_{d,\varepsilon ,X} \cdot [-1,1]^d = P_{X,\varepsilon }$, as desired.$\square$

Note that ${{\,\mathrm{rc}\,}}_\varepsilon (X) \ge {{\,\mathrm{rc}\,}}_{\varepsilon '}(X)$, for every $\varepsilon \ge \varepsilon ' > 0$. Hence, the parameter

$$\begin{aligned} {{\,\mathrm{rc_0}\,}}(X) {:}{=}\min _{\varepsilon > 0} {{\,\mathrm{rc}\,}}_\varepsilon (X) \end{aligned}$$

is well-defined. The main advantage of ${{\,\mathrm{rc_0}\,}}(X)$ and the previously defined ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X)$ is that they are limits of explicitly computable numbers which sandwich ${{\,\mathrm{rc}\,}}(X)$:

Theorem 4

Let $X \in \mathcal {C}(\mathbb {Z}^d)$.

• 1. If $\varepsilon > 0$ is rational, then ${{\,\mathrm{rc}\,}}_\varepsilon (X)$ can be computed in finite time.

• 2. For every $t \in \mathbb {Z}_{>0}$, the number ${{\,\mathrm{rc}\,}}(X,B_t)$ can be computed in finite time.

• 3. One has

  $$\begin{aligned} {{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X) \le {{\,\mathrm{rc}\,}}(X) \le {{\,\mathrm{rc}\,}}_\mathbb {Q}(X) = {{\,\mathrm{rc_0}\,}}(X). \end{aligned}$$

  (1)

Proof

Lemma 3 provides us with a finite and explicitly computable set $Y^\varepsilon \subseteq \mathbb {Z}^d$ such that ${{\,\mathrm{rc}\,}}_\varepsilon (X) = {{\,\mathrm{rc}\,}}_\varepsilon (X,Y^\varepsilon )$. If $\varepsilon \in \mathbb {Q}$, then $X_\varepsilon \subseteq \mathbb {Q}^d$, which means that we ask to minimally separate the finite rational sets $X_\varepsilon$ and $Y^\varepsilon \setminus X_\varepsilon$ from one another. The arguments in [3, Prop. 4.9] show how this can be done via a bounded mixed-integer program and Claim 1 follows. As $B_t$ is finite, the same can be done for ${{\,\mathrm{rc}\,}}(X,B_t)$ implying Claim 2.

For Claim 3, observe that for every $t \in \mathbb {Z}_{>0}$ we clearly have ${{\,\mathrm{rc}\,}}(X,B_t) \le {{\,\mathrm{rc}\,}}(X)$, and thus ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X) \le {{\,\mathrm{rc}\,}}(X) \le {{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$. We thus need to show that ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X) = {{\,\mathrm{rc_0}\,}}(X)$. For one inequality, let $Q \subseteq \mathbb {R}^d$ be a rational relaxation of X having ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$ facets and facet description $Ax \le b$. Since Q is a rational polyhedron, Q is necessarily bounded (otherwise it would contain infinitely many integral points). Thus, there exists $\delta > 0$ such that for each $y \in \mathbb {Z}^d \setminus X$ there exists an inequality ${a}^\intercal {x} \le \beta$ in $Ax \le b$ with ${a}^\intercal {y} \ge \beta + \delta$. Consequently, we can increase $\beta$ slightly to get another relaxation $Q'$ with ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$ facets. This shows that there exists $\varepsilon ' > 0$ such that $X_{\varepsilon '} \subseteq Q'$ and $\mathrm {int}(Q') \cap \mathbb {Z}^d = X$, i.e., $Q'$ is an $\varepsilon '$-relaxation of X, and thus ${{\,\mathrm{rc_0}\,}}(X) \le {{\,\mathrm{rc}\,}}_{\varepsilon '}(X) \le {{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$.

For the reverse inequality, we fix an $\varepsilon > 0$ and note that every $\varepsilon$-relaxation of X is bounded (see the proof of Lemma 3). We may thus perturb any such $\varepsilon$-relaxation slightly into a rational relaxation of X with equally many facets. As a result, we get ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X) \le {{\,\mathrm{rc}\,}}_\varepsilon (X)$, for every $\varepsilon > 0$. $\square$

The main message of Theorem 4 is that, for every rational $\varepsilon > 0$, the number ${{\,\mathrm{rc}\,}}_\varepsilon (X)$ is an explicitly computable upper bound on ${{\,\mathrm{rc}\,}}(X)$ and ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$, and that these upper bounds converge to ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$ with $\varepsilon \rightarrow 0$. However, without further information we do not know whether the computed value ${{\,\mathrm{rc}\,}}_\varepsilon (X)$ agrees with ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$. A situation in which we are sure when to stop computing ${{\,\mathrm{rc}\,}}_\varepsilon (X)$ for decreasing values of $\varepsilon > 0$ is when we can also compute an eventually matching lower bound:

Corollary 5

Let $X \in \mathcal {C}(\mathbb {Z}^d)$. If ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X) = {{\,\mathrm{rc}\,}}_{\mathbb {Q}}(X)$, then there is a finite algorithm that computes ${{\,\mathrm{rc}\,}}(X)$.

Proof

Let $(\varepsilon _t)_{t \in \mathbb {Z}_{>0}} \subseteq \mathbb {Q}$ be a rational strictly decreasing null sequence. Then, by Theorem 4, Part 3, there is $t' \in \mathbb {Z}_{>0}$ such that ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X) = {{\,\mathrm{rc}\,}}_{\varepsilon _{t}}(X)$ for every $t \ge t'$. If ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X) = {{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$, there exists $t^\star \in \mathbb {Z}_{>0}$ such that ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X) = {{\,\mathrm{rc}\,}}(X, B_t)$, for every $t \ge t^\star$. Thus, for $t = \max \{t', t^\star \}$, we get ${{\,\mathrm{rc}\,}}(X, B_t) = {{\,\mathrm{rc}\,}}_{\varepsilon _t}(X)$. Since both quantities can be computed in finite time due to Theorem 4, the assertion follows. $\square$

Our constructions suggest infinite iterative procedures that produce sequences of integer values converging to ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X)$ and ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$ after finitely many steps. However, the existence of such procedures per se does not resolve the computability of ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X)$ or ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$. For computability, one would additionally need to be able to decide when the integer sequence achieves the value it finitely converges to.

2.2 Hiding sets and hiding graphs

Let $X \subseteq \mathbb {Z}^d$ be lattice-convex. To derive a lower bound on ${{\,\mathrm{rc}\,}}(X)$, Kaibel and Weltge [24] introduced the notion of hiding sets. A set $H \subseteq ({{\,\mathrm{aff}\,}}(X) \cap \mathbb {Z}^d) \setminus X$ is called a hiding set for X if, for all distinct $x,y \in H$, we have ${{\,\mathrm{conv}\,}}(\{x,y\}) \cap {{\,\mathrm{conv}\,}}(X) \ne \emptyset$. No valid inequality for X can separate any such x and y simultaneously, showing that

$$\begin{aligned} {{\,\mathrm{rc}\,}}(X) \ge |H |, \quad \text {for every hiding set } H \text { for }X. \end{aligned}$$

(2)

Kaibel and Weltge used this bound to show that the set of integer points in the traveling salesman polytope has exponential relaxation complexity (see [24, Thm. 3]). We refer to the best possible lower bound based on hiding sets as the hiding set bound of X,

$$\begin{aligned} H(X) {:}{=}\max \{ |H | :H \text { is a hiding set for } X\}. \end{aligned}$$

It is not immediate how to explicitly compute this number, because X might admit infinitely many hiding sets. However, a maximum size hiding set can always be found among the observers of X.

Lemma 6

Let $X \subseteq \mathbb {Z}^d$ be lattice-convex, let H be a hiding set for X, and assume that there exists an $h \in H \setminus {{\,\mathrm{Obs}\,}}(X)$. Then, for every $y \in {{\,\mathrm{Obs}\,}}(X) \cap {{\,\mathrm{conv}\,}}(\{h\} \cup X)$, the set $(H \setminus \{h\}) \cup \{y\}$ is a hiding set for X as well.

In particular, there is a maximum size hiding set contained in ${{\,\mathrm{Obs}\,}}(X)$.

Proof

For a point $z \notin {{\,\mathrm{conv}\,}}(X)$, let $C_z {:}{=}z + \left\{ \sum _{x \in X} \lambda _x(x-z) : \lambda _x \ge 0\right\}$ be the smallest convex cone with apex z containing X. Further, $U_z {:}{=}C_z \setminus {{\,\mathrm{conv}\,}}(\{z\} \cup X)$ contains all points $w \notin {{\,\mathrm{conv}\,}}(X)$ such that ${{\,\mathrm{conv}\,}}(\{z,w\}) \cap {{\,\mathrm{conv}\,}}(X) \ne \emptyset$. By the choice of y, we have ${{\,\mathrm{conv}\,}}(\{y\} \cup X) \subseteq {{\,\mathrm{conv}\,}}(\{h\} \cup X)$, since for any $h \in \mathbb {Z}^d \setminus {{\,\mathrm{Obs}\,}}(X)$ one has ${{\,\mathrm{Obs}\,}}(X) \cap {{\,\mathrm{conv}\,}}(X \cup \{ h \}) \ne \emptyset$. So, in particular $U_h \subseteq U_y$. This shows that indeed $H' {:}{=}(H \setminus \{h\}) \cup \{y\}$ is a hiding set for X, and since ${{\,\mathrm{conv}\,}}(\{y,h\}) \cap {{\,\mathrm{conv}\,}}(X) = \emptyset$, we also have $|H' | = |H |$.$\square$

With this in mind, we define the hiding graph $G(X) = (V(X), E(X))$ with node set $V(X) = {{\,\mathrm{Obs}\,}}(X)$, and where two distinct nodes x and y are adjacent if and only if ${{\,\mathrm{conv}\,}}(\{x,y\}) \cap {{\,\mathrm{conv}\,}}(X) \ne \emptyset$. Then, there is a one-to-one correspondence between hiding sets consisting of observers and cliques in G(X). Thus, whenever ${{\,\mathrm{Obs}\,}}(X)$ is finite, we can compute H(X) by solving a maximum size clique problem in G(X).

Using this graph representation, we can strengthen the hiding set bound by using the chromatic number $\chi (G(X))$ of G(X). To this end, observe that any relaxation $Ax \le b$ of X defines a proper coloring of G(X) by associating with each inequality a unique color and assigning $y \in {{\,\mathrm{Obs}\,}}(X)$ the color of one inequality that separates it from X. Hence, the chromatic number of G(X) is a lower bound on ${{\,\mathrm{rc}\,}}(X)$. It is at least as strong as the hiding set bound, because the maximum size of a clique lower bounds the chromatic number in every undirected graph. In the next section, we will see that the chromatic lower bound can be strict, i.e., there exists $X \in \mathcal {C}(\mathbb {Z}^d)$ such that $\chi (G(X)) < {{\,\mathrm{rc}\,}}(X)$, see Remark 12.

Remark 7

Replacing ${{\,\mathrm{Obs}\,}}(X)$ in the definition of G(X) by a general set $Y \subseteq \mathbb {Z}^d$, the chromatic number of the corresponding graph is a lower bound on ${{\,\mathrm{rc}\,}}(X, Y)$.

3 On the existence of finite certificates

In this section, we return to the parameter ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X)$ and, in particular, we focus on the question whether every set $X \in \mathcal {C}(\mathbb {Z}^d)$ admits a finite certificate for ${{\,\mathrm{rc}\,}}(X)$. The already mentioned identity ${{\,\mathrm{rc}\,}}(X) = {{\,\mathrm{rc}\,}}(X, {{\,\mathrm{Obs}\,}}(X))$ does not provide an answer to this question, because ${{\,\mathrm{Obs}\,}}(X)$ can be infinite for $d\ge 3$ (see [35, Sect. 7.5]). An example for which this happens is $\varDelta _3$, but interestingly there exists a finite certificate for the fact that ${{\,\mathrm{rc}\,}}(\varDelta _3) \ge 4$. Indeed, computer experiments led us to a 28-element set $Y \subseteq \mathbb {Z}^3 \setminus \varDelta _3$ such that ${{\,\mathrm{rc}\,}}(\varDelta _3,Y) \ge 4$, and a computer-based proof can be found online¹ using SageMath [34]. A subgraph of the hiding graph of Y that certifies this lower bound is given in Fig. 1; the integer points corresponding to the nodes in this graph can be found in Table 1. The graph has 54 edges, contains no triangles, is not 3-colorable, and the removal of any of its edges leads to a graph that can be colored with 3 colors.

Fig. 1

A non 3-colorable graph that certifies ${{\,\mathrm{rc}\,}}(\varDelta _3) \ge 4$

It turns out that there is nothing special about $\varDelta _3$ concerning such certificates: By [3, Cor. 3.8], every $X \in \mathcal {C}(\mathbb {Z}^3)$ satisfies ${{\,\mathrm{rc}\,}}(X) = {{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$, in particular, every relaxation of X has at least 4 facets. Strengthening this result, we now want to show that there is always a finite certificate for this fact. We need the notion of unimodular equivalence of two sets $X,X' \subseteq \mathbb {R}^d$, which means that there is a unimodular matrix $U \in \mathbb {Z}^{d \times d}$ with $|\det (U)| = 1$ and an integer vector $t \in \mathbb {Z}^d$ such that $X = UX' + t$.

Lemma 8

Let $X \in \mathcal {C}(\mathbb {Z}^3)$ be a subset of the simplex

$$\begin{aligned} T_3 {:}{=}\left\{ (x_1,x_2,x_3) \in \mathbb {R}^3 : x_1 \le 1, x_2 \le 1, x_3 \le 1, x_1 + x_2 + x_3 \ge 0 \right\} , \end{aligned}$$

and which contains $\varDelta _3$. If ${{\,\mathrm{conv}\,}}(\varDelta _3)$ is a simplex of maximal volume in ${{\,\mathrm{conv}\,}}(X)$, then X is unimodularly equivalent to one of the four sets corresponding to the columns of the following matrices:

(3)

Moreover, all these four sets satisfy ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X) = {{\,\mathrm{rc}\,}}(X) = 4$.

Proof

The 20 lattice points in the tetrahedron $T_3$ correspond to the columns of the following matrix:

Every lattice-convex set $X \subseteq T_3$ such that ${{\,\mathrm{conv}\,}}(X)$ contains ${{\,\mathrm{conv}\,}}(\varDelta _3)$ as a simplex of maximal volume corresponds to a subset S of the last 16 columns of that matrix such that every tetrahedron with vertices in $\varDelta _3 \cup S$ has volume 1/6. Using an implementation² in SageMath [34], we find that every such subset S can contain at most 2 elements. This leaves us with $10 + 30 = 40$ candidates for S, each of which corresponding to a lattice polytope $P_S = {{\,\mathrm{conv}\,}}(\varDelta _3 \cup S)$ with 5 or 6 vertices, respectively. Finally, we use that two of the polytopes $P_S$, $P_{S'}$ are unimodularly equivalent if and only if their affine normal form coincides. We refer to [1, Sect. 4] for the definition of this concept and a proof of this statement. The authors of that paper also provide an implementation in SageMath of the affine normal form which is available online³.

The fact that all the four thus obtained sets X satisfy ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X) = {{\,\mathrm{rc}\,}}(X) = 4$ can be certified by (a) corresponding finite certificates Y that prove the necessity of four facets in each relaxation, and (b) an explicit relaxation of X by a tetrahedron. Just as for the standard simplex $\varDelta _3$, our implementation for ${{\,\mathrm{rc}\,}}(X,Y)$ allows us to do just that (see https://github.com/christopherhojny/relaxation_complexity/blob/master/computer-aided-proofs/verify_rc_representatives.sage for the precise data). $\square$

Table 1 A finite certificate for ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(\varDelta _3) \ge 4$

Proposition 9

Any $X \in \mathcal {C}(\mathbb {Z}^3)$ satisfies ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X) \ge 4$.

Proof

Let $b_0,b_1,b_2,b_3 \in X$ be the vertices of a simplex $S_X {:}{=}{{\,\mathrm{conv}\,}}(\{b_0,b_1,b_2,b_3\})$ of maximal volume in ${{\,\mathrm{conv}\,}}(X)$. We may assume w.l.o.g. that $b_0 = \mathbf {0}$ and we write $B=(b_1,b_2,b_3) \in \mathbb {Z}^{3 \times 3}$, so that $S_X = B {{\,\mathrm{conv}\,}}(\varDelta _3)$. It is a well-known result that as $S_X$ is volume-maximal, we have ${{\,\mathrm{conv}\,}}(X) \subseteq B T_3$, where $T_3$ is the simplex in Lemma 8 (cf. [26, Thm. 3]). Now, let $\varLambda = B \mathbb {Z}^3 \subseteq \mathbb {Z}^3$ be the lattice spanned by $b_1,b_2,b_3$. In view of Lemma 8, $\varLambda \cap X = B X'$, where $X'$ is unimodularly equivalent to one of the four sets consisting of the columns of either of the matrices in (3).

Now, if Q is a relaxation of X within $\mathbb {Z}^3$, then $Q \cap \varLambda = (Q \cap \mathbb {Z}^3) \cap \varLambda = X \cap \varLambda = B X'$. Thus, Q is a relaxation of $B X'$ within $\varLambda$. This implies the desired relation ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X) \ge {{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(BX',\varLambda ) = {{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X',\mathbb {Z}^3) = 4$. $\square$

Note that this result does not provide us with a finite certificate for the exact value of ${{\,\mathrm{rc}\,}}(X)$, for a given set $X \subseteq \mathbb {Z}^3$. However, we conjecture that this is always possible, at least in three dimensions:

Conjecture 10

Every set $X \in \mathcal {C}(\mathbb {Z}^3)$ admits a finite set $Y \subseteq \mathbb {Z}^3 \setminus X$ such that ${{\,\mathrm{rc}\,}}(X) = {{\,\mathrm{rc}\,}}(X,Y)$.

A curious phenomenon is that, in contrast to Proposition 9, in any higher dimension $d \ge 4$, the possible necessity of $d+1$ facets in a relaxation of $\varDelta _d$ cannot be certified by a finite subset $Y \subseteq \mathbb {Z}^d \setminus \varDelta _d$. In particular, there is no finite certificate for the fact that ${{\,\mathrm{rc}\,}}(\varDelta _4)=5$, giving a negative answer to a question of Weltge [35, Problem 12] and showing that ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(\varDelta _4) < {{\,\mathrm{rc}\,}}(\varDelta _4)$ (cf. Theorem 4).

Theorem 11

For every $d \in \mathbb {Z}_{>0}$ holds ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(\varDelta _d) \le \left\lceil \frac{d}{2}\right\rceil +2$.

Proof

The assertion is true for $d \le 3$, because ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(\varDelta _d) \le {{\,\mathrm{rc}\,}}(\varDelta _d) \le d+1$ and $d+1 \le \left\lceil \frac{d}{2}\right\rceil +2$ in these cases. Thus, we may assume that $d \ge 4$. Furthermore, if the result holds for even dimensions, we can easily derive it for odd d, because ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(\varDelta _d) \le {{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(\varDelta _{d+1})$, since $\varDelta _d = \left\{ (x_1,\ldots ,x_{d+1}) \in \varDelta _{d+1} \,:\, x_{d+1} = 0 \right\}$ and ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(\varDelta _{d+1}) \le \frac{d+1}{2} + 2 = \lceil \frac{d}{2}\rceil + 2$ for even $d+1$.

For this reason, suppose d is even. Let $k = \frac{d}{2} +1$. Let $N \in \mathbb {Z}_{>0}$ be arbitrary and let $\varepsilon = \frac{1}{3 N}$. Moreover, let $\delta _1, \dots , \delta _{k-2} \in [0,1] \setminus \mathbb {Q}$ be such that $1,\delta _1, \dots , \delta _{k-2}$ are linearly independent over $\mathbb {Q}$ and let $\mathcal {Z} {:}{=}\mathbb {Z}^k \times \{-N,\ldots ,N\}^{k-2} \subseteq \mathbb {Z}^d$. Consider the full-dimensional simplex

$$\begin{aligned} Q = {{\,\mathrm{conv}\,}}\Big \{ \tfrac{1}{1 - \varepsilon } (e_1 + e_k), \tfrac{1}{1 - \varepsilon }e_k, e_2, \dots , e_{k-1}, \mathbf {0}\Big \} \subseteq \mathbb {R}^k \end{aligned}$$

(4)

and the linear map $\pi :\mathbb {R}^d \rightarrow \mathbb {R}^k$, $z \mapsto y$, with

In the following, we show that $P = \{ x \in \mathbb {R}^d : \pi (x) \in Q \}$ is a relaxation of $\varDelta _d$ within $\mathcal {Z}$, that is, $P \cap \mathcal {Z} = \varDelta _d$. Since Q has $k+1$ facets, this shows the relation ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(\varDelta _d) \le {{\,\mathrm{rc}\,}}(\varDelta _d, \mathcal {Z}) \le k+1 = \frac{d}{2} + 2$ as desired.

First, we show $\pi (\varDelta _d) \subseteq Q$. To this end, note that $\pi (\mathbf {0}) = \mathbf {0}$ and $\pi (e_i) = e_i$ for $i \in \{2,\dots ,k-1\}$. Moreover, $\pi (e_1) = e_1 + e_k$ and $\pi (e_k) = e_k$, which are clearly contained in Q since $1 < \frac{1}{1 - \varepsilon }$. For $i \in \{k+1,\dots ,d\}$, $\pi (e_i) = \delta _{i - k} e_{1} + \varepsilon e_{i - k + 1} + e_k$. Since $\pi (e_i)$ is the convex combination of $\frac{1}{1-\varepsilon }(e_1 + e_k)$, $e_{i-k+1}$, and $\frac{1}{1-\varepsilon }e_k$ with coefficients $(1-\varepsilon )\delta _{i-k}$, $\varepsilon$, and $(1 - \varepsilon )(1 - \delta _{i-k})$, respectively, we conclude the first part of the proof because $\pi (\varDelta _d) \subseteq Q$.

In the second part, we show that any $z \in \mathcal {Z}$ with $\pi (z) \in Q$ is necessarily contained in $\varDelta _d$. Then, the theorem’s statement follows, because N can be selected arbitrarily. To show that (4), (5), and $z \in \mathcal {Z}$ imply $z \in \varDelta _d$, we first of all note that (5c) and $z \in \mathcal {Z} \subseteq \mathbb {Z}^d$ yield $y_k \in \mathbb {Z}$. Hence, $0 \le y_k \le \frac{1}{1 - \varepsilon } = \frac{3N}{3N-1} < 2$ by (4). That is, $y_k \in \{0,1\}$. We proceed with a case distinction on the binary value of $y_k$.

On the one hand, if $y_k = 0$, then (4) shows $y \in {{\,\mathrm{conv}\,}}\{e_2, \dots , e_{k-1}, \mathbf {0}\}$, i.e., $y_1 = 0$. Since $1,\delta _1, \dots , \delta _{k-2}$ are linearly independent over $\mathbb {Q}$, we derive from (5a) that $z_1 = z_{k+1} = \dots = z_d = 0$. Then, $0 = y_k = z_1 + \sum \nolimits _{j = k}^d z_j = z_k$ by (5c), and $y_i = z_i$ for all $i \in \{2,\dots ,k-1\}$ by (5b). Consequently, the latter shows $(z_2,\dots ,z_{k-1}) \in \varDelta _{k-2}$ as $y \in {{\,\mathrm{conv}\,}}\{e_2, \dots , e_{k-1}, \mathbf {0}\}$. Since the remaining entries of z equal 0, we get $z \in \varDelta _{d}$.

On the other hand, suppose $y_k = 1$. Since Q can be written as the convex hull of $Q_1 = \frac{1}{1-\varepsilon } {{\,\mathrm{conv}\,}}\{e_1 + e_k, e_k\}$ and $Q_2 = {{\,\mathrm{conv}\,}}\{e_2, \dots , e_{k-1}, \mathbf {0}\}$, the cross-section of Q at height $y_k = 1$ is $(1 - \varepsilon )Q_1 + \varepsilon Q_2$. That is, any y satisfying (4), (5), $z \in \mathcal {Z}$, and $y_k = 1$ also satisfies

$$\begin{aligned} 0 \le y_1 \le 1,&y_2, \dots , y_{k-1} \ge 0,&y_2 + \dots + y_{k-1} \le \varepsilon ,&y_k = 1. \end{aligned}$$

(6)

From $z \in \mathcal {Z}$, we get, for all $i \in \{2, \dots , k-1\}$, $-\frac{1}{3} = -\varepsilon N \le \varepsilon z_{k+i-1} \le \varepsilon N = \frac{1}{3}$. Thus, as $z_i \in \mathbb {Z}$, we conclude from (5b) that $y_i < 0$ or $y_i > \varepsilon$ if $z_i \ne 0$. Consequently, for every $i \in \{2, \dots , k-1\}$, Properties (5b) and (6) imply $z_i = 0$ and ${\tilde{z}} = (z_{k+1}, \dots , z_{d}) \in \varDelta _{k-2}$. The only possible non-zero entries of z are thus $z_1$, $z_k$, and possibly one further entry $z_j$, for some $j \in \{k+1,\dots ,d\}$. To conclude the proof, we distinguish the different cases for ${\tilde{z}}$.

As ${\tilde{z}} \in \varDelta _{k-2}$, either ${\tilde{z}} = \mathbf {0}$ or there exists exactly one $j \in \{2,\dots , k-1\}$ with ${\tilde{z}}_{j-1} = z_{k+j-1} = 1$ and the remaining entries of ${\tilde{z}}$ are 0. In the first case, (5a) shows $y_1 = z_1$. From $z \in \mathcal {Z}$ and $0 \le y_1 \le 1$, we get $y_1 \in \{0,1\}$. If $z_1 = 0$, then (5c) and ${\tilde{z}} = \mathbf {0}$ imply $y_k = z_k$. That is, $z \in \{\mathbf {0}, e_k\} \subseteq \varDelta _d$. If $z_1 = 1$, then $1 = y_k \overset{(5c)}{=} z_1 + z_k = 1 + z_k$ shows $z_k = 0$, i.e., $z = e_1 \in \varDelta _d$. In the second case, let j be the unique index in $\{2,\dots , k-1\}$ with $z_{k+j-1} = 1$. Then, we can derive from (5a) that $[0,1] \ni y_1 = z_1 + \delta _{j-1} z_{k+j-1} = z_1 + \delta _{j-1}$. Since $z_1 \in \mathbb {Z}$ and $\delta _{j-1} \in (0,1)$, we conclude $z_1 = 0$. From (5c), we thus get

$$\begin{aligned} 1 = y_k = z_1 + \sum _{i = k}^d z_i = z_k + z_{k+j-1} = z_k + 1. \end{aligned}$$

Consequently, $z_k = 0$ and hence $z = e_{k+j-1} \in \varDelta _d$, finishing the proof. $\square$

Remark 12

Using this result, we can also show that the chromatic lower bound $\chi (G(X)) \le {{\,\mathrm{rc}\,}}(X)$ introduced in the last section can be strict for certain $X \in \mathcal {C}(\mathbb {Z}^d)$: Theorem 11 gives us that ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(\varDelta _4) \le 4$, so that every finite subgraph of $G(\varDelta _4)$ is four-colorable. Thus, the De Bruijn-Erdős theorem (see [13]) implies $\chi (G(\varDelta _4)) \le 4 < 5 = {{\,\mathrm{rc}\,}}_\mathbb {Q}(\varDelta _4) = {{\,\mathrm{rc}\,}}(\varDelta _4)$, where the last identity holds by [3, Thm. 1.3].

Now, we aim to complement this upper bound on ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(\varDelta _d)$ by providing a lower bound that grows with the dimension d. This is in analogy to the lower bound ${{\,\mathrm{rc}\,}}(X) > \log _2(\dim (X)) - \log _2 \log _2(\dim (X))$, for any $X \subseteq \mathbb {Z}^d$ with $\dim (X) \ge 4$ proven in [3, Thm. 1.2]. To this end, we need a preparatory combinatorial lemma.

We call a tuple $(\alpha _1,\ldots ,\alpha _n) \in \mathbb {R}^n$ monotonic if $\alpha _1 \ge \cdots \ge \alpha _n$ or $\alpha _1 \le \cdots \le \alpha _n$ holds. The following is a special case of De Bruijn’s theorem from combinatorics. Since the result might not be well-known in integer optimization, we reproduce its short proof.

Lemma 13

Let $A \in \mathbb {R}^{m \times n}$ be a matrix with $n = 2^{2^m} +1$ columns. Then, A has an $(m \times 3)$-submatrix whose every row is monotonic.

Proof

We reproduce the argument given in [25, Sect. 5] and [32, Thm. 2]. We use the celebrated Erdős-Szekeres theorem (see [16]) which asserts that each $(N^2+1)$-term sequence of real numbers has an $(N+1)$-term monotonic subsequence.

We argue by induction on m. For $m=1$, the assertion follows from the Erdős-Szekeres theorem with $N=2$. Assume $m \ge 2$ and that the assertion has been verified for matrices with $m-1$ rows. Applying the Erdős-Szekeres theorem with $N=2^{2^{m-1}}$, we see that the matrix $A \in \mathbb {R}^{m \times n}$ has a submatrix B of dimension $m \times (2^{2^{m-1}} + 1)$, whose first row is monotonic. Applying the induction assumption to the matrix obtained from B by removing the first row, we obtain that B has an $(m-1) \times 3$-submatrix $B'$ whose every row is monotonic. Amending this submatrix with the three elements of the first row of B that belong to the three columns of $B'$, we get the desired monotonic $(m \times 3)$-submatrix of A. $\square$

Complementing Theorem 11, we can now prove that ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(\varDelta _d) \rightarrow \infty$ as $d \rightarrow \infty$:

Theorem 14

For each $d \ge 4$, one has ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(\varDelta _d) \ge \log _2\log _2 d$.

Proof

Let $Y {:}{=}\left\{ e_j - e_k + e_\ell \,:\, 1 \le j< k < \ell \le d \right\}$. Since ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(\varDelta _d) \ge {{\,\mathrm{rc}\,}}(\varDelta _d, Y)$, it suffices to show ${{\,\mathrm{rc}\,}}(\varDelta _d, Y) \ge \log _2 \log _2 d$. Consider a system of m linear functionals $U x + v$ such that $\left\{ x \in \varDelta _d \cup Y \,:\, U x + v \ge 0 \right\} = \varDelta _d$. Taking into account that $\mathbf {1}^\intercal x = 1$, for each $x \in \{e_1,\ldots ,e_d\} \cup Y$, we can see that the $(m \times d)$-matrix $A {:}{=}(a_{i,j}) {:}{=}U + v \mathbf {1}^\intercal$ satisfies

$$\begin{aligned} \left\{ x \in \{e_1,\ldots ,e_d\} \cup Y \,:\, A x \ge 0 \right\} = \{e_1,\ldots ,e_d\}. \end{aligned}$$

(7)

We claim that $m \ge \log _2 \log _2 d$. If this were not the case, then we had $d \ge 2^{2^m} + 1$. Hence, Lemma 13 would imply that A has an $(m \times 3)$-submatrix whose every row is monotonic. Let $1 \le j< k < \ell \le d$ be the indices of the columns of A that induce such a submatrix. By construction, $A e_1 \ge 0,\ldots , A e_d \ge 0$, which means that all entries of A are non-negative. For $j,k,\ell$ as chosen above, one has $A (e_j - e_k + e_\ell ) \ge 0$. Indeed, for each $i \in [m]$, one has $a_{i,j} \ge a_{i,k} \ge a_{i,\ell } \ge 0$ or $0 \le a_{i,j} \le a_{i,k} \le a_{i,\ell }$. Independently of which of the two cases occurs, one always has $a_{i,j} - a_{i,k} + a_{i,\ell } \ge 0$. We have thus found the point $y {:}{=}e_j - e_k + e_\ell \in Y$ that satisfies $A y \ge 0$, which is a contradiction to (7). This shows $m \ge \log _2 \log _2 d$ and concludes the proof. $\square$

Remark 15

Since the growth of the doubly logarithmic bound in Theorem 14 is very slow, it is natural to wonder if this bound can be improved. It turns out that our way of estimating ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(\varDelta _d)$ via ${{\,\mathrm{rc}\,}}(\varDelta _d, Y)$, with the particular choice of Y fixed in the proof, cannot give any better estimate, since ${{\,\mathrm{rc}\,}}(\varDelta _d, Y)$ is of order $\log \log d$ up to an absolute multiplicative constant. The upper bound on ${{\,\mathrm{rc}\,}}(\varDelta _d,Y)$ has been clarified in an email communication with M. Bucić and B. Sudakov.

There are two natural open problems related to Theorem 14. The first one is whether the assertion can be extended to ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X) \ge \log _2 \log _2 d$, for an arbitrary set $X \in \mathcal {C}(\mathbb {Z}^d)$. The second is whether the lower bound can be strengthened.

4 Computational complexity and hiding sets in dimension 2

In this section, we confine ourselves with the special case of two-dimensional lattice-convex sets $X \in \mathcal {C}(\mathbb {Z}^2)$. We show that the relaxation complexity of any such X can be computed in weakly polynomial time and that the size H(X) of a maximal hiding set deviates by at most 1 from ${{\,\mathrm{rc}\,}}(X)$.

The main property of two-dimensional lattice-convex sets that leads to these results is that they always have only finitely many observers, a fact that Weltge [35] already realized. Since the set of observers is a certificate for ${{\,\mathrm{rc}\,}}(X)$, this means that ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X) = {{\,\mathrm{rc}\,}}(X,{{\,\mathrm{Obs}\,}}(X)) = {{\,\mathrm{rc}\,}}(X)$. Also, we have ${{\,\mathrm{rc}\,}}(X) = {{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$ for $d=2$, which is a consequence of [35, Sec. 7.5]. Thus, for every $X \in \mathcal {C}(\mathbb {Z}^2)$ the four parameters ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X)$, ${{\,\mathrm{rc}\,}}(X)$, ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$, and ${{\,\mathrm{rc_0}\,}}(X)$ coincide (that is, all inequalities in (1) are identities) and they are computable.

4.1 Computational complexity in the plane

In order to devise an efficient algorithm to compute ${{\,\mathrm{rc}\,}}(X)$ in the plane, we first interpret the relaxation complexity of any lattice-convex set $X \subseteq \mathbb {Z}^d$, such that ${{\,\mathrm{Obs}\,}}(X)$ is finite, as the size of a minimal covering of the observers. More precisely, the separation theorems for convex sets imply that

$$\begin{aligned} \mathcal {I}(X) {:}{=}\left\{ I \subseteq {{\,\mathrm{Obs}\,}}(X) :{{\,\mathrm{conv}\,}}(X) \cap {{\,\mathrm{conv}\,}}(I) = \emptyset \right\} \end{aligned}$$

contains all subsets of observers that can be separated from X by a single hyperplane. By definition of the hiding graph G(X), every $\mathcal {I}\in \mathcal {I}(X)$ is a stable set in G(X). The converse, however, is not true, as the example of the discrete standard triangle $\varDelta _2$ and the set $\mathcal {I}:= \{2e_1, 2e_2, -e_1 - e_2\} \notin \mathcal {I}(\varDelta _2)$ shows.

Let $\mathcal {I}_{\max }(X)$ be the family of all inclusion-wise maximal sets in $\mathcal {I}(X)$.

Observation 16

If $X \subseteq \mathbb {Z}^d$ is such that ${{\,\mathrm{Obs}\,}}(X)$ is finite, then ${{\,\mathrm{rc}\,}}(X)$ is the smallest number k of sets $I_1, \dots , I_k \in \mathcal {I}_{\max }(X)$ such that ${{\,\mathrm{Obs}\,}}(X) = \bigcup _{i = 1}^k I_i$.

In order to computationally make use of this observation, a crucial step is to find the set ${{\,\mathrm{Obs}\,}}(X)$ of observers of any given set $X \in \mathcal {C}(\mathbb {Z}^2)$. This can be done efficiently because ${{\,\mathrm{Obs}\,}}(X)$ is the set of integer points in the boundary of enlarging the lattice polygon $P={{\,\mathrm{conv}\,}}(X)$ by lattice-distance one over each of its edges (see Weltge [35, Prop. 7.5.6]).

To make this precise, let $P = \{x \in \mathbb {R}^2 : a_i x_1 + b_i x_2 \le c_i, i \in [k]\}$ be a polyhedron described by k inequalities with the assumption that $a_i, b_i, c_i \in \mathbb {Z}$ with $a_i$ and $b_i$ coprime, for all $i \in [k]$. Then, following Castryck [6], we write $P^{(-1)} {:}{=}\left\{ x \in \mathbb {R}^2 : a_i x_1 + b_i x_2 \le c_i + 1,\; i \in [k] \right\}$ and say that $P^{(-1)}$ is obtained from P by moving out the edges. Note that neither does $P^{(-1)}$ need to be a lattice polygon again, nor may it have as many edges as P. With this notation the previous discussion can be formulated as

$$\begin{aligned} {{\,\mathrm{Obs}\,}}(X) = {{\,\mathrm{bd}\,}}(P^{(-1)}) \cap \mathbb {Z}^2. \end{aligned}$$

In particular, this means $Y = {{\,\mathrm{Obs}\,}}(X)$ is in convex position, i.e., $Y \subseteq {{\,\mathrm{bd}\,}}({{\,\mathrm{conv}\,}}(Y))$, and that we can efficiently list its elements in counterclockwise order.

Lemma 17

Let $V \subseteq \mathbb {Z}^2$ be a finite two-dimensional set and let $X = {{\,\mathrm{conv}\,}}(V) \cap \mathbb {Z}^2$. There is an algorithm that determines ${{\,\mathrm{Obs}\,}}(X) = \{y_0,\ldots ,y_\ell \}$, with the labeling in counterclockwise order, and which runs in $\mathcal {O}(\ell + k \log k + \gamma k)$ time, where $k = |V |$ and $\gamma$ is an upper bound on the binary encoding size of any point in V.

Proof

Let $P={{\,\mathrm{conv}\,}}(V)$. The algorithm consists of the following four steps:

• 1. Compute an irredundant inequality description $P = \{x \in \mathbb {R}^2 : a_i x_1 + b_i x_2 \le c_i, i \in [k]_0\}$, where $a_i,b_i,c_i \in \mathbb {Z}$ with $a_i$ and $b_i$ coprime, and the outer normal vectors $(a_i,b_i)^\intercal$ labeled in counterclockwise order.

• 2. Move out the edges of P and let $P^{(-1)} = \{x \in \mathbb {R}^2 : a_{i_j} x_1 + b_{i_j} x_2 \le c_{i_j}+1, j \in [m]_0\}$ be such that all redundancies are removed.

• 3. Compute the set of vertices $\{w_0,w_1,\ldots ,w_m\}$ of $P^{(-1)}$ in counterclockwise order.

• 4. For each $i \in [m]_0$, compute the integer points on the segment $[w_i,w_{i+1}]$, with the index taken modulo $m+1$.

Some detailed comments are in order:

In Step 1, we first use a standard convex hull algorithm in the plane, e.g., Graham’s scan (cf. [14, Ch. 8] for details) to compute the vertices $\{v_0,\ldots ,v_r\}$ of P in counterclockwise order. Since P is a lattice polygon, each of its edges, say $[v_j,v_{j+1}]$, corresponds to an integer vector $(\eta _1,\eta _2)^\intercal = v_{j+1}-v_j$. Because of the counterclockwise ordering, $(\eta _2,-\eta _1)^\intercal$ is an outer normal vector to the edge at hand, and dividing out by the greatest common divisor of $\eta _1$ and $\eta _2$ leads to the desired inequality description. When we use Euclid’s Algorithm in this last step, we obtain a running time of $\mathcal {O}(k \log k + \gamma r) \subseteq \mathcal {O}(k \log k + \gamma k)$.

The only thing to do in Step 2, besides increasing all the right hand sides by one unit, is to remove the redundancies. One way to do this is to use duality between convex hulls and intersections of hyperplanes, and again invoke, e.g., Graham’s scan. This can be done in time $\mathcal {O}(k \log k)$.

Step 3 just amounts to an iterative computation of the intersection point $w_j$ of the pair of equations $a_{i_j} x_1 + b_{i_j} x_2 = c_{i_j}+1$ and $a_{i_{j+1}} x_1 + b_{i_{j+1}} x_2 = c_{i_{j+1}}+1$, for $j \in [m]_0$. In this computation we also record the normal vector $(a_{i_{j+1}},b_{i_{j+1}})^\intercal$ that corresponds to the edge with endpoints $w_j$ and $w_{j+1}$. This needs $\mathcal {O}(k)$ steps.

For Step 4 we may use Euclid’s Algorithm on the defining data of the edge of $P^{(-1)}$ that contains $w_i$ and $w_{i+1}$, and determine an affine unimodular transformation $A_i:\mathbb {R}^2 \rightarrow \mathbb {R}^2$ such that $A_i [w_i,w_{i+1}] = [\omega _i e_1, \omega _{i+1} e_1]$, for some $\omega _i < \omega _{i+1}$. The integer points on the latter segment in increasing order of the first coordinate are given by

$$\begin{aligned}&z_j = \frac{\omega _{i+1}-\lceil \omega _i \rceil - j}{\omega _{i+1} - \omega _i} \omega _i e_1 + \frac{\lceil \omega _i \rceil + j - \omega _i}{\omega _{i+1} - \omega _i} \omega _{i+1} e_1,\\&\qquad \text { for }j\in \{0,\ldots ,\lfloor \omega _{i+1} \rfloor - \lceil \omega _i \rceil \}. \end{aligned}$$

Using the inverse transformation $A_i^{-1}$ then leads to the correctly ordered list of integer points on the segment $[w_i,w_{i+1}]$. For a given edge of $P^{(-1)}$ these steps can be performed in time proportional to the number of integer points that it contains, leading to a total running time of $\mathcal {O}(\ell + \gamma k)$ for this step.

Conclusively, we saw that the outlined algorithm terminates with the correctly computed list of observers after $\mathcal {O}(\ell + k \log k + \gamma k)$ iterations.$\square$

We now put things together and show that ${{\,\mathrm{rc}\,}}(X)$ can be computed efficiently.

Theorem 18

Let $V \subseteq \mathbb {Z}^2$ be a finite two-dimensional set, let $X = {{\,\mathrm{conv}\,}}(V) \cap \mathbb {Z}^2$, and let $Y = {{\,\mathrm{Obs}\,}}(X)$. Then, the relaxation complexity ${{\,\mathrm{rc}\,}}(X)$ can be computed in time $\mathcal {O}(|V | \cdot \log |V | + |V | \cdot |Y | \cdot \log |Y | + \gamma \cdot |V |)$, where $\gamma$ is an upper bound on the binary encoding size of any point in V.

Proof

Assume that the set of observers ${{\,\mathrm{Obs}\,}}(X)$ is given in counterclockwise order $y_0,\dots ,y_\ell$. Then, for each $I \in \mathcal {I}_{\max }(X)$, there exist $r,s \in [\ell ]_0$ such that $I = \{y_r,y_{r+1},\dots ,y_{r+s}\}$, where indices are modulo $\ell + 1$. That is, the sets in $\mathcal {I}_{\max }(X)$ form “discrete intervals” of observers, see Fig. 2. In particular, $\mathcal {I}_{\max }(X)$ contains at most $\ell + 1$ intervals.

Because of Observation 16, we can determine ${{\,\mathrm{rc}\,}}(X)$ by finding the smallest number of intervals in $\mathcal {I}_{\max }(X)$ that is sufficient to cover ${{\,\mathrm{Obs}\,}}(X)$. This problem can be solved in $\mathcal {O}(\ell \log \ell )$ time using the minimum circle-covering algorithm by Lee and Lee [27]. Thus, the assertion follows if the observers and the intervals in $\mathcal {I}_{\max }(X)$ can be computed in $\mathcal {O}(|V | \cdot \log |V | + |V | \cdot |Y | \cdot \log |Y | + \gamma \cdot |V |)$ time.

By Lemma 17, the list of observers ${{\,\mathrm{Obs}\,}}(X)$ in counterclockwise order can be found in $\mathcal {O}(|Y | + |V | \cdot \log |V | + \gamma \cdot |V |)$ time. To find the sets $I \in \mathcal {I}_{\max }(X)$, note that we can use binary search on $y_{r+s}$ to find a maximum interval $\{y_r,\dots ,y_{r + s}\}$ such that ${{\,\mathrm{conv}\,}}(\{y_r,y_{r+s}\}) \cap {{\,\mathrm{conv}\,}}(V) = \emptyset$. In each of the $\mathcal {O}(\log |Y |)$ steps of the binary search, we have to check whether the line segment ${{\,\mathrm{conv}\,}}(\{y_r, y_{r+s}\})$ intersects one of the $\mathcal {O}(|V |)$ edges of ${{\,\mathrm{conv}\,}}(V)$. Thus, a single set $I \in \mathcal {I}_{\max }$ can be computed in $\mathcal {O}(|V | \cdot \log |Y |)$ time. Combining these running times concludes the proof.$\square$

Fig. 2

A lattice polygon ${{\,\mathrm{conv}\,}}(X)$ and its observers (left), $\mathcal {I}_{\max }(X)$ (center), and a minimum interval covering (right)

Remark 19

One can show that the number $|Y |$ of observers of X in Theorem 18 can be of order $\varTheta (2^\gamma )$, which means that the presented algorithm is not polynomial in the input size. However, if the points in V are encoded in unary, then the algorithm is indeed polynomial.

A question related to computing the relaxation complexity in the plane has been studied by Edelsbrunner and Preparata [15]: Given two finite sets $X, Y \subseteq \mathbb {R}^2$, they describe an algorithm to find a convex polygon $Q \subseteq \mathbb {R}^2$ with the minimal possible number of edges such that $X \subseteq Q$ and $\mathrm {int}(Q) \cap Y = \emptyset$, or to decide that no such polygon exists. That is, if we apply their algorithm to a lattice-convex set X and its observers Y, we can find a polygon that weakly separates X from Y. Their algorithm runs in $\mathcal {O}(|X \cup Y | \cdot \log |X \cup Y |)$ time.

So, together with Theorem 18, we understand the complexity of these separation problems quite well in the planar case $d=2$. In higher dimensions $d \ge 3$, the situation is much less clear though. The computational problem of strictly separating two finite point sets in any dimension by a given number of hyperplanes has been the focus of Megiddo’s work [29]. He introduces the k-separation problem as the decision problem on whether finite sets $X,Y \subseteq \mathbb {Z}^d$ can be separated by k hyperplanes. The case $k=1$ reduces to the linear separation problem and can be solved by linear programming in polynomial time. Megiddo [29] proves the following for $k \ge 2$:

• (a) If d is arbitrary, but k is fixed, then the k-separation problem is NP-complete. This even holds for $k=2$.

• (b) If $d=2$, but k is arbitrary, then the k-separation problem is NP-complete.

• (c) If both d and k are fixed, then the k-separation problem is solvable in polynomial time.

With regard to computing the relaxation complexity, comparing Theorem 18 with Part (b) shows that (at least in the plane) deciding ${{\,\mathrm{rc}\,}}(X) \le k$ is a computationally easier problem than the general k-separation problem. Part (c) is applicable to computing ${{\,\mathrm{rc}\,}}(X)$ in polynomial time via a binary search whenever $X \subseteq \mathbb {Z}^d$ has finitely many observers whose cardinality is polynomially bounded as a function of $|X |$. Relevant families of lattice-convex sets X with this property have been identified in [3, Thm. 4.4 and Thm. 4.5]: First, if $X \subseteq \mathbb {Z}^d$ contains a representative of every residue class in $(\mathbb {Z}/2\mathbb {Z})^d$, then ${{\,\mathrm{Obs}\,}}(X) \subseteq 2X-X$, i.e., $|{{\,\mathrm{Obs}\,}}(X) | \in \mathcal {O}(|X |^2)$. Second, if ${{\,\mathrm{conv}\,}}(X)$ contains an interior integer point, then $|{{\,\mathrm{Obs}\,}}(X) | \le c_d \cdot |X |$, with $c_d$ a constant only depending on the dimension d.

However, in dimensions $d \ge 3$ not every lattice-convex set has finitely many observers. In general, the following basic questions about computability and computational complexity remain unresolved:

Question 20

For $d = 3$, is it NP-hard to compute ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$ ?

Computability of ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$ is only known for the cases $d=2$ (see Weltge [35, Sec. 7.5] and Theorem 18) and $d=3$ (see [3, Sect. 6]).

Question 21

For $d \ge 4$ fixed, is ${{\,\mathrm{rc}\,}}_\mathbb {Q}(X)$ computable ?

As we remarked several times before, for $d=2$ there are always only finitely many observers. The arguments in [3, Sect. 6] can be used to show that for $d=3$ one can always decide whether ${{\,\mathrm{Obs}\,}}(X)$ is finite. Again, in higher dimensions our knowledge is limited.

Question 22

For $d \ge 4$ fixed, is it decidable whether ${{\,\mathrm{Obs}\,}}(X)$ is finite ?

4.2 Hiding set bound

The interpretation of ${{\,\mathrm{rc}\,}}(X)$ given in Observation 16 as the minimal size of an interval covering of ${{\,\mathrm{Obs}\,}}(X)$ can be used to show that, in the plane, the hiding set bound deviates from ${{\,\mathrm{rc}\,}}(X)$ by at most 1.

Theorem 23

Let $X \in \mathcal {C}(\mathbb {Z}^2)$. Then,

$$\begin{aligned} H(X) \le {{\,\mathrm{rc}\,}}(X) \le H(X) + 1. \end{aligned}$$

Proof

The lower bound is the general hiding set bound from Sect. 2.2. For the upper bound, we construct a suitable hiding set H with the help of Observation 16.

To this end, let $I_1,\dots ,I_k \in \mathcal {I}_{\max }(X)$ be such that ${{\,\mathrm{Obs}\,}}(X) = \bigcup _{i = 1}^k I_i$ is a minimal covering of the observers of X. In particular, ${{\,\mathrm{rc}\,}}(X) = k$. Moreover, we assume that the sets $I_i$, $i \in [k]$, are labeled in clockwise order. The following procedure allows us to extract a hiding set H of size $k-1$ from the sets $I_i$:

First of all, we define $I_i' {:}{=}I_i \setminus I_{i+1}$, for $i \in [k]$, with the indices always understood modulo k. This implies the $I_i'$ to be pairwise distinct. Note that since k is minimal, we must have $I_i' \ne \emptyset$, for all $i \in [k]$. Secondly, we let $y_k$ be the last element (with respect to the clockwise order) in $I_k'$ and let $y_{k-1}$ be the last element in $I_{k-1}'$. If $[y_k,y_{k-1}] \cap {{\,\mathrm{conv}\,}}(X) = \emptyset$, then $I_k' \cup \{y_{k-1}\} \in \mathcal {I}(X)$, since the observers of X are in convex position. We then check whether the line segment connecting the last elements in $I_k' \cup \{y_{k-1}\}$ and $I_{k-1}' \setminus \{y_{k-1}\} \in \mathcal {I}(X)$ has a point in common with ${{\,\mathrm{conv}\,}}(X)$. We repeat this procedure until we arrive at sets $I_k''$ and $I_{k-1}''$ whose respective last elements are endpoints of a line segment that intersects ${{\,\mathrm{conv}\,}}(X)$ non-trivially. This has to happen at some point, as otherwise $I_k' \cup I_{k-1}' \in \mathcal {I}(X)$, yielding a covering of ${{\,\mathrm{Obs}\,}}(X)$ with $k-1$ sets of $\mathcal {I}(X)$.

Now, we iteratively proceed as above and obtain sets $I_k'',I_{k-1}'',\ldots ,I_1'' \in \mathcal {I}(X)$ such that ${{\,\mathrm{Obs}\,}}(X) = \bigcup _{i=1}^k I_i''$ and $[y_i,y_{i-1}] \cap {{\,\mathrm{conv}\,}}(X) \ne \emptyset$, for $2 \le i \le k$, and where $y_i$ is the respective last element in $I_i''$. With this notation we claim that $H= \{y_k,y_{k-1},\ldots ,y_2\}$ is a hiding set for X.

If $k=3$, then there is nothing to show as by construction $[y_3,y_2] \cap {{\,\mathrm{conv}\,}}(X) \ne \emptyset$. So, let $k \ge 4$. Assume for contradiction that $[y_s,y_t] \cap {{\,\mathrm{conv}\,}}(X) = \emptyset$, for some $2 \le s < t \le k$. By construction $t > s+1$, and since ${{\,\mathrm{Obs}\,}}(X)$ is in convex position, this means that $I_s'' \cup I_{s+1}'' \in \mathcal {I}(X)$ or $I_s'' \cup I_{s-1}'' \in \mathcal {I}(X)$, yielding a covering of ${{\,\mathrm{Obs}\,}}(X)$ with only $k-1$ sets of $\mathcal {I}(X)$. $\square$

Already the square $\square _2=\{0,1\}^2$ shows that the upper bound in Theorem 23 is best possible. Indeed, we have ${{\,\mathrm{rc}\,}}(\square _2) = 3$, and a routine check reveals that every hiding set for $\square _2$ has at most two elements. We do not know whether a similar result holds in higher dimensions: The discrete simplex $\varDelta _d$ does not admit hiding sets of size larger than 3 (see [35, Prop. 8.2.4]), but ${{\,\mathrm{rc}\,}}(\varDelta _d)$ grows at least logarithmically with d (see [35, Prop. 8.1.4] and [3, Thm. 1.2]). In fact, Weltge conjectures that ${{\,\mathrm{rc}\,}}(\varDelta _d)=d+1$. However, this example does not exclude an affirmative answer for the following question.

Question 24

Is there a constant $h_d$ only depending on the dimension d such that for every $X \in \mathcal {C}(\mathbb {Z}^d)$ we have

$$\begin{aligned} {{\,\mathrm{rc}\,}}(X) \le H(X) + h_d \,? \end{aligned}$$

5 Relaxation complexity of special families of lattice-convex sets

The exact value of ${{\,\mathrm{rc}\,}}(X)$, or any of its variants, is known only for very few classes of lattice-convex sets $X \subseteq \mathbb {Z}^d$: The discrete unit cube always admits a simplex relaxation, that is, ${{\,\mathrm{rc}\,}}(\{0,1\}^d) = d+1$ (see Weltge [35, Thm. 8.1.3]). The conjectured value ${{\,\mathrm{rc}\,}}(\varDelta _d) = d+1$ for the discrete standard simplex $\varDelta _d$ could only be affirmed for dimensions $d\le 4$ so far (see [3, Cor. 3.8]). Jeroslow [21, Thm. 7] showed that, for every $1 \le k \le 2^{d-1}$ there is a $(2^d-k)$-element subset $X_k$ of $\{0,1\}^d$ such that ${{\,\mathrm{rc}\,}}(X_k,\{0,1\}^d) = k$. Further, for $k=2^{d-1}$ one can choose $X_{2^{d-1}} = X_{\text {even}} {:}{=}\left\{ x \in \{0,1\}^d : \sum _{i=1}^d x_i\text { is even}\right\}$, however the value ${{\,\mathrm{rc}\,}}(X_{\text {even}})$ is not known. Besides some very specific examples that were needed to establish computability of ${{\,\mathrm{rc}\,}}(X)$ for 3-dimensional lattice-convex sets $X \subseteq \mathbb {Z}^3$ (see [3, Sect. 6]), we are not aware of any further classes for which ${{\,\mathrm{rc}\,}}(X)$ is known exactly.

In this section, we determine the relaxation complexity of two classes of lattice-convex sets and thus add to the short list of exact results given above.

5.1 Rectangular boxes

Weltge’s (rational) simplex relaxation of the discrete unit cube $\{0,1\}^d$ can be generalized to the case that one segment of the cube is allowed to have arbitrary length. To this end, we call a set $\{a,\ldots ,b\}$ with $a, b \in \mathbb {Z}$ and $a \le b$ a discrete segment of length $b-a$, and we call the Cartesian product of finitely many discrete segments a discrete rectangular box. These examples contain a representative of every residue class in $(\mathbb {Z}/2\mathbb {Z})^d$, so that the relaxation complexity agrees with the rational relaxation complexity in view of [3, Thm. 1.4].

Lemma 25

For $b \in \mathbb {Z}_{>0}$, let $X_b {:}{=}\{0,1\}^\ell \times \{0,1,\ldots ,b\} \subseteq \mathbb {Z}^{\ell +1}$. Then, the simplex

$$\begin{aligned} P&{:}{=}\left\{ x \in \mathbb {R}^{\ell +1} : x_k \le 1 + \sum _{i=k+1}^{\ell +1} (b+1)^{-i} x_i, \text { for }k \in [\ell ], \right. \\&\quad \left. x_{\ell +1} \le b \quad \text {and} \quad x_1 + \sum _{i=2}^{\ell +1}(b+1)^{-i} x_i \ge 0 \right\} \end{aligned}$$

is a relaxation of $X_b$. In particular, ${{\,\mathrm{rc}\,}}(X_b) \le \ell +2$.

Proof

The proof is just an adjustment of Weltge’s arguments in [35, Lem. 7.2.1]. We need to show that $X_b = P \cap \mathbb {Z}^{\ell +1}$. The inclusion $X_b \subseteq P \cap \mathbb {Z}^{\ell +1}$ is quickly checked, so we give the details for the reverse inclusion.

Let $x \in P \cap \mathbb {Z}^{\ell +1}$. First, we show that $x_i \le 1$, for all $i \in [\ell ]$. To this end, let $k \in [\ell ]$ be the largest index for which $x_k > 1$. Then, the first defining inequalities of P together with basic facts about geometric series give

$$\begin{aligned} 1&< x_k \le 1 + \sum _{i=k+1}^{\ell +1}(b+1)^{-i} x_i \le \sum _{i=k+1}^\ell (b+1)^{-i} + \frac{b}{(b+1)^{\ell +1}}\\&< 1+\frac{1}{b} + \frac{b-1}{(b+1)^{\ell +1}} < 1 + \frac{1}{b} + \frac{b-1}{b} = 2, \end{aligned}$$

in contradiction to $x_k \in \mathbb {Z}$.

Second, we show that $x_1$ is non-negative. Indeed, by the last defining inequality of P, the just established fact that $x_i \le 1$, for $i \in [\ell ]$, and $x_{\ell +1} \le b$, we have

$$\begin{aligned} x_1 \ge -\sum _{i=2}^{\ell +1}(b+1)^{-i} x_i \ge -\sum _{i=2}^\ell (b+1)^{-i} - \frac{b}{(b+1)^{\ell +1}} > -1. \end{aligned}$$

It remains to show that $x_i \ge 0$, for every $i \in \{2,\dots ,\ell +1\}$. To this end, let $j \in \{2,\ldots ,\ell +1\}$ be the smallest index such that $x_j \le -1$. We first show that $x_i = 0$, for every $i < j$: Let $k < j$ be the largest index with $x_k > 0$, which means that $x_k = 1$. Then, by the first defining inequalities of P, we have

$$\begin{aligned} 1&= x_k \le 1 + \sum _{i=k+1}^{\ell +1}(b+1)^{-i} x_i = 1 + (b+1)^{-j} x_j + \sum _{i=j+1}^{\ell +1}(b+1)^{-i} x_i\\&\le 1 - (b+1)^{-j} + \sum _{i=j+1}^\ell (b+1)^{-i} + \frac{b}{(b+1)^{\ell +1}} < 1. \end{aligned}$$

The last inequality follows from the closed form expression for geometric sums and basic algebraic manipulations. Now, knowing that $x_i = 0$, for every $i < j$, we use the last defining inequality of P and get

$$\begin{aligned} 0&\le x_1 + \sum _{i=2}^{\ell +1}(b+1)^{-i} x_i = (b+1)^{-j} x_j + \sum _{i=j+1}^{\ell +1}(b+1)^{-i} x_i\\&\le -(b+1)^{-j} + \sum _{i=j+1}^\ell (b+1)^{-i} + \frac{b}{(b+1)^{\ell +1}} < 0. \end{aligned}$$

This contradiction finishes the proof. $\square$

This simplicial relaxation of $X_b$ quickly leads to an upper bound on the relaxation complexity of general discrete rectangular boxes. To see that these upper bounds are tight, we employ the concept of hiding sets discussed in Sect. 2.2.

Theorem 26

For integers $k>0$ and $\ell \ge 0$, let $S_1, \ldots , S_k$ be discrete segments of length at least 2, and let $T_1,\ldots ,T_\ell$ be discrete segments of length 1. Consider the discrete box $X = S_1 \times \ldots \times S_k \times T_1 \times \ldots \times T_\ell$ in $\mathbb {Z}^{k + \ell }$. Then, ${{\,\mathrm{rc}\,}}(X) = 2 k + \ell$.

Proof

Without loss of generality let $S_i = \{a_i,\ldots ,b_i\}$ be such that $a_i < 0$ and $b_i > 0$, and let $T_j = \{0,1\}$, for all $1 \le j \le \ell$. We form a $2k + \ell$ element set H by attaching to each $S_i$ two points $(a_i-1) e_i$ and $(b_i+1) e_i$ in $\mathbb {Z}^{k + \ell }$, and to each $T_j$ the point $2 e_{k+j}$ in $\mathbb {Z}^{k + \ell }$. It is straightforward to check that H is a hiding set of X. Indeed, by construction the midpoint of any two $p,q \in H$ with $p \ne q$ belongs to ${{\,\mathrm{conv}\,}}(X)$. This shows ${{\,\mathrm{rc}\,}}(X) \ge 2 k + \ell$.

To prove the upper bound ${{\,\mathrm{rc}\,}}(X) \le 2 k + \ell$ it suffices to verify the case $k=1$. Indeed, if $k>0$, a relaxation $Q'$ of $S_k \times T_1 \times \ldots \times T_\ell$ with $2 + \ell$ facets gives rise to the relaxation ${{\,\mathrm{conv}\,}}(S_1 \times \ldots \times S_{k-1}) \times Q'$ of X with $2 k + \ell$ facets. The case $k=1$ is however exactly the content of Lemma 25.$\square$

5.2 Discrete cross-polytopes

Recall that $\lozenge _d = \{ \mathbf {0}, \pm e_1, \dots , \pm e_d\}$ denotes the discrete standard cross-polytope. Weltge [35, Prop. 7.2.2] proved that ${{\,\mathrm{rc}\,}}(\lozenge _d) \le 2d$, for all $d \ge 4$ and wondered whether this bound is best possible. The 2d bound is already quite surprising, as the cross-polytope ${{\,\mathrm{conv}\,}}(\lozenge _d)$ has $2^d$ facets. Using our efficient implementation for ${{\,\mathrm{rc}\,}}(X)$ that we describe in Sect. 6 below, we obtained simplex relaxations of $\lozenge _d$ in small dimensions. In the following we prove that ${{\,\mathrm{rc}\,}}(\lozenge _d)=d+1$ holds for every $d\ge 3$. To this end, we introduce, for two sets X, Y, the notation $X \oplus Y {:}{=}(X \times \{ 0 \}^{\dim (Y)}) \cup (\{ 0 \}^{\dim (X)} \times Y)$. Note that then $\lozenge _{d+1} = \lozenge _d \oplus \lozenge _1$.

Lemma 27

Let d be a positive integer and let $P = \{x \in \mathbb {R}^d :Ax \le \mathbf {1}\}$ be a relaxation of $\lozenge _d$. Then,

$$\begin{aligned} Q = \left\{ (x,y) \in \mathbb {R}^{d+1} : A(x - y e_1) \le \mathbf {1},\; -1 \le x_1 + y \le 1 \right\} \end{aligned}$$

is a relaxation of $\lozenge _{d+1} = \lozenge _d \oplus \lozenge _1$.

Proof

To show $\lozenge _{d+1} \subseteq Q$, let $(x, y) \in \lozenge _{d+1}$. Then, either $x \in \lozenge _d$ and $y = 0$, or $x = \mathbf {0}$ and $y = \pm 1$. In both cases $x - ye^1 \in \lozenge _d$ and $|x_1 + y | \le 1$, thus $(x,y) \in Q$.

For the reverse inclusion assume $(x,y) \in \mathbb {Z}^{d+1}$ is a lattice point in Q. Then, $x - y e_1 \in P \cap \mathbb {Z}^d = \lozenge _d$ and $|x_1 + y| \le 1$. We proceed by a case distinction on y. First, if $y = 0$, then $x = x - y e_1 \in \lozenge _d$ and hence $(x,y) = (x,0) \in \lozenge _d \times \{0\} \subseteq \lozenge _{d+1}$. Second, if $y = 1$, then by $x_1 + 1 = x_1 + y \le 1$ we find that $x_1 \le 0$. Since also $x - e_1 = x - y e_1 \in \lozenge _d$, we conclude $x_1 = 0$ and $(-1,x_2,\ldots ,x_d) \in \lozenge _d$. Hence, $x_2 = \ldots = x_d = 0$, showing $(x,y) = (\mathbf {0},1) = e_{d+1} \in \lozenge _{d+1}$. Third, suppose $y \ge 2$. Note that $A(x - ye_1) \le \mathbf {1}$ implies $|x_1 - y | \le 1$ as P is a relaxation of $\lozenge _d$. Together with $|x_1 + y | \le 1$, we find $|x_1 | \le 1$. Further, from $x_1 + y \le 1$, we get $y \le 1 - x_1 \le 2$. So, in fact $y=2$ and $x_1 = -1$. But then we have $x - y e_1 = (-3,x_2,\ldots ,x_d) \in \lozenge _d$, a contradiction. Finally, the cases of negative values of y are analogous. $\square$

Lemma 28

For any positive integer d, we have ${{\,\mathrm{rc}\,}}(\lozenge _{d+2}) \le {{\,\mathrm{rc}\,}}(\lozenge _{d}) + 2$.

Proof

Let $P = \{x \in \mathbb {R}^d : Ax \le \mathbf {1}\}$ be a bounded relaxation of $\lozenge _d$ with ${{\,\mathrm{rc}\,}}(\lozenge _d)$ facets, which exists by [3, Thm. 1.4]. By Lemma 27, a relaxation of $\lozenge _{d+1}$ is given by $Q = \{(x,y) \in \mathbb {R}^{d+1} : A(x - y e_1) \le \mathbf {1}, |x_1 + y| \le 1\}$. Observe that Q is bounded, since P is. Hence, there is an $\varepsilon > 1$ small enough such that with $A' = \frac{1}{\varepsilon } A$, the polyhedron $Q' = \{(x,y) \in \mathbb {R}^{d+1} : A'(x - y e_1) \le \mathbf {1}, |x_1 + y| \le \varepsilon \}$ is a strict relaxation of $\lozenge _{d+1}$. Since $Q'$ is strict, there are $c_1,\ldots ,c_d \in \mathbb {R}\setminus \mathbb {Q}$ close enough to 0 such that also $Q'' = \{(x,y) \in \mathbb {R}^{d+1} : A'((1+c_1)x - y e_1) \le \mathbf {1}, |(1+c_1)x_1 + c_2 x_2 + \ldots c_d x_d + y| \le \varepsilon \}$ is a relaxation of $\lozenge _{d+1}$. We further assume that $1,c_1,\ldots ,c_d$ are chosen to be linearly independent over $\mathbb {Q}$.

We claim that with this set-up the polyhedron

$$\begin{aligned} R&{:}{=}\big \{(x,y,z) \in \mathbb {R}^{d+2} : A'((1+c_1)x - y e_1) \le (1-z)\mathbf {1},\\&\quad |(1+c_1)x_1 + c_2 x_2 + \ldots + c_d x_d + y| \le (1+z)\varepsilon \big \} \end{aligned}$$

is a relaxation of $\lozenge _{d+2}$, which by construction has at most ${{\,\mathrm{rc}\,}}(\lozenge _d) + 2$ facets. First of all, for $\alpha \in \mathbb {R}$, consider the system $Ax \le \alpha \mathbf {1}$. Since for $\alpha =1$ the system describes the polyhedron P which contains $\mathbf {0}$ in its interior, the system with $\alpha =0$ describes only the origin $\mathbf {0}$, and for any $\alpha < 0$ the corresponding system is infeasible.

Now, the containment $\lozenge _{d+2} \subseteq R \cap \mathbb {Z}^{d+2}$ is quickly checked, since $Q''$ contains $\lozenge _{d+1}$. For the reverse inclusion, let $(x,y,z) \in R \cap \mathbb {Z}^{d+2}$ be an integer point. By the observation on the system $Ax \le \alpha \mathbf {1}$ and the nature of the inequalities $|(1+c_1)x_1 + y| \le (1+z)\varepsilon$, the last coordinate necessarily satisfies $z \in \{0,\pm 1\}$.

If $z=0$, then $(x,y,0) \in \lozenge _{d+2}$, because $(x,y) \in Q'' \cap \mathbb {Z}^{d+1} = \lozenge _{d+1}$. If $z=1$, then by the observation on the system $Ax \le \alpha \mathbf {1}$, we must have $(1+c_1)x = y e_1$, and thus $x = \mathbf {0}$ and $y=0$, as $c_1$ was chosen irrational. Hence, $(x,y,z) = e_{d+2} \in \lozenge _{d+2}$. Finally, if $z = -1$, then $|(1+c_1)x_1 + c_2 x_2 + \ldots c_d x_d + y| \le 0$, which also implies $x = \mathbf {0}$ and $y=0$, since $1,c_1,\ldots ,c_d$ are chosen to be linearly independent over $\mathbb {Q}$. So, $(x,y,z) = -e_{d+2} \in \lozenge _{d+2}$ in this case, finishing the proof. $\square$

Theorem 29

For every positive integer $d \ne 2$, we have ${{\,\mathrm{rc}\,}}(\lozenge _d) = d + 1$.

Proof

Obviously, the result holds for $d=1$. For $d \ge 3$, observe that every relaxation of $\lozenge _d$ is bounded, since ${{\,\mathrm{conv}\,}}(\lozenge _d)$ contains the origin in its interior, cf. [3, Thm. 4.5]. This implies ${{\,\mathrm{rc}\,}}(\lozenge _d) = {{\,\mathrm{rc}\,}}_\mathbb {Q}(\lozenge _d) \ge d+1$ (see [3, Rem. 3.7]). For the upper bound, it is sufficient to prove the statement for $d \in \{3,4\}$, as the remaining cases follow by induction via Lemma 28. One can verify that such relaxations are, e.g.,

for $\lozenge _3$ and $\lozenge _4$, respectively. $\square$

6 Mixed-integer programming formulations for computing the relaxation complexity

In this section, we briefly discuss mixed-integer programming models for the computation of ${{\,\mathrm{rc}\,}}_\varepsilon (X,Y)$ and ${{\,\mathrm{rc}\,}}(X,Y)$, respectively, and evaluate their performance.

Compact model In [3], a mixed-integer programming formulation has been proposed to check whether a finite lattice-convex set $X \subseteq \mathbb {Z}^d$ admits a relaxation w.r.t. a finite set $Y \subseteq \mathbb {Z}^d \setminus X$ with k inequalities. With a minor modification, this model can also be used to compute ${{\,\mathrm{rc}\,}}_\varepsilon (X,Y)$, and thus ${{\,\mathrm{rc}\,}}(X,Y)$, if $\varepsilon$ is sufficiently small. In the following, we just present the idea of this model and refer the reader to [3] for more details.

Given an upper bound k on the number of inequalities needed to separate X and Y, the idea is to introduce variables $a_{ij}$ and $b_i$, $(i,j) \in [k] \times [d]$, to model the k potential inequalities $\sum _{j = 1}^d a_{ij} x_j \le b_i$ needed in a relaxation. By rescaling, one can assume without loss of generality that $a_{ij} \in [-1,1]$ and $b_i \in [-d \rho _X, d \rho _X]$ with $\rho _X =\max \{ \Vert x\Vert _\infty :x \in X\}$. Moreover, for each $i \in [k]$ and $y \in Y$, a binary variable $s_{iy}$ is introduced that indicates whether the i-th inequality is violated by y; binary variables $u_i$, $i \in [k]$, indicate whether the i-th inequality is used in a relaxation. To model that inequality i is violated by at least $\varepsilon$ for $y \in Y$ if $s_{iy} = 1$, a big-M constraint is used.

In its basic version, this model is rather difficult to solve for a black box MIP solver. Reasons for this are (i) the big-M inequalities, (ii) symmetries of the problem formulation, and (iii) combinatorial properties of the relaxation complexity that are not expressed in the model. To overcome these issues, we have used the following enhancements in our implementation. We replace the big-M constraints by so-called indicator constraints which encode the big-M constraint without introducing big-M terms explicitly, cf. Belotti et al. [5]. To handle symmetries, we enforce the used inequalities to be sorted w.r.t. their first coefficient in a; the unused inequalities are fixed to ${0}^\intercal {x} \le d \rho _X$. Finally, we derive additional inequalities that can be used as cutting planes. These inequalities are based on hiding sets $H \subseteq Y$ and encode that each inequality valid for X can cut off at most one point from H. This can be expressed via the hiding set cuts $\sum _{y \in H} s_{iy} \le 1$, $i \in [k]$. Although these cuts are the stronger the bigger the underlying hiding set is, we add these inequalities just for hiding sets of size 2, because the latter can be computed relatively efficiently by a brute-force algorithm.

Column generation based model Based on Observation 16, we can devise an alternative integer programming formulation to compute ${{\,\mathrm{rc}\,}}(X,Y)$. To this end, let

$$\begin{aligned} \mathcal {I}= \mathcal {I}(X, Y) {:}{=}\{ I \subseteq Y :{{\,\mathrm{conv}\,}}(I) \cap {{\,\mathrm{conv}\,}}(X) = \emptyset \}. \end{aligned}$$

Then, ${{\,\mathrm{rc}\,}}(X,Y)$ is the smallest number k of sets $I_1, \dots , I_k \in \mathcal {I}$ with $Y = \bigcup _{i = 1}^k I_i$, which leads immediately to the following integer programming formulation

$$\begin{aligned} \min _{z \in \mathbb {Z}_+^{\mathcal {I}}} \left\{ \sum _{I \in \mathcal {I}} z_I :\sum _{I \in I_y} z_I \ge 1,\; y \in Y \right\} , \end{aligned}$$

(8)

where $I_y = \{ I \in \mathcal {I}:y \in I\}$.

Since this formulation consists of exponentially many variables, we cannot solve it immediately using standard MIP solvers. In our implementation, we thus use a branch-and-price procedure for solving (8), i.e., we use a branch-and-bound procedure in which each LP relaxation is solved by column generation.

The core of the column generation procedure is to solve the pricing problem. Since the dual of the LP relaxation is

$$\begin{aligned} \max _{\alpha \in \mathbb {R}_+^Y} \left\{ \sum _{y \in Y} \alpha _y :\sum _{y \in I} \alpha _y \le 1,\; I \in \mathcal {I}\right\} , \end{aligned}$$

the pricing problem is to find, for fixed shadow prices ${\bar{\alpha }} \in \mathbb {R}_+^Y$, a set $I \in \mathcal {I}$ with $\sum _{y \in I} {\bar{\alpha }}_y > 1$ or to show that no such set exists. Since this problem is NP-hard as it is a generalization of the open hemisphere problem, see Johnson and Preparata [22], we model the pricing problem as a MIP in our implementation. The MIP model that we have used is a variant of the compact model discussed before with $k = 1$.

The branching strategy of our branch-and-price algorithm is to select two sets I and J whose corresponding $\alpha$-variables are fractional in the LP relaxation such that both the intersection $I \cap J$ and the symmetric difference $I \varDelta J$ are non-empty. Then, we select $y_1 \in I \cap J$ and $y_2 \in I \varDelta J$ and create two child nodes. In one, we enforce that $y_1$ and $y_2$ are contained in the same sets $I \in \mathcal {I}$; in the other, $y_1$ and $y_2$ have to be contained in different sets from $\mathcal {I}$ used by the LP relaxation.

Hybrid model Our implementation also allows to use a hybrid model. Instead of solving the full problem using Model (8), we only solve its LP relaxation using column generation. Then, the compact model from [3] is solved after it has been initialized with the best primal solution found during solving the LP relaxation of (8) and the value of the LP relaxation has been transferred as dual bound to the compact model.

An ad-hoc method to compute the relaxation complexity If X has a finite set of observers, we can at least theoretically use the algorithm from [3] to compute ${{\,\mathrm{Obs}\,}}(X)$ and solve the compact or column generation model to find ${{\,\mathrm{rc}\,}}_\varepsilon (X, {{\,\mathrm{Obs}\,}}(X)) = {{\,\mathrm{rc}\,}}_\varepsilon (X)$ and ${{\,\mathrm{rc}\,}}(X, {{\,\mathrm{Obs}\,}}(X)) = {{\,\mathrm{rc}\,}}(X)$, respectively. In practice, however, ${{\,\mathrm{Obs}\,}}(X)$ might be very large or even infinite already in small dimensions and solving these models becomes costly or even impossible. A more ad-hoc fashion for finding ${{\,\mathrm{rc}\,}}_\varepsilon (X)$ and ${{\,\mathrm{rc}\,}}(X)$ is inspired by Theorem 4. Instead of starting with the full set of observers, we select a small set $Y \subseteq {{\,\mathrm{Obs}\,}}(X)$ and compute ${{\,\mathrm{rc}\,}}_\varepsilon (X,Y)$ and ${{\,\mathrm{rc}\,}}(X,Y)$, respectively. If the resulting set of inequalities is already a relaxation of X, we stop. Otherwise, there exists a non-empty set $Y' \subseteq {{\,\mathrm{Obs}\,}}(X) \setminus Y$ that is not separated from X. We can extend the set Y by $Y'$ and iterate this procedure until a relaxation of X has been found.

Of course, this procedure is only guaranteed to work if ${{\,\mathrm{rc_{\scriptscriptstyle \square }}\,}}(X) = {{\,\mathrm{rc_0}\,}}(X)$. Nevertheless, we could use it to compute ${{\,\mathrm{rc}\,}}(X)$ for particular choices of X without computing ${{\,\mathrm{Obs}\,}}(X)$ explicitly.

Numerical experiments Our implementation consists of two separate codes, which are publicly available⁴. The first code essentially consists of one method for computing ${{\,\mathrm{rc}\,}}(X, {{\,\mathrm{Obs}\,}}(X))$ (provided ${{\,\mathrm{Obs}\,}}(X)$ is finite) and another method to compute H(X), which has been used as inspiration for Theorem 23. To find ${{\,\mathrm{Obs}\,}}(X)$, we use a variant of the algorithm suggested in [3], and ${{\,\mathrm{rc}\,}}(X, {{\,\mathrm{Obs}\,}}(X))$ is computed using the compact MIP model. We compute H(X) by finding a maximum cardinality clique in the hiding graph by solving an integer program. We have implemented the aforementioned methods in Python 3.7.8, using SageMath 9.1 [34] for polyhedral computations; all mixed-integer programs were solved using SCIP 7.0.0 [17], which has been called via its Python interface [28]. Note that SCIP is not an exact solver and thus the results reported below are only correct up to numerical tolerances.

We have used this implementation to investigate two properties of ${{\,\mathrm{rc}\,}}(X)$ for lattice polygons in dimension 2. More specifically, we used the representatives of all lattice polygons P with at least one and at most 12 interior integer points provided by Castryck [6]. In our first experiment, we compared the exact value of the relaxation complexity with the hiding set lower bound. Table 2 shows the deviation of these two values parameterized by the number of edges of P, which led us to Theorem 23, i.e., the hiding set bound deviates by at most one. Interestingly, for $k \in \{4,6,7,8,9\}$, the hiding set bound is distributed relatively equally between ${{\,\mathrm{rc}\,}}(X)$ and ${{\,\mathrm{rc}\,}}(X)-1$.

Table 2 Distribution of deviation of maximum hiding set sizes from relaxation complexity in percent

The second experiment compares the relaxation complexity with the number of edges of P, see Table 3. Although P can be rather complex with up to 10 edges (where 10 is realized by a single instance), the relaxation complexity is at most six, where for the majority of all tested instances the relaxation complexity is either four or five. Moreover, there exist polygons with up to nine edges that admit simplicial relaxations. Thus, already in dimension 2, the difference between the number of facets and the relaxation complexity can be very large. In high dimensions this phenomenon does not come as a surprise, since there are knapsack polytopes whose integer hull has super-polynomially many facets (cf. Pokutta and Van Vyve [30, Cor. 3.8]).

Table 3 Distribution of relaxation complexities compared to number of facets in percent

The second code is a C/C++ implementation of the compact, column generation, and the hybrid model to find ${{\,\mathrm{rc}\,}}(X, Y)$. The implementation uses SCIP 7.0.2 as branch-and-bound (-and-price) framework and SoPlex 5.0.2 as LP solver; we use cddlib version 0.94 m⁵ for convex hull computations, e.g., to find hiding set cuts. In the following, we briefly illustrate that selecting the right model is crucial to efficiently find ${{\,\mathrm{rc}\,}}(X, Y)$. As test set, we use the sets $X \in \{\varDelta _3, \varDelta _4, \lozenge _4, \lozenge _5\}$ and Y being all integer points at $\ell _1$-distance at most k from X, where $k \in [9]$ for simplices and $k \in [7]$ for crosspolytopes. These are the test sets that lead us to Theorems 11 and 29. These experiments were run on a Linux cluster with Intel Xeon E5 3.5 GHz quad core processors and 32 GB memory. The code was executed using a single thread. The time limit of all computations is 4 h per instance.

Table 4 Comparison of the different methods for computing ${{\,\mathrm{rc}\,}}(X, Y)$

Table 4 summarizes our experiments and reports running time in shifted geometric mean $\prod _{i = 1}^n (t_i + 10)^{\frac{1}{n}} - 10$ to reduce the impact of outliers. In general, we can see that the compact model performs better than the column generation approach both in terms of running time and number of solved instances (marked as #opt). An explanation for this is the very costly pricing problem. However, we could observe that both the primal and dual bound after solving the root node are in general much tighter than in the compact model. A possible explanation for the better upper bound is that the variables generated by column generation could be well-structured such that SCIP can easily find good integer solutions for the covering formulation of (8). The meaning of the variables of the compact model, however, seems to be less clear to SCIP such that heuristics cannot find feasible solutions easily. This observation also explains why the hybrid model performs much better than the two separate methods: it benefits from the good bounds obtained by column generation and the quick solvability of each node of the branch-and-bound tree in the compact model. In particular for $\lozenge _5$, the hybrid model allows to solve two more instances than the compact model and reduces its running time by 91.9%.