The role of rationality in integer-programming relaxations

For a finite set $X \subset \mathbb{Z}^d$ that can be represented as $X = Q \cap \mathbb{Z}^d$ for some polyhedron $Q$, we call $Q$ a relaxation of $X$ and define the relaxation complexity $rc(X)$ of $X$ as the least number of facets among all possible relaxations $Q$ of $X$. The rational relaxation complexity $rc_\mathbb{Q}(X)$ restricts the definition of $rc(X)$ to rational polyhedra $Q$. In this article, we focus on $X = \Delta_d$, the vertex set of the standard simplex, which consists of the null vector and the standard unit vectors in $\mathbb{R}^d$. We show that $rc(\Delta_d) \leq d$ for every $d \geq 5$. That is, since $rc_{\mathbb{Q}}(\Delta_d)=d+1$, irrationality can reduce the minimal size of relaxations. This answers an open question posed by Kaibel and Weltge (Lower bounds on the size of integer programs without additional variables, Mathematical Programming, 154(1):407-425, 2015). Moreover, we prove the asymptotic statement $rc(\Delta_d) \in O(\frac{d}{\sqrt{\log(d)}})$, which shows that the ratio $rc(\Delta_d)/rc_{\mathbb{Q}}(\Delta_d)$ goes to $0$, as $d\to \infty$.


Introduction
Finding compact representations of a large set of integer points in a high-dimensional space is a recurring theme in combinatorial optimization, where these points usually describe the feasible solutions to a discrete problem.A classical approach is, given a set of points of interest X ⊆ Z d , to find a polyhedron Q ⊆ R d whose integer points are exactly those in X, i.e., such that Q ∩ Z d = X.Such a polyhedron is called a relaxation of X.By feeding a relaxation of X to an integer programming solver, one can optimize a linear objective function over X.The relaxation complexity of X, denoted by rc(X), is the smallest number of facets of a relaxation of X.This concept was introduced in [18] (see also [28]) as a measure of the complexity of a set of integer points.
A basic question is to determine the relaxation complexity of simple sets like the vertices of a hypercube or a simplex.For instance, one has rc({0, 1} d ) ≤ 2d by counting the number of facets of the 0/1 hypercube.However, a relaxation with only d + 1 facets exists [18].Moreover, it is easy to see that any bounded relaxation of a full-dimensional set X ⊆ Z d must have at least d + 1 facets, and one can further show that the 0/1 hypercube does not admit any unbounded relaxation [18], implying rc({0, 1} d ) = d + 1.
In general, however, a finite set of points can admit an unbounded relaxation.Notice that such a relaxation must have irrational rays only (assuming X = ∅): indeed, starting from a point of X and applying a rational ray would produce infinitely many integer points.As a prominent example, the authors of [18] describe a 5-dimensional simplex (that is unimodularly equivalent to the standard simplex) and an unbounded, irrational relaxation of its vertex set, which however has at least six facets.Inspired by this result, they pose the following questions.
Questions.Let ∆ d be the set consisting of the null vector and the canonical unit vectors in R d , and let rc Q (•) be the restriction of rc(•) to rational relaxations.
(Q1) Does rc(∆ d ) = rc Q (∆ d ) = d + 1 hold for all positive integers d? (Q2) If X ⊆ Z d is finite, does rc(X) ≥ dim(X) + 1 hold?Here, dim(X) is the dimension of the affine hull of X.
In this paper, we show that rc(∆ d ) < rc Q (∆ d ) for d ≥ 5.This gives negative answers to all the three questions above.The starting point is an example in dimension 5.
From this example, we can easily show that irrationality helps to reduce the relaxation complexity also in higher dimensions.Thus, Questions (Q1)-(Q3) have a negative answer for every d ≥ 5. where .denotes the rounding-down operation and (d + 1) mod 6 is the remainder of the division of (d + 1) by 6.
Since the relaxation complexity and rational relaxation complexity differ, one might wonder about the ratio rc(∆ d )/ rc Q (∆ d ) of these two quantities.By Corollary 2, this ratio is bounded from above by 5  6 + o(d), as d → ∞.With a significantly more involved construction, we can show that rc(∆ d )/ rc Q (∆ d ) → 0, as d → ∞, by providing the following asymptotic estimate: The remainder of this article is structured as follows.First, we mention related literature (Section 1.1) and introduce the notation and terminology that we use throughout the article (Section 1.2).In Section 2, we prove Theorem 1 and Corollary 2, and, based on the proof of Theorem 1, we derive our strategy for proving Theorem 3. The proof of the latter theorem is then presented in Section 3. Finally, we list some open problems in Section 4.

Related Literature
The relaxation complexity has been formally introduced in [18], where it is also shown how to derive lower bounds for it.Among others, this lower bound shows that any relaxation of a natural encoding of Hamiltonian cycles via binary vectors needs exponentially many inequalities in general.For X ⊆ {0, 1} d , an upper bound on the minimal number of inequalities needed to separate X and {0, 1} d \ X has been derived in [16].Complementing the upper and lower bounds, [5] derived criteria that guarantee that rc(X) can be computed by solving a mixed-integer program.More sophisticated mixed-integer programming techniques to compute rc(X) have been discussed recently in [7].Independently from the criteria by [5] that guarantee computability in arbitrary dimensions, [28] shows that rc(X) for X ⊆ Z d is computable if d = 2, and [5] extended this result to d = 3.In [6] it was shown that, if d = 2, then rc(X) can be found in polynomial time.Furthermore, [6] also addresses issues behind the computability of rc(X) and rc Q (X) for an arbitrary dimension and reveals that having rc(X) < rc Q (X) is an additional obstacle on the way to computing rc(X) and rc Q (X), as certain natural computational procedures fail to stop and so do not give finite algorithms in this case.For more details, see Section 4.
Extension complexity is a related, but more well studied concept than relaxation complexity (see [10] for a survey).The extension complexity of a polyhedron P ⊆ R d is the minimum number of facets of a polyhedron Q ⊆ R d+d that projects down to P .If P is the convex hull of our set of integer points X, the extension complexity of P measures how compactly one can describe X using a linear formulation with additional variables, while the relaxation complexity of X allows to use integer formulations, but with no additional variables.The existence of small extended formulations for a given polyhedron P is related to the existence of small non-negative factorizations of a matrix related to P by the celebrated Yannakakis' Theorem [30].Finding non-negative factorizations of (non-negative) matrices is a problem of independent interest, and in 1993 it was asked [9] whether using irrational numbers could lead to smaller factorizations of rational matrices.This was recently answered in [8] and [24]: irrational numbers indeed help, as there are rational matrices whose minimum size factorizations require irrational numbers.As far as the authors know, the examples given in the aforementioned papers do not correspond to polyhedra, hence it is open whether using irrational numbers leads to smaller extended formulations for rational polyhedra.
Despite the lower bounds and the computability results mentioned above, the existence of small relaxations for a given set of points does not have a general characterization similar to what is known for extended formulations in terms of non-negative factorizations and communication protocols [12,30].Moreover, we lack general approaches to construct relaxations, while several efficient constructions for extended formulations are known [2,3,17,26].This paper can be seen as one step forward in this direction.
Interestingly, sets of points like those corresponding to the spanning trees of a graph do not admit relaxations of subexponential size [18], but do admit polynomial-sized extended formulations [4,20,29]; the opposite happens for other sets, for instance those corresponding to stable sets or matchings in a graph, which admit simple, linear-sized relaxations, but no extended formulation of subexponential size [13,22].

Notation and Terminology
We use ⊆ and ⊂ to denote inclusion and strict inclusion, respectively.As usual, we define [m] := {1, . . ., m} for m ∈ Z >0 and [m] := ∅ for m = 0. We denote by 0 a (resp.0 a,b ) the all-zero vector (resp.all-zero matrix) of size a (resp.a × b), writing only 0 when the dimension is clear from the context.We denote the standard unit vectors in R d by e 1 , . . ., e d .Using this notation, ∆ d = {0, e 1 , . . ., e d }.The dimension dim(X) of X ⊆ R d is the dimension of its affine hull.The cardinality of X is denoted by |X| and its convex hull is denoted by conv(X).For two sets X and Y , we denote by Y X the set of all maps from X to Y .If X is finite, then R X is a |X|-dimensional vector space over R and, thus, it has the standard Euclidean topology.As usual, for φ : X → Z and Y ⊆ X, the notation φ| Y is used to denote the restriction of the map φ to Y .
We call elements of Z d integer points/vectors and we call elements of Z d × R m mixedinteger points/vectors.An affine transformation φ : R d → R d is said to be unimodular if it maps the integer lattice onto itself, that is, φ(Z d ) = Z d .It is known and easy to check (see, e.g., [23]) that φ is unimodular if and only if it can be represented as φ(x) = U x + v where v ∈ Z d and U ∈ Z d×d is a unimodular matrix, which is an integer matrix with determinant equal to one in absolute value.
The recession cone rec(P ) of a nonempty polyhedron P ⊆ R d is the set of all u ∈ R d satisfying p + u ∈ P for each p ∈ P .The recession cone of a polyhedron P = ∅ given by an inequality description Ax ≤ b is a polyhedral cone, which can be given by the inequality description Ax ≤ 0.
For X ⊆ Y ⊆ R k , a polyhedron P satisfying X = P ∩ Y is called a (polyhedral) relaxation of X within Y .By rc(X, Y ) we denote the minimal k such that X = P ∩ Y holds for some polyhedron P with k facets.If such k does not exist, we let rc(X, Y ) = ∞.We call rc(X, Y ) the relaxation complexity of X within Y .It is clear that rc(X, Y ) is monotonically non-decreasing with respect to Y in the sense that X ⊆ Y ⊆ Z implies rc(X, Y ) ≤ rc(X, Z).For X ⊆ Z d , we call rc(X) := rc(X, Z d ) the (integer) relaxation complexity while, for X ⊆ Z d × R n , we call rc(X, Z d × R n ) the (d, n)-mixed-integer relaxation complexity or the mixed-integer relaxation complexity with respect to d integer and n real variables.
We also provide some necessary background from the theory of irrational numbers.The field R is a vector space over Q and it can be seen in different ways that R is infinite-dimensional over Q.One easy non-constructive way to see this is to observe that R is uncountable, whereas Q and by this also every finite-dimensional vector space over Q is countable.This observation shows that for every k ∈ Z >0 there exists numbers a 1 , . . ., a k ∈ R that are linearly independent over Q.As a consequence, we also see that for every k ∈ Z >0 there exist numbers b 1 , . . ., b k ∈ R such that 1, b 1 , . . ., b k are linearly independent over Q.To show the latter, one can start with numbers a 0 , . . ., a k ∈ R linearly independent over Q and fix b i := a i /a 0 for i ∈ [k] 0 .Alternatively, the sequence b 0 = 1, b 1 , . . ., b k of numbers linearly independent over Q can also be obtained by choosing an arbitrary algebraic number c ∈ R of algebraic degree k +1 and fixing b i = c i for i ∈ [k] 0 .See [14,Sect. 4] for more details.

The Projection Approach
In this section, we prove Theorem 1 and Corollary 2.Then, we will discuss how our proof strategy can be generalized to show Theorem 3.
∆ is the vertex set of a 5-dimensional simplex and is unimodularly equivalent to ∆ 5 .We work with ∆ instead of ∆ 5 because this simplifies the calculations: giving a relaxation of ∆ will imply the existence of a relaxation of ∆ 5 of the same size.Let ⊂ R 5 be the line spanned by the vector (0, 0, 0, 1, √ 2).In [18] it is shown that conv(∆) + is an (unbounded) relaxation of ∆.However, such a relaxation has more than five facets.In order to give our relaxation, it is easier to reduce ourselves to a 4dimensional space by projecting on a suitable hyperplane along line , and looking for a relaxation of the projection of ∆ within the projection of Z 5 .In particular, we use the following observation: Observation 4. Let π : R 5 → R 4 be the projection and P ⊆ R 4 be a polyhedron satisfying In other words, one has rc(∆, Z 5 ) ≤ rc(π(∆), Z 3 × R).
Proof.By the choice of π, π(Z 5 ) is a subset of Z 3 × R.Moreover, is the kernel of π and Q is the pre-image of P under π.
We show that The condition π(q) ∈ π(∆) implies q ∈ ∆ + .But since is an irrational line, its only integer point is 0. Hence the points of ∆ are the only integer points in ∆ + .As q ∈ Z 5 , we conclude that q ∈ ∆.Conversely, if q ∈ ∆, then π(q) ∈ π(∆) ∈ P .Thus, q is in the pre-image of P under π, that is, q ∈ Q.This shows Q ∩ Z 5 = ∆ and yields the assertion.
We observe that the sequence of the values rc(∆ d ) is monotonically non-decreasing.
Proof.A relaxation of ∆ d+1 with k inequalities restricted to the coordinate subspace R d × {0} gives a relaxation of ∆ d with at most k inequalities.This shows the desired assertion.
Our use of Proposition 6 is two-fold.We need it as an ingredient in the proof of Theorem 1, on the one hand.On the other hand, since rc(∆ d ) grows with respect to d, we are justified to study the asymptotic growth of this quantity.
Remark 7 (Computer-assisted studies).The system (1)-( 5) has been discovered and validated using a computer.For the separation of ∆ from Z 4 with five inequalities, a mixedinteger programming model has been established, akin to the model suggested in [5].In the established model, instead of Z 4 a finite sufficiently large subset of Z 4 has been used as an "approximation" of Z 4 .The model was represented and solved in the floating-point arithmetic.The values that have been produced as a result have been inspected and manually converted to exact algebraic numbers in the field As a post-processing step, for the validation of the system, along with the manual verification, we also used the computer algebra system SageMath [11].Notably, SageMath can carry out exact polyhedral computations over the field of algebraic numbers.So, in Sage-Math it is possible to carry out all of the computational steps described in the above proof by working with polyhedra over the algebraic numbers.In particular, we used SageMath to determine an appropriate choice of ε > 0 in (1)-(5).

Extensions of the Five-Dimensional Counterexample
To prove Corollary 2, we show that rc(∆ d ) is sub-additive in the cardinality d + 1 of ∆ d .To this end, we introduce the free join of non-empty finite sets X ⊂ Z k and Y ⊂ Z and study its properties.For k, ∈ Z >0 , we define the free join by This operation corresponds to the free join of polytopes, as conv(X * Y ) is the free join of the polytopes conv(X) and conv(Y ), as defined in [15, 15.1.3].It is also useful to define the free join in the case when k or is zero, which means that one takes the free join with ∆ 0 := {0} ⊆ R 0 .This is naturally defined as X * ∆ The analogy for passing from X to X * ∆ 0 within the theory of polyhedra is building a pyramid over a given polytope.Note that (X + u) * (Y + v) coincides with X * Y up to a unimodular transformation for all u ∈ Z k and v ∈ Z .Indeed, (X + u) * (Y + v) is the image of X * Y under the affine unimodular map φ(x, y, z) := (x + (1 − z)u, y + zv, z).
Proof.We may assume w.l.o.g. that both X and Y admit a relaxation, i.e., rc(X) and rc(Y ) are finite, as otherwise the statement holds trivially.We first consider the case k > 0, > 0. Without loss of generality we may assume that both X and Y contain the null vector.Let Ax ≤ b and Cy ≤ d be inequality systems defining relaxations P and Q of X and Y , respectively, having rc(X) and rc(Y ) inequalities, respectively.Since both X and Y contain the origin, b and d are non-negative vectors.In particular, there needs to be at least one strictly positive entry in b and d, as otherwise, P or Q would be a full-dimensional cone and would contain infinitely many integer points.
To verify the inequality, it suffices to show that the polyhedron . If z = 0, then y is contained in the recession cone rec(Q) of Q.Since Q is a relaxation of Y , we conclude y = 0 as otherwise there are infinitely many integer points in Q.Moreover, setting z = 0 gives the original system Ax ≤ b, i.e., x ∈ X follows.Analogously, if z = 1, then x ∈ rec(P ) and y ∈ Y .Thus, (x, y, z) ∈ X * Y follows in both cases.
If z ≤ −1, then {y ∈ R : Cy ≤ d • z} is a subset of rec(Q), since decreasing the right-hand side makes the system more restrictive as d is non-negative.Since there exists at least one entry of d that is positive, Cy ≤ d • z excludes the null vector if z < 0. Hence, {y ∈ R : Cy ≤ d • z} does not contain any integer point.Since we can argue analogously for z ≥ 2 and the system When k or is zero, the argument is similar.Without loss of generality, let = 0. Then Y = ∆ 0 and X * Y = (X × {0}) ∪ {e k+1 }.Again, we can assume that 0 ∈ X and fix a system Ax ≤ b of rc(X) inequalities that defines a relaxation of X.One can see that the system Ax ≤ b • (1 − z), z ≥ 0 of rc(X) + 1 = rc(X) + rc(Y ) inequalities defines a relaxation of X * Y following the proof strategy we used in the case k > 0, > 0.
Let m and r be the quotient and the remainder of the division of d+1 by 6.By the above observation, ∆ d is equivalent to the free join of m copies of ∆ 5 and one copy of ∆ r−1 for r > 0. Thus, using Lemma 8 and Theorem 1, we obtain rc(∆ d ) ≤ m rc(∆ 5 ) + rc(∆ r−1 ) ≤ 5m + r.Remark 9. Combining Proposition 6 and the inequality rc(∆ d+1 ) ≤ rc(∆ d * ∆ 0 ) ≤ rc(∆ d ) + 1, which follows from Lemma 8, we conclude that, for any given d ∈ Z >0 , the value rc(∆ d+1 ) either stays equal to rc(∆ d ) or grows with respect to rc(∆ d ) by exactly one unit.Thus, for the determination of the exact behavior of rc(∆ d ) in d, the frequency of the jumps by one unit would have to be estimated from below and above.
As a consequence of our proof approach for Theorem 1, we also obtain the following.
Corollary 10.For every dimension d ∈ Z ≥5 , up to identification with respect to unimodular transformations, there exist infinitely many sets X ⊂ Z d with dim(X) = d and rc(X) < d + 1.
Proof.We first settle the case d = 5.For each a ∈ Z >0 , we consider the one-parametric generalization X a := {0, e 1 , e 2 , e 3 , (1, 0, 1, 1, 0), (0, 1, 1, 0, a)} of the set ∆, with ∆ = X a for a = 1.The projection of X a along the irrational line a spanned by (0, 0, 0, 1, a √ 2) under the map π a : R 5 → R 4 , is the same set ∆ as the projection of ∆ under the projection map π considered in the analysis of ∆.It is clear that π a maps Z 5 into Z 3 × R. Consequently, in the same way as we showed that π −1 (P1 /8 ) is a five-facet relaxation of ∆ within Z 5 relying on Observation 4 and Proposition 5, we can also show that π −1 a (P1 /8 ) is a five-facet relaxation of X a within Z 5 relying on a straightforward generalization of Observation 4 (with ∆ and π replaced by X a and π a , respectively) and Proposition 5.
Distinct sets X a and X a with a , a ∈ Z >0 and a = a do not coincide up to unimodular transformations when a = a , since X a is the vertex set of a simplex of volume 1  5! a and unimodular transformations preserve volume.This completes the proof of the assertion for d = 5.
For higher dimensions d ≥ 6, we use Y a := X a * ∆ d−6 .The set Y a is the vertex set of a simplex of volume 1 d! a.Thus, all of the sets Y a with a ∈ Z >0 are pairwise distinct up to unimodular transformations.Lemma 8 yields rc(Y a ) ≤ rc(X a ) + rc(

The Core Steps to Find the Counterexample
The core steps to find our counterexample consist of a projection of (a unimodular transformation of) ∆ 5 to a lower dimensional space and a determination of a (3, 1)-mixed-integer relaxation of the projection.As we will see below, the continuous coordinate of such a relaxation can be interpreted as a lifting from a 3-dimensional space.For this reason, to be able to generalize our proof strategy to prove Theorem 3, we provide two lemmas that explain how the relaxation complexity behaves under these two operations: projection and lifting.Afterwards, we will review our approach to prove Theorem 1 in the light of these two lemmas, which will motivate our strategy to prove Theorem 3.
Proof.Let k = rc(π(X), π(Z)).We may assume that k is finite, as otherwise the assertion is trivial.Consider a polyhedron Q with k facets, which is a relaxation of π(X) within π(Z), which means that Q ∩ π(Z) = π(X) holds.It is clear that the polyhedron P = π −1 (Q) has at most k facets.For verifying the asserted inequality, it suffices to show that P is a relaxation of X within Z, which means that P ∩ Z = X.
To check the inclusion P ∩ Z ⊆ X, we consider an arbitrary y ∈ P ∩ Z. From the definition of P , we see that π(y) ∈ Q ∩ π(Z).In view of Q ∩ π(Z) = π(X), we obtain π(y) ∈ π(X).As π| Z is injective and X ⊆ Z, we conclude that y ∈ X.
For showing the converse inclusion X ⊆ P ∩ Z, we consider an arbitrary y ∈ X.Since X ⊆ Z, we have y ∈ Z.The condition y ∈ P , which remains to be verified, is equivalent to π(y) ∈ Q.Since Q ∩ π(Z) = π(X) and π(y) ∈ π(X), we see that π(y) is indeed in Q.
Next, we show how we can derive upper bounds on the (k, 1)-mixed-integer relaxation complexity of finite sets by means of lifting.To formalize this, we introduce, for h : X → R, and call these sets the lift (resp.convex lift) of X with respect to the height function h.
Consider a (k + 1)-dimensional polytope P in R k+1 .The facet-defining inequalities for this polytope are inequalities in (x, y) ∈ R k × R, which can be brought into one of the three forms y ≤ u x + v, y ≥ u x + v or 0 ≤ u x + v, where u ∈ R k and v ∈ R. We call the respective facets of Q the upper facets, the lower facets, and the lateral facets.
Proposition 12. Let P ⊂ R k+1 be a full-dimensional polytope, h : T → R be a function from a finite non-empty set T ⊂ Z k .Then P is a relaxation of lift h (T ) within Z k × R if and only if 1. every p ∈ lift h (T ) is contained in an upper and lower facet of P , and 2. the projection of P onto the first k components is a relaxation of T within Z k .
Proof.First, suppose that P is a relaxation of lift h (T ) within Z k × R. If, by contradiction, there is (x, h(x)) ∈ lift h (T ) that is not contained in any upper facet of P , then there is some y > h(x) such that (x, y) ∈ P , contradicting the fact that P ∩ Z k × R = lift h (T ).Arguing similarly for lower facets, one checks Condition 1. Condition 2 follows immediately from the definition of relaxation.
For the opposite direction, let P satisfy Conditions 1 and 2. Condition 1 easily implies that P ∩ (Z k × R) ⊇ lift h (T ).For the reverse inclusion, let (x, y) ∈ P ∩ (Z k × R).We need to show that (x, y) ∈ lift h (T ).By Condition 2, one has x ∈ T .If (x, y) were not in lift h (T ), one would have y = h(x).Then, if y > h(x) (resp.y < h(x)), we conclude that (x, y) violates every upper facet (resp.lower facet) inequality of P , in particular those that are active on (x, h(x)), a contradiction.Motivated by a free-sum operation for polytopes [15, 15.1.3],we introduce the free sum of X ⊆ Z m and Y ⊆ Z n as

This operation will be useful, because ∆
We combine this map with the map ψ(a, b) := (a, m i=1 h(p i )b i ).Since h(p 1 ), . . ., h(p m ) are linearly independent over Q, the map ψ| Z k+m is injective.Consequently (ψ • φ)| Z k+m is injective as well, so we obtain Having Lemmas 11 and 13 in mind, the summary of the proof of the upper bound rc(∆ 5 ) ≤ 5 is as follows.First, we find it convenient to go from ∆ 5 to ∆.One can represent these two sets as columns of two matrices (see Figures 3a and 3b), which easily allows to check that they are unimodularly equivalent.Second, we apply the projection π on R 4 along line , as defined in Observation 4. This corresponds to applying Lemma 11, where X = ∆, Z = Z 5 .The points in π(∆) are listed as columns of the matrix in Figure 3c.Now, in light of Lemma 13, we see the points in π(∆) as coming from a lift of points in Z 3 .Let X = ∆ 3 and Y = {(1, 0, 1), (0, 1, 1)}, and h such that h(p) = 0 for p ∈ X, h(1, 0, 1) = 1, and h(0, 1, 1) = −1/ √ 2.Then, by Lemma 13, we have that rc In order to obtain Theorem 1, it now remains to show that P , defined as in Section 2.1, is a relaxation of lift h (X ∪ Y ) within Z 3 × R. One can do that by setting T = X ∪ Y and verifying Conditions 1 and 2 in Proposition 12. Condition 1 can be easily verified by the reader (notice that inequalities (1),( 2),(3) define upper facets of P , and inequalities (4),( 5) define lower facets).Verifying Condition 2 can be done directly via Fourier-Motzkin elimination, as mentioned in the proof of Theorem 1.
In summary, our proof of Theorem 1 consists of three steps: project (a unimodular transformation of) the simplex to a lower dimensional space; lift the projection into a space with one continuous coordinate; relax the mixed-integer point set by finding a small relaxation.To prove Theorem 3, we once again utilize this project-lift-relax strategy.That is, the steps of our proof are: (S1) Find a suitable unimodular copy of ∆ d and project this copy onto Z k via a map π.

(S2) Construct a suitable height function
By a variation of Lemma 13, which we formulate below as Lemma 16, the number of facets of P will provide an upper bound on rc( ∆d ) and thus on rc(∆ d ) since ∆ d and ∆d are unimodularly equivalent, provided the height function h satisfies all assumptions.In the following subsections, we discuss each of these three steps in detail.

The Proof of Theorem 3
This section is devoted to proving Theorem 3.

Project: Selecting the Right Simplex and Its Projection
Step (S1) requires us to select a set ∆d that is unimodularly equivalent to ∆ d as well as a projection π : R d → R k .To this end, consider the following d × (d + 1) matrix where I k denotes the identity matrix of order k, and B is the k × (d − k) matrix whose columns are all the vectors in {0, 1} k with at least two 1-entries.Note that this definition requires d = 2 k − 1.Although this assumption might look restrictive, we can reduce the proof of Theorem 3 to this special case as we will see later.
Let ∆d ⊂ R d be the set of columns of M .It is easy to see that ∆d and ∆ d are unimodularly equivalent.We select π to be the orthogonal projection of R d onto the first k coordinates.In the spirit of Lemma 13, we then let X ⊂ Z k be the set containing the first k + 1 columns of M (truncated to the first k coordinates), i.e., X = ∆ k , and Y ⊂ Z k be the set containing the truncation of the other columns of M , so that T := X ∪ Y = {0, 1} k .

Lift: Finding the Right Height Function
Our next goal is to find a suitable height function h for Step (S2).In view of Step (S3), we need to select h in such a way that the we can easily derive a mixed-integer relaxation of lift h (T ).By Proposition 12, such a relaxation is given by taking the 2k box constraints z ∈ [0, 1] k ×R and selecting upper and lower facets of clift h (Y ) that cover all points (z, h(z)) for z ∈ T .Our strategy for finding a small relaxation is thus (S3a) to find a small covering of the points (z, h(z)) for z ∈ T consisting of upper and lower facets, respectively.
To be able to quantify the minimum size of a covering by upper and lower facets, respectively, we need to derive combinatorial properties of the facet structure of clift h (T ).In our analysis in the next subsection, it will thus be convenient to have a function h that satisfies h(z) ∈ Q for all z ∈ T .In view of Lemma 13, however, we require that the values h(y) for all y ∈ Y are linearly independent over Q.To achieve this, we will slightly perturb h, but note that this may change the combinatorial structure of the facets of clift h (T ).As we show next, simplicial facets of clift h (T ) remain simplicial after small perturbations of h.Thus, we can still pursue our strategy if we restrict ourselves to cover the points (z, h(z)) in Step (S3a) by simplicial facets.Therefore, we define the simplicial upper (resp.lower) covering numbers of lift h (T ), denoted by sucn h (T ) (resp.slcn h (T )), as the minimum number of simplicial upper (resp.lower) facets that cover all vertices of the upper (resp.lower) part of clift h (T ).If no covering by simplicial facets exist, the respective number is set to be ∞.Lemma 14 (On semi-continuity).Let T be an m-dimensional subset of {0, 1} m .Then, for every h ∈ R T which is not a restriction of an affine function, there exists an open neighborhood U ⊆ R T of h with the property that for every h ∈ U one has Proof.We only prove the first inequality, since the proof of the second inequality is completely analogous.Every simplicial upper facet of Q := clift h (T ) is given by where v 0 , . . ., v m ∈ T are affinely independent.By standard linear algebra, with an appropriate ordering of v 0 , . . ., v m , the respective facet-defining inequality for Q is given by det By expanding the determinant with respect to the last column, one can see that the fact that F h is an upper facet can be expressed as and the strict validity of the inequality on vert(Q) Since φ h (x) is linear in h, the set is open (it is the interior of a polyhedron) and it satisfies h ∈ H.By construction, for every h ∈ H, the set F h is an upper facet of clift h(T ).Thus, whenever t simplicial upper facets F 1 h , . . ., F t h cover the vertex set of clift h (T ) we see there exist t respective open sets H 1 , . . ., H t with h ∈ H i for every i ∈ {1, . . ., t}.

It follows that, for every h in the open set
. ., F t h are simplicial upper facets of clift h(T ) that cover the vertex set of clift h(T ).In particular, the latter holds for every h in a sufficiently small neighborhood of h.
Another difficulty arises in the tools that we have available for Step (S2), because Lemma 13 requires a height function h with h(x) = 0 for all x ∈ X.We were not able to find such a height function that allows us to find a small relaxation in Step (S3a).For this reason, we need a strengthening of Lemma 13, which we give in Lemma 16 below.In preparation for this, we provide the following proposition, which is actually a classical observation used in the theory of regular polyhedral subdivisions.For the sake of completeness, we give a short proof.
Proposition 15.Let X ⊆ Z m and h, h : X → R be such that h(x) − h(x) = f (x) for every x ∈ X, where f : R m → R is an affine function.Then Proof.The affine map φ(x, y) = (x, y + f (x)) maps both R m × R and Z m × R bijectively onto themselves and it satisfies φ(x, h(x)) = (x, h(x)), which means that φ maps lift h(X ) bijectively onto lift h (X).Thus every relaxation of lift h(X ) within Z m × R is bijectively mapped to a relaxation of lift h (X) within Z m × R and vice versa.This yields the desired assertion.
Lemma 16.Let X be an affinely independent subset of Z k with 0 ∈ X, let Y be a finite m-element subset of Z k disjoint with X, and let h : X ∪ Y → R be such that (a) h(x) ∈ Q for every x ∈ X and (b) the values h(y) with y ∈ Y together with 1 are linearly independent over Q.
Proof.Since X is affinely independent, there exists an affine function f that coincides with h on X.Since h(x) ∈ Q for every x ∈ X, we can choose f with rational coefficients.Then the function h − f is equal to 0 on X.
We now verify that the assumptions of Lemma 13 are fulfilled by checking that the values h(y) − f (y) with y ∈ Y are linearly independent over Q.By the choice of f , we have f (y) ∈ Q for every y ∈ Y .Hence, since the values h(y) with y ∈ Y together with 1 are linearly independent over Q, it follows the values h(y) − f (y) with y ∈ Y are linearly independent over Q.Thus, using Lemma 13 and then Proposition 15, we obtain Equipped with these tools, we are able to provide a statement that allows us to bound the relaxation complexity in terms of simplicial upper and lower covering numbers.Lemma 17.Let X ⊆ {0, 1} k , let Y be an m-element subset of {0, 1} k disjoint with X, and let h : X ∪ Y → R be any function which is not a restriction of an affine function and satisfies h(x) ∈ Q for all x ∈ X.Then Proof.One can approximate h arbitrarily well by h : X ∪Y → R for which h(x) = h(x) ∈ Q for every x ∈ X and the values h(y) with y ∈ Y , together with 1, are linearly independent over Q.Thus, the assertion follows by combining Proposition 12 with T = X ∪ Y and Lemmas 16 and 14.
Recall from the previous subsection that we want to apply a lifting to the set T = X ∪Y with X = ∆ k and Y = {0, 1} k \ X. Due to Lemma 17, we can use in Step (S2) of our strategy any height function h such that h| T is not the restriction of an affine function and such that h(x) ∈ Q for all x ∈ X. Towards proving Theorem 3, we select Remark 18. Function r is the height function that defines the so-called staircase triangulation of [0, 1] k−1 , see, e.g., [19,Sec. 16.7.2].That is, the statements about simplicial facets of lift h (X ∪ Y, Z k × R), which we will make in the following section, can be related to a triangulation of [0, 1] k−1 if we fix x k to 0 or 1.

Relax: Constructing a Small Mixed-Integer Relaxation
To complete Step (S3a), we use the height function h from the preceding subsection and bound sucn h ({0, 1} k ) and slcn h ({0, 1} k ) from above.First, we show that the two values are equal, i.e., it is sufficient to investigate the upper or lower facets.
Proposition 19.One has Proof.The affine map φ(x 1 , . . ., x k , y) := (x 1 , . . ., x k−1 , 1 − x k , −y) is an affine involution on R k+1 .Furthermore, φ can be restricted to an involution on lift h ({0, 1} k ) since ∈ lift h ({0, 1} k ) for each (x 1 , . . ., x k ) ∈ {0, 1} k .This implies that a subset F ⊆ clift h ({0, 1} k ) defines a facet of clift h ({0, 1} k ) if and only if the set φ(F ) does.Now, in order to obtain the identity in the thesis it thus suffices to show that F is an upper facet if and only if φ(F ) is a lower facet.Indeed, let f : R k+1 → R be the affine function defining an upper facet F of clift h ({0, 1} k ), so that f (x) ≤ 0 is valid for each x ∈ clift h ({0, 1} k ), equality is attained for x ∈ F , and the coefficient of x k+1 is positive.Then f (φ(x)) ≤ 0 is valid for each x ∈ clift h ({0, 1} k ), with equality attained for x ∈ φ(F ), and the coefficient of x k+1 is negative, hence the facet φ(F ) given by the affine function f (φ(x)) is a lower facet of clift h ({0, 1} k ).The reverse implication holds by an analogous argument.
Proposition 19 allows us to focus on the upper facets of clift h ({0, 1} k ).We will consider two families of simplicial (upper) facets: one will cover vertices with x k = 0 and the other will cover vertices with x k = 1.For simplicity we denote a vertex of clift h ({0, 1} k ) by (x, x k+1 ), where x ∈ {0, 1} k and x k+1 = h(x).We now define the facets that we will use in our covering.Let π be a permutation of Proof.F π can be easily seen to contain exactly k + 1 affinely independent vectors, hence in order to show that it is a facet, it suffices to exhibit a valid inequality for clift h ({0, 1} k ) that is satisfied at equality precisely by the points of F π .One can argue in the same way for F B,π .Consider the inequality Let (x, x k+1 ) be a vertex of clift h ({0, 1} k ).If x k = 0, then ( 8) is trivially satisfied, as the left-hand side is non-positive and the right-hand side would be non-negative, and it is satisfied with equality only if x = 0. Now, let x k = 1, hence x k+1 = ( k−1 i=1 x i ) 2 = 2 for some ∈ Z ≥0 .Then the right-hand side of ( 8) is the sum of odd numbers, i.e., at least i=1 (2i − 1) = 2 = x k+1 , and equality holds exactly when (x, x k+1 ) ∈ F π .Now, consider the inequality where x(B) = i∈B x i similarly as before.Let h 1 = x(B) and with the sum of the first h 2 odd numbers.The expression above is at least (h 1 + h 2 ) 2 if and only if The latter is easily seen to hold, since h 1 ≤ b.
The case x k = 0 follows from the previous case by noticing that the left-hand side of (9) decreases by 2(h 1 + h 2 ) 2 and the right-hand side decreases by 2b 2 ≤ 2(h 1 + h 2 ) 2 .
Finally, one can easily verify that the inequality is satisfied with equality exactly when (x, x k+1 ) ∈ F B,π , which concludes the proof.
We now give our covering.The idea is to cover all vertices of clift h ({0, 1} k ) that have x k = 1 with facets of the form F π , and the other vertices with facets of the form F B,π .
where for the latter we used the fact that a set of size i−1 points to exactly k −i sets, none of which has to be picked in X .We now set p i = ln(k−i) k−i , and use the fact that 1 − p ≤ e −p for all reals p.We then get where the second equality follows from the known identity ( k ) +1 = ( k+1 +1 ) k+1 , and the last inequality from the binomial expansion.

Putting All Pieces Together
With the preparatory work of the preceding sections, we are able to provide a proof of Theorem 3. ).

Outlook
In (P3) As rc(X) and rc Q (X) do not coincide in general, there are two algorithmic problems that one can ask, for every fixed dimension d as well as for an unspecified d: Determine whether rc(X) and rc Q (X) are algorithmically computable.The existence of sets X satisfying rc(X) < rc Q (X) is an additional complication that makes an algorithmic computation of these two values quite an intricate task, as explained in [7].The authors of [6] introduce a non-increasing sequence (u k ) k∈N of upper bounds on rc Q (X), which are computable for each given k.This sequence stabilizes by attaining the value rc Q (X) after finitely many steps.However, it is open whether one can decide for fixed k whether u k = rc Q (X).One way to decide this is by providing a computable matching lower bound.In [18], a mechanism to derive lower bounds via so-called hiding sets is presented.Deriving a lower bound via hiding sets yields ≤ rc(X) ≤ rc Q (X) ≤ u k .Thus, this approach can only work if rc(X) = rc Q (X).In [5] it is shown that rc(X) = rc Q (X) for any X ⊆ Z d when d ≤ 4, and for other special cases.In this case, one might wonder about a computable lower bound on rc(X) that is tight.Due to Corollary 2 and Theorem 3, however, rc(X) can be strictly smaller than rc Q (X), and thus, matching upper bounds on rc Q (X) and lower bounds on rc(X) do not exist in general.
(P4) Determine the minimum of rc(X) over all sets X ⊆ Z d of dimension d = dim(X).This quantity can be analyzed asymptotically, for d → ∞, as well as for concrete choices of d.It would be interesting to see if the minimum is attained for X = ∆ d .
(P5) Study the relaxation complexity with respect to arbitrary sub-fields of R. One can introduce the relaxation complexity rc F (X) of X ⊆ Z d with respect to relaxations of X described by inequalities with coefficients in F. It is clear that rc F (X) is monotone in F with respect to inclusion.Our arguments show that the estimate rc F (X) ∈ O( d √ log(d) ) holds for every field of dimension at least d + 1 over Q, as we only need d + 1 numbers in R that are linearly independent over Q to prescribe appropriate heights.√ 2] has dimension two over Q.This is the smallest dimension of F over Q with rc F (∆ 5 ) = 5, as the only one-dimensional field over Q is Q itself.It is likely that rc F (∆ 5 ) = 5 holds for every field F ⊂ R of dimension two over Q, which just means that rc Q[r] (∆ 5 ) = 5 should be true for every r ∈ R \ Q.
(P6) When F is the field of real algebraic numbers, determine if rc F (X) is computable for a given finite X ⊂ Z d and if the equality rc F (X) = rc(X) holds for every finite X ⊂ Z d .Indeed, our results show that the choice between rationals and arbitrary reals makes a difference, but the reals in their whole generality include numbers that one cannot even approximate algorithmically, so it makes sense to ask if some reasonably constructive subfield of the reals would be enough for computing the relaxation complexity.The field of algebraic numbers could be a good candidate.

Figure 2 :
Figure 2: The projection on the first three components of ∆ (convex hull in yellow) and the polytope P1 /8 (green).The inequality description of the projection of P1 /8 can be derived from the inequality description (1)-(5) of P1 /8 using Fourier-Motzkin elimination, which combines the three upper bounds (1)-(3) on x 4 with the two lower bounds (4)-(5) on x 4 .As a result, one obtains 3 × 2 = 6 inequalities for x 1 , x 2 , x 3 , and it turns out that each of these inequalities defines a facet of the projected polytope.

Figure 3 :
Figure 3: Visualization of the reductions in the proof of Theorem 1.

Proof of Theorem 3 .
First, assume that there is a positive integer k such that d = 2 k − 1.Let m = d − k.Furthermore, let ∆d as well as T = X ∪ Y be defined as in Section 3.1.Then, X = ∆ k and Y = {0, 1} k \ X.Moreover, as defined at the end of Section 3.2, let h : {0, 1} k → R be given by h(x 1 , . . ., h k ) = (2x k − 1) • r(x 1 , . . ., r k−1 ).By Lemma 17,rc(∆ d ) = rc(∆ k ⊕ ∆ m ) ≤ rc(X ∪ Y ) + sucn h (X ∪ Y ) + slcn h (X ∪ Y ).As X ∪ Y = {0,1} k and rc({0, 1} k ) ≤ 2k, we derive from Proposition 19 and Corollary 24 the upper bound rc(∆ d ) ∈ O( 2 k √ k ).Because k = log 2 (d + 1), the assertion follows.If there is no positive integer k with d = 2 k −1, then let d be the smallest integer greater than d that admits such a representation.Note that d < d ≤ 2d and rc(∆ d ) ≤ rc(∆ d ).Thus, by the above argumentation, rc(∆ d ) ≤ rc(∆ d ) ∈ O( d √ log(d ) ) = O( d √ log(d) light of our new results, we formulate the following open problems:(P1) For given d ∈ Z >0 , characterize all finite subsets X of Z d satisfying rc(X) < rc Q (X).The first open case is d = 5.(P2) Improve the asymptotic upper bound on rc(∆ d ).The best known lower bound on rc(∆ d ) is of asymptotic order Ω(log d), see[5].In contrast, our asymptotic upper bound O( d √ log d ) is only mildly sublinear in d.It is widely open what asymptotic behavior for rc(∆ d ) one should expect, but it is quite likely that our upper bound is not tight.