Constant factor approximations with families of intersection cuts

Transcription

1 Constant factor approximations with families of intersection cuts Gennadiy Averkov, Amitabh Basu 2,, and Joseph Paat 2, Institute of Mathematical Optimization, Faculty of Mathematics, University of Magdeburg, Germany 2 Department of Applied Mathematics and Statistics, Johns Hopkins University, USA Abstract. We study the problem of approximating the corner polyhedron using intersection cuts derived from families of lattice-free sets. In particular, we look that the problem of characterizing families that approximate the corner polyhedron up to a constant factor. The literature already contains several results in this direction. In this paper, we use the maximum number of facets of a lattice-free set in a family as a measure of its complexity and precisely characterize the level of complexity of a family required for constant factor approximations. As one of the main results, we show that for each natural number n, a corner polyhedron for n integer variables is approximated by intersection cuts from lattice-free sets with at most i facets up to a constant factor (depending only on n if i > 2 n and that no such approximation is possible if i 2 n. When the approximation factor is allowed to depend on the bit complexity of the vertex of the corner polyhedron, we show that the threshold is i > n versus i n. The tools introduced for proving such results are of independent interest for studying intersection cuts. Introduction Given a matrix R := (r,..., r k R n k, with k columns r,..., r k R n, a vector f R n \ Z n, the set { C(R, f := conv s R n 0 : f + } k i= s ir i Z n has been studied in the integer programming literature, as a framework for deriving cutting planes (cuts for general mixed-integer programs; see Chapter 6 of [] for a detailed discussion. When both R and f are rational, Meyer s theorem (see [6] implies that C(R, f is a rational polyhedron. We will refer to C(R, f as the corner polyhedron for (R, f. The term corner polyhedron initially constrained s to be integer [3], but recent work allowed for continuous and mixed s-variables (see Chapter 6 of [] and the references therein. A complete inequality description of C(R, f can be obtained via gauge functions of lattice-free sets. A subset B R n is lattice-free if B is n-dimensional, Supported by the NSF grant CMMI

2 closed, convex, and the interior of B does not contain points of Z n. A latticefree set is called maximal if it is not a proper subset of another lattice-free set 3. Let B R n be a closed and convex set with 0 int(b. The gauge function ψ B : R n R of B is ψ B (r := inf {λ > 0 : r λb}. Let R R n k and f R n \ Z n. Given a lattice-free set B with f int(b, the intersection cut for (R, f generated by B (or the B-cut of (R, f is { C B (R, f := s R k 0 : } k i= s iψ B f (r i. We extend this notation to include lattice-free sets B such that f R n \ int(b; in this case we define C B (R, f := R k 0. Given a family B of lattice-free subsets of R n we call the set C B (R, f := B B C B(R, f the B-closure of (R, f. We define C (R, f = R k 0. Intersection cuts can be partially ordered based on set inclusion: if B B 2 are lattice-free sets then C B2 (R, f C B (R, f for all (R, f. Hence maximal lattice-free sets produce the strongest cuts [9, 0]. Furthermore, all lattice-free sets are contained in maximal lattice-free sets, which are polyhedra (see [5]. Therefore focus can be directed towards intersection cuts from polyhedra. Definition (L n i, i-hedral closures, and Ln. For i N, let L n i denote the family of all lattice-free (not necessarily maximal polyhedra in R n with at most i facets; we call C L n i (R, f the i-hedral closure of (R, f. Let L n denote the family of all lattice-free (not necessarily maximal polyhedra in R n. When describing C(R, f using a B-closure, C(R, f C B (R, f for every (R, f, and if L n B then one even has equality C B (R, f = C L n (R, f = C(R, f for every rational (R, f [8]. This implies that one approach to obtain C(R, f for rational (R, f is to classify maximal lattice-free sets and compute cuts using the corresponding gauge functions. Recent work has focused on this classification [, 4, 5, 2]. The classification was given for n = 2 in [2], but a classification is unknown even for n = 3. Moreover, even with a classification, the resulting gauge functions may be difficult to compute. Indeed, for a latticefree polyhedron with k facets, the resulting gauge function is a maximum over k inner products. Therefore, for large values of k (for maximal lattice-free sets in R n, k is at most 2 n, this computation can be expensive. In light of these difficulties, instead of fully describing C(R, f by classifying lattice-free sets, one can aim to find a small and simple family of intersection cuts whose closure closely approximates it [2, 6, 8]. In other words, one can search for a simple family B and a constant α > so that the inclusion C B (R, f C(R, f α holds for all (R, f (note that, when nonempty, C B (R, f and C(R, f are convex sets with recession cone R k 0. This leads to the main question we consider: 3 Some sources omit the condition dim(b = n in the definition of maximal lattice-free sets, but the case dim(b < n is not needed for this paper.

3 Question. Let B and L be families of lattice-free sets in R n. Under what conditions does there exist some α > so that the inclusion C B (R, f α C L(R, f holds for all pairs (R, f? Also, for a fixed f Q n \ Z n, when does there exist α such that C B (R, f α C L(R, f holds for all rational R? If such an α exists, then the B-closure approximates the L-closure within a factor of α, that is the B-closure provides a finite approximation of the L- closure for all choices of (R, f (or for a fixed f and all R. Since computation of the gauge function of a polyhedron depends on the number of facets, there is additional motivation to consider families B made of polyhedra with few facets. In this paper, we focus on answering Question. A consequence of this pursuit is studying the trade-off between the complexity of a family B of lattice-free sets and the quality of approximation of C(R, f by C B (R, f. In fact, our results will be about the quality of approximation of C L n (R, f by C B (R, f. Hence, many results can be stated without any rationality assumption on (R, f. Only when one wants to interpret these results for the corner polyhedron C(R, f, one must consider rationality, which is needed for the equivalence C L n (R, f = C(R, f. 2 Summary of results For α, we call α C B(R, f the α-relaxation of the cut C B (R, f. Analogously, for a family of lattice-free sets B, we call α C B(R, f the α-relaxation of the B- closure C B (R, f. Using α-relaxations, the relative strength of cuts and closures can be quantified naturally as follows. For f R n \ Z n and lattice-free subsets B and L of R n, we define ρ f (B, L := inf { α > 0 : C B (R, f α C L(R, f R }. ( The value ρ f (B, L quantifies up to what extent C B (R, f can replace C L (R, f. For α, the inclusion C B (R, f α C L(R, f says that the cut C B (R, f is at least as strong as the α-relaxation of the cut C L (R, f. For α <, the previous inclusion says that not just C B (R, f but also the α-relaxation of the cut C B (R, f is at least as strong as the cut C L (R, f. Thus, if ρ f (B, L, the B-cuts of (R, f are stronger than the L-cuts of (R, f for every R, and the value ρ f (B, L quantifies how much stronger they are. If < ρ f (B, L <, then the B-cuts of (R, f are not stronger than the L-cuts of (R, f but stronger than α- relaxations of L-cuts for some α > 0 independent of R, where the value ρ f (B, L quantifies up to what extend the L-cuts should be relaxed. If ρ f (B, L =, then C B (R, f cannot replace C L (R, f because there is no α independent of R such that C B (R, f is stronger than the α-relaxation of C L (R, f. In addition to comparing the cuts coming from two lattice-free sets, we want to compare the relative strength of a family B to a single set L, and the relative strength of two families B and L. We consider these comparisons when f is fixed or arbitrary. For the case of a fixed f R n \ Z n we introduce the functional ρ f (B, L := inf { α > 0 : C B (R, f α C L(R, f R },

4 which compares B-closures to L-cuts for a fixed f. We also introduce ρ f (B, L := inf { α > 0 : C B (R, f α C L(R, f R } for comparing B-closures to L-closures for a fixed f. The analysis of ρ f (B, L can be reduced to the analysis of ρ f (B, L for L L, since one obviously has ρ f (B, L = sup {ρ f (B, L : L L}. (2 For the analysis in the case of varying f, we introduce the functionals: Observe that ρ(b, L := sup {ρ f (B, L : f R n \ Z n }, ρ(b, L := sup {ρ f (B, L : f R n \ Z n }. ρ(b, L = sup {ρ f (B, L : f int(l}, (3 ρ(b, L = sup {ρ f (B, L : f int(l, L L}. (4 The functional ρ(b, L was introduced in [6,.2], where the authors initiated a systematic study for the case of n = 2. In the case that R, f is rational, since C L n (R, f = C(R, f, the value ρ(b, L n describes how well C B (R, f approximates C(R, f. As previously mentioned, there is motivation to consider families of latticefree polyhedra with few facets. Our first main result compares i-hedral closures with j-hedral closures using the functional (4. Theorem. Let i {2,..., 2 n }. Then ρ(l n i, Ln i+ = for every i 2n and ρ(l n i, Ln 4 Flt(n for every i > 2 n, where Flt(n is the flatness constant 4. Another way to examine the relative strength of i-hedral closures is with the functional (2 for a fixed f R n \ Z n. Theorem immediately implies that ρ f (L n i, Ln < for i > 2 n. If approximation is allowed to depend on f, then this can be improved to ρ f (L n i, Ln < for i > n. Theorem 2. Fix f Q n \ Z n. Let i {2,..., 2 n }. Then ρ f (L n i, Ln i+ = for every i n and ρ f (L n i, Ln < for every i > n. An important point to note is that Theorem 2 assumes rationality of f, and if ρ f (L n i, Ln is finite, it depends on the bit complexity of f and the dimension n. Rationality on f or R is not required for the other results in this paper. The proofs of Theorems and 2 are outlined in subsections 5. and 5.2, respectively. Our main tool used in proving Theorems and 2 is Theorem 3, which is geometric in nature. We set up some notation necesssary to state the result. 4 The flatness constant Flt(n is a number depending only on n and is known to be upper bounded by γn 3/2 for some universal constant γ > 0 [7, Chapter VII].

5 Definition 2 (C f and B f. For f R n \Z n, let C f be the collection of all closed, full-dimensional, convex sets with f in the interior. For a family of lattice-free sets B, define B f := C f B. Theorem 3 requires a metric on the space C f. We refer to this as the f-metric, whose definition appears in Section 3. In that section, we motivate the necessity of such topological concepts for the analysis of relative strength. We believe that this topology on the collection of lattice-free sets could be useful in future research. We use cl f to denote the closure operator with respect to the f-metric. Theorem 3. Let B be a family of lattice-free sets of R n and let L be a family of lattice-free polyhedra such that L L n i for some i N. Then the following hold: (a Suppose cl f (B f = B f for a fixed f R n \ Z n. Then ρ f (B, L < if and only if there exists µ (0, such that for every L L with f int(l, there exists some B B satisfying B µl + ( µf. (b Suppose cl f (B f = B f for all f R n \ Z n. Then ρ(b, L < if and only if there exists µ (0, such that for every L L and every f int(l, there exists some B B satisfying B µl + ( µf. Theorem 3 establishes a relationship between the functional ρ f (B, L (similarly, ρ(b, L, and individual lattice-free sets B B and L L. It is clear from the definitions that if B µl + ( µf then the cut C B (R, f is contained in the cut C µl+( µf (R, f for all R. If such a µ can be found so that this homothetical inclusion holds for all L (where B is allowed to depend on L, then the inclusion of intersection cuts implies the inclusion C B (R, f µc L (R, f. This implies that ρ f (B, L µ < (similarly, ρ(b, L µ <, establishing the sufficiency of the conditions stated in Theorem 3. The non-trivial direction is proving the necessity of these conditions when ρ f (B, L and ρ(b, L are finite. This forms the content of Section 4. 3 The f-metric One nice property of ρ f (B, L is that it can be analyzed using individual sets B and L (in fact, this idea is one way to interpret Theorem 3. However it is not always sufficient to only consider sets B B, as seen in the following example. Example. Let L R 2 be a maximal lattice-free split. Choose f int(l and let B L 2 3 C f be the set of all maximal lattice-free triangles containing f in the interior. Since L is a split, there is a nonzero vector r in lin(l. Set R = (r, r. The intersection cut C L (R, f = while C B (R, f for each B B. Hence ρ f (B, L = for each B B. However, it was shown in [8] that ρ f (B, L 2. The issue in Example is that the set L is a limit point of the family B, but L B. Examples such as this one motivates the use of a metric so that limit sets may be considered.

6 For f R n \ Z n, recall from Definition 2 the collection C f to be all closed, full-dimensional, convex sets in R n that contain f in their interior. We define the f-metric d f : C f C f R 0 on C f to be d f (B, B 2 := d H ((B f, (B 2 f, (5 where B, B 2 C f, (B i f denotes the polar of B i for i =, 2, and d H denotes the Hausdorff metric 5. Note that d f (, is always finite. This follows since f int(b for each B C f, and the polar of a convex set is bounded if and only if the set contains the origin in its interior. 4 One-for-all theorems and proof of Theorem 3 Let f R n \ Z n and B, L be families of lattice-free sets in R n. Theorem 3 states that the finiteness of ρ f (B, L and ρ(b, L reduces to understanding the relationship between individual sets in B and L. Theorem 4 argues that this is true in the case when f is fixed and L consists of a single set. Theorem 4 (One-for-all Theorem for a family B and a set L. Let B be a family of lattice-free subsets of R n and L = conv(v + cone(w a lattice-free polyhedron in R n with V. Let f R n and suppose cl f (B f = B f. Then ( V + W + inf ρ f (B, L ρ f (B, L inf ρ f (B, L. (6 B B B B Proof (Sketch. Note that for every B B, ρ f (B, L ρ f (B, L and so the right inequality of (6 holds. For the left inequality of (6, first suppose that L is a polytope, i.e. W =. We may assume f int(l, otherwise all the functionals evaluate to 0. We may also assume ρ f (B, L <, which implies B f. Using the fact that ρ f (B, L <, one can show that for ɛ := /(( V + ρ f (B, L, there is some B B f B satisfying ɛ L + ( ɛ f B. Therefore C B (R, f ɛ C L (R, f for all R. From (, ρ f (B, L ( V + ρ f (B, L. This yields ( V + inf ρ f (B, L B B ( V + ρ f (B, L ρ f (B, L. If L is not a polytope, we can restrict ourselves to the polytope case by representing L using a sequence of polytopes and employing a limiting argument. A complete rigorous proof appears in Appendix A.4. An immediate corollary of Theorem 4 handles the case of arbitrary f. Corollary. Let B and L be as in Theorem 4. Then ( sup inf V + W + ρ f (B, L ρ(b, L B B f R n \Z n sup f R n \Z n inf ρ f (B, L. (7 B B 5 The Hausdorff metric is defined on the family of compact sets of R n as follows: d H(A, B for compact sets A, B is the minimum ɛ > 0 such that A B + N(0, ɛ and B A + N(0, ɛ, where N(0, ɛ is the ball of radius ɛ around the origin.

7 Corollary and Theorem 4 together give a more general One-for-all type result where L is a family of sets. That is, in order to understand when a B- closure provides a finite approximation for an L-closure, it is enough to compare individual cuts coming from B B and L L. Theorem 5 (One-for-all Theorem for two families. Let B be a family of lattice-free subsets of R n and let L L n i for some i N. Since L L n i, there is some N N so that every L L can be written as L = conv(v + cone(w with V + W + N. Let f R n \ Z n and suppose cl f (B f = B f. Then N sup inf ρ f (B, L ρ f (B, L sup inf ρ f (B, L. (8 B B B B L L If cl f (B f = B f for all f R n \ Z n, one even has N sup L L,f int(l inf ρ f (B, L ρ(b, L B B L L sup L L,f int(l inf ρ f (B, L. (9 B B Note that (8 follows from (6, and (9 follows from (7. Theorem 5 characterizes when ρ f (B, L < and ρ(b, L < for two families B and L. In particular, finiteness depends on the individual values ρ f (B, L for sets B B and L L. For two sets B and L and µ (0,, it can be shown that ρ f (B, L /µ if and only if B µl + ( µf (see Proposition 0 in Appendix A.3. Therefore, with this equivalence, Theorem 5 implies Theorem 3. 5 The relative strength of i-hedral closures Theorem 3 provides a characterization for when a B-closure finitely approximates a L-closure, but it does not immediately provide any specific (and interesting families for when this happens. However, we can use this characterization to determine the relative strength of i-hedral closures. This yields Theorems and 2, which we show in subsections 5. and 5.2, respectively. From Theorem 3, identifying which values of i yield ρ(l n i, Ln < or ρ f (L n i, Ln < can be done by analyzing the structure of polyhedra with at most i facets. In order to help with this geometric analysis, we make use of the flatness constant. For every nonempty subset X of R n, the width function w(x, : R n [0, ] of X is defined to be w(x, u := sup x u inf x u. x X x X The value w(x := inf u Z n \{0} w(x, u is called the lattice width of X. Theorem 6 (Flatness Theorem. For every choice of n N, the flatness constant Flt(n := sup {w(b : B lattice free set in R n } is finite.

8 The literature contains many upper bounds on Flt(n (see [7, Chapter VII]. A final note before proving Theorems and 2 is that Theorem 3 requires knowledge of cl f (L n i C f for f R n \ Z n. However, in light of the following result, understanding the f-metric is not required to understand the relative strength of i-hedral closures; the proof appears in Appendix A.. Proposition. Let i, n N and f R n \ Z n. Then cl f (L n i C f = L n i C f. 5. The relative strength of i-hedral closures for arbitrary f We divide the proof of Theorem into Propositions 2 and 3. For Proposition 2, most of the proof is contained in Lemma 2 stated below. This lemma, in turn, relies on the next result whose proof appears in Appendix A.6. Lemma. Let L L n be such that L (R n [γ, λ] for γ [, 0], λ [0, ], and λ γ. Suppose there is some k N and M 0 L n k so that L (R n {0} M 0 {0}. Then for each f int(l, there is some B L n k+ such that 4 L f B. Lemma 2. Let L R n be a lattice-free set such that w(l. Then for each f int(l, there exists a lattice-free set B L n 2 n + so that 4 L f B. Proof (Sketch. After applying a unimodular transformation, we may assume that w(l, e n, where e n is the n-th standard basis vector in R n. We may also assume 4 L f Rn [, ]. If 4 L f is in Rn [, 0] or R n [0, ], then setting B to be the split R n [, 0] or R n [0, ], respectively, yields the desired result. So assume that this is not the case. Note that L R n {0} is a lattice-free set in R n, and is therefore contained in some maximal lattice-free set M 0 R n with at most 2 n facets. By Lemma, there exists B L n 2 n + such that 4 L + 3 4f B. A complete, rigorous proof appears in Appendix A.7. Proposition 2. ρ(l n i, Ln 4 Flt(n for i > 2 n. Proof. From Theorem 3 and Proposition, it is enough to show that for each f R n \ Z n and each L L n C f, there is some polyhedron B L n i so that 4 Flt(n L + ( 4 Flt(n f B. To this end, let f Rn \ Z n. Note that w( Flt(n L + ( Flt(n f, and so Lemma 2 gives the desired result. In order to use Theorem 3 to show that ρ(l n i, Ln i+ = for a given i N, for every µ (0, we must find a set L(µ = L L n that cannot be scaled by µ and placed inside a polyhedron B L n i. To this end, we start by showing every lattice-free polyhedron with k facets must be scaled enough to fit inside a lattice-free polyhedron with k facets. The proof of Lemma 3 appears in Appendix A.7. Lemma 3. Let M R n be a maximal lattice-free polyhedron. Suppose M has k facets and take c int(m. Then there exists an ε (0, so that every lattice-free polyhedron in R n containing ( εm + εc has at least k facets.

9 Proposition 3. ρ(l n i, Ln i+ = for i 2n. Proof. Assuming ρ(l n i, Ln i+ <, we derive a contradiction. Since i 2n, Lemma 5 yields that there is some M L n i \L n i that is maximal lattice-free for R n. We consider M as embedded in R n R n. For any z relint(m, let ε(z denote the value obtained from Lemma 3 by setting c = z. Fix z relint(m. For each ε > 0, we consider the pyramid L ε := conv ( {z + εe n } (( + ε (M z + z en with apex z + εe n and base ( + ε (M z + z en. The pyramid L ε L n i+ is maximal lattice-free with exactly i + facets. Theorem 3(ii and Proposition imply the existence of µ (0, independent of ε > 0 such that for every f L ε some B ɛ L n i satisfies µl ε + ( µf B ɛ. Let ɛ > 0 and γ (0, ɛ be chosen so that (a ε( ( µ ( µ < 0, and (b µ + γ µ ε µ > ε(z. For example, one can choose ɛ = 2 ( µ µ and γ = max{ ɛ 2, ɛ ε(z µ 2 (+ µ }. Now choose f = z + γe n. With this choice of f, L := µl ε + ( µf {z } ( (( = conv( + (µε + ( µγen µ + ε (M z + z + (γ( µ µen is a pyramid with apex z +(µε+( µγe n and base µ (( + ε (M z + z + (γ( µ µe n. From (a and the fact that γ < ɛ, we obtain that γ( µ < µ, i.e., γ( µ µ < 0. Thus the base of L is below the hyperplane R n {0}. For λ R, define L ε λ := L ε (R n {λ} and L λ := L (R n {λ}. Claim. Let λ [, 0]. Then L ε λ = ( λ ε (M z + z + λen. Proof of Claim. Using the definitions of L ε λ and L ε it follows that x L ε λ iff x n = λ and x L ε ( iff x = +λ +ε ( (z + εe n + +λ +ε iff x = ( λ ε (y z + z + λen, for y M iff x ( λ ε (M z + z + λen. (( + ε (y z + z en, for y M Define β := γ( µ µ. From (a and the fact that γ < ɛ, we see that β (, 0. Claim 2. L 0 = µl ɛ β + ( µf.

10 Proof of Claim. L 0 is the set of points y L with y n = 0. L is exactly the set of points that can be written as µx + ( µf for some x L ɛ. All points of this form that have 0 in the last coordinate must satisfy x n = γ( µ µ = β. Thus L is the set of points that can be written as µx + ( µf for some x L ɛ β. We now follow this sequence of equalities: L 0 = µl ε β + ( µf from Claim 2 (( = µ β ε (M z + z + βe n + ( µ(z + γe n from Claim ( ( ( = µ β ε M + µ β ε z. ( From (a and (b, µ β ε > ε(z. Thus the definition of ε(z implies that any lattice-free polyhedron containing L 0 requires i facets. In particular, B ε must have at least i facets since the cross-section of B ε by the hyperplane R n {0} contains C 0. Since B ε L n i, B ε must have exactly i facets. However, for small enough ε, the base of L can be made to have lattice-width bigger than the flatness constant Flt(n. This would imply that B ɛ is not a cylinder, because it must contain the base of L. Therefore B ε must have a full-dimensional recession cone. However, this contradicts that B ε is lattice-free. 5.2 The relative strength of i-hedral closures for a fixed f We divide the proof of Theorem 2 into Propositions 4 and 5. Proposition 4. ρ f (L n i, Ln < for each f Q n \ Z n and every i n +. Proof. We use induction on n. As a base case, take n =. Note that i {n +,..., 2 n } implies i = 2 n = 2. Using Theorem 3(a, the result holds since L 2 contains all maximal lattice-free sets in R. So suppose ρ f (L n i, L n < for each f Q n \ Z n and every i {n,..., 2 n }. Fix f Q n \Z n and take i {n+,..., 2 n }. From Theorem 3(a, it is enough to find some µ (0, so that for each L L n C f, there exists a B L n i so that µl + ( µf B. From here, we consider cases on f. In what follows, L L n C f. Since L is lattice-free, L := Flt(n (L f + f satisfies w(l. After a unimodular transformation, we may assume L (R n [, ] =: C. f Case : Assume f n > 0. We claim µ = n (+ f n Flt(n (0, gives the desired result. Note that x L implies x n and so x n f n + f n. For each x L it follows that ( fn + f (x n n f n + f n f n = f n + f x n n f n f n. fn Hence + f n (L f+f is contained in the split D := R n [f n f n, f n + f n ]. Since f int(l and L f is convex, it holds that n + f n (L f + f C. Hence f n + f n (L f+f = µl+( µf is contained in the lattice-free split B := C D.

11 Case 2: Assume f n = 0. Define f = (f,..., f n Q n \Z n. From the induction hypothesis, ρ f (L n i, Ln <. We claim µ := /(4ρ f (L n i, Ln (0, gives the desired result. Define L 0 := L (R n {0}. When restricted to R n, notice that L 0 L n C f. Hence there is some B 0 L n i so that ( ( L ρ f (L n i,ln 0 + f B ρ f (L n i,ln 0. From Lemma, there is some B L n i so that ( µl + ( µf = L + 4ρ f (L n i,ln ( 4ρ f (L n i,ln f B. Similar to Proposition 3, proving ρ f (L n i, Ln = requires us to identify, for every choice of µ (0,, some polyhedron L L n k with k > i satisfying B µl+( µf for every B L n i. However, unlike in the proof of Proposition 3 where we chose a specific value for f to construct L, proving ρ f (L n i, Ln = requires L be dependent on some fixed f. The next result helps us overcome this complication. The proof of Lemma 4 appears in Appendix A.8. Lemma 4. Let n N, f Q n \ Z n and µ (0,. Then (a There exists a maximal lattice-free simplex L L n n+ C f such that, for some choice of n+ integer points z,..., z n+ in the relative interior of the n+ distinct facets of L, the following is fulfilled: every closed halfspace disjoint from int(µl+( µf contains at most one point of the set {z,..., z n+ }. (b For every simplex L in (a, B µl + ( µf for every B L n+ n C f. Proposition 5. ρ f (L n i, Ln i+ = for each f Qn \ Z n and every i n. Proof. From Theorem 3, it suffices to show that for all i {,..., n}, f Q n \Z n and µ (0,, there exists L L n i+ C f satisfying B µl + ( µf for all B L n i C f. For i = n, the assertion follows by choosing L as in Lemma 4(b. Consider the case i < n. After applying an appropriate unimodular transformation we may assume that f = (f, 0,..., 0 R n for some f Q i \ Z i. Application of Lemma 4 in dimension i yields the existence of a maximal latticefree simplex L L i i+ such that B µl +( µf holds for every B L i i. We choose L = L R n i and show that B µl + ( µf for every B L n i. The homothetical copy µl+( µf of L contains the affine space A := {f } R n i. If B A, we get B µl + ( µf. Otherwise, B A and thus B can be represented as B = B R n i with B L i i. In this case, B µl + ( µf since (B R i µl + ( µf. References. Andersen, K., Louveaux, Q., Weismantel, R., Wolsey, L.: Inequalities from two rows of a simplex tableau. In: Fischetti, M., Williamson, D. (eds. Integer Programming and Combinatorial Optimization. 2th International IPCO Conference, Ithaca, NY, USA, June 25 27, Proceedings, Lecture Notes in Computer Science, vol. 453, pp. 5. Springer Berlin / Heidelberg (2007

12 2. Andersen, K., Wagner, C., Weismantel, R.: On an analysis of the strength of mixedinteger cutting planes from multiple simplex tableau rows. SIAM Journal on Optimization 20(2, ( Averkov, G.: A proof of Lovász s theorem on maximal lattice-free sets. Beitr. Algebra Geom. 54(, (203, 4. Averkov, G., Krümpelmann, J., Weltge, S.: Notions of maximality for integral lattice-free polyhedra: the case of dimension three. ( Averkov, G., Wagner, C., Weismantel, R.: Maximal lattice-free polyhedra: Finiteness and an explicit description in dimension three. Math. Oper. Res. 36(4, (Nov 20, 6. Awate, Y., Cornuejols, G., Guenin, B., Tuncel, L.: On the relative strength of families of intersection cuts arising from pairs of tableau constraints in mixed integer programs. Mathematical Programming 50, ( Barvinok, A.: A course in convexity. American Mathematical Society ( Basu, A., Bonami, P., Cornuéjols, G., Margot, F.: On the relative strength of split, triangle and quadrilateral cuts. Mathematical Programming Ser. A 26, ( Basu, A., Conforti, M., Cornuéjols, G., Zambelli, G.: Maximal lattice-free convex sets in linear subspaces. Mathematics of Operations Research 35, ( Borozan, V., Cornuéjols, G.: Minimal valid inequalities for integer constraints. Mathematics of Operations Research 34, (2009, Conforti, M., Cornuéjols, G., Zambelli, G.: Integer programming, vol. 27. Springer ( Dey, S.S., Wolsey, L.A.: Lifting integer variables in minimal inequalities corresponding to lattice-free triangles. In: Lodi, A., Panconesi, A., Rinaldi, G. (eds. Integer Programming and Combinatorial Optimization. 3th International Conference, IPCO 2008, Bertinoro, Italy, May 26 28, Proceedings, Lecture Notes in Computer Science, vol. 5035, pp Springer Berlin / Heidelberg ( Gomory, R.E.: Some polyhedra related to combinatorial problems. Linear Algebra and its Applications 2(4, ( Hoffman, A.: Binding constraints and helly numbers. Annals of the New York Academy of Sciences 39, ( Lovász, L.: Geometry of numbers and integer programming. In: Iri, M., Tanabe, K. (eds. Mathematical Programming: State of the Art, pp Mathematical Programming Society ( Meyer, R.: On the existence of optimal solutions to integer and mixed-integer progamming problems. Mathematical Programming 7, ( Schneider, R.: Convex Bodies: The Brunn-Minkowski Theory, vol. 44. Cambridge University Press ( Zambelli, G.: On degenerate multi-row Gomory cuts. Operations Research Letters 37(, 2 22 (Jan 2009,

13 A Appendix A. Properties of the f-metric Here we collect some notes on the f-metric, many of which follow from the next f result from [7]. In what follows, we use B t B to denote convergence in the H f-metric and C t C to denote convergence in the Hausdorff metric. Theorem 7 (Theorem.8.7 in [7]. Let (K t t= be a sequence of nonempty, convex, compact sets in R n and K R n be nonempty, convex, and compact. Suppose that K t K. This is equivalent to the following two conditions H together: (a each point x K is a limit of points (x t t=, where x t K t for t N; (b the limit of any convergent sequence (x t t=, where x t K t for t N, is contained in K. When dealing with intersection cuts, our interest in the f-metric will restrict to lattice-free sets. The following result shows that lattice-free sets form a closed subset under the f-metric, allowing us to freely consider sequences of such sets. Proposition 6. Let (B t t= C f and B C f so that B t f B. Then Proof. (a if x B t for each t N, then x B; (b if x int(b t for each t N, then x int(b. (a It is well known that B f = ((B f. Thus it is enough to show r (x f for each r (B f. From Theorem 7 (a, each r (B f is a limit of points (r t t=, where r t (B t f. Observe r t (x f for all t, and so, by continuity of the inner product, r (x f. (b This follows using an argument similar to that of part (a. In Example, the collection B is not a closed subset of C f. In general, the collection (L n i \Ln i C f of polyhedra in R n with exactly i-facets is not closed in C f under d f. However, the set L n i C f is closed under d f. This is Proposition. Proof (of Proposition. Let (B t t= be a sequence in L n i C f and B C f so f that B t B. For each t N, let kt denote the number of facets of B t. Since 0 int(b t f there is some A t R n kt so that B t = {x R n : A t x }. Note that, as each B t f is a polyhedron with at most i facets, k t i for each t N. Writing each B t using such a system of inequalities, it follows that (B t f = conv{v : v row(a t }. It follows from Theorem 7(a that any extreme point v of (B f is a limit point of the form v = lim t v t, where v t row(a t. Hence there is some A R n k so that k i and (B f = conv{v row(a}. This implies that B is a polyhedron with at most i facets.

14 A.2 Background on lattice-free sets and gauge functions A geometric characterization of maximal lattice-free sets was given by Lovász [5]; see also [3] and [9]. We say a transformation T : R n R n is affine unimodular if T (x = Ux + v for a unimodular matrix U R n n and some v Z n. Theorem 8. Let B be a lattice-free subset of R n. Then the following hold (a B is maximal lattice-free if and only if B is a lattice-free polyhedron and the relative interior of each facet of B contains a point of Z n. (b If B is maximal lattice-free, then it is a polyhedron with at most 2 n facets. (c If B is an unbounded maximal lattice-free set, then B coincides up to an affine unimodular transformation with B R k, where k n and B is a bounded maximal lattice-free set. The bound 2 n in Theorem 8(b is tight, and moreover there is a maximal lattice free set with i facets for every i between 2 and 2 n. Lemma 5. Let i, n N so that 2 i 2 n. There exists a maximal lattice free set with exactly i facets. Proof. We begin with the following claim. Claim 3. Let i, n N so that 2 i 2 n. There exists a family G of i disjoint faces of the unit cube [0, ] n that cover all of its vertices {0, } n. Proof of Claim. We provide a constructive argument for the existence of G. Begin with G = {F 0, F }, where F 0 = {x [0, ] n : x n = 0} and F = {x [0, ] n : x n = }. Iteratively update G as follows. If G = i, then we are done. If G < i 2 n then choose H G such that H is not 0-dimensional. Since H is not a vertex, there is some k [n] so that neither H 0 = {x H : x k = 0} nor H = {x H : x k = } are empty. Recursively update G to G \ {H} {H 0, H }. After i 2 recursive steps, G = i. From Claim 3, there exist disjoint faces F,..., F i of the unit cube [0, ] n that cover the points {0, } n. For j [i], let u j R n and c j R be such that the inequality u j x c j is valid for [0, ] n and defines the face F j. Then we claim that B = {x R n : u j x c j, j [i]} is lattice-free with i facets. Moreover, by construction, v {0, } n satisfies u j v = c j if and only if v F j. Because the F j were chosen to be disjoint, the vertices of F j are contained in the relative interior of the facet of B corresponding to the inequality u j x c j. It remains to show that B is lattice-free indeed, the result then follows from Theorem 8. Suppose int(b Z n. Let z int(b Z n be such that it is the closest to [0, ] n amongst all points in int(b Z n. This implies that {z} {0, } n are in Z n - convex position, i.e., conv({z} {0, } n Z n = vert(conv({z} {0, } n Z n = {z} {0, } n. This contradicts a classical theorem due to Doignon-Bell-Scarf stating one can have at most 2 n integer points in Z n -convex position [4].

15 A gauge function ψ B satisfies the following simple properties ψ B (r 0 r R n, ψ B (r + r 2 ψ B (r + ψ B (r 2 r, r 2 R n, λψ B (r = ψ B (λr r R n λ 0, λψ B (r = ψ λ B (r r R n λ > 0, and the equivalence ψ B (r = 0 if and only if r rec(b is straightforward to check. The definition of an intersection cut is in terms of the gauge function of a lattice-free set, while the f-metric defined in Section 3 is on convex sets and not functions. Proposition 7 states that the f-metric acts nicely when transitioning from lattice-free sets to their gauges (and consequently, to intersection cuts. Proposition 7. Fix f R n \ Z n. Let B C f, and take (B t t N C f f B. Then ψbt f ψ B f pointwise. B t so that Proposition 7 follows from the fact that the gauge function of a set containing the origin in its interior is equal to the so-called support function of the polar 6 (see [7, Theorem.7.6], and the fact that if a family of compact, convex sets converge to a limit in the Hausdorff topology, then their support functions converge pointwise to the support function of the limit (see [7, Theorem.8.]. Proposition 7 implies that it is possible to restrict considerations to topologically closed families. Proposition 8. Let B be a family lattice-free convex sets such that B f. Then C B (R, f = C Bf (R, f = C clf (B f (R, f for every (R, f. Proof. Choose a (R, f pair with R R n k and f R n \ Z n. The equality C B (R, f = C Bf (R, f holds because C B (R, f = R k 0 for each B B with f int(b. So we focus our attention on the second equality. Since B f cl f (B f, it follows that C clf (B f (R, f C Bf (R, f. In order to show C Bf (R, f C clf (B f (R, f, we derive C Bf (R, f C B (R, f for every B cl f (B f. Consider a sequence (B t t N of elements of B f converging to B in the f-metric. Consider an arbitrary s C Bf (R, f. For each t N we have k i= ψ B t f (r i s i since B t B f. In view of Proposition 7, in the latter inequality we can pass to the limit as t obtaining k i= ψ B f (r i. Thus, s C B (R, f. A.3 Properties of ρ f (B, L, ρ f (B, L, and ρ f (B, L The following proposition shows that, for α, the α-relaxation α C B(R, f of the cut C B (R, f is also an intersection cut (the claim follows from the basic properties of the gauge functions. 6 For a given set S R n, the support function σ S is defined as σ S(r = sup x S r x.

16 Proposition 9. Let B be a lattice-free subset of R n. Consider the intersection cut C B (R, f of some (R, f with f int(b and let α. Then α C B(R, f is an intersection cut for (R, f too, since α C B(R, f = C B (R, f, where B is a lattice-free subset of B defined by B := ( α B + f. α We characterize the condition C B (R, f C L (R, f, for all R and a fixed f, which can also be formulated as ρ f (B, L. Proposition 0. Let B and L be lattice-free subsets of R n and let f int(b int(l. Then the following conditions are equivalent: (i C B (R, f C L (R, f holds for every R, (ii B L, (iii ψ B f ψ L f Proof. (i (ii: Take f + r L. This implies that ψ L f (r and int(c L (r, f. By (i, this implies that int(c B (r, f, or equivalently, that ψ B f (r. Hence r B f and so f + r B. (ii (iii: Let r R n and take λ > 0 such that r λ(l f. By (ii, r λ(b f and so ψ B f (r λ. Taking the infimum over λ > 0 gives ψ B f (r ψ L f (r. (iii (i: Choose R R n k and take s C B (R, f. Note that k ψ L f (r i s i i= k ψ B f (r i s i, i= where the first inequality follows from (iii and the second since s C B (R, f. Hence s C L (R, f. Propositions 9 and 0 give the following description of ρ f (B, L. Proposition. Let B and L be maximal lattice-free subsets of R n f R n \ Z n. Then one has ρ f (B, L = inf {α > 0 : B α L + ( α } f and let = inf {α > 0 : αb + ( αf L} = inf {α > 0 : ψ B f αψ L f } (0 if f int(l and ρ f (B, L = 0, otherwise. Observe that whenever the infimum in (0 is finite, it is attained for some α.

17 A.4 Proof of Theorem 4 The inequality ρ f (B, L inf B B ρ f (B, L holds because ρ f (B, L ρ f (B, L for every B B. So we need to check the lower inequality in (6. First consider the case when L is a polytope. For the lower inequality in (6, it suffices to consider the case inf B B ρ f (B, L > 0 and ρ f (B, L <. This implies that f int(l and B f. For α, α > 0 we consider the conditions and R : C B (R, f αc L (R, f ( B B so that R : C B (R, f α C L (R, f. (2 We want to find a relation between α and α under which ( implies (2. Propositions 9 and 0 together imply that C B (R, f α C L (R, f holds for every R if and only if B f α (L f holds. Thus, (2 is equivalent to B B : B f α (L f, (3 where B f α (L f can also be reformulated as B f α vert(l f. Thus we want to determine α and α under which one has the implication R : C B (R, f αc L (R, f B B : B f α vert(l f. Taking the contrapositive we arrive at the equivalent implication B B : B f α vert(l f R : C B (R, f αc L (R, f. (4 So assume that the premise of (4 is fulfilled. We fix R = (r,..., r k to be the matrix with k = vert(l columns in which all vertices of L f are listed in some order. By our assumption, for every B B there exists i [k] with α r i B f. The latter means that ψ B f (r i > /α. This implies that α (,..., C B (R, f. Indeed, one needs to check that α k i= ψ B f (r i holds for every B B. Since for every B B, there exists r i such that ψ B f (r i is at least /α, the latter inequality is clearly true. The point α (,..., does not belong to αc L (R, f if α α (,..., does not belong to C L(R, f. Since, by construction, ψ L f (r i = for every i [k], the latter condition amounts to the inequality α k α <. This inequality is fulfilled if say α = α (k +. Summarizing, we conclude that for α = α/(k +, condition ( implies (2. Now let α > 0 be so that ρ f (B, L < /α. For all R, it follows that C B (R, f αc L (R, f. From the previous discussion, there exists some B B so that for all R, ( α C B (R, f C L (R, f. + vert(l Hence ρ f (B, L ( + vert(l/α ( + V /α. Since ρ f (B, L < /α was arbitrarily chosen, dividing this inequality by + V and taking the infimum gives the lower inequality of (6.

18 Now consider the case when L is not necessarily a polytope. The proof idea is to approximate L by a polytopes, use the proof of the polytope case and pass to the limit. For this, we pick v 0 V and consider the lattice-free polytopes L t = conv(v (v 0 + tw with t N. Let α > 0 be so that ρ f (B, L /α. Hence for every R one has C B (R, f αc L (R, f. Set α = α/( V + W +. From the proof of (a, we get that the implication R : C B (R, f αc Lt (R, f B B : B f α (L t f. (5 For each t, we have L t L, which implies C L (R, f C Lt (R, f. Thus, the premise of (5 holds for every t. It follows that there exists B t B with B t f α (L t f. Dualization of the latter inequality gives (B t f (α (L t f for every t N. Now the sets (B t f with t N form a bounded sequence of compact convex sets. We conclude using Blaschke s selection theorem [7, Theorem.8.6] that there is a convegerent subsequent, whose limit we represent as (B f. Since the family B is closed under the f-metric, we get B B. It is not hard to check that the limit of (α (L t f is (α (L f. Indeed, for each t N, α (L t f α (L f and so lim t (α (L t f (α (L f. For the inclusion, suppose that y R n so that y (α (L f. Hence there is some x L f so that (/α y x >. For large enough values of t, x L t f and so y (α (L t f. This implies that lim t (α (L t f (α (L f. Thus passing t to the limit, we get the inclusion (B f (α (L f. Dualization gives the inequality B f α (L f. This containment implies inf ρ f (B, L ρ f (B, L V + W + =. B B α α Dividing through by V + W +, we get ( inf ρ f (B, L V + W + B B α. Since α > 0 was arbitrarily chosen so that ρ f (B, L /α, the lower inequality in (6 follows. A.5 Truncated Pyramids Definition 3 (Truncated pyramid. Let M R n be a closed, convex set, and let M = ( + αm + p, for α 0 and p R n. Suppose that aff(m aff(m. Then P := conv(m M is a truncated pyramid with bases M and M. Lemma 6. Let P be a truncated pyramid with bases M and M = (+αm +p. Then

19 (a P can be given by P = 0 λ In particular every point t P can be given as ( ( + λαm + λp. (6 t = ( + λαx + λp with 0 λ and x M. (b If f P can be given as f = ( + µαx + µp with x M and /3 µ, then 4 P f conv({x} M P. (7 Proof. (a Equality (6 can be shown directly: P = conv(m M = (( λm + λm = 0 λ 0 λ ( ( + λαm + λp. (b conv({x} M P is clear, since x M. We show 4 P f conv({x} M. Since P = conv(m M, it suffices to check the two inclusions 4 M f conv({x} M 4 M f conv({x} M. This is equivalent to showing the following inclusions obtained by translating the right and the left hand sides by x: where 4 (M x (f x conv({0} (M x 4 (M x (f x conv({0} (M x, M x = ( + α(m x + αx + p, f x = µ(αx + p. This shows that for proving the two inclusions we can assume x = 0 (this corresponds to substitution of M for M x and p for αx + p. So we assume x = 0 and need to verify 4 M f conv({0} M 4 M f conv({0} M, where f = µp and M = ( + αm + p. Since conv({0} M = 0 λ λm it suffices to verify to show that the left hand sides are subsets of λm for

20 some appropriate choices of λ. For the first inclusion we choose λ = 3 4 µ. We have to check 4 M f µm that is 4 M µp 3 4 µ( + αm µp. The latter is fulfilled since µ 3. For the second inclusion we choose λ = 3 4 µ + 4. We have to check 4 M + 3 ( 3 4 f 4 µ + M 4 that is 4 ( + αm + 4 p µp The latter is obviously fulfilled. ( 3 4 µ + ( 3 ( + αm µ + p. 4 Proposition 2. Suppose P R n is a truncated pyramid with bases M and M. Let f int(p and assume (i M (R n {λ} and M (R n {γ}, for λ (0, and γ (, 0; (ii λ γ ; (iii for some k N, M = S U with S a k-simplex and U a linear space; (iv 4 (P f + f is not contained in a split. Then there exists a polyhedron B with at most k + 2 facets such that and 4 P f B (Rn [, ], (8 B (R n {0} P (R n {0}. (9 Proof. Since M = S U has k + facets, P can be written as P = {x R n : a i x b i, i [k + ]} (R n [γ, λ]. (20 We consider two cases on the facet structure of P. Case : Suppose that neither P (R n {λ} nor P (R n {γ} is a facet of P. Set B := P. Note (9 follows immediately. Also B has at most k +2 facets as seen from (20. Moreover, since f int(p and P is convex, (8 follows. Case 2: Suppose that P (R n {λ} and P (R n {γ} are facets of P. Define the polyhedron P := {x R n : a i x b i, i [k + ]} (R n [, ], (2 and consider the faces M = P (R n {} and M = P (R n { } of P. If either M or M is not a facet of P, then set B = P. Equation (9 holds immediately and (2 implies that B has at most k + 2 facets. Also, since f int(p P (R n [, ], Equation (8 follows. In Case 2, it is left to consider when both M and M are facets of P.

21 Claim 4. P is a truncated pyramid with bases M and M = ( + α M + p. Proof of Claim. Note that truncated pyramids are preserved under invertible linear transformations and translations. Therefore we may apply an invertible linear transformation so that assume M = S U R k R n k and p = e n. The projection of P onto R k is a translated cone, since S is a simplex. Furthermore, the projections of P (R n {λ} and P (R n {γ} are two cross-section of this cone, as are the projections of M and M. Since γ < λ and α > 0, it follows that M is of the form ( + α M + βp for some α > 0 and β R. Using Claim 4 and Lemma 6(a, we can write f = (+µα x+µp for x M and µ [0, ]. Observe that f n = (+µα x n +µp n = µ[(+α x n +p n]+( µx n = µ+( µ = 2µ. (22 From (i, we get the inclusion 4 P f (Rn [, ]. From this inclusion and assumption (iv, there exist x, y 4 P f so that x n (0, and y n (, 0. This implies the inequalities 4 λ+ 3 4 f n > 0 and 4 γ f n < 0. Using λ [0, ] and γ [, 0] from (i, these inequalities can be rearranged to show f n ( 3, 3. Applying these bounds on of f n to Equation (22, we get µ ( 3, 3 2. From Lemma 6(b, the set B := conv({x} (( + α M + p satisfies Equation (8. Note B has at most k + 2 facets since ( + α M + p has at most k + facets. Equation (9 holds as P P is convex. A.6 Proof of Lemma We begin with the following technical lemma. Lemma 7. Let P = {x R n : a i x b i, i I} be a lattice-free polyhedron. Then there exists a nonempty subset I I and some k N so that {x R n : a i x b i, i I } = S U, where S is a k-simplex and U a linear space. Proof. We may assume that a i 0 for each i I and I = k N. Let z P. Since P is lattice-free, the linear subspace L = aff(rec(p z is not fulldimensional. Therefore dim(l. Let u L be nonzero so that u x and ( u x for each x P. By Farkas Lemma, there exists nonnegative scalars {λ i } k i= so that k i= λ ia i = u and 0 < k i= λ i. Set M = k i= λ i and note that M u conv{a i : i I }. A similar argument with u shows that αu conv{a i : i I } for some α < 0. Hence, 0 conv{ M u, αu} conv{a i : i I}. Let I I be a minimal set so that 0 conv{a i : i I }. Then 0 relint(conv{a i : i I }. Since I is minimal, Caratheodory s Theorem implies that conv{a i : i I } is a k-simplex for k = I. Therefore, {x R n : a i x b i, i I } is of the form S U for U a n k dimensional linear space.

22 Proof (of Lemma. Fix f int(l. In the case that 4 L + 3 4f is contained in a split, we may choose B to be such a split. So we can focus on the case when 4 L + 3 4f is not contained in a split. We may write M 0 {0} using the inequality description M 0 {0} = {(x, x n R n R : (a i, 0 (x, x n b i, i [k]} (R n {0}. Define L 0 := L (R n {0}. Since L 0 M 0 {0} and L is not contained in a split, the facets of M 0 can be extended in R n to create a polyhedron M containing L. Extending those facets and truncating at the hyperplanes R n {λ} and R n {γ}, we obtain the polyhedron M := {(x, x n R n R : (a i, α i (x, x n b i, i [k]} (R n [γ, λ]. From Lemma 7, the cross-section M (R n {0} is contained in M 0 = {(x, x n R n R : (a i, α i (x, x n b i, i I } (R n {0}, where M 0 = S U for S a k-simplex and U a linear space. Define M := {(x, x n R n R : (a i, α i (x, x n b i, i I } (R n [γ, λ]. Note B M and M 0 := M (R n {0} = M 0. Proposition 2 gives a set B = {x R n : c j x d j, j J}, where J I +, so that 4 L f B (R n [, ], (23 and B (R n {0} M 0. (24 Finally, define the set B R n to be B := {x R n : c j x d j, j J} {(x, x n R n R : (a i, α i (x, x n b i, i [k]\i }. Both sets defining B contain 4 L + 3 4f, so B does as well. Also, B has at most k + facets. Finally, from Equations (23 and (24, B is lattice-free. A.7 Lemmas used in the proof of Theorem Proof (of Lemma 2. Since w(l, by performing a unimodular transformation we may assume that L (R n [γ, λ], where γ [, 0], λ [0, ], and λ γ. Define L 0 := L R n {0}. Note L 0 is lattice-free in R n. Thus there is a maximal lattice free set M 0 R n containing L 0. Theorem 8(b implies that M 0 L n 2 n. The result follows from Lemma.

23 Proof (of Lemma 3. Let F,..., F k denote the facets of M. From Theorem 8(a, for each i [k] there is a z i Z n so that z i relint(f i. We will show that for each pair of indices {i, j} [k], there is some α i,j (0, so that no single valid inequality for ( α i,j M + α i,j c separates both z i and z j. Then for ε = min{α i,j : {i, j} [k]}, any lattice-free polyhedron containing ( ɛm + ɛc must contain a separate facet for each z i. This will give the desired result. Let {i, j} [k]. Since z i relint(f i and z j relint(f j, the point m i,j = 2 (z i + z j is contained in int(m. Therefore, there is some α i,j > 0 so that m i,j int(( α i,j M +α i,j c. Assume to the contrary that some valid inequality a x b for ( α i,j M + α i,j c separates both z i and z j. Then b 2 (a z i + a z j = a m b. Hence a m = b and so m int(( α i,j M + α i,j c which is a contradiction. A.8 Proof of Lemma 4 (a: We argue by induction on n. In the case n =, we can choose L, z and z 2 such that L = [z, z 2 ] and z < f < z 2 = z + for some z Z. Assume the assertion is true for some dimension n N and let us verify the assertion in dimension n +. Let f Q n+ \ Z n+ and µ (0,. Applying a unimodular transformation, we can assume that f has the form f = (f, 0 with f Q n \Z n. By the induction assumption there exists a simplex L L n n+ such that, for some choice of n + integer points z,..., z n+ in the relative interior of the n + distinct facets of L, every closed halfspace disjoint with L µ := µl + ( µf contains at most one point of the set {z,..., z n+}. For an α > that will be fixed in what follows, we choose L L n+ n+2 to be the pyramid L := conv({a} F with the apex ( a := f, α and the base F := (αl + ( αf { }. The cross-section L (R n {0} of L coincides with L {0}. The apex a lies above the points (f, 0 L {0} and (f, F. We now fix z i = (z i, 0 for i [n + ]. Fix z n+2 := (z,. Since α >, one has L { } relint(f. Thus, z n+2 relint(f and the chosen L is indeed a maximal lattice-free simplex with the integer points z,..., z n+2 lying in the relative interior of the n + 2 distinct facets of L. Set L µ := µl + ( µf and consider the facet F µ := µf + ( µ f = (αµl + ( αµ f { µ} (25 of L µ. We also consider the polytope conv(z,..., z n+2 with n + vertices z,..., z n+ above aff(f µ and one vertex z n+2 below aff(f µ. The cross-section