ON MAXIMAL S-FREE CONVEX SETS 1 ... - Optimization Online

ON MAXIMAL S-FREE CONVEX SETS ´ R.∗ AND SANTANU S. DEY† DIEGO A. MORAN Abstract. Let S ⊆ Zn satisfy the property that conv(S) ∩ Zn = S. Then a convex set K is called an S-free convex set if int(K) ∩ S = ∅. A maximal S-free convex set is an S-free convex set that is not properly contained in any S-free convex set. We show that maximal S-free convex sets are polyhedra. This result generalizes a result of Basu et al. [6] for the case where S is the set of integer points in a rational polyhedron and a result of Lov´ asz [18] and Basu et al. [5] for the case where S is the set of integer points in some affine subspace of Rn . Key words. Integer nonlinear programming, Cutting planes, Maximal lattice-free convex sets AMS subject classifications. 90C11, 90C57

1. Introduction. A convex set is called lattice-free if it contains no integer points in its interior. A maximal lattice-free convex set is a lattice-free convex set not strictly contained in any lattice-free convex set. Lov´asz [18] and Basu et al. [5] proved that maximal lattice-free convex sets are polyhedra. Given a convex set P ⊆ Rn , let S be the set of integer points contained in P, that is S = P ∩ Zn . A set K is called S-free convex set if int(K) ∩S = ∅. Hence the concept of S-free convex sets represents a generalization of the concept of lattice-free convex sets. Maximal S-free convex sets are defined analogously to maximal lattice-free convex sets. Dey and Wolsey [12] show that maximal S-free convex sets are polyhedra under restrictive conditions. Fukasawa and G¨ unl¨ uk [14] show that maximal S-free convex sets are polyhedra where P is a rational polyhedron in R2 . Basu et al. [6] prove that if P is any general rational polyhedron, then maximal S-free convex sets are polyhedra. A natural question then is to ask whether the polyhedrality of maximal S-free convex sets carries through to the case where P is an arbitrary convex set, that is the case where S ⊆ Zn is assumed only to satisfy the property that conv(S) ∩ Zn = S. In this paper we verify that maximal S-free convex sets are polyhedra, where S is the set of integer points in any convex set P ⊆ Rn . Since the proof in Basu et al. [6] explicitly uses that fact that conv(S) is a rational polyhedron when S is the set of integer points contained in a rational polyhedron, the result in this paper also provides an alternative proof to the result in [6]. Finally, we establish the polyhedrality of maximal S-free convex sets in some cases where S does not satisfy conv(S) ∩ Zn = S. We now briefly describe one motivation for the study of maximal S-free convex sets. A key algorithmic technique in solving mixed integer optimization problems is to sequentially obtain tighter approximations of the convex hull of integer feasible solutions. This is achieved by the addition of cutting planes, that is inequalities that separate fractional point from the convex hull of integer feasible points. See for example Marchand et al. [19] and Johnson et al. [17] for description of cutting plane methods. A connection between S-free convex sets and cutting planes for mixed integer linear programming was first discovered by Balas [3]. The main idea is the following. Consider the mixed integer set T := {(x, y) ∈ Zp × Rq : (x, y) ∈ C},

(1.1)

∗ H. Milton Stewart School of Industrial and Systems Engineering, Georgia Institute of Technology, Atlanta, GA, USA. ([email protected]). † H. Milton Stewart School of Industrial and Systems Engineering, Georgia Institute of Technology, Atlanta, GA, USA. ([email protected]).

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´ R. AND S. S. DEY D. A. MORAN

where C is a convex set. Now suppose that we are able to identify a set S ⊆ Zn such ˆ = B × Rq that S ⊇ Projx (T ). If B ⊆ Rp is an S-free convex set, then by letting B ˆ we can construct a valid relaxation of conv(T ) as conv(C \ int(B)). Often by a good ˆ is a much better approximation of the convex hull of T choice of B, conv(C \ int(B)) than C. A classical example of this procedure is that of split disjunctions; see Cook et al. [10]. Notice that if B 1 and B 2 are S-free convex sets such that B 1 ⊇ B 2 , then conv(C \ int(Bˆ1 )) ⊆ conv(C \ int(Bˆ2 )). This motivates the search for maximal S-free convex sets. Various families of cutting planes based on maximal S-free convex sets (for different S) have been proposed. See for example Andersen et al. [1], Andersen et al. [2], Basu et al. [6], Borozan and Cornu´ejols [9], Cornu´ejols and Margot [11], Dey and Wolsey [12], Fukasawa and G¨ unl¨ uk [14], Johnson [16], Zambelli [23]. Since we verify that maximal S-free convex sets are polyhedra, one possible way to compute ˆ is by computing the convex hull of the following disjunction conv(C \ int(B)) t [

(C ∩ {(x, y) : hai , xi ≥ bi }),

(1.2)

i=1

where B = {x ∈ Rp : hai , xi ≤ bi , i ∈ {1, ..., t}}. For many classes of convex sets C, the convex hull of the union of sets in (1.2) can be described ‘conveniently’; see for example the case of second order conic representable sets in Ben-Tal and Nemirovski [8]. The rest of the paper is organized as follows. In Section 2, we review some standard results from convex analysis and present a key result about maximal latticefree convex sets in affine subspaces due to Basu et al. [5]. In Section 3, we present the characterization of maximal S-free convex sets. In Section 4, we discuss some differences between the properties of maximal S-free convex sets in the general case to the case where S is the set of integer points contained in a rational polyhedron and point out some generalizations of the results presented in Section 3. 2. Preliminaries. Let W be an affine subspace of Rn . We use intW (A) to denote the interior of the set A ⊆ W with respect to the topology induced by Rn on W . Therefore rel.int(A) = intaff.hull(A) (A). We will call a set H a half-space (resp. hyperplane) of W if H is the intersection of W with a half-space (resp. hyperplane) of Rn and W * H. For u ∈ Rn and ε > 0, B(u, ε) is the open ball around u of radius ε > 0, that is B(u, ε) := {x ∈ Rn : ||x − u|| < ε}. We will frequently use the following basic result from convex analysis that we prove for completeness. Lemma 2.1. Let U ⊆ V ⊆ Rn be affine subspaces and let A ⊆ V be a convex set such that intV (A) ∩ U 6= ∅. Then intV (A) ∩ U = intU (A ∩ U ). Proof. The inclusion intV (A) ∩ U ⊆ intU (A ∩ U ) is straightforward. For the other inclusion, assume that y ∈ intU (A ∩ U ). Then there exists ε > 0 such that B(y, ε) ∩ U ⊆ A ∩ U . Since intV (A) ∩ U 6= ∅, there exists x ∈ intV (A) ∩ U . If x = y, then the proof is complete. Otherwise, since U is a affine subspace, we obtain that ε z = y + 2kx−yk (y − x) ∈ B(y, ε) ∩ U . It follows that y is a strict convex combination of z ∈ A and x ∈ intV (A), so y ∈ intV (A), and then y ∈ intV (A) ∩ U . Using Lemma 2.1, we restate a version of separation theorem for convex sets that we use later. Theorem 2.2 (Separation Theorem). Let W ⊆ Rn be an affine subspace. If A, B ⊆ W , A and B are convex sets, and rel.int(A) ∩ rel.int(B) = ∅, then there exists half-space H of W such that A ⊆ H and intW (H) ∩ B = ∅.

On Maximal S-free Convex Sets

3

Proof. By Theorem 4.14 in Hiriart-Urrut and Lemar´echal [15], there exists s ∈ Rn such that supy∈A hs, yi ≤ infy∈B hs, yi infy∈A hs, yi < supy∈B hs, yi.

(2.1) (2.2)

˜ = {x ∈ Rn : hs, xi ≤ supy∈A hs, yi} and H = H ˜ ∩ W . By (2.2), s is not Let H orthogonal to W . Therefore H is a halfspace of W . Finally, by Lemma 2.1 we obtain ˜ ∩ W = intW (H), which completes the proof. intRn (H) Listed below are some properties of interiors and relative interiors of convex sets that are also used frequently. See Hiriart-Urrut and Lemar´echal [15] and Rockafellar [20] for proofs. Proposition 2.3. Let W be an affine subspace and let A, B ⊆ W be convex sets. Then, 1. intW (A) ∩ intW (B) = intW (A ∩ B). 2. rel.int(A) ∩ rel.int(B) ⊆ rel.int(A ∩ B). 3. If rel.int(A) ∩ rel.int(B) 6= ∅, then rel.int(A) ∩ rel.int(B) = rel.int(A ∩ B). 4. A ⊆ rel.int(A), where C represent the closure of C. 5. rel.int(A) + rel.int(B) = rel.int(A + B). Definition 2.4 (S-free and Lattice-free Convex sets). Let W ⊆ Rn be an affine subspace of dimension dim(W ) ≥ 1, P ⊆ W be a convex set and S = P ∩ Zn . A set K is called S-free (resp. lattice-free) convex set of W if K ⊆ W , K is convex and intW (K) ∩ S = ∅ (resp. intW (K) ∩ Zn = ∅). A convex set K ⊆ W is called maximal S-free (resp. lattice-free) convex set of W if K is S-free (resp. lattice-free) convex set and there does not exists a set K 0 ⊆ W such that K 0 6= K, K 0 ⊇ K and K 0 is an S-free (resp. lattice-free) convex set of W . Basu et al. [5] proved that maximal lattice-free convex sets of affine subspaces are polyhedra. This is a crucial ingredient in the proof presented in this paper. We present this result next. Theorem 2.5 (Basu et al. [5]). Let W ⊆ Rn be an affine subspace containing an integral point and V be the affine hull of W ∩ Zn . A set K ⊆ W is a maximal lattice-free convex set of W if and only if one of the following holds: 1. K is a polyhedron in W whose dimension equals dim(W ), K ∩V is a maximal lattice-free convex set of V whose dimension equals dim(V ), and for every facet F of K, F ∩ V is a facet of K ∩ V , 2. K is an affine hyperplane of W such that K ∩ V is an irrational hyperplane of V , 3. K is a half-space of W that contains V on its boundary. The proof of Theorem 2.5 in Basu et al. [5] involves showing that if K is a latticefree convex set of W , then K must be contained in maximal lattice-free convex set of W . We will therefore use the following simplified version of Theorem 2.5. Theorem 2.6 (Basu et al. [5]). Let W ⊆ Rn be an affine subspace containing an integral point. If K ⊆ W is a lattice-free convex set of W , then it is contained in a maximal lattice-free convex set B of W , where B is a polyhedron. 3. Maximal S-free Convex Sets. As observed in Basu et al. [5], the existence of maximal S-free convex sets is a consequence of Zorn’s Lemma. We will verify the following results in this section. Theorem 3.1. Let W ⊆ Rn be an affine subspace of dimension dim(W ) ≥ 1, V the affine hull of W ∩ Zn , P ⊆ W a convex set and S = P ∩ Zn . If K ⊆ W is a maximal S-free convex set, then one of the following holds:

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1. dim(K) = dim(W ) and rel.int(K) ∩ rel.int(P ) 6= ∅: Then K is a polyhedron in W , K ∩ V is a maximal S-free convex set of V whose dimension equals dim(V ), and for every facet F of K, F ∩ V is a facet of K ∩ V , 2. dim(K) < dim(W ): Then K is an hyperplane of W , 3. dim(K) = dim(W ) and rel.int(K) ∩ rel.int(P ) = ∅: Then K is a halfspace of W, 4. S = ∅ and K = W . If S is the set of integer points contained in an rational polyhedron, then every facet of maximal S-free polyhedron contains a point of S in its relative interior. A slightly weaker result holds in the general case. Before we present this result, we require some additional notation. Let K be a polyhedron of W , that is let K = {x ∈ W : hai , xi ≤ bi ∀i ∈ {1, ..., N }}. Then we denote the ith facet of K as Fi (K). Also for all ε > 0, let Fiε (K) = {x ∈ W : haj , xi < bj ∀ j 6= i and bi < hai , xi < bi + ε}. Theorem 3.2. Let W ⊆ Rn be an affine subspace of dimension dim(W ) ≥ 1, P ⊆ W a convex set and S = P ∩ Zn . Let K be a S-free convex set such that dim(K) = dim(W ). If K is a maximal S-free convex set, then K is a polyhedron with N facets such that 1. (rel.int(Fi ) ∪ Fiε (K)) ∩ S 6= ∅ ∀ ε > 0 ∀ i ∈ {1, . . . , N } and 2. N ≤ 2dim(conv(S)) . The rest of the section is organized as follows. In Section 3.1, we prove that maximal S-free convex sets are polyhedra. In Section 3.2, we verify properties of facets of maximal S-free convex sets. In particular, if dim(K) = dim(W ) and rel.int(K) ∩ rel.int(P ) 6= ∅, then we show that K ∩ V is a maximal S-free convex set of V whose dimension equals dim(V ), and for every facet F of K, F ∩V is a facet of K ∩V (where V = aff.hull(W ∩Zn )). We also verify part (1.) of Theorem 3.2 in this section. Finally in Section 3.3, we obtain an upper bound on the number of facets of maximal S-free convex sets, completing the proof of Theorem 3.2. This upper bound is obtained using the upper bound result for maximal lattice-free sets (Doignon [13], Bell [7], Scarf [21]) but involves a little more work as facets of maximal S-free convex sets do not in general contain points of S in their relative interior. 3.1. Polyhedrality of Maximal S-free Convex Sets. To show that a maximal S-free convex set is a polyhedron, it is sufficient to show that every S-free convex set is contained in a polyhedral S-free convex set. This is verified next. Proposition 3.3. Let W ⊆ Rn be an affine subspace of dimension dim(W ) ≥ 1, P ⊆ W be a convex set and S = P ∩ Zn . Let K ⊆ W be an S-free convex set of W . Then one of the following holds: 1. dim(K) = dim(W ) and rel.int(K) ∩ rel.int(P ) 6= ∅: K is contained in an S-free convex set B ⊆ W such that B is a polyhedron, 2. dim(K) < dim(W ): Then K is contained in an S-free hyperplane of W , 3. rel.int(K) ∩ rel.int(P ) = ∅: Then K is contained in an S-free halfspace of W , 4. S = ∅ and W is an S-free convex set (K is contained in W ). Proof. Consider first the case where dim(W ) = 1. Then since K ⊆ W is a convex set and dim(K) ≤ 1, we conclude that K is an S-free polyhedron. Therefore the cases (1.), (2.), (3.) and (4.) are easily verifiable when dim(W ) = 1. The proof is now by induction on the dimension of W . If S ∩W = ∅, then W is an S-free convex set and this completes the proof. Hence, we can assume S ∩ W 6= ∅. Denote P = conv(S). Observe that since P is convex, S = P ∩ Zn .

On Maximal S-free Convex Sets

5

If dim(K) < dim(W ), then K is contained in a hyperplane H of W . Since intW (H) ∩ S = ∅, the proof is complete. If rel.int(K) ∩ rel.int(P ) = ∅, then by separation theorem there exists a half-space H ≤ of W such that K ⊆ H ≤ and P ∩intW (H ≤ ) = ∅. Thus, K is contained in an S-free half-space. Therefore we can assume dim(K) = dim(W ) and rel.int(K) ∩ rel.int(P ) 6= ∅. Since K is a S-free convex set, we obtain that intW (K ∩ P ) ∩ Zn = intW (K) ∩ intW (P ) ∩ Zn ⊆ intW (K) ∩ (P ∩ Zn ) = intW (K) ∩ S = ∅. So we conclude K ∩ P is a lattice-free convex set of W . Also since S ∩ W 6= ∅, we obtain that W ∩ Zn 6= ∅. Hence, by Theorem 2.6, there exists a maximal lattice-free convex set of W , Q ⊆ W such that K ∩ P ⊆ Q and Q is a polyhedron. Observe also that since S 6= ∅, Q ( W . We therefore obtain that

Q=

m \

Hi≤ ,

i=1

where each Hi≤ , i = 1, . . . , m is a half-space of W defining a facet of Q. For i ∈ I = {1, . . . , m}, denote Hi≥ = W \ (intW (Hi≤ )) and Hi= = Hi≤ ∩ Hi≥ . We partition the set I into four disjoints subsets. 1. I1 := {i ∈ I : rel.int(P ) ∩ rel.int(Hi≥ ) 6= ∅}. If i ∈ I1 , then rel.int(P ∩ Hi≥ ) = rel.int(P ) ∩ rel.int(Hi≥ ). Therefore rel.int(K) ∩ rel.int(P ∩ Hi≥ ) = rel.int(K) ∩ rel.int(P ) ∩ rel.int(Hi≥ ) = rel.int(K ∩ P ) ∩ rel.int(Hi≥ ) ⊆ Q ∩ rel.int(Hi≥ ) = ∅. The second equality holds because rel.int(K)∩rel.int(P ) 6= ∅. Since rel.int(K)∩ rel.int(P ∩ Hi≥ ) = ∅, we obtain that there exists Gi , a half-space of W , such that

intW (Gi ) ∩ P ∩

K ⊆ Gi ,

(3.1)

Hi≥

(3.2)

= ∅.

2. Consider I2 := {i ∈ I \ I1 : intW (K) ∩ Hi= = ∅, S ∩ Hi= 6= ∅}. In this case, we verify that K ⊆ Hi≤ . Since K ⊆ rel.int(K) = intW (K) and Hi≤ is a closed convex set, it is sufficient to verify that intW (K) ⊆ Hi≤ . Assume by contradiction, that there exists x ∈ intW (K) ∩ intW (Hi≥ ). Since ∅ 6= rel.int(K) ∩ rel.int(P ) ⊆ K ∩ P ⊆ Hi≤ , we obtain that rel.int(K) ∩ Hi≤ 6= ∅. However, note that since the affine hull of rel.int(K) and Hi≤ is W , rel.int(K) is an open set (wrt W ) and Hi≤ is a half space of W , we obtain that rel.int(K) ∩ rel.int(Hi≤ ) 6= ∅. (Choose y˜ ∈ rel.int(K) ∩ Hi≤ . If y˜ ∈ rel.int(H ≤ ), then we are done. Otherwise, since y˜ ∈ rel.int(K) and affine hull of rel.int(K) is W , it is possible to choose a neighborhood B(˜ y , ε) ∩ W of y˜ contained in rel.int(K). However, since y˜ ∈ / rel.int(Hi≤ ), we obtain that y˜ ∈ Hi= . Therefore, B(˜ y , ε) ∩ W ∩ rel.int(Hi≤ ) 6= ∅ and thus there exists y ∈ ≤ rel.int(K)∩rel.int(Hi )). Let y ∈ rel.int(K)∩rel.int(Hi≤ ). Hence every convex

´ R. AND S. S. DEY D. A. MORAN

6

Case I1= H S

Case I2

Case I3 Gi1 Gi2

G S

Q

S Q

Q

H= H=

K

K

K

Fig. 3.1. Illustration of case I1 , I2 , and I3 .

combination of x and y belongs to intW (K) since x, y ∈ intW (K). Moreover, since x ∈ intW (Hi≥ ) and y ∈ intW (Hi≤ ), there exists a convex combination of x and y that belongs to intW (K) ∩ Hi= . Therefore intW (K) ∩ Hi= 6= ∅, which contradicts the fact that i ∈ I2 . If i ∈ I2 , then define Gi = Hi≤ . Therefore from the above, we obtain Hi=

K ⊆ Gi , ∩ intW (Gi ) = ∅.

(3.3) (3.4)

3. Consider I3 := {i ∈ I \ I1 : intW (K) ∩ Hi= 6= ∅, S ∩ Hi= 6= ∅}. If i ∈ I3 , then by Lemma 2.1 we obtain that intHi= (K ∩ Hi= ) = intW (K) ∩ Hi= . Since K is an S-free convex set and intHi= (K∩Hi= ) = intW (K)∩Hi= we obtain that intHi= (K ∩ Hi= ) ∩ S = ∅. Hence, K ∩ Hi= is a (S ∩ Hi= )-free convex set of Hi= . Note that since Hi≤ ( W , we obtain that dim(Hi= ) = dim(W ) − 1. By the induction hypothesis, there exists a polyhedron Ti ⊆ Hi= such that K ∩ Hi= ⊆ Ti , Ti is a (S ∩ Hi= )-free convex set of Hi= . Note that since Tmi ≤ Fij , where (S ∩ Hi= ) 6= ∅, we obtain that Ti 6= Hi= . Therefore T = j=1 Fij≤ ⊆ Hi= , j = 1, . . . mi are the half-spaces of Hi= defining the facets of Ti . Denote Fij≥ = Hi= \ (intHi= (Fij≤ )). We have: rel.int(Fij≥ ) ∩ rel.int(K) = rel.int(Fij≥ ) ∩ Hi= ∩ intW (K) ⊆ Fij≥ ∩ intHi= (K ∩ Hi= ) ⊆ Fij≥ ∩ intHi= (Ti ) = ∅. Therefore there exists Gij , half-spaces of W , such that: Fij≥ Tm i

K ⊆ Gij ,

(3.5)

∩ intW (Gij ) = ∅.

(3.6)

In this case, define Gi = j=1 Gij . We verify a property that we require later. Note that since intW (K) ⊆ intW (Gij ) and intW (K)∩Hi= 6= ∅, we obtain that intW (Gij )∩Hi= 6= ∅. Therefore using Lemma 2.1 we obtain the equality intHi= (Gij ∩ Hi= ) = intW (Gij ) ∩

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On Maximal S-free Convex Sets

Hi= . Therefore, (3.6) is equivalent to Fij≥ ∩ intHi= (Gij ∩ Hi= ) = ∅.

(3.7)

4. Finally define I4 := {i ∈ I \ I1 : S ∩ Hi= = ∅}. Note that I = I1 ∪ I2 ∪ I3 ∪ I4 . Before the final step in the proof, we verify that if rel.int(P ) ∩ rel.int(Hi≥ ) = ∅, then P ∩ rel.int(Hi≥ ) = ∅.

(3.8)

As rel.int(P ) ∩ rel.int(Hi≥ ) = ∅, we obtain rel.int(P ) ⊆ Hi≤ . Since Hi≤ is a closed set, we obtain rel.int(P ) ⊆ Hi≤ . Finally, note that since P is a convex set, we obtain ≥ P ⊆ rel.int(P ) ⊆ Hi≤ (see Proposition 2.3). Thus P ∩ rel.int(H T i ) = ∅. To complete the proof we will verify that the set B = i∈I\I4 Gi ⊆ W is an Sfree convex set of W containing K. By (3.1), (3.3) and (3.5), we obtain that K ⊆ B. We need to prove that B is an S-free convex set to complete the proof. Assume by contradiction that there exists y ∈ S ∩ intW (B), that is y ∈ S and y ∈ intW (Gi ) ∀i ∈ I \ I4 . We have P =W " ∩P # [ ≥ = Hi ∪ intW (Q) ∩ P " ⊆ " =

i∈I

[

# (Hi≥

∩ P ) ∪ intW (Q)

i∈I

[

(Hi≥

∩ P) ∪

i∈I1

∪ intW (Q).

[ i∈I2

(Hi=

∩ P) ∪

[ i∈I3

(Hi=

∩ P) ∪

[

# (Hi=

∩ P)

i∈I4

(3.9)

The last equality is a consequence of (3.8), since for i ∈ I2 ∪ I3 ∪ I4 we have that ∩ P ) = (rel.int(Hi≥ ) ∩ P ) ∪ (Hi= ∩ P ) = (Hi= ∩ P ). Since y ∈ S = P ∩ Zn and S ∩ Hi= = ∅ for i ∈ I4 , we obtain that y ∈ / Hi= ∩ P for i ∈ I4 . Moreover Q is lattice-free. Therefore, using (3.9) there are three cases: 1. For i ∈ I1 , if y ∈ Hi≥ ∩ P and y ∈ intW (Gi ), then we obtain a contradiction to (3.2). 2. For i ∈ I2 , if y ∈ Hi= and y ∈ intW (Gi ), then we obtain a contradiction to (3.4). Tm i Tm i intW (Gij )∩Hi= = j=1 intHi= (Gij ∩ 3. For i ∈ I3 , if y ∈ Hi= , then since y ∈ j=1 T T mi mi (Hi= \ Fij ≥ ) = j=1 intHi= (Fij ≤ ) (the last inclusion is a conseHi= ) ⊆ j=1 quence of (3.7)), we obtain that y ∈ intHi= (Ti ) which is in contradiction with the fact that Ti is an S ∩ Hi= -free convex set of Hi= . Therefore B is S-free polyhedron containing K. Cases (2.), (3.), (4.) of Theorem 3.1 follow from maximality of K and cases (2.), (3.), (4.) of Proposition 3.3 respectively. Case (1.) of Proposition 3.3 shows that maximal S-free convex sets are polyhedra when dim(K) = dim(W ) and rel.int(K) ∩ rel.int(P ) 6= ∅. To complete the proof of Theorem 3.1, we need to show that K ∩ V is a maximal S-free convex set of V whose dimension equals dim(V ), and for every facet F of K, F ∩ V is a facet of K ∩ V (where V = aff.hull(W ∩ Zn )). This is verified in the next section. (Hi≥

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3.2. Structure of Facets of Maximal S-free Convex Sets. For convenience we repeat the definition of some notation. Let K be the polyhedron of W , that is let K = {x ∈ W : hai , xi ≤ bi ∀i ∈ {1, ..., N }}. Then we denote the ith facet of K as Fi (K). Also for all ε > 0, let Fiε (K) = {x ∈ W : haj , xi < bj ∀j 6= i and bi < hai , xi < bi + ε}. Before completing the proof of Theorem 3.1, we prove part (1.) of Theorem 3.2. Proposition 3.4. Let K = {x ∈ W : hai , xi ≤ bi , ∀ i ∈ {1, . . . , N }} ⊆ Rn be a maximal S-free convex set, such that dim(K) = dim(W ). Then (rel.int(Fi (K)) ∪ Fiε (K)) ∩ S 6= ∅

∀ ε > 0 ∀ i ∈ {1, . . . , N }.

Proof. For ε > 0 and i ∈ {1, ..., N } consider the set Kiε = {x ∈ W : haj , xi ≤ bj ∀j 6= i and hai , xi ≤ bi + ε}. Since K ( Kiε we obtain that Kiε is not an S-free convex set, that is, intW (Kiε )∩S 6= ∅. Hence, the set intW (Kiε ) \ intW (K) must contain a point of S. Observe that: intW (Kiε ) \ intW (K) = {x ∈ W : haj , xi < bj ∀ j 6= i and bi ≤ hai , xi < bi + ε} = Fiε (K) ∪ rel.int(Fi (K)). Therefore ∅ 6= intW (Kiε ) \ intW (K) ∩ S = (Fiε (K) ∪ rel.int(Fi (K))) ∩ S. Note that Proposition 3.4 highlights an important difference between maximal S-free convex sets for general S and for the case where S is the set of integer points contained in a rational polyhedron. In the case of general S, there is no guarantee that every facet of a maximal S-free convex set contains points belonging to S in its relative interior. This is illustrated in the next √ example. 2 Example 3.5. Let S = {x ∈ Z : 2x1 + x2 ≤ 0, x1 ≥ 1}. Then the set √ K = {x ∈ R2 : 2x1√+ x2 ≥ 0} is a maximal S-free convex set, but the facet of K defined by {x ∈ R2 : 2x1 + x2 = 0} contains no point belonging to S. Now we complete the proof of Theorem 3.1. The proof of the first part of the next proposition is similar to the proof of a similar property for maximal lattice-free convex sets, appearing in Basu et al. [5]. Proposition 3.6. Let K = {x ∈ W : hai , xi ≤ bi , ∀ i ∈ {1, . . . , N }} ⊆ W be a maximal S-free convex set such that dim(K) = dim(W ) and intW (K) ∩ rel.int(P) 6= ∅. Let V be the affine hull of W ∩ Zn . Then K ∩ V is a maximal S-free convex set of V such that dim(K ∩ V ) = dim(V ) and F ⊆ V is a facet of K ∩ V if and only if F = Fi (K) ∩ V for some i ∈ {1, . . . , N }. Proof. Since intW (K)∩rel.int(P) 6= ∅, we obtain that intW (K)∩V 6= ∅. Therefore Lemma 2.1 implies that intW (K) ∩ V = intV (K ∩ V ). We first verify that K ∩ V is a maximal S-free convex set of V . Since K is an S-free convex set and intW (K) ∩ S = intW (K) ∩ S ∩ V = intV (K ∩ V ) ∩ S, we obtain that K ∩ V is an S-free convex set of V . If K ∩ V is not maximal, then there exist B ⊆ V , an S-free convex set, such that B ) K ∩ V . Consider K 0 = conv(K ∪ B). Then K 0 ∩ V = B. We have intW (K 0 ) ∩ S = intW (K 0 ) ∩ S ∩ V = intV (B) ∩ S = ∅. Therefore K 0 is an S-free convex set of W . Since K 0 ⊇ K and B ) K ∩ V , we obtain K 0 ) K which is in contradiction with the fact that K is maximal S-free convex set. Therefore, K ∩ V is a maximal S-free convex set of V .

On Maximal S-free Convex Sets

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Since intV (K ∩ V ) 6= ∅, we obtain dim(K ∩ V ) = dim(V ). Finally, we verify that F ⊆ V is a facet of K ∩ V if and only if F = Fi (K) ∩ V for some i ∈ {1, . . . , N }. (⇒) If F is a facet of K ∩ V = {x ∈ V : hai , xi ≤ bi , ∀ i ∈ {1, . . . , N }}, then F = Fi (K) ∩ V for some i ∈ {1, . . . , N }. (⇐) Given ε > 0, by the use of Proposition 3.4 we obtain that for all i = 1, . . . , N , ∅ 6= (relint(Fi (K)) ∪ Fiε (K)) ∩ S = (rel.int(Fi (K)) ∪ Fiε (K)) ∩ V ∩ S ⊆ intW ({x ∈ W : hak , xi ≤ bk , ∀ k ∈ {1, . . . , N } \ {i}}) ∩ V ∩ S = intV ({x ∈ V : hak , xi ≤ bk , ∀ k ∈ {1, . . . , N } \ {i}}) ∩ S. Note that the last equality is obtained as a consequence of Lemma 2.1, {x ∈ W : hak , xi ≤ bk , ∀ k ∈ {1, . . . , N }\{i}} ⊇ K and intW (K)∩V 6= ∅. As K ∩ V is an S-free convex set of V and dim(K ∩ V ) = dim(V ), we conclude that hai , xi ≤ bi must define a facet of K ∩ V , that is Fi (K) ∩ V is a facet of K ∩ V for all i ∈ {1, ..., N }. 3.3. Upper Bound on the Number of Facets of Maximal S-free Convex Sets. When S is the set of integer points contained in a general convex set, maximal S-free convex sets do not need to have points of S in the relative interior of each facet. Therefore, proving upper bound on the number of facets is slightly more involved than the case of maximal lattice-free convex sets. We first begin with a corollary of Theorem 3.1 and Proposition 3.4. Alternatively, this also follows from results in Basu et al. [6]. Corollary 3.7. Let K 0 ⊆ W , P 0 a rational polytope, S 0 = P 0 ∩ Zn , dim(K 0 ) = dim(W ), then K 0 is a maximal S 0 -free convex set of W if and only if K 0 is a S-free polyhedron with a point of S 0 in the relative interior of each of its facets. The next lemma is a standard result. See Schrijver [22] for a proof. Lemma 3.8. If L ⊆ Rn is a rational affine subspace (i.e. aff.hull(L ∩ Zn ) = L), then there exists an affine transformation T : Rdim(L) 7−→ L such that T is invertible and T (Zdim(L) ) = L ∩ Zn . Using Corollary 3.7 and Lemma 3.8 and a proof similar to that in Doignon [13], Bell [7], Scarf [21] we obtain the following result. Lemma 3.9. Let K 0 ⊆ W , P 0 a polytope, S 0 = P 0 ∩ Zn , dim(K 0 ) = dim(W ). If K 0 is a maximal S 0 -free convex set of W , then K 0 is a polyhedron with at most 0 2dim(P ) facets. We now have all the tools needed to verify the upper bound on the number of facets of maximal S-free convex sets. Proposition 3.10. If K = {x ∈ W : hai , xi ≤ bi , ∀ i ∈ {1, . . . , N }} ⊆ W is a maximal S-free convex set such that dim(K) = dim(W ), then N ≤ 2dim(conv(S)) . Proof. Let ε > 0. Consider the sets I1 and I2 defined as: 1. ∀ i ∈ I1 , S ∩ rel.int(Fi (K)) 6= ∅. 2. ∀ i ∈ I2 , S ∩ rel.int(Fi (K)) = ∅ and, Fiε (K) ∩ S 6= ∅. By Proposition 3.4, these sets are well defined and I1 ∪ I2 = {1, ..., N }. If I1 6= ∅, then for each i ∈ I1 take a point xi ∈ rel.int(Fi (K)) ∩ S and if I2 6= ∅, then for each i ∈ I2 , take a point xi ∈ Fiε (K) ∩ S. Define P 0 = conv({x1 , . . . , xN }) and S 0 = P 0 ∩ Zn . Since P 0 ⊆ P, we have that 0 S ⊆ S. This implies that intW (K) ∩ S 0 ⊆ intW (K) ∩ S = ∅, so K is an S 0 -free convex set. Note that P 0 is a rational polytope, so by Corollary 3.7 and Lemma 3.9 every 0 maximal S 0 -free convex set K 0 is a polyhedron that has at most 2dim(P ) facets and 0 contains an integer point of S in the relative interior of each of its facets. Observe that dim(P 0 ) ≤ dim(conv(S)).

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If I2 = ∅, then K is a maximal S 0 -free convex set. Therefore N ≤ 2dim(P ) ≤ dim(conv(S)) 2 . If I2 6= ∅, then consider i ∈ I2 . We will construct a polyhedron K1 with N facets such that: 1. K1 is an S 0 -free convex set. 2. Fjε (K) ⊆ Fjε (K1 ), ∀ j ∈ I2 \ {i}. 3. K1 has N facets. 4. K1 has at least |I1 | + 1 facets with a point of S 0 in the relative interior. We construct K1 from K by changing the right hand side of the ith inequality. Since P 0 is bounded and S 0 ⊆ Zn , we obtain that |S 0 | is finite. So S 0 ∩ Fiε (K) is also finite. Moreover, since xi ∈ S 0 ∩ Fiε (K), we have S 0 ∩ Fiε (K) 6= ∅. Hence, there exists zi ∈ S 0 ∩ Fiε (K) such that min{hai , xi : x ∈ S 0 ∩ Fiε (K)} = hai , zi i. Denote b0i = hai , zi i. Consider the polyhedron K1 = {x ∈ W : hai , xi ≤ bi , ∀ i ∈ I1 ∪ I2 \ {i}, hai , xi ≤ b0i } We verify that K1 satisfies (1.), (2.), (3.), and (4.): 1. Since K is S 0 -free, we only need to show that int(K1 )\int(K) does not contain points of S 0 . Since b0i ≤ bi + ε we have: int(K1 ) \ int(K) = {x ∈ W : hai , xi < bi , ∀ i ∈ I1 ∪ I2 \ {i}, bi ≤ hai , xi < b0i , } ⊆ {x ∈ W : haj , xi < bj ∀ j 6= i and bi ≤ hai , xi < bi + ε} = Fiε (K) ∪ rel.int(Fi (K)) Since i ∈ I2 , S ∩ rel.int(Fi (K)) = ∅ so (int(K1 ) \ int(K)) ∩ S 0 ⊆ Fiε (K) ∩ S 0 . On the other hand, since ∀ x ∈ S 0 ∩ Fiε (K), ha, xi ≥ b0i , we conclude (int(K1 ) \ int(K)) ∩ S 0 = ∅. Therefore, K1 is S 0 -free convex set. 2. Using the fact that bi ≤ b0i and j ∈ I2 \ {i} we have, Fjε (K) = {x ∈ W : hak , xi < bk ∀k 6= j and bj < haj , xi < bj + ε} ⊆ {x ∈ W : hak , xi < bk ∀k = 6 i, j, hai , xi < b0i and bj < haj , xi < bj + ε} = Fjε (K1 ). 3. By definition, the point zi satisfies hai , zi i = b0i and ∀ j 6= i, haj , zi i < bj . Therefore the inequality hai , xi ≤ b0i is a facet defining inequality for K1 . Observe also that the inequalities haj , xi ≤ bj for j 6= i are facet-defining inequality for K1 . Therefore K1 has N facets. 4. As verified earlier, zi belongs to the relative interior of the facet of K1 defined by {x ∈ K1 : hai , xi = b0i }. Also for j ∈ I1 , since bi ≤ b0i , we have: rel.int(Fj (K)) = {x ∈ W : hak , xi < bk ∀k 6= j and haj , xi = bj } ⊆ {x ∈ W : hak , xi < bk ∀k = 6 i, j, hai , xi < b0i and haj , xi = bj }. Therefore, since xj ∈ rel.int(Fj ) ∩ S 0 , the facet of K1 defined by {x ∈ K1 : haj , xi = bj } has a point of S 0 in its relative interior. In conclusion, K1 has at least |I1 | + 1 facets with a point of S 0 in its relative interior.

On Maximal S-free Convex Sets

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So we have a procedure to construct, from K, an S 0 -free polyhedron K1 with N facets that has at least one more facet that contains a point of S 0 in its relative interior than K has. If K1 is not a maximal S 0 -free convex set, then we can choose a facet of K1 without a point of S 0 in its relative interior and by properties (1.) and (2.), repeat the above procedure. We can repeat this a finite number of times, obtaining a sequence K1 , K2 , . . . , KT of polyhedra with the same number of facets as K and such that KT does not have any facet without a point of S 0 in its relative interior. So KT 0 is a maximal S 0 -free convex set. Thus, N ≤ 2dim(P ) ≤ 2dim(conv(S)) . 4. Notes. 4.1. Differences Between General Maximal S-free Convex Sets and the Case Where S is the Set of Integer Points Contained in a Rational Polyhedron. As discussed in Section 3.2, in the case of general S, maximal S-free convex sets do not necessarily have integer points in the relative interior of each facet. There is another difference between maximal S-free convex sets in the general case and the case when S is the set of integer points contained in a rational polyhedron. The result of Lov´asz [18] states that maximal lattice-free convex sets are in the form of a polytope plus a cylinder, that is, if r is a recession direction of maximal lattice-free convex set, then so is −r. Similarly, Basu et al. [6] prove the following result. Proposition 4.1 (Basu et al. [6]). If S is a nonempty set of integer points contained in a rational polyhedron P, K is a maximal S-free convex set such that K ∩ conv(S) has nonempty interior, and r belongs to the recession cone of P and K, then −r belongs to the recession cone of K. Proposition 4.1 is an important property and is useful in verifying many results. See for example Basu et al. [4]. The following example shows that this result is not true in general. √ √ Example 4.2. Let P = {u ∈ R3 : u3 ≥ u2 − 2u1 , u3 ≥ 2u1 − u2 }. √ Then K = {v ∈ R3 : v3 ≤ 1, v2 ≥ 0} is a maximal S-free convex set. Also r = (1, 2, 0) ∈ rec(K ∩ P). However, −r ∈ / rec(K). We present some sufficient conditions on P for a property like the one presented in Proposition 4.1 to hold. For simplicity we restrict the discussion to the case where W = Rn . Given a closed convex set P ⊆ Zn , and a non-zero vector r ∈ rec(P), let Z(P, r) = {x ∈ Zn : ∃λ ≥ 0, s.t. x + λr ∈ P}. Definition 4.3 (Convex Set with Dirichlet Property). We say a closed convex set P ⊆ Rn satisfies Dirichlet Property if P satisfies the following conditions: For all r ∈ rec(P) \ {0} and for all x ∈ Z(P, r), given any ² > 0 and γ ≥ 0 there exists x ¯ ∈ P ∩ Zn such that the distance between the integer point x ¯ and the half line {x + λr : λ ≥ γ} is less than ². We first show that the Dirichlet property indeed implies a property similar to that presented in Proposition 4.1. Proposition 4.4. Let P ⊆ Rn be a closed convex set satisfying Dirichlet Property, let S = P ∩ Zn , and let K be a full-dimensional maximal S-free convex set. If r belongs to the recession cone of P and K, then −r belongs to the recession cone of K. Proof. By part (5.) of Proposition 2.3, we obtain int(K + {λr | λ ∈ R}) = int(K) + rel.int({λr | λ ∈ R}) = int(K) + {λr | λ ∈ R}. To prove the result of this proposition we need to verify that int(K + {λr | λ ∈ R}) ∩ S = ∅. Assume by contradiction that x ∈ int(K + {λr | λ ∈ R}) ∩ S. By the ¯ ∈ R+ such that x + λr ¯ ∈ int(K). Therefore B(0, δ) + previous claim, there exists λ

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¯ is contained in the interior of K for some suitable small and positive {x + λr : λ ≥ λ} δ. Since r is a recession direction of P, we obtain that x ∈ Z(P, r). As P satisfies Dirichlet property, there exists an integer point z belonging to P in the interior of the ¯ However, this set lies in the interior of K. Therefore set B(0, δ) + {x + λr : λ ≥ λ}. z ∈ P and lies in interior of K, a contradiction. Our motivation for the name ‘Dirichlet property’ is due to the fact that we often use the following consequence of Dirichlet Theorem to prove the ‘Dirichlet property’. Lemma 4.5 (Basu et al. [5]). If x ∈ Zn and r ∈ Rn , then for all ² > 0 and γ ≥ 0, there exists a point of Zn at a distance less than ² from the half line {x+λr : λ ≥ γ}. 4.1.1. Some Convex Sets Satisfying Dirichlet Property. Next we give examples of convex sets with Dirichlet Property. Proposition 4.6. Every full dimensional closed convex set in R2 is a convex set satisfying Dirichlet Property. Proof. Let P be a full-dimensional, closed convex set in R2 . Consider any ² > 0 and γ ≥ 0. Let r be a non zero vector in the recession cone of P and x ∈ Z(P, r). Denote η = min{λ ∈ R+ : x + λr ∈ P}. 1. Since P is a closed set, we obtain that if r is a rational vector, then there exists x ¯ ∈ P ∩Zn such that the distance between x ¯ and the half line {x+λr : λ ≥ γ} is 0. 2. Suppose now that r is not a rational vector (i.e. λr ∈ / Z2 ∀λ > 0). Assume by n contradiction that there exists no x ¯ ∈ P ∩ Z such that the distance between x ¯ and the half line {x + λr : λ ≥ γ} is less than ². Since P is full dimensional and L := {x + λr : λ ≥ η} ⊆ P, the set M := {y ∈ P : distance(y, L) ≤ ²} is a full-dimensional convex subset of P. Note now that M is lattice-free and r belongs to the recession cone of M . Therefore M is contained in a maximal lattice-free convex set. Since the only class of unbounded fulldimensional maximal lattice-free convex set in R2 is the split set (a set of form {y ∈ R2 : b ≤ ha, yi ≤ b + 1} where a ∈ Z2 and b ∈ Z), we obtain that r is a rational vector which is a contradiction. We next verify that all the examples which do not satisfy the property similar to that presented in Proposition 4.1, also satisfy the condition that conv(S) is not closed. Note that Proposition 4.6 can be used to construct examples where conv(S) is not closed and yet the property similar to that presented in Proposition 4.1 is satisfied. Proposition 4.7. If S ⊆ Zn and conv(S) is full-dimensional and closed, then conv(S) satisfies the Dirichlet property. Proof. Let P := conv(S). If n = 1, then the result is straightforward to verify. The proof is now by induction on n. Consider any ² > 0 and γ ≥ 0. Let r be a non zero vector in the recession cone of P and x ∈ Z(P, r). Denote η = min{λ ∈ R+ : x + λr ∈ P}. 1. Since P is a closed set, we obtain that if r is a rational vector, then there exists x ¯ ∈ P ∩Zn such that the distance between x ¯ and the half line {x+λr : λ ≥ γ} is 0. 2. Suppose now that r is not a rational vector. There are two cases: (a) Suppose ∃ δ > 0 and ζ > 0 such that B(x + ζr, δ) ⊆ P. Let ²0 = min{δ, ²}. Now as a consequence of the Dirichlet Theorem there exists an integer point x ¯ at a distance less that ²0 from the half line {x + λr : λ ≥ max{ζ, γ}}. Then x ¯ belongs to P, completing the proof.

On Maximal S-free Convex Sets

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(b) The half-line {x + λr : λ ≥ η} lies on the boundary of conv(S). Let F be a proper face of conv(S) containing this half-line. (Recall that a face F of a closed convex set P is a closed convex subset such that if x ∈ F and x can be written as convex combination of x1 , x2 ∈ P, then x1 , x2 ∈ F ). Then we claim: i. F = conv(S ∩ F ): Since S ∩ F ⊆ F and F is a convex set, we obtain that conv(S ∩ F ) ⊆ F . For the other inclusion, observe that if x ∈ F , then x can be written as a convex combination of a finite number of vectors in S (since F ⊆ conv(S)). However by definition of a face, each of these vectors belong to F . Thus x ∈ conv(S ∩ F ), or equivalently F ⊆ conv(S ∩ F ). ii. aff.hull(F ) is a rational affine half-space containing an integer point: By the above claim S ∩ F 6= ∅ and aff.hull(F ) can be generated by vectors in S, which are integral vectors. Thus, aff.hull(F ) is a rational affine half-space. Now, we apply the invertible affine transformation A : aff.hull(F ) 7−→ Rdim(aff.hull(F )) to F , where A is the function described in Lemma 3.8. Observe now that A(F ) is a closed convex set, A(F ) = conv(A(S ∩ F )), Ar is a recession direction of A(F ), Ax ∈ Z(A(F ), Ar) and 1 ≤ dim(aff.hull(F )) < n. Therefore by the induction hypothesis, for all δ > 0, there exists an integer point x ˆ belonging to A(conv(S ∩ F )) that lies within a distance of δ from the half-line {u : u = Ax + λAr, λ ≥ max{η, γ}}. However, this implies that there exists a point x ¯ ∈ S such that x ¯ lies at a distance less than ² to the half line {u : u = x + λr, λ ≥ max{η, γ}}. We remark here that by the use of Lemma 3.8, the result of Proposition 4.7 can be extended to the case where aff.hull(conv(S)) is not full-dimensional since aff.hull(conv(S)) is a rational affine subspace. Next we show an interesting class of non-polyhedral convex sets that satisfy Dirichlet Property. Definition 4.8. A set P is called a strictly convex set if it satisfies the following property: If u ∈ P such that u is not an extreme point, then u belongs to the relative interior of P. Proposition 4.9. Every full dimensional, closed, strictly convex set satisfies Dirichlet Property. Proof. Let P be a full dimensional, closed, strictly convex set. Consider any ² > 0 and γ ≥ 0. Let r be a non zero vector in the recession cone of P and x ∈ Z(P, r). Denote η = min{λ ∈ R+ : x + λr ∈ P}. 1. Since P is a closed set, we obtain that if r is a rational vector, then there exists x ¯ ∈ P ∩Zn such that the distance between x ¯ and the half line {x+λr : λ ≥ γ} is 0. 2. Suppose now that r is not a rational vector. Let γ¯ := max{γ, η + 1}. The point x+ γ¯ r belongs to P and is not an extreme point of P. Since P is strictly convex, the point x + γ¯ r lies in the interior of P. Therefore, there exists a ball of radius δ > 0 around the point x + γ¯ r that lies in P. Set ²0 := min{², 2δ }. Now as a consequence of the Dirichlet Theorem there exists an integer point x ¯ at a distance less that ²0 from the half line {x + λr : λ ≥ γ¯ }. Since x ¯ lies at a distance less than δ from the half line {x + λr : λ ≥ γ¯ } and the

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set B(0, δ) + {x + λr : λ ≥ γ¯ } belongs to P, we obtain that x ¯ belongs to P. Moreover, the point x ¯ lies at a distance less that ² from the half line {x + λr : λ ≥ γ}. 4.2. Other Extensions. 1. Instead of defining S ⊆ Zn , we can define S as a subset of points belonging to a general lattice. All the results in Section 3 carry through. 2. The condition that S ⊆ Zn such that conv(S) ∩ Zn = S is not necessary for the polyhedrality of maximal S-free convex sets. For example, the following corollary of Theorem 3.1 can be proven. Corollary 4.10. Let Si ⊆ Zn ∩ W , i = 1, . . . , N such that Si = conv(Si ) ∩ Zn . SN Denote S = i=1 Si . Let K ⊆ W be an S-free convex set of W . Then there exists an S-free polyhedron B ⊆ W such that K ⊆ B. Proof. For all i = 1, . . . , N , intW (K) ∩ Si ⊆ intW (K) ∩ S = ∅. Therefore K is an Si -free convex set of W . By Theorem 3.1,Tthere exists an Si -free polyhedron Bi ⊆ W N such that K ⊆ Bi . The polyhedron B = i=1 Bi satisfies h i KS⊆ B and intW (B) ∩ S = TN SN SN TN N i=1 intW (Bi ) ∩ j=1 Sj = j=1 Sj ∩ i=1 intW (Bi ) ⊆ j=1 [Sj ∩ intW (Bj )] = ∅. Thus B is an S-free polyhedron containing K. An analysis similar to that in Section 3.1 and 3.2 can be carried out for this more general class of S. The upper bound on the number of facets of maximal Sfree convex sets presented in Section 3.3 is not valid for this class of more general S. For example, if S = {(0, 0), (1, 0), (−1, 1), (2, 1), (0, 2), (1, 2)}, then the hexagon with vertices {(0.5, −0.25), (2, 0.5), (2, 1.5), (0.5, 2.25), (−1, 1.5), (−1, 0.5)} is a maximal Sfree convex set. REFERENCES [1] K. Andersen, Q. Louveaux, R. Weismantel, and L. Wolsey, Cutting planes from two rows of a simplex tableau, Proceedings 12th Conference on Integer Programming and Combinatorial Optimization, LNCS 4513 (M. Fischetti and D. P. Williamson, eds.), Springer-Verlag, 2007, pp. 1–15. [2] K. Andersen, C. Wagner, and R. Weismantel, On an analysis of the strength of mixed integer cutting planes from multiple simplex tableau rows, SIAM Journal on Optimization 20 (2009), 967–982. [3] E. Balas, Intersection cuts - a new type of cutting planes for integer programming, Operations Research 19 (1971), 19–39. [4] A. Basu, M. Campelo, M. Conforti, G. Cornu´ ejols, and G. Zambelli, On lifting integer variables in minimal inequalities, To appear in IPCO 2010 (2010). [5] A. Basu, M. Conforti, G. Cornu´ ejols, and G. Zambelli, Maximal lattice-free convex sets in linear subspaces, Manuscript, 2009. [6] , Minimal inequalities for an infinite relaxation of integer programs, SIAM Journal on Discrete Mathematics 24 (2010), 158–168. [7] D. E. Bell, A theorem concerning the integer lattice, Studies in Applied Mathematics 56 (1977), 187–188. [8] A. Ben-Tal and A. Nemirovski, Lectures on modern convex optimization: analysis, algorithms, and engineering applications, Society for Industrial and Applied Mathematics, Philadelphia, PA, 2001. [9] V. Borozan and G. Cornu´ ejols, Minimal valid inequalities for integer constraints, Mathematics of Operations Research 34 (2009), 538–546. [10] W.J. Cook, R. Kannan, and A. Schrijver, Chv´ atal closures for mixed integer programming problems, Mathematical Programming 58 (1990), 155–174. [11] G. Cornu´ ejols and F. Margot, On the facets of mixed integer programs with two integer variables and two constraints, Mathematical Programming 120 (2009), 429–456.

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[12] S. S. Dey and L. A. Wolsey, Constrained infinite group relaxations of MIPs, Tech. Report CORE DP 33, Universit´ e catholique de Louvain, Louvain-la-Neuve, Belgium, 2009. [13] J. P. Doignon, Convexity in crystallographic lattices, Journal of Geometry 3 (1973), 71–85. [14] R. Fukasawa and O. G¨ unl¨ uk, Strengthening lattice-free cuts using non-negativity, http://www.optimization-online.org/DB HTML/2009/05/2296.html, 2009. [15] J.-P. Hiriart-Urrut and C. Lemar´ echal, Fundamentals of convex analysis, Springer-Verlag, Berlin, 2000. [16] E. L. Johnson, Characterization of facets for multiple right-hand side choice linear programs, Mathematical Programming Study 14 (1981), 112–142. [17] E. L. Johnson, G. L. Nemhauser, and M. W. P. Savelsbergh, Progress in linear programmingbased algorithms for integer programming: an exposition, INFORMS Journal of Computing 12 (2000), 2–23. [18] L. Lov´ asz, Geometry of numbers and integer programming, Mathematical Programming: Recent Developments and Applications (M. Iri and K. Tanabe, eds.), Kluwer, Dordrecht, 1989, pp. 177–210. [19] H. Marchand, A. Martin, R. Weismantel, and L. A. Wolsey, Cutting planes in integer and mixed integer programming, Discrete Applied Mathematics 123 (2002), 397–446. [20] G. T. Rockafeller, Convex analysis, Princeton University Press, New Jersey, NJ, 1970. [21] H. E. Scarf, An observation on the structure of production sets with indivisibilities, Proceedings of the National Academy of Sciences USA 74 (1977), 3637–3641. [22] A. Schrijver, Theory of linear and integer programming, John Wiley & Sons, Inc., New York, NY, 1986. [23] G. Zambelli, On degenerate multi-row Gomory cuts, Operations Research Letters 37 (2009), 21–22.