Intersection Cuts with Infinite Split Rank

# Intersection Cuts with Infinite Split Rank

Amitabh Basu,   Gérard Cornuéjols
François Margot
April 2010
###### Abstract

We consider mixed integer linear programs where free integer variables are expressed in terms of nonnegative continuous variables. When this model only has two integer variables, Dey and Louveaux characterized the intersection cuts that have infinite split rank. We show that, for any number of integer variables, the split rank of an intersection cut generated from a bounded convex set is finite if and only if the integer points on the boundary of satisfy a certain “2-hyperplane property”. The Dey-Louveaux characterization is a consequence of this more general result.

11footnotetext: Tepper School of Business, Carnegie Mellon University, Pittsburgh, PA 15213.22footnotetext: Supported by a Mellon Fellowship.33footnotetext: LIF, Faculté des Sciences de Luminy, Université de Marseille, France.44footnotetext: Supported by NSF grant CMMI0653419, ONR grant N00014-09-1-0033 and ANR grant ANR06-BLAN-0375.55footnotetext: Supported by ONR grant N00014-09-1-0033.

## 1 Introduction.

In this paper, we consider mixed integer linear programs with equality constraints expressing free integer variables in terms of nonnegative continuous variables.

 x=f+k∑j=1rjsjx∈Zms∈Rk+. (1)

The convex hull of the solutions to (1) is a corner polyhedron (Gomory [11], Gomory and Johnson [12]). In the remainder we assume , and . Hence is not a solution of (1). To avoid discussing trivial cases, we assume that , and this implies that . The facets of are the nonnegativity constraints and intersection cuts (Balas [2]), namely inequalities

 k∑j=1ψ(rj)sj≥1 (2)

obtained from lattice-free convex sets containing in their interior, where denotes the gauge of (Borozan and Cornuéjols [4]). By lattice-free convex set, we mean a convex set with no point of in its interior. By gauge of a convex set containing the origin in its interior, we mean the function . Intersection cut (2) for a given convex set is called -cut for short.

When , Andersen, Louveaux, Weismantel and Wolsey [1] showed that the only intersection cuts needed arise from splits (Cook et al. [6]), triangles and quadrilaterals in the plane and a complete characterization of the facet-defining inequalities was obtained in Cornuéjols and Margot [7]. More generally, Borozan and Cornuéjols [4] showed that the only intersection cuts needed in (1) arise from full-dimensional maximal lattice-free convex sets . Lovász [13] showed that these sets are polyhedra with at most facets and that they are cylinders, i.e., their recession cone is a linear space. These cylinders can be written in the form where is a polytope of dimension at least one and is a linear space such that . We say that is a cylinder over . A split in is a maximal lattice-free cylinder over a line segment.

Let be a polytope containing in its interior. For , let the boundary point for the ray be the intersection of the half-line with the boundary of . We say that has rays going into its corners if each vertex of is the boundary point for at least one of the rays , .

The notions of split closure and split rank were introduced by Cook et al. [6] (precise definitions are in Section 2). They gave an example of (1) with and that has an infinite split rank. Specifically, there is a facet-defining inequality for that cannot be deduced from a finite recursive application of the split closure operation. This inequality is an intersection cut generated from a maximal lattice-free triangle with integer vertices and rays going into its corners. That triangle has vertices , and and has six integer points on its boundary. It is a triangle of Type 1 according to Dey and Wolsey [9]. Dey and Louveaux [8] showed that, when , an intersection cut has an infinite split rank if and only if it is generated from a Type 1 triangle with rays going into its corners. In this paper we prove a more general theorem, whose statement relies on the following definitions.

A set of points in is 2-partitionable if either or there exists a partition of into nonempty sets and and a split such that the points in are on one of its boundary hyperplanes and the points in are on the other. We say that a polytope is 2-partitionable if its integer points are 2-partitionable.

Let be a rational lattice-free polytope in and let be the convex hull of the integer points in . We say that has the 2-hyperplane property if every face of that is not contained in a facet of is 2-partitionable. Note that one of the faces of is itself and thus if has the 2-hyperplane property and is not contained in a facet of , then there exists a split containing all the integer points of on its boundary hyperplanes, with at least one integer point of on each of the hyperplanes.

We illustrate the 2-hyperplane property by giving an example in . Consider the tetrahedron given in Figure 1. We will show that has the 2-hyperplane property. Let be the shaded triangle with corners and and let be the shaded triangle with corners and . For any point with , let be the tetrahedron obtained as the intersection of the half-space with the cone with apex and three extreme rays joining it to the corners of . Observe that for , the vertices of are and the corners of . The tetrahedron depicted in Figure 1 is for . As is in , and the intersection of with the plane is the triangle , it follows that the intersection of with the plane contains . To check whether has the 2-hyperplane property or not, we need to check if some of the faces of the convex hull of the integer points in are 2-partitionable or not. Note that is the convex hull of and and that a face of that is not contained in a facet of is either the triangle (which is 2-partitionable, using for example the split with boundary hyperplanes and ) or it contains an integer point in the plane and an integer point in the plane (and thus is 2-partitionable using these two planes as boundary for the split). As a result, has the 2-hyperplane property. We now give an example of a polytope that does not have the 2-hyperplane property. Define as above, except that we change the half-space to . Let be obtained by truncating by , where as earlier. Then is again the convex hull of and . However is a face of not contained in a facet of . Furthermore, is not 2-partitionable because it is not possible to find a split and a partition of the six integer points of into two nonempty sets with the property that lies on one boundary of the split and on the other. Therefore does not have the 2-hyperplane property.

The main result of this paper is the following theorem.

###### Theorem 1.1.

Let be a rational lattice-free polytope in containing in its interior and having rays going into its corners. The -cut has finite split rank if and only if has the 2-hyperplane property.

Given a polytope containing in its interior, let be the convex hull of the boundary points for the rays . We assume here that nonnegative combinations of the rays in (1) span . This implies that is in the interior of and that has rays going into its corners. Moreover, the -cut and -cut are identical. Therefore the previous theorem implies the following.

###### Corollary 1.2.

Assume that nonnegative combinations of the rays in (1) span . Let be a rational lattice-free polytope in containing in its interior. The -cut has finite split rank if and only if has the 2-hyperplane property.

This corollary is a direct generalization of the characterization of Dey and Louveaux for , as triangles of Type 1 do not have the 2-hyperplane property whereas all other lattice-free polytopes in the plane do, as one can check using Lovász’ [13] characterization of maximal lattice-free convex sets in the plane.

The paper is organized as follows. In Section 2, we give a precise definition of split inequalities and split rank, as well as useful related results. In Section 3 we give an equivalent formulation of (1) that proves convenient to compute the split rank of -cuts. We prove one direction of Theorem 1.1 in Section 4 and we prove the other direction in Section 5.

## 2 Split inequalities and split closure.

Consider a mixed integer set , where and are respectively and rational matrices, and . Let be its linear relaxation.

Let and . Note that all points in satisfy the split disjunction induced by , i.e.,

 πx≤π0orπx≥π0+1 .

The hyperplanes in defined by and are the boundary hyperplanes of the split. Conversely, two parallel hyperplanes and with rational equations, both containing points with integer and such that no points with integer are between them, define a valid split . Indeed, only variables can have nonzero coefficients in the equation of the planes and we can assume that they are relatively prime integers. An application of Bézout’s Theorem [15] shows that if, for , the equation of the hyperplane is given as , then . Define

 Q≤=Q∩{(x,y)∈Rp+q | πx≤π0} ,Q≥=Q∩{(x,y)∈Rp+q | πx≥π0+1} Q(π,π0)=conv(Q≤ ∪ Q≥) .

As , any inequality that is valid for is valid for . The facets of that are not valid for are split inequalities obtained from the split . As shown by Cook et al. [6], the intersection of all for all possible splits yields a polyhedron called the split closure of . Let the rank- split closure of be itself. For , the rank- split closure of is obtained by taking the split closure of the rank- split closure of .

Note that since is the convex hull of with , any point in is a (possibly trivial) convex combination of a point and a point . Moreover, if is neither in nor in , then the segment intersects in for . By convexity of , we have that and . In summary, we have the following.

###### Observation 2.1.
• If or , then and ;

• If , then is a convex combination of and .

Let be a valid inequality for and let be the smallest nonnegative integer such that the inequality is valid for the rank- split closure of a polyhedron or if no such integer exists. The value is the split rank of the inequality with respect to . It is known that valid inequalities for may have infinite split rank with respect to (Cook et al. [6]).

The following lemma gives three useful properties of split ranks of inequalities.

###### Lemma 2.2.

Let be the linear relaxation of .

• Let . The split rank with respect to of a valid inequality for is at most its split rank with respect to ;

• Let be a subset of the variables and let be the orthogonal projection of onto the variables . Consider a valid inequality for whose coefficients for the variables not in are all 0. The split rank of inequality with respect to is greater than or equal to its split rank with respect to .

• Assume that all points in satisfy an equality. Adding any multiple of this equality to a valid inequality for does not change the split rank of the inequality with respect to .

###### Proof.

(i) Let be a split on the variables. We have

 Q≤1⊆Q≤,andQ≥1⊆Q≥ .

It follows that the split closure of is contained in the split closure of and that, for each , the rank- split closure of is contained in the rank- split closure of .

(ii) Let proj be the operation of projecting orthogonally onto the variables . It follows from the definitions of projection and convex hull that the operations of taking the projection and taking the convex hull commute. Therefore we have, for any split on the variables,

 proj(conv(Q≤∪Q≥))=conv(proj(Q≤)∪proj(Q≥))=conv(Q(x,y′)≤∪Q(x,y′)≥) ,

with the validity of the last equality coming from the fact that none of the variables involved in the disjunction are projected out. Hence, for all , the projection proj of the rank- split closure of is contained in the rank- split closure of as the projection of an intersection of polyhedra is contained in the intersection of their projections. The result then follows from the fact that inequality is valid for a polyhedron in the -space if and only if it is valid for .

(iii) For any , all points in the rank- split closure of satisfy the equality. An inequality is valid for the rank- split closure of if and only if the inequality obtained by adding to it any multiple of the equality is. ∎

Let be a polyhedron where is the space of integer variables of the mixed integer set . Assume that is rational and full-dimensional. For each facet of there exists a split with boundary hyperplanes and parallel to , with between and and with some points in strictly between and . (Note that, if the hyperplane supporting contains an integer point, then supports .) Let the width of the split be the Euclidean distance between and .

As earlier, let . Performing a round of splits around on means generating the intersection of for all facets of . Note that if contains the rank- split closure of an arbitrary polytope , then contains the rank- split closure of . Define the width of a round of splits around as the minimum of the width of the splits for all facets of .

## 3 Changing space to compute the split rank.

In the remainder of this paper, we will use to denote the linear relaxation of (1). Given a lattice-free polytope containing in its interior, our goal is to compute the split rank of the -cut with respect to . We show that this rank can be computed in another space that we find convenient.

Let be the polyhedron obtained from by adding one equation corresponding to the -cut (2) with a free continuous variable representing the difference between its left and right-hand sides:

 x=f+k∑j=1rjsjz=1−k∑j=1ψ(rj)sjx∈Rms∈Rk+z∈R . (3)

Clearly, is the orthogonal projection of onto the -space. As the relation between and is a bijection projecting or adding a single continuous variable , the split rank of (2) with respect to or are identical. Let be the orthogonal projection of onto the -space. By Lemma 2.2 (iii), inequalities (2) and have the same split rank with respect to . By Lemma 2.2 (ii), the split rank of the inequality for is smaller than or equal to its rank for . We thus have the following:

###### Observation 3.1.

Let be a lattice-free polytope containing in its interior and let be the linear relaxation of (1). The split rank of the -cut (2) with respect to is smaller than or equal to the split rank of the inequality with respect to .

Let be the zero vector and let be the unit vector in direction . Observe that is a cone with apex and extreme rays . As is a vertex of , the latter is also a pointed cone. Its apex is and its extreme rays are among . Note also that if we embed in the hyperplane , then, for all , the point is on the boundary of . The intersection of with the hyperplane is the convex hull of the points for .

Consider the pointed cone with apex and extreme rays joining to the vertices of embedded in the hyperplane (Figure 2). Note that and have the same apex and that all the extreme rays of are convex combinations of those of . Thus, and if and only if has rays going into its corners.

Let be a polyhedron in the -space. For any , define the height of with respect to as , with the convention that this number may be if is unbounded in the direction of the -unit vector or if the maximum is taken over an empty set. Define the height of as the maximum height of in . Observe that if the height of is then the height with respect to is a concave function over . Note that the split rank of the inequality with respect to is at most if and only if the rank- split closure of has height at most zero.

By Lemma 2.2 (i), the split rank of with respect to is at most its split rank with respect to . With the help of Observation 3.1, we get:

###### Observation 3.2.

Let be a lattice-free polytope containing in its interior. If the rank- split closure of has height at most zero, then the split rank of the -cut (2) with respect to is at most .

## 4 Proof of necessity.

In this section, we prove the “only if” part of Theorem 1.1. For a polyhedron , we denote by (resp. ) the interior (resp. relative interior) of .

###### Theorem 4.1.

Let be a rational lattice-free polytope in containing in its interior and having rays going into its corners. If the -cut has finite split rank with respect to , then every face of that is not contained in a facet of is 2-partitionable.

###### Proof.

Assume that the -cut has split rank for some finite . Thus, the height of the rank- split closure of is at most zero. By a theorem of Cook et al. [6], the split closure of a polyhedron is a polyhedron. Therefore there exists a finite number of splits such that applying these splits to in that order reduces its height to zero or less. Let and for .

Suppose for a contradiction that there exists a face of that is not 2-partitionable and not contained in a facet of . As is not 2-partitionable, it must contain at least two integer points and thus . We claim that any point has positive height with respect to for , a contradiction.

We prove the claim by induction on . For , the result follows from the fact that and every point in has a positive height with respect to . Indeed, any point can be written as where is a point on the boundary of with . By convexity of , the height of with respect to is at least , as the height of (resp. ) with respect to is 1 (resp. 0).

Suppose now that and that the claim is true for . If is contained in one of the boundary hyperplanes of then is contained in or and Observation 2.1 (i) shows that the height of any with respect to and is identical, proving the claim for . Otherwise, as is not 2-partitionable, there exists an integer point of that is not on the boundary hyperplanes of . Since is integer, it is strictly on one of the two sides of the split disjunction implying that there exists a point that is also strictly on one of the two sides of the split disjunction. The height of with respect to and is identical and positive by induction hypothesis. All points on the relative boundary of have non-negative height with respect to as they are convex combinations of vertices of that have height zero, and the height is a concave function. As any point is a convex combination of and a point on the boundary of with a positive coefficient for , the height of with respect to is positive. This proves the claim.∎

## 5 Proof of sufficiency.

Recall that denotes the linear relaxation of (1). In this section, we prove the following theorem.

###### Theorem 5.1.

Let be a rational lattice-free polytope in containing in its interior. If has the 2-hyperplane property, then the -cut has finite split rank with respect to .

Before giving the details of the proof, we present the main ideas. The first one is presented in Section 5.1, where we prove that Theorem 5.1 holds when there is a sequence of “intersecting splits” followed by an “englobing split”. These notions are defined as follows.

Let be a polytope in and let be a split. The part of contained between or on the boundary hyperplanes and is denoted by . The split is a -intersecting split if both of its boundary hyperplanes have a nonempty intersection with . The split is -englobing if . Note that a split can be simultaneously englobing and intersecting, as illustrated in Figure 3.

Let be a finite sequence of splits. Recall the definition of from Section 2. The polytopes , and for are the polytopes obtained from the sequence and is the polytope at the end of the sequence. When denotes the sequence of splits, we will use the notation to denote the polytope . We say that this sequence is a sequence of -intersecting splits if, for all , we have that is -intersecting.

Our approach to proving Theorem 5.1 is to work with , as introduced at the end of Section 2, instead of with directly. By Observation 3.2, to show that an -cut has finite split rank with respect to , it is sufficient to show that the height of can be reduced to at most 0 in a finite number of split operations. Lemma 5.7 guarantees a reduction of the height of when applying repeatedly a sequence of intersecting splits followed by an englobing split. A key result to proving this lemma is Lemma 5.5, showing that, for an intersecting split, we can essentially guarantee a constant reduction in height for the points cut off by the split. A consequence of these results is Corollary 5.8 that provides a sufficient condition for proving that the -cut obtained from a bounded rational lattice-free set has finite split rank. It is indeed enough to exhibit a finite sequence of -intersecting cuts followed by a final englobing cut.

Sections 5.2-5.4 deal with the unfortunate fact that it is not obvious that a sequence of intersecting splits followed by an englobing split always exists. However, since it is possible to reduce to using Chvátal cuts, the result is proved by replacing each of the Chvátal cuts by a finite collection of intersecting splits for enlarged polytopes, and using the 2-hyperplane property for proving that a final englobing split exists. Section 5.2 shows how to enlarge polytopes so that they have desirable properties. Section 5.3 proves a technical lemma about Chvátal cuts. In Section 5.4, induction on the dimension is used to prove that the region removed by a Chvátal cut can also be removed by a finite number of intersecting splits for enlarged polytopes.

We now present details of the proof.

### 5.1 Intersecting splits.

The Euclidean distance between two points is denoted by . For a point and a set , we define . The diameter of a polytope , denoted by , is the maximum Euclidean distance between two points in .

Let be a rational polytope in of dimension at least 1, and let be two finite numbers. Define (see Figure 4) as the nonconvex region in the -space containing all points such that is in the affine subspace spanned by and

 ¯z≤⎧⎪⎨⎪⎩Mif \ ¯x∈relint(Qx)M0− d(¯x,Qx)diam(Qx) (M−M0)% otherwise.

Note that this definition implies that if is on the boundary of .

The purpose of is to provide an upper bound on the height of points outside with respect to any polyhedron of height having the property that the height with respect to of any is at most , as shown in the next lemma.

###### Lemma 5.2.

Let be two finite numbers, let be a rational polyhedron of height in the -space, and let be a rational polytope containing in its relative interior. Then .

###### Proof.

Let . As the height of is , we have . If then it follows that . Otherwise and therefore by definition of . Let be a point in with height with respect to . Let be the intersection of the half-line starting at and going through with the boundary of . Notice that this intersection is on the segment . As the height of a point with respect to is a concave function, the height of is , and the height of is at most by choice of , we have that

 ¯z≤M0−(M−M0)d(xM,x0) d(¯x,x0) .

The result follows from and . ∎

We define a triplet as follows: is a full-dimensional rational polytope in with ; is a hyperplane with ; is a hyperplane in supporting a nonempty face of and is a closed half-space bounded by not containing (Figure 5). Such a triplet naturally occurs when using a -intersecting split when both and are full dimensional: is one of the two boundary hyperplanes of the split, is a hyperplane supporting a facet of that is not a facet of , and is the half-space bounded by that does not contain .

Given a triplet, we claim that there exists a hyperplane separating from with and maximizing the angle between and . This maximum value is denoted by .

To see that the claim holds, let be a point in the relative interior of . As is a hyperplane of separating from , we can rotate around to get a hyperplane supporting a face of . Then, we can possibly rotate around to increase the angle between and while keeping the resulting hyperplane supporting a face of . This rotation is stopped either if an angle of is obtained, or if gains a point of outside of .

###### Lemma 5.3.
 d(¯x,Qx)≥d(¯x,H∩H1)⋅sinθ(Qx,H1,¯H) . (4)
###### Proof.

The observation follows from

 d(¯x,Qx)≥d(¯x,H∗)=d(¯x,H∗∩H1)⋅sinθ(Qx,H1,¯H)=d(¯x,H∩H1)⋅sinθ(Qx,H1,¯H) . (5)

Indeed, the first inequality comes from the fact that separates from , the first equality is pictured in Figure 6, and the last equality follows from .

###### Lemma 5.4.

Consider a triplet. Let be two finite numbers and let be a rational polyhedron of height . Let . The height of with respect to satisfies

 ¯z≤M∗0−sinθ(Qx,H1,¯H)⋅(M∗−M∗0)diam(Qx)⋅d(¯x,H∩H1) .
###### Proof.

As and , we have that

 ¯z≤M∗0− d(¯x,Qx)diam(Qx) (M∗−M∗0) . (6)

If is on the boundary of , then , and the result holds since . Otherwise, the result follows from using Lemma 5.3 to replace the term in (6) by . ∎

Let be a full-dimensional rational polytope in with and let be a -intersecting split with boundary hyperplanes and (see Figure 7). Assume first that is full-dimensional and strictly contained in , and that the width of a round of splits around satisfies . Recall that the width of a round of splits is defined at the end of Section 2. Let be a hyperplane supporting a facet of that is not a facet of and let be the closed half-space not containing bounded by . As mentioned earlier, we have that, for , is a triplet. Let

 δ(Qx,(π,π0),¯HF)=wdiam(Qx)⋅min{sinθ(Qx,H1,¯HF), sinθ(Qx,H2,¯HF)} .

Define the reduction coefficient for , denoted by , as the minimum of taken over all hyperplanes supporting a facet of that is not a facet of . As has a finite number of facets, this minimum is well-defined and its value is positive and at most one. Assume now that is not full-dimensional or that or that . The reduction coefficient for is then defined as the value 1. Note that the reduction coefficient depends only on and and always has a positive value smaller than or equal to 1.

Lemma 5.4 can be used to prove a bound on the height of some points after applying an intersecting split. Given a set we denote its closure by . (We mean the topological closure here, not to be confused with the split closure.)

###### Lemma 5.5.

Let be a full-dimensional rational polytope in with . Let be a -intersecting split and let be the sequence of a round of splits around followed by .

Then, for any two finite numbers and for any rational polyhedron , the height with respect to of any point in is at most , where is the height of .

###### Proof.

Let be the width of the round of splits around . Observe that if then all the splits used during the round of splits around are -englobing. It follows that the height of is at most and the result holds (recall that if then its height is