Maximal lattice-free convex sets in linear subspaces
We consider a model that arises in integer programming, and show that all irredundant inequalities are obtained from maximal lattice-free convex sets in an affine subspace. We also show that these sets are polyhedra. The latter result extends a theorem of Lovász characterizing maximal lattice-free convex sets in .
The study of maximal lattice-free convex sets dates back to Minkowski’s work on the geometry of numbers. Connections between integer programming and the geometry of numbers were investigated in the 1980s starting with the work of Lenstra . See Lovász  for a survey. Recent work in cutting plane theory ,,,,,,,,,,,, has generated renewed interest in the study of maximal lattice-free convex sets. In this paper we further pursue this line of research. In the first part of the paper we consider convex sets in an affine subspace of that are maximal with the property of not containing integral points in their relative interior. When this affine subspace is rational, these convex sets are characterized by a result of Lovász . The extension to irrational subspaces appears to be new, and it has already found an application in the proof of a key result in . It is also used to prove the main result in the second part of this paper: We consider a model that arises in integer programming, and show that all irredundant inequalities are obtained from maximal lattice-free convex sets in an affine subspace.
Let be an affine subspace of . Assume that contains an integral point, i.e. . We say that a set is a maximal lattice-free convex set in if , is convex, has no integral point in its interior with respect to the topology induced on by , and is inclusionwise maximal with these three properties. This definition implies that either contains no integral point in its relative interior or has dimension strictly less than .
The subspace is said to be rational if it is generated by the integral points in . So, if we denote by the affine hull of the integral points in , if and only if is rational. If is not rational, then the inclusion is strict. When is not rational, we will also say that is irrational. An example of an irrational affine subspace is the set of points satisfying the equation . The affine hull of is the set of points satisfying the equations .
Let be an affine space containing an integral point and the affine hull of . A set is a maximal lattice-free convex set of if and only if one of the following holds:
is a polyhedron in whose dimension equals , is a maximal lattice-free convex set of whose dimension equals , the facets of and are in one-to-one correspondence and for every facet of , is the facet of corresponding to ;
is an hyperplane of of the form , where and is an irrational hyperplane of ;
is a half-space of that contains on its boundary.
A characterization of maximal lattice-free convex sets of , needed in of the previous theorem, is given by the following.
(Lovász ) Let be a rational affine subspace of containing an integral point. A set is a maximal lattice-free convex set of if and only if one of the following holds:
is a polyhedron of the form where is a polytope, is a rational linear space, , does not contain any integral point in its relative interior and there is an integral point in the relative interior of each facet of ;
is an affine hyperplane of of the form , where and is an irrational hyperplane of ;
Theorem 1 is new and it is used in the proof of our main result about integer programming, Theorem 3 below. It is also used to prove the last theorem in . Theorem 2 is due to Lovász ( Proposition 3.1). Lovász only gives a sketch of the proof and it is not clear how case (ii) in Theorem 2 arises in his sketch or in the statement of his proposition. Therefore in Section 2 we will prove both theorems.
Figure 1 shows examples of maximal lattice-free convex sets in a 2-dimensional affine subspace of . We denote by the affine space generated by . In the first picture is rational, so , while in the second one is a subspace of of dimension 1.
We now give an example of Theorem 1(ii). Let be the set of points satisfying the equation . The affine hull of is the set of points satisfying the equations . The set defined by the equations , satisfies Theorem 1(ii). Indeed, . Furthermore, and is the line satisfying the equations and it is an irrational subspace since the only integral point it contains is .
Next we highlight the relation between lattice-free convex sets and valid inequalities in integer programming. This was first observed by Balas .
Suppose we consider rows of the optimal tableau of the LP relaxation of a given MILP, relative to basic integer variables . Let be the nonbasic variables, and be the vector of components of the optimal basic feasible solution. The tableau restricted to these rows is of the form
where , , and denotes the set of integer nonbasic variables. Gomory  proposed to consider the relaxation of the above problem obtained by dropping the nonnegativity conditions . This gives rise to the so called corner polyhedron. A further relaxation is obtained by also dropping the integrality conditions on the nonbasic variables, obtaining the mixed-integer set
Note that, since is completely determined by , the above is equivalent to
We denote by the set of points satisfying (1). The above relaxation was studied by Andersen et al.  in the case of two rows and Borozan and Cornuéjols  for the general case. In these papers they showed that the irredundant valid inequalities for correspond to maximal lattice-free convex sets in . In [1, 10] data are assumed to be rational. Here we consider the case were may have irrational entries.
Let be the linear space generated by . Note that, for every , the point , hence we assume contains an integral point. Let be the affine hull of . Notice that and coincide if and only if is a rational space. Borozan and Cornuéjols  proposed to study the following semi-infinite relaxation, which is a special case of Gomory and Johnson’s group problem . Let be the set of points of satisfying
where is the set of all with finite support, i.e. the set has finite cardinality. Notice that .
Given a function and , the linear inequality
is valid for if it is satisfied by every .
Note that, given a valid inequality (3) for , the inequality
is valid for . Hence a characterization of valid linear inequalities for provides a characterization of valid linear inequalities for .
Next we observe how maximal lattice-free convex sets in give valid linear inequalities for . Let be a maximal lattice-free convex set in containing in its interior. Since, by Theorem 1, is a polyhedron and since is in its interior, there exist such that . We define the function by
Note that the function is subadditive, i.e. , and positively homogeneous, i.e. for every . We claim that
is valid for .
Indeed, let , and . Since and is lattice-free, . Then
where the first equation follows from positive homogeneity, the first inequality follows from subadditivity of and the last one follows from the fact that .
We will show that all nontrivial irredundant valid linear inequalities for are indeed of the type described above. Furthermore, if is irrational, we will see that is contained in a proper affine subspace of , so each inequality has infinitely many equivalent forms. Note that, by definition of , if is not in the recession cone of , when is in the interior of the recession cone of , while when is on the boundary of the recession cone of . We will show that one can always choose a form of the inequality so that is a nonnegative function. We make this more precise in the next theorem.
Given a point , then . Recall that we denote by the affine hull of . Thus is contained in the affine subspace of defined as
Observe that, given and such that , we have
Given two valid inequalities and for , we say that they are equivalent if there exist and such that and . Note that, if two valid inequalities and for are equivalent, then
A linear inequality that is satisfied by every element in is said to be trivial.
We say that inequality dominates inequality if for all . Note that, for any such that for all , if satisfies the first inequality, then also satisfies the second. A valid inequality for is minimal if it is not dominated by any valid linear inequality for such that . It is not obvious that nontrivial valid linear inequalities are dominated by minimal ones. We will show that this is the case. Note that it is not even obvious that minimal valid linear inequalities exist.
We will show that, for any maximal lattice-free convex set of with in its interior, the inequality is a minimal valid inequality for . The main result is a converse, stated in the next theorem. We need the notion of equivalent inequalities, which define the same region in .
Every nontrivial valid linear inequality for is dominated by a nontrivial minimal valid linear inequality for .
Every nontrivial minimal valid linear inequality for is equivalent to an inequality of the form
such that for all and is a maximal lattice-free convex set in with in its interior.
This theorem generalizes earlier results of Borozan and Cornuéjols . In their setting it is immediate that all valid linear inequalities are of the form with nonnegative. From this, it follows easily that must be equal to for some maximal lattice-free convex set . The proof is much more complicated for the case of when is an irrational space. In this case, valid linear inequalities might have negative coefficients. For minimal inequalities, however, Theorem 3 shows that there always exists an equivalent one where all coefficients are nonnegative. The function is nonnegative if and only if the recession cone of has empty interior. Although there are nontrivial minimal valid linear inequalities arising from maximal lattice-free convex sets whose recession cone is full dimensional, Theorem 3 states that there always exists a maximal lattice-free convex set whose recession cone is not full dimensional that gives an equivalent inequality. A crucial ingredient in showing this is a new result about sublinear functions proved in .
In light of Theorem 3, it is a natural question to ask what is the subset of obtained by intersecting the set of nonnegative elements of with all half-spaces defined by inequalities as in Theorem 3. In a finite dimensional space, the intersection of all half-spaces containing a given convex set is the closure of . Things are more complicated in infinite dimension. First of all, while in finite dimension all norms are topologically equivalent, and thus the concept of closure does not depend on the choice of a specific norm, in infinite dimension different norms may produce different topologies. Secondly, in finite dimensional spaces linear functions are always continuous, while in infinite dimension there always exist linear functions that are not continuous. In particular, half-spaces (i.e. sets of points satisfying a linear inequality) are not always closed in infinite dimensional spaces (see Conway  for example).
To illustrate this, note that if is endowed with the Euclidean norm, then belongs to the closure of with respect to this norm, as shown next. Let be an integral point in and let be defined by
Clearly, for every choice of , , and for that goes to infinity the point is arbitrarily close to with respect to the Euclidean
distance. Now, given a valid linear inequality for , since the hyperplane separates strictly from even though is in the closure of . This implies that is not a closed hyperplane of , and in particular the function is not continuous with respect to the Euclidean norm on .
A nice answer to our question is given by considering a different norm on . We endow with the norm defined by
It is straightforward to show that is indeed a norm. Given , we denote by the closure of with respect to the norm .
Let be the family of all maximal lattice-free convex sets of with in their interior.
A valid inequality for is said to be extreme if there do not exist distinct functions and satisfying , such that , , are both valid for . The above definition is due to Gomory and Johnson . Note that, if an inequality is not extreme, then it is not necessary to define .
The next theorem exhibits a correspondence between extreme inequalities for the infinite model and extreme inequalities for some finite problem where . The theorem is very similar to a result of Dey and Wolsey .
Let be a maximal lattice-free convex set in with in its interior. Let and let . Then , is a rational space, and is a polytope. Let be the vertices of , and be a rational basis of . Define for .
Then the inequality is extreme for if and only if the inequality is extreme for .
Even though the data in integer programs are typically rational and studying the infinite relaxation (1) for seems natural [10, 13], some of its extreme inequalities arise from maximal lattice-free convex sets that are not rational polyhedra .
For example, the irrational triangle defined by the inequalities is a maximal lattice-free convex set in the plane, and it gives rise to an extreme valid inequality for for any rational in the interior of . In fact, every maximal lattice-free triangle gives rise to an extreme valid inequality for . Therefore, even when in (1), irrational coefficients are needed to describe some of the extreme inequalities for . Indeed, it follows from Theorem 3 and from  that the extreme inequalities for are precisely the restrictions to of the extreme inequalities for . This suggests that the more natural setting for (1) might in fact be .
2 Maximal lattice-free convex sets
Given , we denote by the linear space generated by the vectors in . The underlying field is in this paper. The purpose of this section is to prove Theorems 1 and 2. For this, we will need to work with general lattices.
An additive group of is said to be finitely generated if there exist vectors such that .
If a finitely generated additive group of can be generated by linearly independent vectors , then is called a lattice of the linear space . The set of vectors is called a basis of the lattice .
Let be a lattice of a linear space of . Given a linear subspace of , we say that is a -subspace of if there exists a basis of contained in .
For example, in , consider the lattice generated by vectors and . The line is a -subspace, whereas the line is not.
Given and , we will denote by the open ball centered at of radius . Given an affine space of and a set , we denote by the interior of with respect to the topology induced on by , namely is the set of points such that for some . We denote by the relative interior of , that is .
Let be a lattice of a linear space of , and let be a linear space of containing . A set is said to be a -free convex set of if , is convex and , and is said to be a maximal -free convex set of if it is not properly contained in any -free convex set.
Let be a lattice of a linear space of , and let be a linear space of containing . A set is a maximal -free convex set of if and only if one of the following holds:
is a polyhedron in , , is a maximal -free convex set of , the facets of and are in one-to-one correspondence and for every facet of , is the facet of corresponding to ;
is an affine hyperplane of of the form where and is a hyperplane of that is not a lattice subspace of ;
is a half-space of that contains on its boundary.
Let be a lattice of a linear space of . A set is a maximal -free convex set of if and only if one of the following holds:
is a polyhedron of the form where is a polytope, is a -subspace of , , does not contain any point of in its interior and there is a point of in the relative interior of each facet of ;
is an affine hyperplane of of the form where and is not a -subspace of .
2.1 Proof of Theorem 9
We assume Theorem 10 holds. Its proof will be given in the next section.
() Let be a maximal -free convex set of . We show that one of holds. If , then cannot occur and either or follows from Theorem 10. Thus we assume .
Assume first that . Then there exists a hyperplane
of containing , and since , then
by maximality of . Since is a hyperplane of , then
either or is a hyperplane of . If
, then let be one of the two half spaces of
separated by . Then ,
contradicting the maximality of . Hence is a
hyperplane of . We show that is a maximal
-free convex set of . Indeed, let be a convex set in
such that and .
Since , then . By maximality of , ,
Given , for some hyperplane of , and . Applying Theorem 10 to , we get that is not a lattice subspace of , and case holds.
So we may assume . Since is convex, then . We consider two cases.
Case 1. .
Since and are nonempty disjoint convex sets, there exists a hyperplane separating them, i.e. there exist and such that for every and for every . Since is a linear space, then for every , hence . Then the half space contains and lies on the boundary of . Hence is a maximal -free convex set of containing , therefore by the maximality assumption, so holds.
Case 2. .
We claim that
To prove this claim, notice that the direction is straightforward. Conversely, let . Then there exists such that . Since , there exists . Then there exists such that . We may assume since otherwise the result holds. Since , there exists such that is in the relative interior of the segment . Since , and therefore . The ball with radius is contained in . By convexity of , and therefore . Thus . Since , it follows that .
Since , by Theorem 10 applied to , is a polyhedron with a point of in the relative interior of each of its facets. Let be the facets of . For , let be a point in . By (6), . By the separation theorem, there exists a half-space of containing such that . Notice that is on the boundary of . Then . By construction , hence by maximality of , . For every , contains . Therefore defines a facet of for .
() Let be a set in satisfying one of . Clearly is a convex set in and , so we only need to prove maximality. If satisfies , then this is immediate. This is also immediate when satisfies and . So we may assume that either satisfies (i) and , or satisfies .
Suppose that there exists a closed convex set strictly containing such that . Let . Then . To conclude the proof of the theorem, it suffices to prove that is strictly contained in . Indeed, by maximality of , this claim implies that the set contains a point in . Now implies that contains a point of , a contradiction.
It only remains to prove that . This is clear when is a hyperplane satisfying (ii). The statement is also clear if . Assume now that is a polyhedron satisfying (i) and and that . Let be a facet of that separates from . If is contained in a proper face of , then is contained in at least two facets of , a contradiction to the one-to-one correspondence property. So is not contained in proper face of . Therefore, there exists . Choose such that . Note that since otherwise but this is a contradiction since , and , .
Let . Note that and are distinct affine hyperplanes of . Let be the two open half-spaces of defined by , and assume w.l.o.g. that . Since , contains some point . Since , it follows that . Let be the line segment joining and . Since and it follows that contains exactly one point, say . Note that . Since and , we have that but . ∎
2.2 Proof of Theorem 10
To simplify notation, given , we denote simply by .
The following standard result in lattice theory provides a useful equivalent definition of lattice (see Barvinok , p. 284 Theorem 1.4).
Let be the additive group generated by vectors . Then is a lattice of the linear space if and only if there exists such that for every .
Let be a lattice of a linear space of . Then every bounded set in contains a finite number of points in .
Throughout this section, will be a lattice of a linear space of . The following lemma proves the “only if” part of Theorem 10 when is bounded and full-dimensional.
Let be a bounded maximal -free convex set with . Then is a polytope with a point of in the relative interior of each of its facets.
Since is bounded, there exist integers such that is contained in the box . For each , since is convex there exists a closed half-space of such that and . By Corollary 12, is finite, therefore is a polyhedron. Thus is a polytope and by construction . Since and for every , it follows that . By maximality of , , therefore is a polytope. We only need to show that has a point of in the relative interior of each of its facets. Let be the facets of , and let be the closed half-space defining , . Then . Suppose, by contradiction, that one of the facets of , say , does not contain a point of in its relative interior. Given , the polyhedron is a polytope since it has the same recession cone as . The polytope contains points of in its interior by the maximality of . By Corollary 12, has a finite number of points in , hence there exists one minimizing , say . By construction, the polytope does not contain any point of in its interior and properly contains , contradicting the maximality of . ∎
We will also need the following famous theorem of Dirichlet.
Theorem 14 (Dirichlet).
Given real numbers with , there exist integers and such that
The following is a consequence of Dirichlet’s theorem.
Given and , then for every and , there exists a point of at distance less than from the half line
First we show that, if the statement holds for , then it holds for arbitrary . Given , let be the set of points of at distance less than from . Suppose, by contradiction, that no point in has distance less than from . Then is contained in . By Corollary 12, is finite, thus there exists an such that every point in has distance greater than from , a contradiction. So we only need to show that, given , there exists at least one point of at distance at most from . We may assume .
Without loss of generality, assume . Let and be a basis of . Then there exists such that . Denote by the matrix with columns , and define where, for a vector , denotes the Euclidean norm of . Choose such that and . By Dirichlet’s theorem, there exist and such that
Let . Since , then . Note that since , where the first inequality follows from the definition of and the last one follows from the assumptions on , and . Therefore, . Furthermore
Let be a -free convex set, and let . Then also is -free.
Let , . We only need to show that is -free. Suppose there exists . We show that . Suppose not. Then