1 Maximal Lattice-free Convex Sets

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1 47-831: Advanced Integer Programming Lecturer: Amitabh Basu Lecture 3 Date: 03/23/2010 In this lecture, we explore the connections between lattices of R n and convex sets in R n. The structures will prove to be useful for developing a general theory of cutting planes for general Mixed-Integer Linear Programs (MILPs). 1 Maximal Lattice-free Convex Sets Consider some lattice Λof R n. Definition 1 Let Λ be a lattice of R n. A set S R n is said to be a Λ-free convex set of R n if S is convex, Λ int(s) =, and S is said to be a maximal Λ-free convex set of R n if it is not properly contained in any Λ-free convex set. When the lattice is clear from the context, we will often use the term maximal lattice-free convex sets. A characterization of maximal lattice-free convex sets, is given by the following. Theorem 2 Let Λ be a lattice of a linear space V of R n. A set S R n is a maximal Λ-free convex set of V if and only if one of the following holds: (i) S is a polyhedron of the form S = P + L where P is a polytope, L is a lattice-subspace of R n, dim(s) = dim(p ) + dim(l) = n, S does not contain any point of Λ in its interior and there is a point of Λ in the relative interior of each facet of S; (ii) dim(s) < n, S is an affine hyperplane of R n of the form S = v + L where v S and L is not a lattice-subspace of R n. The polyhedron S = P + L in Theorem 2(i) is called a cylinder over the polytope P and can be shown to have at most 2 dim(p ) facets. You proved the following results in the homeworks. Proposition 3 A bounded set Π R n contains only finitely many points from Λ. Lemma 4 Given y Λ and r V, then for every ε > 0 and λ 0, there exists a point of Λ at distance less than ε from the half line {y + λr λ λ}. Lemma 5 Given a lattice-subspace L of V, the orthogonal projection of Λ onto L is a lattice of L V. The following is a converse to Lemma 7 from Lecture 1. Lemma 6 If a linear subspace L of V is not a lattice-subspace of V, then for every ε > 0 there exists y Λ \ L at distance less than ε from L. 1

2 Proof: The proof is by induction on k = dim(l). Assume L is a linear subspace of V that is not a latticesubspace, and let ε > 0. If k = 1, then, since the origin 0 is contained in Λ, by Lemma 4 there exists y Λ at distance less than ε from L. If y L, then L = y, thus L is a lattice-subspace of V, contradicting our assumption. Hence we may assume that k 2 and the statement holds for spaces of dimension k 1. Suppose L contains a nonzero vector r Λ. Let L = proj r (L), Λ = proj r (Λ). By Lemma 5, Λ is a lattice of r V. Also, L is not a lattice subspace of r V with respect to Λ, because if there exists a basis a 1,..., a k 1 of L contained in Λ, then there exist scalars µ 1,..., µ k 1 such that a 1 + µ 1 r,..., a k 1 + µ k 1 r Λ, but then r, a 1 + µ 1 r,..., a k 1 + µ k 1 r is a basis of L contained in Λ, a contradiction. By induction, there exists a point y Λ \ L at distance less than ε from L. Since y Λ, there exists a scalar µ such that y = y + µr Λ, and y has distance less than ε from L. Thus L Λ = {0}. By Lemma 4, there exists a nonzero vector y Λ at distance less than ε from L. Since L does not contain any point in Λ other than the origin, y / L. The following lemma proves the only if part of Theorem 2 when S is bounded and full-dimensional. Lemma 7 Let Λ be a lattice of a linear space V of R n. Let S V be a bounded maximal Λ-free convex set with dim(s) = dim(v ). Then S is a polytope with a point of Λ in the relative interior of each of its facets. Proof: Since S is bounded, there exist integers L, U such that S is contained in the box B = {x R d L x i U}. For each y Λ B, since S is convex there exists a closed half-space H y of V such that S H y and y / int(h y ). By Proposition 3, B Λ is finite, therefore y B Λ Hy is a polyhedron. Thus P = y B Λ Hy B is a polytope and by construction Λ int(p ) =. Since S B and S H y for every y B Λ, it follows that S P. By maximality of S, S = P, therefore S is a polytope. We only need to show that S has a point of Λ in the relative interior of each of its facets. Let F 1,..., F t be the facets of S, and let H i = {x V α i x β i } be the closed half-space defining F i, i = 1,..., t. Then S = t i=1 H i. Suppose, by contradiction, that one of the facets of S, say F t, does not contain a point of Λ in its relative interior. Given ε > 0, the polyhedron S = {x V α i x β i, i = 1,..., t 1, α t x β t + ε} contains points of Λ in its interior by the maximality of S. By Proposition 3, int(s ) has a finite number of points in Λ, hence there exists one minimizing α t x, say z. By construction, the polytope S = {x V α i x β i, i = 1,..., t 1, α t x α t z} does not contain any point of Λ in its interior and properly contains S, contradicting the maximality of S. Lemma 8 Let L be a linear subspace of V with dim(l) = dim(v ) 1, and let v V. Then v + L is a maximal Λ-free convex set if and only if L is not a lattice subspace of V. Proof: ( ) Let S = v + L and assume that S is a maximal Λ-free convex set. Suppose by contradiction that L is a lattice-subspace. Then there exists a basis a 1,..., a m of Λ such that a 1,..., a m 1 is a basis of L. Thus S = { m i=1 x ia i x m = β} for some β R. Then, K = { m i=1 x ia i β 1 x m β } strictly contains S and int(k) Λ =, contradicting the maximality of S. ( ) Assume L is not a lattice-subspace of V. Since S = v+l is an affine hyperplane of V, int(s) =, thus int(s) Λ =, hence we only need to prove that S is maximal with such property. Suppose not, and 2

3 let K be a maximal convex set in V such that int(k) Λ = and S K. Then by maximality K is closed. Let w K \ S. Since K is convex and closed, then K conv({v, w}) + L. Let ε be the distance between v + L and w + L, and δ be the distance of conv({v, w}) + L from the origin. By Lemma 6, since L is not a lattice-subspace of V, there exists a vector y Λ \ L at distance ε < ε from L. Let z = ( δ ε + 1)y. By definition, z is strictly between v + L and w + L, hence z int(k). Since z is an integer multiple of y Λ, then z Λ, a contradiction. We are now ready to prove Lovász s Theorem. Proof:[Proof of Theorem 2.] ( ) If S satisfies (ii), then by Lemma 8, S is a maximal Λ-free convex set. If S satisfies (i), then, since int(s) Λ =, we only need to show that S is maximal. Suppose not, and let K be a convex set in V such that int(k) Λ = and S K. Given y K \ S, there exists a hyperplane H separating y from S such that F = S H is a facet of S. Since K is convex and S K, then conv(s {y}) K. Since dim(s) = dim(v ), F S hence the relint(f ) int(k). By assumption, there exists x Λ relint(f ), so x int(k), a contradiction. ( ) Let S be a maximal Λ-free convex set. We show that S satisfies either (i) or (ii). Observe that, by maximality, S must be closed. If dim(s) < dim(v ), then S is contained in some affine hyperplane H. Since int(h) =, we have S = H by maximality of S, therefore S = v + L where v S and L is a hyperplane in V. By Lemma 8, (ii) holds. Therefore we may assume that dim(s) = dim(v ). In particular, since S is convex, int(s). By Lemma 7, if S is bounded, (i) holds. Hence we may assume that S is unbounded. Let C be the recession cone of S and L the lineality space of S. Since S is unbounded, C 0. Claim 1. L = C. By the definition of L and C, L C. We show the opposite inclusion. Let r C, r 0. We only need to show that S + r is Λ-free; by maximality of S this will imply that S = S + r. Suppose there exists y int(s + r ) Λ. We show that y int(s) + r. Suppose not. Then (y + r ) int(s) =, which implies that there is a hyperplane H separating the line y+ r and S+ r. This contradicts y int(s+ r ). This shows y int(s) + r. Thus there exists λ such that ȳ = y + λr int(s), i.e. there exists ε > 0 such that B ε (ȳ) V S. Since y Λ, then y / int(s), and thus, since ȳ int(s) and r C, we must have λ > 0. Since r C, then B ε (ȳ) + {λr λ 0} S. Since y Λ, by Lemma 4 there exists z Λ at distance less than ε from the half line {y + λr λ λ}. Thus z B ε (ȳ) + {λr λ 0}, hence z int(s), a contradiction. Let P = proj L (S) and Λ = proj L (Λ). By Claim 1, S = P + L and P L V is a bounded set. Furthermore, dim(s) = dim(p ) + dim(l) = dim(v ) and dim(p ) = dim(l V ). Notice that int(s) = relint(p ) + L, hence relint(p ) Λ =. Furthermore P is inclusionwise maximal among the convex sets of L V without points of Λ in the relative interior: if not, given a convex set K L V strictly containing P and with no point of Λ in its relative interior, we have S = P + L K + L, and K + L does not contain any point of Λ in its interior, contradicting the maximality of S. You are asked to prove the following Claim in the homework. Claim 2. L is a lattice-subspace of V. 3

4 Since L is a lattice-subspace of V, Λ is a lattice of L V by Lemma 5. Since P is a bounded maximal Λ -free convex set, it follows from Lemma 7 that P is a polytope with a point of Λ in the relative interior of each of its facets, therefore S = P + L has a point of Λ in the relative interior of each of its facets, and (i) holds. Exercise 1 Prove rigorously that any lattice-free convex set is contained in a maximal lattice-free convex set. There exists both a non-constructive and a constructive proof of this fact. We now show a bound on the number of facets for a maximal lattice-free convex set. Theorem 9 Any maximal Z n -free convex set B in R n has at most 2 n facets. Proof: If B is an affine hyperplane, then we are done. So we can assume it is full dimensional with nonempty interior by Theorem 2. Suppose to the contrary that B has at least 2 n + 1 facets. By Theorem 2, every facet has a point from Z n in its relative interior. We denote this set of integral points by I, so that I 2 n + 1. Consider the set of vectors Ī = {p mod 2 p I}, where p mod 2 represents reducing each coordinate of p modulo 2. Note that each element in Ī is a vector in {0, 1}n. Since I 2 n + 1, this implies that there exist p, q I with p q and p mod 2 = q mod 2. This implies that p+q 2 is in fact integral and moreover lies in the interior of B, because p and q lie in distinct facets of B. This is a contradiction. 2 Corner Polyhedron and Cutting Planes 2.1 Corner Polyhedron Corner polyhedra were introduced by Gomory as a generalization of his original cutting plane method and this structure was studied extensively by Gomory and Johnson. We introduce this concept and the notation in this section. Consider a general MILP in the following standard form. Let A be a rational m n matrix, b be a rational m 1 column vector. We are interested in x R n such that Ax = b, x 0, x j Z for j I {1,... n} (1) We assume, without loss of generality, that A has full row rank. A basis D is a maximal linearly independent set of columns in A; so D is an invertible m m matrix. Corresponding to any basis D, we define the corner polyhedron with respect to D, denoted by corner(d) as the convex hull of all x R n such that x D = D 1 b D 1 Nx N, x N 0, x j Z for j I {1,... n} (2) where x D refers to the variables corresponding to the columns of D (a.k.a the basic variables) and x N refers to the remaining variables (a.k.a the non-basic variables). To simplify notation, we will denote the set of indices of the basic variables by D, and the non-basic variables by N. The point x = (D 1 b, 0) is called the basic solution corresponding to the basis D. Let LP (corner(d)) be the relaxation of corner(d) where the integrality constraints are dropped on all the variables in (2). It is well-known that corner(d) 4

5 LP (corner(d)) if and only if x does not satisfy the integrality constraints. Furthermore, corner(d) is a relaxation of (1) because the non-negativity constraints on the basic variables have been dropped. This implies that valid inequalities for corner(d) are also valid inequalities for (1). The significance of this corner polyhedron lies in the fact that most cutting planes for general MILPs (e.g. GMI cuts, Split cuts, Reduce-and-Split cuts, MIR cuts and so forth) are obtained by solving the linear relaxation using linear programming methods and then using integrality arguments to find an inequality which cuts off the fractional solution of the linear relaxation. Many families of such cutting planes, including the four examples listed above, are valid inequalities for corner(d) for some appropriate basis D, which cut off the corresponding basic solution x. D usually corresponds to the optimal basis in the standard simplex method for solving the LP relaxation. Moreover, for the derivation of these cutting planes, one needs to only consider a relaxation of the corner polyhedra obtained by retaining only one equality constraint. Therefore, having a characterization of polyhedra in R n defined using constraints like (2) would be an automatic generalization of most cutting planes used in practice. In addition, corner polyhedra have facets which cannot be obtained by applying integrality arguments to a single constraint from the system. So the analysis of corner polyhedra provides us with new families of cutting planes, derived from multiple constraints, which may be useful for solving general MILPs. 2.2 Continuous Relaxation Andersen et al. introduced a relaxation of the corner polyhedron, where the integrality conditions on the non-basic variables are dropped. We can formulate this relaxation in the following notation, which provides a little more geometric intuition. x = f + k i=1 ri s i x Z m s i 0 for all i. In the above, f R m corresponds to D 1 b in (2) and r i s are rays in R m which correspond to the non-basic columns ( D 1 N i ) in (2) (N i denotes the i-th column of the matrix N). x corresponds to x D and s corresponds to x N. Note that to describe the solutions of (3), one only needs to record the values of the s i variables. We use R f (r 1,..., r k ) to denote the set of all points s such that (3) is satisfied. Since conv(r f (r 1,..., r k )) is a relaxation of the corner polyhedron, any valid inequality for conv(r f (r 1,..., r k )) is valid for the corner polyhedron and hence for our initial MILP problem from which the corner polyhedron is derived. 2.3 Intersection Cuts and the Minkowski functional In this section, we will develop the ideas for deriving valid inequalities for conv(r f (r 1,..., r k )). As discussed before, this will provide cutting planes for our original MILP, derived using multiple constraints. The key observation is that one can use lattice-free convex sets to derive valid inequalities for conv(r f (r 1,..., r k )) using the following classical tool from convex analysis. Definition 10 Let K R n be a closed convex set containing the origin in its interior. The gauge or the Minkowski functional, is the function γ K defined by By definition γ K is nonnegative. γ K (x) = inf{t > 0 t 1 x K} for all x R n. (3) 5

6 We now collect some key properties of Minkowski functionals. Proposition 11 Let K R n be a closed, convex set containing the origin in its interior. Then the Minkowski functional γ K satisfies : i γ K (r 1 + r 2 ) γ K (r 1 ) + γ K (r 2 ), for all r 1, r 2 R n. ii γ K (sr) = sγ K (r) for every s R and r R n. iii K = {r γ K (r) 1}. Proof: We will prove only the subadditivity condition (condition i) here. You prove the other two conditions in the homework. Let s 1 > 0 be any real number such that s 1 1 r 1 K and s 2 > 0 be a real number such that s 1 2 r 2 K. Since K is convex, the point s 1 s 1 + s 2 (s 1 1 r 1) + s 2 s 1 + s 2 (s 1 2 r 2) K. Therefore (s 1 + s 2 ) 1 (r 1 + r 2 ) K. Hence, γ K (r 1 + r 2 ) s 1 + s 2. Taking an infimum over all s 1 on both sides of the inequality, we obtain that γ K (r 1 + r 2 ) γ K (r 1 ) + s 2 and then taking an infimum over s 2, we obtain γ K (r 1 + r 2 ) γ K (r 1 ) + γ K (r 2 ). We now show how to derive valid inequalities for conv(r f (r 1,..., r k )) using the Minkowski functional. This idea was first presented by Egon Balas under the name of Intersection Cuts. Theorem 12 (Intersection Cuts) Consider any closed, convex set B containing the point f in its interior, but no integer point in its interior. Let K = B f. Then the inequality k i=1 γ K(r i )s i 1 is valid for conv(r f (r 1,..., r k )). Proof: Consider any point s that satisfies (3). Let x = f + k i=1 ri s i. Note that since x Z n, x is not in the interior of B and therefore x f is not in the interior of K. Therefore, γ K ( x f) 1 by condition iii) in Proposition 11. Therefore, k k k 1 γ K ( x f) = γ K ( r i s i ) γ K (r i s i ) γ K (r i )s i i=1 i=1 i=1 where the second and third inequalities follow from conditions i) and ii) of Proposition 11 respectively. You prove the following proposition in HW2. Proposition 13 Consider two closed convex sets C 1 and C 2 containing the origin in their interiors, such that C 1 C 2. Show that the associated Minkowski functionals satisfy the inequality γ C2 (r) γ C1 (r) for all r R n. Note that if we have two valid inequalities k i=1 γ is i 1 and k i=1 γ i s i 1 such that γ i γ i, then the inequality k i=1 γ i s i 1 dominates the inequality k i=1 γ is i 1, i.e. every point in R n + satisfying k i=1 γ i s i 1 also satisfies k i=1 γ is i 1. Hence, the inequality k i=1 γ is i 1 is redundant. Combining 6

7 this simple observation with Proposition 13, we conclude that we only need to consider maximal latticefree convex sets to derive our valid inequalities. In other words, the inequalities derived from non-maximal lattice-free convex sets are redundant and are dominated by inequalities from maximal lattice-free convex sets. One can further show that every valid inequality for conv(r f (r 1,..., r k )) comes from an intersection cut. Theorem 14 Assume that all data in the problem (3) is rational. Then every valid inequality for conv(r f (r 1,..., r k )) is dominated by an intersection cut obtained from a closed, convex set B with f in its interior, and no integer point in its interior. Proof:[Sketch of Proof] You show in your homework that one can assume that every valid inequality for conv(r f (r 1,..., r k )) is of the form k i=1 γ is i 1. We next consider the following set in R n. B = conv{f + rj γ j γ j 0} + cone{r j γ j = 0}. Observe that γ j = γ K (r j ), where γ K is the Minkowski functional of B f. Claim 1 No face of B containing f contains an integer point in its relative interior. Proof: Suppose to the contrary that a face F containing f contains an integer point x in its relative interior. Then x = αf + (1 α)p for some p in F and α > 0 (since x is in the relative interior of F ). Now p C and therefore p can be expressed as p = βf + n j:γ j 0 α j(f + rj γ j ) + j:γ j =0 β jr j with β + α j = 1. Therefore, x = αf + (1 α)p = αf + (1 α)(βf + n j:γ j 0 α j(f + rj γ j ) + j:γ j =0 β jr j ) = f + n j:γ j 0 (1 α)α j rj γ j + (4) j:γ j =0 (1 α)β jr j One observes that α j 1 and 1 α < 1. Therefore (1 α) α j < 1. We consider the solution { α (1 α) j γ s j = j γ j 0 (1 α)β j γ j = 0 From the last equation in (4), one sees that s j is a valid solution to (3). But γ j s j < 1 because (1 α) α j < 1. This is a contradiction to the fact that γ j s j 1 is a valid inequality. Firstly note that the claim implies that there is no integer point in the relative interior of B, since B is a face of itself and B contains f. Further, if f lies on any facet F of B, then one can push this facet out a little bit so that f does not lie on F any more. Note that this modification uses the fact that the affine hull of F is a lattice-subspace, which comes from the fact that the data is rational and the fact that F does not have any integer point in its relative interior. By doing this pushing for every facet that contains f, we can ensure that f is actually in the relative interior of B. One finally observes, that enlarging the lattice-free set B like this only improves the inequality one derives using the Minkowski functional, using Proposition 13. 7

8 If B happens to be full-dimensional, then we would be done at this point, because we have ensured that f is in the interior of B and the Minkowski functional dominates the given inequality k j=1 γ js j 1. The case where B is not full-dimensional can also be handled, but we will not give the details here. In conclusion, we studied how one can use lattice-free convex sets to derive cutting planes for general MILPs. In this context, the maximal lattice-free convex sets are the important ones because all valid inequalities for conv(r f (r 1,..., r k )) are dominated by valid inequalities derived using the Minkowski functional of maximal lattice-free convex sets. Note further that the Minkowski functional is a function that depends only on the point f and the convex set used. It does not see the actual rays in the system (3). In this sense, it gives generic formulas dependent only on f, such that one can simply plug in the actual rays of the problem and derive the inequalities. This is very much in the spirit of the GMI cuts that you studied last Mini. We will explore this in more detail in the subsequent lectures, where we will derive even more explicit formulas for the Minkowski functionals of maximal lattice-free sets. This will use Lovasz s structural theorem for maximal lattice-free sets, which essentially says that they are polyhedra whose recession cones are not full dimensional. 8