aa r X i v : . [ m a t h . O C ] O c t Reverse split rank ∗ Michele Conforti † Alberto Del Pia ‡ Marco Di Summa § Yuri Faenza ¶ Abstract
The reverse split rank of an integral polyhedron P is defined as the supremumof the split ranks of all rational polyhedra whose integer hull is P . Already in R there exist polyhedra with infinite reverse split rank. We give a geometriccharacterization of the integral polyhedra in R n with infinite reverse split rank. Keywords : Integer programming, Cutting planes, Split inequalities, Split rank,Integer hull.
The problem of finding or approximating the convex hull of the integer points in arational polyhedron is crucial in Integer Programming (see, e.g., [17, 22]). In thispaper we consider one of the most well-known procedures used for this purpose: thesplit inequalities.Given an integral polyhedron P ⊆ R n , a relaxation of P is a rational polyhedron Q ⊆ R n such that P ∩ Z n = Q ∩ Z n , i.e., conv ( Q ∩ Z n ) = P , where “conv” denotes theconvex hull operator. A split S ⊆ R n is a set of the form S = { x ∈ R n : b ≤ ax ≤ b + } for some integer number b and some primitive vector a ∈ Z n (i.e., an integer vectorwhose entries have greatest common divisor equal to 1). Note that a split does notcontain any integer point in its interior int S . Therefore, if Q is a rational polyhedronand S is a split, then the set conv ( Q \ int S ) contains the same integer points as Q . The split closure SC ( Q ) of Q is defined as SC ( Q ) = \ S split conv ( Q \ int S ) . As shown in [9], if Q is a rational polyhedron, its split closure SC ( Q ) is a rationalpolyhedron, and it clearly contains the same integer points as Q . For k ∈ N , the k-thsplit closure of Q is SC k ( Q ) = SC ( SC k − ( Q )) , with SC ( Q ) = Q . If Q is a rational ∗ This work was supported by the
Progetto di Eccellenza 2008–2009 of Fondazione Cassa diRisparmio di Padova e Rovigo . † Dipartimento di Matematica, Universit`a degli Studi di Padova, Italy. ‡ Department of Industrial and Systems Engineering & Wisconsin Institute for Discovery, Universityof Wisconsin-Madison, United States, [email protected] . § Dipartimento di Matematica, Universit`a degli Studi di Padova, Italy. ¶ DISOPT, Institut de math´ematiques d’analyse et applications, EPFL, Switzerland. k such that SC k ( Q ) = conv ( Q ∩ Z n ) (see [9]); theminimum k for which this happens is called the split rank of Q , and we denote it by s ( Q ) .While one can verify that the split rank of all rational polyhedra in R is boundedby a constant, there is no bound for the split rank of all rational polyhedra in R .Furthermore, even if the set of integer points in Q is fixed, there might be no con-stant bounding the split rank of Q . For instance, let P ⊆ R be the convex hullof the points ( , , ) , ( , , ) and ( , , ) . For every t ≥
0, the polyhedron Q t = conv ( P , ( / , / , t )) is a relaxation of P . As shown in [7], s ( Q t ) → + ¥ as t → + ¥ .In this paper we aim at understanding which polyhedra admit relaxations witharbitrarily high split rank. For this purpose, given an integral polyhedron P , we definethe reverse split rank of P , denoted s ∗ ( P ) , as the supremum of the split ranks of allrelaxations of P : s ∗ ( P ) = sup { s ( Q ) : Q is a relaxation of P } . For instance, the polyhedron P given in the above example satisfies s ∗ ( P ) = + ¥ .In order to state our main result, given a subset K ⊆ R n , we denote by int K itsinterior and by relint K its relative interior. We say that K is (relatively) lattice-freeif there are no integer points in its (relative) interior. We denote by lin P the linealityspace of a polyhedron P . Furthermore, given to sets A , B ⊆ R n , A + B denotes theMinkowski sum of A and B , defined by A + B = { a + b : a ∈ A , b ∈ B } . (See, e.g.,[19, 22].) Theorem 1.
Let P ⊆ R n be an integral polyhedron. Then s ∗ ( P ) = + ¥ if and only ifthere exist a nonempty face F of P and a rational linear subspace L lin P such that (i) relint ( F + L ) is not contained in the interior of any split, (ii) G + L is relatively lattice-free for every face G of P that contains F.
Note that for the polyhedron P given in the example above, conditions (i)–(ii) aresatisfied by taking F = P and L equal to the line generated by the vector ( , , ) .We also remark that the condition L lin P in the theorem implies in particular that L = { } . Furthermore, we observe that the dimension of any face F as in the statementof the theorem is at least two.The analogous concept of reverse Chv´atal–Gomory (CG) rank of an integral poly-hedron P was introduced in [6]. We recall that an inequality cx ≤ ⌊ d ⌋ is a CG inequal-ity for a polyhedron Q ⊆ R n if c is an integer vector and cx ≤ d is valid for Q . Alter-natively, a CG inequality is a split inequality in which the split S = { x ∈ R n : b ≤ ax ≤ b + } is such that one of the half-spaces { x ∈ R n : ax ≤ b } and { x ∈ R n : ax ≥ b + } does not intersect Q . The CG closure , the
CG rank r ( Q ) , and the reverse CG rankr ∗ ( Q ) of Q are defined as for the split inequalities. The facts that the CG closure of arational polyhedron is a rational polyhedron and that the CG rank of a rational polyhe-dron is finite were shown in [21]. In [6] the following characterization was proved. Theorem 2 ([6]) . Let P ⊆ R n be an integral polyhedron. Then r ∗ ( P ) = + ¥ if and onlyif P = ∅ and there exists a one-dimensional rational linear subspace L lin P suchthat P + L is relatively lattice-free. P , a nonempty face F of P and a rational linearsubspace L lin P satisfy conditions (i)–(ii) of Theorem 1, then P = ∅ and P + L isrelatively lattice-free. Thus the conditions of Theorem 1 are a strengthening of thoseof Theorem 2. This is not surprising, as every CG inequality is a split inequality,thus s ( Q ) ≤ r ( Q ) for every rational polyhedron Q and s ∗ ( P ) ≤ r ∗ ( P ) for every integralpolyhedron P . Indeed, there are examples of integral polyhedra with finite reverse splitrank but infinite reverse CG rank: for instance, the polytope defined as the convex hullof points ( , ) and ( , ) in R (see [6]).The comparison between Theorem 1 and Theorem 2 suggests that there is some“gap” between the CG rank and the split rank. This is not surprising, as the literaturealready offers results in this direction. For instance, if we consider a rational polytopecontained in the cube [ , ] n , it is known that its split rank is at most n [2], while its CGrank can be as high as W ( n ) (see [20]; weaker results were previously given in [13,18]). Some more details about the differences between the statements of Theorem 1and Theorem 2 will be given at the end of the paper.We remark that, despite the similarity between the statements of Theorem 1 andTheorem 2, the proof of the former result (which we give here) needs more sophisti-cated tools and is more involved.The rest of the paper is organized as follows. In Sect. 2 we recall some knownfacts. In Sect. 3 we present two results on the position of integer points close to linearor affine subspaces: these results, beside being used in the proof of Theorem 1, seemto be of their own interest. The sufficiency of conditions (i)–(ii) of Theorem 1 isproved in Sect. 4, while the necessity of the conditions is shown in Sect. 5 for boundedpolyhedra, and in Sect. 6 for unbounded polyhedra. In Sect. 7 we discuss a connectionbetween the concept of reverse split rank in the pure integer case and that of split rankin the mixed-integer case. We conclude with some observations in Sect. 8. In this section we introduce some notation and present some basic facts that will beused in the proof of Theorem 1. We refer the reader to a textbook, e.g. [22], forstandard preliminaries that do not appear here.Given a point x ∈ R n and a number r >
0, we denote by B ( x , r ) the closed ball ofradius r centered at x . We write aff P to indicate the affine hull of a polyhedron P ⊆ R n ,rec P for the recession cone of P , and we recall that by lin P we denote the linealityspace of P . The angle between two vectors v , w ∈ R n is denoted by f ( v , w ) . TheEuclidean norm of a vector v ∈ R n is denoted by k v k , while dist ( A , B ) : = inf {k a − b k : a ∈ A , b ∈ B } is the Euclidean distance between two subsets A , B ⊆ R n . (If A = { a } ,we write dist ( a , B ) instead of dist ( { a } , B ) .) Given subsets S , . . . , S k of R n , we indicatewith h S , . . . , S k i the linear subspace of R n generated by the union of S , . . . , S k . (If S isa subset of R n and v ∈ R n , we write h S , v i instead of h S , { v }i and h v i instead of h{ v }i .)Further, L ⊥ is the orthogonal complement of a linear subspace L ⊆ R n . Finally, wedenote by cone ( v , . . . , v k ) the set of conic combinations of vectors v , . . . , v k ∈ R n .3 .1 Unimodular transformations A unimodular transformation u : R n → R n maps a point x ∈ R n to u ( x ) = U x + v ,where U is an n × n unimodular matrix (i.e., a square integer matrix with | det ( U ) | = v ∈ Z n . It is well-known (see e.g. [22]) that U is a unimodular matrix if and onlyif so is U − . Furthermore, a unimodular transformation is a bijection of both R n and Z n . It follows that if Q ⊆ R n is a rational polyhedron and u : R n → R n is a unimodulartransformation, then the split rank of Q coincides with the split rank of u ( Q ) .The following basic fact will prove useful: if L ⊆ R n is a rational linear subspace ofdimension d , then there exists a unimodular transformation that maps L to the subspace { x ∈ R n : x d + = · · · = x n = } ; in other words, L is equivalent to R d up to a unimodulartransformation. We will use the following result (see [1, Lemma 10]) and its easy corollary.
Lemma 3.
For every n ∈ N there exists a number q ( n ) such that the following holds:for every rational polyhedron Q ⊆ R n , c ∈ Z n and d , d ′ ∈ R with d ′ ≥ d , where cx ≤ d is valid for conv ( Q ∩ Z n ) and cx ≤ d ′ is valid for Q, the inequality cx ≤ d is valid forthe p-th CG closure of Q, where p = ( ⌊ d ′ ⌋ − ⌊ d ⌋ ) q ( n ) + . Corollary 4.
Given an integral polytope P ⊆ R n and a bounded set B containing P,there exists an integer N such that r ( Q ) ≤ N for all relaxations Q of P contained in B.
We also need the following lemma.
Lemma 5.
Let Q ⊆ R n be a rational polyhedron contained in a split S, where S = { x ∈ R n : b ≤ ax ≤ b + } . Let Q (resp., Q ) be the face of Q induced by the inequalityax ≥ b (resp., ax ≤ b + ). Then s ( Q ) ≤ max { s ( Q ) , s ( Q ) } + .Proof. For j = ,
1, since Q j is a (possibly non-proper) face of Q , we have SC ( Q j ) = SC ( Q ) ∩ Q j (see [9]). Then, for k = max { s ( Q ) , s ( Q ) } , both SC k ( Q ) ∩ Q and SC k ( Q ) ∩ Q are integral polyhedra. It follows that after another application of the split closure(actually, the split S is sufficient) we obtain an integral polyhedron. A maximal lattice-free convex set is a convex set that is not strictly contained in anylattice-free convex set. A result in [4, Theorem 2] (see also [16]) states that a maximallattice-free convex set in R n is either an irrational hyperplane or a polyhedron P + L ,where P is a polytope and L is a rational linear subspace. In particular, since a full-dimensional set is never contained in a hyperplane, every full-dimensional lattice-freeconvex set is contained in a set of the form P + L , where P is a polytope and L is arational linear subspace. 4 .4 Lattice width The lattice width w ( K ) of a closed convex set K ⊆ R n (with respect to the integerlattice Z n ) is defined by w ( K ) : = inf c ∈ Z n \{ } (cid:26) sup x ∈ K cx − inf x ∈ K cx (cid:27) . If K is full-dimensional and w ( K ) < + ¥ , then there exists a nonzero integer vector c for which w ( K ) = max x ∈ K cx − min x ∈ K cx . Furthermore, c is a primitive vector. (See, e.g., [3].)We will use the following extension of the well-known Flatness Theorem of Khint-chine [15] (see also [3, Chapter 7]), which is taken from [1, Corollary 5] (see also [14,Theorem (4.1)]). Lemma 6.
For every k ∈ N and every convex body K ⊆ R n with | K ∩ k Z n | = , onehas w ( K ) ≤ w ( n , k ) , where w is a function depending on n and k only. The proof of Theorem 1 exploits the notion of compactness and sequential compact-ness, which we recall here. A subset K of a topological space is compact if everycollection of open sets covering K contains a finite subcollection which still covers K .It is well-known that a subset of R n is compact (with respect to the usual topologyof R n ) if and only if it is closed and bounded. For a normed space (such as R n ) thenotion of compactness coincides with that of sequential compactness: a set K is se-quentially compact if every sequence ( x i ) i ∈ N of elements of K admits a subsequencethat converges to an element of K . A result given in [4], based on Dirichlet’s approximation theorem (see, e.g., [22]),shows that for each line passing through the origin there are integer points arbitrarilyclose to the line and arbitrarily far from the origin. (Note that if the line is not rational,then the origin is the only integer point lying on in it.) We give here a strengthening ofthat result, showing that for every line passing through the origin the integer points thatare “very close” to the line are not too far from each other. Furthermore, this result ispresented in a more general version, valid for every linear subspace. This lemma willbe used in the proof of Theorem 1, but we find it interesting in its own right.
Lemma 7.
Let L ⊆ R n be a linear subspace and fix d > . Then there exists R > suchthat, for every x ∈ L, there is an integer point y satisfying k y − x k ≤ R and dist ( y , L ) ≤ d .Proof. The proof is a double induction on n and d : = dim L . The statement is easilyverified for n =
1; so we fix n ≥ n . The proof is now by induction on d .5 r w r z ′ r z ′ − wC ( z ) B ′ z r Figure 1: Illustration of the proof of the claim.B
ASE STEP
The statement is trivial if d =
0. We show that it is correct for d =
1, aswell. So we now assume that L = h v i for some v ∈ R n \ { } . Claim 1.
If there is no row-vector a ∈ Z n \ { } such that av = , then the result of thelemma holds.Proof . Define the ( n − ) -dimensional ball B ′ = B ( , d ) ∩ h v i ⊥ . Note that B ( , d ) + h v i = B ′ + h v i . Let us denote by relbd B ′ the relative boundary of B ′ . For every z ∈ relbd B ′ , denote by C ( z ) the intersection of B ′ with the open cone of revolution ofdirection z and angle p /
6. We claim that for every z ∈ relbd B ′ and z ′ , w ∈ C ( z ) , z ′ − w ∈ B ′ (see Fig. 1). To see this, assume wlog that k z ′ k ≤ k w k ≤ d . Then k z ′ − w k = k z ′ k + k w k − z ′ · w = k z ′ k + k w k − k z ′ kk w ′ k≤ k z ′ k + d k w k − k z ′ kk w ′ k = k z ′ k ( k z ′ k − k w k ) + d ( k w k − d ) + d ≤ d , where the second equality holds because the angle between z ′ and w is at most p / z ′ · w k z ′ kk w ′ k = cos ( p / ) = /
2. This proves that z ′ − w ∈ B ′ whenever z ′ , w ∈ C ( z ) .Note that B ′ is a compact set, B ′ = S z ∈ relbd B ′ C ( z ) , and each C ( z ) is an open set in B ′ . Then there exist z , . . . , z m ∈ relbd B ′ such that B ′ = S mi = C ( z i ) . Define C i = C ( z i ) for i = , . . . , m .We claim that for every i = , . . . , m , the set C i + h v i is not lattice-free. To seethis, assume by contradiction that C i + h v i is lattice-free. Then it is contained in afull-dimensional maximal lattice-free polyhedron, i.e., a set of the form P + L ′ forsome polytope P ⊆ R n and some rational linear subspace L ′ ⊆ R n (recall Sect. 2.3).Note that v ∈ L ′ . However, this is not possible, as we assumed that there is no row-vector a ∈ Z n \ { } such that av = v is not contained in any rational subspace).Therefore C i + h v i is not lattice-free for i = , . . . , m . This implies that C i + cone ( v ) isnot lattice-free for i = , . . . , m .For i = , . . . , m , let w i be an integer point in C i + cone ( v ) ; note that w , . . . , w m / ∈h v i ⊥ because of the hypothesis of the claim. Observe that for every z ∈ B ′ + h v i , atleast one of the points z − w , . . . , z − w m is still in B ′ + h v i (just choose i such that6 ∈ C i + h v i ). We define M = max i = ,..., m dist ( w i , B ′ ) > , m = min i = ,..., m dist ( w i , B ′ ) > , and show that the statement of the lemma holds by choosing R = M + d .Take any x ∈ h v i . Since v is contained in no rational subspace, the set x + B ′ + cone ( v ) contains an integer point z (see, e.g., [4, Lemma 2.2]). If dist ( z , x + B ′ ) > M ,we choose i ∈ { , . . . , m } such that the point z ′ = z − w i is still in B ( x , d ) + cone ( v ) .Since 0 < dist ( z ′ , x + B ′ ) ≤ dist ( z , x + B ′ ) − m , and m >
0, by iterating this procedurea finite number of times we arrive at an integer point y ∈ B ( x , d ) + cone ( v ) such thatdist ( y , x + B ′ ) ≤ M . Then k y − x k ≤ M + d = R and dist ( y , h v i ) ≤ d . This concludesthe proof of the claim. (cid:4) We can now prove the lemma for d =
1. Let S ⊆ R n be a minimal rational subspacecontaining v . If S = R n , then the hypothesis of the claim is satisfied and we are done.So assume that dim S < n . In this case, by applying a unimodular transformation wecan reduce ourselves to the case in which S = R dim S × { } n − dim S and use induction,as S is now equivalent to an ambient space of dimension smaller than n . Thoughunimodular transformations do not preserve distances, there exist positive constants c ≤ c (depending only on the transformation) such that the distance between any twopoints (or sets) is scaled by a factor between c and c , so the arguments can be easilyadapted.I NDUCTIVE STEP
Fix d ≥ R n of dimension smaller than d . Fix any v ∈ L and define L ′ = L ∩ h v i ⊥ . By the basestep of the induction, there exists R > x ∈ h v i , there is an integerpoint y satisfying k y − x k ≤ R and dist ( y , h v i ) ≤ d /
2. Furthermore, by induction,there exists R > x ∈ L ′ , there is an integer point y satisfying k y − x k ≤ R and dist ( y , L ′ ) ≤ d /
2. Note that this remains true also if we replace L ′ with an affine subspace L obtained by translating L ′ by an integer vector. We showthat the result of the lemma holds with R = R + R ; see Fig. 2 to follow the proof.Take any x ∈ L and decompose it by writing x = x + x , where x ∈ h v i and x ∈ L ′ .Let y be an integer point satisfying k y − x k ≤ R and dist ( y , h v i ) ≤ d /
2. Define L = y + L ′ . Let x ′ = y + x ; note that k x ′ − x k = k y − x k ≤ R . Since L is atranslation of L ′ by an integer vector, there is an integer point y satisfying k y − x ′ k ≤ R and dist ( y , L ) ≤ d /
2. Now, dist ( y , L ) ≤ dist ( y , L ) + dist ( L , L ) ≤ d / + d / ≤ d . Furthermore, k y − x k ≤ k y − x ′ k + k x ′ − x k ≤ R + R = R . This concludes theproof of Lemma 7.We now prove a result that gives sufficient conditions guaranteeing that a non-full-dimensional simplex of a special type is “very close” to an integer point. Lemma 8.
For a given k ∈ N , let L ⊆ L ⊆ · · · ⊆ L k ⊆ H be a sequence of linearsubspaces, where dim L t = t for t = , . . . , k, and H is a rational subspace. Then forevery d > there exists M > such that the following holds: for every x ∈ L k and forevery y , . . . , y k satisfying y t ∈ x + L t and dist ( y t , x + L t − ) ≥ M for t = , . . . , k, onehas dist ( conv ( x , y , . . . , y k ) , H ∩ Z n ) ≤ d . L ′ h v i r x r x r x r y L r x ′ r y Figure 2: Illustration of the inductive step in the proof of Lemma 7. The space L isrepresented. Underlined symbols indicate points that do not necessarily belong to L ;in other words, their orthogonal projection onto L is represented. The circle on the lefthas radius R , the one on the right has radius R . Proof.
We first observe that it is enough to show the result for H = R n . Indeed, if H is a d -dimensional rational subspace of R n with d < n , we can apply a unimodulartransformation mapping H to R d × { } n − d to reduce ourselves to the case in which H coincides with the ambient space. As in the proof of Lemma 7, there exist posi-tive constants c ≤ c (depending only on the transformation) such that the distancebetween any two points is scaled by a factor between c and c , and this is enough toconclude.Therefore in the following we assume H = R n . We proceed by induction on k .B ASE STEP
The case k = k =
1. Apply Lemma 7 with L = L and d = d , and define M = R . Pick y ∈ x + L at distance at least M from x .Because of Lemma 7, we know that there exists an integer point z at distance at most d from L and at most M / x and y . The latter conditionimplies that the orthogonal projection of z onto L lies between x and y , hence z is atdistance at most d from conv ( x , y ) .I NDUCTIVE STEP
We prove the lemma when subspaces L ⊆ L ⊆ · · · ⊆ L k are given( k ≥ R > z ∈ L k , there is an integer point u satisfying k u − z k ≤ R and dist ( u , L k ) ≤ d / M ′ > x ′ ∈ L k − and y , . . . , y k − satisfy y t ∈ x ′ + L t and dist ( y t , x ′ + L t − ) ≥ M ′ for t = , . . . , k −
1, then dist ( conv ( x ′ , y , . . . , y k − ) , Z n ) ≤ d / M = max { M ′ , R } (see Fig. 3 to followthe proof). So fix x ∈ L k and y , . . . , y k satisfying y t ∈ x + L t and dist ( y t , x + L t − ) ≥ M for t = , . . . , k . Let v be the unit-norm vector in L k ∩ L ⊥ k − such that y k ∈ x + L k − + a v for some a ≥ x + L k − , the point y k lies on the side8ointed by v ). If we define z = x + Rv ∈ L k , then there exists an integer point u suchthat k u − z k ≤ R and dist ( u , L k ) ≤ d /
2. Let ˜ u be the orthogonal projection of u onto L k . Note that dist ( ˜ u , x + L k − ) ≤ R .Define ˜ x as the unique point in [ x , y k ] ∩ ( ˜ u + L k − ) ; for t = , . . . , k −
1, define ˜ y t asthe unique point in [ y t , y k ] ∩ ( ˜ u + L k − ) . Sincedist ( y k , x + L k − ) ≥ M ≥ R ≥ ( ˜ u , x + L k − ) and since y t ∈ x + L k − for t = , . . . , k −
1, we have, for t = , . . . , k − ( ˜ y t , ˜ x + L t − ) ≥
12 dist ( y t , x + L t − ) ≥ M ≥ M ′ . Now define x ′ , y ′ , . . . , y ′ k − as the points obtained by projecting ˜ x , ˜ y , . . . , ˜ y k − or-thogonally onto u + L k − , which is a translation of L k − by an integer vector. Note thatthe points x ′ , y ′ , . . . , y ′ k − are obtained by translating ˜ x , ˜ y , . . . , ˜ y k − by the vector u − ˜ u ,whose norm is at most d /
2. Since we still havedist ( y ′ t , x ′ + L t − ) ≥ M ′ , t = , . . . , k − , by induction we obtain an integer point ¯ z such that dist ( conv ( x ′ , y ′ , . . . , y ′ k − ) , ¯ z ) ≤ d / ( conv ( x , y . . . , y k ) , ¯ z ) ≤ dist ( conv ( ˜ x , ˜ y . . . , ˜ y k − ) , ¯ z ) ≤ dist ( conv ( x ′ , y ′ . . . , y ′ k − ) , ¯ z ) + d / ≤ d , where the first inequality holds because conv ( ˜ x , ˜ y . . . , ˜ y k − ) ⊆ conv ( x , y . . . , y k ) Thisconcludes the proof.
In this section we prove that if F and L satisfying conditions (i)–(ii) of Theorem 1exist, then P has infinite reverse split rank.By hypothesis, F and P are nonempty. Since L is a rational subspace, it admits abasis v , . . . , v k ∈ Z n . Fix ¯ x ∈ relint F , and for l ≥ Q l F = conv ( F , ¯ x ± l v , . . . , ¯ x ± l v k ) , Q l P = conv ( P , ¯ x ± l v , . . . , ¯ x ± l v k ) , where conv denotes the closed convex hull. Clearly Q l F ⊆ Q l P for every l ≥
0. As¯ x ∈ relint F and F + L is relatively lattice-free, it follows that Q l F is a relaxation of F for every l ≥
0. We now show that also Q l P is a relaxation of P for every l ≥ Claim 9. Q l P is a relaxation of P for every l ≥ .Proof. Fix l ≥ Q l P contains an integer point z / ∈ P .Since Q l P ⊆ P + L , z ∈ P + L . Let G ′ be a minimal face of P + L containing z ; thus z ∈ relint G ′ . If ax ≤ b is an inequality that defines face G ′ of P + L , then ax ≤ b defines a face G of P such that G ′ = G + L . Then z is an integer point in relint ( G + L ) ,9 h v i L k − r xx + L k − r zR r ˜ u ˜ u + L k − r y t r y k r ˜ y t r ˜ x r ¯ z Figure 3: Illustration of the inductive step in the proof of Lemma 8. The space L k isrepresented. Symbol ¯ z is underlined to indicate that ¯ z does not necessarily belong to L k ; in other words, its orthogonal projection onto L k is represented. Points x ′ , y ′ t and u ′ are not depicted; however, they project down to ˜ x , ˜ y t and ˜ u respectively, and theirdistance from the corresponding projected point is at most d / G + L is not relatively lattice-free. By condition (ii) of Theorem 1, this impliesthat G is a face of P not containing F , and thus ¯ x / ∈ G . Recall that z ∈ Q l P = conv ( P , ¯ x ± l v , . . . , ¯ x ± l v k ) . Note that all points in P ∪ { ¯ x ± l v , . . . , ¯ x ± l v k } satisfy ax ≤ b , andevery point of the form ¯ x ± l v i , i = , . . . , k satisfies ax < b , as ¯ x / ∈ G . Since az = b , itfollows that z is a convex combination of points in P . Then z ∈ P , a contradiction.Let r > F centered at ¯ x and contained in F . Clearly r is finite, because otherwise F = aff F and so F + L would be an integralaffine subspace of R n , thus not relatively lattice-free, contradicting condition (ii) ofTheorem 1.Since F is an integral polyhedron, it can be written in the form F = conv { g , . . . , g p } + cone { h , . . . , h q } , (1)where g i , i = , . . . , p , are integer points (one in each minimal face of F ), and h i , i = , . . . , q , are integer vectors; here p ≥ q ≥
0. Let R g = max {k ¯ x − g i k : i = , . . . , p } , R h = max { , k h i k : i = , . . . , q } . (Note that the 0 in the latter definition makes R h well defined.) Both R g and R h arefinite, and so is R = max { R g , R h } .We will show below that, for each l ≥ SC ( Q l F ) contains the two points¯ x ± min (cid:26) ( l − ) , r ( r + R ) l (cid:27) v i (2)for every i = , . . . , k . As Q l F ⊆ Q l P , we have SC ( Q l F ) ⊆ SC ( Q l P ) . As l was chosenarbitrarily, and at least one of the 2 k points in (2) is not in P for l large enough10because L is not contained in lin P ), this implies that P ( SC ( Q l P ) , i.e., s ( Q l P ) >
1. If l is large then the argument can be iterated, showing that s ( Q l P ) → + ¥ as l → + ¥ ,hence s ∗ ( P ) = + ¥ .It remains to prove that SC ( Q l F ) contains the two points given in (2) for every i = , . . . , k . To do so, we prove that for every split S , the set conv ( Q l F \ int S ) containsthe two points ¯ x ± ( l − ) v i or the two points ¯ x ± l r ( r + R ) v i , for every i = , . . . , k . Notethat the lineality space of every minimal face of Q l F is lin F = lin P , thus for every split S that satisfies lin P * lin S , we have conv ( Q l F \ int S ) = Q l F . Hence we now consideronly splits with lin P ⊆ lin S . To simplify notation, for fixed S and l we define T = conv ( Q l F \ int S ) , omitting the dependence on S and l . Case 1. Let S be a split such that there exists a vector ¯ v ∈ { v , . . . , v k } not in lin S . In this case we show that T contains the point ¯ x + ( l − ) v i for every i = , . . . , k .Symmetrically, T will also contain the point ¯ x − ( l − ) v i for every i = , . . . , k .Let i ∈ { , . . . , k } be such that v i / ∈ lin S . As v i ∈ Z n , it is easy to check that int S can contain at most one of the points ¯ x + l v i and ¯ x + ( l − ) v i . Thus T contains thepoint ¯ x + l v i or the point ¯ x + ( l − ) v i . If T contains ¯ x + l v i , then it also contains¯ x + ( l − ) v i , since the latter can be written as a convex combination of the points¯ x + l v i and ¯ x , which are both in T .Now let i ∈ { , . . . , k } be such that v i ∈ lin S . If ¯ x + ( l − ) v i / ∈ int S we are done,thus we assume that ¯ x + ( l − ) v i ∈ int S . Since the three points ¯ x + l v i , ¯ x ± l ¯ v are in Q l F , also their convex combinations ¯ x + ( l − ) v i ± ¯ v are in Q l F . As ¯ v ∈ Z n , ¯ v / ∈ lin S ,and ¯ x + ( l − ) v i ∈ int S , both points ¯ x + ( l − ) v i ± ¯ v are not in int S , and thus are in T . Therefore also their convex combination ¯ x + ( l − ) v i is in T . Case 2. Let S be a split such that v i ∈ lin S for every i = , . . . , k . In this case weshow that T contains the point ¯ x + l r ( r + R ) v i for every i = , . . . , k . Symmetrically, T will also contain the point ¯ x − l r ( r + R ) v i for every i = , . . . , k .Let ˜ v ∈ { v , . . . , v k } . If ¯ x / ∈ int S , then also ¯ x + l ˜ v / ∈ S , and the statement followstrivially, as ¯ x ∈ F . Thus we now assume that ¯ x ∈ int S , which implies that also ¯ x + l ˜ v ∈ int S .Since, by (i), relint ( F + L ) is not contained in int S , and since F ∩ int S = ∅ , F + L is not contained in S . As v i ∈ lin S for every i = , . . . , k , this implies that F is notcontained in S . Therefore wlog ax ≥ b is not valid for F , where a ∈ Z n , b ∈ Z aresuch that S = { x ∈ R n : b ≤ ax ≤ b + } . Since F is integral, there exists a point in F that satisfies ax ≤ b −
1. By (1), such point can be written as p (cid:229) i = l i g i + q (cid:229) i = m i h i , for nonnegative scalars l , . . . , l p and m , . . . , m q with (cid:229) pi = l i = w in F that satisfies the inequality ax ≤ b − R from ¯ x . If there exists a point in { g , . . . , g p } that satisfies ax ≤ b −
1, let w be such point. Otherwise, for every i = , . . . , p , the integral point g i satisfies ax ≥ b , and so does the convex combination (cid:229) pi = l i g i . Therefore the scalarproduct of vectors (cid:229) qi = m i h i and a is strictly negative, implying that there exists a11ector h ∈ { h , . . . , h q } such that the scalar product of h and a is strictly negative.Define w = ¯ x + h . As h is in the recession cone of F , it follows that w is in F .Moreover, since h is integral, aw ≤ b − S ax = b ax = b + Fww ′ ¯ xw ′′ r ≤ RR ′ Figure 4: Illustration of Case 2.Since aw ≤ b −
1, and a ¯ x > b , we can define w ′ as the unique point in the inter-section of the hyperplane { x ∈ R n : ax = b } with the segment [ w , ¯ x ] . (See Fig. 4.) As a ¯ x < b +
1, it follows that the segment [ w , ¯ x + l ˜ v ] ⊆ Q l F contains a point w ′ + l ′ ˜ v , with l ′ > l . Thus the point w ′ + l ˜ v is in Q l F , and in T .We finally show that T contains the point ¯ x + l r ( r + R ) ˜ v . Let w ′′ be the intersectionpoint of the line aff { w , ¯ x } with the boundary of B ( ¯ x , r ) that does not lie in the segment [ w , ¯ x ] . The point w ′′ is in F , and the distance between ¯ x and w ′′ is r . Let R ′ be thedistance between w ′ and w ′′ , and note that r < R ′ ≤ R + r . Both points w ′ + l / v and w ′′ are in T , thus also is their convex combination rR ′ (cid:16) w ′ + l ˜ v (cid:17) + R ′ − rR ′ w ′′ = ¯ x + l r R ′ ˜ v .As R ′ ≤ R + r , the point ¯ x + l r ( r + R ) ˜ v is a convex combination of the latter point and ¯ x ,implying that ¯ x + l r ( r + R ) ˜ v ∈ T . In this section we prove that if an integral polytope P has infinite reverse split rank,then F and L satisfying conditions (i)–(ii) of Theorem 1 exist, while the case of anunbounded polyhedron will be considered in Sect. 6. We remark that if P = ∅ then itsreverse split rank is finite, as this is the case even for the reverse CG rank (see [8, 6]).Therefore in this section we assume that P = ∅ . Also, we recall that for a polytope thecondition L lin P is equivalent to L = { } .In order to prove the necessity of conditions (i)–(ii), we need to extend the notionof relaxation and reverse split rank to rational polyhedra. Indeed, when dealing witha non-full-dimensional integral polytope P in Sect. 5.7, we will approximate P with anon-integral full-dimensional polytope containing the same integer points as P .Given a rational polyhedron P ⊆ R n , a relaxation of P is a rational polyhedron Q ⊆ R n such that P ⊆ Q and P ∩ Z n = Q ∩ Z n . The reverse split rank of a rational12olyhedron P is defined as follows: s ∗ ( P ) = sup { s ( Q ) : Q is a relaxation of P } . In the following we prove that if a nonempty rational polytope has infinite reversesplit rank, then F and L satisfying conditions (i)–(ii) of Theorem 1 exist. Given a full-dimensional rational polytope P ⊆ R n with s ∗ ( P ) = + ¥ , we prove condi-tions (i)–(ii) of Theorem 1 under the assumption that the result holds for all ( possiblynon-full-dimensional ) rational polytopes in R n − . (The case of a non-full-dimensionalpolytope in R n will be treated in Sect. 5.7.) Note that the theorem holds for n = s ∗ ( P ) is always finite. We also remark that if P is bounded then everyrelaxation of P is bounded, as every rational polyhedron has the same recession coneas its integer hull.So let P ⊆ R n be a full-dimensional rational polytope with s ∗ ( P ) = + ¥ . We nowgive a procedure that returns F and L satisfying the conditions of the theorem. Wejustify it and prove its correctness in the rest of this section. We remark that at thisstage the linear subspace returned by the procedure might be non-rational, but we willshow in Sect. 5.6 how to choose a rational subspace. Also, we point out that theprocedure below is not an “executable algorithm”, but only a theoretical proof of theexistence of F and L .1. Fix a point ¯ x ∈ int P ; choose a sequence ( Q i ) i ∈ N of relaxations of P with sup i s ( Q i ) =+ ¥ ; initialize k = L = { } , and S = P .2. Choose a sequence of points ( x i ) i ∈ N such that x i ∈ Q i for all i ∈ N and sup i dist ( x i , S ) =+ ¥ ; let v i be the projection of x i − ¯ x onto L ⊥ k − ; define ¯ v as the limit of somesubsequence of the sequence (cid:16) v i k v i k (cid:17) i ∈ N and assume wlog that this subsequencecoincides with the original sequence; define L k = h L k − , ¯ v i .3. If P + L k is not contained in any split, then return F = P and L = L k , and stop;otherwise, let S be a split such that P + L k ⊆ S , where S = { x ∈ R n : b ≤ ax ≤ b + } .4. If there exists M ∈ R such that Q i ⊆ { x ∈ R n : b − M ≤ ax ≤ b + M } for every i ∈ N , then choose j ∈ { , } such that P j : = P ∩ { x ∈ R n : ax = b + j } hasinfinite reverse split rank when viewed as a polytope in the affine space H = { x ∈ R n : ax = b + j } ; since H is a rational subspace and we assumed that theresult holds in dimension n −
1, there exist F and L satisfying conditions (i)–(ii)of the theorem with respect to the space H ; return F and L , and stop. Otherwise,if no M as above exists, set k ← k + ( x i ) i ∈ N and a vector ¯ v as required can be found;13b) the procedure terminates (either in step 3 or step 4);(c) if the procedure terminates in step 4, then there exists j ∈ { , } such that P j hasinfinite reverse split rank in the affine space H = { x ∈ R n : ax = b + j } , and theoutput is correct;(d) if the procedure terminates in step 3, then the output is correct. We prove that at every execution of step 2 a sequence ( x i ) i ∈ N and a vector ¯ v as requiredcan be found.Consider first the iteration k =
1; in this case, S = P . Since sup i s ( Q i ) = + ¥ , wealso have sup i r ( Q i ) = + ¥ . By Corollary 4 applied to the integral polytope conv ( P ∩ Z n ) , there is no bounded set containing every Q i for i ∈ N . Then there is a sequence ofpoints ( x i ) i ∈ N such that x i ∈ Q i for every i ∈ N and sup i dist ( x i , P ) = + ¥ . As L = { } ,for k = v i given in step 2 reduces to v i = x i − ¯ x for i ∈ N . Since everyvector v i k v i k belongs to the unit sphere, which is a compact set, the sequence (cid:16) v i k v i k (cid:17) i ∈ N has a subsequence converging to some unit-norm vector ¯ v .Assume now that we are at the k -th iteration ( k ≥ S ⊆ R n such that P + L k − ⊆ S = { x ∈ R n : b ≤ ax ≤ b + } . Furthermore,we know that there is no M ∈ R such that Q i ⊆ { x ∈ R n : b − M ≤ ax ≤ b + M } forevery i ∈ N (see step 4). This implies that there is a sequence of points ( x i ) i ∈ N suchthat x i ∈ Q i for i ∈ N and sup i dist ( x i , S ) = + ¥ . For i ∈ N , let v i be the projection ofthe vector x i − ¯ x onto the space L ⊥ k − . Note that, for i large enough, x i − ¯ x / ∈ L k − ,as ¯ x + L k − ⊆ S and sup i dist ( x i , S ) = + ¥ ; thus v i = i large enough. Since theelements of the sequence (cid:16) v i k v i k (cid:17) i ∈ N belong to the intersection of L ⊥ k − with the unitsphere, and this intersection gives a compact set, there is a subsequence converging tosome unit-norm vector belonging to L ⊥ k − , which we call ¯ v . In order to show that the procedure terminates after a finite number of iterations, it issufficient to observe that at every iteration in step 2 we select a vector ¯ v ∈ L ⊥ k − , thusthe dimension of L k = h L k − , ¯ v i is k . In particular, the procedure terminates after atmost n iterations, as for k = n no split S can be found in step 3. We now prove that if the procedure terminates in step 4, then there exists j ∈ { , } such that P j has infinite reverse split rank when viewed as a polytope in the affinespace { x ∈ R n : ax = b + j } , and the output is correct.Since Q i ⊆ { x ∈ R n : b − M ≤ ax ≤ b + M } for every i ∈ N , by Lemma 3 thereexists a number N such that, for each i ∈ N , N iterations of the CG closure operator(hence, also of the split closure operator) applied to Q i are sufficient to obtain a relax-ation of P contained in S . For i ∈ N , let e Q i be the relaxation of P obtained this way.Then we have sup i s ( e Q i ) = + ¥ . 14ecall that P and P are the faces of P induced by equations ax = b and ax = b +
1, respectively. Similarly, for i ∈ N , let e Q i and e Q i be the faces of e Q i induced byequations ax = b and ax = b +
1, respectively. Since e Q i ⊆ S , by Lemma 5 we have s ( e Q i ) ≤ max { s ( e Q i ) , s ( e Q i ) } +
1. Then there exists j ∈ { , } such that sup i s ( e Q ji ) =+ ¥ . Since every relaxation e Q ji is contained in the affine space H = { x ∈ R n : ax = b + j } , we have s ∗ ( P j ) = + ¥ with respect to the ambient space H (which is equivalent to R n − under some unimodular transformation). Let H ∗ be the translation of H passingthrough the origin. Since H is a rational space of dimension n −
1, by induction thereexist a face F of P j and a nonzero linear subspace L ⊆ H ∗ satisfying conditions (i)–(ii)of Theorem 1 for P j : specifically, relint ( F + L ) is not contained in the interior of any ( n − ) -dimensional split in the affine space H , and G + L is relatively lattice-free forevery face G of P j containing F .We show that F and L satisfy conditions (i)–(ii) for P , too. First, note that F is a face of P and L is a nonzero linear subspace of R n . To prove (i), assume bycontradiction that there is an n -dimensional split T such that relint ( F + L ) ⊆ int T .Then, as F + L is contained in the boundary of S , we have lin T = lin S . This impliesthat T ∩ H is contained in some ( n − ) -dimensional split U living in H . But then, withrespect to the ambient space H , we would have relint ( F + L ) ⊆ int U , a contradiction.To prove (ii), let G be a face of P containing F . If G ⊆ P j , then G is a face of P j and thus G + L is relatively lattice-free by induction. So we assume that G P j . Since G ⊆ P ⊆ S , this implies that G contains some points in int S , and thus relint G ⊆ int S .Since L ⊆ lin S , this yields relint ( G + L ) ⊆ int S , hence G + L is relatively lattice-free. We now prove that if the procedure terminates in step 3, then the output is correct.Note that it is sufficient to prove that P + L k is lattice-free at every iteration of thealgorithm.The subspace L is constructed following the same procedure as in the proof ofTheorem 2 given in [6, Sect. 3.2]. Therefore, with the same arguments as in [6], oneproves that P + L is lattice-free.We now assume that k ∈ { , . . . , n } . Recall that L k = h L k − , ¯ v i and ¯ v ∈ L ⊥ k − . Sup-pose by contradiction that there is an integer point ¯ z ∈ int ( P + L k ) = int P + L k . Firstof all we show that ¯ z can be taken sufficiently far from P (we will specify later howfar it should be taken). To see this, choose any integer point ¯ z ∈ int P + L k and applyLemma 7 to the affine subspace ¯ z + h ¯ v i (which is a translation of a linear subspace byan integer point), with d small enough. This guarantees the existence of integer pointsin int P + L k that lie arbitrarily far from P .Since ¯ z ∈ int P + L k , there exists a vector u ∈ L k − such that z : = ¯ z + u ∈ int P + h ¯ v i .Let x ∈ int P be such that x = z − d ¯ v , where wlog we can assume d >
0. Note thatsince k ¯ v k = d = k z − x k . Let r > B ( x , r ) ⊆ P . Furthermore, denoteby p : R n → L ⊥ k − the orthogonal projection onto L ⊥ k − .Recall that there is a split S such that P + L k − ⊆ S (step 3 of the previous iteration).Define H = lin S and H = z + H . By starting the above construction with a point ¯ z sufficiently far from P , we can assume wlog that H does not intersect P .We need the following lemma (recall that x ∈ int P ).15 laim 10. For every M ′ > and e > , there exist an index i ∈ N and points y , . . . , y k satisfying y t ∈ Q i ∩ ( x + L t ) , dist ( y t , x + L t − ) ≥ M ′ for t = , . . . , k, and f ( p ( y k − x ) , ¯ v ) ≤ e .Proof. We proceed by induction on k : we assume that the property of the lemma holdswhen k is replaced with k − k = M ′ > e >
0. By induction, there exist an index i and points y , . . . , y k − such that y t ∈ Q i ∩ ( x + L t ) and dist ( y t , x + L t − ) ≥ M ′ for t = , . . . , k −
1. Note thatthe existence of such an index i implies the existence of infinitely-many such indices.(To see this, one has to reapply the lemma with a sufficiently large M ′ : since Q i isbounded, if M ′ is large enough then a different index i ′ must exist.) Now we need tofind an additional point y k ∈ Q i ∩ ( x + L k ) such that dist ( y k , x + L k − ) ≥ M ′ .Let r > B ( x , r ) ⊆ P and define d = k x − ¯ x k . Denote again by p : R n → L ⊥ k − the orthogonal projection onto L ⊥ k − (recall that ¯ v ∈ L ⊥ k − ). The choicesof the sequence ( x i ) i ∈ N and the vector ¯ v made in step 2 of the algorithm imply that, for i ∈ N large enough, the norm of p ( x i − ¯ x ) can be made arbitrarily large and the angle f ( p ( x i − ¯ x ) , ¯ v ) = f ( v i , ¯ v ) can be made arbitrarily small. Thus we can assume that D : = k p ( x i − ¯ x ) k ≥ r ( M ′ ( r + d + ) + d ) , (3) a : = f ( p ( x i − ¯ x ) , ¯ v ) = f ( v i , ¯ v ) ≤ min (cid:26) p , arcsin (cid:18) r ( M ′ ( r + d + ) + d ) (cid:19)(cid:27) . By replacing x i with a suitable point in the line segment [ ¯ x , x i ] , we can assume that (3)holds at equality: D = ( M ′ ( r + d + )+ d ) r .Let w i be the orthogonal projection of x i onto the affine space x + L ⊥ k − . We claimthat there exists x ′ ∈ B ( x , r ) ∩ ( x + L ⊥ k ) such that k x ′ − x k = r and [ x ′ , w i ] ∩ ( x + h ¯ v i ) contains a (single) point, which we call z . To see this, consider the set obtained asthe convex hull of x i and B ( x , r ) ∩ ( x + L ⊥ k ) . This set intersects the line x + h ¯ v i ina segment. The point of this segment at maximum distance from x is the point x ′ satisfying the requirements.Observe that the orthogonal projections of x and x ′ onto w i + h ¯ v i coincide; let uscall u this common projected point. Then k u − x ′ k ≤ r + d + D sin a ≤ r + d + Dr ( M ′ ( r + d + ) + d ) = r + d + k u − w i k ≥ D cos a − d ≥ D / − d . Since the two triangles with vertices respectively w i , u , x ′ and z , x , x ′ are similar, wededuce that k z − x k ≥ D / − dr + d + · r ≥ M ′ ( r + d + ) r + d + = M ′ . (4)The construction of z shows that there exists y k ∈ Q i ∩ ( x + L k ) whose projection onto x + L ⊥ k − is z , hence dist ( y k , x + L k − ) ≥ dist ( z , x ) ≥ M ′ . ¯ x ¯ x + h ¯ v i r x x + h ¯ v i r x ′ r r w i , x i a r z , y k w i + h ¯ v i r u Figure 5: Illustration of the proof of Claim 10. The affine space x + L ⊥ k − is repre-sented. Underlined symbols indicate points that do not necessarily belong to x + L ⊥ k − ;in other words, their orthogonal projection onto x + L ⊥ k − is represented. Note that x i may also lie “above” the line ¯ x + h ¯ v i (and even “above” the line x + h ¯ v i ).To show the last part of Claim 10, note that if we choose i sufficiently large thenthe angle f ( p ( y k − ¯ x ) , ¯ v ) can be made arbitrarily small. Since the norm of p ( y x − ¯ x ) can be made arbitrarily large, this implies that also the angle f ( p ( y k − x ) , ¯ v ) can bemade arbitrarily small. (In other words, the angles f ( p ( y k − ¯ x ) , ¯ v ) and f ( p ( y k − x ) , ¯ v ) are almost the same for large i .)We first apply Lemma 8 with d = r / k − k , and obtain M > x ∈ L k − and forevery y , . . . , y k − satisfying y t ∈ x + L t and dist ( y t , x + L t − ) ≥ M for t = , . . . , k − ( conv ( x , y , . . . , y k − ) , H ∩ Z n ) ≤ d . Define M ′ = max { M , d } . Now,by Claim 10, there exist i ∈ N and points y , . . . , y k satisfying y t ∈ Q i ∩ ( x + L t ) anddist ( y t , x + L t − ) ≥ M ′ for t = , . . . , k (see Fig. 6). Now let w denote the unit-normvector which is orthogonal to H and forms an acute angle with ¯ v (recall that ¯ v / ∈ H , thus¯ v and w cannot be orthogonal) and define a = f ( ¯ v , w ) . Again because of Claim 10, wecan enforce the condition b : = f ( p ( y k − x ) , ¯ v ) ≤ arctan (cid:16) tan a + r d (cid:17) − a (5)(see Fig. 7). Note that the value on the right-hand-side of (5) is nonnegative, as 0 ≤ a < p / r >
0, define B ′ ( r ) = B ( , r ) ∩ L ⊥ k − ∩ H . For t = , . . . , k −
1, let ˜ y t be themidpoint of the segment [ x , y t ] . Note that Q i ⊇ conv ( B ( x , r ) ∪ { y , . . . , y k − } ) ⊇ C : = conv (cid:0) x + B ′ ( r / ) , ˜ y + B ′ ( r / ) , . . . , ˜ y k − + B ′ ( r / ) (cid:1) . Let x ′ be the unique point in [ x , y k ] ∩ H , and, for i = , . . . , k −
1, let y ′ t be theunique point in [ ˜ y t , y k ] ∩ H . Since dist ( y k , x + L k − ) ≥ M ′ ≥ d ≥ ( x + L k − , x ′ + L k − ) ,conv ( C , y k ) ∩ H ⊇ C ′ : = conv ( x ′ + B ′ ( r / ) , y ′ + B ′ ( r / ) , . . . , y ′ k + B ′ ( r / )) . (6)17 + H q x q y t r q ˜ y t C H ¯ vL k − L ⊥ k − ∩ H q z q ¯ zu q y k qq C ′ qq x ′ x ′′ y ′ t y ′′ t ¯ z + L k − d ¯ v Figure 6: Illustration of the proof of (d).Moreover, as B ( x , r ) ⊆ P ⊆ Q i and B ( ˜ y t , r / ) ⊆ Q i for t = , . . . , k −
1, we have B ( x ′ , r / ) ⊆ Q i and B ( y ′ t , r / ) ⊆ Q i for t = , . . . , k − . (7)Let x ′′ be the projection of x ′ onto the space z + L k − . We claim that k x ′′ − x ′ k = k p ( x ′′ − x ′ ) k ≤ d tan ( a + b ) − d tan a ≤ r / x ′′ − x ′ ∈ L ⊥ k − by construction; the firstinequality describes the worst case (which is the one depicted in the figure), i.e., when k p ( x ′′ − x ′ ) k is as large as possible; the last bound follows from (5).Now define y ′′ , . . . , y ′′ k − as the orthogonal projections of y ′ , . . . , y ′ k − onto z + L k − = ¯ z + L k − . Note that y ′′ , . . . , y ′′ k − are obtained by translating y ′ , . . . , y ′ k − byvector x ′′ − x ′ . By the definition of C ′ given in (6), y ′′ , . . . , y ′′ k − still belong to C ′ . Oneverifies that y ′′ t ∈ x ′′ + L t and dist ( y ′′ t , x ′′ + L t − ) ≥ M ′ / ≥ M for t = , . . . , k −
1. Since¯ z + L k − is a translation of L k − by an integer vector, by the choice of M given byLemma 8 there is an integer point p ∈ ¯ z + H = H at distance at most d = r / ( x ′′ , y ′′ , . . . , y ′′ k − ) .We claim that p ∈ Q i . To see this, first observe thatdist ( p , conv ( x ′ , y ′ , . . . , y ′ k − )) ≤ dist ( p , conv ( x ′′ , y ′′ , . . . , y ′′ k − )) + k x ′′ − x ′ k≤ r + r = r . (8)Now from (7) we obtain that conv ( x ′ , y ′ , . . . , y ′ k − ) + B ( , r / ) ⊆ Q i and thus, by (8), p ∈ Q i . This is a contradiction, as p is an integer point in Q i \ P ( p does not belong to P because p ∈ H and H ∩ P = ∅ by assumption).18 qq q x + h ¯ v i w y k x x ′ H ∩ ( x + L ⊥ k − ) a b z , x ′′ d Figure 7: Illustration of the proof of (d). The space x + L ⊥ k − is represented. Un-derlined symbols indicate points that do not necessarily belong to x + L ⊥ k − ; in otherwords, their orthogonal projection onto x + L ⊥ k − is represented. L As mentioned in Sect. 5.1, our procedure might return a non-rational linear subspace L . Note that this cannot be the case if the procedure terminates in step 4, as in thiscase the rationality of L follows from the fact that we assumed that the theorem holdsin R n − . Therefore we now assume that the procedure terminates in step 3, and showthat we can replace L with a suitable nonzero rational linear subspace e L and still haveconditions (i)–(ii) fulfilled.Since P + L is full-dimensional, as discussed in Sect. 2.3 we have P + L ⊆ e P + e L ,where e P is a polytope and e L is a rational linear subspace. Moreover, e L = { } , asit contains L . Since we are assuming that the procedure terminates in step 3 (thus F = P ), conditions (i)–(ii) are satisfied if L is replaced with e L . The proof of the necessity of Theorem 1 given above covers the case of a full-dimensionalrational polytope P ⊆ R n , assuming the result true both for full-dimensional and non-full-dimensional rational polytopes in R n − . We now deal with the case of a non-full-dimensional polytope in R n . For this purpose, we will take a non-full-dimensionalpolytope P and make it full-dimensional by “growing” it along directions orthogonalto its affine hull. This will be done in such a way that no integer point is added to P .The idea is then to use the proof of the full-dimensional case given above. We remarkthat even if we start from an integral polytope P , the new polytope that we constructwill not be integral. This is why at the beginning of Sect. 5 we extended the notion ofreverse split rank to rational polyhedra.Note that if P I is the convex hull of integer points in a rational polytope P , it isnot true (in general) that s ∗ ( P I ) = + ¥ implies s ∗ ( P ) = + ¥ . However, the key factunderlying our approach is the following: Given a non-full-dimensional rational polytope P with s ∗ ( P ) = + ¥ , it ispossible to “enlarge” P and obtain a full-dimensional polytope P ′ con-taining the same integer points as P, in such a way that s ∗ ( P ′ ) = + ¥ . Now, let P be a d -dimensional rational polytope P , where d < n . Assume that19 ∗ ( P ) = + ¥ . By applying a suitable unimodular transformation, we can assume thataff P = R d × { } n − d .Given a rational basis { b d + , . . . , b n } of the subspace ( aff P ) ⊥ = { } d × R n − d , arational point ¯ x ∈ relint P , and a rational number e >
0, we define P ( ¯ x , e ) = conv ( P , ¯ x + e b d + , . . . , ¯ x + e b n ) ;we do not write explicitly the dependence on vectors b d + , . . . , b n , as they will be soonfixed. Note that P ( ¯ x , e ) is a full-dimensional rational polytope.We can now present the procedure that finds F and L as required. Recall thatwe are assuming by induction that the theorem is true for both full-dimensional andnon-full-dimensional rational polytopes in R n − .0. Let w : = w ( n − d , ) be the constant of Lemma 6. Choose a sequence ( Q i ) i ∈ N ofrelaxations of P such that sup i s ( Q i ) = + ¥ . For i ∈ N , let b Q i be the orthogonalprojection of Q i onto the space ( aff P ) ⊥ , which we identify with R n − d . If thereexists an infinite subsequence of indices i , i , . . . such that the lattice width ofevery polyhedron b Q i t in R n − d is at most w , then s ∗ ( P ) = + ¥ also when we view P as a polyhedron in R n − . In this case, return F and L by induction, and stop.1. Fix a rational point ¯ x ∈ relint P ; choose a rational basis { b d + , . . . , b n } of ( aff P ) ⊥ ,a rational number e >
0, and redefine the sequence of rational polyhedra ( Q i ) i ∈ N so that:(a) P ( ¯ x , e ) has the same integer points as P ,(b) Q i is a relaxation of P ( ¯ x , e ) (and thus of P ) for every i ∈ N ,(c) sup i s ( Q i ) = + ¥ ;initialize k = L = { } , and S = P ( ¯ x , e ) .2. Choose a sequence of points ( x i ) i ∈ N such that x i ∈ Q i for all i ∈ N and sup i dist ( x i , S ) =+ ¥ ; let v i be the projection of x i − ¯ x onto L ⊥ k − ; define ¯ v as the limit of some sub-sequence of the sequence (cid:16) v i k v i k (cid:17) i ∈ N and assume that this subsequence coincideswith the original sequence; define L k = h L k − , ¯ v i .3. If, for every strictly positive rational number e ′ ≤ e , P ( ¯ x , e ′ )+ L k is not containedin any split, then choose a rational subspace L ⊇ L k such that P ( ¯ x , e ) + L islattice-free, return F = P and L , and stop; otherwise, let S = { x ∈ R n : b ≤ ax ≤ b + } be a split such that P ( ¯ x , e ′ ) + L k ⊆ S for some strictly positive rationalnumber e ′ ≤ e , and update e ← e ′ .4. If there exists M ∈ R such that Q i ⊆ { x ∈ R n : b − M ≤ ax ≤ b + M } for every i ∈ N , then choose j ∈ { , } such that P j : = P ∩ { x ∈ R n : ax = b + j } hasinfinite reverse split rank (when viewed as a polytope in the affine space { x ∈ R n : ax = b + j } ), then F and L exist by induction; return F and L , and stop.Otherwise, set k ← k +
1, and go to 2.The fact that step 2 can be executed follows from the same argument given for thefull dimensional case (see Sect. 5.2). The following additional facts, which we provebelow, imply the correctness of the procedure:20 in step 0, if there exists an infinite subsequence of indices i , i , . . . such that thelattice width of every polyhedron b Q i t in R n − d is at most w , then s ∗ ( P ) = + ¥ also when we view P as a polyhedron in R n − (Claim 11); • in step 1, a basis { b d + , . . . , b n } , a number e , and a sequence ( Q i ) i ∈ N satisfying(a)–(c) do exist (Claim 12); • if we stop in step 3, then F and L are correctly determined (Claim 13); • if the condition of step 4 is true, then there exists j ∈ { , } such that P j hasinfinite reverse split rank when viewed as a polytope in the affine space { x ∈ R n : ax = b + j } (Claim 14).In the next four claims we prove the correctness of the procedure. Recall thataff ( P ) = R d × { } n − d . Also, we identify the space ( aff P ) ⊥ with R n − d ; we will denoteits variables by x d + , . . . , x n . We denote by e j the unit vector in R n with its only 1 inposition j , for j = , . . . , n . Claim 11.
If there exists an infinite subsequence of indices i , i , . . . such that thelattice width of every polyhedron b Q i t in R n − d is at most w , then s ∗ ( P ) = + ¥ also whenwe view P as a polyhedron in R n − .Proof. If the lattice width of every polyhedron b Q i t in R n − d is at most w , then for every t ∈ N there is a primitive direction c t ∈ Z n − d such that every polyhedron b Q i t has widthat most w with respect to c t . For each t ∈ N , we can find a unimodular transformation u t that maps c t to e n and keeps the subspace aff P unchanged. The resulting polyhedra u ( Q i ) , u ( Q i ) , . . . are still relaxations of P , and they have the same split rank as Q i , Q i , . . . , respectively (see Sect. 2.1). Thus sup { s ( u ( Q i )) , s ( u ( Q i )) , . . . } = + ¥ .By Lemma 3, there is an integer N such that N iterations of the CG closure operatorare sufficient to reduce each u t ( Q i t ) to a polyhedron contained in { x ∈ R n : x n = } .Then s ∗ ( P ) = + ¥ also when we view P as a polyhedron in R n − .Under the hypothesis of the above claim, by induction there are F and L satisfyingthe conditions of the theorem when P is viewed as a polytope in R n − . It is immediateto check that with this choice of F and L the conditions of theorem are also satisfiedwhen P is viewed as a polytope in R n .From now on we can assume that the hypothesis of the previous lemma is not satis-fied. Wlog, we assume that every polyhedron in the sequence ( b Q i ) i ∈ N has (minimum)lattice width larger than w . Claim 12.
For every ¯ x ∈ relint P, there exist a rational basis { b d + , . . . , b n } of ( aff P ) ⊥ ,a rational number e > , and a sequence of rational polyhedra ( Q i ) i ∈ N such that: (a) P ( ¯ x , e ) has the same integer points as P; (b) Q i is a relaxation of P ( ¯ x , e ) (and thus of P) for every i ∈ N ; (c) sup i s ( Q i ) = + ¥ . ¯ x aff P aff P + e d + r r r r / / ⋆ Figure 8: Illustration of the proof of Claim 12. Each point in the higher part of thefigure is a point of the type p i + e d + for some p i ∈ aff P such that k p i − ¯ x k ¥ ≤ / B ( ¯ x , r ) . The pyramids have a common point of theform ¯ x + e e d + for some e >
0, e.g., the one marked with an asterisk.
Proof.
Since every b Q i has lattice width larger than w , because of Lemma 6 every b Q i contains a nonzero integer point ˆ y i . Since the origin belongs to b Q i , we can assume wlogthat ˆ y i is a primitive vector in Z n − d . For every i ∈ N , there exists a unimodular lineartransformation of R n − d that maps ˆ y i to e d + . Furthermore, each of these transforma-tions can be extended to a unimodular linear transformation of R n that maps aff P toitself. To simplify notation, we assume that every Q i coincides with its image via thelatter transformation. Then every polyhedron in the sequence ( Q i ) i ∈ N is a relaxation of P that contains a point y i of the form y i = x i + e d + for some x i ∈ aff P = R d × { } n − d .For every i ∈ N , define z i = y i − ¯ x . Let ¯ z i be the vector obtained from z i by roundingeach entry to the closest integer. Note that ¯ z i ∈ Z d + × { } n − d − and has its ( d + ) -th component equal to 1. Then ¯ z i is a primitive vector and therefore there exists aunimodular linear transformation u i such that u i ( x ) = x for every x ∈ aff P and u i ( ¯ z i ) = e d + . We then have u i ( y i ) = u i ( ¯ x + z i ) = u i ( ¯ x ) + u i ( z i − ¯ z i ) + u i ( ¯ z i ) = ¯ x + ( z i − ¯ z i ) + e d + , where the equality u i ( z i − ¯ z i ) = z i − ¯ z i holds because z i − ¯ z i ∈ aff P and u i ( x ) = x forevery x ∈ aff P . Since z i − ¯ z i ∈ aff P = R d × { } n − d and has its components in theinterval [ − / , / ] , we obtain that every u i ( Q i ) is a relaxation of P that containsa point of the type p i + e d + for some p i ∈ aff P such that k p i − ¯ x k ¥ ≤ / x ∈ relint P , there exists r > B : = B ( ¯ x , r ) ∩ aff P ⊆ P . Thenconv ( B , p i + e d + ) ⊆ u i ( Q i ) for i ∈ N . This implies that there exists a point of the type¯ x + e e d + that belongs to u i ( Q i ) for every i ∈ N (for some rational e > e sufficiently small, the polyhedron e P = conv ( P , ¯ x + e e d + ) will have the same integerpoints as P . Note that e P is a polyhedron of dimension d + e Q i = u i ( Q i ) is arelaxation of e P . We take b d + = e d + .The conclusion now follows by iterating the arguments used in this proof until afull-dimensional polytope is obtained. Note that we cannot use the same ¯ x , since itdoes not lie in relint e P . However, we can slightly perturb it, e.g. by taking the middlepoint of ¯ x and ¯ x + e e d + . After the last iteration, we will determine a rational basis { b d + , . . . , b n } of ( aff P ) ⊥ and a rational number e > ( Q i ) i ∈ N consists of relaxations of P ( ¯ x , e ′ ) for every strictly22ositive rational number e ′ ≤ e .The next claim shows that if we stop in step 3 at some iteration k , then F and L arecorrectly determined by the algorithm. The fact that P ( ¯ x , e ) + L k is lattice-free can beproved as in Sect. 5.5. Then the existence of a rational subspace L containing L k suchthat P ( ¯ x , e ) + L is lattice-free (which is required in step 3) follows from the discussionmade in Sect. 2.3. Claim 13.
Assume that, for every strictly positive rational number e ′ ≤ e , P ( ¯ x , e ′ ) + L k is not contained in any split. Let L be a rational subspace containing L k such thatP ( ¯ x , e ) + L is lattice-free. Then P + L is relatively lattice-free and relint ( P + L ) is notcontained in the interior of any split (i.e., the conditions of the theorem are satisfiedwith F = P).Proof.
Assume by contradiction that there is a split S such that relint ( P + L ) ⊆ int S .Then ¯ x ∈ int S . This implies that for e ′ > P ( ¯ x , e ′ ) + L k ⊆ P ( ¯ x , e ′ ) + L ⊆ S , a contradiction.We now prove that P + L is relatively lattice-free. Note that since P is a face of P ( ¯ x , e ) , P + L is a face of P ( ¯ x , e ) + L . Then either relint ( P + L ) is contained in theboundary of P ( ¯ x , e ) + L , or relint ( P + L ) ⊆ int ( P ( ¯ x , e ) + L ) . The latter case immedi-ately implies that P + L is relatively lattice-free, as P ( ¯ x , e ) + L is lattice-free. So weassume that relint ( P + L ) is contained in the boundary of P ( ¯ x , e ) + L .Let H be a rational hyperplane containing P + L and not containing any interiorpoint of P ( ¯ x , e ) + L ; note that H is a supporting hyperplane for P ( ¯ x , e ) + L . We denoteby ax = b an equation defining H , where a ∈ Z n is a primitive vector and b ∈ Z .Assume wlog that ax ≥ b is a valid inequality for P ( ¯ x , e ) + L . Define the split S = { x ∈ R n : b ≤ ax ≤ b + } . For e ′ > P ( ¯ x , e ′ ) + L k ⊆ P ( ¯ x , e ′ ) + L ⊆ S , acontradiction. Claim 14.
In step 4, if there exists M ∈ R such that Q i ⊆ { x ∈ R n : b − M ≤ ax ≤ b + M } for every i ∈ N , then there exists j ∈ { , } such that P j has infinite reversesplit rank when viewed as a polytope in the affine space { x ∈ R n : ax = b + j } .Proof. Denote by H and H the hyperplanes defining the split S . For j ∈ { , } , let P j ( ¯ x , e ) be the face of P ( ¯ x , e ) induced by H j . The proof of the full-dimensional caseshows that there exists j ∈ { , } such that P j ( ¯ x , e ) has infinite reverse split rank whenviewed as a polytope in the ( n − ) -dimensional space H j . Since P j ⊆ P j ( ¯ x , e ) andthese two polytopes have the same integer points, P j has also infinite reverse split rankwhen viewed as a polytope in H j . F We conclude this section with an observation that gives some more information on theface F in the statement of Theorem 1. Observation 15.
Let P ⊆ R n be an integral polytope with s ∗ ( P ) = + ¥ , and let F andL be the output of the procedure of Sect. 5.1 or Sect. 5.7. Then F + L is a face of P + L.Proof.
If the procedure terminates at the first iteration, then F = P and the statementis trivial. Therefore we assume that the procedure ends at some iteration k >
1. In23his case F and L are determined by induction on a face P j of P , for some for some j ∈ { , } , where P j is viewed as a polytope in a rational hyperplane H j = { x ∈ R n : ax = b + j } . Define H ∗ = lin H j . Then, assuming the statement true by induction, F + L is a face of P j + L . Since P j ⊆ H j , and L ⊆ H ∗ , P j + L is a face of P + L . Then F + L is a face of P + L . We prove here the necessity of conditions (i)–(ii) of Theorem 1 for unbounded poly-hedra. We assume that P ( R n , as for P = R n we have s ∗ ( P ) = Lemma 16.
Let P ⊆ R n be an integral polyhedron with lin P = h e k + , . . . , e n i for somek ∈ { , . . . , n } . Let P ′ be the polyhedron P ∩ h e , . . . , e k i viewed as a convex set in thespace h e , . . . , e k i (which is equivalent to R k ). Then s ∗ ( P ) = s ∗ ( P ′ ) .Proof. Let p : R n → R k be the map that drops the last n − k components of every vector.Note that p ( P ) = P ′ . Furthermore, p maps integer points to integer points. Since everyrelaxation Q of P is such that lin Q ⊇ h e k + , . . . , e n i , p induces a bijection between therelaxations of P and those of P ′ . Also, p induces a bijection between the splits of R n whose lineality space contains h e k + , . . . , e n i and the splits of R k . We remark thatif S is a split of R n whose lineality space contains h e k + , . . . , e n i , then p ( conv ( Q \ int S )) = conv ( p ( Q ) \ int ( p ( S ))) , while if lin S does not contain h e k + , . . . , e n i thenconv ( Q \ int ( S )) = Q (i.e., S has no effect when applied to a relaxation of P ), as in thiscase S does not contain any minimal face of Q . We conclude that if Q is a relaxationof P then p ( Q ) is a relaxation of P ′ with the same split rank. The lemma follows.Let P ( R n be an integral polyhedron with s ∗ ( P ) = + ¥ . We now show that thanksto the above lemma we can reduce to the case lin P = { } . Indeed, if this is not thecase, we can assume wlog that lin P = h e k + , . . . , e n i for some k ∈ { , . . . , n } . If wedefine P ′ as in Lemma 16, P ′ is an integral polyhedron satisfying lin P ′ = { } and s ∗ ( P ′ ) = + ¥ . Given a face F ′ of P ′ and a nonzero rational subspace L ′ ⊆ R k such that(i)–(ii) are satisfied for P ′ , we have that by setting F = F ′ × R n − k and L = L ′ × { } n − k (which is not contained in lin P ), conditions (i)–(ii) of Theorem 1 are satisfied for P .Therefore in the following we assume that lin P = { } (but rec P = { } ). Note thatin this case the condition “ L lin P ” of Theorem 1 simplifies to “ L = { } ”.A second useful lemma is now stated. Lemma 17.
Let Q ⊆ R n be a rational polyhedron, and define e Q = Q + h rec Q i . ThenQ is relatively lattice-free if and only if e Q is relatively lattice-free.Proof.
Since aff e Q = aff Q and Q ⊆ e Q , if e Q is relatively lattice-free then Q is relativelylattice-free as well.To show the reverse implication, assume that there is an integer point ˜ x ∈ relint e Q .Since Q is a rational polyhedron, we can write rec Q = cone { r , . . . , r k } , where r , . . . , r k are integer vectors. Then h rec Q i = h± r , . . . , ± r k i . This implies that we can write˜ x = x + (cid:229) ki = l i r i , where x ∈ relint Q and l , . . . , l k ∈ R . Define x = x + (cid:229) ki = ( + i − ⌈ l i ⌉ ) r i = ˜ x + (cid:229) ki = ( − ⌈ l i ⌉ ) r i . We claim that x is an integer point in relint Q . Theintegrality of x follows from the fact that x is a translation of ˜ x by an integer combi-nation of the integer vectors r , . . . , r k . Furthermore, x ∈ relint Q as x ∈ relint Q and1 + l i − ⌈ l i ⌉ ≥ i = , . . . , k . Therefore x is an integer point in relint Q , and thus Q is not relatively lattice-free.Since s ∗ ( P ) = + ¥ , there is a sequence ( Q i ) i ∈ N of relaxations of P such that sup s ( Q i ) =+ ¥ . Define L = h rec P i . Note that L = { } . Wlog, L = h e k + , . . . , e n i for some k ∈ { , . . . , n } . Also, define e P = P + L and e Q i = Q i + L for i ∈ N .Since the reverse split rank of P is infinite, the same is true for its reverse CG rank;by Theorem 2, this implies that P is relatively lattice-free. Then, by Lemma 17, e P isalso relatively lattice-free. If relint e P is not contained in the interior of any split, then(i)–(ii) hold with F = P and L = L . Therefore in the remainder of the proof we assumethat relint e P is contained in the interior of some split S . Claim 2. e Q i is a relaxation of e P for every i ∈ N .Proof. Fix i ∈ N and assume that e Q i contains some integer point ˜ x ; we prove that ˜ x ∈ e P .By using arguments that are similar to those in the proof of Lemma 17, Q i contains aninteger point x of the form x = ˜ x + r , where r is an integer vector in L . Since Q i is arelaxation of P , we have x ∈ P . But then ˜ x = x − r is in e P .Assume that s ∗ ( e P ) < + ¥ , say s ∗ ( e P ) = t . Then, for every i ∈ N , applying t timesthe split closure operator to e Q i yields e P . If the same splits are applied to Q i , we obtaina relaxation of P which is contained in e P , which in turn is contained in S . In otherwords, t rounds of the split closure operator are sufficient to make Q i contained in S for every i ∈ N . As in the proof for polytopes (Sect. 5.4), this implies that at leastone of the two faces of P induced by the boundary of S ( P , say) has infinite reversesplit rank. By induction, there exist a face F of P and a nonzero rational subspace L satisfying (i)–(ii). The same choice of F and L is also good for P .Therefore we now assume that s ∗ ( e P ) = + ¥ . Since lin e P = { } , we can replicatethe argument in the discussion following Lemma 16 and conclude that there exist aface e F of e P and a rational subspace e L lin e P such that (i)–(ii) are fulfilled for e P . Let H be any supporting hyperplane for e F . We now verify that the face F of P supportedby H and the space L = e L satisfy the conditions for P . To show that condition (i)is satisfied, observe that a split contains relint ( F + L ) in its interior if and only if itcontains relint ( e F + L ) in its interior, as e F = F + h rec F i ; to check condition (ii), onecan use Lemma 17. This concludes the proof of Theorem 1 for unbounded polyhedra. Remark 18.
Using Observation 15, one verifies that if F and L are obtained as abovethen F + L is a face of P + L. In this section we discuss a link between the concept of infinite reverse split rank inthe pure integer case and that of infinite split rank in the mixed-integer case.Fix k ∈ { , . . . , n } , and consider x , . . . , x k as integer variables and x k + , . . . , x n ascontinuous variables. A split S ⊆ R n is now defined as a set of the form S = { x ∈ n : b ≤ ax ≤ b + } for some primitive vector a ∈ Z k × { } n − k and some integernumber b . Note that every set of this type is also a split in the pure integer sense.The split closure of Q is defined as in the pure integer case. The split rank of Q is the minimum integer k such that the k -th split closure of Q coincides with Q I = conv ( Q ∩ ( Z k × R n − k )) . Unlike the pure integer case, in the mixed-integer case sucha number k does not always exist; in other words, there are rational polyhedra withinfinite split rank, see e.g. [9]. Note in fact that the example given in [9] is obtainedfrom the polytope presented in Sect. 1 by considering x as the unique continuousvariable and “enlarging” it along x . (We will develop this idea below.) We remark,however, that the split closure of a rational polyhedron Q asymptotically converges to Q I (with respect to the Hausdorff distance), as shown in [11].Given a rational polyhedron Q and a valid inequality cx ≤ d for its mixed-integerhull Q I , we say that the split rank of cx ≤ d is k if the inequality is valid for the k -thsplit closure of Q but not for the ( k − ) -th split closure of Q . The following theorem,which was proven in [10] and extends results presented in [5], characterizes the validinequalities for Q I that have infinite split rank. Theorem 19.
Let Q ⊆ R n be a rational polyhedron. For some fixed k ∈ { , . . . , n } ,define Q I = conv ( Q ∩ ( Z k × R n − k )) and let p denote the orthogonal projection ontothe space h e , . . . , e k i . Let cx ≤ d be a valid inequality for Q I . Then cx ≤ d has infinitesplit rank for Q if and only if there exists a face M of p ( { x ∈ Q I : cx = d } ) such thatM ∩ p ( { x ∈ Q : cx > d } ) = ∅ and relint M is not contained in the interior of any split.
The above result, compared with Theorem 1, suggests that there is a connectionbetween the integral polyhedra with infinite reverse split rank and the rational polyhe-dra with infinite split rank in the mixed-integer case. We propose such a connectionbelow.
Proposition 20.
Let P ⊆ R n be an integral polyhedron with s ∗ ( P ) = + ¥ . Let Fand L be as in Theorem 1, where we assume wlog L = h e k + , . . . , e n i for some k ∈{ , . . . , n } . Denote by p the orthogonal projection onto the space h e , . . . , e k i , and de-fine e P = p ( P ) . Choose ¯ x ∈ relint F and define ˜ x = p ( ¯ x ) . Then the rational polyhedronQ = conv ( e P , ˜ x + e k + , . . . , ˜ x + e n ) has infinite split rank, where variables x , . . . , x k areinteger and variables x k + , . . . , x n are continuous.Proof. Note that e P is an integral polyhedron. We claim that Q I = e P , where Q I = conv ( Q ∩ ( Z k × R n − k )) . If x is an integer point in e P , then clearly x ∈ Q I ; since e P is an integral polyhedron, this implies that e P ⊆ Q I . Assume by contradiction that e P ( Q I . Then Q contains a point w ∈ ( Z k × R n − k ) \ e P . Note that z : = w + (cid:229) ni = k + l i e i ∈ ( P + L ) ∩ Z n for some l k + , . . . , l n ∈ R . We can then proceed as in the proof of Claim 9to obtain a contradiction.Define e F = p ( F ) and let F be the minimal face of e P containing e F . Let cx ≤ d bean inequality defining face F of e P , where wlog c / ∈ L ⊥ . Using Theorem 19, we showbelow that the inequality cx ≤ d has infinite split rank for Q , thus implying that Q hasinfinite split rank.Note that p ( { x ∈ Q I : cx = d } ) = F ; we choose M to be this set. Since c / ∈ L ⊥ ,the set M ∩ p ( { x ∈ Q : cx > d } ) contains ˜ x and thus it is nonempty. In order to applyTheorem 19, it remains to show that relint M (i.e., relint F ) is not contained in the26nterior of any split. Assume by contradiction that there is a split S such that relint F ⊆ int S . Since F ⊆ L ⊥ , we can assume that L ⊆ lin S . Since F and e F have the samedimension, and e F ⊆ F , we have that relint ( e F ) ⊆ int S . Then relint ( F + L ) ⊆ int S , acontradiction to condition (i) of Theorem 1.One might wonder why in Proposition 20 the polyhedron Q is not defined simplyas conv ( P , ¯ x + e k + , . . . , ¯ x + e n ) , or perhaps conv ( P , ¯ x + l e k + , . . . , ¯ x + l e n ) for some l >
0. In fact, with this definition Q might have finite split rank. For instance, let P ⊆ R be defined as the convex hull of the points ( , , ) , ( , , ) , ( , , ) , ( , , ) , ( , , − ) . P is an integral polyhedron with infinite reverse split rank, as shown by the face F = P and the linear space L = h e i . Take k =
2. We claim that if we choose any ¯ x ∈ int P and any l >
0, then the polyhedron Q = conv ( P , ¯ x + l e ) = conv ( P , ¯ x + l e ) has finitesplit rank. To see this, observe that Q I = P has five facets, defined by the followinginequalities: − x + x ≤ − x − x ≤ − x + x ≤ − x − x ≤ x + x ≤ . The last inequality is valid for Q , thus its split rank is zero. One verifies that for eachof the first four inequalities there is no M satisfying the conditions of Theorem 19;therefore all these inequalities have finite split rank. It follows that Q has finite splitrank.We now present a result which is, in a sense, the inverse of Proposition 20. In orderto prove it, we will use of the following lemma, shown in [10, Lemma 2.1]. Lemma 21.
Let Q ⊆ R n be a rational polyhedron. For some fixed k ∈ { , . . . , n } ,let p denote the orthogonal projection onto the space h e , . . . , e k i . Let cx ≤ d be aninequality, and let M be a polyhedron contained in p ( { x ∈ Q : cx ≥ d } ) . If M ∩ p ( { x ∈ Q : cx > d } ) = ∅ , then relint M ⊆ p ( { x ∈ Q : cx > d } ) . Proposition 22.
Let Q ⊆ R n be a rational polyhedron. For some fixed k ∈ { , . . . , n } ,define Q I = conv ( Q ∩ ( Z k × R n − k )) and let p denote the orthogonal projection ontothe space h e , . . . , e k i . Let cx ≤ d be a valid inequality for Q I with infinite split rankfor Q. Then p ( { x ∈ Q I : cx = d } ) has infinite reverse split rank in the space R n , whereall variables are integer.Proof. By Theorem 19, there exists a face M of P : = p ( { x ∈ Q I : cx = d } ) such that M ∩ p ( { x ∈ Q : cx > d } ) = ∅ and relint M is not contained in the interior of any split(in the mixed-integer sense). Let L = h e k + , . . . , e n i and note that L lin P ; moreover,relint ( M + L ) is not contained in the interior of any split (in the pure integer sense).Let G be a face of P that contains M . Note that G is contained in p ( { x ∈ Q : cx ≥ d } ) . As G ∩ p ( { x ∈ Q : cx > d } ) = ∅ , it follows by Lemma 21 that relint G ⊆ p ( { x ∈ : cx > d } ) . The set { x ∈ Q : cx > d } contains no point with the first k componentsinteger, and thus its projection contains no integer point, implying that both G and G + L are relatively lattice-free. Hence by Theorem 1 (with F = M ), s ∗ ( P ) = + ¥ . L As illustrated in the introduction, Theorem 1 has strong similarities with Theorem 2,which characterizes the integral polyhedra with infinite reverse CG rank. One of thedifferences between the two statements is that in Theorem 2 the subspace L has di-mension one. We show below that L cannot be assumed to have dimension one inTheorem 1.Consider the integral polytope P in R defined by P = conv { ( , , , ) , ( , , , ) , ( , , , ) } . Note that P lives in the linear subspace R × { } .First we show that P has infinite reverse split rank. In order to do so, by our mainresult, it is sufficient to give a nonzero rational linear subspace L ⊆ R such that P + L is relatively lattice-free and relint ( P + L ) is not contained in the interior of any split.Let L be the linear subspace of R generated by vectors v = ( / , , , ) , v = ( , / , , ) . Consider the polytope P ′ obtained from P by projecting out variables x and x , i.e., P ′ = conv { ( , ) , ( , ) , ( , ) } . (See Fig. 9(a) for the drawings of P ′ and the lattice Z .) Consider also the lattice Y in R obtained as the projection of Z onto R × { } by means of L . More formally, a point y ∈ R is in Y if and only if there exists ℓ ∈ L such that ( y , , ) + ℓ ∈ Z . It can be checked that Y is the lattice Z . (See Fig. 9(b)for the drawing of P ′ and the lattice Y .) s ss s (a) P ′ and lattice Z s s ss s ss s s (b) P ′ and lattice Y s s ss s s (c) P ′ and a possible lattice Y ′ Figure 9: Illustrations of P ′ and different lattices Z , Y , and Y ′ .We say that a set Q is Y -free if it contains no point of Y in its interior, and a Y -split is the convex hull of two parallel hyperplanes containing points in Y that is Y -free. As P ′ is Y -free, one checks that P + L is lattice-free. Moreover, since P ′ is notcontained in the interior of any Y -split, one verifies that P + L is not contained in anysplit. Therefore L satisfies the desired conditions and thus P has infinite reverse splitrank. 28e now show that for every face F of P , there is no nonzero rational vector v ∈ R n such that conditions (i)–(ii) of Theorem 1 hold for L = h v i . We already observed in theintroduction that F must have dimension at least two, thus we only consider the case F = P .Assume that P + h v i is relatively lattice-free. If v = v =
0, it is easy to checkthat relint ( P + h v i ) is always contained in the interior of a split. Therefore assume nowthat ( v , v ) = ( , ) and, by scaling, that v and v are coprime integers. Considerthe lattice Y ′ in R obtained as the projection of Z onto R × { } by means of v .More formally, a point y ∈ R is in Y ′ if and only if there exists l ∈ R such that ( y , , ) + l v ∈ Z . Y ′ is the lattice generated by the vectors ( , ) , ( , ) , and ( v , v ) .(See Fig. 9(c) for a drawing of P ′ and a possible lattice Y ′ .) Note that the lattice Y ′ can contain at most one of the three points ( / , ) , ( , / ) , and ( / , / ) (and inparticular Y ′ is different from the lattice Y ). Since P + h v i is relatively lattice-free, thepolytope P ′ is Y ′ -free. Hence P ′ is a Y ′ -free triangle with vertices in Y ′ and at mostone of the three middle points of its edges is in Y ′ . This is well known to imply that P ′ is contained in the interior of a Y ′ -split, which in turn shows that relint ( P + h v i ) iscontained in the interior of a split. Remark 23.
The previous example also shows that there exist 0/1 polytopes with infi-nite reverse split rank. (The example can be made full-dimensional, if one is interestedin this further condition.) This contrasts with the fact that the split rank (and even theCG rank) of 0/1 polytopes in dimension n is bounded by a function of n (see [2, 13]).
In order to determine whether a polyhedron has infinite reverse split rank, all facesneed to be considered in Theorem 1, while this is not the case for the reverse CG rank( F = P is the only interesting face in that case). We now show that this “complication”is necessary.Let P ⊆ R be defined as the convex hull of points ( , , , ) , ( , , , ) , ( , , , ) ,and ( , , , ) . If F is the face of P induced by equation x =
1, and L = h e i , thenthe conditions of the theorem are satisfied; thus s ∗ ( P ) = + ¥ . However, the conditionsare not fulfilled if we choose F = P and the same L , as relint ( P + L ) is contained inthe interior of the split { x ∈ R : 0 ≤ x ≤ } . Indeed one can verify that there is nosubspace L ′ such that the conditions are satisfied with F = P . Acknowledgements
This work was supported by the
Progetto di Eccellenza 2008–2009 of FondazioneCassa di Risparmio di Padova e Rovigo . Yuri Faenza was supported by the GermanResearch Foundation (DFG) within the Priority Programme 1307 Algorithm Engineer-ing. The authors are grateful to two anonymous referees, whose detailed commentshelped us to improve the paper. 29 eferences [1] Averkov, G., Conforti, M., Del Pia, A., Di Summa, M., Faenza, Y.: On the con-vergence of the affine hull of the Chv´atal-Gomory closures. SIAM J. DiscreteMath. 27, 1492–1502 (2013)[2] Balas, E.: Disjunctive programming: Properties of the convex hull of feasiblepoints. Discrete Appl. Math. 89, 3–44 (1998).[3] Barvinok, A.: A Course in Convexity. Grad. Stud. Math. 54. AMS, Providence(2002)[4] Basu, A., Conforti, M., Cornu´ejols, G., Zambelli, G.: Maximal lattice-free con-vex sets in linear subspaces. Math. Oper. Res. 35, 704–720 (2010)[5] Basu, A., Cornu´ejols, G., Margot, F.: Intersection cuts with infinite split rank.Math. Oper. Res. 37, 21–40 (2012)[6] Conforti, M., Del Pia, A., Di Summa, M., Faenza, Y., Grappe R.: ReverseChv´atal–Gomory rank. In Goemans M., Correa, J. (eds.) Proceedings of the XVIInternational Conference on Integer Programming and Combinatorial Optimiza-tion (IPCO), Lecture Notes in Computer Science 7801, pp. 133–144, Springer-Verlag, Heidelberg (2013)[7] Conforti, M., Del Pia, A., Di Summa, M., Faenza, Y., Grappe, R.: ReverseChv´atal–Gomory rank. To appear in SIAM J. Discrete Math.[8] Cook, W., Coullard, C.R., T ´uran, G.: On the complexity of cutting-plane proofs.Discrete Appl. Math. 18, 25–38 (1987)[9] Cook, W., Kannan, R., Schrijver, A.: Chv´atal closures for mixed integer pro-gramming problems. Math. Program. 47, 155–174 (1990)[10] Del Pia, A.: On the rank of disjunctive cuts. Math. Oper. Res. 37, 372–378 (2012)[11] Del Pia, A., Weismantel, R.: On convergence in mixed integer programming.Math. Program. 135, 397–412 (2012)[12] Dey, S.S., Louveaux, Q.: Split rank of triangle and quadrilateral inequalities.Math. Oper. Res. 36, 432–461 (2011)[13] Eisenbrand, F., Schulz, A.S.: Bounds on the Chv´atal rank of polytopes in the 0 //