Two-halfspace closure

Abstract

We define a new cutting plane closure for pure integer programs called the two-halfspace closure. It is a natural generalization of the well-known Chvátal-Gomory closure. We prove that the two-halfspace closure is polyhedral. We also study the corresponding two-halfspace rank of any valid inequality and show that it is at most the split rank of the inequality. Moreover, while the split rank can be strictly larger than the two-halfspace rank, the split rank is at most twice the two-halfspace rank. A key step of our analysis shows that the split closure of a rational polyhedron can be obtained by considering the split closures of all k-dimensional (rational) projections of the polyhedron, for any fixed $k \ge 2$. This result may be of independent interest.

Appendices

A Integer hull of two dimensional simplicial cones

Definition 6

Given a simplicial cone $C{\subseteq } \mathbb R^2$, let $F^1,\dots , F^n$ be the facets of $C_I$. For every $i\in \{1,\dots ,n\}$ we denote by $H^i$ the halfspace defining $F^i$, which can be described as $\{x\in \mathbb R^2:\langle a_i,x \rangle \le \delta _i\}$, for some $a_i\in {\mathbb Z}^2$ of which the two entries are coprime and $\delta _i\in {\mathbb Z}$. Furthermore, we define $\widehat{\ell ^i}$ as $\{x\in \mathbb R^2:\langle a_i,x \rangle = \delta _i+1\}$. Also, let $\widehat{H^i}$ denote the halfspace described by $\{x\in \mathbb R^2:\langle a_i,x \rangle \le \delta _i+1\}$.

Definition 7

Given a line $\ell \subseteq \mathbb R^2$ containing integer points, we call each closed segment whose endpoints are two consecutive integer points of $\ell$ as a unit interval of $\ell$.

Theorem 5

Given a simplicial cone $C\in \mathbb R^2$, the integer hull of C can be derived by taking the split closure of C, and then taking the CG closure of this split closure.

Proof

We use the same notations as in Definition 6. We first verify that $\mathcal {S}(C)\subseteq \widehat{H^i}$ for every $i=1, \ldots , n$. Given $i\in \{1,\ldots ,n\}$, if $\widehat{l^i}$ does not intersect with C, then C is contained in $\widehat{H^i}$ and we are done. If $\widehat{l^i}$ intersects with C. Then there exists a unique unit interval U of $\widehat{l^i}$ intersecting C. By definition of $\widehat{l^i}$, the two integer points of U are outside of C. Also, there exists at least one unit interval on $F^i$. Let $U'$ be one of them. Then we take the unique split set S containing the apex of C and ${{\,\mathrm{conv}\,}}(U\cup U')$. The two intersection points between the boundaries of S and the boundaries of C determine a split cut H produced by S, which cuts off $U\cap C$. Thus $\mathcal {S}(C)\subseteq H\cap C\subseteq \widehat{H^i}{\backslash \widehat{l^i}}$.

Thus by the definition of $\widehat{H^i}$, the facet defining halfspace $H^i$ of $F^i$ is a CG cut for $\mathcal {S}(C)$, for $i\in \{1,\ldots ,n\}$. This finishes the proof.$\square$

General integer hulls in two dimensions

We give a new proof of the following result.

Theorem 6

[7, 9] Let $P\subseteq \mathbb R^2$ be any two-dimensional rational polyhedron. Then $P_I$ is equal to the intersection of the integer hulls of all the simplicial cones (or split sets) constructed from pairs of facets of P.

We break the proof into two cases: $P_I$ is nonempty or empty. These are dealt with in Theorems 9 and 10 below.

Definition 8

A convex set $B\subseteq \mathbb R^n$ is a lattice-free convex set if there is no integer point in its interior, and it is maximal if for any lattice-free set $B'\supseteq B$, we have $B'=B$.

Theorem 7

In $\mathbb R^2$, a convex set B is a maximal lattice-free convex set if and only if it satisfies one of the following properties.

• 1. B is a split set, and each of its facets contains integer points.

• 2. B is a triangle such that each of its facets contains at least one integer point in its relative interior.

• 3. B is a four-facet lattice-free set and each of its facet contains exactly one integer point in its relative interior. Moreover, the convex hull of the union of these four integer points is a parallelogram with area 1.

Lemma 2

Let $Q\in \mathbb R^2$ be a rational polyhedron such that ${{\,\mathrm{int}\,}}(Q)\cap {\mathbb Z}^2=\emptyset$, and Q has four facets $F_1$, $F_2$, $F_3$ and $F_4$ in clockwise order. Let $H_i$ and $l_i$ denote the corresponding facet-defining halfspace and hyperplane for $i=1,2,3,4$. Assume $F_1$ contains at least one integer point in its relative interior, and $l_i\cap {\mathbb Z}^2=\emptyset$ for $i=2,3,4$. Let the corresponding facet defining halfspace $H_i$ be $\{x\in \mathbb R^2:\langle a_i,x\rangle \le \delta _i\}$ for $i=1,2,3,4$. Then $H:=\{x\in \mathbb R^2:\langle a_1,x\rangle \ge \delta _1\}\supseteq (H_2\cap H_3)_I\cap (H_2\cap H_4)_I\cap (H_3\cap H_4)_I$.

Proof

We will attempt to construct a maximal lattice-free set containing Q by “pushing its facets out”. More formally, we do the following.

If $(H_1 \cap H_2 \cap H_4) \cap {\mathbb Z}^2 = (H_1 \cap H_2 \cap H_3 \cap H_4) \cap {\mathbb Z}^2$, i.e., removing $H_3$ does not change the set of integer points, then $H \supseteq (H_2\cap H_4)_I$ and we are done. Suppose then that $(H_1 \cap H_2 \cap H_4) \cap {\mathbb Z}^2$ contains integer points that are not in $(H_1 \cap H_2 \cap H_3 \cap H_4) \cap {\mathbb Z}^2$. All such integer points must be in the interior of $H_1 \cap H_2 \cap H_4$, since $\ell _2$ and $\ell _4$ do not contain integer points. Therefore, there exists $\delta '_3 > \delta _3$ such that if one defines $H'_3 = \{x \in \mathbb R^2: \langle a_3, x \rangle \le \delta '_3\}$, then $H_1 \cap H_2 \cap H'_3 \cap H_4$ is also lattice free, but the facet corresponding to $H'_3$ contains integer points in its relative interior. One now checks if removing $H_2$ introduces new integer points in $H_1 \cap H_2 \cap H'_3 \cap H_4$. If not, then we observe that $H \supseteq (H_3\cap H_4)_I$ since $\delta '_3 > \delta _3$ and we are done. Otherwise, we find $\delta '_2 > \delta _2$ such that if one defines $H'_2 = \{x \in \mathbb R^2: \langle a_2, x \rangle \le \delta '_2\}$, then $H_1 \cap H'_2 \cap H'_3 \cap H_4$ is also lattice free, but the facet corresponding to $H'_2$ contains integer points in its relative interior. Finally, we “push out” $H_4$ and either realize that $H \supseteq (H_2\cap H_3)_I$, or end up with a maximal lattice-free quadrilateral satisfying Case 3. in Theorem 7.

Let us make the notation uniform and use $H_k'$ and $\ell _k'$ denote the corresponding facet defining halfspace and hyperplane, and $v_j$ be the integer point located on the corresponding facet, for $j=1,2,3,4$. By assumption, $\ell _k\ne \ell _k'$ for $k=2,3,4$. Let $\ell _5$, $\ell _5'$ and $\ell _6$ be the lines such that $\{v_1,v_2\}\subseteq \ell _5$, $\{v_3,v_4\}\subseteq \ell _5'$, and $\{v_1,v_4\}\subseteq \ell _6$. Furthermore, let $H_5$ be the halfspace defined by $\ell _5$ such that $v_4\notin H_5$. Similarly $H_6$ be the halfspace defined by $\ell _6$ such that $v_2\notin H_6$. Since there are no integer points between $\ell _5$ and $\ell _5'$ (because $v_1, v_2, v_3, v_4$ form a fundmanetal parallelopiped of the integer lattice), all integer points in $H'_3 \cap H'_4$ are contained in $H_5$, except for the points $v_3$ and $v_4$. Since these points are not contained in $H_3\cap H_4$, we must have $(H_3\cap H_4)_I\subseteq H_5$. Similarly, we have $(H_2\cap H_3)_I\subseteq H_6$. Therefore, $(H_2\cap H_3)_I\cap (H_4\cap H_3)_I\subseteq H_5\cap H_6\subseteq H$. $\square$

Theorem 8

(Integer Helly’s Theorem [8, 21, 26, 30]) Let $\mathcal {I}$ be a finite family of convex sets in $\mathbb R^n$ such that $\bigcap _{C\in \mathcal {I}}C\cap {\mathbb Z}^n=\emptyset$, then there exists $\mathcal {I}'\subseteq \mathcal {I}$ such that $\bigcap _{C\in \mathcal {I}'}C\cap {\mathbb Z}^n=\emptyset$ and $|\mathcal {I}'| \le 2^n$.

Corollary 3

Given a full dimensional polyhedron $P\in \mathbb R^2$ with at least four facets, assume only one of its facets F contains integer points in its relative interior. Let H be the facet defining halfspace of F. Furthermore, we assume ${{\,\mathrm{int}\,}}(H)\cap P\cap {\mathbb Z}^2=\emptyset$. Then there exists three facet defining halfspaces for P denoted by $H_1$, $H_2$ and $H_3$ other than H, such that ${{\,\mathrm{int}\,}}(H)\cap H_1\cap H_2\cap H_3\cap {\mathbb Z}^2=\emptyset$ and H is irredundant to ${{\,\mathrm{int}\,}}(H)\cap H_1\cap H_2\cap H_3$.

Theorem 9

Given a polyhedron $P\in \mathbb R^2$ such that $P\cap {\mathbb Z}^2\ne \emptyset$. Then $P_I$ is the intersection of the integer hulls of all the simplicial cones (or split sets) constructed from pairs of facets of P.

Proof

Let H be a halfspace containing $P_I$ and described by $\{x\in \mathbb R^2:\langle a,x \rangle \ge \delta \}$. We wish to show that H is valid for the intersection of the integer hulls of all the simplicial cones (or split sets) constructed from pairs of facets of P. For this purpose, we may strengthen H such that its bounding hyperplane has a nonempty intersection with $P_I$, and show that this strengthening has the desired property. Let $H'$ be the halfspace $\{x\in \mathbb R^2:\langle a,x \rangle \le \delta \}$. Let P be the intersection of halfspaces $H_i:=\{x\in \mathbb R^2:\langle a_i,x \rangle \le \delta _i\}$, and the two entries of $a_i$ be coprime for $i=1,\dots ,m$. If $\delta _i\in {\mathbb Z}$, then let $H_i'$ be the halfspace $\{x\in \mathbb R^2:\langle a_i,x \rangle \le \delta _i+\frac{1}{2}\}$. Otherwise, $H_i'=H_i$. Let $P'$ be the intersection of $H_i'$ for $i=1,\ldots ,m$. Note that $P_I=P'_I\subseteq {{\,\mathrm{int}\,}}(P')$, and there is no integer point on any facet defining line of $P'$.

Claim: H is valid for the intersection of the integer hulls of all the simplicial cones (or split sets) constructed from pairs of facets of $P'$.

Proof

By our assumption, the bounding hyperplane of H contains integer points from $P_I$. Thus, $H'\cap P_I\ne \emptyset$. Then $P'\cap H'$ is a lattice-free set with one facet defined by $H'$ and containing at least one integer point in its relative interior since $P'_I=P_I\subseteq {{\,\mathrm{int}\,}}(P)$. If $P'\cap H'$ has only two or three facets, the proof is trivial. Otherwise, by Corollary 3, there exist three facet defining halfspaces of $P'$, say $H'_1$, $H'_2$ and $H'_3$, such that $Q:=H'_1\cap H'_2\cap H'_3\cap H'$ is lattice free and nonempty since $H'\cap P_I\ne \emptyset$, and $H'$ is irredundant to Q. Moreover, the facet of Q defined by $H'$ contains a integer point in its relative interior since $P_I\subseteq {{\,\mathrm{int}\,}}(P)$. If Q only has two or three facets, then the proof is trivial. If Q has four facets, then by Lemma 2, we can finish the proof.$\square$

The claim immediately implies that $P_I$ is the intersection of the integer hulls of all the simplicial cones (or split sets) constructed from pairs of facets of $P'$. The proof can be finished by the fact that $H_i\subseteq H_i'$ for $i=1,\ldots , m$.$\square$

Theorem 10

Given a polyhedron P such that $P\cap {\mathbb Z}^2= \emptyset$, we have that the intersection, denoted by U, of the integer hulls of all the simplicial cones (or split sets) constructed from pairs of facets of P, is empty.

Proof

By Theorem 8, we can assume P has at most four facets. If P has two or three facets, then the proof is trivial. Therefore, assume P has four facets and let $H_i$ for $i=1,\ldots ,4$ denote the facet defining halfspaces in clockwise order. If $H_1\cap H_3$ or $H_2\cap H_4$ forms a split set, then the proof is trivial. So we assume both $H_1\cap H_3$ and $H_2\cap H_4$ contain integer points. Since $H_1\cap H_3 \cap {\mathbb Z}^2 \ne \emptyset$, $(H_1\cap H_3\cap H_2)_I$ or $(H_1\cap H_3\cap H_4)_I$ is not empty. Without loss of generality, assume $(H_1\cap H_3\cap H_2)_I\ne \emptyset$. By Theorem 9, we have $(H_1\cap H_3\cap H_2)_I=(H_3\cap H_2)_I\cap (H_1\cap H_2)_I\cap (H_1\cap H_3)_I$. Therefore $U\subseteq (H_1\cap H_3\cap H_2)_I$. Similarly, using the fact that $H_2\cap H_4 \cap {\mathbb Z}^2 \ne \emptyset$, we can assume $(H_2\cap H_4\cap H_1)_I\ne \emptyset$ and have $U\subseteq (H_2\cap H_4\cap H_1)_I$. Hence $U\subseteq (H_1\cap H_3\cap H_2)_I \cap (H_2\cap H_4\cap H_1)_I\subseteq (H_1 \cap H_2\cap H_3)_I \cap H_4=\emptyset$.$\square$