Branching with a pre-specified finite list of $k$ -sparse split sets for binary MILPs

Santanu S. Dey santanu.dey@isye.gatech.edu School of Industrial and Systems Engineering, Georgia Institute of Technology Diego Morán morand@rpi.edu Department of Industrial and Systems Engineering, Rensselaer Polytechnic Institute Jingye Xu jxu673@gatech.edu School of Industrial and Systems Engineering, Georgia Institute of Technology

Abstract

When branching for binary mixed integer linear programs with disjunctions of sparsity level $2$ , we observe that there exists a finite list of $2$ -sparse disjunctions, such that any other $2$ -sparse disjunction is dominated by one disjunction in this finite list. For sparsity level greater than $2$ , we show that a finite list of disjunctions with this property cannot exist. This leads to the definition of covering number for a list of splits disjunctions. Given a finite list of split sets $\mathcal{F}$ of $k$ -sparsity, and a given $k$ -sparse split set $S$ , let $\mathcal{F}(S)$ be the minimum number of split sets from the list $\mathcal{F}$ , whose union contains $S\cap[0,\ 1]^{n}$ . Let the covering number of $\mathcal{F}$ be the maximum value of $\mathcal{F}(S)$ over all $k$ -sparse split sets $S$ . We show that the covering number for any finite list of $k$ -sparse split sets is at least $\lfloor k/2\rfloor$ for $k\geq 4$ . We also show that the covering number of the family of $k$ -sparse split sets with coefficients in $\{-1,0,1\}$ is upper bounded by $k-1$ for $k\leq 4$ .

1 Introduction

Land and Doig [19] invented the branch-and-bound procedure to solve mixed integer linear programs (MILP). Today, all state-of-the-art MILP solvers use the branch-and-bound procedure at its core. An important decision in formalizing a branch-and-bound algorithm is to decide the method to partition the feasible region of the linear program corresponding to a node in the branch-and-bound tree. Given $\pi\in\mathbb{Z}^{n}$ and $\eta\in\mathbb{Z}$ , a general way to partition a feasible region where all variables are binary is to the use the following disjunction for $x\in\{0,1\}^{n}$ : $\left(\pi^{\top}x\leq\eta\right)\vee\left(\pi^{\top}x\geq\eta+1\right),$ in order to create two child nodes. The open set

S(\pi,\eta):=\left\{x\in\mathbb{R}^{n}\,|\,\eta<\pi^{\top}x<\eta+1\right\},

is called split set and the associated disjunction is called split disjunction. We say a split set $S(\pi,\eta)$ is $k$ -sparse if the number of non-zero entries of $\pi$ , denoted by $\|\pi\|_{0}$ , is at most $k$ , that is, $\|\pi\|_{0}\leq k.$

Most state-of-the-art MILP solvers are based on branch-and-bound trees built using $1$ -sparse split disjunctions; such branch-and-bound trees are called simple branch-and-bound trees [11]. One rationale for using $1$ -sparse split disjunctions is to maintain the sparsity of linear programs solved at child nodes; see discussion in [12, 13]. Recently, [10, 5] showed that on random instances, using $1$ -sparse split disjunctions is sufficient to obtain a polynomial size branch-and-bound tree when the number of constraints are fixed. However, several papers have shown the power of constructing branch-and-bound trees with dense disjunctions. See, for example, the papers [24, 1, 20, 22, 18, 7, 21, 26, 23] which present several evidences of dramatic reduction in the number of nodes in a branch-and-bound tree when using dense disjunctions in comparison to branch-and-bound trees based on $1$ -sparse disjunctions. Moreover, the papers [17, 6] present examples of MILPs where every $1$ -sparse branching scheme leads to exponential size branch-and-bound trees, although these instances can be solved using polynomial-size branch-and-bound trees when using denser inequalities see [26, 4]. While the worst-case size of a branch-and-tree may be exponential even when using dense disjunctions [8, 11, 14, 16], the papers [25, 4] present other compelling theoretical evidence on the importance of branching using dense disjunctions.

The papers [24, 20, 22, 26] show significant improvement in the size of the branch-and-bound tree by using split disjunctions of a specified sparsity level together which having the coefficients of the associated split sets being in $\{-1,0,1\}$ . One perspective to view this line of work, is that they explore the paradigm of expanding the list of disjunctions used to build the branch-and-bound tree, from the typically used $1$ -sparse disjunctions, to a finite list of pre-specified denser disjunctions. In this paper, we explore a geometric problem motivated by the use of such pre-specified finite lists of dense disjunctions to solve binary MILPs.

2 Main results

2.1 Dominance result for $2$ -sparse disjunctions

Consider two split sets $S(\pi^{1},\eta^{1})$ and $S(\pi^{0},\eta^{0})$ in $\mathbb{R}^{n}$ . We say that $S(\pi^{1},\eta^{1})$ dominates $S(\pi^{0},\eta^{0})$ if

\displaystyle S(\pi^{1},\eta^{1})\cap[0,1]^{n}\supseteq S(\pi^{0},\eta^{0})\cap[0,1]^{n}.

(1)

If (1) holds, then in any branch-and-bound tree that solves a binary MILP using the disjunction corresponding to $S(\pi^{0},\eta^{0})$ , we may replace this disjunction by the disjunction corresponding to $S(\pi^{1},\eta^{1})$ , resulting in a branch-and-bound tree that cannot increase in size in comparison to the original branch-and-bound tree.

Let $\mathcal{F}_{k}$ be the finite list of $k$ -sparse split sets, such that $S(\pi,\eta)\in\mathcal{F}_{k}$ if $\pi\in\{-1,0,1\}^{n}$ , $\|\pi\|_{0}\leq k$ , and $\eta\in\{-k,\dots,-1,0,1,\dots,k\}$ . If we only use $1$ -sparse disjunctions, then clearly there are only $n$ possible split sets in $\mathcal{F}_{1}$ , none of which dominate each other. Next let us consider the case of $2$ -sparse disjunctions.

Proposition 1.

Consider any arbitrary $2$ -sparse split set $S(\pi,\eta)\subseteq\mathbb{R}^{n}$ , that is, $\pi\in\mathbb{Z}^{n}$ and $\|\pi\|_{0}\leq 2$ . Then there exists a split set in $\mathcal{F}_{2}$ that dominates $S(\pi,\eta)$ .

Proposition 1 shows that if one decides to branch using $2$ -sparse disjunctions only for solving binary MILPs, there is no reason to use general $2$ -sparse split disjunctions – in particular, one may restrict the use of disjunctions to the finite list described in Proposition 1. Indeed, the paper [26] shows the importance of branching using $2$ -sparse disjunctions by employing exactly the split sets described in Proposition 1 and shows significant improvement over sizes of tree constructed using the $1$ -sparse disjunctions. See Section 4 for a proof of Proposition 1.

Generalizing the result of Proposition 1, we would like to fix the level of sparsity $k$ of the split disjunctions used to build branch-and-bound tree and ask the question: Does there exist a finite list of $k$ -sparse disjunctions, such that it is sufficient to restrict attention to this finite list in order to get the full power of branching with $k$ -sparse disjunctions. Unfortunately, as shown in the next result, such finite lists do not exist for $k$ -sparse disjunctions with $k\geq 3$ .

Theorem 2.

Let $k\geq 3$ . There does not exist any finite list $\mathcal{F}$ of $k$ -sparse split sets such that any arbitrary $k$ -sparse split set is dominated by exactly one of the split sets from $\mathcal{F}$ .

This negative result in the context of the use of split disjuctions in branch-and-bound is in striking contrast to the case of cutting planes computed using split sets: for any rational polyhedral there exists a finite list of split sets such that cutting planes derived from an arbitrary split set are dominated by cutting planes derived using one split set from this finite list [2, 3, 9]. See Section 5 for a proof of Theorem 2.

2.2 Lower bound on covering number for general finite list of dense disjunctions with $k\geq 4$

Given the negative result of Theorem 2, the next natural question to ask is if there exists finite list of $k$ -sparse split sets, such that any other arbitrary $k$ -sparse split set is a subset of an union of a small number of split sets from the list. Formally, given split sets $\{S(\pi^{i},\eta^{i})\}_{i=0}^{p}\subseteq\mathbb{R}^{n}$ , we say that $\{S(\pi^{i},\eta^{i})\}_{i=1}^{p}$ dominates $S(\pi^{0},\eta^{0})$ if:

\displaystyle\left(\bigcup_{i=1}^{p}S(\pi^{i},\eta^{i})\right)\cap[0,1]^{n}\supseteq S(\pi,\eta)\cap[0,1]^{n}.

(2)

If (2) holds, then in any branch-and-bound tree that solves a binary MILP using the disjunction corresponding to $S(\pi^{0},\eta^{0})$ , we may replace this disjunction by the disjunctions corresponding to $\{S(\pi^{i},\eta^{i})\}_{i=0}^{p}$ resulting in a branch-and-bound tree whose size is no more than $2^{p-1}$ times the original branch-and-bound tree.

Definition 1 (Covering number for a finite list of $k$ -sparse split sets).

Let $\mathcal{F}$ be a finite list of $k$ -sparse split sets. Given an arbitrary $k$ -sparse split set $S$ , let $\mathcal{F}(S)$ be the smallest number of split sets from $\mathcal{F}$ that dominates $S$ . We define the covering number of $\mathcal{F}$ , denoted as $C(\mathcal{F})$ , as:

C(\mathcal{F}):=\textup{max}\{\mathcal{F}(S)\,|\,S\ \textup{ is a }k\textup{-sparse split set}\}.

If one can show that a finite list of $k$ -sparse disjunctions has a small covering number, then it could be considered a theoretical justification for using just this finite list of pre-specified $k$ -sparse disjunctions instead of general $k$ -sparse disjunctions.

The covering number of $\mathcal{F}_{1}$ is $k$ , since, for example, in order to dominate the split set $\{x\in\mathbb{R}^{k}\,|\,k-1<\sum_{i=1}^{k}x_{i}<k\}$ we require all the $k$ disjunctions $0<x_{i}<1$ for $i\in\{1,\ldots,k\}$ . Unfortunately, the next result indicates that it is not possible to find a finite list of disjunctions with significantly smaller covering number.

Theorem 3.

Let $\mathcal{F}$ be any finite list of $k$ -sparse split sets. Then $C(\mathcal{F})\geq\left\lfloor\frac{k}{2}\right\rfloor$ .

See Section 6 for a proof of Theorem 3.

2.3 Covering number of $\{-1,0,1\}$ -disjunctions

Finally, since a number of papers have successfully employed the very natural list of disjunctions with coefficients only in $\{-1,0,1\}$ , we explore the covering number of such finite list of disjunctions for sparsity level less or equal than $4$ .

Proposition 4.

For $k=2,3,4$ we have that $C(\mathcal{F}_{k})\leq k-1$ .

See Section 7 for a proof of Proposition 4.

3 Conclusions

The results of this paper justify the use of pre-specified list of disjunctions with coefficients in $\{-1,0,1\}$ for low levels of sparsity. For $k=2$ , Proposition 1 provides this justification. For a branch-and-bound tree using $3$ -sparse disjunctions, Theorem 3 and Proposition 4 imply that any finite list has a covering number of at least $2$ and $\mathcal{F}_{3}$ also has a covering number of $2$ . Thus with respect to covering number, it is optimal to limit the use of disjunctions from $\mathcal{F}_{3}$ . It is an open question if $\mathcal{F}_{k}$ is optimal for higher values of $k$ with respect to covering number. In order to answer this question, results of both Theorem 3 and Proposition 4 may need to be tightened and generalized.

More generally, Theorem 3 may also be an indication that the use of pre-specified list of disjunctions may not be the best way to generate small branch-and-bound trees. While using disjunctions in $\mathcal{F}_{k}$ already produces smaller branch-and-bound trees than those produced using $1$ -sparse disjunctions [24, 22, 15, 26], in order to truly obtain significantly smaller branch-and-bound trees, one may need to further develop and expand on methods to select problem-specific dense disjunctions that are not pre-specified [1, 20, 18, 7, 21, 26, 23].

4 Proof of Proposition 1

In order to prove Proposition 1 ( $k=2$ ) and Proposition 4 ( $k=3,4$ ) in Section 7, we have to show that for any given arbitrary split set $S=\{x\in\mathbb{R}^{k}\,|\,\eta<\pi^{\top}x<\eta+1\}$ at most $k-1$ split sets from $\mathcal{F}_{k}$ are needed to dominate it. Without loss of generality, we may assume that $0\leq\pi_{1}\leq\pi_{2}\leq\ldots\leq\pi_{k}$ . This is because, if $\pi_{i}<0$ we can change $x_{i}$ to $1-x_{i}$ , and then permute the order of the variables. Note that this is fine because $\mathcal{F}_{k}$ is closed under taking the same operations.

Proof for $k=2$ .

We assume $\|\pi\|_{0}=2$ , since otherwise the result is trivial. Let $x\in S\cap[0,1]^{2}$ . We consider the following cases.

•

$0\leq\eta<\eta+1\leq\pi_{1}$ : For $x\in S$ , we have $x_{1}+x_{2}\leq x_{1}+\frac{\pi_{2}}{\pi_{1}}x_{2}=\frac{\pi_{1}x_{1}+\pi_{2}x_{2}}{\pi_{1}}<\frac{\eta+1}{\pi_{1}}\leq 1.$ Since $(0,0)\notin S$ , we obtain $S\cap[0,1]^{2}\subseteq\{x\in[0,1]^{2}\,|\,0<x_{1}+x_{2}<1\}$ .
•

$0\leq\pi_{1}\leq\eta;\ \eta+1\leq\pi_{2}$ : We have $0\leq\frac{\eta-\pi_{1}}{\pi_{2}}\leq\frac{\eta-\pi_{1}x_{1}}{\pi_{2}}<x_{2}$ for $x\in S$ . On the other hand, for $x\in S$ we have $x_{2}\leq\frac{\pi_{1}}{\pi_{2}}x_{1}+x_{2}<\frac{\eta+1}{\pi_{2}}\leq 1$ . Thus, $S\cap[0,1]^{2}\subseteq\{x\in[0,1]^{2}\,|\,0<x_{2}<1\}$ .
•

$0\leq\pi_{1}\leq\eta;\ 0\leq\pi_{2}\leq\eta$ : We have $1<\frac{\pi_{1}}{\eta}x_{1}+\frac{\pi_{2}}{\eta}x_{2}\leq x_{1}+x_{2}$ for $x\in S$ . Since $(1,1)\notin S$ , we obtain $S\cap[0,1]^{2}\subseteq\{x\in[0,1]^{2}\,|\ 1<x_{1}+x_{2}<2\}$ .

∎

5 Proof of Theorem 2

We will prove Theorem 2 for $k=3$ . A similar proof can be given for $k\geq 4$ , but the result in this case is implied by Theorem 3 so we do not consider it in this section.

Proof of Theorem 2.

In order to prove Theorem 2, we show that for the infinite family of split sets

S_{\gamma}=\{x\in\mathbb{R}^{3}\,|\,\gamma<x_{1}+\gamma x_{2}+(\gamma+1)x_{3}<\gamma+1\},

where $\gamma\in\mathbb{Z}_{+},\gamma\geq 1$ , there is no split set in $\mathbb{R}^{3}$ that contains $S_{\gamma}\cap[0,1]^{3}$ for infinitely many values of $\gamma$ . Assume for a contradiction that there exists an split set $S=\{x\in\mathbb{R}^{3}\,|\,\eta<\pi^{T}x<\eta+1\}$ , where $\pi\in(\pi_{1},\pi_{2},\pi_{3})\in\mathbb{Z}^{3},\eta\in\mathbb{Z}$ , such that $S$ dominates $S_{\gamma}$ for infinitely many $\gamma\in\mathbb{Z}_{+}$ , that is,

S_{\gamma}\cap[0,1]^{3}\subseteq S\cap[0,1]^{3}\ \forall\gamma\in\Gamma\quad(\Rightarrow\overline{S}_{\gamma}\cap[0,1]^{3}\subseteq\overline{S}\cap[0,1]^{3},\ \forall\gamma\in\Gamma),

(3)

where $\overline{S}_{\gamma}$ and $\overline{S}$ are closure of ${S}_{\gamma}$ and ${S}$ respectively, and $\Gamma\subseteq\mathbb{Z}_{+}$ is an infinite set.

We first show that:

	$\displaystyle H^{0}:=\{x\in[0,1]^{3}\,\|\,\pi^{T}x\leq\eta\}$	$\displaystyle\subseteq\{x\in[0,1]^{3}\,\|\,x_{1}+\gamma x_{2}+(\gamma+1)x_{3}\leq\gamma\}=:H_{\gamma}^{0},$		(4)
	$\displaystyle H^{1}:=\{x\in[0,1]^{3}\,\|\,\pi^{T}x\geq\eta+1\}$	$\displaystyle\subseteq\{x\in[0,1]^{3}\,\|\,x_{1}+\gamma x_{2}+(\gamma+1)x_{3}\geq\gamma+1\}=:H_{\gamma}^{1}.$		(5)

Notice that we must have $H_{\gamma}^{j}\cap\{0,1\}^{3}\neq\emptyset$ for all $j\in\{0,1\}$ ; otherwise if, for instance $H_{\gamma}^{0}\cap\{0,1\}^{3}=\emptyset$ , then we would have $\{0,1\}^{3}\subseteq H_{\gamma}^{1}$ which implies $[0,1]^{3}\subseteq H_{\gamma}^{1}$ , a contradiction with the fact $S_{\gamma}\cap[0,1]^{3}\neq\emptyset$ . On the other hand, observe that $S_{\gamma}$ being dominated by $S$ is equivalent to: for all $i\in\{0,1\}$ there exists $j\in\{0,1\}$ such that $H^{i}\subseteq H_{\gamma}^{j}$ . Since $(H^{0}\cup H^{1})\cap\{0,1\}^{3}=\{0,1\}^{3}$ , we conclude that it cannot happen that $H^{0}\subseteq H_{\gamma}^{j}$ and $H^{1}\subseteq H_{\gamma}^{j}$ for the same $j$ since $H_{\gamma}^{i}\cap\{0,1\}^{3}\neq\emptyset$ for $i\neq j$ . Therefore, (4) and (5) hold (we may assume that we have $H^{i}\subseteq H_{\gamma}^{i}$ for $i=\in\{0,1\}$ by considering $S$ to be defined by $\hat{\pi}=-\pi$ and $\hat{\eta}=-\eta-1$ if necessary).

Since $(0,1,0)$ satisfies the equation $x_{1}+\gamma x_{2}+(\gamma+1)x_{3}=\gamma$ and by (4) we have $H^{0}\cap\{0,1\}^{3}=H_{\gamma}^{0}\cap\{0,1\}^{3}$ , we must have that $(0,1,0)$ satisfies the inequality $\pi^{T}x\leq\eta$ . We now show that $(0,1,0)$ must satisfy $\pi^{T}x=\eta$ . Assume for a contradiction that it satisfies $\pi^{T}x<\eta$ . Let $x_{0}\in S_{\gamma}\cap[0,1]^{3}$ be an arbitrary point. For $\lambda>0$ small enough we have that the point $x_{\lambda}=(0,1,0)+\lambda(x_{0}-(0,1,0))$ satisfies $\pi^{T}x_{\lambda}<\eta$ and, by convexity of $\overline{S}_{\gamma}\cap[0,1]^{3}$ , that $x_{\lambda}\in S_{\gamma}\cap[0,1]^{3}$ . Since $S_{\gamma}\cap[0,1]^{3}\subseteq S\cap[0,1]^{3}$ , it follows that $x_{\lambda}\in S\cap[0,1]^{3}$ , a contradiction with the fact that $\pi^{T}x_{\lambda}<\eta$ . Thus, we must have that $\pi_{2}=\eta$ . By a similar argument, since $(1,1,0)$ and $(0,0,1)$ satisfy $x_{1}+\gamma x_{2}+(\gamma+1)x_{3}=\gamma+1$ , it follows from (5) that we must have that these points satisfy $\pi^{T}x=\eta+1$ , and therefore $\pi_{1}+\pi_{2}=\eta+1$ and $\pi_{3}=\eta+1$ . Therefore, we obtain that $\pi_{1}=1$ , $\pi_{2}=\eta$ and $\pi_{3}=\eta+1$ .

Since $(1,0,\gamma/(\gamma+1))\in\overline{S}_{\gamma}\cap[0,1]^{3}$ , by (3) we obtain $1+(\eta+1)\frac{\gamma}{\gamma+1}\leq\eta+1\Leftrightarrow 1\leq\frac{\eta+1}{\gamma+1}.$ Since this inequality holds for any $\gamma\in\Gamma$ , we obtain $1\leq 0$ , a contradiction.

∎

6 Proof of Theorem 3

We first prove the result when the sparsity level is an even positive integer $2k$ .

Consider the following family of split sets parameterized by a positive integer $\theta$ :

\displaystyle\begin{array}[]{rl}&S^{\theta}=\left\{(x,y)\in\mathbb{R}^{k}\times\mathbb{R}^{k}\,\left|\,\sum\limits_{i=1}^{k}\theta^{i}<\sum\limits_{i=1}^{k}\theta^{i}(x_{i}+y_{i})<1+\sum\limits_{i=1}^{k}\theta^{i}\right.\right\}.\end{array}

In order to prove Theorem 3 it is sufficient to prove the following result:

Lemma 1.

For every finite collection of split sets $\mathcal{F}$ , there exists $\theta\in\mathbb{Z}_{+}$ with $\theta\geq 1$ , such that one needs at least an union of $k$ split sets from $\mathcal{F}$ to dominate $S^{\theta}\cap[0,1]^{2k}$ .

Before presenting the proof of Lemma 1, we introduce some notation. Consider a list of split sets $A^{(i)}:=\left\{(x,y)\,|\,c^{(i)}<a^{(i)}x+b^{(i)}y^{(i)}<c^{(i)}+1\right\},\ \textup{for}\ i=1,\ldots,p.$ For each $i$ , we denote the two connected components of the complement set to $A^{(i)}$ by

A^{(i)}_{0}:=\left\{(x,y)\,|\,a^{(i)}x+b^{(i)}y^{(i)}\leq c^{(i)}\right\}\quad\textup{and}\quad A^{(i)}_{1}:=\left\{(x,y)\,|\,a^{(i)}x+b^{(i)}y^{(i)}\geq c^{(i)}+1\right\}.

Given a binary vector $u\in\{0,1\}^{p}$ , we further define $A_{u}=\bigcap_{i=1}^{p}A^{(i)}_{u_{i}}$ . Note that the fact that $\bigcup_{i=1}^{p}A^{(i)}$ dominates $S^{\theta}$ can be written as:

S^{\theta}\cap[0,1]^{2k}\subseteq\left(\bigcup_{i\in[g]}A^{(i)}\right)\cap[0,1]^{2k}\ \Leftrightarrow\ [0,1]^{2k}\setminus S^{\theta}\supseteq[0,1]^{2k}\setminus\left(\bigcup_{i\in[g]}A^{(i)}\right).

So dominance of the given list of split sets is equivalent to:

\forall u\in\{0,1\}^{p},\text{ either }A_{u}\cap[0,1]^{2k}\subseteq S^{\theta}_{0}\cap[0,1]^{2k}\text{ or }A_{u}\cap[0,1]^{2k}\subseteq S^{\theta}_{1}\cap[0,1]^{2k}.

(6)

Now we present a proof of Lemma 1.

Proof.

We argue by contradiction. Suppose one needs at most $k-1$ split sets from $\mathcal{F}$ to dominate $S^{\theta}$ for all $\theta$ . Since $\mathcal{F}$ is finite, but there are infinitely many choices of $S^{\theta}$ , there must exist $p$ split sets from $\mathcal{F}$ , where $p\leq k-1$ , and an infinite set $\Theta\subseteq\mathbb{Z}_{+}$ such that those $p$ split sets dominate $S^{\theta}\cap[0,1]^{2k}$ for all $\theta\in\Theta$ . We denote those split sets by

\displaystyle A^{(i)}:=\left\{(x,y)\,|\,c^{(i)}<a^{(i)}x+b^{(i)}y^{(i)}<c^{(i)}+1\right\},\ \textup{for}\ i=1,\ldots,p.

We will show that $(\ref{key_relationship})$ fails for sufficiently large $\theta\in\Theta$ . Our main idea is to construct a certain point $z\in[0,1]^{2k}$ such that $z\in A_{u}\cap[0,1]^{2k}$ for some $u$ but $z$ violates $(\ref{key_relationship})$ .

Consider the following linear system:

	$\displaystyle a^{(i)}x+b^{(i)}y=0\$	$\displaystyle\quad\textup{for}\ i=1,\ldots,p$		(7)
	$\displaystyle y_{i}=0\$	$\displaystyle\quad\textup{for}\ i=1,\ldots,k.$		(8)

This linear system has $2k$ variables and $k+p$ constraints. Since $k+p<2k$ it has at least one non-zero solution $(x^{*},y^{*})$ . Without loss of generality, we may assume that $\|{(x^{*},y^{*})}\|_{2}=1$ and $x^{*}_{j}>0$ where $j$ is the largest index $i=1,\ldots,k$ such that $x^{*}_{i}\neq 0$ .

By (8) and for sufficiently large $\theta\in\Theta$ we have that

\displaystyle\sum_{i=1}^{k}\theta^{i}(x_{i}^{*}+y_{i}^{*})=\sum_{i=1}^{j}\theta^{i}x_{i}^{*}>0.

(9)

We now construct a binary vector $(s,t)\in\{0,1\}^{2k}$ in the following way:

s_{i}=0,t_{i}=1\text{ if }x^{*}_{i}\geq 0\quad\textup{and}\quad s_{i}=1,t_{i}=0\text{ if }x^{*}_{i}<0.

Notice that since $(s,t)$ is a integer vector, it must belong to either $A^{(i)}_{0}$ or $A^{(i)}_{1}$ for all $i=1,\ldots,p$ and therefore $(s,t)\in A_{u^{*}}\cap[0,1]^{2k}$ for some $u^{*}$ .

We now verify that $(s,t)+\lambda(x^{*},y^{*})\in A_{u^{*}}\cap[0,1]^{2k}$ for some sufficiently small $\lambda>0$ . Indeed, $(s,t)+\lambda(x^{*},y^{*})$ stays in $A_{u^{*}}$ for any $\lambda>0$ because of (7). On the other hand, $(s,t)+\lambda(x^{*},y^{*})$ stays in $[0,1]^{2k}$ for sufficiently small $\lambda>0$ because $t_{i}+\lambda y^{*}_{i}$ does not change due to (8), components associated to $s_{i}=1$ decrease a little and components associated to $s_{i}=0$ increase a little.

Now observe that $\sum\limits_{i\in[k]}\theta^{i}(s_{i}+t_{i})=\sum\limits_{i\in[k]}\theta^{i}$ and $\sum\limits_{i\in[n]}\theta^{i}(x_{i}^{*}+y_{i}^{*})>0$ by (9), hence we obtain that

\displaystyle\sum\limits_{i=1}^{k}\theta^{i}<\sum\limits_{i=1}^{k}\theta^{i}(s_{i}+\lambda x_{i}^{*}+t_{i}+\lambda y_{i}^{*})<1+\sum\limits_{i=1}^{k}\theta^{i},

for sufficiently small $\lambda>0$ . In other words, for sufficiently small $\lambda>0$ and large enough $\theta\in\Theta$ we have that $(s,t)+\lambda(x^{*},y^{*})\in S^{\theta}\cap(A_{u^{*}}\cap[0,1]^{2k})$ . We conclude that $(\ref{key_relationship})$ is not satisfied for the point $(s,t)+\lambda(x^{*},y^{*})$ , a contradiction. ∎

In order to prove Theorem 3 for odd sparsity levels of split disjunctions, a similar proof can be presented using the family of split sets:

\displaystyle\begin{array}[]{rl}&S^{\theta}:=\left\{(x,y)\in\mathbb{R}^{k}\times\mathbb{R}^{k+1}:\sum\limits_{i=1}^{k}\theta^{i}<\sum\limits_{i=1}^{k}\theta^{i}(x_{i}+y_{i})+y_{k+1}<1+\sum\limits_{i=1}^{k}\theta^{i}\right\}.\end{array}

7 Proof of Proposition 4

The case $k=2$ is proven in Proposition 1. We now consider the cases $k=3,4$ .

Proof for $k=3$ .

Let $S=\{x\in\mathbb{R}^{3}\,|\,\eta<\pi^{\top}x<\eta+1\}$ , recall that we may assume that $0\leq\pi_{1}\leq\pi_{2}\leq\pi_{3}$ (see Section 4). We have to show that at most $2$ split sets from $\mathcal{F}_{3}$ are needed to dominate it. There are three cases:

•

$\pi_{3}\geq\eta+1$ : In this case, observe that $x\in S$ , implies that $x_{3}<1$ . By Proposition 1, we know there exist $1$ split set of sparsity $2$ (or lesser) from $\mathcal{F}_{3}$ whose union contains the set $\{x\in[0,1]^{3}\,|\,x\in S,x_{3}=0\}$ . The set of points in $\{x\in[0,1]^{3}\,|\,x\in S,0<x_{3}<1\}$ is contained in the split set $0<x_{3}<1$ .
•

$\pi_{1}+\pi_{2}\leq\eta$ : In this case, observe that $x\in S$ , implies that $x_{3}>0$ . By Proposition 1, we know there exist $1$ split set of sparsity $2$ (or lesser) from $\mathcal{F}_{3}$ whose union contains the set $\{x\in[0,1]^{3}\,|\,x\in S,x_{3}=1\}$ . The set of points in $\{x\in[0,1]^{3}\,|\,x\in S,0<x_{3}<1\}$ is contained in the split set $0<x_{3}<1$ .
•

$\pi_{3}\leq\eta$ and $\pi_{1}+\pi_{2}\geq\eta+1$ : Since $\pi_{1}\leq\pi_{2}\leq\pi_{3}\leq\eta$ , if $\sum_{j=1}^{3}x_{j}\leq 1$ , then $\sum_{j=1}^{3}\pi_{j}x_{j}\leq\eta$ . Thus, $x\in S$ implies that $\sum_{j=1}^{3}x_{j}>1$ . Moreover, if $\sum_{j=1}^{3}x_{j}\geq 2$ , then $\sum_{j=1}^{3}\pi_{j}x_{j}\geq\pi_{1}+\pi_{2}\geq\eta+1$ . Thus $x\in S$ implies that $\sum_{j=1}^{3}x_{j}<2$ . Therefore, $S$ is dominated by the split set $\{x\in\mathbb{R}^{3}\,|\,1<\sum_{j=1}^{3}x_{j}<2\}$ .

∎

Proof for $k=4$ .

Let $S=\{x\in\mathbb{R}^{4}\,|\,\eta<\pi^{\top}x<\eta+1\}$ with $0\leq\pi_{1}\leq\pi_{2}\leq\pi_{3}\leq\pi_{4}$ . There are ten cases:

•

$\pi_{4}\geq\eta+1$ : In this case, observe that $x\in S$ , implies that $x_{4}<1$ . By Proposition 4 for $k=3$ case, we know there at most $2$ split set of sparsity $3$ (or lesser) from $\mathcal{F}_{4}$ whose union contains the set $\{x\in[0,1]^{4}\,|\,x\in S,x_{4}=0\}$ . The set of points in $\{x\in[0,1]^{4}\,|\,x\in S,0<x_{4}<1\}$ is contained in the split set $0<x_{4}<1$ .
•

$\pi_{1}+\pi_{2}+\pi_{3}\leq\eta$ : In this case, observe that $x\in S$ , implies that $x_{4}>0$ . By Proposition 4 for $k=3$ , we know there at most $2$ split set of sparsity $3$ (or lesser) from $\mathcal{F}_{4}$ whose union contains the set $\{x\in[0,1]^{4}\,|\,x\in S,x_{4}=1\}$ . The set of points in $\{x\in[0,1]^{4}\,|\,x\in S,0<x_{4}<1\}$ is contained in the split set $0<x_{4}<1$ .
•

$\pi_{1}+\pi_{2}\geq\eta+1$ : We may assume that $\pi_{4}\leq\eta$ . Thus, we have $x\in S$ implies $1<x_{1}+x_{2}+x_{3}+x_{4}$ . On the other hand, we also must have $x_{1}+x_{2}+x_{3}+x_{4}<2$ , since otherwise, $\sum_{j=1}^{4}\pi_{j}x_{j}\geq\pi_{1}+\pi_{2}\geq\eta+1$ . Thus $S\cap[0,1]^{4}$ is contained in $1<x_{1}+x_{2}+x_{2}+x_{3}+x_{4}<2$ .
•
$\pi_{1}+\pi_{3}\geq\eta+1$ : We may assume $\pi_{4}\leq\eta$ and $\pi_{1}+\pi_{2}\leq\eta$ . We claim that $S$ is contained in the union of $1<x_{1}+x_{2}+x_{3}+x_{4}<2$ and $0<x_{3}+x_{4}<1$ . Consider the following cases for $x\in S$ :
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  If $x_{1}+x_{2}\leq 1$ : We claim that that $x_{1}+x_{2}+x_{3}+x_{4}<2$ . Assume by contradiction $x_{1}+x_{2}+x_{3}+x_{4}\geq 2$ . Then we have $x_{3}+x_{4}\geq 1$ and thus $\sum_{j=1}^{4}\pi_{j}x_{j}\geq\pi_{1}\cdot\textup{min}\{1,2-x_{3}-x_{4}\}+\pi_{3}\cdot\textup{max}\{1,x_{3}+x_{4}\}\geq\pi_{1}+\pi_{3}\geq\eta+1.$ On the other hand, since $\pi_{4}\leq\eta$ , we have $1<x_{1}+x_{2}+x_{3}+x_{4}$ . Thus, in this case $x$ belongs to $1<x_{1}+x_{2}+x_{2}+x_{3}+x_{4}<2$ .
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  If $x_{1}+x_{2}>1$ : Then note that $x_{3}+x_{4}<1$ , since otherwise $\sum_{j=1}^{4}\pi_{j}x_{j}\geq\pi_{1}+\pi_{3}\geq\eta+1$ . Also note that if $x_{3}+x_{4}=0$ , then $\sum_{j=1}^{4}\pi_{j}x_{j}\leq\pi_{1}+\pi_{2}\leq\eta$ . Thus, in this case we have that $x$ belongs $0<x_{3}+x_{4}<1$ .
•

$\pi_{3}+\pi_{4}\leq\eta$ : We assume that $\pi_{1}+\pi_{2}+\pi_{3}\geq\eta+1$ . First, note that $S$ is contained in $x_{1}+x_{2}+x_{3}+x_{4}>2$ . Also, since $\pi_{1}+\pi_{2}+\pi_{3}\geq\eta+1$ , we have that $S$ is contained in $x_{1}+x_{2}+x_{3}+x_{4}<3$ . Thus, $S$ is contained in $2<x_{1}+x_{2}+x_{3}+x_{4}<3$ .
•
$\pi_{2}+\pi_{4}\leq\eta$ : We may assume $\pi_{1}+\pi_{2}+\pi_{3}\geq\eta+1$ and $\pi_{3}+\pi_{4}\geq\eta+1$ . We claim that $S$ is contained in the union of $2<x_{1}+x_{2}+x_{3}+x_{4}<3$ and $1<x_{3}+x_{4}<2$ . Consider the following cases for $x\in S$ :
- –
  
  If $x_{3}+x_{4}\leq 1$ : We claim that that $x_{1}+x_{2}+x_{3}+x_{4}>2$ . Assume by contradiction $x_{1}+x_{2}+x_{3}+x_{4}\leq 2$ . Thus, $\sum_{j=1}^{4}\pi_{j}x_{j}\leq\pi_{2}\cdot\textup{max}\{1,2-x_{3}-x_{4}\}+\pi_{4}\cdot\textup{min}\{1,x_{3}+x_{4}\}\leq\pi_{2}+\pi_{4}\leq\eta.$ On the other hand, since $\pi_{1}+\pi_{2}+\pi_{3}\geq\eta+1$ , we have $3>x_{1}+x_{2}+x_{3}+x_{4}$ . Thus, in this case $x$ belongs to $2<x_{1}+x_{2}+x_{2}+x_{3}+x_{4}<3$ .
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  Also note that $x_{3}+x_{4}=2$ is not possible, since then $\sum_{j}\pi_{j}x_{j}\geq\pi_{3}+\pi_{4}\geq\eta+1$ .
Thus, $S\cap[0,1]^{4}$ is contained in the union of $2<x_{1}+x_{2}+x_{3}+x_{4}<3$ and $1<x_{3}+x_{4}<2$ .
•
$\pi_{2}+\pi_{3}\geq\eta+1$ and $\pi_{1}+\pi_{4}\geq\eta+1$ : We may assume $\pi_{4}\leq\eta$ and $\pi_{1}+\pi_{3}\leq\eta$ . We claim that $S$ is contained in the union of $1<x_{2}+x_{3}+x_{4}<2$ , $0<x_{1}<1$ and $0<x_{4}<1$ . Consider the following cases:
- –
  
  $x_{1}=0$ : In this case, note that because of $\pi_{2}+\pi_{3}\geq\eta+1$ and $\pi_{4}\leq\eta$ , we have that $x$ belongs to $1<x_{2}+x_{3}+x_{4}<2$ .
- –
  
  $0<x_{1}<1$ : In this case, note that $x$ belongs to $0<x_{1}<1$ .
- –
  
  $x_{1}=1$ : Clearly, due to $\pi_{2}+\pi_{3}\geq\eta+1$ we have that $x_{2}+x_{3}+x_{4}<2$ . If $1<x_{2}+x_{3}+x_{4}$ , then $x$ belongs to $1<x_{2}+x_{3}+x_{4}<2$ .
  
  Otherwise suppose, $x_{2}+x_{3}+x_{4}\leq 1$ . We claim that $0<x_{4}<1$ . By contradiction, if $x_{4}=0$ , then note that $\sum_{j}\pi_{j}x_{j}\leq\pi_{1}+\pi_{3}\leq\eta$ . If $x_{4}=1$ , then note that $\sum_{j}\pi_{j}x_{j}\geq\pi_{1}+\pi_{4}\geq\eta+1.$
•
$\pi_{2}+\pi_{3}\leq\eta$ and $\pi_{1}+\pi_{4}\leq\eta$ : We may assume $\pi_{1}+\pi_{2}+\pi_{3}\geq\eta+1$ and $\pi_{2}+\pi_{4}\geq\eta+1$ . We claim that $S$ is contained in the union of $0<x_{2}+x_{3}<1$ , $2<x_{1}+x_{2}+x_{3}<3$ and $0<x_{4}<1$ . Consider the following cases:
- –
  
  $x_{4}=0$ : In case, note that due to $\pi_{1}+\pi_{2}+\pi_{3}\geq\eta+1$ , we have that $x_{1}+x_{2}+x_{3}<3$ . Also, since $\pi_{2}+\pi_{3}\leq\eta$ , we have that $x_{1}+x_{2}+x_{3}>2$ . Thus, $x$ is contained in $2<x_{1}+x_{2}+x_{3}<3$ .
- –
  
  $0<x_{4}<1$ : In this case, note that $x$ belongs to $0<x_{4}<1$ .
- –
  
  $x_{4}=1$ : In this case note that $x_{2}+x_{3}\geq 1$ is not possible, since $\sum_{j}\pi_{j}x_{j}\geq\pi_{2}+\pi_{4}\geq\eta+1$ . Also note that $x_{2}+x_{3}=0$ is not possible, since that $\sum_{j}\pi_{j}x_{j}\leq\pi_{1}+\pi_{4}\leq\eta$ . Thus, in this case, $x$ belongs to $0<x_{2}+x_{3}<1$ .
•
$\pi_{2}+\pi_{3}\geq\eta+1$ and $\pi_{1}+\pi_{4}\leq\eta$ : We may assume $\pi_{1}+\pi_{2}+\pi_{3}\geq\eta+1$ , $\pi_{1}+\pi_{3}\leq\eta$ and $\pi_{4}\leq\eta$ . We claim that $S$ is contained in the union of $1<x_{2}+x_{3}+x_{4}<2$ , $0<x_{1}<1$ , and $0<x_{2}+x_{3}<1$ . Consider the following cases:
- –
  
  $x_{1}=0$ : In this case, note that because of $\pi_{2}+\pi_{3}\geq\eta+1$ and $\pi_{4}\leq\eta$ , we have that $x$ belongs to $1<x_{2}+x_{3}+x_{4}<2$ .
- –
  
  $0<x_{1}<1$ : In this case, note that $x$ belongs to $0<x_{1}<1$ .
- –
  
  $x_{1}=1$ : Clearly, due to $\pi_{2}+\pi_{3}\geq\eta+1$ we have that $x_{2}+x_{3}+x_{4}<2$ . If $1<x_{2}+x_{3}+x_{4}$ , then $x$ belongs to $1<x_{2}+x_{3}+x_{4}<2$ .
  
  Otherwise suppose, $x_{2}+x_{3}+x_{4}\leq 1$ . In this case, note that $x_{2}+x_{3}=1$ is not possible, since that $x_{4}=0$ and we have $\sum_{j}\pi_{j}x_{j}\leq\pi_{1}+\pi_{3}\leq\eta$ . Also note that $x_{2}+x_{3}=0$ is not possible, since that $\sum_{j}\pi_{j}x_{j}\leq\pi_{1}+\pi_{4}\leq\eta$ . Thus, in this case, $x$ belongs to $0<x_{2}+x_{3}<1$ .
•
$\pi_{2}+\pi_{3}\leq\eta$ and $\pi_{1}+\pi_{4}\geq\eta+1$ : We may assume $\pi_{1}+\pi_{2}+\pi_{3}\geq\eta+1$ and $\pi_{4}\leq\eta$ . We claim that $S$ is contained in the union of $1<x_{1}+x_{2}+x_{3}+x_{4}<2$ , $2<x_{1}+x_{2}+x_{3}<3$ and $0<x_{4}<1$ . Consider the following cases:
- –
  
  $x_{4}=0$ : In case, note that due to $\pi_{1}+\pi_{2}+\pi_{3}\geq\eta+1$ , we have that $x_{1}+x_{2}+x_{3}<3$ . Also, since $\pi_{2}+\pi_{3}\leq\eta$ , we have that $x_{1}+x_{2}+x_{3}>2$ . Thus, $x$ is contained in $2<x_{1}+x_{2}+x_{3}<3$ .
- –
  
  $0<x_{4}<1$ : In this case, note that $x$ belongs to $0<x_{4}<1$ .
- –
  
  $x_{4}=1$ : In this case note that since $\pi_{1}+\pi_{4}\geq\eta+1$ , we have that $x_{1}+x_{2}+x_{3}+x_{4}<2$ . Also note that since $\pi_{4}\leq\eta$ , we have that $x_{1}+x_{2}+x_{3}+x_{4}>1$ . Thus, in this case, $x$ belongs to $1<x_{1}+x_{2}+x_{3}+x_{4}<2$ .

∎

Acknowledgements

We would like to thank Diego Cifuentes, Amitabh Basu, Antoine Deza, and Lionel Pournin for various discussions. We would also like to thank the support from AFOSR grant # F9550-22-1-0052 and from the ANID grant Fondecyt # 1210348 .

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	$\displaystyle H^{0}:=\{x\in[0,1]^{3}\,\|\,\pi^{T}x\leq\eta\}$	$\displaystyle\subseteq\{x\in[0,1]^{3}\,\|\,x_{1}+\gamma x_{2}+(\gamma+1)x_{3}\leq\gamma\}=:H_{\gamma}^{0},$		(4)
	$\displaystyle H^{1}:=\{x\in[0,1]^{3}\,\|\,\pi^{T}x\geq\eta+1\}$	$\displaystyle\subseteq\{x\in[0,1]^{3}\,\|\,x_{1}+\gamma x_{2}+(\gamma+1)x_{3}\geq\gamma+1\}=:H_{\gamma}^{1}.$		(5)

Branching with a pre-specified finite list of kk-sparse split sets for binary MILPs