Distances between optimal solutions of mixed-integer programs

Abstract

A classic result of Cook et al. (Math. Program. 34:251–264, 1986) bounds the distances between optimal solutions of mixed-integer linear programs and optimal solutions of the corresponding linear relaxations. Their bound is given in terms of the number of variables and a parameter $\varDelta$, which quantifies sub-determinants of the underlying linear inequalities. We show that this distance can be bounded in terms of $\varDelta$ and the number of integer variables rather than the total number of variables. To this end, we make use of a result by Olson (J. Number Theory 1:8–10, 1969) in additive combinatorics and demonstrate how it implies feasibility of certain mixed-integer linear programs. We conjecture that our bound can be improved to a function that only depends on $\varDelta$, in general.

Appendix: A proof of Theorem 3

The original proof of Theorem 3 in [10] uses the notion of group algebras. We provide a variation of the original proof that avoids explicitly defining group algebras.

Let $d,p \in \mathbb {Z}_{\ge 1}$ with p prime and consider the additive group $G := \mathbb {Z}^d/p\mathbb {Z}^d$. We denote the all-zero vector by $\mathbf {0}\in G$, and the standard unit vectors by $e_1,\cdots ,e_d \in G$. Let $r \ge pd -d +1$ and $g_1, \cdots , g_r \in G$. For a group element $g \in G$, we can uniquely write g as $g = \sum _{i=1}^d z_i^g e_i$, where $z_1^g, \cdots , z_d^g \in \{0,\cdots ,p-1\}$, and the coefficient vector of g as

$$\begin{aligned} \alpha (g) := (z_1^g, \cdots , z_d^g). \end{aligned}$$

(2)

Let us first provide some intuition for the proof of Theorem 3. Consider polynomials of the form

$$\begin{aligned} p_{\lambda }(x_1, \cdots , x_d) := \sum _{g \in G} \lambda _g x^{\alpha (g)}, \end{aligned}$$

(3)

where $\lambda \in \mathbb {Z}^G$, $\alpha (g)$ is from (2), and $x^{\alpha (g)} := \prod _{i=1}^d x_i^{z_i^g}$. For $\lambda , \tilde{\lambda } \in \mathbb {Z}^G$, we define the product of $p_{\lambda }$ and $p_{\tilde{\lambda }}$ to be the polynomial

$$\begin{aligned} p_{\lambda }(x) p_{\tilde{\lambda }}(x) := \sum _{g_1, g_2 \in G} \lambda _{g_1}\tilde{\lambda }_{g_2} x^{\alpha (g_1) + \alpha (g_2) \ {\mathrm {mod}}\ p}. \end{aligned}$$

(4)

Notice that the exponents are taken modulo p, rather than normal polynomial multiplication. This extra condition on polynomial multiplication reflects the group behavior of G. We illustrate the polynomial multiplication in (4) with the following example.

Example 2

Let $g_1, g_2 \in G \setminus \{\mathbf {0}\}$ and set $g_0 := \mathbf {0}$. Using the notation introduced in (2),

$$\begin{aligned} \alpha (g_0) = (0, \cdots , 0) \quad \text { and } \quad \alpha (g_i) = (z_1^{g_i} \cdots , z_d^{g_j}) ~~ \forall ~ i \in \{1,2\}. \end{aligned}$$

For $i \in \{0,1,2\}$, define $\lambda ^i \in \mathbb {Z}^G$ component-wise to be

$$\begin{aligned} \lambda ^i_h := {\left\{ \begin{array}{ll} 1, &{} \text {if } h = g_i\\ 0, &{} \text {else} \end{array}\right. }. \end{aligned}$$

(5)

The product $(p_{\lambda ^0} - p_{\lambda ^1}) ( p_{\lambda ^0} - p_{\lambda ^2})$ evaluates to

$$\begin{aligned}&(p_{\lambda ^0} - p_{\lambda ^1}) ( p_{\lambda ^0} - p_{\lambda ^2})(x) = \left( 1 - x^{\alpha (g_1)} \right) \left( 1 -x^{\alpha (g_2)} \right) \nonumber \\&\quad = 1- x^{\alpha (g_1)} - x^{\alpha (g_2)} +x^{\alpha (g_1)+\alpha (g_2)\ {\mathrm {mod}}\ p}. \end{aligned}$$

(6)

Now, note that

$$\begin{aligned} \alpha (g_1 + g_2) = \sum _{i=1}^d (z_i^{g_1}+z_i^{g_2} \ {\mathrm {mod}}\ p) e_i, \end{aligned}$$

so $\alpha (g_1+g_2) = \alpha (g_1)+\alpha (g_2)\ {\mathrm {mod}}\ p$. Therefore, (6) becomes

$$\begin{aligned} (p_{\lambda ^0} - p_{\lambda ^1}) ( p_{\lambda ^0} - p_{\lambda ^2}) (x) = 1- x^{\alpha (g_1)} - x^{\alpha (g_2)} +x^{\alpha (g_1+g_2)}. \end{aligned}$$

(7)

$\diamond$

Observe that (7) encodes information about $\mathbf {0}, g_1, g_2$, and $g_1 + g_2$, that is, (7) encodes information about the partial sums of the sequence $g_1, g_2$. Therefore, Example 2 shows that multiplying suitably chosen polynomials yields information about partial sums of elements in G. More generally, if we are given a sequence $g_1, \dots , g_r \in G$, then we can determine information about the partial sums of $g_1, \dots , g_r$ by looking at the coefficients of the polynomial $(p_{\lambda ^0} - p_{\lambda ^1}) \cdot (p_{\lambda ^0} - p_{\lambda ^2}) \cdot \cdots \cdot (p_{\lambda ^0} - p_{\lambda ^r})$.

This relationship between polynomial multiplication and partial sums is the main idea behind the proof of Theorem 3. Upon closer inspection, it can be seen that the partial sum information can also be determined by looking only at a discrete convolution of the vectors of coefficients $(\lambda _g)_{g \in G}$, which can be viewed as functions from G to $\mathbb {Z}$. Therefore, we formally prove Theorem 3 by examining convolutions of functions from G to $\mathbb {Z}$. For $f_1,f_2 : G \rightarrow \mathbb {Z}$, the convolution of $f_1$ and $f_2$ is the function $f_1 \otimes f_2 : G \rightarrow \mathbb {Z}$ defined by

$$\begin{aligned} (f_1 \otimes f_2)(g) = \sum _{h \in G} f_1(h) f_2(g - h) \end{aligned}$$

for $g \in G$. This operation is associative, commutative, and distributive, and satisfies

$$\begin{aligned} (f_1 \otimes \cdots \otimes f_k)(g) = \sum _{\begin{array}{c} (g_1,\cdots ,g_k) \in G^k,\\ g_1+\cdots +g_k = g \end{array}} \prod _{j=1}^k f_j(g_j). \end{aligned}$$

(8)

For $g \in G$, define $\chi _g : G \rightarrow \mathbb {Z}$ by $\chi _g(g) = 1$ and $\chi _g(h) = 0$ for all $h \in G \setminus \{g\}$.

Lemma 4

Let $r \in \mathbb {Z}_{\ge 1}$ and $g_1,\cdots ,g_r \in G$, and consider the function $\pi := (\chi _{\mathbf {0}} - \chi _{g_1}) \otimes \dots \otimes (\chi _{\mathbf {0}} - \chi _{g_r})$. Then $\pi$ can be written as a finite sum of functions of the form

$$\begin{aligned} f \otimes \bigotimes _{i=1}^d \bigotimes _{j=1}^{t_i} (\chi _\mathbf {0}- \chi _{e_i}) \end{aligned}$$

(9)

for some $f : G \rightarrow \mathbb {Z}$ and $t_1,\cdots ,t_d \in \mathbb {Z}_{\ge 0}$ with $\sum _{i=1}^d t_i = r$.

Proof

Since $\otimes$ is associative, commutative, and distributive, it suffices to prove the claim for $r = 1$. Recall from (2) that every element $h \in G$ has unique numbers $z_1^h,\cdots ,z_d^h \in \{0,\cdots ,p-1\}$ such that $h = \sum _{i=1}^d z_i^h e_i$. Thus, we can define $w(h) := z_1^h + \cdots + z_d^h$. We prove the claim via induction over $w(g_1)$. If $w(g_1) = 0$, then $g_1 = \mathbf {0}$ and $\pi \equiv 0$, which satisfies the claim by choosing the empty sum. If $w(g_1) \ge 1$, then there exists some $i \in [ d ]$ with $z_i^{g_1} \ge 1$. It is straightforward to verify that

$$\begin{aligned} \chi _\mathbf {0}- \chi _{g_1} = \chi _\mathbf {0}\otimes (\chi _\mathbf {0}- \chi _{e_i}) + \chi _g \otimes (\chi _\mathbf {0}- \chi _{g_1 - e_i}). \end{aligned}$$

Since $w(g_1 - e_i) = w(g_1) - 1$, the latter term has a representation as in (9), and therefore, so does $\pi = \chi _\mathbf {0}- \chi _{g_1}$. $\square$

Lemma 5

(Olson (1969), see [10, Theorem 1]) Let r, $g_1,\cdots ,g_r$, and $\pi$ be defined as in Lemma 4. If $r \ge pd-d+1$, then $\pi (h) \in p \mathbb {Z}$ for each $h \in G$.

Proof

By Lemma 4, we know that $\pi$ is a finite sum of functions of the form (9). Thus, it suffices to show that every function $\pi '$ of the form

$$\begin{aligned} \pi ' = f \otimes \bigotimes _{i=1}^d \bigotimes _{j=1}^{t_i} (\chi _\mathbf {0}- \chi _{e_i}), \end{aligned}$$

with $\sum _{i=1}^d t_i = r$ satisfies $\pi '(h) \in p \mathbb {Z}$ for every $h \in G$.

Since, $\sum _{i=1}^d t_i = r \ge pd-d+1 > d(p-1)$, there exists some $i \in \{1, \cdots , d\}$ such that $t_i \ge p$. Define $\pi '' := \bigotimes _{j=1}^p (\chi _\mathbf {0}- \chi _{e_i})$, and note that $\pi ' = f' \otimes \pi ''$ for some $f' : G \rightarrow \mathbb {Z}$. Let $h \in G$ and consider $\pi ''(h)$. If $h = \mathbf {0}$, then using (8) it follows that $\pi ''(h) = 2 \in p \mathbb {Z}$ when $p = 2$ and $\pi ''(h) = 0 \in p \mathbb {Z}$ when $p > 2$. Now, suppose that $h \ne \mathbf {0}$. Using (8), it follows that if h is not a multiple of $e_i$, then $\pi ''(h) = 0 \in p \mathbb {Z}$. In the case that if h is a multiple of $e_i$, say $h = k e_i$ for some $k \in \{1,\cdots ,p - 1\}$, one can verify that $\pi ''(h) = (-1)^k \left( {\begin{array}{c}p\\ k\end{array}}\right)$ holds. Since p divides $\left( {\begin{array}{c}p\\ k\end{array}}\right)$, we have that $\pi ''(h) \in p \mathbb {Z}$. Thus, we obtain that $\pi ''(h) \in p \mathbb {Z}$ holds for all $h \in G$, which implies that $\pi '(h) = (f' \otimes \pi '')(h) \in p \mathbb {Z}$ also holds for all $h \in G$. $\square$

Proof of Theorem 3

Consider the function $\pi$ as defined in Lemma 4. By Lemma 5, $\pi (\mathbf {0}) \in p \mathbb {Z}$. Since $p \ge 2$, it follows that $\pi (\mathbf {0}) \ne 1$. Express $\pi (\mathbf {0})$ using the sum in (8) and note that one of the summands is

$$\begin{aligned} \prod _{i=1}^r (\chi _{\mathbf {0}} - \chi _{g_i})(\mathbf {0}) = \prod _{i=1}^r 1 = 1. \end{aligned}$$

Since $\pi (\mathbf {0}) \ne 1$, there must exist another non-zero summand. This means that there exist $h_1,\cdots ,h_r \in G$ not all zero with $h_1+\cdots +h_r = \mathbf {0}$ such that

$$\begin{aligned} \prod _{i=1}^r (\chi _{\mathbf {0}} - \chi _{g_i})(h_i) \ne 0. \end{aligned}$$

For each $i \in \{1, \dots , r\}$, the function $\chi _{\mathbf {0}} - \chi _{g_i}$ is zero everywhere except at the points $\mathbf {0}$ and $g_i$. So, the latter implies that $h_i \in \{\mathbf {0}, g_i \}$ for each $i \in \{ 1,\cdots ,r \}$. Defining $I := \{i \in \{1, \dots , r\} : h_i = g_i \}$, we obtain

$$\begin{aligned} \sum _{i \in I} g_i = \sum _{i \in I} h_i = \sum _{i=1}^r h_i = \mathbf {0}. \end{aligned}$$

Note that I is not empty since not all $h_i$ are zero. $\square$