The Integrality Number of an Integer Program

Abstract

We introduce the integrality number of an integer program (IP) in inequality form. Roughly speaking, the integrality number is the smallest number of integer constraints needed to solve an IP via a mixed integer (MIP) relaxation. One notable property of this number is its invariance under unimodular transformations of the constraint matrix. Considering the largest minor $\varDelta$ of the constraint matrix, our analysis allows us to make statements of the following form: there exist numbers $\tau (\varDelta )$ and $\kappa (\varDelta )$ such that an IP with $n\ge \tau (\varDelta )$ many variables and $n + \kappa (\varDelta )\cdot \sqrt{n}$ many inequality constraints can be solved via a MIP relaxation with fewer than n integer constraints. A special instance of our results shows that IPs defined by only n constraints can be solved via a MIP relaxation with $O(\sqrt{\varDelta })$ many integer constraints.

Appendix

Proof

(of Lemma 2). Let $W \in \mathbb {Z}^{k\times n}$ be a matrix with minimal k such that all vertices of $W\!\text {-}\text {MIP}_{A,c}(b)$ are integral. Set $\overline{A} := AU$, $\overline{c} := c^\intercal U$, and $\overline{W} := WU$. The matrix $U^{-1}$ maps the vertices of $W\!\text {-}\text {MIP}_{A,c}(b)$ to those of $\overline{W}\!\text {-}\text {MIP}_{\overline{A}, \overline{c}}(b)$, and $U^{-1}$ maps $\mathbb {Z}^n$ to $\mathbb {Z}^n$. Thus, $\overline{W} \in \mathbb {Z}^{k\times n}$ and the vertices of $\overline{W}\!\text {-}\text {MIP}_{\overline{A}, \overline{c}}(b)$ are integral. Hence, $i_{A}(b) \ge i_{\overline{A}}(b)$. To see why the reverse inequality holds, it is enough to notice that $U^{-1}$ is also unimodular.    $\square$

Proof

(of Lemma 4). By Ghouila-Houri (see, e.g.,[17, §19]) it is enough to show

$$\begin{aligned} \textstyle y := \sum _{w \in \widehat{W} \cap {W}^B} -w ~ + \sum _{w \in \widehat{W} \cap {W}^T} w \in \{-1,0,1\}^{n} \end{aligned}$$

for a subset $\widehat{W}$ of the rows of W. Recall that every column u of $C = (B ~~ T ) W$ can be written as $u = v+t$ for some $v \in B$ and $t \in T \cup \{0\}$. Hence, a column of W has at most two non-zero entries, where a non-zero entry equals 1. One of these entries is in the rows of $W^B$ while the other is in $W^T$. This shows $y \in \{-1,0,1\}^n$.    $\square$

Proof

(of Lemma 5). Let z be a column of $A_I^1$. If $z \in \{z^1, \dotsc , z^{\ell }\}$, then $z = z+0 \in B+(T \cup \{0\})$. Else, $z \in \{0,...,\alpha _1-1\} \times ... \times \{0,...,\alpha _\ell -1\}$. Define

$$\begin{aligned} t_i = \bigg \lfloor \frac{z_i}{k_i+1} \bigg \rfloor \cdot (k_i+1) \text { and } v_i = z_i -t_i \text { for all } i \in \{0,\ldots ,\ell \}. \end{aligned}$$

We have $v_i \in \{0,\ldots ,k_i\}$ for each $i \in \{0,\ldots ,\ell \}$, so $v := (v_0, \ldots , v_\ell )^\intercal \in B$. To see that $(t_1,\ldots ,t_\ell )^\intercal \in T$ note that $t_i \le \beta _i\cdot (k_i+1)$ for each $i \in \{1,\ldots , \ell \}$.

It is left to choose $k_1, \dotsc , k_{\ell }$ such that $|B|+|T| \le 6{\delta }^{1/2}+\log _2({\delta })$. Note $\ell \le \log _2({\delta })$ as $\alpha _1, \dotsc , \alpha _{\ell } \ge 2$. By permuting rows and columns we assume $\alpha _1 \le \alpha _2\le \ldots \le \alpha _{\ell }$. We consider two cases.

Case 1. Assume $\alpha _\ell = {\delta }^\tau$ for $\tau \ge 1/2$. This implies $\prod _{i=1}^{\ell -1} \alpha _i = {\delta }^{1 - \tau } \le {\delta }^{1/2}$. Let $\sigma \ge 0$ such that $1-\tau + \sigma = 1/2$. For each $i \in \{1, \dotsc , \ell -1\}$ define $k_i := \alpha _i-1$ and set $k_{\ell } := \lceil {\delta }^\sigma \rceil$. The value $\beta _{\ell }$ in (9) satisfies

$$ \beta _{\ell } = \bigg \lfloor \frac{\alpha _{\ell }-1}{\big \lceil {\delta }^\sigma \big \rceil +1} \bigg \rfloor = \bigg \lfloor \frac{{\delta }^\tau -1}{\big \lceil {\delta }^\sigma \big \rceil +1} \bigg \rfloor \le \big \lceil {\delta }^{1/2}\big \rceil \le {\delta }^{1/2} + 1. $$

Define $B = \overline{B} \cup \{z^1, \dotsc , z^{\ell }\}$, where $\overline{B} := \{0, \dotsc , k_1\} \times \dotsc \times \{0, \dotsc , k_{\ell }\},$ and the set T via (9). A direct computation reveals that $|B| + |T| = |\overline{B}|+|T|+\ell$ is upper bounded by

$$ {\delta }^{1-\tau }(\lceil {\delta }^\sigma \rceil +1) + (\beta _\ell +1) + \log _2({\delta }) \le 6{\delta }^{1/2}+ \log _2({\delta }). $$

Case 2. Assume $\alpha _{\ell } < {\delta }^{1/2}$, which implies ${\delta }^{1/2} < \prod _{i=1}^{\ell -1} \alpha _i$. Let $j \in \{1, \dotsc , \ell -2\}$ be the largest index with $\gamma := \prod _{i = 1}^j \alpha _i \le {\delta }^{1/2}$. Let $\sigma \ge 0$ be such that $\gamma \cdot {\delta }^ \sigma = {\delta } ^ {1/2}$ and $\tau < 1/2$ be such that $\alpha _{j+1} = {\delta }^\tau$. Note that $0 \le \sigma < \tau$ and ${\delta }^{\tau - \sigma } \cdot \prod _{i = j+2}^\ell \alpha _i = {\delta }^{1/2}$. For each $i \in \{1, \dotsc , j\}$ define $k_i := 0$, for each $i \in \{j+2, \dotsc , \ell \}$ define $k_i := \alpha _i - 1$, and set $k_{j+1} := \lceil {\delta }^{\tau - \sigma } \rceil$. The value $\beta _{j+1}$ in (9) satisfies

$$ \beta _{j+1} = \bigg \lfloor \frac{\alpha _{j+1}-1}{\big \lceil {\delta }^{\tau - \sigma } \big \rceil +1} \bigg \rfloor = \bigg \lfloor \frac{{\delta }^\tau -1}{\big \lceil {\delta }^{\tau - \sigma } \big \rceil +1} \bigg \rfloor \le \big \lceil {\delta }^{\sigma }\big \rceil . $$

Define $B = \overline{B} \cup \{z^1, \dotsc , z^{\ell }\}$, where $\overline{B} := \{0, \dotsc , k_1\} \times \dotsc \times \{0, \dotsc , k_{\ell }\},$ and the set T via (9). In this case we have $|B| + |T| = |\overline{B}|+|T|+l$ is upper bounded by

$$ \textstyle \left( \prod _{i = j+2}^\ell \alpha _i\right) ({\delta }^{\tau - \sigma } +2 ) + \gamma ({\delta }^\sigma +2) + \log _2({\delta }) \le 6{\delta }^{1/2}+ \log _2({\delta }). $$

   $\square$

Proof

(of Lemma 7). Let $x^* := A_I^{-1} b_I$ be the feasible vertex solution to ${\text {LP}}_{A,c}(b)$ with respect to the basis I. Applying Theorem 1 in [6] to the simplicial problems ${\text {LP}}_{A_I,c}(b)$ and ${\text {IP}}_{A_I,c}(b)$ shows that $z^*$ satisfies $\Vert z^* - x^*\Vert _{\infty } \le n\varDelta ^{\max }$. Thus, for every $j \in \{1, \dotsc , m\}\setminus I$ we have

$$ |A_j z^*-A_jx^*| \le \Vert A_j\Vert _1 \cdot \Vert z^* - x^*\Vert _\infty \le n^2\Vert A_j\Vert _\infty \cdot \varDelta ^{\max } \le (n\varDelta ^{\max })^2. $$

The assumption $A_jA_I^{-1} b_I+ (n\varDelta ^{\max })^2 < b_j$ implies

$$\begin{aligned} A_j z^*\le A_jx^* +|A_jz^*-A_jx^*| \le A_jx^* + (n\varDelta ^{\max })^2 = A_jA_I^{-1} b_I + (n\varDelta ^{\max })^2 < b_j. \end{aligned}$$

Thus, $z^*$ is feasible for $W\!\text {-}\text {MIP}_{A,c}(b)$.    $\square$

Proof

(of Lemma 8). Let $b \in \mathbb {Z}^m \setminus \mathcal {G}$. Therefore, there exists a feasible ${\text {LP}}_{A,c}(b)$ basis $I \subseteq \{1,\ldots ,m\}$ and $j \in \{1,\ldots ,m\} \setminus I$ such that $b_j \le A_jA_I^{-1} b_I + (n\varDelta ^{\max })^2$. Recall $A_jA_I^{-1} b_I \le b_j$ because $A_I^{-1} b_I$ is feasible for ${\text {LP}}_{A,c}(b)$. Thus, $\mathbb {Z}^m\setminus \mathcal {G}$ is in

$$\begin{aligned} \{b \in \mathbb {Z}^m:~\exists \text { a basis } I \text { and } j \in \{1,\ldots ,m\}\setminus I \text { with } b_j \le A_jA_I^{-1} b_I + (n\varDelta ^{\max })^2\}. \end{aligned}$$

Cramer’s Rule implies that $\varDelta _I \cdot A_jA_I^{-1} b_I \in \mathbb {Z}$ for all $j \in \{1,\ldots ,m\} \setminus I$. Thus,

$$\begin{aligned}&\{b\in \mathbb {Z}^m: \exists \text { a basis } I \text { and } j \not \in I \text { with } \varDelta _Ib_j \le \varDelta _IA_jA_I^{-1} b_I + \varDelta _I(n\varDelta ^{\max })^2\} \\ \subseteq&\bigcup _{\begin{array}{c} I \subseteq \{1,\ldots ,m\} \\ I \text { basis} \end{array}} \bigcup _{j \not \in I} \bigcup _{r = 0}^{\varDelta _I(n\varDelta ^{\max })^2}\{b \in \mathbb {Z}^m : \varDelta _I b_j = \varDelta _IA_jA_I^{-1} b_I + r\}. \end{aligned}$$

   $\square$