Sparsity of Integer Solutions in the Average Case

Abstract

We examine how sparse feasible solutions of integer programs are, on average. Average case here means that we fix the constraint matrix and vary the right-hand side vectors. For a problem in standard form with m equations, there exist LP feasible solutions with at most m many nonzero entries. We show that under relatively mild assumptions, integer programs in standard form have feasible solutions with O(m) many nonzero entries, on average. Our proof uses ideas from the theory of groups, lattices, and Ehrhart polynomials. From our main theorem we obtain the best known upper bounds on the integer Carathéodory number provided that the determinants in the data are small.

Appendices

A Proof of Theorem 2

We construct both matrices A and B using a submatrix $\tilde{A}$, which we construct first. Let $d \in \mathbb {Z}_{\ge 1}$ and $p_1< \cdots < p_d$ be prime. For $i \in \{1, \cdots , d\}$, define $q_i := \prod _{j=1, j \ne i}^d p_i$ and $\delta := \prod _{j=1}^d p_i$. Define the matrix $\tilde{A} := \begin{bmatrix} q_1,&\cdots&q_d,&- \delta \end{bmatrix}.$ The matrix $\tilde{A}$ has $d+1$ columns, so $\sigma ^{\text {asy}}(\tilde{A}) \le 1+d$. The matrix $\tilde{A}$ is similar to the example in [1, Theorem 2] and the theory of so-called primorials. We claim

$$\begin{aligned} \text {if } b \in \mathbb {Z}_{< 0} \text { and } b \equiv 1 \text { mod } \delta , \text { then } P(\tilde{A},b) \ne \emptyset \text { and } \sigma (\tilde{A},b) = 1 + d. \end{aligned}$$

(20)

Note that $\gcd (q_1, \cdots , q_d) = 1$. The Frobenius number of $\{q_1, \cdots , q_d\}$ is the largest integer that cannot be written as a positive integer linear combination of $q_1, \cdots ,$ and $q_d$. Hence, if we choose $\bar{b} \in \mathbb {Z}_{\ge 1}$ to be the Frobenius number of $\{q_1, \cdots , q_d\}$, then $b \ge \bar{b}+1$ implies $P(\tilde{A},b) \ne \emptyset$. If $b \equiv 1 \text { mod } \delta$, then b is not divisible by $p_i$ for any $i \in \{1, \cdots , d\}$. Thus, if $b \ge \bar{b}+1$ and $b \equiv 1 \text { mod } \delta$, then $\sigma (\tilde{A}, b) = d$. Finally, observe that if $b < 0$, then $b + k\delta >\bar{b}$ for large enough $k \in \mathbb {Z}_{\ge 1}$. The only negative column of $\tilde{A}$ is $-\delta$, so $\sigma (\tilde{A},b) = 1+d$. This proves (20).

Now we define the matrix A. Let $m \in \mathbb {Z}_{\ge 1}$ and define

$$ A := \begin{bmatrix} I^{m-1} ~&0^{(m-1)\times (d+1)} \\ 0^{1 \times (m-1)}&\tilde{A} \end{bmatrix} \in \mathbb {Z}^{m \times (m+d)}, $$

where $I^k \in \mathbb {Z}^{k \times k}$ is the identity matrix and $0^{k \times s} \in \mathbb {Z}^{k \times s}$ is the all zero matrix for $k,s\in \mathbb {Z}_{\ge 1}$. Note that $\phi ^{\max }(A) = d$. If $b \in \mathbb {Z}^{m-1}_{> 0} \times \mathbb {Z}_{<0}$ is such that the last component is equivalent to $1 \text { mod } \delta$, then $\sigma (A,b) = m+d$ by the arguments above. Now, the set of $b\in \mathbb {Z}^m$ such that $P(A,b) \ne \emptyset$ is contained in $\mathbb {Z}^{m-1}_{\ge 0} \times \mathbb {Z}$. So, for every $t \in \mathbb {Z}_{\ge 1}$, the set of feasible solutions in $\{-t\delta , \cdots , t\delta \}^m$ contains $t(t\delta -1)^{m-1}$ points b such that $\sigma (A,b) = m+d$. Moreover, if $t \in \mathbb {Z}_{\ge \bar{b}}$, then $P(A,b) \ne \emptyset$ for every $b \in \{0, \cdots , t\delta \}^{m-1}\times \{-t\delta , \cdots , t\delta \}$. Therefore,

$$ \begin{array}{rclcl} &{}\displaystyle \lim _{t \rightarrow \infty } \frac{|\{b \in \{-t, ..., t\} : \sigma (A,b) \le (m-1) + d\}|}{|\{b \in \{-t, ..., t\} : P(A,b) \ne \emptyset \}|} \\ = &{}\displaystyle \lim _{t \rightarrow \infty } \frac{|\{b \in \{-t\delta , ..., t\delta \} : \sigma (A,b) \le (m-1) + d\}|}{|\{b \in \{-t\delta , ..., t\delta \} : P(A,b) \ne \emptyset \}|} \\ \le &{} \displaystyle \lim _{t \rightarrow \infty } \frac{(2t\delta +1)(t\delta +1)^{m-1} - t(t\delta +1)^{m-1}}{(2t\delta +1)(t\delta +1)^{m-1}} &{} <&{} 1. \end{array} $$

Using this and the fact that A has $m + d$ columns, we have $\sigma ^{\text {asy}}(A) = m+d$.

Now we define the matrix B. Let $A \in \mathbb {Z}^{m\times (m+d)}$ be as above. Let $e^{1 \times (m+1)} \in \mathbb {Z}^{1 \times (m+1)}$ be the all ones matrix and $U \in \mathbb {Z}^{m\times (m+1)}$. Assume

$$ \bigg | \det \bigg (\begin{bmatrix} U \\ e^{1 \times (m+1)} \end{bmatrix} \bigg )\bigg | =1 $$

and set

$$ B := \begin{bmatrix} U ~&A \\ e^{1 \times (m+1)} ~&0^{1 \times (m+d)} \end{bmatrix} \in \mathbb {Z}^{(m+1) \times (2m+1+d)}. $$

Note that $\phi ^{\min }(B) = 0$, so Theorem 1 (ii) implies that $\sigma ^{\text {asy}}(B) \le 2m+2$. Let $b \in \mathbb {Z}^{m}\times \{0\}$ be such that $P(B,b) \ne \emptyset$. If $z \in P(B,b)$, then the first $m+1$ components of z are zero. So, similarly to above, there are $b \in \mathbb {Z}^{m+1}$ such that $\sigma (B,b) = m+d$. Hence, $\sigma ^{asy}(B) \le 2m+2 < m+d = \sigma (B)$.    $\square$

B Proof of Lemma 3

Assume that $t = 2$. Let $x := x^1$ and $y := x^2$. First, we show that $K \cap (K + x) \cap (K+y) \ne \emptyset$. Since $v^1, \cdots , v^m$ are linearly independent, K is a full-dimensional simplicial cone. Hence, there exist linearly independent vectors $a^1, \dots , a^m \in \mathbb {R}^m$ such that $K = \{ w \in \mathbb {R}^m : (a^i)^\intercal w \le 0 ~ \forall ~ i \in \{1, \dots , m\}\}$ and linearly independent vectors $r^1, \dots , r^m \in K$ such that $(a^i)^\intercal r^i < 0$ for each $i \in \{1, \dots , m\}$.

There is a set $J \subseteq \{1, \cdots , m\}$ such that $(a^j)^\intercal (x-y) > 0$ for each $j \in J$ and $(a^j)^\intercal (x-y) \le 0$ for each $j \in \{1, \dots , m\}\setminus J$. For $j \in \{1, \dots , m\}$, set

$$ \lambda _j := {\left\{ \begin{array}{ll} \max \left\{ 0, -\frac{(a^j)^\intercal x }{(a^j)^\intercal r^j}\right\} , &{} \text {if } j \in \{1, \cdots , m\} \setminus J\\ \\ \max \left\{ -\frac{(a^j)^\intercal (x-y) }{ (a^j)^\intercal r^j}, -\frac{(a^j)^\intercal x }{(a^j)^\intercal r^j}\right\} , &{} \text {if } j \in J. \end{array}\right. } $$

Note that $\lambda _1, \dots , \lambda _m \in \mathbb {R}_{\ge 0}$, so $x +\sum _{j=1}^m \lambda _j r^j \in K + x.$ For each $i \in \{1, \dots , m\}$, it follows that

$$ (a^i)^\intercal \bigg (x +\sum _{j=1}^m \lambda _j r^j - y \bigg ) \le (a^i)^\intercal (x-y) + \lambda _i (a^i)^\intercal r^i \le 0. $$

So, $x +\sum _{j=1}^m \lambda _j r^j -y \in K$ and $x +\sum _{j=1}^m \lambda _j r^j \in K + y$. Finally, for each $i \in \{1, \dots , m\}$, it follows that

$$ (a^i)^\intercal \bigg (x +\sum _{j=1}^m \lambda _j r^j \bigg ) \le (a^i)^\intercal x + \lambda _i (a^i)^\intercal r^i \le 0. $$

Hence, $x +\sum _{j=1}^m \lambda _j r^j \in K$ and $K \cap (K+x) \cap (K+y) \ne \emptyset$.

Let $w \in K \cap (K+x) \cap (K+y)$. Then $K+ w \subseteq K \cap (K+x) \cap (K+y)$. Because K is full-dimensional, there exists a point $z \in (K+w) \cap \mathbb {Z}^m$ such that $z = \sum _{i=1}^m k_i v^i$ for $k_i \in \mathbb {Z}_{\ge 0}$. Note that $z \in K + w \subseteq K$ and

$$ K + z \subseteq K + w \subseteq K + \left( K \cap (K+x) \cap (K+y)\right) \subseteq K \cap (K+x) \cap (K+y). $$

For $t \ge 3$, the result follows by induction.    $\square$