In our work, we consider the problem of computing a vector $x \in Z^n$ of minimum $\|\cdot\|_p$-norm such that $a^\top x \not= a_0$, for any vector $(a,a_0)$ from a given subset of $Z^n$ of size $m$. In other words, we search for a vector of minimum norm that avoids a given finite set of hyperplanes, which is natural to call as the $\textit{Hyperplanes Avoiding Problem}$. This problem naturally appears as a subproblem in Barvinok-type algorithms for counting integer points in polyhedra. More precisely, it appears when one needs to evaluate certain rational generating functions in an avoidable critical point. We show that: 1) With respect to $\|\cdot\|_1$, the problem admits a feasible solution $x$ with $\|x\|_1 \leq (m+n)/2$, and show that such solution can be constructed by a deterministic polynomial-time algorithm with $O(n \cdot m)$ operations. Moreover, this inequality is the best possible. This is a significant improvement over the previous randomized algorithm, which computes $x$ with a guaranty $\|x\|_{1} \leq n \cdot m$. The original approach of A.~Barvinok can guarantee only $\|x\|_1 = O\bigl((n \cdot m)^n\bigr)$; 2) The problem is NP-hard with respect to any norm $\|\cdot\|_p$, for $p \in \bigl(R_{\geq 1} \cup \{\infty\}\bigr)$. 3) As an application, we show that the problem to count integer points in a polytope $P = \{x \in R^n \colon A x \leq b\}$, for given $A \in Z^{m \times n}$ and $b \in Q^m$, can be solved by an algorithm with $O\bigl(\nu^2 \cdot n^3 \cdot \Delta^3 \bigr)$ operations, where $\nu$ is the maximum size of a normal fan triangulation of $P$, and $\Delta$ is the maximum value of rank-order subdeterminants of $A$. It refines the previous state-of-the-art $O\bigl(\nu^2 \cdot n^4 \cdot \Delta^3\bigr)$-time algorithm.

翻译：在我们的工作中，我们考虑计算一个向量 $x \in Z^n$，使其在给定的大小为 $m$ 的 $Z^n$ 子集中的任意向量 $(a,a_0)$ 下满足 $a^\top x \not= a_0$，并且其 $\|\cdot\|_p$-范数最小。换言之，我们寻找一个最小范数向量以规避给定的有限超平面集合，这自然可称为 $\textit{超平面规避问题}$。该问题自然地作为 Barvinok 类型算法中用于计数多面体内整数点的子问题出现。更准确地说，它出现在需要评估某个有理生成函数在一个可规避的临界点时。我们证明：1) 对于 $\|\cdot\|_1$，该问题存在一个可行解 $x$ 满足 $\|x\|_1 \leq (m+n)/2$，并证明这样的解可以通过一个确定性多项式时间算法以 $O(n \cdot m)$ 次操作构造。此外，该不等式是最优的。这相对于先前的随机算法（其计算出的 $x$ 保证 $\|x\|_{1} \leq n \cdot m$）是一个显著改进。A.~Barvinok 的原始方法仅能保证 $\|x\|_1 = O\bigl((n \cdot m)^n\bigr)$；2) 对于任意范数 $\|\cdot\|_p$，其中 $p \in \bigl(R_{\geq 1} \cup \{\infty\}\bigr)$，该问题是 NP-难的。3) 作为一个应用，我们证明了对于给定的 $A \in Z^{m \times n}$ 和 $b \in Q^m$，计算多面体 $P = \{x \in R^n \colon A x \leq b\}$ 中整数点个数的问题可以通过一个算法以 $O\bigl(\nu^2 \cdot n^3 \cdot \Delta^3 \bigr)$ 次操作解决，其中 $\nu$ 是 $P$ 的法扇三角剖分的最大规模，$\Delta$ 是 $A$ 的秩阶子式的最大值。这改进了先前最先进的 $O\bigl(\nu^2 \cdot n^4 \cdot \Delta^3\bigr)$ 时间复杂度算法。