This paper introduces the \emph{$d$-distance $b$-matching problem}, in which we are given a bipartite graph $G=(S,T;E)$ with $S=\{s_1,\dots,s_n\}$, a weight function on the edges, an integer $d\in\mathbb{Z}_+$ and a degree bound function $b:S\cup T\to\mathbb{Z}_+$. The goal is to find a maximum-weight subset $M\subseteq E$ of the edges satisfying the following two conditions: 1) the degree of each node $v\in S\cup T$ is at most $b(v)$ in $M$, 2) if $s_it,s_jt\in M$, then $|i-j|\geq d$. In the cyclic version of the problem, the nodes in $S$ are considered to be in cyclic order. We get back the \emph{(cyclic) $d$-distance matching problem} when $b(s) = 1$ for $s\in S$ and $b(t) = \infty$ for $t\in T$. We prove that the $d$-distance matching problem is APX-hard even in the unweighted case. We show that $(2-\frac{1}{d})$ is a tight upper bound on the integrality gap of the natural integer programming model for the cyclic $d$-distance $b$-matching problem provided that $(2d-1)$ divides the size of $S$. For the non-cyclic case, the integrality gap is shown to be at most $(2-\frac{2}{d})$. The proofs give approximation algorithms with guarantees matching these bounds, and also improve the best known algorithms for the (cyclic) $d$-distance matching problem. In a related problem, our goal is to find a permutation of $S$ maximizing the weight of the optimal $d$-distance $b$-matching. This problem can be solved in polynomial time for the (cyclic) $d$-distance matching problem -- even though the (cyclic) $d$-distance matching problem itself is NP-hard and also hard to approximate arbitrarily. For (cyclic) $d$-distance $b$-matchings, however, we prove that finding the best permutation is NP-hard even if $b\equiv 2$ or $d=2$, and we give $e$-approximation algorithms.

翻译：本文介绍了\emph{$d$-距离 $b$-匹配问题}，其中给定一个二分图 $G=(S,T;E)$，$S=\{s_1,\dots,s_n\}$，边上的权函数，整数 $d\in\mathbb{Z}_+$ 以及度界函数 $b:S\cup T\to\mathbb{Z}_+$。目标是找到一个最大权子集 $M\subseteq E$，满足以下两个条件：1) 在 $M$ 中，每个节点 $v\in S\cup T$ 的度数不超过 $b(v)$；2) 若 $s_it,s_jt\in M$，则 $|i-j|\geq d$。在循环版本的问题中，$S$ 中的节点被视为按循环顺序排列。当 $b(s) = 1$（对于 $s\in S$）且 $b(t) = \infty$（对于 $t\in T$）时，我们得到\emph{(循环) $d$-距离匹配问题}。我们证明了即使是无权情况，$d$-距离匹配问题也是 APX-难的。我们表明，当 $(2d-1)$ 整除 $S$ 的大小时，$(2-\frac{1}{d})$ 是循环 $d$-距离 $b$-匹配问题的自然整数规划模型整数性间隙的紧上界。对于非循环情况，整数性间隙至多为 $(2-\frac{2}{d})$。这些证明给出了与这些界相匹配的近似算法，并改进了 (循环) $d$-距离匹配问题的最佳已知算法。在一个相关的问题中，我们的目标是找到 $S$ 的一个排列，使得最优 $d$-距离 $b$-匹配的权值最大化。对于(循环) $d$-距离匹配问题，该问题可在多项式时间内求解——尽管(循环) $d$-距离匹配问题本身是 NP-难的，且难以任意近似。然而，对于(循环) $d$-距离 $b$-匹配，我们证明即使当 $b\equiv 2$ 或 $d=2$ 时，寻找最佳排列也是 NP-难的，并给出了 $e$-近似算法。