arXiv:2501.06506cs.GTcs.AI2025-01中稿 · AAMAS 2025 as an e…被引 3

在拉丁方约束下优化资源分配,兼顾公平与效率

Resource Allocation under the Latin Square Constraint

  • 基于拉丁方结构设计分配规则,确保每轮每人最多得一件,每件仅分配一次
  • 提出(1-1/e)和(1-1/e)/4近似算法,分别用于部分与完整分配场景
  • 证明多个公平性指标判定为NP难,适合资源调度与实验设计研究者

拉丁方是填满n个不同符号的n×n矩阵,使每个符号在每行每列恰好出现一次。我们提出一个在n轮内将n个不可分物品分配给n个代理的问题,满足拉丁方约束:每个代理每轮至多获得一个物品,且每个物品至多被分配一次。每个代理对物品-轮次组合具有可加效用。现实中的排程、资源管理与实验设计需此约束以保障分配的公平或均衡。目标是最大化代理效用总和(效用最大)或最小效用(平等最大)。我们证明,即使效用为二元可加,最大化效用最大仍为NP难。针对部分与完整分配,分别给出(1-1/e)与(1-1/e)/4近似算法。同时,针对拉丁方阶数和最优值提供固定参数可解(FPT)算法。对于平等最大问题,即使在二元效用下,判断最优值是否≤1或≥2均为NP难。此外,验证是否存在满足无嫉妒、比例、等价、无嫉妒到任意物品、比例到任意物品或等价到任意物品的完整分配,即使在相同效用下也均为NP难。

原文摘要 · Abstract (English)

A Latin square is an $n \times n$ matrix filled with $n$ distinct symbols, each of which appears exactly once in each row and exactly once in each column. We introduce a problem of allocating $n$ indivisible items among $n$ agents over $n$ rounds while satisfying the Latin square constraint. This constraint ensures that each agent receives no more than one item per round and receives each item at most once. Each agent has an additive valuation on the item--round pairs. Real-world applications like scheduling, resource management, and experimental design require the Latin square constraint to satisfy fairness or balancedness in allocation. Our goal is to find a partial or complete allocation that maximizes the sum of the agents' valuations (utilitarian social welfare) or the minimum of the agents' valuations (egalitarian social welfare). For the problem of maximizing utilitarian social welfare, we prove NP-hardness even when the valuations are binary additive. We then provide $(1-1/e)$ and $(1-1/e)/4$-approximation algorithms for partial and complete settings, respectively. Additionally, we present fixed-parameter tractable (FPT) algorithms with respect to the order of Latin square and the optimum value for both partial and complete settings. For the problem of maximizing egalitarian social welfare, we establish that deciding whether the optimum value is at most $1$ or at least $2$ is NP-hard for both the partial and complete settings, even when the valuations are binary. Furthermore, we demonstrate that checking the existence of a complete allocation that satisfies each of envy-free, proportional, equitable, envy-free up to any good, proportional up to any good, or equitable up to any good is NP-hard, even when the valuations are identical.

资源分配拉丁方近似算法公平性

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