Envy-freeness up to any good (EFX) is a central fairness notion for allocating indivisible goods, yet its existence is unresolved in general. In the setting with few surplus items, where the number of goods exceeds the number of agents by a small constant (at most three), EFX allocations are guaranteed to exist, shifting the focus from existence to efficiency and computation. We study how EFX interacts with generalized-mean ($p$-mean) welfare, which subsumes commonly-studied utilitarian ($p=1$), Nash ($p=0$), and egalitarian ($p \rightarrow -\infty$) objectives. We establish sharp complexity dichotomies at $p=0$: for any fixed $p \in (0,1]$, both deciding whether EFX can attain the global $p$-mean optimum and computing an EFX allocation maximizing $p$-mean welfare are NP-hard, even with at most three surplus goods; in contrast, for any fixed $p \leq 0$, we give polynomial-time algorithms that optimize $p$-mean welfare within the space of EFX allocations and efficiently certify when EFX attains the global optimum. We further quantify the welfare loss of enforcing EFX via the price of fairness framework, showing that for $p > 0$, the loss can grow linearly with the number of agents, whereas for $p \leq 0$, it is bounded by a constant depending on the surplus (and for Nash welfare it vanishes asymptotically). Finally we show that requiring Pareto-optimality alongside EFX is NP-hard (and becomes $Σ_2^P$-complete for a stronger variant of EFX). Overall, our results delineate when EFX is computationally costly versus structurally aligned with welfare maximization in the setting with few surplus items.

翻译：在分配不可分割物品时，任意物品无嫉妒（EFX）是一个核心的公平性概念，但其一般存在性仍未解决。在具有少量剩余物品的场景中（物品数量超过智能体数量一个小的常数，至多为三），EFX分配被保证存在，这使得研究重点从存在性转向效率和计算问题。我们研究了EFX如何与广义均值（$p$-均值）福利相互作用，后者包含了通常研究的功利主义（$p=1$）、纳什（$p=0$）和平等主义（$p \rightarrow -\infty$）目标。我们在$p=0$处建立了尖锐的复杂性二分：对于任意固定的$p \in (0,1]$，即使最多只有三个剩余物品，判定EFX是否能达到全局$p$-均值最优以及计算最大化$p$-均值福利的EFX分配都是NP难的；相反，对于任意固定的$p \leq 0$，我们给出了多项式时间算法，可以在EFX分配的空间内优化$p$-均值福利，并有效地验证EFX何时达到全局最优。我们进一步通过公平性代价框架量化了强制执行EFX带来的福利损失，表明对于$p > 0$，损失可能随智能体数量线性增长，而对于$p \leq 0$，损失受限于一个依赖于剩余物品数量的常数（对于纳什福利，它会渐近地消失）。最后，我们证明了同时要求帕累托最优和EFX是NP难的（对于一个更强的EFX变体，则变为$Σ_2^P$完全的）。总体而言，我们的结果刻画了在具有少量剩余物品的场景中，EFX何时在计算上是昂贵的，何时在结构上与福利最大化相一致。