We study the problem of fairly allocating a set of indivisible goods among agents with bivalued submodular valuations -- each good provides a marginal gain of either $a$ or $b$ ($a < b$) and goods have decreasing marginal gains. This is a natural generalization of two well-studied valuation classes -- bivalued additive valuations and binary submodular valuations. We present a simple sequential algorithmic framework, based on the recently introduced Yankee Swap mechanism, that can be adapted to compute a variety of solution concepts, including leximin, max Nash welfare (MNW) and $p$-mean welfare maximizing allocations when $a$ divides $b$. This result is complemented by an existing result on the computational intractability of leximin and MNW allocations when $a$ does not divide $b$. We further examine leximin and MNW allocations with respect to two well-known properties -- envy freeness and the maximin share guarantee. On envy freeness, we show that neither the leximin nor the MNW allocation is guaranteed to be envy free up to one good (EF1). This is surprising since for the simpler classes of bivalued additive valuations and binary submodular valuations, MNW allocations are known to be envy free up to any good (EFX). On the maximin share guarantee, we show that MNW and leximin allocations guarantee each agent $\frac14$ and $\frac{a}{b+3a}$ of their maximin share respectively when $a$ divides $b$. This fraction improves to $\frac13$ and $\frac{a}{b+2a}$ respectively when agents have bivalued additive valuations.
翻译:我们研究了在具有双值次模估值(每个物品提供$a$或$b$($a<b$)的边际增益且边际增益递减)的智能体间公平分配一组不可分割物品的问题。这是两类被广泛研究的估值类别——双值可加估值和二元次模估值——的自然推广。我们提出了一种基于近期引入的Yankee Swap机制的简单顺序算法框架,该框架可适用于计算多种解概念,包括当$a$整除$b$时的leximin、最大纳什福利(MNW)和$p$-均值福利最大化分配。该结果与已有关于当$a$不整除$b$时leximin和MNW分配计算难解性的结果互为补充。我们进一步研究了leximin和MNW分配关于两个众所周知的性质——无嫉妒性和最大最小份额保证。在无嫉妒性方面,我们证明leximin和MNW分配均不能保证达到至多一个物品的无嫉妒性(EF1)。这令人惊讶,因为对于更简单的双值可加估值和二元次模估值类别,已知MNW分配可达到至多任意物品无嫉妒性(EFX)。在最大最小份额保证方面,我们证明当$a$整除$b$时,MNW和leximin分配分别保证每个智能体获得其最大最小份额的$\frac14$和$\frac{a}{b+3a}$。当智能体具有双值可加估值时,该比例分别提升至$\frac13$和$\frac{a}{b+2a}$。