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：我们研究在拥有双值次模估值的智能体间公平分配一组不可分割物品的问题——每件物品提供边际增益$a$或$b$（$a<b$），且物品的边际增益递减。这是两类被广泛研究的估值类别——双值加性估值与二元次模估值的自然推广。我们提出一种基于近期引入的Yankee Swap机制的简单顺序算法框架，该框架可适配计算包括leximin、最大纳什福利（MNW）及$p$-均值福利最大化配置在内的多种解概念（当$a$整除$b$时）。该成果与现有关于当$a$不整除$b$时leximin与MNW配置的计算难解性结果形成互补。我们进一步从两个经典性质——无嫉妒性与极大极小份额保证——角度考察leximin与MNW配置。关于无嫉妒性，我们证明leximin与MNW配置均不能保证达到至多一件物品无嫉妒（EF1）。这令人惊讶，因为对于更简单的双值加性估值与二元次模估值类别，MNW配置已知能保证达到至多任意物品无嫉妒（EFX）。关于极大极小份额保证，我们证明当$a$整除$b$时，MNW与leximin配置分别保证每个智能体获得其极大极小份额的$\frac14$和$\frac{a}{b+3a}$。当智能体具有双值加性估值时，该比例分别提升至$\frac13$与$\frac{a}{b+2a}$。