We study fair division of indivisible chores among $n$ agents with additive disutility functions. Two well-studied fairness notions for indivisible items are envy-freeness up to one/any item (EF1/EFX) and the standard notion of economic efficiency is Pareto optimality (PO). There is a noticeable gap between the results known for both EF1 and EFX in the goods and chores settings. The case of chores turns out to be much more challenging. We reduce this gap by providing slightly relaxed versions of the known results on goods for the chores setting. Interestingly, our algorithms run in polynomial time, unlike their analogous versions in the goods setting. We introduce the concept of $k$ surplus which means that up to $k$ more chores are allocated to the agents and each of them is a copy of an original chore. We present a polynomial-time algorithm which gives EF1 and PO allocations with $(n-1)$ surplus. We relax the notion of EFX slightly and define tEFX which requires that the envy from agent $i$ to agent $j$ is removed upon the transfer of any chore from the $i$'s bundle to $j$'s bundle. We give a polynomial-time algorithm that in the chores case for $3$ agents returns an allocation which is either proportional or tEFX. Note that proportionality is a very strong criterion in the case of indivisible items, and hence both notions we guarantee are desirable.

翻译：我们研究在$n$个具有可加负效用函数的智能体之间对不可分家务的公平分配问题。针对不可分物品，两个被深入研究的概念是至多一个/任意物品下的无妒忌性（EF1/EFX），而经济效率的标准概念则是帕累托最优（PO）。在商品与家务情境中，关于EF1和EFX的现有结果之间存在显著差距。家务情形被证明更具挑战性。我们通过提供已知商品结果在家务情境中的宽松版本，缩小了这一差距。有趣的是，我们的算法可在多项式时间内运行，而其在商品情境中的对应版本则无法实现。我们引入了$k$剩余的概念，即最多额外分配$k$件家务给智能体，且每件家务均为原始家务的副本。我们提出了一种多项式时间算法，可在具有$(n-1)$剩余的情况下实现EF1与PO分配。我们略微放宽了EFX的定义，提出tEFX要求：若从智能体$i$的捆绑中转移任意一件家务到智能体$j$的捆绑，则消除$i$对$j$的妒忌。我们给出一个多项式时间算法，在家务情境下针对$3个$智能体，返回一个满足比例性或tEFX的分配。需注意，在不可分物品情形中比例性是非常严格的标准，因此我们保证的这两个性质均具有理想性。