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$n$个具有可加负效用的智能体之间公平分配不可分杂务的问题。两个广为人知的不可分物品公平性概念是单个/任意物品免嫉妒性（EF1/EFX），而经济效率的标准概念是帕累托最优性（PO）。在物品与杂务场景中，关于EF1和EFX的已知结果存在显著差距。杂务情形被证明显得更具挑战性。我们通过为杂务场景提供已知物品结果的略微松弛版本缩小了这一差距。值得注意的是，与物品场景中的对应版本不同，我们的算法运行于多项式时间。我们引入$k$剩余的概念，即最多可向智能体额外分配$k$件杂务，且每件均为原始杂务的副本。我们提出一种多项式时间算法，可在$(n-1)$剩余条件下实现EF1与PO分配。我们略微放松EFX概念，定义tEFX要求：从智能体$i$向智能体$j$转移任意杂务后，$i$对$j$的嫉妒即可消除。我们给出一种多项式时间算法，在3个智能体的杂务场景中返回满足比例性或tEFX的分配。需注意，比例性在不可分物品情形中是非常严格的标准，因此我们保障的两种概念均具有理想性质。