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Obtaining a Proportional Allocation by Deleting Items

机译:通过删除项目获得比例分配

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We consider the following control problem on fair allocation of indivisible goods. Given a set I of items and a set of agents, each having strict linear preferences over the items, we ask for a minimum subset of the items whose deletion guarantees the existence of a proportional allocation in the remaining instance; we call this problem Proportionality by Item Deletion (PID). Our main result is a polynomial-time algorithm that solves PID for three agents. By contrast, we prove that PID is computationally intractable when the number of agents is unbounded, even if the number k of item deletions allowed is small-we show that the problem is W[3]-hard with respect to the parameter k. Additionally, we provide some tight lower and upper bounds on the complexity of PID when regarded as a function of vertical bar I vertical bar and k. Considering the possibilities for approximation, we prove a strong inapproximability result for PID. Finally, we also study a variant of the problem where we are given an allocation pi in advance as part of the input, and our aim is to delete a minimum number of items such that pi is proportional in the remainder; this variant turns out to be NP-hard for six agents, but polynomial-time solvable for two agents, and we show that it is W[2]-hard when parameterized by the number k of delections.
机译:我们认为以下控制问题在公平分配不可分割的商品。给定项目和一组代理,每个代理都具有严格的线性偏好,我们要求删除剩余实例中的比例分配的项目的最小项目的最小子集;我们按项目删除(PID)称为此问题比例。我们的主要结果是一种多项式算法,可以解决三个代理的PID。相比之下,当代理的数量无界时,我们证明了PID在计算上难以解决,即使允许的项目删除的数量k是小的,我们表明问题是W [3] - 相对于参数k。另外,当视为PID的复杂性时,我们提供了一些紧密的下限和上限,当垂直条I垂直条和k的函数被视为PID时。考虑到近似的可能性,我们证明了PID的强烈不可识别结果。最后,我们还研究了一个问题的变体,我们提前给予分配PI作为输入的一部分,我们的目标是删除最少数量的项目,使得PI在其余部分中比例;该变体对六种剂量产生NP - 硬,但两种代理的多项式溶解,我们表明它是W [2] - 当由分层的数量k参数化时,它是W [2]。

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