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 \({\mathsf {W}}[3]\)-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 |I| 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 \({{\mathsf {N}}}{{\mathsf {P}}}\)-hard for six agents, but polynomial-time solvable for two agents, and we show that it is \(\mathsf {W[2]}\)-hard when parameterized by the number k of

我们考虑以下关于不可分割物品公平分配的控制问题。给定一组我的项目和一组代理，在项目每个都具有严格的线性的偏好，我们会要求其删除，保证在剩余的实例的比例分配存在的项目的最低子集; 我们将这个问题称为按项目删除 (PID) 的比例。我们的主要结果是一个多项式时间算法，它解决了三个代理的 PID。相比之下，我们证明了当代理的数量是无限的时，PID 在计算上是难以处理的，即使允许的项目删除数量k很小——我们证明问题是\({\mathsf {W}}[3]\) -hard 相对于参数k. 此外，当将 PID 视为 | 我| 和k。考虑到逼近的可能性，我们证明了 PID 的强不可逼近性结果。最后，我们还研究了这个问题的一个变体，其中我们预先给定了一个分配\(\pi \)作为输入的一部分，我们的目标是删除最小数量的项目，使得\(\pi \)是余数成正比；这个变体对于六个代理来说是\({{\mathsf {N}}}{{\mathsf {P}}}\) -hard，但是对于两个代理来说多项式时间是可解的，我们证明它是\ (\mathsf {W[2]}\) -hard 当由数字k参数化时  的