Envy-free relaxations for goods, chores, and mixed items

被引:1
|
作者
Berczi, Kristof [1 ,2 ]
Berczi-Kovacs, Erika R. [1 ,2 ]
Boros, Endre [3 ,4 ]
Gedefa, Fekadu Tolessa [5 ]
Kamiyama, Naoyuki [6 ]
Kavitha, Telikepalli [7 ]
Kobayashi, Yusuke [8 ]
Makino, Kazuhisa [8 ]
机构
[1] ELTEE Eotvos Lorand Univ, MTA ELTE Matroid Optimizat Res Grp, Budapest, Hungary
[2] Eotvos Lorand Univ, Dept Operat Res, HUN REN ELTE Egervary Res Grp, Budapest, Hungary
[3] Rutgers State Univ, MSIS Dept, Camden, NJ USA
[4] Rutgers State Univ, RUTCOR, Camden, NJ USA
[5] Eotvos Lorand Univ, Dept Operat Res, Budapest, Hungary
[6] Kyushu Univ, Inst Math Ind, Fukuoka, Japan
[7] Tata Inst Fundamental Res, Sch Technol & Comp Sci, Mumbai, India
[8] Kyoto Univ, Res Inst Math Sci RIMS, Kyoto, Japan
基金
欧洲研究理事会;
关键词
Fair division; Indivisible items; Envy-freeness; Non-monotone utility function; Non-additive utility function; FAIR; ASSIGNMENT; COMPLEXITY; DIVISION;
D O I
10.1016/j.tcs.2024.114596
中图分类号
TP301 [理论、方法];
学科分类号
081202 ;
摘要
In fair division problems, we are given a set S of m items and a set N of n agents with individual preferences, and the goal is to find an allocation of items among agents so that each agent finds the allocation fair. There are several established fairness concepts and envy -freeness is one of the most extensively studied ones. However envy -free allocations do not always exist when items are indivisible and this has motivated relaxations of envy -freeness: envy -freeness up to one item (EF1) and envy -freeness up to any item (EFX) are two well -studied relaxations. We consider the problem of finding EF1 and EFX allocations for utility functions that are not necessarily monotone, and propose four possible extensions of different strength to this setting. In particular, we present a polynomial time algorithm for finding an EF1 allocation for two agents with arbitrary utility functions. An example is given showing that EFX allocations need not exist for two agents with non -monotone, non -additive, identical utility functions. However, when all agents have monotone (not necessarily additive) identical utility functions, we give a pseudo -polynomial time algorithm that always finds an EFX allocation of chores. As a step toward understanding the general case, we discuss two subclasses of utility functions: Boolean utilities that are {0, +1} -valued functions, and negative Boolean utilities that are {0, -1}-valued functions. For the latter, we give a polynomial time algorithm that finds an EFX allocation when the utility functions are identical.
引用
收藏
页数:14
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