The problem of lifting a preference order on a set of objects to a preference order on a family of subsets of this set is a fundamental problem with a wide variety of applications in AI. The process is often guided by axioms postulating properties the lifted order should have. Well-known impossibility results by Kannai and Peleg and by Barber\`a and Pattanaik tell us that some desirable axioms - namely dominance and (strict) independence - are not jointly satisfiable for any linear order on the objects if all non-empty sets of objects are to be ordered. On the other hand, if not all non-empty sets of objects are to be ordered, the axioms are jointly satisfiable for all linear orders on the objects for some families of sets. Such families are very important for applications as they allow for the use of lifted orders, for example, in combinatorial voting. In this paper, we determine the computational complexity of recognizing such families. We show that it is $\Pi_2^p$-complete to decide for a given family of subsets whether dominance and independence or dominance and strict independence are jointly satisfiable for all linear orders on the objects if the lifted order needs to be total. Furthermore, we show that the problem remains coNP-complete if the lifted order can be incomplete. Additionally, we show that the complexity of these problem can increase exponentially if the family of sets is not given explicitly but via a succinct domain restriction. Finally, we show that it is NP-complete to decide for family of subsets whether dominance and independence or dominance and strict independence are jointly satisfiable for at least one linear orders on the objects.
翻译:取消对一组物品的偏好命令而取消对一组物品的偏好命令的问题,是该组物品的多种应用都存在的一个根本问题。 这一过程往往以对取消命令所应具备的属性的暗喻为指南。 Kannai和Peleg以及Barberçáa和Pattanaik的众所周知的不可能结果告诉我们, 某些可取的暗喻( 即支配地位和( 严格) 独立性), 对于任何对象的直线命令来说, 如果要订购所有非空的物品组, 则是一个根本性的问题。 另一方面, 如果不是所有非空的物品组的固定独立, 则该暗喻对于某些组物品的直线性命令是共同可置信的。 这些家庭对于使用解除命令非常重要, 例如在轮椅投票中, 我们决定承认这些家庭的计算复杂性是不可合并的。 我们表明, 在一个子类物品的直线性命令是否直线性、 独立或直线性命令是完全的直线性命令, 我们才能显示, 平坦的顺序是完全的。