The Tournament Equilibrium Set Is PSPACE-Complete, Even under Hereditary Uniqueness
This paper proves that the problem of deciding whether a given alternative belongs to the tournament equilibrium set (TEQ) is PSPACE-complete, and studies the complexity of the problem under different conditions.
This paper provides the first PSPACE-completeness result for membership in a tournament solution.
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Abstract
More Like ThisThe tournament equilibrium set (TEQ), proposed by Schwartz in 1990, is a classical voting rule based on pairwise majority comparisons. Its definition is recursive: an alternative is judged by the alternatives that beat it, called its dominators, and by which of those dominators TEQ itself selects. Deciding whether a given alternative belongs to TEQ was known to be NP-hard and to lie in PSPACE. We close this gap by proving that the problem is PSPACE-complete. To our knowledge, this is the first PSPACE-completeness result for membership in a tournament solution. The hardness persists even when every nonempty subtournament has a unique minimal retentive set, the property conjectured by Schwartz to hold for all tournaments. Moreover, every such subtournament in our hard instances contains an alternative with at most three dominators. We prove that a single alternative with at most four dominators suffices both to guarantee uniqueness and to find a TEQ winner in $O(N^2)$ time. Thus, on our hard instances, finding one winner is efficient even though recognizing a specified winner is PSPACE-complete. We also study how the complexity changes with the dominator bound. If every nonempty subtournament has an alternative with at most one dominator, the entire TEQ can be computed in quadratic time; allowing two dominators makes membership NP-hard. Under the same condition with four dominators, all $t$ winners can be computed in $O(2^t N^2)$ time, where $N$ is the number of alternatives.