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A counterexample to a conjecture of Schwartz

Author(s): Brandt, Felix; Chudnovsky, Maria; Kim, Ilhee; Liu, Gaku; Norin, Sergey; et al

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dc.contributor.authorBrandt, Felix-
dc.contributor.authorChudnovsky, Maria-
dc.contributor.authorKim, Ilhee-
dc.contributor.authorLiu, Gaku-
dc.contributor.authorNorin, Sergey-
dc.contributor.authorScott, Alex-
dc.contributor.authorSeymour, Paul D.-
dc.contributor.authorThomassé, Stéphan-
dc.date.accessioned2018-07-20T15:08:24Z-
dc.date.available2018-07-20T15:08:24Z-
dc.date.issued2013-03en_US
dc.identifier.citationBrandt, Felix, Chudnovsky, Maria, Kim, Ilhee, Liu, Gaku, Norin, Sergey, Scott, Alex, Seymour, Paul, Thomassé, Stephan. (2013). A counterexample to a conjecture of Schwartz. Social Choice and Welfare, 40 (3), 739 - 743. doi:10.1007/s00355-011-0638-yen_US
dc.identifier.issn0176-1714-
dc.identifier.urihttp://arks.princeton.edu/ark:/88435/pr1h96s-
dc.description.abstractIn 1990, motivated by applications in the social sciences, Thomas Schwartz made a conjecture about tournaments which would have had numerous attractive consequences. In particular, it implied that there is no tournament with a partition A, B of its vertex set, such that every transitive subset of A is in the out-neighbour set of some vertex in B, and vice versa. But in fact there is such a tournament, as we show in this article, and so Schwartz’ conjecture is false. Our proof is non-constructive and uses the probabilistic method.en_US
dc.format.extent739 - 743en_US
dc.language.isoen_USen_US
dc.relation.ispartofSocial Choice and Welfareen_US
dc.rightsAuthor's manuscripten_US
dc.titleA counterexample to a conjecture of Schwartzen_US
dc.typeJournal Articleen_US
dc.identifier.doidoi:10.1007/s00355-011-0638-y-
dc.date.eissued2012-01-17en_US
dc.identifier.eissn1432-217X-
pu.type.symplectichttp://www.symplectic.co.uk/publications/atom-terms/1.0/journal-articleen_US

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