2009
DOI: 10.1007/s10479-009-0648-7
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An updated survey on the linear ordering problem for weighted or unweighted tournaments

Abstract: In this paper, we survey some results, conjectures and open problems dealing with the combinatorial and algorithmic aspects of the linear ordering problem. This problem consists in finding a linear order which is at minimum distance from a (weighted or not) tournament. We show how it can be used to model an aggregation problem consisting of going from individual preferences defined on a set of candidates to a collective ranking of these candidates.

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Cited by 62 publications
(44 citation statements)
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References 201 publications
(223 reference statements)
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“…The extension of the method of Kemeny to rankings with ties (and to tournaments and to binary relations) is a common problem in social choice [8,21,30,31], usually referred to as the search for 'median orders' or 'median relations'. These extensions are obtained by allowing the relations in the profile to be rankings with ties (or tournaments, or binary relations) or by allowing the winning relations to be rankings with ties (or tournaments, or binary relations).…”
Section: Definition 841mentioning
confidence: 99%
“…The extension of the method of Kemeny to rankings with ties (and to tournaments and to binary relations) is a common problem in social choice [8,21,30,31], usually referred to as the search for 'median orders' or 'median relations'. These extensions are obtained by allowing the relations in the profile to be rankings with ties (or tournaments, or binary relations) or by allowing the winning relations to be rankings with ties (or tournaments, or binary relations).…”
Section: Definition 841mentioning
confidence: 99%
“…There are many publications addressing this problem, for a review of the literature see, for example, Charon & Hudry (2010).…”
Section: Two Measures Of the Dependence Of Preferential Rankings On Cmentioning
confidence: 99%
“…Finding these linear orders is equivalent to solving an instance of the NP-complete problem feedback arc set [1,14,20], which implies that checking membership in the Slater set is NP-hard [16]. Yet, there are implementations that are sufficiently fast on small instances (e.g., [17]).…”
Section: Top Cycle a Nonempty Subset Of Alternativesmentioning
confidence: 99%