2007
DOI: 10.1145/1273340.1273346
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Improved approximation results for the stable marriage problem

Abstract: The stable marriage problem has recently been studied in its general setting, where both ties and incomplete lists are allowed. It is NP-hard to find a stable matching of maximum size, while any stable matching is a maximal matching and thus trivially we can obtain a 2-approximation algorithm.In this article, we give the first nontrivial result for approximation of factor less than two. Our algorithm achieves an approximation ratio of 2/(1 + L −2 ) for instances in which only men have ties of length at most L.… Show more

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Cited by 62 publications
(56 citation statements)
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“…For an instance I of SMTI, let opt(I) denote the maximum size of a weakly stable matching in I. Halldorsson et al [6] showed [in the proof of Corollary 3.4] that given an instance I of SMTI of size n, where only one side of the market has agents with indifferences and each of these agents has a single tie of size two, and any arbitrary small positive , it is NP-hard to distinguish between the following two cases: (1) opt(I) ≥ 21− 27 n (2) opt(I) < 19+ 27 n. When choosing so that 0 < < 1 2 we can simplify the above cases to (1) opt(I) > 41 54 n, since opt(I) ≥ 21− 27 n > 41 54 n and (2) opt(I) < 39 54 n, since opt(I) < 19+ 27 n < 39 54 n. Therefore, the number of agents left unmatched on either side of the market is less than 13 54 n in the first case and more than 15 54 n in the second case. Let us now extend instance I to a larger instance of SMTI I as follows.…”
Section: Theoremmentioning
confidence: 99%
See 1 more Smart Citation
“…For an instance I of SMTI, let opt(I) denote the maximum size of a weakly stable matching in I. Halldorsson et al [6] showed [in the proof of Corollary 3.4] that given an instance I of SMTI of size n, where only one side of the market has agents with indifferences and each of these agents has a single tie of size two, and any arbitrary small positive , it is NP-hard to distinguish between the following two cases: (1) opt(I) ≥ 21− 27 n (2) opt(I) < 19+ 27 n. When choosing so that 0 < < 1 2 we can simplify the above cases to (1) opt(I) > 41 54 n, since opt(I) ≥ 21− 27 n > 41 54 n and (2) opt(I) < 39 54 n, since opt(I) < 19+ 27 n < 39 54 n. Therefore, the number of agents left unmatched on either side of the market is less than 13 54 n in the first case and more than 15 54 n in the second case. Let us now extend instance I to a larger instance of SMTI I as follows.…”
Section: Theoremmentioning
confidence: 99%
“…For the purpose of this paper we assume that the preference lists are complete, i.e., each agent finds each member of the opposite side acceptable. 6 In the stable marriage problem the goal is to compute a stable matching; a matching where no two agents prefer to be matched to each other rather than be matched to their current partners. Unlike most of the literature on stable matching problems [5,10,12], we assume that men and women may have uncertainty in their preferences which can be captured by various probabilistic uncertainty models.…”
Section: Introductionmentioning
confidence: 99%
“…Predictably, however, this is an intractable problem. Irving and Leather [12] have shown that the maximum number of stable matches for the simple marriage matching problem grows exponentially in n (see also [10,11]). Further, they provide a lower bound on the maximum by problem size.…”
Section: Stop Eachmentioning
confidence: 99%
“…When ties are allowed on both sides, the problem is NP-hard to approximate within 33/29 [23] and the currently best known ration is 3/2 [20]. When ties are only allowed on the side of the women the problem is NP-hard to approximate within 21/19 [11] and the currently best known ratio is 25/17 [14].…”
Section: Introductionmentioning
confidence: 99%