given is the following algorithm: we choose an initial assignment of man^woman, so that every man is engaged to exactly one woman and vice versa. now if there's a man m that prefers a woman f' to his actual partner f, and a woman f that prefers man m to her actual partner m', then m and f get engaged, and m' and f' get engaged. this is repeated until no changes occur. now i'm looking for a case with 3 men and 3 women that never ends. what would the preferences of every man and woman has to be so that the algorithm never ends? Thanks for your time!
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What did you try? It seems like there's not many possibilities to try. – Paul Hankin Sep 23 '17 at 07:02
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i've been using brute force but I don't seem to come to a result. what have you been thinking of? – Cherry Sep 23 '17 at 07:03
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I can't imagine how it should work, for me it seems logical it to end after at most 2 changes, because one a person is with their preferred person, they won't accept any other partner. any change in partner includes the person to go "one person up in their preferences" so this can only happen at most twice until they reach their top choice... – Cherry Sep 23 '17 at 08:20
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If each man prefers woman `a` to woman `b` and woman `b` to woman `c` AND prefers woman `c` to woman `a`, and if each woman has a similar set of preferences it is trivial to find unstable arrangements which never settle to a stable arrangement. The problem statement (real life too) does not require that partner-preferences is a *transitive* relation. – High Performance Mark Sep 23 '17 at 10:21
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I think I misunderstand, if I do it your way I don't get a loop.. could you explain a bit further please? Would be really kind! – Cherry Sep 23 '17 at 22:13