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I was reading up on the stable marriage problem and chanced upon this question: Is it possible that a man1 has woman1 on the top of his preference list, and woman1 has man1 on top of hers, but still there exists a stable matching (not necessarily man optimal or woman optimal) where they are not paired together?

Anannya Uberoi
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  • what order are pairs matched in? i think either may get "claimed" (put in a marriage with non-optimal mate) if the ordering of the pairings allows for that to happen – SlimsGhost Jan 27 '17 at 18:20
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    Reread the definition of a stable matching. You'll find that the answer to your question follows immediately from the definition. – user2357112 Jan 27 '17 at 18:20
  • this is not really a programming problem, by the way. maybe on http://stats.stackexchange.com/ or another SE forum, but i like the question! I also realize now that i am not too familiar with this particular problem and didn't realize there's a tag for it on SO, so maybe don't listen to me ;) – SlimsGhost Jan 27 '17 at 18:21
  • Actually from the definition it sounds like she's asking are there some solutions that are stable but not strongly stable – Gonen I Jan 27 '17 at 18:26

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The case you are describing sounds like one that is stable but not strongly stable.

From "The structure of stable marriage with indifference": A matching M is strongly stable if there is no couple (x,y) such that x strictly prefers y to his/her partner in M, and y either strictly prefers x to his/her partner ... For a given instance I of SMP, the existence of a weakly stable matching is guaranteed: ... On the other hand, it is straightforward to construct instances of SMT which admit no strongly stable matching.

So it sounds like the answer is yes... it is possible.

Gonen I
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