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Jack Edmonds

Jack Edmonds. Jack Edmonds. Stable marriage. Network Algorithms 2005. The stable marriage problem. Story: there are n men and n women, which are unmarried. Each has a preference list on the persons of the opposite sex

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Jack Edmonds

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  1. Jack Edmonds

  2. Jack Edmonds

  3. Stable marriage Network Algorithms 2005

  4. The stable marriage problem • Story: there are n men and n women, which are unmarried. Each has a preference list on the persons of the opposite sex • Does there exist and can we find a stable matching (stable marriage): a matching of men and women, such that there is no pair of a man and a woman who both prefer each other above their partner in the matching?

  5. Application • Origin: assignment of medical students to hospitals (for internships) • Students list hospitals in order of preference • Hospitals list students in order of preference

  6. M1: 2 1 3 M2 1 3 2 M3 1 3 2 W1 2 1 3 W2 1 3 2 W3 2 1 3 Stable matching: (1,2), (2,1), (3,3) Matching (1,1), (2,2), (3,3) is not stable, e.g., M1 and W2 prefer each other above given partner Example

  7. Result • Gale/Stanley algorithm: finds always a stable matching • Input: list of men, women, and their preference list • Output: stable matching

  8. The algorithm • Fix some ordering on the men • Repeat until everyone is matched • Let X be the first unmatched man in the ordering • Find woman Y such that Y is the most desirable woman in X’s list such that Y is unmatched, or Y is currently matched to a Z and X is more preferable to Y than Z. • Match X and Y; possible this turns Z to be unmatched Questions: Does this terminate? How fast? Does this give a stable matching?

  9. Termination and number of steps • Once a woman is matched, she stays matched (her partner can change). • When the partner of a woman changes, this is to a more preferable partner for her: at most n – 1times. • Every step, either an unmatched woman becomes matched, or a matched woman changes partner: at most n2steps.

  10. Suppose final matching is not stable. Take: Mx is matched to Wx, My is matched to Wy, Mx prefers Wy to Wx, Wy prefers Mx to My. When Wy is before Wx in the preference list of Mx, but Mx is not matched to Wy. Two cases: When Mx considers Wy, she has a partner Mz preferable to Mx: Mz is also preferable to My, but in the algorithm woman can get only more preferable partners, contradiction. When Mx considers Wy, she is free, but Mx is later replaced by someone preferable to Mx. Again, Wy can never end up with My. Stability of final matching

  11. Final comments • A stable matching exists and can be found in polynomial time • Controversy: the algorithm is better for the men: hospitals in the application

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