Stable matching problem: Difference between revisions

Revision as of 23:15, 6 November 2004

Revision as of 07:55, 19 December 2004

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In 1962, Gale and Shapley proved that, for any equal number of men and women, it is always possible to solve the SMP and make all marriages stable. They presented an [[algorithm]] to do so.

The Gale-Shapley algorithm involves a number of "rounds" (or "[[iteration]]s") where each unengaged man "proposes" to the most-preferred woman thatto whom he has not yet proposed to, and she accepts or rejects him based on whether she is already engaged to someone she prefers. If she is unengaged, or engaged to a man lower down her preference list than her new suitor, she accepts the proposal (and in the latter case, the other man becomes unengaged again).

This algorithm guarantees that:

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'''The marriages are stable:''' At the end, it is not possible for a man and a woman (call them Alice and Bob) to both prefer each other to their current partners. This is easy to show. If Bob prefers Alice to his current partner, he must have proposed to Alice before he proposed to his current partner. If Alice accepted his proposal, yet is not married to him at the end, she must have dumped him for someone she likes more (since women only switch partners to increase their happiness). If Alice rejected his proposal, she was already with someone she liked more than Bob.

==Properties of the Gale-Shapley pairing==

The pairing generated by the Gale-Shapley algorithm has a number of interesting properties. In particular, we can show that, in some sense, the Gale-Shapley pairing is '''male-optimal''' and '''female-pessimal'''. To see this, consider the definition of a '''feasible marriage'''. We say that the marriage between man A and woman B is feasible if there exists a stable pairing in which A and B are married. When we say a pairing is male-optimal, we mean that every man is paired with his highest ranked feasible partner. Similarly, a female-pessimal pairing is one in which each female is paired with her lowest ranked feasible partner.

'''Proof that the Gale-Shapley pairing is male-optimal:''' We go by contradiction. Suppose the pairing generated by the Gale-Shapley algorithm were not male-optimal. Let T be the first iteration where a man is rejected by his optimal partner. Let man M be one man who is rejected by his optimal partner, woman W, during this iteration. Then woman W must have rejected man M for some other man, man Z. Further, since this is the first iteration where a man is rejected by his optimal partner, woman W must be at least as high on man Z's preference list as his optimal partner. Also note that, since woman W is man M's optimal partner, there exists some stable pairing S where man M is paired with woman W. Then, in pairing S, woman W prefers man Z to man M (since she rejected M for Z in the Gale-Shapley algorithm) and man Z prefers woman W to his partner in S, since he likes woman W at least as much as his optimal partner and, by our definition of optimal, he could not be paired with anyone better than this. But then, by definition, this pairing is unstable, since man Z and woman W prefer each other to their partners in S. We have reached a contradiction, so the Gale-Shapley algorithm must produce a male-optimal pairing. <math>\Box</math>

==Real world applications==