Stable marriage problem

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In mathematics, the stable marriage problem (SMP) is the problem of finding a stable matching — a matching in which no element of the first matched set prefers an element of the second matched set that also prefers the first element.

It is commonly stated as:

Given n men and n women, where each person has ranked all members of the opposite sex with a unique number between 1 and n in order of preference, marry the men and women off such that there are no two people of opposite sex who would both rather have each other than their current partners. If there are no such people, all the marriages are "stable".

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[edit] Solution

In 1962, David Gale and Lloyd 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 "iterations") where each unengaged man "proposes" to the most-preferred woman to whom he has not yet proposed, 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). Note that only women can switch partners to increase their happiness.

This algorithm guarantees that:

  • Everyone gets married: Once a woman becomes engaged, she is always engaged to someone. So, at the end, there cannot be a man and a woman both unengaged, as he must have proposed to her at some point (since a man will eventually propose to everyone, if necessary) and, being unengaged, she would have to have said yes.
  • The marriages are stable: Let Alice be a woman and Bob be a man. They are each paired/partnered/married, but not to each other. Upon completion of the algorithm, it is not possible for both Alice and Bob to prefer each other over their current partners. 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, and therefore doesn't like Bob more than her current partner. If Alice rejected his proposal, she was already with someone she liked more than Bob.

[edit] Algorithm

function stableMatching {
    Initialize all m \in M and w \in W to free
    while \exists free man m who still has a woman w to propose to {
       w = m's highest ranked such woman
       if w is free
         (m, w) become engaged
       else some pair (m', w) already exists
         if w prefers m to m'
           (m, w) become engaged
           m' becomes free
         else
           (m', w) remain engaged
    }
}

[edit] Gale-Shapley pairing

The pairing generated by the Gale-Shapley algorithm has a number of interesting properties. In particular, one may prove that, in some sense, the Gale-Shapley pairing, in the form presented here, is male-optimal and female-pessimal (it would be the reverse, of course, if the roles of "male" and "female" participants in the algorithm were interchanged). 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.

Proof that the Gale-Shapley pairing is female-pessimal: We can show this using the fact a male-optimal pairing must be female-pessimal. Consider any woman W paired to a man M in the male-optimal stable pairing G. Also consider an arbitrary stable pairing S, where M and W are not paired. Because G is male-optimal, M prefers W to his match in S. Because S is stable, W must prefer her match in S to M. Thus women will always prefer another match in any stable pairing over their match in a male-optimal pairing. Thus, the Gale-Shapley pairing is female-pessimal because it is male-optimal (see above).

[edit] Applications

There are a number of real world applications for the Gale-Shapley solution. For example, their algorithm has been used to pair graduating medical students with hospital jobs. [1]

[edit] Similar problems

The weighted matching problem is to find a matching in a weighted bipartite graph that has maximum weight. Maximum weighted matchings do not have to be stable, but in some applications a maximum weighted matching is better than a stable one.

[edit] References

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