Stable matching problem: Difference between revisions

Revision as of 15:37, 6 May 2006

Revision as of 15:38, 6 May 2006

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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.

Proof of stability: 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.