The Stable Marriage Problem was originally described by David Gale and Lloyd Shapley in their 1962 paper, "College Admissions and the Stability of Marriage". They describe the problem as follows:
A certain community consists of n men and n women. Each person ranks those of the opposite sex in accordance with his or her preferences for a marriage partner. We seek a satisfactory way of marrying off all member of the community. We call a set of marriage unstable if under it there are a man and a woman who are not married to each other, but prefer each other to their actual mates.
Gale and Shapley shows that for any pattern of preferences it's possible to find a stable set of marriages.
On its own, this doesn't sound very interesting. However, bringing together resources is an important economic principle and this work formed part of the puzzle of Cooperative Game Theory and Shapley was jointly awarded the Nobel Prize for economics in 2012.
So how does the algorithm for Stable Marriages work?
Let's start by defining the problem. Given two lists of preferences, find the match such that there is no unstable match (that is two pairs that would cooperatively trade partners to make each other better off). The only constraint the types have is that they have is that they are equatable. This isn't the ideal representation (to put it mildly) in a strongly typed language (it doesn't enforce any invariants about the structure of the lists), but it's probably the simplest representation for explaining the algorithm.
stableMatch :: (Eq m, Eq w) => [(m,[w])] -> [(w,[m])] -> [(m,w)]
The algorithm continues whilst there are any unmarried men. If there are no unmarried men, then the algorithm terminates.
stableMatch :: (Eq m, Eq w) => [(m,[w])] -> [(w,[m])] -> [(m,w)] stableMatch ms ws = stableMatch'  where stableMatch' ps = case unmarried ms ps of Just unmarriedMan -> stableMatch' (findMatch unmarriedMan ws ps) Nothing -> ps unmarried :: Eq m => [(m,[w])] -> [(m,w)] -> Maybe (m,[w]) unmarried ms ps = find (\(m,_) -> m `notElem` engagedMen) ms where engagedMen = map fst ps
If there is at least one unmarried man, then we need to find a match. We do this by proposing to each of his preferences in turn. If his first preference is not engaged, then we propose. Otherwise, if his potential partner is already engaged and would prefer him then this violates the stable marriage principle and we breakup the engagement and re-engage with our first choice.
findMatch :: (Eq m,Eq w) => (m,[w]) -> [(w,[m])] -> [(m,w)] -> [(m,w)] findMatch (m,w:rest) ws ps = case isEngaged w ps of -- w is already engaged to m' - is there a better match? Just m' -> if prefers (getPrefs ws w) m m' then engage (breakup m' ps) m w else findMatch (m,rest) ws ps -- can match with first choice Nothing -> engage ps m w
You can see the full code at Stable Marriage Problem. As always flames, comments and critiques gratefully received.