I have a data set of players' skill ranking, age and sex and would like to create evenly matched teams.
I would like to try this in Haskell but the choice of coding language is the least important aspect of this problem.
This is a bin packing problem, or a multi-dimensional knapsack problem. Björn B. Brandenburg has made a bin packing heuristics library in Haskell that you may find useful.
You need something like...
data Player = P { skill :: Int, gender :: Bool, age :: Int }
Decide on a number of teams n (I'm guessing this is a function of the total number of players).
Find the desired total skill per team:
teamSkill n ps = sum (map skill ps) / n
Find the ideal gender ratio:
genderRatio ps = sum (map (\x -> if gender x then 1 else 0)) / length ps
Find the ideal age variance (you'll want the Math.Statistics package):
ageDist ps = pvar (map age ps)
And you must assign the three constraints some weights to come up with a scoring for a given team:
score skillW genderW ageW team = skillW * sk + genderW * g + ageW * a
where (sk, (g, a)) = (teamSkill 1 &&& genderRatio &&& ageDist) team
The problem reduces to the minimization of the difference in scores between teams. A brute force approach will take time proportional to Θ(nk−1). Given the size of your problem (8 teams of 12 players each), this translates to about 6 to 24 hours on a typical modern PC.
EDIT
An approach that may work well for you (since you don't need an exact solution in practise) is simulated annealing, or continual improvement by random permutation:
Given the number of players per team and the gender ration (which you can easily compute). The remaining problem is called n-partition problem, which is unfortunately NP-complete and thus very hard to solve exactly. You will have to use approximative or heuristic allgorithms (evolutionary algorithms), if your problem size is too big for a brute force solution. A very simple approximation would be sorting by age and assign in an alternating way.
You can tweak the skill level, gender, and age point values to change the distribution of each.
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