Group list by equivalence relation

Question

I have a equivalence relation R on a set A. How can I build equivalence classes on A? It's something like groupBy do, but between all the elements, not only neighbors.

For example, equal is equivalence relation (it is reflexive, symmetric and transitive binary relation):

type Sometuple = (Int, Int, Int)

equal :: Sometuple -> Sometuple -> Bool
equal (_, x, _) (_, y, _) = x == y

It is actually a predicate that connect 2 Sometuple elements.

λ> equal (1,2,3) (1,2,2)
True

So, how can I build all equivalence classes on [Sometuple] based on equal predicate? Something like that:

equivalenceClasses :: (Sometuple -> Sometuple -> Bool) -> [Sometuple] -> [[Sometuple]]
λ> equivalenceClasses equal [(1,2,3), (2,1,4), (0,3,2), (9,2,1), (5,3,1), (1,3,1)]
[[(1,2,3),(9,2,1)],[(2,1,4)],[(0,3,2),(5,3,1),(1,3,2)]]

Answer 1

If you can define a compatible ordering relation, you can use

equivalenceClasses equal comp = groupBy equal . sortBy comp

which would give you O(n*log n) complexity. Without that, I don't see any way to get better complexity than O(n^2), basically

splitOffFirstGroup :: (a -> a -> Bool) -> [a] -> ([a],[a])
splitOffFirstGroup equal xs@(x:_) = partition (equal x) xs
splitOffFirstGroup _     []       = ([],[])

equivalenceClasses _     [] = []
equivalenceClasses equal xs = let (fg,rst) = splitOffFirstGroup equal xs
                              in fg : equivalenceClasses equal rst