[Haskell-cafe] Fixed points
alex.solla at gmail.com
Sat Jun 11 21:21:47 CEST 2011
On Sat, Jun 11, 2011 at 3:19 AM, Max Rabkin <max.rabkin at gmail.com> wrote:
> On Fri, Jun 10, 2011 at 21:05, Alexander Solla <alex.solla at gmail.com>
> > equivalenceClosure :: (Ord a) => Relation a -> Relation a
> > equivalenceClosure = fix (\f -> reflexivity . symmetry . transitivity)
> If you want to learn about fix, this won't help you, but if you're
> just want the best way to calculate equivalence closures of relations,
> then it's probably
> equivalenceClosure = transitivity . symmetry . reflexivity
> assuming those are the transitive, symmetric and reflexive closure
> functions. You still need some kind of iteration to get the transitive
> closure. The algorithm I know of for that is Warshall's Algorithm,
> which is O(N^3) (possibly with a log N factor for pure data
Cool, thanks for the suggestion. I was iterating all of them, since an
iteration of "transitive" introduces new pairs to the relation (which are
not guaranteed to have symmetric "complements" in my implementation). I
suppose I can get away with not iterating "reflexive", for something like an
O(n) speed up for each iteration.
This is a summary of the code. I haven't done order analysis on it.
Relation is a newtype over a Set of pairs:
-- | Iterate 'transitivity' to compute the transitive closure for a
transitivity :: (Ord a) => Relation a -> Relation a
transitivity (Relation set) = Relation $ (Set.fold _joinOn set) (set)
-- | Compute the reflexive closure for a relation. In other words, take a
-- containing @(a,b)@, @(c,d)@, ... into one containing the originals and
-- @(b,a)@, @(d,c)@, and so on.
reflexivity :: (Ord a) => Relation a -> Relation a
reflexivity (Relation set) = Relation $ Set.unions [ set
, (Set.map (\(x,_) ->
, (Set.map (\(_,y) ->
-- | Compute the symmetric closure for a relation.
symmetry :: (Ord a) => Relation a -> Relation a
symmetry (Relation set) = Relation $ Set.union set (Set.map _symmetry set)
_symmetry :: (a, a) -> (a, a)
_symmetry (a, b) = (b, a)
_joinOn :: (Ord a) => (a,a) -> Set (a,a) -> Set (a,a)
_joinOn (a,b) set =
let fst' = Set.filter ((b ==) . fst) $ set
snd' = Set.filter ((a ==) . snd) $ set
in Set.unions [ set
, Set.map (\(x,y) -> (a,y)) fst'
, Set.map (\(x,y) -> (x,b)) snd'
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