permute package:combinatorial

permute :: [a] -> [[a]]
combinatorial Combinatorics
Generate list of all permutations of the input list. The list is sorted lexicographically. 
>>> Comb.permute "abc"
["abc","acb","bac","bca","cab","cba"]

>>> Comb.permute "aabc"
["aabc","aacb","abac","abca","acab","acba","aabc","aacb","abac","abca","acab","acba","baac","baca","baac","baca","bcaa","bcaa","caab","caba","caab","caba","cbaa","cbaa"]

QC.forAll (take 6 <$> QC.arbitrary :: QC.Gen [Int]) $ \xs -> allEqual $ map (\p -> sort (p xs)) $ Comb.permute : Comb.permuteFast : Comb.permuteShare : []
permuteFast :: [a] -> [[a]]
combinatorial Combinatorics
Generate list of all permutations of the input list. It is not lexicographically sorted. It is slightly faster and consumes less memory than the lexicographical ordering permute.
permuteRep :: [(a, Int)] -> [[a]]
combinatorial Combinatorics
>>> Comb.permuteRep [('a',2), ('b',1), ('c',1)]
["aabc","aacb","abac","abca","acab","acba","baac","baca","bcaa","caab","caba","cbaa"]

QC.forAll (genPermuteRep  7) $ \xs -> let perms = Comb.permuteRep $ Key.nub fst xs in perms == nub perms

QC.forAll (genPermuteRep 10) $ \xs -> let perms = Comb.permuteRep $ Key.nub fst xs in List.sort perms == Set.toList (Set.fromList perms)

QC.forAll (genPermuteRep 10) $ isAscending . Comb.permuteRep . Key.nub fst . sort

QC.forAll (QC.choose (0,10)) $ \n k -> Comb.choose n k == Comb.permuteRep [(False, n-k), (True, k)]
permuteShare :: [a] -> [[a]]
combinatorial Combinatorics
All permutations share as much suffixes as possible. The reversed permutations are sorted lexicographically.