Consider the following Haskell function which enumerates permutations of a given length:
> {-# LANGUAGE GADTs #-}
> {-# LANGUAGE PolyKinds #-}
> import Data.List
> perms 0 = [[]]
> perms n = [ insertAt i n p | p <- perms (n - 1), i <- [0..n-1] ]
> insertAt 0 x xs = x:xs
> insertAt n x (x':xs) = x':xs'
> where xs' = insertAt (n - 1) x xs
The goal of this post is to derive a partial inverse indexOfPerm to the indexing function perms n !! as an exercise in equational reasoning in Haskell. That is, we seek a function such that for all i:
indexOfPerm :: Int -> [Int] -> Int
indexOfPerm n (perms n !! i) = i
This will serve as a specification of the function indexOfPerm.
Expanding the definition of perms at zero gives the following:
indexOfPerm 0 []
{ Definition of perms }
= indexOfPerm 0 (perms 0 !! 0)
{ By assumption }
= 0
Expanding the recursive definition of perms gives the following:
indexOfPerm n (perms n !! (j * n + k))
{ Definition of perms }
= indexOfPerm n (insertAt j n (perms (n - 1) !! k))
{ Let xs = insertAt j n (perms (n - 1) !! k) }
= indexOfPerm n xs
{ By assumption }
= j * n + k
Suppose we can find a function extract which satisfies the following:
extract :: Int -> [Int] -> (Int, [Int])
extract n (insertAt i n xs) = (i, xs)
Now calculate as follows:
extract n xs
{ Definition of xs }
= extract n (insertAt j n (perms (n - 1) !! k) )
{ Definition of extract }
= (j, perms (n - 1) !! k)
{ Let xs' = perms (n - 1) !! k) }
so that k = indexOfPerm (n - 1) xs' }
= (j, xs')
indexOfPerm n xs
{ From earlier }
= j * n + k
{ Expressing k in terms of xs' }
= j * n + indexOfPerm (n - 1) xs'
We can now define indexOfPerm as follows:
> indexOfPerm 0 [] = 0
> indexOfPerm n xs = n * (indexOfPerm (n - 1) xs') + j
> where (j, xs') = extract n xs
It remains to compute the function extract. Expanding the definition of insertAt at zero gives:
extract x (x:xs)
{ Definition of insertAt }
= extract x (insertAt 0 xs)
{ By assumption }
= (0, xs)
Expanding the recursive definition of insertAt gives:
extract x (x':xs)
{ Assume xs = insertAt i x xs'
so that (i, xs') = extract x xs }
= extract x (x':insertAt i x xs')
{ Definition of insertAt }
= extract x (insertAt (i + 1) x x':xs'))
{ By assumption }
= (i + 1, x':xs')
Now we can define extract as follows:
> extract x (x':xs) | x == x' = (0, xs)
> | otherwise = (i + 1, x':xs')
> where (i, xs') = extract x xs
One can check that the relation that we are interested in between insertAt and extract actually holds.
Generating Permutations
We can now combine perms and indexOf to give a function nextPerm which generates the next permutation in the list perms n:
> fact 0 = 1
> fact n = n * fact (n - 1)
> nextPerm' n xs = perms n !! ((1 + indexOfPerm n xs) `mod` (fact n))
However, we can rewrite this function by fusing the definition of perms with the definition of indexOfPerm:
nextPerm 0 []
{ Definition of nextPerm }
= perms 0 !! ((1 + indexOfPerm 0 []) mod (fact 0))
{ Definition of indexOfPerm }
= perms 0 !! (1 mod 1)
{ Definition of perms }
= []
The recursive case is only slightly more tricky. We divide into two cases.
nextPerm n xs
{ Definition of nextPerm }
= perms n !! ((1 + indexOfPerm n xs) mod (fact n))
{ Let (j, xs') = extract n xs }
= [ insertAt i n p | p <- perms (n - 1), i <- [0..n- 1] ] !! ((n * (indexOfPerm (n - 1) xs') + j + 1) mod (n * fact (n - 1)))
The value j is the index of n in xs, so that 0 < j < n. The first case is j < n - 1:
nextPerm n xs
{ Assume j < n - 1 }
= insertAt (j + 1) n (perms (n - 1) !! (indexOfPerm (n - 1) xs'))
{ By earlier assumption }
= insertAt (j + 1) n xs'
The second case is when j = n - 1:
nextPerm n xs
{ Assume j = n - 1 }
= [ insertAt i n p | p <- perms (n - 1), i <- [0..n-1] ] !! ((n * (1 + indexOfPerm (n - 1) xs')) mod (n * fact (n - 1)))
{ By earlier assumption }
= insertAt 0 n (perms (n - 1) !! (1 + indexOfPerm (n - 1) xs'))
{ Definition of nextPerm }
= insertAt 0 n (nextPerm (n - 1) xs')
Thus we arrive at our final definition of nextPerm:
> nextPerm 0 [] = []
> nextPerm n xs | j == n - 1 = insertAt 0 n (nextPerm (n - 1) xs')
> | otherwise = insertAt (j + 1) n xs'
> where (j, xs') = extract n xs
A Better Data Structure
The inverse indexOfPerm is only a partial function, because perms n returns a collection of lists of size n. In addition, the types of lists does not enforce the invariant that each element perms n is a permutation of [1..n].
Using the -XPolyKinds GHC extension, we can express a type of permutations, indexed by size, allowing us to strengthen the type of nextPerm, specifying that nextPerm preserves the size of a permutation.
The following type definition will be lifted to the kind level, generating two constructors Z :: Nat and S :: Nat -> Nat
> data Nat = Z | S Nat
> _1 = S Z
> _2 = S $ S Z
> _3 = S $ S $ S Z
> _4 = S $ S $ S $ S Z
> showNat :: Nat -> String
> showNat n = show (show' n 0) where
> show' :: Nat -> Int -> Int
> show' Z m = m
> show' (S n) m = show' n (m + 1)
> instance Show Nat where
> show = showNat
The type Leq n of natural numbers less than or equal to n. The type is parameterised over the kind Nat.
> data Leq :: Nat -> * where
> LeqZero :: Leq n
> LeqSucc :: Leq n -> Leq (S n)
We can embed numbers less than or equal to n into numbers less than or equal to n + 1 for every n:
> embed :: Leq n -> Leq (S n)
> embed LeqZero = LeqZero
> embed (LeqSucc n) = LeqSucc (embed n)
We can convert to and from regular integers:
> leqToInt :: Leq n -> Int
> leqToInt LeqZero = 0
> leqToInt (LeqSucc n) = 1 + leqToInt n
> intToLeq :: Int -> EqNat n -> Leq n
> intToLeq 0 n = LeqZero
> intToLeq n (EqSucc m) = LeqSucc (intToLeq (n - 1) m)
> showLeq :: Leq n -> String
> showLeq n = show (show' n 0) where
> show' :: Leq n -> Int -> Int
> show' LeqZero m = m
> show' (LeqSucc n) m = show' n (m + 1)
> instance Show (Leq n) where
> show = showLeq
The type of natural numbers equal to n, that is, a singleton type for each natural number:
> data EqNat :: Nat -> * where
> EqZero :: EqNat Z
> EqSucc :: EqNat n -> EqNat (S n)
> eq1 = EqSucc EqZero
> eq2 = EqSucc $ EqSucc EqZero
> eq3 = EqSucc $ EqSucc $ EqSucc EqZero
> eq4 = EqSucc $ EqSucc $ EqSucc $ EqSucc EqZero
We can convert the sole inhabitant of each singleton type to its natural number representation:
> eqToInt :: EqNat n -> Int
> eqToInt EqZero = 0
> eqToInt (EqSucc n) = 1 + eqToInt n
> showEq :: EqNat n -> String
> showEq n = show (show' n 0) where
> show' :: EqNat n -> Int -> Int
> show' EqZero m = m
> show' (EqSucc n) m = show' n (m + 1)
> instance Show (EqNat n) where
> show = showEq
We will need the following helper method, which returns the value of n in the type of numbers less than or equal to n:
> maxLeq :: EqNat n -> Leq n
> maxLeq EqZero = LeqZero
> maxLeq (EqSucc n) = LeqSucc (maxLeq n)
We can turn collect the list of all numbers in Leq n recursively:
> for :: EqNat n -> [Leq n]
> for EqZero = [LeqZero]
> for (EqSucc n) = LeqZero : map LeqSucc (for n)
Finally, we define the type of permutations, again parameterised by the kind Nat and containing two type constructors: the empty permutation and the permutation obtained by inserting the
value n + 1 into a permutation of the list [1..n]:
> data Perm :: Nat -> * where
> Empty :: Perm Z
> Insert :: Leq n -> Perm n -> Perm (S n)
> showPerm :: Perm n -> String
> showPerm p = "(" ++ concat (intersperse "," (map show (toList p 0))) ++ ")" where
> toList :: Perm n -> Int -> [Int]
> toList Empty m = []
> toList (Insert n p) m = let (l, r) = splitAt (leqToInt n) (toList p (m + 1)) in l ++ [m] ++ r
> instance Show (Perm n) where
> show = showPerm
Note now that invalid permutations are no longer inhabitants of the type Perm n for any n: to insert value n into a permutation of [1..n-1], we have to specify a position to insert which is in the range [0..n], and this is enforced by the type Perm n! One cannot, for example, represent a list with a duplicate element - the elements are not even mentioned explicitly.
The rank of a permutation is the size of the set it permutes:
> rank :: Perm n -> EqNat n
> rank Empty = EqZero
> rank (Insert n p) = EqSucc (rank p)
The identity permutation is easily defined by recursion:
> identity :: EqNat n -> Perm n
> identity EqZero = Empty
> identity (EqSucc n) = Insert LeqZero (identity n)
The method perms translates easily to this new setting:
> perms1 :: EqNat n -> [Perm n]
> perms1 EqZero = [Empty]
> perms1 (EqSucc n) = [ Insert i xs | xs <- perms1 n, i <- for n ]
We can create a permutation from its list representation by repeatedly extracting the highest element:
> fromList :: EqNat n -> [Int] -> Perm n
> fromList EqZero [] = Empty
> fromList (EqSucc n) xs = Insert (intToLeq i n) (fromList n (map (flip (-) 1) (delete 0 xs)))
> where Just i = elemIndex 0 xs
We can also translate the function indexOfPerm without difficulty:
> indexOfPerm1 :: Perm n -> EqNat n -> Int
> indexOfPerm1 Empty EqZero = 0
> indexOfPerm1 (Insert n p) (EqSucc m) = (indexOfPerm1 p m) * (1 + eqToInt m) + leqToInt n
The following function emulates the indexing function perms r !!, returning the nth permutation in the set of permutations of a given rank:
> nth :: Int -> EqNat n -> Perm n
> nth 0 EqZero = Empty
> nth m (EqSucc n) = Insert (intToLeq k n) (nth j n)
> where (j, k) = divMod m (1 + eqToInt n)
As before, we can combine nth with indexOfPerm1 to step to the next permutation:
> nextPerm1' :: Perm n -> Perm n
> nextPerm1' p = let r = rank p in nth (indexOfPerm1 p r - 1) r
Finally, we can perform the same fusion as before, and express nextPerm1 directly without the need for helper functions nth and indexOfPerm1.
> nextPerm1 :: Perm n -> Perm n
> nextPerm1 Empty = Empty
> nextPerm1 (Insert LeqZero p) = Insert (maxLeq (rank p)) (nextPerm1 p)
> nextPerm1 (Insert (LeqSucc n) p) = Insert (embed n) p
Note here that we have also removed the dependence on the intermediate type Int, representing the index of the permutation, and we are left with a type which conveys some valuable information about the function nextPerm1:
nextPerm1 :: forall (n :: Nat). Perm n -> Perm n
That is, nextPerm1 preserves the rank of its argument.
Download the source or copy this literate Haskell file into a text editor, and compile with GHC 7.4.
Edit (12/29/2011) - Corrected the definition of fact.
Interesting. Unfortunately, the derivation derails from the point it defines fact 0 = 0. As it is, fact n == 0 for all n. For me, the giveaway that something is amiss was using mod 0 in the derivation.
ReplyDeleteThanks, I've made the correction.
ReplyDelete