Calculating permutations in F#

Question

Inspired by this question and answer, how do I create a generic permutations algorithm in F#? Google doesn't give any useful answers to this.

EDIT: I provide my best answer below, but I suspect that Tomas's is better (certainly shorter!)

Answer 1

My latest best answer

//mini-extension to List for removing 1 element from a list
module List = 
    let remove n lst = List.filter (fun x -> x <> n) lst

//Node type declared outside permutations function allows us to define a pruning filter
type Node<'a> =
    | Branch of ('a * Node<'a> seq)
    | Leaf of 'a

let permutations treefilter lst =
    //Builds a tree representing all possible permutations
    let rec nodeBuilder lst x = //x is the next element to use
        match lst with  //lst is all the remaining elements to be permuted
        | [x] -> seq { yield Leaf(x) }  //only x left in list -> we are at a leaf
        | h ->   //anything else left -> we are at a branch, recurse 
            let ilst = List.remove x lst   //get new list without i, use this to build subnodes of branch
            seq { yield Branch(x, Seq.map_concat (nodeBuilder ilst) ilst) }

    //converts a tree to a list for each leafpath
    let rec pathBuilder pth n = // pth is the accumulated path, n is the current node
        match n with
        | Leaf(i) -> seq { yield List.rev (i :: pth) } //path list is constructed from root to leaf, so have to reverse it
        | Branch(i, nodes) -> Seq.map_concat (pathBuilder (i :: pth)) nodes

    let nodes = 
        lst                                     //using input list
        |> Seq.map_concat (nodeBuilder lst)     //build permutations tree
        |> Seq.choose treefilter                //prune tree if necessary
        |> Seq.map_concat (pathBuilder [])      //convert to seq of path lists

    nodes

The permutations function works by constructing an n-ary tree representing all possible permutations of the list of 'things' passed in, then traversing the tree to construct a list of lists. Using 'Seq' dramatically improves performance as it makes everything lazy.

The second parameter of the permutations function allows the caller to define a filter for 'pruning' the tree before generating the paths (see my example below, where I don't want any leading zeros).

Some example usage: Node<'a> is generic, so we can do permutations of 'anything':

let myfilter n = Some(n)  //i.e., don't filter
permutations myfilter ['A';'B';'C';'D'] 

//in this case, I want to 'prune' leading zeros from my list before generating paths
let noLeadingZero n = 
    match n with
    | Branch(0, _) -> None
    | n -> Some(n)

//Curry myself an int-list permutations function with no leading zeros
let noLZperm = permutations noLeadingZero
noLZperm [0..9]

(Special thanks to Tomas Petricek, any comments welcome)

Answer 2

Hi, you can also write something like this:

let rec permutations list taken = 
  seq { if Set.count taken = List.length list then yield [] else
        for l in list do
          if not (Set.mem l taken) then 
            for perm in permutations list (Set.add l taken)  do
              yield l::perm }

The 'list' argument contains all the numbers that you want to permute and 'taken' is a set that contains numbers already used. The function returns empty list when all numbers all taken. Otherwise, it iterates over all numbers that are still available, gets all possible permutations of the remaining numbers (recursively using 'permutations') and appends the current number to each of them before returning (l::perm).

To run this, you'll give it an empty set, because no numebrs are used at the beginning:

permutations [1;2;3] Set.empty;;

Answer 3

http://fsharpcode.blogspot.com/2010/01/f-permutations.html

Answer 4

I like this implementation (but can't remember the source of it):

let rec insertions x = function
    | []             -> [[x]]
    | (y :: ys) as l -> (x::l)::(List.map (fun x -> y::x) (insertions x ys))

let rec permutations = function
    | []      -> seq [ [] ]
    | x :: xs -> Seq.concat (Seq.map (insertions x) (permutations xs))

Answer 5

Take a look at this one:

http://fsharpcode.blogspot.com/2010/04/permutations.html

Answer 6

In Tomas Petricek's solution I found I had to change 'Set.mem l taken' to 'taken.Contains l'

Answer 7

Tomas' solution is quite elegant: it's short, purely functional, and lazy. I think it may even be tail-recursive. Also, it produces permutations lexicographically. However, we can improve performance two-fold using an imperative solution internally while still exposing a functional interface externally.

The function permutations takes a generic sequence e as well as a generic comparison function f : ('a -> 'a -> int) and lazily yields immutable permutations lexicographically. The comparison functional allows us to generate permutations of elements which are not necessarily comparable as well as easily specify reverse or custom orderings.

The inner function permute is the imperative implementation of the algorithm described here. The conversion function let comparer f = { new System.Collections.Generic.IComparer<'a> with member self.Compare(x,y) = f x y } allows us to use the System.Array.Sort overload which does in-place sub-range custom sorts using an IComparer.

let permutations f e =
    ///Advances (mutating) perm to the next lexical permutation.
    let permute (perm:'a[]) (f: 'a->'a->int) (comparer:System.Collections.Generic.IComparer<'a>) : bool =
        try
            //Find the longest "tail" that is ordered in decreasing order ((s+1)..perm.Length-1).
            //will throw an index out of bounds exception if perm is the last permuation,
            //but will not corrupt perm.
            let rec find i =
                if (f perm.[i] perm.[i-1]) >= 0 then i-1
                else find (i-1)
            let s = find (perm.Length-1)
            let s' = perm.[s]

            //Change the number just before the tail (s') to the smallest number bigger than it in the tail (perm.[t]).
            let rec find i imin =
                if i = perm.Length then imin
                elif (f perm.[i] s') > 0 && (f perm.[i] perm.[imin]) < 0 then find (i+1) i
                else find (i+1) imin
            let t = find (s+1) (s+1)

            perm.[s] <- perm.[t]
            perm.[t] <- s'

            //Sort the tail in increasing order.
            System.Array.Sort(perm, s+1, perm.Length - s - 1, comparer)
            true
        with
        | _ -> false

    //permuation sequence expression 
    let c = f |> comparer
    let freeze arr = arr |> Array.copy |> Seq.readonly
    seq { let e' = Seq.toArray e
          yield freeze e'
          while permute e' f c do
              yield freeze e' }

Now for convenience we have the following where let flip f x y = f y x:

let permutationsAsc e = permutations compare e
let permutationsDesc e = permutations (flip compare) e

Answer 8

If you need distinct permuations (when the original set has duplicates), you can use this:

let rec insertions pre c post =
    seq {
        if List.length post = 0 then
            yield pre @ [c]
        else
            if List.forall (fun x->x<>c) post then
                yield pre@[c]@post
            yield! insertions (pre@[post.Head]) c post.Tail
        }

let rec permutations l =
    seq {
        if List.length l = 1 then
            yield l
        else
            let subperms = permutations l.Tail
            for sub in subperms do
                yield! insertions [] l.Head sub
        }

This is a straight-forward translation from this C# code. I am open to suggestions for a more functional look-and-feel.