I'm trying to write a function in F# to get the powersets of a set. So far I have written :
let rec powerset = function
|[] -> [[]]
| [x] -> [[x]; []]
|x::xs -> [x] :: (List.map (fun n -> [x; n]) xs) # powerset xs;;
but this isn't returning the cases that have 3 or more elements, only the pairs, the single elements, and the empty set.
You are on the right track, here is a working solution:
let rec powerset =
function
| [] -> [[]]
| (x::xs) ->
let xss = powerset xs
List.map (fun xs' -> x::xs') xss # xss
See you only have to use this trick:
for each element x you there half of the elements of the powerset will include x and half will not
so you recursively generate the powerset of the remaining elements xss and concat the two parts (List.map (fun xs' -> x::xs') xss will prepend the x to each of those)
But please note that this is not tail recursive and will blow the stack for bigger lists - you can take this idea and try to implement it with seq or make a tail-recursive version if you like
Using seq
Here is a version that uses seq and the bijection between the binary representation of natural numbers (a subset of those) and the subsets of a set (you map the elements to digits and set 1 if the corresponding element is in the subset and 0 if not):
let powerset (xs : 'a seq) : 'a seq seq =
let digits (n : bigint) : bool seq =
Seq.unfold (fun n ->
if n <= 0I
then None
else Some (n &&& 1I = 1I, n >>> 1))
n
let subsetBy (i : bigint) : 'a seq =
Seq.zip xs (digits i)
|> Seq.choose (fun (x,b) -> if b then Some x else None)
seq { 0I .. 2I**(Seq.length xs)-1I }
|> Seq.map subsetBy
this will work for things like powerset [1..100] but it might take a long time to enumerate them all ;) (but it should not take to much memory...)
Related
So I have a function which computes the powerset of a set:
let rec powerset = function
| [] -> [[]]
| x::xs -> List.collect (fun subset -> [subset; x::subset]) (powerset xs)
This returns 'a list list.
Now I'm trying to make a function which gives me all the subsets with a certain length k.
let rec allSubsets n k =
let powersetN = Set.ofList (powerset [1..n])
let subsets ps =
Set.filter (fun e -> Set.count e = k) ps
subsets powersetN
The thing is, powersetN is a Set<int list>, whereas subsets expect a Set<Set<'a>>.
The obvious solution of course would be to create the powerset as an actual set instead of a list, but I haven't been able to come up with a way to do this.
Any hints?
I am interested to implement fold3, fold4 etc., similar to List.fold and List.fold2. e.g.
// TESTCASE
let polynomial (x:double) a b c = a*x + b*x*x + c*x*x*x
let A = [2.0; 3.0; 4.0; 5.0]
let B = [1.5; 1.0; 0.5; 0.2]
let C = [0.8; 0.01; 0.001; 0.0001]
let result = fold3 polynomial 0.7 A B C
// 2.0 * (0.7 ) + 1.5 * (0.7 )^2 + 0.8 * (0.7 )^3 -> 2.4094
// 3.0 * (2.4094) + 1.0 * (2.4094)^2 + 0.01 * (2.4094)^3 -> 13.173
// 4.0 * (13.173) + 0.5 * (13.173)^2 + 0.001 * (13.173)^3 -> 141.75
// 5.0 * (141.75) + 0.2 * (141.75)^2 + 0.0001 * (141.75)^3 -> 5011.964
//
// Output: result = 5011.964
My first method is grouping the 3 lists A, B, C, into a list of tuples, and then apply list.fold
let fold3 f x A B C =
List.map3 (fun a b c -> (a,b,c)) A B C
|> List.fold (fun acc (a,b,c) -> f acc a b c) x
// e.g. creates [(2.0,1.5,0.8); (3.0,1.0,0.01); ......]
My second method is to declare a mutable data, and use List.map3
let mutable result = 0.7
List.map3 (fun a b c ->
result <- polynomial result a b c // Change mutable data
// Output intermediate data
result) A B C
// Output from List.map3: [2.4094; 13.17327905; 141.7467853; 5011.963942]
// result mutable: 5011.963942
I would like to know if there are other ways to solve this problem. Thank you.
For fold3, you could just do zip3 and then fold:
let polynomial (x:double) (a, b, c) = a*x + b*x*x + c*x*x*x
List.zip3 A B C |> List.fold polynomial 0.7
But if you want this for the general case, then you need what we call "applicative functors".
First, imagine you have a list of functions and a list of values. Let's assume for now they're of the same size:
let fs = [ (fun x -> x+1); (fun x -> x+2); (fun x -> x+3) ]
let xs = [3;5;7]
And what you'd like to do (only natural) is to apply each function to each value. This is easily done with List.map2:
let apply fs xs = List.map2 (fun f x -> f x) fs xs
apply fs xs // Result = [4;7;10]
This operation "apply" is why these are called "applicative functors". Not just any ol' functors, but applicative ones. (the reason for why they're "functors" is a tad more complicated)
So far so good. But wait! What if each function in my list of functions returned another function?
let f1s = [ (fun x -> fun y -> x+y); (fun x -> fun y -> x-y); (fun x -> fun y -> x*y) ]
Or, if I remember that fun x -> fun y -> ... can be written in the short form of fun x y -> ...
let f1s = [ (fun x y -> x+y); (fun x y -> x-y); (fun x y -> x*y) ]
What if I apply such list of functions to my values? Well, naturally, I'll get another list of functions:
let f2s = apply f1s xs
// f2s = [ (fun y -> 3+y); (fun y -> 5+y); (fun y -> 7+y) ]
Hey, here's an idea! Since f2s is also a list of functions, can I apply it again? Well of course I can!
let ys = [1;2;3]
apply f2s ys // Result: [4;7;10]
Wait, what? What just happened?
I first applied the first list of functions to xs, and got another list of functions as a result. And then I applied that result to ys, and got a list of numbers.
We could rewrite that without intermediate variable f2s:
let f1s = [ (fun x y -> x+y); (fun x y -> x-y); (fun x y -> x*y) ]
let xs = [3;5;7]
let ys = [1;2;3]
apply (apply f1s xs) ys // Result: [4;7;10]
For extra convenience, this operation apply is usually expressed as an operator:
let (<*>) = apply
f1s <*> xs <*> ys
See what I did there? With this operator, it now looks very similar to just calling the function with two arguments. Neat.
But wait. What about our original task? In the original requirements we don't have a list of functions, we only have one single function.
Well, that can be easily fixed with another operation, let's call it "apply first". This operation will take a single function (not a list) plus a list of values, and apply this function to each value in the list:
let applyFirst f xs = List.map f xs
Oh, wait. That's just map. Silly me :-)
For extra convenience, this operation is usually also given an operator name:
let (<|>) = List.map
And now, I can do things like this:
let f x y = x + y
let xs = [3;5;7]
let ys = [1;2;3]
f <|> xs <*> ys // Result: [4;7;10]
Or this:
let f x y z = (x + y)*z
let xs = [3;5;7]
let ys = [1;2;3]
let zs = [1;-1;100]
f <|> xs <*> ys <*> zs // Result: [4;-7;1000]
Neat! I made it so I can apply arbitrary functions to lists of arguments at once!
Now, finally, you can apply this to your original problem:
let polynomial a b c (x:double) = a*x + b*x*x + c*x*x*x
let A = [2.0; 3.0; 4.0; 5.0]
let B = [1.5; 1.0; 0.5; 0.2]
let C = [0.8; 0.01; 0.001; 0.0001]
let ps = polynomial <|> A <*> B <*> C
let result = ps |> List.fold (fun x f -> f x) 0.7
The list ps consists of polynomial instances that are partially applied to corresponding elements of A, B, and C, and still expecting the final argument x. And on the next line, I simply fold over this list of functions, applying each of them to the result of the previous.
You could check the implementation for ideas:
https://github.com/fsharp/fsharp/blob/master/src/fsharp/FSharp.Core/array.fs
let fold<'T,'State> (f : 'State -> 'T -> 'State) (acc: 'State) (array:'T[]) =
checkNonNull "array" array
let f = OptimizedClosures.FSharpFunc<_,_,_>.Adapt(f)
let mutable state = acc
for i = 0 to array.Length-1 do
state <- f.Invoke(state,array.[i])
state
here's a few implementations for you:
let fold2<'a,'b,'State> (f : 'State -> 'a -> 'b -> 'State) (acc: 'State) (a:'a array) (b:'b array) =
let mutable state = acc
Array.iter2 (fun x y->state<-f state x y) a b
state
let iter3 f (a: 'a[]) (b: 'b[]) (c: 'c[]) =
let f = OptimizedClosures.FSharpFunc<_,_,_,_>.Adapt(f)
if a.Length <> b.Length || a.Length <> c.Length then failwithf "length"
for i = 0 to a.Length-1 do
f.Invoke(a.[i], b.[i], c.[i])
let altIter3 f (a: 'a[]) (b: 'b[]) (c: 'c[]) =
if a.Length <> b.Length || a.Length <> c.Length then failwithf "length"
for i = 0 to a.Length-1 do
f (a.[i]) (b.[i]) (c.[i])
let fold3<'a,'b,'State> (f : 'State -> 'a -> 'b -> 'c -> 'State) (acc: 'State) (a:'a array) (b:'b array) (c:'c array) =
let mutable state = acc
iter3 (fun x y z->state<-f state x y z) a b c
state
NB. we don't have an iter3, so, implement that. OptimizedClosures.FSharpFunc only allow up to 5 (or is it 7?) params. There are a finite number of type slots available. It makes sense. You can go higher than this, of course, without using the OptimizedClosures stuff.
... anyway, generally, you don't want to be iterating too many lists / arrays / sequences at once. So I'd caution against going too high.
... the better way forward in such cases may be to construct a record or tuple from said lists / arrays, first. Then, you can just use map and iter, which are already baked in. This is what zip / zip3 are all about (see: "(array1.[i],array2.[i],array3.[i])")
let zip3 (array1: _[]) (array2: _[]) (array3: _[]) =
checkNonNull "array1" array1
checkNonNull "array2" array2
checkNonNull "array3" array3
let len1 = array1.Length
if len1 <> array2.Length || len1 <> array3.Length then invalidArg3ArraysDifferent "array1" "array2" "array3" len1 array2.Length array3.Length
let res = Microsoft.FSharp.Primitives.Basics.Array.zeroCreateUnchecked len1
for i = 0 to res.Length-1 do
res.[i] <- (array1.[i],array2.[i],array3.[i])
res
I'm working with arrays at the moment, so my solution pertained to those. Sorry about that. Here's a recursive version for lists.
let fold3 f acc a b c =
let mutable state = acc
let rec fold3 f a b c =
match a,b,c with
| [],[],[] -> ()
| [],_,_
| _,[],_
| _,_,[] -> failwith "length"
| ahead::atail, bhead::btail, chead::ctail ->
state <- f state ahead bhead chead
fold3 f atail btail ctail
fold3 f a b c
i.e. we define a recursive function within a function which acts upon/mutates/changes the outer scoped mutable acc variable (a closure in functional speak). Finally, this gets returned.
It's pretty cool how much type information gets inferred about these functions. In the array examples above, mostly I was explicit with 'a 'b 'c. This time, we let type inference kick in. It knows we're dealing with lists from the :: operator. That's kind of neat.
NB. the compiler will probably unwind this tail-recursive approach so that it is just a loop behind-the-scenes. Generally, get a correct answer before optimising. Just mentioning this, though, as food for later thought.
I think the existing answers provide great options if you want to generalize folding, which was your original question. However, if I simply wanted to call the polynomial function on inputs specified in A, B and C, then I would probably do not want to introduce fairly complex constructs like applicative functors with fancy operators to my code base.
The problem becomes a lot easier if you transpose the input data, so that rather than having a list [A; B; C] with lists for individual variables, you have a transposed list with inputs for calculating each polynomial. To do this, we'll need the transpose function:
let rec transpose = function
| (_::_)::_ as M -> List.map List.head M :: transpose (List.map List.tail M)
| _ -> []
Now you can create a list with inputs, transpose it and calculate all polynomials simply using List.map:
transpose [A; B; C]
|> List.map (function
| [a; b; c] -> polynomial 0.7 a b c
| _ -> failwith "wrong number of arguments")
There are many ways to solve this problem. Few are mentioned like first zip3 all three list, then run over it. Using Applicate Functors like Fyodor Soikin describes means you can turn any function with any amount of arguments into a function that expects list instead of single arguments. This is a good general solution that works with any numbers of lists.
While this is a general good idea, i'm sometimes shocked that so few use more low-level tools. In this case it is a good idea to use recursion and learn more about recursion.
Recursion here is the right-tool because we have immutable data-types. But you could consider how you would implement it with mutable lists and looping first, if that helps. The steps would be:
You loop over an index from 0 to the amount of elements in the lists.
You check if every list has an element for the index
If every list has an element then you pass this to your "folder" function
If at least one list don't have an element, then you abort the loop
The recursive version works exactly the same. Only that you don't use an index to access the elements. You would chop of the first element from every list and then recurse on the remaining list.
Otherwise List.isEmpty is the function to check if a List is empty. You can chop off the first element with List.head and you get the remaining list with the first element removed by List.tail. This way you can just write:
let rec fold3 f acc l1 l2 l3 =
let h = List.head
let t = List.tail
let empty = List.isEmpty
if (empty l1) || (empty l2) && (empty l3)
then acc
else fold3 f (f acc (h l1) (h l2) (h l3)) (t l1) (t l2) (t l3)
The if line checks if every list has at least one element. If that is true
it executes: f acc (h l1) (h l2) (h l3). So it executes f and passes it the first element of every list as an argument. The result is the new accumulator of
the next fold3 call.
Now that you worked on the first element of every list, you must chop off the first element of every list, and continue with the remaining lists. You achieve that with List.tail or in the above example (t l1) (t l2) (t l3). Those are the next remaining lists for the next fold3 call.
Creating a fold4, fold5, fold6 and so on isn't really hard, and I think it is self-explanatory. My general advice is to learn a little bit more about recursion and try to write recursive List functions without Pattern Matching. Pattern Matching is not always easier.
Some code examples:
fold3 (fun acc x y z -> x + y + z :: acc) [] [1;2;3] [10;20;30] [100;200;300] // [333;222;111]
fold3 (fun acc x y z -> x :: y :: z :: acc) [] [1;2;3] [10;20;30] [100;200;300] // [3; 30; 300; 2; 20; 200; 1; 10; 100]
I'm attempting to rewrite List.filter manually
so far I have this:
let rec filter f = function
|[] -> []
|x::xs -> if f x = true then x # filter f xs
else filter f xs;;
I'd add to the accepted answer that recognizing and applying functional patterns may be as important as mastery of recursion and pattern matching. And probably the first of such patterns is folding.
Implementing your task with folding takes a terse:
let filter p ls = List.foldBack (fun l acc -> if p l then l::acc else acc) ls []
Operator # appends 2 lists so x in if ... then ... else expression is supposed to be of type list.
You probably meant to use list cons operator ::. Also you don't need to compare the result of function f application to true.
let rec filter f = function
|[] -> []
|x::xs -> if f x then x :: filter f xs
else filter f xs
[1;2;3;4;5;6;7;8;9] |> filter (fun x -> x % 2 = 0)
val it : int list = [2; 4; 6; 8]
Note: this function is not tail recursive so you'll get stack overflow exception with big lists.
I've found this question on hubFS, but that handles a splitting criteria based on individual elements. I'd like to split based on a comparison of adjacent elements, so the type would look like this:
val split = ('T -> 'T -> bool) -> 'T list -> 'T list list
Currently, I am trying to start from Don's imperative solution, but I can't work out how to initialize and use a 'prev' value for comparison. Is fold a better way to go?
//Don's solution for single criteria, copied from hubFS
let SequencesStartingWith n (s:seq<_>) =
seq { use ie = s.GetEnumerator()
let acc = new ResizeArray<_>()
while ie.MoveNext() do
let x = ie.Current
if x = n && acc.Count > 0 then
yield ResizeArray.to_list acc
acc.Clear()
acc.Add x
if acc.Count > 0 then
yield ResizeArray.to_list acc }
This is an interesting problem! I needed to implement exactly this in C# just recently for my article about grouping (because the type signature of the function is pretty similar to groupBy, so it can be used in LINQ query as the group by clause). The C# implementation was quite ugly though.
Anyway, there must be a way to express this function using some simple primitives. It just seems that the F# library doesn't provide any functions that fit for this purpose. I was able to come up with two functions that seem to be generally useful and can be combined together to solve this problem, so here they are:
// Splits a list into two lists using the specified function
// The list is split between two elements for which 'f' returns 'true'
let splitAt f list =
let rec splitAtAux acc list =
match list with
| x::y::ys when f x y -> List.rev (x::acc), y::ys
| x::xs -> splitAtAux (x::acc) xs
| [] -> (List.rev acc), []
splitAtAux [] list
val splitAt : ('a -> 'a -> bool) -> 'a list -> 'a list * 'a list
This is similar to what we want to achieve, but it splits the list only in two pieces (which is a simpler case than splitting the list multiple times). Then we'll need to repeat this operation, which can be done using this function:
// Repeatedly uses 'f' to take several elements of the input list and
// aggregate them into value of type 'b until the remaining list
// (second value returned by 'f') is empty
let foldUntilEmpty f list =
let rec foldUntilEmptyAux acc list =
match f list with
| l, [] -> l::acc |> List.rev
| l, rest -> foldUntilEmptyAux (l::acc) rest
foldUntilEmptyAux [] list
val foldUntilEmpty : ('a list -> 'b * 'a list) -> 'a list -> 'b list
Now we can repeatedly apply splitAt (with some predicate specified as the first argument) on the input list using foldUntilEmpty, which gives us the function we wanted:
let splitAtEvery f list = foldUntilEmpty (splitAt f) list
splitAtEvery (<>) [ 1; 1; 1; 2; 2; 3; 3; 3; 3 ];;
val it : int list list = [[1; 1; 1]; [2; 2]; [3; 3; 3; 3]]
I think that the last step is really nice :-). The first two functions are quite straightforward and may be useful for other things, although they are not as general as functions from the F# core library.
How about:
let splitOn test lst =
List.foldBack (fun el lst ->
match lst with
| [] -> [[el]]
| (x::xs)::ys when not (test el x) -> (el::(x::xs))::ys
| _ -> [el]::lst
) lst []
the foldBack removes the need to reverse the list.
Having thought about this a bit further, I've come up with this solution. I'm not sure that it's very readable (except for me who wrote it).
UPDATE Building on the better matching example in Tomas's answer, here's an improved version which removes the 'code smell' (see edits for previous version), and is slightly more readable (says me).
It still breaks on this (splitOn (<>) []), because of the dreaded value restriction error, but I think that might be inevitable.
(EDIT: Corrected bug spotted by Johan Kullbom, now works correctly for [1;1;2;3]. The problem was eating two elements directly in the first match, this meant I missed a comparison/check.)
//Function for splitting list into list of lists based on comparison of adjacent elements
let splitOn test lst =
let rec loop lst inner outer = //inner=current sublist, outer=list of sublists
match lst with
| x::y::ys when test x y -> loop (y::ys) [] (List.rev (x::inner) :: outer)
| x::xs -> loop xs (x::inner) outer
| _ -> List.rev ((List.rev inner) :: outer)
loop lst [] []
splitOn (fun a b -> b - a > 1) [1]
> val it : [[1]]
splitOn (fun a b -> b - a > 1) [1;3]
> val it : [[1]; [3]]
splitOn (fun a b -> b - a > 1) [1;2;3;4;6;7;8;9;11;12;13;14;15;16;18;19;21]
> val it : [[1; 2; 3; 4]; [6; 7; 8; 9]; [11; 12; 13; 14; 15; 16]; [18; 19]; [21]]
Any thoughts on this, or the partial solution in my question?
"adjacent" immediately makes me think of Seq.pairwise.
let splitAt pred xs =
if Seq.isEmpty xs then
[]
else
xs
|> Seq.pairwise
|> Seq.fold (fun (curr :: rest as lists) (i, j) -> if pred i j then [j] :: lists else (j :: curr) :: rest) [[Seq.head xs]]
|> List.rev
|> List.map List.rev
Example:
[1;1;2;3;3;3;2;1;2;2]
|> splitAt (>)
Gives:
[[1; 1; 2; 3; 3; 3]; [2]; [1; 2; 2]]
I would prefer using List.fold over explicit recursion.
let splitOn pred = function
| [] -> []
| hd :: tl ->
let (outer, inner, _) =
List.fold (fun (outer, inner, prev) curr ->
if pred prev curr
then (List.rev inner) :: outer, [curr], curr
else outer, curr :: inner, curr)
([], [hd], hd)
tl
List.rev ((List.rev inner) :: outer)
I like answers provided by #Joh and #Johan as these solutions seem to be most idiomatic and straightforward. I also like an idea suggested by #Shooton. However, each solution had their own drawbacks.
I was trying to avoid:
Reversing lists
Unsplitting and joining back the temporary results
Complex match instructions
Even Seq.pairwise appeared to be redundant
Checking list for emptiness can be removed in cost of using Unchecked.defaultof<_> below
Here's my version:
let splitWhen f src =
if List.isEmpty src then [] else
src
|> List.foldBack
(fun el (prev, current, rest) ->
if f el prev
then el , [el] , current :: rest
else el , el :: current , rest
)
<| (List.head src, [], []) // Initial value does not matter, dislike using Unchecked.defaultof<_>
|> fun (_, current, rest) -> current :: rest // Merge temporary lists
|> List.filter (not << List.isEmpty) // Drop tail element
I need to generate permutations on a given list. I managed to do it like this
let rec Permute (final, arr) =
if List.length arr > 0 then
for x in arr do
let n_final = final # [x]
let rest = arr |> List.filter (fun a -> not (x = a))
Permute (n_final, rest)
else
printfn "%A" final
let DoPermute lst =
Permute ([], lst)
DoPermute lst
There are obvious issues with this code. For example, list elements must be unique. Also, this is more-less a same approach that I would use when generating straight forward implementation in any other language. Is there any better way to implement this in F#.
Thanks!
Here's the solution I gave in my book F# for Scientists (page 166-167):
let rec distribute e = function
| [] -> [[e]]
| x::xs' as xs -> (e::xs)::[for xs in distribute e xs' -> x::xs]
let rec permute = function
| [] -> [[]]
| e::xs -> List.collect (distribute e) (permute xs)
For permutations of small lists, I use the following code:
let distrib e L =
let rec aux pre post =
seq {
match post with
| [] -> yield (L # [e])
| h::t -> yield (List.rev pre # [e] # post)
yield! aux (h::pre) t
}
aux [] L
let rec perms = function
| [] -> Seq.singleton []
| h::t -> Seq.collect (distrib h) (perms t)
It works as follows: the function "distrib" distributes a given element over all positions in a list, example:
distrib 10 [1;2;3] --> [[10;1;2;3];[1;10;2;3];[1;2;10;3];[1;2;3;10]]
The function perms works (recursively) as follows: distribute the head of the list over all permutations of its tail.
The distrib function will get slow for large lists, because it uses the # operator a lot, but for lists of reasonable length (<=10), the code above works fine.
One warning: if your list contains duplicates, the result will contain identical permutations. For example:
perms [1;1;3] = [[1;1;3]; [1;1;3]; [1;3;1]; [1;3;1]; [3;1;1]; [3;1;1]]
The nice thing about this code is that it returns a sequence of permutations, instead of generating them all at once.
Of course, generating permutations with an imperative array-based algorithm will be (much) faster, but this algorithm has served me well in most cases.
Here's another sequence-based version, hopefully more readable than the voted answer.
This version is similar to Jon's version in terms of logic, but uses computation expressions instead of lists. The first function computes all ways to insert an element x in a list l. The second function computes permutations.
You should be able to use this on larger lists (e.g. for brute force searches on all permutations of a set of inputs).
let rec inserts x l =
seq { match l with
| [] -> yield [x]
| y::rest ->
yield x::l
for i in inserts x rest do
yield y::i
}
let rec permutations l =
seq { match l with
| [] -> yield []
| x::rest ->
for p in permutations rest do
yield! inserts x p
}
It depends on what you mean by "better". I'd consider this to be slightly more elegant, but that may be a matter of taste:
(* get the list of possible heads + remaining elements *)
let rec splitList = function
| [x] -> [x,[]]
| x::xs -> (x, xs) :: List.map (fun (y,l) -> y,x::l) (splitList xs)
let rec permutations = function
| [] -> [[]]
| l ->
splitList l
|> List.collect (fun (x,rest) ->
(* permute remaining elements, then prepend head *)
permutations rest |> List.map (fun l -> x::l))
This can handle lists with duplicate elements, though it will result in duplicated permutations.
In the spirit of Cyrl's suggestion, here's a sequence comprehension version
let rec permsOf xs =
match xs with
| [] -> List.toSeq([[]])
| _ -> seq{ for x in xs do
for xs' in permsOf (remove x xs) do
yield (x::xs')}
where remove is a simple function that removes a given element from a list
let rec remove x xs =
match xs with [] -> [] | (x'::xs')-> if x=x' then xs' else x'::(remove x xs')
IMHO the best solution should alleviate the fact that F# is a functional language so imho the solution should be as close to the definition of what we mean as permutation there as possible.
So the permutation is such an instance of list of things where the head of the list is somehow added to the permutation of the rest of the input list.
The erlang solution shows that in a pretty way:
permutations([]) -> [[]];
permutations(L) -> [[H|T] H<- L, T <- permutations( L--[H] ) ].
taken fron the "programming erlang" book
There is a list comprehension operator used, in solution mentioned here by the fellow stackoverflowers there is a helper function which does the similar job
basically I'd vote for the solution without any visible loops etc, just pure function definition
I'm like 11 years late, but still in case anyone needs permutations like I did recently. Here's Array version of permutation func, I believe it's more performant:
[<RequireQualifiedAccess>]
module Array =
let private swap (arr: _[]) i j =
let buf = arr.[i]
arr.[i] <- arr.[j]
arr.[j] <- buf
let permutations arr =
match arr with
| null | [||] -> [||]
| arr ->
let last = arr.Length - 1
let arr = Array.copy arr
let rec perm arr k =
let arr = Array.copy arr
[|
if k = last then
yield arr
else
for i in k .. last do
swap arr k i
yield! perm arr (k + 1)
|]
perm arr 0