For cartesian production there is a good enough function - sequence which defined like that:
let rec sequence = function
| [] -> Seq.singleton []
| (l::ls) -> seq { for x in l do for xs in sequence ls do yield (x::xs) }
but look at its result:
sequence [[1..2];[1..10000]] |> Seq.skip 1000 ;;
val it : seq = seq [[1; 1001]; [1; 1002]; [1; 1003]; [1; 1004]; ...]
As we can see the first "coordinate" of the product alters very slowly and it will change the value when the second list is ended.
I wrote my own sequence as following (comments below):
/// Sum of all producted indeces = n
let rec hyper'plane'indices indexsum maxlengths =
match maxlengths with
| [x] -> if indexsum < x then [[indexsum]] else []
| (i::is) -> [for x in [0 .. min indexsum (i-1)] do for xs in hyper'plane'indices (indexsum-x) is do yield (x::xs)]
| [] -> [[]]
let finite'sequence = function
| [] -> Seq.singleton []
| ns ->
let ars = [ for n in ns -> Seq.toArray n ]
let length'list = List.map Array.length ars
let nmax = List.max length'list
seq {
for n in [0 .. nmax] do
for ixs in hyper'plane'indices n length'list do
yield (List.map2 (fun (a:'a[]) i -> a.[i]) ars ixs)
}
The key idea is to look at (two) lists as at (two) orthogonal dimensions where every element marked by its index in the list. So we can enumerate all elements by enumerating every element in every section of cartesian product by hyper plane (in 2D case this is a line). In another words imagine excel's sheet where first column contains values from [1;1] to [1;10000] and second - from [2;1] to [2;10000]. And "hyper plane" with number 1 is the line that connects cell A2 and cell B1. For the our example
hyper'plane'indices 0 [2;10000];; val it : int list list = [[0; 0]]
hyper'plane'indices 1 [2;10000];; val it : int list list = [[0; 1]; [1; 0]]
hyper'plane'indices 2 [2;10000];; val it : int list list = [[0; 2]; [1; 1]]
hyper'plane'indices 3 [2;10000];; val it : int list list = [[0; 3]; [1; 2]]
hyper'plane'indices 4 [2;10000];; val it : int list list = [[0; 4]; [1; 3]]
Well if we have indeces and arrays that we are producing from the given lists than we can now define sequence as {all elements in plane 0; than all elements in plane 1 ... and so on } and get more volatile function than original sequence.
But finite'sequence turned out very gluttonous function. And now the question. How I can improve it?
With best wishes, Alexander. (and sorry for poor English)
Can you explain what exactly is the problem - time or space complexity or performance? Do you have a specific benchmark in mind? I am not sure how to improve on the time complexity here, but I edited your code a bit to remove the intermediate lists, which might help a bit with memory allocation behavior.
Do not do this:
for n in [0 .. nmax] do
Do this instead:
for n in 0 .. nmax do
Here is the code:
let rec hyper'plane'indices indexsum maxlengths =
match maxlengths with
| [] -> Seq.singleton []
| [x] -> if indexsum < x then Seq.singleton [indexsum] else Seq.empty
| i :: is ->
seq {
for x in 0 .. min indexsum (i - 1) do
for xs in hyper'plane'indices (indexsum - x) is do
yield x :: xs
}
let finite'sequence xs =
match xs with
| [] -> Seq.singleton []
| ns ->
let ars = [ for n in ns -> Seq.toArray n ]
let length'list = List.map Array.length ars
let nmax = List.max length'list
seq {
for n in 0 .. nmax do
for ixs in hyper'plane'indices n length'list do
yield List.map2 Array.get ars ixs
}
Does this fare any better? Beautiful problem by the way.
UPDATE: Perhaps you are more interested to mix the sequences fairly than in maintaining the exact formula in your algorithm. Here is a Haskell code that mixes a finite number of possibly infinite sequences fairly, where fairness means that for every input element there is a finite prefix of the output sequence that contains it. You mention in the comment that you have a 2D incremental solution that is hard to generalize to N dimensions, and the Haskell code does exactly that:
merge :: [a] -> [a] -> [a]
merge [] y = y
merge x [] = x
merge (x:xs) (y:ys) = x : y : merge xs ys
prod :: (a -> b -> c) -> [a] -> [b] -> [c]
prod _ [] _ = []
prod _ _ [] = []
prod f (x:xs) (y:ys) = f x y : a `merge` b `merge` prod f xs ys where
a = [f x y | x <- xs]
b = [f x y | y <- ys]
prodN :: [[a]] -> [[a]]
prodN [] = [[]]
prodN (x:xs) = prod (:) x (prodN xs)
I have not ported this to F# yet - it requires some thought as sequences do not match to head/tail very well.
UPDATE 2:
A fairly mechanical translation to F# follows.
type Node<'T> =
| Nil
| Cons of 'T * Stream<'T>
and Stream<'T> = Lazy<Node<'T>>
let ( !! ) (x: Lazy<'T>) = x.Value
let ( !^ ) x = Lazy.CreateFromValue(x)
let rec merge (xs: Stream<'T>) (ys: Stream<'T>) : Stream<'T> =
lazy
match !!xs, !!ys with
| Nil, r | r, Nil -> r
| Cons (x, xs), Cons (y, ys) -> Cons (x, !^ (Cons (y, merge xs ys)))
let rec map (f: 'T1 -> 'T2) (xs: Stream<'T1>) : Stream<'T2> =
lazy
match !!xs with
| Nil -> Nil
| Cons (x, xs) -> Cons (f x, map f xs)
let ( ++ ) = merge
let rec prod f xs ys =
lazy
match !!xs, !!ys with
| Nil, _ | _, Nil -> Nil
| Cons (x, xs), Cons (y, ys) ->
let a = map (fun x -> f x y) xs
let b = map (fun y -> f x y) ys
Cons (f x y, a ++ b ++ prod f xs ys)
let ofSeq (s: seq<'T>) =
lazy
let e = s.GetEnumerator()
let rec loop () =
lazy
if e.MoveNext()
then Cons (e.Current, loop ())
else e.Dispose(); Nil
!! (loop ())
let toSeq stream =
stream
|> Seq.unfold (fun stream ->
match !!stream with
| Nil -> None
| Cons (x, xs) -> Some (x, xs))
let empty<'T> : Stream<'T> = !^ Nil
let cons x xs = !^ (Cons (x, xs))
let singleton x = cons x empty
let rec prodN (xs: Stream<Stream<'T>>) : Stream<Stream<'T>> =
match !!xs with
| Nil -> singleton empty
| Cons (x, xs) -> prod cons x (prodN xs)
let test () =
ofSeq [
ofSeq [1; 2; 3]
ofSeq [4; 5; 6]
ofSeq [7; 8; 9]
]
|> prodN
|> toSeq
|> Seq.iter (fun xs ->
toSeq xs
|> Seq.map string
|> String.concat ", "
|> stdout.WriteLine)
Related
I need some help with my hometask: to express one function (sort) through others (smallest, delete, insert). If you know how, please, tell me, how I can do running my recursion cicle? it doing now only one step. maybe something like this: val4 -> head :: tail |> sort tail on line 25 (val4)?
let rec smallest = function
| x :: y :: tail when x <= y -> smallest (x :: tail)
| x :: y :: tail when x > y -> smallest (y :: tail)
| [x] -> Some x
| _ -> None
let rec delete (n, xs) =
match (n, xs) with
| (n, x :: xs) when n <> x -> x :: delete (n, xs)
| (n, x :: xs) when n = x -> xs
| (n, _) -> []
let rec insert (xs, n) =
match (xs, n) with
| ([x], n) when x < n -> [x]#[n]
| (x :: xs, n) when x < n -> x :: insert (xs, n)
| (x :: xs, n) when x >= n -> n :: x :: xs
| (_, _) -> []
let rec sort = function
| xs -> let val1 = smallest xs
let val2 = val1.[0]
let val3 = delete (val2, xs)
let val4 = insert (val3, val2)
val4
let res = sort [5; 4; 3; 2; 1; 1]
printfn "%A" res
This is sort of like insertion sort, but since you're always finding the smallest number in the whole list instead of the next highest number, it will recurse forever unless you skip whatever you've already found to be the smallest.
Furthermore, your insert and delete functions act not on the item index, but on equality to the value, so it won't be able to handle repeated numbers.
Keeping most of your original code the same, usually you have an inner recursive function to help you keep track of state. This is a common FP pattern.
let sort lst =
let size = lst |> List.length
let rec sort' xs = function
| index when index = size -> xs
| index ->
let val1 = smallest (xs |> List.skip index)
let val2 = val1.[0]
let val3 = delete (val2, xs)
let val4 = insert (val3, val2)
sort' val4 (index + 1)
sort' lst 0
let res = sort [5; 3; 2; 4; 1; ]
printfn "%A" res
Needless to say, this isn't correct or performant, and each iteration traverses the list multiple times. It probably runs in cubic time.
But keep learning!
I found it... I only had changed 4 & 5 lines above in the "smallest" on this: | [x] -> Some x
| _ -> None, when there was: | [x] -> [x]
| _ -> []
let rec sort = function
| xs -> match xs with
| head :: tail -> let val1 = smallest xs
match val1 with
| Some x -> let val2 = delete (x, xs)
let val3 = insert (val2, x)
let val4 = (fun list -> match list with head :: tail -> head :: sort tail | _ -> [])
val4 val3
| None -> []
| _ -> []
// let res = sort [5; 4; 3; 2; 1]
// printfn "%A" res
Given a set with n elements {1, 2, 3, ..., n}, I want to declare a function which returns the set containing the sets with k number of elements such as:
allSubsets 3 2
Would return [[1;2];[1;3];[2;3]] since those are the sets with 2 elements in a set created by 1 .. n
I've made the initial create-a-set-part but I'm a little stuck on how to find out all the subsets with k elements in it.
let allSubsets n k =
Set.ofList [1..n] |>
UPDATE:
I managed to get a working solution using yield:
let allSubsets n k =
let setN = Set.ofList [1..n]
let rec subsets s =
set [
if Set.count s = k then yield s
for e in s do
yield! subsets (Set.remove e s) ]
subsets setN
allSubsets 3 2
val it : Set<Set<int>> = set [set [1; 2]; set [1; 3]; set [2; 3]]
But isn't it possible to do it a little cleaner?
What you have is pretty clean, but it's also pretty inefficient. Try running allSubsets 10 3 and you'll know what I mean.
This is what I came up with:
let input = Set.ofList [ 1 .. 15 ]
let subsets (size:int) (input: Set<'a>) =
let rec inner elems =
match elems with
| [] -> [[]]
| h::t ->
List.fold (fun acc e ->
if List.length e < size then
(h::e)::e::acc
else e::acc) [] (inner t)
inner (Set.toList input)
|> Seq.choose (fun subset ->
if List.length subset = size then
Some <| Set.ofList subset
else None)
|> Set.ofSeq
subsets 3 input
The inner recursive function is a modified power set function from here. My first hunch was to generate the power set and then filter it, which would be pretty elegant, but that proved to be rather inefficient as well.
If this was to be production-quality code, I'd look into generating lists of indices of a given length, and use them to index into the input array. This is how FsCheck generates subsets, for example.
You can calculate the powerset and then filter in order to get only the ones with the specified length":
let powerset n k =
let lst = Set.toList n
seq [0..(lst.Length |> pown 2)-1]
|> Seq.map (fun i ->
set ([0..lst.Length-1] |> Seq.choose (fun x ->
if i &&& (pown 2 x) = 0 then None else Some lst.[x])))
|> Seq.filter (Seq.length >> (=) k)
However this is not efficient for large sets (n) of where k is close to n. But it's easy to optimize, you'll have to filter out early based on the digit count of the binary representation of each number.
This function implements the popular n-choose-k function:
let n_choose_k (arr: 'a []) (k: int) : 'a list list =
let len = Array.length arr
let rec choose lo x =
match x with
| 0 -> [[]]
| i -> [ for j in lo..(len-1) do
for ks in choose (j+1) (i-1) do
yield arr.[j]::ks ]
choose 0 k
> n_choose_k [|1..3|] 2;;
val it : int list list = [[1; 2]; [1; 3]; [2; 3]]
You can use Set.toArray and Set.ofList to convert to and from Set.
You can consider the following approach:
get powerset
let rec powerset xs =
match xs with
| [] -> [ [] ]
| h :: t -> List.fold (fun ys s -> (h :: s) :: s :: ys) [] (powerset t)
filter all subsets with a neccessary number of elements
let filtered xs k = List.filter (fun (x: 'a list) -> x.Length = k) xs
finally get the requested allSubsets
let allSubsets n k = Set.ofList (List.map (fun xs -> Set.ofList xs) (filtered (powerset [ 1 .. n ]) k))
Just to check and play with you can use:
printfn "%A" (allSubsets 3 2) // set [ set [1; 2]; set [1; 3]; set [2; 3] ]
I have to write a program which give in output a tuple with: min and max of a not-empty list and the value that appears most often.
In particular:
min_max [1;0;-1;2;0;-4] ==> (-4; 2)
min_max: int list -> (int * int)
mode [-1;2;1;2;5;-1;5;5;2] ==> 2
mode: int list -> int
This is the code that I wrote for max (min is almost equal) but how can I do to receive as output a tuple with the two values?
let rec max_list xs =
match xs with
| [] -> failwith "xs" "Empty list"
| [x] -> x
| x1::x2::xs' -> max_list((max2 x1 x2)::xs');;
I'll take the first suggestion from #Mark Seemann's answer and run with it, in order to make it generic, working with any collection type, and handle the case of the empty collection sensibly.
let tryMinMax xs =
Seq.fold (function
| Some(mn, mx) -> fun i -> Some(min mn i, max mx i)
| None -> fun i -> Some(i, i) ) None xs
[1;0;-1;2;0;-4]
|> tryMinMax
// val it : (int * int) option = Some (-4, 2)
For the most frequent part of the question:
let mostFrequent xs =
xs
|> Seq.countBy id
|> Seq.maxBy snd
|> fst
[1;0;-1;2;0;-4]
|> mostFrequent
// val it : int = 0
let minMax xs =
xs
|> List.fold
(fun (mn, mx) i -> min mn i, max mx i)
(System.Int32.MaxValue, System.Int32.MinValue)
Not particularly efficient, but fun to write:
let mode xs =
xs
|> List.groupBy id
|> List.map (fun (i, is) -> i, Seq.length is)
|> List.maxBy snd
|> fst
Option without the use of standard modules:
open System
let tryMinMax xs =
let rec minMax xs mn mx =
match xs with | [] -> mn, mx | h::t -> minMax t (min mn h) (max mx h)
match xs with | [] -> None | _ -> Some(minMax xs Int32.MaxValue Int32.MinValue)
dotnetfiddle
On the second question - show their attempts to solve.
I want to calculate powerset of a set. Because I don't need the whole powerset at a time, it's better to generate it lazily.
For example:
powerset (set ["a"; "b"; "c"]) =
seq {
set [];
set ["a"];
set ["b"];
set ["c"];
set ["a"; "b"];
set ["a"; "c"];
set ["b"; "c"];
set ["a";"b"; "c"];
}
Since the result is a sequence, I prefer it in the above order. How can I do it in an idomatic way in F#?
EDIT:
This is what I'm going to use (based on BLUEPIXY's answer):
let powerset s =
let rec loop n l =
seq {
match n, l with
| 0, _ -> yield []
| _, [] -> ()
| n, x::xs -> yield! Seq.map (fun l -> x::l) (loop (n-1) xs)
yield! loop n xs
}
let xs = s |> Set.toList
seq {
for i = 0 to List.length xs do
for x in loop i xs -> set x
}
Thanks everyone for excellent input.
let rec comb n l =
match n, l with
| 0, _ -> [[]]
| _, [] -> []
| n, x::xs -> List.map (fun l -> x ::l) (comb (n - 1) xs) # (comb n xs)
let powerset xs = seq {
for i = 0 to List.length xs do
for x in comb i xs -> set x
}
DEMO
> powerset ["a";"b";"c"] |> Seq.iter (printfn "%A");;
set []
set ["a"]
set ["b"]
set ["c"]
set ["a"; "b"]
set ["a"; "c"]
set ["b"; "c"]
set ["a"; "b"; "c"]
val it : unit = ()
From F# for Scientists, slightly modified to be lazy
let rec powerset s =
seq {
match s with
| [] -> yield []
| h::t -> for x in powerset t do yield! [x; h::x]
}
Here's another approach, using maths instead of recursion:
let powerset st =
let lst = Set.toList st
seq [0..(lst.Length |> pown 2)-1]
|> Seq.map (fun i ->
set ([0..lst.Length-1] |> Seq.choose (fun x ->
if i &&& (pown 2 x) = 0 then None else Some lst.[x])))
I would like to start some questions about simplifying different expressions in F#.
Anyone have ideas for better and/or simpler implementation of insertAt (parameters could be reordered, too). Lists or Sequences could be used.
Here is some start implementation:
let insertAt x xs n = Seq.concat [Seq.take n xs; seq [x]; Seq.skip n xs]
The implementation dannyasher posted is a non-tail-recursive one. In order to make the function more efficient, we'll have to introduce an explicit accumulator parameter which makes the function tail-recursive and allows the compiler to optimize the recursion overhead away:
let insertAt =
let rec insertAtRec acc n e list =
match n, list with
| 0, _ -> (List.rev acc) # [e] # list
| _, x::xs -> insertAtRec (x::acc) (n - 1) e xs
| _ -> failwith "Index out of range"
insertAtRec []
Tail-recursive using Seqs:
let rec insertAt = function
| 0, x, xs -> seq { yield x; yield! xs }
| n, x, xs -> seq { yield Seq.hd xs; yield! insertAt (n-1, x, Seq.skip 1 xs) }
Here's an F# implementation of the Haskell list insertion:
let rec insertAt x ys n =
match n, ys with
| 1, _
| _, [] -> x::ys
| _, y::ys -> y::insertAt x ys (n-1)
let a = [1 .. 5]
let b = insertAt 0 a 3
let c = insertAt 0 [] 3
>
val a : int list = [1; 2; 3; 4; 5]
val b : int list = [1; 2; 0; 3; 4; 5]
val c : int list = [0]
My Haskell isn't good enough to know whether the case of passing an empty list is correctly taken care of in the Haskell function. In F# we explicitly take care of the empty list in the second match case.
Danny
For case you really want to work with sequence:
let insertAt x ys n =
let i = ref n
seq {
for y in ys do
decr i
if !i = 0 then yield x
yield y
}
For all other cases dannyasher's answer is definitly nicer and faster.
From the Haskell Wiki - http://www.haskell.org/haskellwiki/99_questions/21_to_28
insertAt :: a -> [a] -> Int -> [a]
insertAt x ys 1 = x:ys
insertAt x (y:ys) n = y:insertAt x ys (n-1)
I'm not an F# programmer so I don't know the equivalent syntax for F# but this is a nice recursive definition for insertAt