F# continuation recursion bug - recursion

I'm having an issue with a recursive function that runs into a stack overflow on larger data sets so I've attempted to rewrite the function to use continuous recursion but to say I'm new to this would be an understatement. In the below example the first function, processList, gives the desired results on a small data set. The second function, processListCont, seems to work however I know there must be a bug since when I run the same small data set through it I get different results. Would processListCont be the correct way to express the processList function or am I missing something?
open System
type Something(id) =
member val id = id with get, set
member val children : list<Something> = [] with get, set
member val processed : bool = false with get, set
let rec processList (item:Something, itemList:list<Something>) =
for child in item.children do
let parent = itemList |> Seq.find (fun (i:Something) -> i.id = child.id)
if parent.processed = false then
parent.processed <- true
processList(parent, itemList)
let processListCont (item:Something, itemList:list<Something>) =
let rec _processListCont (item:Something, itemList:list<Something>, f) =
for child in item.children do
let parent = itemList |> Seq.find (fun (i:Something) -> i.id = child.id)
if parent.processed = false then
parent.processed <- true
f(parent, itemList)
_processListCont(item, itemList, (fun (item:Something, itemList:list<Something>) -> ()))
[<EntryPoint>]
let main argv =
// generate some data
let count = 10000
let idList = List.init count (fun index -> index)
let items = [for (id) in idList -> Something id]
let rnd = System.Random()
for i in items do
i.children <- List.init 100 (fun _ -> Something (rnd.Next(0, count - 1)))
// process the list
for i in items do
processList(i, items)
Console.WriteLine("Processing completed successfully")
Console.ReadKey()
|> ignore
0

The main issue is that you are calling the continuation f in the body of the for loop, but your non-tail-recursive version makes a recurisve call here.
This is tricky because you want to make a recursive call and the continuation should be "run the rest of the for loop". To express this, you'll need to use pattern matching instead of for loop.
I did not have a small example to test this, but I think something like this should do the trick:
let rec processListCont (item:Something, itemList:list<Something>) cont =
let rec loop (children:list<Something>) cont =
match children with
| child::tail ->
let parent = itemList |> Seq.find (fun (i:Something) -> i.id = child.id)
if parent.processed = false then
parent.processed <- true
processListCont (parent, itemList) (fun () -> loop tail cont)
| [] -> cont ()
loop item.children cont

Your code is unidiomatic in F# nonetheless consider the following example.
Suppose you want to add a list of numbers. You could write a function like this:
let rec add (l:int list) :int =
match l with
| [] -> 0
| x::xs -> x + (add xs)
but this would overflow the stack very quickly. Instead you could use cps to allow the code to become tail recursive:
type
cont = int -> int
let rec add2 (l:int list) (k:cont):int =
match l with
| [] -> k 0
| x::xs -> add2 xs (fun a -> k (a + x))
which you can use like this:
printfn "%i" (add2 [1..10000] id)
In a similar fashion you could rewrite your function like this:
type cont2 = Something list->unit
let rec p (item:Something, itemList:list<Something>) (k:cont2) =
match item.children with
| [] -> k []
| child::xs ->
let parent = itemList |> Seq.find (fun (i:Something) -> i.id = child.id)
if parent.processed = false then
parent.processed <- true
p (parent, itemList) (fun _ ->k xs)
else
k xs
let p2 (item:Something,itemList:Something list) = p (item,itemList) ignore
and you can call it like this:
for i in items do
p2(i, items)

Related

Removing from a list of a tuples that contains an empty element in the second projection F#

I need to learn the right way to do pattern matching on Pair types:
let pairToBeFiltered = Ok ([(1,[]);(2,[3;4]);(5,[6;7;8]);(9,[]);(10,[])])
let filterEmpty (pair: int * int list) =
match pair with
| (x,y) when y <> [] -> (x,y) //This gives error because of incomplete pattern matching!
let filtering = List.map(filterEmpty) pairToBeFiltered
Desired output:
Ok([(2,[3;4]);(5,[6;7;8])])
This should do it:
let pairsToBeFiltered = Ok ([(1,[]);(2,[3;4]);(5,[6;7;8]);(9,[]);(10,[])])
let filterEmpty pairs =
List.where (fun (_, y) -> y <> []) pairs // pattern match on pair occurs here
let filtering : Result<_, string> =
pairsToBeFiltered
|> Result.map filterEmpty
printfn "%A" filtering // Ok [(2, [3; 4]); (5, [6; 7; 8])]
There are a number of issues here:
For clarity, I modified filterEmpty so it processes the entire list, rather than a single pair. This is where we apply the filtering function, List.where, using pattern matching. (In your code, note that List.map with a match expression doesn't filter anything.)
Since your list is wrapped in a Result, you need to unwrap it via Result.map in order to process it. (Since you didn't specify a 'TError type, I assumed string to pacify the compiler.)
Three more versions:
(* using match statement *)
module Version1 =
let pairsToBeFiltered : Result<_, string> =
Ok [(1,[]);(2,[3;4]);(5,[6;7;8]);(9,[]);(10,[])]
let myWhere (pair : int * List<int>) =
match pair with
| _, [] -> false
| _, _ -> true
let myFilter l0 = l0 |> Result.map (List.filter myWhere)
let result = pairsToBeFiltered |> myFilter
(* using lambda functions and List.isEmpty *)
module Version2 =
let pairsToBeFiltered : Result<_, string> =
Ok [(1,[]);(2,[3;4]);(5,[6;7;8]);(9,[]);(10,[])]
let myFilter l0 =
l0
|> Result.map (fun l1 ->
l1 |> List.filter (fun (_, l2) ->
l2 |> List.isEmpty |> not))
let result = pairsToBeFiltered |> myFilter
(* shortening Version2 (point free style - take care, can be confusing) *)
module Version3 =
let pairsToBeFiltered : Result<_, string> =
Ok [(1,[]);(2,[3;4]);(5,[6;7;8]);(9,[]);(10,[])]
let myFilter = Result.map (List.filter (snd >> List.isEmpty >> not))
let result = pairsToBeFiltered |> myFilter

F# rewrite computation expression

I'm studying continuations because I want to make some interesting use of coroutines... anyway, I want to better understand one implementation I found.
To do so I want to rewrite the implementation without using the computation expression (continuation Monad), but I'm not quite able to do it.
I have this:
type K<'T,'r> = (('T -> 'r) -> 'r)
let returnK x = (fun k -> k x)
let bindK m f = (fun k -> m (fun a -> f a k))
let runK (c:K<_,_>) cont = c cont
let callcK (f: ('T -> K<'b,'r>) -> K<'T,'r>) : K<'T,'r> =
fun cont -> runK (f (fun a -> (fun _ -> cont a))) cont
type ContinuationBuilder() =
member __.Return(x) = returnK x
member __.ReturnFrom(x) = x
member __.Bind(m,f) = bindK m f
member this.Zero () = this.Return ()
let K = new ContinuationBuilder()
/// The coroutine type from http://fssnip.net/7M
type Coroutine() =
let tasks = new System.Collections.Generic.Queue<K<unit,unit>>()
member this.Put(task) =
let withYield = K {
do! callcK (fun exit ->
task (fun () ->
callcK (fun c ->
tasks.Enqueue(c())
exit ())))
if tasks.Count <> 0 then
do! tasks.Dequeue() }
tasks.Enqueue(withYield)
member this.Run() =
runK (tasks.Dequeue()) ignore
// from FSharpx tests
let ``When running a coroutine it should yield elements in turn``() =
// This test comes from the sample on http://fssnip.net/7M
let actual = System.Text.StringBuilder()
let coroutine = Coroutine()
coroutine.Put(fun yield' -> K {
actual.Append("A") |> ignore
do! yield' ()
actual.Append("B") |> ignore
do! yield' ()
actual.Append("C") |> ignore
do! yield' ()
})
coroutine.Put(fun yield' -> K {
actual.Append("1") |> ignore
do! yield' ()
actual.Append("2") |> ignore
do! yield' ()
})
coroutine.Run()
actual.ToString() = "A1B2C"
``When running a coroutine it should yield elements in turn``()
So, I want rewrite the Put member of the Coroutine class without using the computation expression K.
I have read of course this and this and several other articles about catamorphisms but it is not quite easy to rewrite this continuation monand as it is to rewrite the Write Monad for example...
I try several ways, this is one of them:
member this.Put(task) =
let withYield =
bindK
(callcK (fun exit ->
task (fun () ->
callcK (fun c ->
tasks.Enqueue(c())
exit ()))))
(fun () ->
if tasks.Count <> 0
then tasks.Dequeue()
else returnK ())
tasks.Enqueue(withYield)
Of course it does not work :(
(By the way: there is some extensive documentation of all rules the compiler apply to rewrite the computation in plain F#?)
Your version of Put is almost correct. Two issues though:
The bindK function is being used backwards, the parameters need to be swaped.
task should be passed a Cont<_,_> -> Cont<_,_>, not a unit -> Cont<_,_> -> Cont<_,_>.
Fixing those issues it could look like this:
member this.Put(task) =
let withYield =
bindK
(fun () ->
if tasks.Count <> 0
then tasks.Dequeue()
else returnK ())
(callcK (fun exit ->
task (
callcK (fun c ->
tasks.Enqueue(c())
exit ()))))
tasks.Enqueue(withYield)
Of course it is not too elegant.
When using bind it is better to declare an operator >>=:
let (>>=) c f = bindK f c
that way
do! translates to putting >>= fun () -> after
let! a = translates to putting >>= fun a -> after
and then your code will look a little bit better:
member this.Put2(task) =
let withYield =
callcK( fun exit ->
task( callcK (fun c ->
tasks.Enqueue(c())
exit())
)
) >>= fun () ->
if tasks.Count <> 0 then
tasks.Dequeue()
else returnK ()
tasks.Enqueue withYield

F# adding lists

How would I go about adding sub-lists.
For example, [ [10;2;10]; [10;50;10]] ----> [20;52;20] that is 10+10, 2+50 and 10+10. Not sure how to start this.
Fold is a higher order function:
let input = [[10;2;10]; [10;50;10]]
input |> Seq.fold (fun acc elem -> acc + (List.nth elem 1)) 0
val it : int = 52
Solution 1: Recursive version
We need a helper function to add two lists by summing elements one-to-one. It is recursive and assumes that both lists are of the same length:
let rec sum2Lists (l1:List<int>) (l2:List<int>) =
match (l1,l2) with
| ([],[]) -> []
| (x1::t1, x2::t2) -> (x1+x2)::sum2Lists t1 t2
Then the following recursive function can process a list of lists, using our helper function :
let rec sumLists xs =
match xs with
| [] -> [] // empty list
| x1::[] -> x1 // a single sublist
| xh::xt -> sum2Lists xh (sumLists xt) // add the head to recursion on tail
let myres = sumLists mylist
Solution 2: higher order function
Our helper function can be simplified, using List.map2:
let sum2hfLists (l1:List<int>) (l2:List<int>) = List.map2 (+) l1 l2
We can then use List.fold to create an on the flow accumulator using our helper function:
let sumhfList (l:List<List<int>>) =
match l with
| [] -> [] // empty list of sublist
| h::[] -> h // list with a single sublist
| h::t -> List.fold (fun a x -> sum2hfLists a x) h t
The last match case is applied only for lists of at least two sublists. The trick is to take the first sublist as starting point of the accumulator, and let fold execute on the rest of the list.

Remove duplicates from a list in SML

I just started learning functional programming in SML and I want to know how I can combine the following two functions into a single function. The function isolate deletes the duplicates of a list of any type ('a) using the helper function 'removes'.
fun isolate [] = []
| isolate (l as x::xs) = x::isolate(remove(x,xs))
fun remove (x,[]) = []
| remove (x,l as y::ys) = if x = y then remove(x,ys) else y::remove(x,ys)
So, for the purpose of better understanding the constructs in the SML, how would you include the function remove within isolate? This may seem trivial, but I have thinking about it and can't figure it out. Thank you for your help!
One method would be to just define remove inside isolate.
fun isolate [] = []
| isolate (l as x::xs) =
let fun remove (x,[]) = []
| remove (x,l as y::ys) = if x = y
then remove(x,ys)
else y::remove(x,ys)
in
x::isolate(remove(x,xs))
end
Alternately, to make deduplication one function, though all this really does is use the library function List.filter to do the same thing that remove does.
fun isolate [] = []
| isolate (x::xs) = x::isolate(List.filter (fn y => y <> x) xs)
My idea: define a nested function to check if there are duplicated elements in the list:
fun set(nums:int list)=
let fun duplicate(x:int, l:int list)=
if null l
then false
else hd l=x orelse duplicate(x,tl l)
in
if null nums
then []
else
let val s=set(tl nums)
in if duplicate(hd nums,s)
then s
else hd nums::s
end
end
But it will give a list that only remains the last one for every duplicated elements.
I want to propound the following solutions of this problem:
fun remove_duplicates(xs: int list) =
let
fun check(xs: int list, item: int) =
if null xs
then false
else if hd xs = item
then true
else check (tl xs, item)
fun go_through_list(xs: int list) =
if null xs
then []
else if check(tl xs, hd xs)
then go_through_list(tl xs)
else hd xs :: go_through_list(tl xs)
in
go_through_list(xs)
end
It's more lines of code than in the solution propounded by #qaphla
My idea is to first sort the list, then recursively return a new list without duplicates:
fun remove_duplicates(l: int list) =
if null(l)
then []
else if null(tl l)
then l
else
let
fun compare(x: int, y: int) = x > y
fun sort(l: int list) = ListMergeSort.sort(compare) l
val l_sorted = sort(l)
in
if (hd l_sorted) = (hd (tl l_sorted))
then remove_duplicates(tl l_sorted)
else (hd l_sorted)::remove_duplicates(tl l_sorted)
end

F# Cutting a list in half using functional programming

I am trying to input a list into the function and it send me a list with the first half of the elements taken away using f# with the below recursion but I keep running into a base case problem that I just cant figure out. any thoughts? I am using the second shadow list to count how far I need to go until I am half way into the list (by removing two elements at a time)
let rec dropHalf listToDrop shadowList =
match shadowList with
| [] -> listToDrop
| shadowHead2::shadowHead1::shadowTail -> if shadowTail.Length<=1 then listToDrop else
match listToDrop with
|[] -> listToDrop
|listToDropHead::listToDropTail -> dropHalf listToDropTail shadowTail
let rec dropHalf listToDrop shadowList =
match shadowList with
| [] -> listToDrop
| shadowHead2::[] -> listToDrop (* odd number! *)
| shadowHead1::shadowHead2::shadowTail ->
match listToDrop with
| [] -> listToDrop (* should never happen? *)
| listToDropHead::listToDropTail -> dropHalf listToDropTail shadowTail
i'm afraid i don't use F#, but it's similar to ocaml, so hopefully the following is close to what you're looking for (maybe the comment format has changed?!). the idea is that when you exhaust the shadow you're done. your code was almost there, but the test for length on the shadow tail made no sense.
i want to emphasize that this isn't anything like anyone would write "in real life", but it sounds like you're battling with some weird requirements.
Because you use the shadow list with the same length as the original list and remove elements from these lists with different rates, it's better to create an auxiliary function:
let dropHalf xs =
let rec dropHalf' ys zs =
match ys, zs with
| _::_::ys', _::zs' -> dropHalf' ys' zs'
| _, zs' -> zs' (* One half of the shadow list ys *)
dropHalf' xs xs
If you don't care to traverse the list twice, the following solution is simpler:
let rec drop n xs =
match xs, n with
| _ when n < 0 -> failwith "n should be greater or equals to 0"
| [], _ -> []
| _, 0 -> xs
| _::xs', _ -> drop (n-1) xs'
let dropHalf xs =
xs |> drop (List.length xs/2)
and another simple solution needs some extra space but doesn't have to use recursion:
let dropHalf xs =
let xa = Array.ofList xs
xa.[xa.Length/2..] |> List.ofArray
As a general rule of thumb, if you're calling Length on a list, then there is most likely a better way to do what you're doing. Length has to iterate the entire list and is therefore O(n).
let secondHalf list =
let rec half (result : 'a list) = function
| a::b::sx -> half result.Tail sx
// uncomment to remove odd value from second half
// | (a::sx) -> result.Tail
| _ -> result
half list list
Here is a sample does what you described.
open System
open System.Collections.Generic
let items = seq { 1 .. 100 } |> Seq.toList
let getBackOfList ( targetList : int list) =
if (targetList.Length = 0) then
targetList
else
let len = targetList.Length
let halfLen = len / 2
targetList |> Seq.skip halfLen |> Seq.toList
let shortList = items |> getBackOfList
("Len: {0}", shortList.Length) |> Console.WriteLine
let result = Console.ReadLine()
Hope this helps

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