I am trying to convert the following normal-recursive code to tail-recursive in F#, but I am failing miserably.
let rec insert elem lst =
match lst with
| [] -> [elem]
| hd::tl -> if hd > elem then
elem::lst
else
hd::(insert elem tl)
let lst1 = []
let lst2 = [1;2;3;5]
printfn "\nInserting 4 in an empty list: %A" (insert 4 lst1)
printfn "\nInserting 4 in a sorted list: %A" (insert 4 lst2)
Can you guys help? Unfortunately I am a beginner in f#. Also, can anyone point me to a good tutorial to understand tail-recursion?
The point of tail recursion is the following: the last operation before returning from a function is a call to itself; this is called a tail call, and is where tail recursion gets its name from (the recursive call is in last, i.e. tail position).
Your function is not tail recursive because at least one of its branches has an operation after the recursive call (the list cons operator).
The usual way of converting a recursive function into a tail-recursive function is to add an argument to accumulate intermediate results (the accumulator). When it comes to lists, and when you realize that the only elementary list operation is prepending an element, this also means that after you are through with processing your list, it will be reversed, and thus the resulting accumulator will usually have to be reversed again.
With all these points in mind, and given that we do not want to change the function's public interface by adding a parameter that is superfluous from the caller's point of view, we move the real work to an internal subfunction. This particular function is slightly more complicated because after the element has been inserted, there is nothing else to do but concatenate the two partial lists again, one of which is now in reverse order while the other is not. We create a second internal function to handle that part, and so the whole function looks as follows:
let insert elm lst =
let rec iter acc = function
| [] -> List.rev (elm :: acc)
| (h :: t) as ls ->
if h > elm then finish (elm :: ls) acc
else iter (h :: acc) t
and finish acc = function
| [] -> acc
| h :: t -> finish (h :: acc) t
iter [] lst
For further studying, Scott Wlaschin's F# for Fun and Profit is a great resource, and tail recursion is handled in a larger chapter about recursive types and more: https://fsharpforfunandprofit.com/posts/recursive-types-and-folds
Related
I was trying to write a generic mapFoldWhile function, which is just mapFold but requires the state to be an option and stops as soon as it encounters a None state.
I don't want to use mapFold because it will transform the entire list, but I want it to stop as soon as an invalid state (i.e. None) is found.
This was myfirst attempt:
let mapFoldWhile (f : 'State option -> 'T -> 'Result * 'State option) (state : 'State option) (list : 'T list) =
let rec mapRec f state list results =
match list with
| [] -> (List.rev results, state)
| item :: tail ->
let (result, newState) = f state item
match newState with
| Some x -> mapRec f newState tail (result :: results)
| None -> ([], None)
mapRec f state list []
The List.rev irked me, since the point of the exercise was to exit early and constructing a new list ought to be even slower.
So I looked up what F#'s very own map does, which was:
let map f list = Microsoft.FSharp.Primitives.Basics.List.map f list
The ominous Microsoft.FSharp.Primitives.Basics.List.map can be found here and looks like this:
let map f x =
match x with
| [] -> []
| [h] -> [f h]
| (h::t) ->
let cons = freshConsNoTail (f h)
mapToFreshConsTail cons f t
cons
The consNoTail stuff is also in this file:
// optimized mutation-based implementation. This code is only valid in fslib, where mutation of private
// tail cons cells is permitted in carefully written library code.
let inline setFreshConsTail cons t = cons.(::).1 <- t
let inline freshConsNoTail h = h :: (# "ldnull" : 'T list #)
So I guess it turns out that F#'s immutable lists are actually mutable because performance? I'm a bit worried about this, having used the prepend-then-reverse list approach as I thought it was the "way to go" in F#.
I'm not very experienced with F# or functional programming in general, so maybe (probably) the whole idea of creating a new mapFoldWhile function is the wrong thing to do, but then what am I to do instead?
I often find myself in situations where I need to "exit early" because a collection item is "invalid" and I know that I don't have to look at the rest. I'm using List.pick or Seq.takeWhile in some cases, but in other instances I need to do more (mapFold).
Is there an efficient solution to this kind of problem (mapFoldWhile in particular and "exit early" in general) with functional programming concepts, or do I have to switch to an imperative solution / use a Collections.Generics.List?
In most cases, using List.rev is a perfectly sufficient solution.
You are right that the F# core library uses mutation and other dirty hacks to squeeze some more performance out of the F# list operations, but I think the micro-optimizations done there are not particularly good example. F# list functions are used almost everywhere so it might be a good trade-off, but I would not follow it in most situations.
Running your function with the following:
let l = [ 1 .. 1000000 ]
#time
mapFoldWhile (fun s v -> 0, s) (Some 1) l
I get ~240ms on the second line when I run the function without changes. When I just drop List.rev (so that it returns the data in the other order), I get around ~190ms. If you are really calling the function frequently enough that this matters, then you'd have to use mutation (actually, your own mutable list type), but I think that is rarely worth it.
For general "exit early" problems, you can often write the code as a composition of Seq.scan and Seq.takeWhile. For example, say you want to sum numbers from a sequence until you reach 1000. You can write:
input
|> Seq.scan (fun sum v -> v + sum) 0
|> Seq.takeWhile (fun sum -> sum < 1000)
Using Seq.scan generates a sequence of sums that is over the whole input, but since this is lazily generated, using Seq.takeWhile stops the computation as soon as the exit condition happens.
As per the MSDN documentation, while writing recursive function, The use of the accumulator argument makes the function tail recursive, which saves stack space.
I'm using two example given on the MSDN site to calculate of sum of all the numbers in a list-
first without tail recursion-
let rec Sum myList =
match myList with
| [] -> 0
| h::t -> h + Sum t
and now with tail recursion-
let Sumtail list =
let rec loop list acc =
match list with
| h::t -> loop t acc + h
| [] -> acc
loop list 0
and running both the functions with input [1..100000].
Function Sum successfully calculates the sum of this list but gives stackoverflow exception if I pass [1..1000000]
but the second function Sumtail fails at [1..100000] while it should give better performance then the first function since it uses tail recursion.
Are there any other factors which affects the recursive function?
Your second function isn't tail-recursive since loop t acc + h is parsed as (loop t acc) + h which makes + become the last operation on loop.
Change loop t acc + h to loop t (acc + h) in order that the function becomes tail-recursive.
As I'm reading the Programming F# book, I found the example code snippet on page 195 as follows:
type ContinuationStep<'a> =
| Finished
| Step of 'a * (unit -> ContinuationStep<'a>)
let iter f binTree =
let rec linearize binTree cont =
match binTree with
| Empty -> cont()
| Node(x, l, r) ->
Step(x, (fun () -> linearize l (fun() -> linearize r cont)))
let steps = linearize binTree (fun () -> Finished)
let rec processSteps step =
match step with
| Finished -> ()
| Step(x, getNext)
-> f x
processSteps (getNext())
processSteps steps
By using continuation, the binary recursion of traversing a binary has been transformed to tail-recursive function processSteps. My question is that the other function, linearize seems to be non-tail-recursive. Does that mean we are not able to transform a binary-recursion to a tail-recursion completely even using continuation?
linearize is tail-recursive: you don't need to come back from the recursive call to continue the computation.
fun () -> linearize l (fun() -> linearize r cont)
doesn't call linearize. The computation is suspended until processSteps calls getNext ().
The example is a bit subtle because it does not use ordinary continuations, but instead builds a structure that can be evaluated step-by-step. In a place where you would normally make a recursive call, it returns a value Step that contains the function that you'd (recursively) call.
In the second case, the linearize function returns a Step containing a function that will call linearize recursively, but it does not immediately make the recursive call. So the function does not make a recursive call (it just stores a recursive reference).
It only makes sense to consider whether the program is tail-recursive when you look at processSteps, because that does the actual looping - and that is tail-recursive, because it runs a Step by Step without keeping stack space for each Step.
If you wanted to construct a list instead of a chain of lazy steps then you'd have to make the recursive call to linearize immediately inside the continuation:
let rec linearize binTree cont =
match binTree with
| Empty -> cont []
| Node(x, l, r) ->
linearize l (fun l -> linearize r (fun v -> cont (x::v)))
This is essentially the same as the previous function, but it actually calls linearize instead of building Step containing a function that will call linearize.
I was reading this post While or Tail Recursion in F#, what to use when? were several people say that the 'functional way' of doing things is by using maps/folds and higher order functions instead of recursing and looping.
I have this function that returns the item at position x in a list:
let rec getPos l c = if c = 0 then List.head l else getPos (List.tail l) (c - 1)
how can it be converted to be more functional?
This is a primitive list function (also known as List.nth).
It is okay to use recursion, especially when creating the basic building blocks. Although it would be nicer with pattern matching instead of if-else, like this:
let rec getPos l c =
match l with
| h::_ when c = 0 -> h
| _::t -> getPos t (c-1)
| [] -> failwith "list too short"
It is possible to express this function with List.fold, however the result is less clear than the recursive version.
I'm not sure what you mean by more functional.
Are you rolling this yourself as a learning exercise?
If not, you could just try this:
> let mylist = [1;2;3;4];;
> let n = 2;;
> mylist.[n];;
Your definition is already pretty functional since it uses a tail-recursive function instead of an imperative loop construct. However, it also looks like something a Scheme programmer might have written because you're using head and tail.
I suspect you're really asking how to write it in a more idiomatic ML style. The answer is to use pattern matching:
let rec getPos list n =
match list with
| hd::tl ->
if n = 0 then hd
else getPos tl (n - 1)
| [] -> failWith "Index out of range."
The recursion on the structure of the list is now revealed in the code. You also get a warning if the pattern matching is non-exhaustive so you're forced to deal with the index too big error.
You're right that functional programming also encourages the use of combinators like map or fold (so called points-free style). But too much of it just leads to unreadable code. I don't think it's warranted in this case.
Of course, Benjol is right, in practice you would just write mylist.[n].
If you'd like to use high-order functions for this, you could do:
let nth n = Seq.take (n+1) >> Seq.fold (fun _ x -> Some x) None
let nth n = Seq.take (n+1) >> Seq.reduce (fun _ x -> x)
But the idea is really to have basic constructions and combine them build whatever you want. Getting the nth element of a sequence is clearly a basic block that you should use. If you want the nth item, as Benjol mentioned, do myList.[n].
For building basic constructions, there's nothing wrong to use recursion or mutable loops (and often, you have to do it this way).
Not as a practical solution, but as an exercise, here is one of the ways to express nth via foldr or, in F# terms, List.foldBack:
let myNth n xs =
let step e f = function |0 -> e |n -> f (n-1)
let error _ = failwith "List is too short"
List.foldBack step xs error n
Hey guys, I'm trying to get cozy with functional programming (particularly with F#), and I've hit a wall when it comes to building tail-recursive functions. I'm pretty good with turning basic recursion (where the function basically calls itself once per invocation), into tail recursion, but I now have a slightly more complicated situation.
In my case, the function must accept a single list as a parameter. When the function is called, I have to remove the first element from the list, and then recur using the remainder of the list. Then I need to apply the first element which I removed in some way to the result of the recursion. Next, I remove the second element and do the same thing (Note: when I say "remove the seond element", that is from the original list, so the list passed at the recursion includes the first element as well). I do the same for the third, fourth, etc. elements of the list.
Is there a way to convert the above situation into a tail-recursive function? Maybe nested tail-recursive functions??? Thank you for any answers.
Okay, so here's my basic code. This particular one is a permutation generator (I'm not too concern with the permutation part, though - it's the recursion I'd like to focusing on):
let permutationsOther str =
match str with
| value :: [] ->
[[value]]
| _ ->
let list = (List.map (fun a -> // This applies the remove part for every element a
let lst = (List.filter (fun b -> b <> a) str) // This part removes element a from the list
let permutedLst = permutations lst // recursive call
consToAll a permutedLst // constToAll this is my own function which performs "cons" operation with a and every element in the list permutedLst
) str)
List.reduce (fun acc elem -> elem # acc) list // flatten list of lists produce by map into a single list
I hope this is clear enough - I'll be happy to provide clarifications if needed.
By the way, I have found just a way to rewrite this particular function so that it only uses a single recursion, but it was a fluke more than an informed decision. However, this has encouraged me that there may be a general method of turning multiple recursion into single recursion, but I have not yet found it.
Conversion to CPS should do the trick:
NOTE 1: Source of the sample is typed directly in browser, so may contain errors :(. But I hope it can demonstrate the general idea.
NOTE 2: consToAll function should be converted to CPS too: consToAll: 'T -> 'T list list -> ('T list list -> 'R) -> 'R
let remove x l = List.filter ((<>) x) l // from original post: should duplicates also be removed ???
let permute l =
let rec loop k l =
match l with
| [] -> k []
| [value] -> k [[value]]
| _ -> filter l [] l (fun r -> r |> List.reduce (fun acc elem -> elem # acc) |> k )
and filter l acc orig fk =
match l with
| [] -> fk acc
| x::xs ->
remove x orig
|> loop (fun res ->
consToAll x res (fun rs -> filter xs (rs::acc) orig fk)
)
loop id l