I'm working through the book Real-World Functional Programming, and I tried to come up with my own example of tail recursion before reading the book's example (listing 10.2, p. 265). The book's example works; mine causes a stack overflow.
I've found if I use a tuple argument or pre-calculate a + accum then mine will work. I want to understand why.
let rnd = new System.Random()
let test2 = List.init 1000000 (fun _ -> rnd.Next(-50, 51))
let rec sum2 list accum =
match list with
| [] -> accum
| a::b -> sum2 b a + accum
let result = sum2 test2 0
printfn "%d" result
sum2 b a + accum
Note that this is parsed as (sum2 b a) + accum, not sum2 b (a + accum).
So this calls sum2 b a. Then it takes the result of that call and adds accum to it. So the last expression evaluated is the addition, not the call to sum2. Thus the call to sum2 is not a tail call.
Maybe the compiler is reading
a::b -> (sum2 b a) + accum
instead of
a::b -> sum2 b (a + accum)
Related
I was told that this first piece of code is, in terms of performance, worse than the second.
But, I honestly cannot figure out how they can be different, if in the end the same calls are made.
Am I missing something?
First example with explicit calls:
#let rec max l =
match l with
x::[]->x
| x::xs -> if(x > max xs)
then x else max xs;;
Second example with variable usage:
#let rec max l =
match l with
x::[]->x
| x::xs -> let m = max xs
in
if (x>m) then x else m;;
The key is that the ocaml compiler does not know that max xs and max xs are the same thing, so your first example is equivalent to something like:
let rec max l =
match l with
| x::[]-> x
| x::xs ->
let m1 = max xs in (* first call *)
if (x > m1) then
x
else
let m2 = max xs in
m2 (* second call *)
;;
Making only one call is a valid optimization, but is not correct in the general cases. For example:
let f () =
print_endline "hello";
print_endline "hello";
3
is not equivalent to:
let g () =
let x = print_endline "hello" in
x;
x;
3
I'm trying to write a function that would take an input like :
repeat 3 [1;2] ;;
and display something like:
[1;2;1;2;1;2]
Now the code I have is:
let repeat ls n =
let rec helper acc n l =
if n = 0 then acc else helper (l :: acc) (n-1) l in
let rec helper2 acc = function
| [] -> acc
| h :: t -> helper2 (helper acc n h) t in helper2 [] (List.rev ls);;
which gives me an output of:
[1;1;1;2;2;2]
for the same input. What can I do to fix this?
You are almost at the end ;)
Just modify the first helper :
let rec helper acc n l =
if n = 0 then acc else helper (l # acc) (n-1) l ;;
And you will be close to the solution.
(you just want to replicate the input list so # is ok to concatenate this list to the acc, you do not want to parse each and every element of the list, so :: is not what you need)
I think this solution may be a little bit faster in term of complexity (and simplicity):
let repeat ls n =
let rec f l = function
| 0 -> l
| n -> f (List.rev_append ls l) (n-1) in
List.rev (f [] n)
Also I always forget if List.rev is a tail-recursive or not, so this may be even better:
let repeat ls n =
let rec rev l = function
| [] -> l
| a::t -> rev (a::l) t in
let rec f l = function
| 0 -> l
| n -> f (List.rev_append ls l) (n-1) in
rev [] (f [] n)
Note: in my opinion Pierre's answer is good enough, my post is more like remark.
What do you usually do with a list in purely functional programming?
Obviously we have map f [x0; x1; x2] which produces [f x0; f x1; f x2] and fold f acc [x0; x1; x2] which produces f (f (f acc x0) x1) x2).
For map, no information is transmitted between calls to f; for fold, all the information produced by f is re-injected in the next call through the accumulator acc.
I've also seen things like flatmap that concatenates the lists produced by map f when f returns a list and for_all that applies a predicate to all elements of the lists, but these are just special cases of map and fold.
I can think about something intermediate that would produce a list but also keep an accumulator during iteration (OCaml syntax):
let rec map_acc ~f acc = function
| [] -> []
| x::xs -> let (y, acc) = f x acc in y::(map_acc ~f acc xs)
Questions:
is map_acc a usual concept in functional programming?
if yes, what's its canonical name?
is it implemented in standard libraries?
what are the other usual higher-order functions?
Notes:
Here is an example where map_acc is used to produce running sums:
let sums xs = map_acc ~f:(fun x sum -> let sum = sum + x in (sum, sum)) 0 xs
And sums [3; 5; 6; 9] produces [3; 8; 14; 23]. Granted sums can be implemented using fold and a well-tailored accumulator, but it is simpler with mac_acc.
map_acc looks like a slightly more general scan_left:
let scan_left f init list =
let _, xs = List.fold_left
(fun (acc, coll) elt ->
let acc' = f acc elt in acc', acc'::coll)
(init, []) list in
List.rev xs
scan_left (+) 0 [3; 5; 6; 9] => [3; 8; 14; 23]
This function and variants appear under various names in the standard libraries of a few functional languages, although not OCaml.
None of these functions really deserve being called primitives.
I am have some problems with recursion in Lazy Computations. I need calculation the square root by Newton Raphson method. I do not know how to apply a lazy evaluation. This is my code:
let next x z = ((x + z / x) / 2.);
let rec iterate f x =
List.Cons(x, (iterate f (f x)));
let rec within eps list =
let a = float (List.head list);
let b = float (List.head (List.tail list));
let rest = (List.tail (List.tail (list)));
if (abs(a - b) <= eps * abs(b))
then b
else within eps (List.tail (list));
let lazySqrt a0 eps z =
within eps (iterate (next z) a0);
let result2 = lazySqrt 10. Eps fvalue;
printfn "lazy approach";
printfn "result: %f" result2;
Of course, stack overflow exception.
You're using F# lists which has eager evaluation. In your example, you need lazy evaluation and decomposing lists, so F# PowerPack's LazyList is appropriate to use:
let next z x = (x + z / x) / 2.
let rec iterate f x =
LazyList.consDelayed x (fun () -> iterate f (f x))
let rec within eps list =
match list with
| LazyList.Cons(a, LazyList.Cons(b, rest)) when abs(a - b) <= eps * abs(b) -> b
| LazyList.Cons(a, res) -> within eps res
| LazyList.Nil -> failwith "Unexpected pattern"
let lazySqrt a0 eps z =
within eps (iterate (next z) a0)
let result2 = lazySqrt 10. Eps fvalue
printfn "lazy approach"
printfn "result: %f" result2
Notice that I use pattern matching which is more idiomatic than head and tail.
If you don't mind a slightly different approach, Seq.unfold is natural here:
let next z x = (x + z / x) / 2.
let lazySqrt a0 eps z =
a0
|> Seq.unfold (fun a ->
let b = next z a
if abs(a - b) <= eps * abs(b) then None else Some(a, b))
|> Seq.fold (fun _ x -> x) a0
If you need lazy computations, then you have to use appropriate tools. List is not lazy, it is computed to the end. Your iterate function never ends, so the entire code stack overflows in this function.
You may use Seq here.
Note: Seq.skip almost inevitably leads you to an O(N^2) complexity.
let next N x = ((x + N / x) / 2.);
let rec iterate f x = seq {
yield x
yield! iterate f (f x)
}
let rec within eps list =
let a = Seq.head list
let b = list |> Seq.skip 1 |> Seq.head
if (abs(a - b) <= eps * abs(b))
then b
else list |> Seq.skip 1 |> within eps
let lazySqrt a0 eps z =
within eps (iterate (next z) a0);
let result2 = lazySqrt 10. 0.0001 42.;
printfn "lazy approach";
printfn "result: %f" result2;
// 6.4807406986501
Yet another approach is to use LazyList from F# PowerPack. The code is available in this article. Copying it to my answer for sake of integrity:
open Microsoft.FSharp.Collections.LazyList
let next N (x:float) = (x + N/x) / 2.0
let rec repeat f a =
LazyList.consDelayed a (fun() -> repeat f (f a))
let rec within (eps : float) = function
| LazyList.Cons(a, LazyList.Cons(b, rest)) when (abs (a - b)) <= eps -> b
| x -> within eps (LazyList.tail x)
let newton_square a0 eps N = within eps (repeat (next N) a0)
printfn "%A" (newton_square 16.0 0.001 16.0)
Some minor notes:
Your next function is wrong;
The meaning of eps is relative accuracy while in most academic books I've seen an absolute accuracy. The difference between the two is whether or not it's measured against b, here: <= eps * abs(b). The code from FPish treats eps as an absolute accuracy.
This is actually a solution to Project Euler Problem 14 in F#. However, I'm running into a System.OutOfMemory exception when attempting to calculate an iterative sequence for larger numbers. As you can see, I'm writing my recursive function with tail calls.
I was running into a problem with StackOverFlowException because I was debugging in visual studio (which disables the tail calls). I've documented that in another question. Here, I'm running in release mode--but I'm getting out of memory exceptions when I run this as a console app (on windows xp with 4gb ram).
I'm really at a loss to understand how I coded myself into this memory overflow & hoping someone can show my the error in my ways.
let E14_interativeSequence x =
let rec calc acc startNum =
match startNum with
| d when d = 1 -> List.rev (d::acc)
| e when e%2 = 0 -> calc (e::acc) (e/2)
| _ -> calc (startNum::acc) (startNum * 3 + 1)
let maxNum pl=
let rec maxPairInternal acc pairList =
match pairList with
| [] -> acc
| x::xs -> if (snd x) > (snd acc) then maxPairInternal x xs
else maxPairInternal acc xs
maxPairInternal (0,0) pl
|> fst
// if I lower this to like [2..99999] it will work.
[2..99999]
|> List.map (fun n -> (n,(calc [] n)))
|> List.map (fun pair -> ((fst pair), (List.length (snd pair))))
|> maxNum
|> (fun x-> Console.WriteLine(x))
EDIT
Given the suggestions via the answers, I rewrote to use a lazy list and also to use Int64's.
#r "FSharp.PowerPack.dll"
let E14_interativeSequence =
let rec calc acc startNum =
match startNum with
| d when d = 1L -> List.rev (d::acc) |> List.toSeq
| e when e%2L = 0L -> calc (e::acc) (e/2L)
| _ -> calc (startNum::acc) (startNum * 3L + 1L)
let maxNum (lazyPairs:LazyList<System.Int64*System.Int64>) =
let rec maxPairInternal acc (pairs:seq<System.Int64*System.Int64>) =
match pairs with
| :? LazyList<System.Int64*System.Int64> as p ->
match p with
| LazyList.Cons(x,xs)-> if (snd x) > (snd acc) then maxPairInternal x xs
else maxPairInternal acc xs
| _ -> acc
| _ -> failwith("not a lazylist of pairs")
maxPairInternal (0L,0L) lazyPairs
|> fst
{2L..999999L}
|> Seq.map (fun n -> (n,(calc [] n)))
|> Seq.map (fun pair -> ((fst pair), (Convert.ToInt64(Seq.length (snd pair)))))
|> LazyList.ofSeq
|> maxNum
which solves the problem. I'd also look at Yin Zhu's solution which is better, though.
As mentioned by Brian, List.* operations are not appropriate here. They cost too much memory.
The stackoverflow problem comes from another place. There are two possible for you to have stackoverflow: calc and maxPairInternal. It must be the first as the second has the same depth as the first. Then the problem comes to the numbers, the number in 3n+1 problem could easily go to very large. So you first get a int32 overflow, then you get a stackoverflow. That's the reason. After changing the numbers to 64bit, the program works.
Here is my solution page, where you can see a memoization trick.
open System
let E14_interativeSequence x =
let rec calc acc startNum =
match startNum with
| d when d = 1L -> List.rev (d::acc)
| e when e%2L = 0L -> calc (e::acc) (e/2L)
| _ -> calc (startNum::acc) (startNum * 3L + 1L)
let maxNum pl=
let rec maxPairInternal acc pairList =
match pairList with
| [] -> acc
| x::xs -> if (snd x) > (snd acc) then maxPairInternal x xs
else maxPairInternal acc xs
maxPairInternal (0L,0) pl
|> fst
// if I lower this to like [2..99999] it will work.
[2L..1000000L]
|> Seq.map (fun n -> (n,(calc [] n)))
|> Seq.maxBy (fun (n, lst) -> List.length lst)
|> (fun x-> Console.WriteLine(x))
If you change List.map to Seq.map (and re-work maxPairInternal to iterate over a seq) that will probably help tons. Right now, you're manifesting all the data at once in a giant structure before processing the whole structure to get a single number result. It is much better to do this lazily via Seq, and just create one row, and compare it with the next row, and create a single row at a time and then discard it.
I don't have time to code my suggestion now, but let me know if you are still having trouble and I'll revisit this.
Stop trying to use lists everywhere, this isn't Haskell! And stop writing fst pair and snd pair everywhere, this isn't Lisp!
If you want a simple solution in F# you can do it directly like this without creating any intermediate data structures:
let rec f = function
| 1L -> 0
| n when n % 2L = 0L -> 1 + f(n / 2L)
| n -> 1 + f(3L * n + 1L)
let rec g (li, i) = function
| 1L -> i
| n -> g (max (li, i) (f n, n)) (n - 1L)
let euler14 n = g (0, 1L) n
That takes around 15s on my netbook. If you want something more time efficient, reuse previous results via an array:
let rec inside (a : _ array) n =
if n <= 1L || a.[int n] > 0s then a.[int n] else
let p =
if n &&& 1L = 0L then inside a (n >>> 1) else
let n = 3L*n + 1L
if n < int64 a.Length then inside a n else outside a n
a.[int n] <- 1s + p
1s + p
and outside (a : _ array) n =
let n = if n &&& 1L = 0L then n >>> 1 else 3L*n + 1L
1s + if n < int64 a.Length then inside a n else outside a n
let euler14 n =
let a = Array.create (n+1) 0s
let a = Array.Parallel.init (n+1) (fun n -> inside a (int64 n))
let i = Array.findIndex (Array.reduce max a |> (=)) a
i, a.[i]
That takes around 0.2s on my netbook.
Found this looking for Microsoft.FSharp.Core.Operators.Checked.
I'm just learning F#, so I thought I'd take the Project Euler 14 Challenge.
This uses recursion but not tail-recursion.
Takes about 3.1 sec for me, but has the advantage that I can almost understand it.
let Collatz (n:int64) = if n % 2L = 0L then n / 2L else n * 3L + 1L
let rec CollatzLength (current:int64) (acc:int) =
match current with
| 1L -> acc
| _ -> CollatzLength (Collatz current) (acc + 1)
let collatzSeq (max:int64) =
seq{
for i in 1L..max do
yield i, CollatzLength i 0
}
let collatz = Seq.toList(collatzSeq 1000000L)
let result, steps = List.maxBy snd collatz