I recently started solving Project Euler problems in Scala, however when I got to problem 14, I got the StackOverflowError, so I rewrote my solution in F#, since (I am told) the F# compiler, unlike Scala's (which produces Java bytecode), translates recursive calls into loops.
My question to you therefore is, how is it possible that the code below throws the StackOverflowException after reaching some number above 113000? I think that the recursion doesn't have to be a tail recursion in order to be translated/optimized, does it?
I tried several rewrites of my code, but without success. I really don't want to have to write the code in imperative style using loops, and I don't think I could turn the len function to be tail-recursive, even if that was the problem preventing it from being optimized.
module Problem14 =
let lenMap = Dictionary<'int,'int>()
let next n =
if n % 2 = 0 then n/2
else 3*n+1
let memoize(num:int, lng:int):int =
lenMap.[num]<-lng
lng
let rec len(num:int):int =
match num with
| 1 -> 1
| _ when lenMap.ContainsKey(num) -> lenMap.[num]
| _ -> memoize(num, 1+(len (next num)))
let cand = seq{ for i in 999999 .. -1 .. 1 -> i}
let tuples = cand |> Seq.map(fun n -> (n, len(n)))
let Result = tuples |> Seq.maxBy(fun n -> snd n) |> fst
NOTE: I am aware that the code below is very far from optimal and several lines could be a lot simpler, but I am not very proficient in F# and did not bother looking up ways to simplify it and make it more elegant (yet).
Thank you.
Your current code runs without error and gets the correct result if I change all the int to int64 and append an L after every numeric literal (e.g. -1L). If the actual problem is that you're overflowing a 32-bit integer, I'm not sure why you get a StackOverflowException.
module Problem14 =
let lenMap = System.Collections.Generic.Dictionary<_,_>()
let next n =
if n % 2L = 0L then n/2L
else 3L*n+1L
let memoize(num, lng) =
lenMap.[num]<-lng
lng
let rec len num =
match num with
| 1L -> 1L
| _ when lenMap.ContainsKey(num) -> lenMap.[num]
| _ -> memoize(num, 1L+(len (next num)))
let cand = seq{ for i in 999999L .. -1L .. 1L -> i}
let tuples = cand |> Seq.map(fun n -> (n, len(n)))
let Result = tuples |> Seq.maxBy(fun n -> snd n) |> fst
Related
I'm trying to concatenate a string with a certain amount of times, but I feel like I've cheated a bit (or at least not actually understood how it's supposed to be done) by using a higher-order function:
let repeat s n =
String.replicate n s |> printfn "%s"
repeat "a" 10
Obviously gives me "aaaaaaaaaa", but how could I do this without a higher-order function? I feel like it's a very simple problem but I can't seem to wrap my head around it, the F# syntax, or way of thinking, is still troublesome for me.
If you just want a recursive solution, how about this?
let rec repeat s n =
match n with
| _ when n <= 0 -> ""
| _ -> s + (repeat s (n-1))
repeat "a" 10
or in a more "classic" style with an if-expression:
let rec repeat s n =
if n <= 0 then
""
else
s + (repeat s (n-1))
repeat "a" 10
And here's one way using list comprehension and fold, which is the go to function for recursion:
[for i in 1..10 -> "a"] |> List.fold (+) ""
Tail Recursive version
let repeat2 s n =
let rec loop acc n =
match n with
| _ when n > 0 -> loop (acc + s) (n - 1)
| _ -> acc
loop "" n
repeat "oijdfsaoijdoyasjd" 100000 // Process is terminated due to StackOverflowException.
[for i in 1..100000 -> "oijdfsaoijdoyasjd"] |> List.fold (+) "" // no stack overflow
repeat2 "oijdfsaoijdoyasjd" 100000 // no stack overflow
But prepared for massive amounts of gen2 GC and a few min. of runtime.
I'm trying to implement Kosaraju's algorithm on a large graph
as part of an assignment [MOOC Algo I Stanford on Coursera]
https://en.wikipedia.org/wiki/Kosaraju%27s_algorithm
The current code works on a small graph, but I'm hitting Stack Overflow during runtime execution.
Despite having read the relevant chapter in Expert in F#, or other available examples on websites and SO, i still don't get how to use continuation to solve this problem
Below is the full code for general purpose, but it will already fail when executing DFSLoop1 and the recursive function DFSsub inside. I think I'm not making the function tail recursive [because of the instructions
t<-t+1
G.[n].finishingtime <- t
?]
but i don't understand how i can implement the continuation properly.
When considering only the part that fails, DFSLoop1 is taking as argument a graph to which we will apply Depth-First Search. We need to record the finishing time as part of the algo to proceed to the second part of the algo in a second DFS Loop (DFSLoop2) [of course we are failing before that].
open System
open System.Collections.Generic
open System.IO
let x = File.ReadAllLines "C:\Users\Fagui\Documents\GitHub\Learning Fsharp\Algo Stanford I\PA 4 - SCC.txt";;
// let x = File.ReadAllLines "C:\Users\Fagui\Documents\GitHub\Learning Fsharp\Algo Stanford I\PA 4 - test1.txt";;
// val x : string [] =
let splitAtTab (text:string)=
text.Split [|'\t';' '|]
let splitIntoKeyValue (A: int[]) =
(A.[0], A.[1])
let parseLine (line:string)=
line
|> splitAtTab
|> Array.filter (fun s -> not(s=""))
|> Array.map (fun s-> (int s))
|> splitIntoKeyValue
let y =
x |> Array.map parseLine
//val it : (int * int) []
type Children = int[]
type Node1 =
{children : Children ;
mutable finishingtime : int ;
mutable explored1 : bool ;
}
type Node2 =
{children : Children ;
mutable leader : int ;
mutable explored2 : bool ;
}
type DFSgraphcore = Dictionary<int,Children>
let directgraphcore = new DFSgraphcore()
let reversegraphcore = new DFSgraphcore()
type DFSgraph1 = Dictionary<int,Node1>
let reversegraph1 = new DFSgraph1()
type DFSgraph2 = Dictionary<int,Node2>
let directgraph2 = new DFSgraph2()
let AddtoGraph (G:DFSgraphcore) (n,c) =
if not(G.ContainsKey n) then
let node = [|c|]
G.Add(n,node)
else
let c'= G.[n]
G.Remove(n) |> ignore
G.Add (n, Array.append c' [|c|])
let inline swaptuple (a,b) = (b,a)
y|> Array.iter (AddtoGraph directgraphcore)
y|> Array.map swaptuple |> Array.iter (AddtoGraph reversegraphcore)
for i in directgraphcore.Keys do
if reversegraphcore.ContainsKey(i) then do
let node = {children = reversegraphcore.[i] ;
finishingtime = -1 ;
explored1 = false ;
}
reversegraph1.Add (i,node)
else
let node = {children = [||] ;
finishingtime = -1 ;
explored1 = false ;
}
reversegraph1.Add (i,node)
directgraphcore.Clear |> ignore
reversegraphcore.Clear |> ignore
// for i in reversegraph1.Keys do printfn "%d %A" i reversegraph1.[i].children
printfn "pause"
Console.ReadKey() |> ignore
let num_nodes =
directgraphcore |> Seq.length
let DFSLoop1 (G:DFSgraph1) =
let mutable t = 0
let mutable s = -1
let mutable k = num_nodes
let rec DFSsub (G:DFSgraph1)(n:int) (cont:int->int) =
//how to make it tail recursive ???
G.[n].explored1 <- true
// G.[n].leader <- s
for j in G.[n].children do
if not(G.[j].explored1) then DFSsub G j cont
t<-t+1
G.[n].finishingtime <- t
// end of DFSsub
for i in num_nodes .. -1 .. 1 do
printfn "%d" i
if not(G.[i].explored1) then do
s <- i
( DFSsub G i (fun s -> s) ) |> ignore
// printfn "%d %d" i G.[i].finishingtime
DFSLoop1 reversegraph1
printfn "pause"
Console.ReadKey() |> ignore
for i in directgraphcore.Keys do
let node = {children =
directgraphcore.[i]
|> Array.map (fun k -> reversegraph1.[k].finishingtime) ;
leader = -1 ;
explored2= false ;
}
directgraph2.Add (reversegraph1.[i].finishingtime,node)
let z = 0
let DFSLoop2 (G:DFSgraph2) =
let mutable t = 0
let mutable s = -1
let mutable k = num_nodes
let rec DFSsub (G:DFSgraph2)(n:int) (cont:int->int) =
G.[n].explored2 <- true
G.[n].leader <- s
for j in G.[n].children do
if not(G.[j].explored2) then DFSsub G j cont
t<-t+1
// G.[n].finishingtime <- t
// end of DFSsub
for i in num_nodes .. -1 .. 1 do
if not(G.[i].explored2) then do
s <- i
( DFSsub G i (fun s -> s) ) |> ignore
// printfn "%d %d" i G.[i].leader
DFSLoop2 directgraph2
printfn "pause"
Console.ReadKey() |> ignore
let table = [for i in directgraph2.Keys do yield directgraph2.[i].leader]
let results = table |> Seq.countBy id |> Seq.map snd |> Seq.toList |> List.sort |> List.rev
printfn "%A" results
printfn "pause"
Console.ReadKey() |> ignore
Here is a text file with a simple graph example
1 4
2 8
3 6
4 7
5 2
6 9
7 1
8 5
8 6
9 7
9 3
(the one which is causing overflow is 70Mo big with around 900,000 nodes)
EDIT
to clarify a few things first
Here is the "pseudo code"
Input: a directed graph G = (V,E), in adjacency list representation. Assume that the vertices V are labeled
1, 2, 3, . . . , n.
1. Let Grev denote the graph G after the orientation of all arcs have been reversed.
2. Run the DFS-Loop subroutine on Grev, processing vertices according to the given order, to obtain a
finishing time f(v) for each vertex v ∈ V .
3. Run the DFS-Loop subroutine on G, processing vertices in decreasing order of f(v), to assign a leader
to each vertex v ∈ V .
4. The strongly connected components of G correspond to vertices of G that share a common leader.
Figure 2: The top level of our SCC algorithm. The f-values and leaders are computed in the first and second
calls to DFS-Loop, respectively (see below).
Input: a directed graph G = (V,E), in adjacency list representation.
1. Initialize a global variable t to 0.
[This keeps track of the number of vertices that have been fully explored.]
2. Initialize a global variable s to NULL.
[This keeps track of the vertex from which the last DFS call was invoked.]
3. For i = n downto 1:
[In the first call, vertices are labeled 1, 2, . . . , n arbitrarily. In the second call, vertices are labeled by
their f(v)-values from the first call.]
(a) if i not yet explored:
i. set s := i
ii. DFS(G, i)
Figure 3: The DFS-Loop subroutine.
Input: a directed graph G = (V,E), in adjacency list representation, and a source vertex i ∈ V .
1. Mark i as explored.
[It remains explored for the entire duration of the DFS-Loop call.]
2. Set leader(i) := s
3. For each arc (i, j) ∈ G:
(a) if j not yet explored:
i. DFS(G, j)
4. t + +
5. Set f(i) := t
Figure 4: The DFS subroutine. The f-values only need to be computed during the first call to DFS-Loop, and
the leader values only need to be computed during the second call to DFS-Loop.
EDIT
i have amended the code, with the help of an experienced programmer (a lisper but who has no experience in F#) simplifying somewhat the first part to have more quickly an example without bothering about non-relevant code for this discussion.
The code focuses only on half of the algo, running DFS once to get finishing times of the reversed tree.
This is the first part of the code just to create a small example
y is the original tree. the first element of a tuple is the parent, the second is the child. But we will be working with the reverse tree
open System
open System.Collections.Generic
open System.IO
let x = File.ReadAllLines "C:\Users\Fagui\Documents\GitHub\Learning Fsharp\Algo Stanford I\PA 4 - SCC.txt";;
// let x = File.ReadAllLines "C:\Users\Fagui\Documents\GitHub\Learning Fsharp\Algo Stanford I\PA 4 - test1.txt";;
// val x : string [] =
let splitAtTab (text:string)=
text.Split [|'\t';' '|]
let splitIntoKeyValue (A: int[]) =
(A.[0], A.[1])
let parseLine (line:string)=
line
|> splitAtTab
|> Array.filter (fun s -> not(s=""))
|> Array.map (fun s-> (int s))
|> splitIntoKeyValue
// let y =
// x |> Array.map parseLine
//let y =
// [|(1, 4); (2, 8); (3, 6); (4, 7); (5, 2); (6, 9); (7, 1); (8, 5); (8, 6);
// (9, 7); (9, 3)|]
// let y = Array.append [|(1,1);(1,2);(2,3);(3,1)|] [|for i in 4 .. 10000 do yield (i,4)|]
let y = Array.append [|(1,1);(1,2);(2,3);(3,1)|] [|for i in 4 .. 99999 do yield (i,i+1)|]
//val it : (int * int) []
type Children = int list
type Node1 =
{children : Children ;
mutable finishingtime : int ;
mutable explored1 : bool ;
}
type Node2 =
{children : Children ;
mutable leader : int ;
mutable explored2 : bool ;
}
type DFSgraphcore = Dictionary<int,Children>
let directgraphcore = new DFSgraphcore()
let reversegraphcore = new DFSgraphcore()
type DFSgraph1 = Dictionary<int,Node1>
let reversegraph1 = new DFSgraph1()
let AddtoGraph (G:DFSgraphcore) (n,c) =
if not(G.ContainsKey n) then
let node = [c]
G.Add(n,node)
else
let c'= G.[n]
G.Remove(n) |> ignore
G.Add (n, List.append c' [c])
let inline swaptuple (a,b) = (b,a)
y|> Array.iter (AddtoGraph directgraphcore)
y|> Array.map swaptuple |> Array.iter (AddtoGraph reversegraphcore)
// définir reversegraph1 = ... with....
for i in reversegraphcore.Keys do
let node = {children = reversegraphcore.[i] ;
finishingtime = -1 ;
explored1 = false ;
}
reversegraph1.Add (i,node)
for i in directgraphcore.Keys do
if not(reversegraphcore.ContainsKey(i)) then do
let node = {children = [] ;
finishingtime = -1 ;
explored1 = false ;
}
reversegraph1.Add (i,node)
directgraphcore.Clear |> ignore
reversegraphcore.Clear |> ignore
// for i in reversegraph1.Keys do printfn "%d %A" i reversegraph1.[i].children
printfn "pause"
Console.ReadKey() |> ignore
let num_nodes =
directgraphcore |> Seq.length
So basically the graph is (1->2->3->1)::(4->5->6->7->8->....->99999->10000)
and the reverse graph is (1->3->2->1)::(10000->9999->....->4)
here is the main code written in direct style
//////////////////// main code is below ///////////////////
let DFSLoop1 (G:DFSgraph1) =
let mutable t = 0
let mutable s = -1
let rec iter (n:int) (f:'a->unit) (list:'a list) : unit =
match list with
| [] -> (t <- t+1) ; (G.[n].finishingtime <- t)
| x::xs -> f x ; iter n f xs
let rec DFSsub (G:DFSgraph1) (n:int) : unit =
let my_f (j:int) : unit = if not(G.[j].explored1) then (DFSsub G j)
G.[n].explored1 <- true
iter n my_f G.[n].children
for i in num_nodes .. -1 .. 1 do
// printfn "%d" i
if not(G.[i].explored1) then do
s <- i
DFSsub G i
printfn "%d %d" i G.[i].finishingtime
// End of DFSLoop1
DFSLoop1 reversegraph1
printfn "pause"
Console.ReadKey() |> ignore
its not tail recursive, so we use continuations, here is the same code adapted to CPS style:
//////////////////// main code is below ///////////////////
let DFSLoop1 (G:DFSgraph1) =
let mutable t = 0
let mutable s = -1
let rec iter_c (n:int) (f_c:'a->(unit->'r)->'r) (list:'a list) (cont: unit->'r) : 'r =
match list with
| [] -> (t <- t+1) ; (G.[n].finishingtime <- t) ; cont()
| x::xs -> f_c x (fun ()-> iter_c n f_c xs cont)
let rec DFSsub (G:DFSgraph1) (n:int) (cont: unit->'r) : 'r=
let my_f_c (j:int)(cont:unit->'r):'r = if not(G.[j].explored1) then (DFSsub G j cont) else cont()
G.[n].explored1 <- true
iter_c n my_f_c G.[n].children cont
for i in maxnum_nodes .. -1 .. 1 do
// printfn "%d" i
if not(G.[i].explored1) then do
s <- i
DFSsub G i id
printfn "%d %d" i G.[i].finishingtime
DFSLoop1 reversegraph1
printfn "faré"
printfn "pause"
Console.ReadKey() |> ignore
both codes compile and give the same results for the small example (the one in comment) or the same tree that we are using , with a smaller size (1000 instead of 100000)
so i don't think its a bug in the algo here, we've got the same tree structure, just a bigger tree is causing problems. it looks to us the continuations are well written. we've typed the code explicitly. and all calls end with a continuation in all cases...
We are looking for expert advice !!! thanks !!!
I did not try to understand the whole code snippet, because it is fairly long, but you'll certainly need to replace the for loop with an iteration implemented using continuation passing style. Something like:
let rec iterc f cont list =
match list with
| [] -> cont ()
| x::xs -> f x (fun () -> iterc f cont xs)
I didn't understand the purpose of cont in your DFSub function (it is never called, is it?), but the continuation based version would look roughly like this:
let rec DFSsub (G:DFSgraph2)(n:int) cont =
G.[n].explored2 <- true
G.[n].leader <- s
G.[n].children
|> iterc
(fun j cont -> if not(G.[j].explored2) then DFSsub G j cont else cont ())
(fun () -> t <- t + 1)
Overflowing the stack when you recurse through hundreds of thousands of entries isn't bad at all, really. A lot of programming language implementations will choke on much shorter recursions than that. You're having serious programmer problems — nothing to be ashamed of!
Now if you want to do deeper recursions than your implementation will handle, you need to transform your algorithm so it is iterative and/or tail-recursive (the two are isomorphic — except that tail-recursion allows for decentralization and modularity, whereas iteration is centralized and non-modular).
To transform an algorithm from recursive to tail-recursive, which is an important skill to possess, you need to understand the state that is implicitly stored in a stack frame, i.e. those free variables in the function body that change across the recursion, and explicitly store them in a FIFO queue (a data structure that replicates your stack, and can be implemented trivially as a linked list). Then you can pass that linked list of reified frame variables as an argument to your tail recursive functions.
In more advanced cases where you have many tail recursive functions each with a different kind of frame, instead of simple self-recursion, you may need to define some mutually recursive data types for the reified stack frames, instead of using a list. But I believe Kosaraju's algorithm only involves self-recursive functions.
OK, so the code given above was the RIGHT code !
the problem lies with the compiler of F#
here is some words about it from Microsoft
http://blogs.msdn.com/b/fsharpteam/archive/2011/07/08/tail-calls-in-fsharp.aspx
Basically, be careful with the settings, in default mode, the compiler may NOT make automatically the tail calls. To do so, in VS2015, go to the Solution Explorer, right click with the mouse and click on "Properties" (the last element of the scrolling list)
Then in the new window, click on "Build" and tick the box "Generate tail calls"
It is also to check if the compiler did its job looking at the disassembly using
ILDASM.exe
you can find the source code for the whole algo in my github repository
https://github.com/FaguiCurtain/Learning-Fsharp/blob/master/Algo%20Stanford/Algo%20Stanford/Kosaraju_cont.fs
on a performance point of view, i'm not very satisfied. The code runs on 36 seconds on my laptop. From the forum with other fellow MOOCers, C/C++/C# typically executes in subsecond to 5s, Java around 10-15, Python around 20-30s.
So my implementation is clearly not optimized. I am now happy to hear about tricks to make it faster !!! thanks !!!!
Closed. This question does not meet Stack Overflow guidelines. It is not currently accepting answers.
This question does not appear to be about programming within the scope defined in the help center.
Closed 7 years ago.
Improve this question
I'm an experienced C# developer trying to teach myself F#. I spent a day or 3 reading throught the F# wikibook trying to get to know the syntax and F# fundamentals.
As an exercise I'm trying to go through the Project Euler problems to get a better feeling of the syntax and working with the language.
I've just solved problem 5. But I'm not so happy about the hoops I had to jump through to get a data structure that represents my solution.
I've used this blogpost to get to the algorithm for solving the problem.
I was wondering if anyone could give me some pointers as to how this code could be improved? My guess is that the inherent immutability of F# values is causing me to have to perform a lot of steps to get the exact data I want...
This is my code:
let main argv =
//calculates the prime factors of a number
let findPrimeFactors x =
let primes = [|2I;3I;5I;7I;11I;13I;17I;19I|]
let rec loop acc counter = function
| x when x = 1I -> failwith "A PRIME IS BY DEFINITION GREATER THAN 1"
| x when primes |> Array.contains x -> x :: acc
| x when counter = primes.Length -> failwith "MY LIST OF KNOWN PRIMES IS NOT BIG ENOUGH"
| x when x%primes.[counter]=0I-> loop (primes.[counter]::acc) (counter) (x/primes.[counter])
| x -> loop acc (counter + 1) x
let primeFactor = loop [] 0 x |> List.rev
primeFactor
//calculates the prime factors for each of the numbers between 2 and n
//then, for each of the prime factorizations it tries to find the highest power for each occurring prime factor
let findPrimeFactorsPowers n =
//builds a map of all the prime factor powers for all prime factorizations
let rec addCounterFactorPowers factorPowers = function
| counter when counter = n -> factorPowers
| (counter : int) -> addCounterFactorPowers ((findPrimeFactors (counter|>bigint) |> List.countBy (fun x-> x)) # factorPowers) (counter + 1)
let allFactorPowers = addCounterFactorPowers [] 2
//group all the powers per prime factor
let groupedFactorPowers = allFactorPowers |> List.groupBy (fun (factor, power) -> factor)
//get the highest power per prime factor
let maxFactorPowers = groupedFactorPowers |> List.map (fun (key, powers) -> (key, powers |> List.map (fun (factor, power) -> power) |> List.max))
//return the result
maxFactorPowers
let n = 20;
let primeFactorSet = findPrimeFactorsPowers n
printfn "%A" primeFactorSet
let smallestNumberDivisableByAllNumbersBelown = (primeFactorSet |> List.fold (fun state (factor, power) -> state * pown factor power) 1I)
printfn "Result %A" smallestNumberDivisableByAllNumbersBelown
System.Console.ReadKey(true)|>ignore
0 // return an integer exit code
There are many direct simplifications you can apply to your code, but I don't think that's the best approach.
This is the way I would solve that problem in F#:
let rec tryDivide n m =
if m = 1 then true
else
if n % m = 0 then tryDivide n (m-1)
else false
let rec findIt i m =
if tryDivide i m then i
else findIt (i+1) m
findIt 1 20
It's a bit slower than yours because it doesn't use hardcoded primes, they're calculated on the fly, but still it takes less than 2 secs on my computer to get the right answer.
Note that I'm not using any list-like data structure and no need to rely on big integers in this specific problem.
UPDATE
Here's a better approach based on Kvb's proposed solution:
let rec gcd x y = if y = 0 then abs x else gcd y (x % y)
let lcm a b =
match (a, b) with
| (_, 0) | (0, _) -> 0
| (x, y) -> abs ((x / (gcd x y)) * y)
Seq.fold lcm 1 {2..20}
I write some code to learning F#.
Here is a example:
let nextPrime list=
let rec loop n=
match n with
| _ when (list |> List.filter (fun x -> x <= ( n |> double |> sqrt |> int)) |> List.forall (fun x -> n % x <> 0)) -> n
| _ -> loop (n+1)
loop (List.max list + 1)
let rec findPrimes num=
match num with
| 1 -> [2]
| n ->
let temp = findPrimes <| n-1
(nextPrime temp ) :: temp
//find 10 primes
findPrimes 10 |> printfn "%A"
I'm very happy that it just works!
I'm totally beginner to recursion
Recursion is a wonderful thing.
I think findPrimes is not efficient.
Someone help me to refactor findPrimes to tail recursion if possible?
BTW, is there some more efficient way to find first n primes?
Regarding the first part of your question, if you want to write a recursive list building function tail-recursively you should pass the list of intermediate results as an extra parameter to the function. In your case this would be something like
let findPrimesTailRecursive num =
let rec aux acc num =
match num with
| 1 -> acc
| n -> aux ((nextPrime acc)::acc) (n-1)
aux [2] num
The recursive function aux gathers its results in an extra parameter conveniently called acc (as in acc-umulator). When you reach your ending condition, just spit out the accumulated result. I've wrapped the tail-recursive helper function in another function, so the function signature remains the same.
As you can see, the call to aux is the only, and therefore last, call to happen in the n <> 1 case. It's now tail-recursive and will compile into a while loop.
I've timed your version and mine, generating 2000 primes. My version is 16% faster, but still rather slow. For generating primes, I like to use an imperative array sieve. Not very functional, but very (very) fast.
An alternative is to use an extra continuation argument to make findPrimes tail recursive. This technique always works. It will avoid stack overflows, but probably won't make your code faster.
Also, I put your nextPrime function a little closer to the style I'd use.
let nextPrime list=
let rec loop n = if list |> List.filter (fun x -> x*x <= n)
|> List.forall (fun x -> n % x <> 0)
then n
else loop (n+1)
loop (1 + List.head list)
let rec findPrimesC num cont =
match num with
| 1 -> cont [2]
| n -> findPrimesC (n-1) (fun temp -> nextPrime temp :: temp |> cont)
let findPrimes num = findPrimesC num (fun res -> res)
findPrimes 10
As others have said, there's faster ways to generate primes.
Why not simply write:
let isPrime n =
if n<=1 then false
else
let m = int(sqrt (float(n)))
{2..m} |> Seq.forall (fun i->n%i<>0)
let findPrimes n =
{2..n} |> Seq.filter isPrime |> Seq.toList
or sieve (very fast):
let generatePrimes max=
let p = Array.create (max+1) true
let rec filter i step =
if i <= max then
p.[i] <- false
filter (i+step) step
{2..int (sqrt (float max))} |> Seq.iter (fun i->filter (i+i) i)
{2..max} |> Seq.filter (fun i->p.[i]) |> Seq.toArray
BTW, is there some more efficient way to find first n primes?
I described a fast arbitrary-size Sieve of Eratosthenes in F# here that accumulated its results into an ever-growing ResizeArray:
> let primes =
let a = ResizeArray[2]
let grow() =
let p0 = a.[a.Count-1]+1
let b = Array.create p0 true
for di in a do
let rec loop i =
if i<b.Length then
b.[i] <- false
loop(i+di)
let i0 = p0/di*di
loop(if i0<p0 then i0+di-p0 else i0-p0)
for i=0 to b.Length-1 do
if b.[i] then a.Add(p0+i)
fun n ->
while n >= a.Count do
grow()
a.[n];;
val primes : (int -> int)
I know that this is a bit late, and an answer was already accepted. However, I believe that a good step by step guide to making something tail recursive may be of interest to the OP or anyone else for that matter. Here are some tips that have certainly helped me out. I'm going to use a strait-forward example other than prime generation because, as others have stated, there are better ways to generate primes.
Consider a naive implementation of a count function that will create a list of integers counting down from some n. This version is not tail recursive so for long lists you will encounter a stack overflow exception:
let rec countDown = function
| 0 -> []
| n -> n :: countDown (n - 1)
(* ^
|... the cons operator is in the tail position
as such it is evaluated last. this drags
stack frames through subsequent recursive
calls *)
One way to fix this is to apply continuation passing style with a parameterized function:
let countDown' n =
let rec countDown n k =
match n with
| 0 -> k [] (* v--- this is continuation passing style *)
| n -> countDown (n - 1) (fun ns -> n :: k ns)
(* ^
|... the recursive call is now in tail position *)
countDown n (fun ns -> ns)
(* ^
|... and we initialize k with the identity function *)
Then, refactor this parameterized function into a specialized representation. Notice that the function countDown' is not actually counting down. This is an artifact of the way the continuation is built up when n > 0 and then evaluated when n = 0. If you have something like the first example and you can't figure out how to make it tail recursive, what I'm suggesting is that you write the second one and then try to optimize it to eliminate the function parameter k. That will certainly improve the readability. This is an optimization of the second example:
let countDown'' n =
let rec countDown n ns =
match n with
| 0 -> List.rev ns (* reverse so we are actually counting down again *)
| n -> countDown (n - 1) (n :: ns)
countDown n []
Background:
I have a sequence of contiguous, time-stamped data. The data-sequence has gaps in it where the data is not contiguous. I want create a method to split the sequence up into a sequence of sequences so that each subsequence contains contiguous data (split the input-sequence at the gaps).
Constraints:
The return value must be a sequence of sequences to ensure that elements are only produced as needed (cannot use list/array/cacheing)
The solution must NOT be O(n^2), probably ruling out a Seq.take - Seq.skip pattern (cf. Brian's post)
Bonus points for a functionally idiomatic approach (since I want to become more proficient at functional programming), but it's not a requirement.
Method signature
let groupContiguousDataPoints (timeBetweenContiguousDataPoints : TimeSpan) (dataPointsWithHoles : seq<DateTime * float>) : (seq<seq< DateTime * float >>)= ...
On the face of it the problem looked trivial to me, but even employing Seq.pairwise, IEnumerator<_>, sequence comprehensions and yield statements, the solution eludes me. I am sure that this is because I still lack experience with combining F#-idioms, or possibly because there are some language-constructs that I have not yet been exposed to.
// Test data
let numbers = {1.0..1000.0}
let baseTime = DateTime.Now
let contiguousTimeStamps = seq { for n in numbers ->baseTime.AddMinutes(n)}
let dataWithOccationalHoles = Seq.zip contiguousTimeStamps numbers |> Seq.filter (fun (dateTime, num) -> num % 77.0 <> 0.0) // Has a gap in the data every 77 items
let timeBetweenContiguousValues = (new TimeSpan(0,1,0))
dataWithOccationalHoles |> groupContiguousDataPoints timeBetweenContiguousValues |> Seq.iteri (fun i sequence -> printfn "Group %d has %d data-points: Head: %f" i (Seq.length sequence) (snd(Seq.hd sequence)))
I think this does what you want
dataWithOccationalHoles
|> Seq.pairwise
|> Seq.map(fun ((time1,elem1),(time2,elem2)) -> if time2-time1 = timeBetweenContiguousValues then 0, ((time1,elem1),(time2,elem2)) else 1, ((time1,elem1),(time2,elem2)) )
|> Seq.scan(fun (indexres,(t1,e1),(t2,e2)) (index,((time1,elem1),(time2,elem2))) -> (index+indexres,(time1,elem1),(time2,elem2)) ) (0,(baseTime,-1.0),(baseTime,-1.0))
|> Seq.map( fun (index,(time1,elem1),(time2,elem2)) -> index,(time2,elem2) )
|> Seq.filter( fun (_,(_,elem)) -> elem <> -1.0)
|> PSeq.groupBy(fst)
|> Seq.map(snd>>Seq.map(snd))
Thanks for asking this cool question
I translated Alexey's Haskell to F#, but it's not pretty in F#, and still one element too eager.
I expect there is a better way, but I'll have to try again later.
let N = 20
let data = // produce some arbitrary data with holes
seq {
for x in 1..N do
if x % 4 <> 0 && x % 7 <> 0 then
printfn "producing %d" x
yield x
}
let rec GroupBy comp (input:LazyList<'a>) : LazyList<LazyList<'a>> =
LazyList.delayed (fun () ->
match input with
| LazyList.Nil -> LazyList.cons (LazyList.empty()) (LazyList.empty())
| LazyList.Cons(x,LazyList.Nil) ->
LazyList.cons (LazyList.cons x (LazyList.empty())) (LazyList.empty())
| LazyList.Cons(x,(LazyList.Cons(y,_) as xs)) ->
let groups = GroupBy comp xs
if comp x y then
LazyList.consf
(LazyList.consf x (fun () ->
let (LazyList.Cons(firstGroup,_)) = groups
firstGroup))
(fun () ->
let (LazyList.Cons(_,otherGroups)) = groups
otherGroups)
else
LazyList.cons (LazyList.cons x (LazyList.empty())) groups)
let result = data |> LazyList.of_seq |> GroupBy (fun x y -> y = x + 1)
printfn "Consuming..."
for group in result do
printfn "about to do a group"
for x in group do
printfn " %d" x
You seem to want a function that has signature
(`a -> bool) -> seq<'a> -> seq<seq<'a>>
I.e. a function and a sequence, then break up the input sequence into a sequence of sequences based on the result of the function.
Caching the values into a collection that implements IEnumerable would likely be simplest (albeit not exactly purist, but avoiding iterating the input multiple times. It will lose much of the laziness of the input):
let groupBy (fun: 'a -> bool) (input: seq) =
seq {
let cache = ref (new System.Collections.Generic.List())
for e in input do
(!cache).Add(e)
if not (fun e) then
yield !cache
cache := new System.Collections.Generic.List()
if cache.Length > 0 then
yield !cache
}
An alternative implementation could pass cache collection (as seq<'a>) to the function so it can see multiple elements to chose the break points.
A Haskell solution, because I don't know F# syntax well, but it should be easy enough to translate:
type TimeStamp = Integer -- ticks
type TimeSpan = Integer -- difference between TimeStamps
groupContiguousDataPoints :: TimeSpan -> [(TimeStamp, a)] -> [[(TimeStamp, a)]]
There is a function groupBy :: (a -> a -> Bool) -> [a] -> [[a]] in the Prelude:
The group function takes a list and returns a list of lists such that the concatenation of the result is equal to the argument. Moreover, each sublist in the result contains only equal elements. For example,
group "Mississippi" = ["M","i","ss","i","ss","i","pp","i"]
It is a special case of groupBy, which allows the programmer to supply their own equality test.
It isn't quite what we want, because it compares each element in the list with the first element of the current group, and we need to compare consecutive elements. If we had such a function groupBy1, we could write groupContiguousDataPoints easily:
groupContiguousDataPoints maxTimeDiff list = groupBy1 (\(t1, _) (t2, _) -> t2 - t1 <= maxTimeDiff) list
So let's write it!
groupBy1 :: (a -> a -> Bool) -> [a] -> [[a]]
groupBy1 _ [] = [[]]
groupBy1 _ [x] = [[x]]
groupBy1 comp (x : xs#(y : _))
| comp x y = (x : firstGroup) : otherGroups
| otherwise = [x] : groups
where groups#(firstGroup : otherGroups) = groupBy1 comp xs
UPDATE: it looks like F# doesn't let you pattern match on seq, so it isn't too easy to translate after all. However, this thread on HubFS shows a way to pattern match sequences by converting them to LazyList when needed.
UPDATE2: Haskell lists are lazy and generated as needed, so they correspond to F#'s LazyList (not to seq, because the generated data is cached (and garbage collected, of course, if you no longer hold a reference to it)).
(EDIT: This suffers from a similar problem to Brian's solution, in that iterating the outer sequence without iterating over each inner sequence will mess things up badly!)
Here's a solution that nests sequence expressions. The imperitave nature of .NET's IEnumerable<T> is pretty apparent here, which makes it a bit harder to write idiomatic F# code for this problem, but hopefully it's still clear what's going on.
let groupBy cmp (sq:seq<_>) =
let en = sq.GetEnumerator()
let rec partitions (first:option<_>) =
seq {
match first with
| Some first' -> //'
(* The following value is always overwritten;
it represents the first element of the next subsequence to output, if any *)
let next = ref None
(* This function generates a subsequence to output,
setting next appropriately as it goes *)
let rec iter item =
seq {
yield item
if (en.MoveNext()) then
let curr = en.Current
if (cmp item curr) then
yield! iter curr
else // consumed one too many - pass it on as the start of the next sequence
next := Some curr
else
next := None
}
yield iter first' (* ' generate the first sequence *)
yield! partitions !next (* recursively generate all remaining sequences *)
| None -> () // return an empty sequence if there are no more values
}
let first = if en.MoveNext() then Some en.Current else None
partitions first
let groupContiguousDataPoints (time:TimeSpan) : (seq<DateTime*_> -> _) =
groupBy (fun (t,_) (t',_) -> t' - t <= time)
Okay, trying again. Achieving the optimal amount of laziness turns out to be a bit difficult in F#... On the bright side, this is somewhat more functional than my last attempt, in that it doesn't use any ref cells.
let groupBy cmp (sq:seq<_>) =
let en = sq.GetEnumerator()
let next() = if en.MoveNext() then Some en.Current else None
(* this function returns a pair containing the first sequence and a lazy option indicating the first element in the next sequence (if any) *)
let rec seqStartingWith start =
match next() with
| Some y when cmp start y ->
let rest_next = lazy seqStartingWith y // delay evaluation until forced - stores the rest of this sequence and the start of the next one as a pair
seq { yield start; yield! fst (Lazy.force rest_next) },
lazy Lazy.force (snd (Lazy.force rest_next))
| next -> seq { yield start }, lazy next
let rec iter start =
seq {
match (Lazy.force start) with
| None -> ()
| Some start ->
let (first,next) = seqStartingWith start
yield first
yield! iter next
}
Seq.cache (iter (lazy next()))
Below is some code that does what I think you want. It is not idiomatic F#.
(It may be similar to Brian's answer, though I can't tell because I'm not familiar with the LazyList semantics.)
But it doesn't exactly match your test specification: Seq.length enumerates its entire input. Your "test code" calls Seq.length and then calls Seq.hd. That will generate an enumerator twice, and since there is no caching, things get messed up. I'm not sure if there is any clean way to allow multiple enumerators without caching. Frankly, seq<seq<'a>> may not be the best data structure for this problem.
Anyway, here's the code:
type State<'a> = Unstarted | InnerOkay of 'a | NeedNewInner of 'a | Finished
// f() = true means the neighbors should be kept together
// f() = false means they should be split
let split_up (f : 'a -> 'a -> bool) (input : seq<'a>) =
// simple unfold that assumes f captured a mutable variable
let iter f = Seq.unfold (fun _ ->
match f() with
| Some(x) -> Some(x,())
| None -> None) ()
seq {
let state = ref (Unstarted)
use ie = input.GetEnumerator()
let innerMoveNext() =
match !state with
| Unstarted ->
if ie.MoveNext()
then let cur = ie.Current
state := InnerOkay(cur); Some(cur)
else state := Finished; None
| InnerOkay(last) ->
if ie.MoveNext()
then let cur = ie.Current
if f last cur
then state := InnerOkay(cur); Some(cur)
else state := NeedNewInner(cur); None
else state := Finished; None
| NeedNewInner(last) -> state := InnerOkay(last); Some(last)
| Finished -> None
let outerMoveNext() =
match !state with
| Unstarted | NeedNewInner(_) -> Some(iter innerMoveNext)
| InnerOkay(_) -> failwith "Move to next inner seq when current is active: undefined behavior."
| Finished -> None
yield! iter outerMoveNext }
open System
let groupContigs (contigTime : TimeSpan) (holey : seq<DateTime * int>) =
split_up (fun (t1,_) (t2,_) -> (t2 - t1) <= contigTime) holey
// Test data
let numbers = {1 .. 15}
let contiguousTimeStamps =
let baseTime = DateTime.Now
seq { for n in numbers -> baseTime.AddMinutes(float n)}
let holeyData =
Seq.zip contiguousTimeStamps numbers
|> Seq.filter (fun (dateTime, num) -> num % 7 <> 0)
let grouped_data = groupContigs (new TimeSpan(0,1,0)) holeyData
printfn "Consuming..."
for group in grouped_data do
printfn "about to do a group"
for x in group do
printfn " %A" x
Ok, here's an answer I'm not unhappy with.
(EDIT: I am unhappy - it's wrong! No time to try to fix right now though.)
It uses a bit of imperative state, but it is not too difficult to follow (provided you recall that '!' is the F# dereference operator, and not 'not'). It is as lazy as possible, and takes a seq as input and returns a seq of seqs as output.
let N = 20
let data = // produce some arbitrary data with holes
seq {
for x in 1..N do
if x % 4 <> 0 && x % 7 <> 0 then
printfn "producing %d" x
yield x
}
let rec GroupBy comp (input:seq<_>) = seq {
let doneWithThisGroup = ref false
let areMore = ref true
use e = input.GetEnumerator()
let Next() = areMore := e.MoveNext(); !areMore
// deal with length 0 or 1, seed 'prev'
if not(e.MoveNext()) then () else
let prev = ref e.Current
while !areMore do
yield seq {
while not(!doneWithThisGroup) do
if Next() then
let next = e.Current
doneWithThisGroup := not(comp !prev next)
yield !prev
prev := next
else
// end of list, yield final value
yield !prev
doneWithThisGroup := true }
doneWithThisGroup := false }
let result = data |> GroupBy (fun x y -> y = x + 1)
printfn "Consuming..."
for group in result do
printfn "about to do a group"
for x in group do
printfn " %d" x