I'm writing some thing really easy, a program that finds all factors of an int. Here is what I have
let factor n=
let ls=[]
for i=1 to n do
if i % n =0 then ls = i::ls
l
If I do this then it pops an error This expression was expected to have the type unit. But I tried to put an expression that prints something after if..then.., which is suppose to return type unit, but it still gives the same error. I am lost about this. Can someone help please? Thanks
You are trying to make ls into a mutable variable nd assign it with =. While this is possible, by using mutable (1) or ref (2) along with <- or := assignment operators, this is generally discouraged in the functional world.
A possibly more idiomatic implementation of the naive algorithm could be:
let factor n =
let rec factorLoop curr divs =
if curr > n then divs
else
if n % curr = 0
then factorLoop (curr+1) (curr::divs)
else factorLoop (curr+1) divs
factorLoop 1 [] |> List.rev
> factor 12;;
val it : int list = [1; 2; 3; 4; 6; 12]
Here the main function defines an inner factorLoop function that is recursive. Recursion is the way we can avoid many uses of mutable variables in functional languages. The recursive inner function threads along a curr variable that is the current divisor to be tested and a list divs of currently found divisors. The result includes 1 and n. This can be altered respectively by changing the initial value of curr and the terminating condition in the first line of factorLoop.
It is worth noting that it can all be shrunk down to one line by making use of the F# library:
let factor n =
[1..n] |> List.filter (fun x -> n % x = 0)
Here we build a list of values 1..n and feed them to List.filter which applies the given predicate (at the end of the line) to select only divisors on n. If n is large, however, the temp list will grow very large. We can use a lazily evaluated sequence instead, which won't blow the memory usage:
let factor n =
{1..n} |> Seq.filter (fun x -> n % x = 0) |> Seq.toList
Here we filter on a 'lazy' sequence and only convert the (much smaller) sequence of results to a list at the end:
> factor 10000000;;
val it : int list =
[1; 2; 4; 5; 8; 10; 16; 20; 25; 32; ... etc
= is comparison, not assignment. You want either
let factor n =
let mutable ls = []
for i = 1 to n do
if n % i = 0 then ls <- i::ls
ls
or
let factor n =
let ls = ref []
for i = 1 to n do
if n % i = 0 then ls := i::(!ls)
!ls
Note, however, that both of these solutions are highly unidiomatic, as there are equally easy immutable solutions to this problem.
Related
Creating a Parallel.ForEach expression of this form:
let low = max 1 (k-m)
let high = min (k-1) n
let rangesize = (high+1-low)/(PROCS*3)
Parallel.ForEach(Partitioner.Create(low, high+1, rangesize), (fun j ->
let i = k - j
if x.[i-1] = y.[j-1] then
a.[i] <- b.[i-1] + 1
else
a.[i] <- max c.[i] c.[i-1]
)) |> ignore
Causes me to receive the error: No overloads match for method 'ForEach'. However I am using the Parallel.ForEach<TSource> Method (Partitioner<TSource>, Action<TSource>) and it seems right to me. Am I missing something?
Edited: I am trying to obtain the same results as the code below (that does not use a Partitioner):
let low = max 1 (k-m)
let high = min (k-1) n
let rangesize = (high+1-low)/(PROCS*3)
let A = [| low .. high |]
Parallel.ForEach(A, fun (j:int) ->
let i = k - j
if x.[i-1] = y.[j-1] then
a.[i] <- b.[i-1] + 1
else
a.[i] <- max c.[i] c.[i-1]
) |> ignore
Are you sure that you have opened all necessary namespaces, all the values you are using (low, high and PROCS) are defined and that your code does not accidentally redefine some of the names that you're using (like Partitioner)?
I created a very simple F# script with this code and it seems to be working fine (I refactored the code to create a partitioner called p, but that does not affect the behavior):
open System.Threading.Tasks
open System.Collections.Concurrent
let PROCS = 10
let low, high = 0, 100
let p = Partitioner.Create(low, high+1, high+1-low/(PROCS*3))
Parallel.ForEach(p, (fun j ->
printfn "%A" j // Print the desired range (using %A as it is a tuple)
)) |> ignore
It is important that the value j is actually a pair of type int * int, so if the body uses it in a wrong way (e.g. as an int), you will get the error. In that case, you can add a type annotation to j and you would get a more useful error elsewhere:
Parallel.ForEach(p, (fun (j:int * int) ->
printfn "%d" j // Error here, because `j` is used as an int, but it is a pair!
)) |> ignore
This means that if you want to perform something for all j values in the original range, you need to write something like this:
Parallel.ForEach(p, (fun (loJ, hiJ) ->
for j in loJ .. hiJ - 1 do // Iterate over all js in this partition
printfn "%d" j // process the current j
)) |> ignore
Aside, I guess that the last argument to Partitioner.Create should actually be (high+1-low)/(PROCS*3) - you probably want to divide the total number of steps, not just the low value.
I have this code written for a Project Euler problem in c++:
int sum = 0;
for(int i =0; i < 1000; i++)
{
//Check if multiple of 3 but not multiple of 5 to prevent duplicate
sum += i % 3 == 0 && i % 5 != 0 ? i: 0;
//check for all multiple of 5, including those of 3
sum += i % 5 == 0 ? i: 0;
}
cout << sum;
I'm trying to learn f# and rewriting this in f#. This is what I have so far:
open System
//function to calculate the multiples
let multiple3v5 num =
num
//function to calculate sum of list items
let rec SumList xs =
match xs with
| [] -> 0
| y::ys -> y + SumList ys
let sum = Array.map multiple3v5 [|1 .. 1000|]
What I have may be complete nonsense...so help please?
Your sumList function is a good start. It already iterates (recursively) over the entire list, so you don't need to wrap it in an additional Array.map. You just need to extend your sumList so that it adds the number only when it matches the specified condition.
Here is a solution to a simplified problem - add all numbers that are divisible by 3:
open System
let rec sumList xs =
match xs with
| [] -> 0 // If the list is empty, the sum is zero
| y::ys when y % 3 = 0 ->
// If the list starts with y that is divisible by 3, then we add 'y' to the
// sum that we get by recursively processing the rest of the list
y + sumList ys
| y::ys ->
// This will only execute when y is not divisible by 3, so we just
// recursively process the rest of the list and return
/// that (without adding current value)
sumList ys
// For testing, let's sum all numbers divisble by 3 between 1 and 10.
let sum = sumList [ 1 .. 10 ]
This is the basic way of writing the function using explicit recursion. In practice, the solution by jpalmer is how I'd solve it too, but it is useful to write a few recursive functions yourself if you're learning F#.
The accumulator parameter mentioned by sashang is a more advanced way to write this. You'll need to do that if you want to run the function on large inputs (which is likely the case in Euler problem). When using accumulator parameter, the function can be written using tail recursion, so it avoids stack overflow even when processing long lists.
The idea of a accumulator-based version is that the function takes additional parameter, which represents the sum calculated so far.
let rec sumList xs sumSoFar = ...
When you call it initially, you write sumList [ ... ] 0. The recursive calls will not call y + sumList xs, but will instead add y to the accumulator and then make the recursive call sumList xs (y + sumSoFar). This way, the F# compiler can do tail-call optimization and it will translate code to a loop (similar to the C++ version).
I'm not sure if translating from an imperative language solution is a good approach to developing a functional mindset as instrument (C++ in your case) had already defined an (imperative) approach to solution, so it's better sticking to original problem outlay.
Overall tasks from Project Euler are excellent for mastering many F# facilities. For example, you may use list comprehensions like in the snippet below
// multipleOf3Or5 function definition is left for your exercise
let sumOfMultiples n =
[ for x in 1 .. n do if multipleOf3Or5 x then yield x] |> List.sum
sumOfMultiples 999
or you can a bit generalize the solution suggested by #jpalmer by exploiting laziness:
Seq.initInfinite id
|> Seq.filter multipleOf3Or5
|> Seq.takeWhile ((>) 1000)
|> Seq.sum
or you may even use this opportunity to master active patterns:
let (|DivisibleBy|_) divisior num = if num % divisor = 0 the Some(num) else None
{1..999}
|> Seq.map (fun i ->
match i with | DivisibleBy 3 i -> i | DivisibleBy 5 i -> i | _ -> 0)
|> Seq.sum
All three variations above implement a common pattern of making a sequence of members with sought property and then folding it by calculating sum.
F# has many more functions than just map - this problem suggests using filter and sum, my approach would be something like
let valid n = Left as an exercise
let r =
[1..1000]
|> List.filter valid
|> List.sum
printfn "%i" r
I didn't want to do the whole problem, but filling in the missing function shouldn't be too hard
This is how you turn a loop with a counter into a recursive function. You do this by passing an accumulator parameter to the loop function that holds the current loop count.
For example:
let rec loop acc =
if acc = 10 then
printfn "endloop"
else
printfn "%d" acc
loop (acc + 1)
loop 0
This will stop when acc is 10.
In F#, imagine we have an array of bytes representing pixel data with three bytes per pixel in RGB order:
[| 255; 0; 0; //Solid red
0; 255; 0; //Solid green
0; 0; 255; //Solid blue
1; 72; 9;
34; 15; 155
... |]
I'm having a hard time knowing how to functionally operate on this data as-is, since a single item is really a consecutive block of three elements in the array.
So, I need to first group the triples in the array into something like this:
[|
[| 255; 0; 0 |];
[| 0; 255; 0 |];
[| 0; 0; 255 |];
[| 1; 72; 9 |];
[| 34; 15; 155 |]
... |]
Now, gathering up the triples into sub-arrays is easy enough to do with a for loop, but I'm curious--is there a functional way to gather up groups of array elements in F#? My ultimate goal is not simply to convert the data as illustrated above, but to solve the problem in a more declarative and functional manner. But I have yet to find an example of how to do this without an imperative loop.
kvb's answer may not give you what you want. Seq.windowed returns a sliding window of values, e.g., [1; 2; 3; 4] becomes [[1; 2; 3]; [2; 3; 4]]. It seems like you want it split into contiguous chunks. The following function takes a list and returns a list of triples ('T list -> ('T * 'T * 'T) list).
let toTriples list =
let rec aux f = function
| a :: b :: c :: rest -> aux (fun acc -> f ((a, b, c) :: acc)) rest
| _ -> f []
aux id list
Here's the inverse:
let ofTriples triples =
let rec aux f = function
| (a, b, c) :: rest -> aux (fun acc -> f (a :: b :: c :: acc)) rest
| [] -> f []
aux id triples
EDIT
If you're dealing with huge amounts of data, here's a sequence-based approach with constant memory use (all the options and tuples it creates have a negative impact on GC--see below for a better version):
let (|Next|_|) (e:IEnumerator<_>) =
if e.MoveNext() then Some e.Current
else None
let (|Triple|_|) = function
| Next a & Next b & Next c -> Some (a, b, c) //change to [|a;b;c|] if you like
| _ -> None
let toSeqTriples (items:seq<_>) =
use e = items.GetEnumerator()
let rec loop() =
seq {
match e with
| Triple (a, b, c) ->
yield a, b, c
yield! loop()
| _ -> ()
}
loop()
EDIT 2
ebb's question about memory use prompted me to test and I found toSeqTriples to be slow and cause surprisingly frequent GCs. The following version fixes those issues and is almost 4x faster than the list-based version.
let toSeqTriplesFast (items:seq<_>) =
use e = items.GetEnumerator()
let rec loop() =
seq {
if e.MoveNext() then
let a = e.Current
if e.MoveNext() then
let b = e.Current
if e.MoveNext() then
let c = e.Current
yield (a, b, c)
yield! loop()
}
loop()
This has relatively constant memory usage vs a list or array-based approach because a) if you have a seq to start with the entire sequence doesn't have to be slurped into a list/array; and b) it also returns a sequence, making it lazy, and avoiding allocating yet another list/array.
I need to first group the triples in the array into something like this:
If you know they will always be triples then representing then as a tuple int * int * int is more "typeful" than using an array because it conveys the fact that there are only ever exactly three elements.
Other people have described various ways to massage the data but I would actually recommend not bothering (unless there is more to this than you have described). I would opt for a function to destructure your array as-is instead:
let get i = a.[3*i], a.[3*i+1], a.[3*i+2]
If you really want to change the representation then you can now do:
let b = Array.init (a.Length/3) get
The answer really depends upon what you want to do next though...
(Hat tip: Scott Wlaschin) As of F# 4.0, you can use Array.chunkBySize(). It does exactly what you want:
let bs = [| 255; 0; 0; //Solid red
0; 255; 0; //Solid green
0; 0; 255; //Solid blue
1; 72; 9;
34; 15; 155 |]
let grouped = bs |> Array.chunkBySize 3
// [| [|255; 0; 0|]
// [| 0; 255; 0|]
// [| 0; 0; 255|]
// [| 1; 72; 9|]
// [| 34; 15; 155|] |]
The List and Seq modules also have chunkBySize() in F# 4.0. As of this writing, the docs at MSDN don't show chunkBySize() anywhere, but it's there if you're using F# 4.0.
UPDATE: As pointed out by Daniel, this answer is wrong because it creates a sliding window.
You can use the Seq.windowed function from the library. E.g.
let rgbPix = rawValues |> Seq.windowed 3
This returns a sequence rather than an array, so if you need random access, you could follow that with a call to Seq.toArray.
Another approach, that takes and yields arrays directly:
let splitArrays n arr =
match Array.length arr with
| 0 ->
invalidArg "arr" "array is empty"
| x when x % n <> 0 ->
invalidArg "arr" "array length is not evenly divisible by n"
| arrLen ->
let ret = arrLen / n |> Array.zeroCreate
let rec loop idx =
ret.[idx] <- Array.sub arr (idx * n) n
match idx + 1 with
| idx' when idx' <> ret.Length -> loop idx'
| _ -> ret
loop 0
Or, yet another:
let splitArray n arr =
match Array.length arr with
| 0 ->
invalidArg "arr" "array is empty"
| x when x % n <> 0 ->
invalidArg "arr" "array length is not evenly divisible by n"
| arrLen ->
let rec loop idx = seq {
yield Array.sub arr idx n
let idx' = idx + n
if idx' <> arrLen then
yield! loop idx' }
loop 0 |> Seq.toArray
I've trying to learn F#. I'm a complete beginner, so this might be a walkover for you guys :)
I have the following function:
let removeEven l =
let n = List.length l;
let list_ = [];
let seq_ = seq { for x in 1..n do if x % 2 <> 0 then yield List.nth l (x-1)}
for x in seq_ do
let list_ = list_ # [x];
list_;
It takes a list, and return a new list containing all the numbers, which is placed at an odd index in the original list, so removeEven [x1;x2;x3] = [x1;x3]
However, I get my already favourite error-message: Incomplete construct at or before this point in expression...
If I add a print to the end of the line, instead of list_:
...
print_any list_;
the problem is fixed. But I do not want to print the list, I want to return it!
What causes this? Why can't I return my list?
To answer your question first, the compiler complains because there is a problem inside the for loop. In F#, let serves to declare values (that are immutable and cannot be changed later in the program). It isn't a statement as in C# - let can be only used as part of another expression. For example:
let n = 10
n + n
Actually means that you want the n symbol to refer to the value 10 in the expression n + n. The problem with your code is that you're using let without any expression (probably because you want to use mutable variables):
for x in seq_ do
let list_ = list_ # [x] // This isn't assignment!
list_
The problematic line is an incomplete expression - using let in this way isn't allowed, because it doesn't contain any expression (the list_ value will not be accessed from any code). You can use mutable variable to correct your code:
let mutable list_ = [] // declared as 'mutable'
let seq_ = seq { for x in 1..n do if x % 2 <> 0 then yield List.nth l (x-1)}
for x in seq_ do
list_ <- list_ # [x] // assignment using '<-'
Now, this should work, but it isn't really functional, because you're using imperative mutation. Moreover, appending elements using # is really inefficient thing to do in functional languages. So, if you want to make your code functional, you'll probably need to use different approach. Both of the other answers show a great approach, although I prefer the example by Joel, because indexing into a list (in the solution by Chaos) also isn't very functional (there is no pointer arithmetic, so it will be also slower).
Probably the most classical functional solution would be to use the List.fold function, which aggregates all elements of the list into a single result, walking from the left to the right:
[1;2;3;4;5]
|> List.fold (fun (flag, res) el ->
if flag then (not flag, el::res) else (not flag, res)) (true, [])
|> snd |> List.rev
Here, the state used during the aggregation is a Boolean flag specifying whether to include the next element (during each step, we flip the flag by returning not flag). The second element is the list aggregated so far (we add element by el::res only when the flag is set. After fold returns, we use snd to get the second element of the tuple (the aggregated list) and reverse it using List.rev, because it was collected in the reversed order (this is more efficient than appending to the end using res#[el]).
Edit: If I understand your requirements correctly, here's a version of your function done functional rather than imperative style, that removes elements with odd indexes.
let removeEven list =
list
|> Seq.mapi (fun i x -> (i, x))
|> Seq.filter (fun (i, x) -> i % 2 = 0)
|> Seq.map snd
|> List.ofSeq
> removeEven ['a'; 'b'; 'c'; 'd'];;
val it : char list = ['a'; 'c']
I think this is what you are looking for.
let removeEven list =
let maxIndex = (List.length list) - 1;
seq { for i in 0..2..maxIndex -> list.[i] }
|> Seq.toList
Tests
val removeEven : 'a list -> 'a list
> removeEven [1;2;3;4;5;6];;
val it : int list = [1; 3; 5]
> removeEven [1;2;3;4;5];;
val it : int list = [1; 3; 5]
> removeEven [1;2;3;4];;
val it : int list = [1; 3]
> removeEven [1;2;3];;
val it : int list = [1; 3]
> removeEven [1;2];;
val it : int list = [1]
> removeEven [1];;
val it : int list = [1]
You can try a pattern-matching approach. I haven't used F# in a while and I can't test things right now, but it would be something like this:
let rec curse sofar ls =
match ls with
| even :: odd :: tl -> curse (even :: sofar) tl
| even :: [] -> curse (even :: sofar) []
| [] -> List.rev sofar
curse [] [ 1; 2; 3; 4; 5 ]
This recursively picks off the even elements. I think. I would probably use Joel Mueller's approach though. I don't remember if there is an index-based filter function, but that would probably be the ideal to use, or to make if it doesn't exist in the libraries.
But in general lists aren't really meant as index-type things. That's what arrays are for. If you consider what kind of algorithm would require a list having its even elements removed, maybe it's possible that in the steps prior to this requirement, the elements can be paired up in tuples, like this:
[ (1,2); (3,4) ]
That would make it trivial to get the even-"indexed" elements out:
thelist |> List.map fst // take first element from each tuple
There's a variety of options if the input list isn't guaranteed to have an even number of elements.
Yet another alternative, which (by my reckoning) is slightly slower than Joel's, but it's shorter :)
let removeEven list =
list
|> Seq.mapi (fun i x -> (i, x))
|> Seq.choose (fun (i,x) -> if i % 2 = 0 then Some(x) else None)
|> List.ofSeq
I am trying to learn F# so I paid a visit to Project Euler and I am currently working on Problem 3.
The prime factors of 13195 are 5, 7,
13 and 29.
What is the largest prime
factor of the number 600851475143?
Some things to consider:
My first priority is to learn good functional habits.
My second priority is I would like it to be fast and efficient.
Within the following code I have marked the section this question is regarding.
let isPrime(n:int64) =
let rec check(i:int64) =
i > n / 2L or (n % i <> 0L && check(i + 1L))
check(2L)
let greatestPrimeFactor(n:int64) =
let nextPrime(prime:int64):int64 =
seq { for i = prime + 1L to System.Int64.MaxValue do if isPrime(i) then yield i }
|> Seq.skipWhile(fun v -> n % v <> 0L)
|> Seq.hd
let rec findNextPrimeFactor(number:int64, prime:int64):int64 =
if number = 1L then prime else
//************* No variable
(fun p -> findNextPrimeFactor(number / p, p))(nextPrime(prime))
//*************
//************* Variable
let p = nextPrime(prime)
findNextPrimeFactor(number / p, p)
//*************
findNextPrimeFactor(n, 2L)
Update
Based off some of the feedback I have refactored the code to be 10 times faster.
module Problem3
module private Internal =
let execute(number:int64):int64 =
let rec isPrime(value:int64, current:int64) =
current > value / 2L or (value % current <> 0L && isPrime(value, current + 1L))
let rec nextPrime(prime:int64):int64 =
if number % prime = 0L && isPrime(prime, 2L) then prime else nextPrime(prime + 1L)
let rec greatestPrimeFactor(current:int64, prime:int64):int64 =
if current = 1L then prime else nextPrime(prime + 1L) |> fun p -> greatestPrimeFactor(current / p, p)
greatestPrimeFactor(number, 2L)
let execute() = Internal.execute(600851475143L)
Update
I would like to thank everyone for there advice. This latest version is a compilation of all the advice I received.
module Problem3
module private Internal =
let largestPrimeFactor number =
let rec isPrime value current =
current > value / 2L || (value % current <> 0L && isPrime value (current + 1L))
let rec nextPrime value =
if number % value = 0L && isPrime value 2L then value else nextPrime (value + 1L)
let rec find current prime =
match current / prime with
| 1L -> prime
| current -> nextPrime (prime + 1L) |> find current
find number (nextPrime 2L)
let execute() = Internal.largestPrimeFactor 600851475143L
Functional programming becomes easier and more automatic with practice, so don't sweat it if you don't get it absolutely right on the first try.
With that in mind, let's take your sample code:
let rec findNextPrimeFactor(number:int64, prime:int64):int64 =
if number = 1L then prime else
//************* No variable
(fun p -> findNextPrimeFactor(number / p, p))(nextPrime(prime))
//*************
//************* Variable
let p = nextPrime(prime)
findNextPrimeFactor(number / p, p)
//*************
Your no variable version is just weird, don't use it. I like your version with the explicit let binding.
Another way to write it would be:
nextPrime(prime) |> fun p -> findNextPrimeFactor(number / p, p)
Its ok and occasionally useful to write it like this, but still comes across as a little weird. Most of the time, we use |> to curry values without needing to name our variables (in "pointfree" style). Try to anticipate how your function will be used, and if possible, re-write it so you can use it with the pipe operator without explicit declared variables. For example:
let rec findNextPrimeFactor number prime =
match number / prime with
| 1L -> prime
| number' -> nextPrime(prime) |> findNextPrimeFactor number'
No more named args :)
Ok, now that we have that out of the way, let's look at your isPrime function:
let isPrime(n:int64) =
let rec check(i:int64) =
i > n / 2L or (n % i <> 0L && check(i + 1L))
check(2L)
You've probably heard to use recursion instead of loops, and that much is right. But, wherever possible, you should abstract away recursion with folds, maps, or higher order functions. Two reasons for this:
its a little more readable, and
improperly written recursion will result in a stack overflow. For example, your function is not tail recursive, so it'll blow up on large values of n.
I'd rewrite isPrime like this:
let isPrime n = seq { 2L .. n / 2L } |> Seq.exists (fun i -> n % i = 0L) |> not
Most of the time, if you can abstract away your explicit looping, then you're just applying transformations to your input sequence until you get your results:
let maxFactor n =
seq { 2L .. n - 1L } // test inputs
|> Seq.filter isPrime // primes
|> Seq.filter (fun x -> n % x = 0L) // factors
|> Seq.max // result
We don't even have intermediate variables in this version. Coolness!
My second priority is I would like it
to be fast and efficient.
Most of the time, F# is going to be pretty comparable with C# in terms of speed, or it's going to be "fast enough". If you find your code takes a long time to execute, it probably means you're using the wrong data structure or a bad algorithm. For a concrete example, read the comments on this question.
So, the code I've written is "elegant" in the sense that its concise, gives the correct results, and doesn't rely on any trickery. Unfortunately, its not very fast. For start:
it uses trial division to create a sequence of primes, when the Sieve of Eratosthenes would be much faster. [Edit: I wrote a somewhat naive version of this sieve which didn't work for numbers larger than Int32.MaxValue, so I've removed the code.]
read Wikipedia's article on the prime counting function, it'll give you pointers on calculating the first n primes as well as estimating the upper and lower bounds for the nth prime.
[Edit: I included some code with a somewhat naive implementation of a sieve of eratosthenes. It only works for inputs less than int32.MaxValue, so it probably isn't suitable for project euler.]
Concerning "good functional habit" or rather good practice I see three minor things. Using the yield in your sequence is a little harder to read than just filter. Unnecessary type annotations in a type inferred language leads to difficult refactoring and makes the code harder to read. Don't go overboard and try to remove every type annotation though if you're finding it difficult. Lastly making a lambda function which only takes a value to use as a temp variable reduces readability.
As far as personal style goes I prefer more spaces and only using tupled arguments when the data makes sense being grouped together.
I'd write your original code like this.
let isPrime n =
let rec check i =
i > n / 2L || (n % i <> 0L && check (i + 1L))
check 2L
let greatestPrimeFactor n =
let nextPrime prime =
seq {prime + 1L .. System.Int64.MaxValue}
|> Seq.filter isPrime
|> Seq.skipWhile (fun v -> n % v <> 0L)
|> Seq.head
let rec findNextPrimeFactor number prime =
if number = 1L then
prime
else
let p = nextPrime(prime)
findNextPrimeFactor (number / p) p
findNextPrimeFactor n 2L
Your updated code is optimal for your approach. You would have to use a different algorithm like Yin Zhu answer to go faster. I wrote a test to check to see if F# makes the "check" function tail recursive and it does.
the variable p is actually a name binding, not a variable. Using name binding is not a bad style. And it is more readable. The lazy style of nextPrime is good, and it actually prime-test each number only once during the whole program.
My Solution
let problem3 =
let num = 600851475143L
let rec findMax (n:int64) (i:int64) =
if n=i || n<i then
n
elif n%i=0L then
findMax (n/i) i
else
findMax n (i+1L)
findMax num 2L
I basically divides num from 2, 3, 4.. and don't consider any prime numbers. Because if we divides all 2 from num, then we won't be able to divide it by 4,8, etc.
on this number, my solution is quicker:
> greatestPrimeFactor 600851475143L;;
Real: 00:00:01.110, CPU: 00:00:00.702, GC gen0: 1, gen1: 1, gen2: 0
val it : int64 = 6857L
>
Real: 00:00:00.001, CPU: 00:00:00.000, GC gen0: 0, gen1: 0, gen2: 0
val problem3 : int64 = 6857L
I think that the code with the temporary binding is significantly easier to read. It's pretty unusual to create an anonymous function and then immediately apply it to a value as you do in the other case. If you really want to avoid using a temporary value, I think that the most idiomatic way to do that in F# would be to use the (|>) operator to pipe the value into the anonymous function, but I still think that this isn't quite as readable.