I am working on some homework and we are supposed to be making a combination function in F#. I have got the factorial function down, but it seems to overflow once I get a big number to use factorial on. (Let's say 20) I understand I can use an int64 or a float, but that would change all the inputs on the code. What data type should I use?
let rec Fact (a:int)=
if (a = 0) then 1 else a*Fact(a-1);;
let combo (n:int) (k:int)=
if (n = 0) then 0 else (Fact n)/((Fact k)*(Fact (n-k)));;
On the code right now, when I do combo 20 5;; it gives me 2147. Which is clearly the wrong answer. I looked at the factorial function and when I put 20 in there it gave me a big negative number. Any help would be much appreciated. Thanks in advance.
First of all, if you want to avoid surprises, you can open the Checked module at the top of your file. This will redefine the numerical operators so that they perform overflow checks - and you'll get an exception rather than unexpected number:
open Microsoft.FSharp.Core.Operators.Checked
As Fyodor points out in the comment, you cannot fit factorial of 20 in int and you need int64. However, your combo function then performs division which will make the result of combo 20 5 small enough to fit into int.
One option is to change Fact to use int64, but keep combo as a function that takes and returns integers - you'll need to convert them to int64 before calling Fact and then back to int after performing the division:
let rec Fact (a:int64) =
if (a = 0L) then 1L else a * Fact(a-1L)
let combo (n:int) (k:int) =
if (n = 0) then 0 else int (Fact (int64 n) / (Fact (int64 k) * Fact (int64 (n-k))))
Now you can call combo 20 5 and you'll get 15504 as the result.
EDIT: As noted by #pswg in the other answer, int64 is also quite limited and so you'll need BigInteger for larger factorials. However, the same method should work for you with BigInteger. You can keep the combo function as a function that returns int by converting back from BigInteger to int.
You simply won't be able to do that with an 32-bit integer (int). A 64-bit integer will get you up to 20!, but will fail at 21!. The numbers just get too big, too quickly. To go any further than that you'll need to use System.Numerics.BigInteger (abbreviated bigint in F#).
The parameter can probably stay as an int to be reasonable, but you need to return a bigint:
let rec Fact (n : int) =
if n = 0 then bigint.One else (bigint n) * Fact (n - 1)
Or to be a little more idiomatic:
let rec Fact = function | 0 -> bigint.One | n -> (bigint n) * Fact (n - 1)
And now, in your Combo function, you'll need to use these bigint's internally for all math (thankfully the integer division is all you need in this case).
let Combo (n : int) (k : int) =
if n = 0 then bigint.Zero else (Fact n) / ((Fact k) * (Fact (n - k)))
If you really wanted to make Combo return an int, you can do that conversion here:
let Combo (n : int) (k : int) =
if n = 0 then 0 else (Fact n) / ((Fact k) * (Fact (n - k))) |> int
Examples:
Combo 20 5 // --> 15504
Combo 99 5 // --> 71523144 (would break if you used int64)
Edit: By rethinking your implementation of Combo you can get some big performance improvements out of this. See this question on Math.SE for the basis of this implementation:
let ComboFast (n : int) (k : int) =
let rec Combo_r (n : int) = function
| 0 -> bigint.One
| k -> (bigint n) * (Combo_r (n - 1) (k - 1)) / (bigint k)
Combo_r n (if (2 * k) > n then n - k else k)
A quick benchmark showed this to be significantly faster than the Fact-based version above:
Function Avg. Time (ms)
Combo 99 5 30.12570
ComboFast 99 5 0.72364
Related
What is the preferable style for F# definitions?
The book I am studying makes regular use the following style:
let foo = fun x y ->
let aux1 = fun z -> z * 2
in aux1 x +
let aux2 = fun q -> q * 3
in aux2 y;;
On the other side, when I look for information on the web, it is most likely to meet something like:
let foo (x: int) (y: int) =
let aux1 (z:int) = z * 2
in aux1 x +
let aux2 (q: int) = q * 3
in aux2 y;;
On the Guide I failed to find a reference about it. Is it a matter that goes beyond "mere style"? There are efficiency implications behind these two approaches?
What does your experience suggest?
As a general rule, F# function definitions tend to do one of two things:
Define as few types as possible (let foo x y = ...). This is the case for most functions. Or...
Explicitly define the types of each argument and the return type (let foo (x : int) (y : int) : int = ....
Style #2 is rare, and I've usually seen it for functions that are explicitly part of the API of a module, and that have /// comments to provide documentation as well. For internal functions, though, the typeless variant is usually used, since F#'s type inference works so well.
Also, as s952163 pointed out in a comment, the in keyword is almost never used anymore, since the #light style makes it unnecessary. I'd expect to see your sample code written as follows in modern F# style:
let foo x y =
let aux1 z = z * 2
let aux2 q = q * 3
(aux1 x) + (aux2 y)
No ;; necessary, either, unless you're typing into the F# Interactive console. If you're using VS Code + Ionide, and highlighting segments of code and pressing Alt + Enter to send them to F# Interactive, then you don't need any ;; separators because Ionide adds them automatically.
I found evidence suggesting that the first style, even if today unconventional, is intrinsically connected to currying and anonymous functions.
Currying is a powerful characteristic of F#, where, I remember, every function could take only one parameter. For example:
let add x y = x + y
val add: int -> int -> int
The signature is interpreted as add is a function that takes two integers as input and return an integer.
When compile time comes, the function is interpreted like:
let add2 = fun x -> fun y -> x + y
val add2: int -> int -> int
where val add2: int -> int -> int is semantically equivalent to val add: (int -> (int -> int))
By providing an argument to add2, such as 6, it returns fun y -> 6 + y, which is another function waiting for its argument, while x is replaced by 6.
Currying means that every argument actually returns a separate function: that's why when we call a function with only few of its parameters returns another function.
If I got it correctly, the more common F# syntax of the second example, let add x y = x + y, could be thought like syntactic sugar for the explicit currying style shown above.
My understanding of the difference between List.fold and List.foldBack is that foldBack iterates over its list in a reverse order. Both functions accumulate a result from the items in the list.
I'm having trouble coming up with a good example where it is preferable to foldBack over a list. In the examples I have come up with, the results are the same for both fold and foldBack, if the function logic does the same thing.
[<Fact>]
let ``List.foldBack accumulating a value from the right to the left``() =
let list = [1..5]
let fFoldBack x acc =
acc - x
let fFold acc x =
acc - x
let foldBackResult = List.foldBack fFoldBack list 0
let foldResult = List.fold fFold 0 list
Assert.Equal( -15, foldBackResult ) // 0 - 5 - 4 - 3 - 2 - 1
Assert.Equal( -15, foldResult ) // 0 - 1 - 2 - 3 - 4 - 5
You don't see a difference in your example because you chose a function such that for any x1 and x2:
(acc - x1) - x2 = (acc - x2) - x1
So it doesn't matter in what order you go through the list, you will get the same result.
List construction is an example of function for which it is not the case:
x1 :: (x2 :: acc) <> x2 :: (x1 :: acc)
So the following will yield different results:
List.fold (fun acc x -> x :: acc) [] [1; 2; 3; 4; 5]
// val it : int list = [5; 4; 3; 2; 1]
List.foldBack (fun x acc -> x :: acc) [1; 2; 3; 4; 5] [];;
// val it : int list = [1; 2; 3; 4; 5]
List.fold starts with an empty result list and goes forward through the input, adding each element to the front of the result list; therefore the final result is in the reverse order.
List.foldBack, on the other hand, goes backward through the input; so each element newly added to the front of the result list was itself to the front in the original list. So the final result is the same list as the original.
Tarmil's answer has already demonstrated the difference between the two in a good, concise manner. I'm going to give an example that uses a somewhat more complex data type. (Actually, if you ignore the naming then my example is a linked list, but you can imagine how it could be expanded to something much more complex.)
The purpose of fold vs. foldBack isn't necessarily obvious when you are computing a scalar value, but when you start using it to build data structures, it becomes clear that most such structures must be built in a particular direction. This is especially true if you use immutable data structures, since you don't have the option of constructing a node and then updating it to point to another node.
In this example, I've defined a structure for a trivial programming language that does nothing but print numbers. It recognizes two instructions, Print and End. Each print instruction stores a pointer to the next instruction. Thus, a program consists of a chain of instructions, each pointing to the next. (The reason I've used this particular example is because, by executing the program, you demonstrate its structure.)
The program uses three different methods of constructing the program from a list of integers. The first, make_list_printer, is defined recursively with no fold, to demonstrate what we're trying to achieve. The second, make_list_printer_foldBack, uses foldBack to achieve the same result. The third, make_list_printer_fold, uses fold to demonstrate how it produces the wrong result.
I've mostly coded in OCaml, not F#, so I apologize if some of the coding conventions used below aren't really considered proper style in F#. I did test it in the F# interpreter, though, and it works.
(* Data structure of our mini-language. *)
type prog =
| End
| Print of int * prog
(* Builds a program recursively. *)
let rec make_list_printer = function
| [] -> End
| i :: program -> Print (i, make_list_printer program)
(* Builds a program using foldBack. *)
let make_list_printer_foldBack ints =
List.foldBack (fun i p -> Print (i, p)) ints End
(* Builds a program in the wrong direction. *)
let make_list_printer_fold ints =
List.fold (fun p i -> Print (i, p)) End ints
(* The interpreter for our mini-language. *)
let rec run_list_printer = function
| End ->
printfn ""
| Print (i, program) ->
printf "%d " i;
run_list_printer program
(* Build and run three different programs based on the same list of numbers. *)
let () =
let base_list = [1; 2; 3; 4; 5] in
let a = make_list_printer base_list in
let b = make_list_printer_foldBack base_list in
let c = make_list_printer_fold base_list in
run_list_printer a;
run_list_printer b;
run_list_printer c
The output that I get when I run this is:
1 2 3 4 5
1 2 3 4 5
5 4 3 2 1
I am currently experimenting with F#. The articles found on the internet are helpful, but as a C# programmer, I sometimes run into situations where I thought my solution would help, but it did not or just partially helped.
So my lack of knowledge of F# (and most likely, how the compiler works) is probably the reason why I am totally flabbergasted sometimes.
For example, I wrote a C# program to determine perfect numbers. It uses the known form of Euclids proof, that a perfect number can be formed from a Mersenne Prime 2p−1(2p−1) (where 2p-1 is a prime, and p is denoted as the power of).
Since the help of F# states that '**' can be used to calculate a power, but uses floating points, I tried to create a simple function with a bitshift operator (<<<) (note that I've edit this code for pointing out the need):
let PowBitShift (y:int32) = 1 <<< y;;
However, when running a test, and looking for performance improvements, I also tried a form which I remember from using Miranda (a functional programming language also), which uses recursion and a pattern matcher to calculate the power. The main benefit is that I can use the variable y as a 64-bit Integer, which is not possible with the standard bitshift operator.
let rec Pow (x : int64) (y : int64) =
match y with
| 0L -> 1L
| y -> x * Pow x (y - 1L);;
It turns out that this function is actually faster, but I cannot (yet) understand the reason why. Perhaps it is a less intellectual question, but I am still curious.
The seconds question then would be, that when calculating perfect numbers, you run into the fact that the int64 cannot display the big numbers crossing after finding the 9th perfectnumber (which is formed from the power of 31). I am trying to find out if you can use the BigInteger object (or bigint type) then, but here my knowledge of F# is blocking me a bit. Is it possible to create a powerfunction which accepts both arguments to be bigints?
I currently have this:
let rec PowBigInt (x : bigint) (y : bigint) =
match y with
| bigint.Zero -> 1I
| y -> x * Pow x (y - 1I);;
But it throws an error that bigint.Zero is not defined. So I am doing something wrong there as well. 0I is not accepted as a replacement, since it gives this error:
Non-primitive numeric literal constants cannot be used in pattern matches because they
can be mapped to multiple different types through the use of a NumericLiteral module.
Consider using replacing with a variable, and use 'when <variable> = <constant>' at the
end of the match clause.
But a pattern matcher cannot use a 'when' statement. Is there another solution to do this?
Thanks in advance, and please forgive my long post. I am only trying to express my 'challenges' as clear as I can.
I failed to understand why you need y to be an int64 or a bigint. According to this link, the biggest known Mersenne number is the one with p = 43112609, where p is indeed inside the range of int.
Having y as an integer, you can use the standard operator pown : ^T -> int -> ^T instead because:
let Pow (x : int64) y = pown x y
let PowBigInt (x: bigint) y = pown x y
Regarding your question of pattern matching bigint, the error message indicates quite clearly that you can use pattern matching via when guards:
let rec PowBigInt x y =
match y with
| _ when y = 0I -> 1I
| _ -> x * PowBigInt x (y - 1I)
I think the easiest way to define PowBigInt is to use if instead of pattern matching:
let rec PowBigInt (x : bigint) (y : bigint) =
if y = 0I then 1I
else x * PowBigInt x (y - 1I)
The problem is that bigint.Zero is a static property that returns the value, but patterns can only contain (constant) literals or F# active patterns. They can't directly contain property (or other) calls. However, you can write additional constraints in where clause if you still prefer match:
let rec PowBigInt (x : bigint) (y : bigint) =
match y with
| y when y = bigint.Zero -> 1I
| y -> x * PowBigInt x (y - 1I)
As a side-note, you can probably make the function more efficent using tail-recursion (the idea is that if a function makes recursive call as the last thing, then it can be compiled more efficiently):
let PowBigInt (x : bigint) (y : bigint) =
// Recursive helper function that stores the result calculated so far
// in 'acc' and recursively loops until 'y = 0I'
let rec PowBigIntHelper (y : bigint) (acc : bigint) =
if y = 0I then acc
else PowBigIntHelper (y - 1I) (x * acc)
// Start with the given value of 'y' and '1I' as the result so far
PowBigIntHelper y 1I
Regarding the PowBitShift function - I'm not sure why it is slower, but it definitely doesn't do what you need. Using bit shifting to implement power only works when the base is 2.
You don't need to create the Pow function.
The (**) operator has an overload for bigint -> int -> bigint.
Only the second parameter should be an integer, but I don't think that's a problem for your case.
Just try
bigint 10 ** 32 ;;
val it : System.Numerics.BigInteger =
100000000000000000000000000000000 {IsEven = true;
IsOne = false;
IsPowerOfTwo = false;
IsZero = false;
Sign = 1;}
Another option is to inline your function so it works with all numeric types (that support the required operators: (*), (-), get_One, and get_Zero).
let rec inline PowBigInt (x:^a) (y:^a) : ^a =
let zero = LanguagePrimitives.GenericZero
let one = LanguagePrimitives.GenericOne
if y = zero then one
else x * PowBigInt x (y - one)
let x = PowBigInt 10 32 //int
let y = PowBigInt 10I 32I //bigint
let z = PowBigInt 10.0 32.0 //float
I'd probably recommend making it tail-recursive, as Tomas suggested.
What is the most elegant way to implement dynamic programming algorithms that solve problems with overlapping subproblems? In imperative programming one would usually create an array indexed (at least in one dimension) by the size of the problem, and then the algorithm would start from the simplest problems and work towards more complicated once, using the results already computed.
The simplest example I can think of is computing the Nth Fibonacci number:
int Fibonacci(int N)
{
var F = new int[N+1];
F[0]=1;
F[1]=1;
for(int i=2; i<=N; i++)
{
F[i]=F[i-1]+F[i-2];
}
return F[N];
}
I know you can implement the same thing in F#, but I am looking for a nice functional solution (which is O(N) as well obviously).
One technique that is quite useful for dynamic programming is called memoization. For more details, see for example blog post by Don Syme or introduction by Matthew Podwysocki.
The idea is that you write (a naive) recursive function and then add cache that stores previous results. This lets you write the function in a usual functional style, but get the performance of algorithm implemented using dynamic programming.
For example, a naive (inefficient) function for calculating Fibonacci number looks like this:
let rec fibs n =
if n < 1 then 1 else
(fibs (n - 1)) + (fibs (n - 2))
This is inefficient, because when you call fibs 3, it will call fibs 1 three times (and many more times if you call, for example, fibs 6). The idea behind memoization is that we write a cache that stores the result of fib 1 and fib 2, and so on, so repeated calls will just pick the pre-calculated value from the cache.
A generic function that does the memoization can be written like this:
open System.Collections.Generic
let memoize(f) =
// Create (mutable) cache that is used for storing results of
// for function arguments that were already calculated.
let cache = new Dictionary<_, _>()
(fun x ->
// The returned function first performs a cache lookup
let succ, v = cache.TryGetValue(x)
if succ then v else
// If value was not found, calculate & cache it
let v = f(x)
cache.Add(x, v)
v)
To write more efficient Fibonacci function, we can now call memoize and give it the function that performs the calculation as an argument:
let rec fibs = memoize (fun n ->
if n < 1 then 1 else
(fibs (n - 1)) + (fibs (n - 2)))
Note that this is a recursive value - the body of the function calls the memoized fibs function.
Tomas's answer is a good general approach. In more specific circumstances, there may be other techniques that work well - for example, in your Fibonacci case you really only need a finite amount of state (the previous 2 numbers), not all of the previously calculated values. Therefore you can do something like this:
let fibs = Seq.unfold (fun (i,j) -> Some(i,(j,i+j))) (1,1)
let fib n = Seq.nth n fibs
You could also do this more directly (without using Seq.unfold):
let fib =
let rec loop i j = function
| 0 -> i
| n -> loop j (i+j) (n-1)
loop 1 1
let fibs =
(1I,1I)
|> Seq.unfold (fun (n0, n1) -> Some (n0 , (n1, n0 + n1)))
|> Seq.cache
Taking inspiration from Tomas' answer here, and in an attempt to resolve the warning in my comment on said answer, I propose the following updated solution.
open System.Collections.Generic
let fib n =
let cache = new Dictionary<_, _>()
let memoize f c =
let succ, v = cache.TryGetValue c
if succ then v else
let v = f c
cache.Add(c, v)
v
let rec inner n =
match n with
| 1
| 2 -> bigint n
| n ->
memoize inner (n - 1) + memoize inner (n - 2)
inner n
This solution internalizes the memoization, and while doing so, allows the definitions of fib and inner to be functions, instead of fib being a recursive object, which allows the compiler to (I think) properly reason about the viability of the function calls.
I also return a bigint instead of an int, as int quickly overflows with even a small of n.
Edit: I should mention, however, that this solution still runs into stack overflow exceptions with sufficiently large values of n.
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.