Convert procedural solution to functional one. F# - f#

Question

I'm trying to wrap my head around functional programming using F#. I'm working my way through the Project Euler problems, and I feel like I am just writing procedural code in F#. For instance, this is my solution to #3.
let Calc() =
let mutable limit = 600851475143L
let mutable factor = 2L // Start with the lowest prime
while factor < limit do
if limit % factor = 0L then
begin
limit <- limit / factor
end
else factor <- factor + 1L
limit
This works just fine, but all I've really done is taken how I would solve this problem in c# and converted it to F# syntax. Looking back over several of my solutions, this is becoming a pattern. I think that I should be able to solve this problem without using mutable, but I'm having trouble not thinking about the problem procedurally.

Answer 1

Why not with recursion?
let Calc() =
let rec calcinner factor limit =
if factor < limit then
if limit % factor = 0L then
calcinner factor (limit/factor)
else
calcinner (factor + 1L) limit
else limit
let limit = 600851475143L
let factor = 2L // Start with the lowest prime
calcinner factor limit

Answer 2

For algorithmic problems (like project Euler), you'll probably want to write most iterations using recursion (as John suggests). However, even mutable imperative code sometimes makes sense if you are using e.g. hashtables or arrays and care about performance.
One area where F# works really well which is (sadly) not really covered by the project Euler exercises is designing data types - so if you're interested in learning F# from another perspective, have a look at Designing with types at F# for Fun and Profit.
In this case, you could also use Seq.unfold to implement the solution (in general, you can often compose solutions to sequence processing problems using Seq functions - though it does not look as elegant here).
let Calc() =
// Start with initial state (600851475143L, 2L) and generate a sequence
// of "limits" by generating new states & returning limit in each step
(600851475143L, 2L)
|> Seq.unfold (fun (limit, factor) ->
// End the sequence when factor is greater than limit
if factor >= limit then None
// Update limit when divisible by factor
elif limit % factor = 0L then
let limit = limit / factor
Some(limit, (limit, factor))
// Update factor
else
Some(limit, (limit, factor + 1L)) )
// Take the last generated limit value
|> Seq.last

Answer 3

In functional programming when I think mutable I think heap and when trying to write code that is more functional, you should use the stack instead of the heap.
So how do you get values on to the stack for use with a function?
Place the value in the function's parameters.
let result01 = List.filter (fun x -> x % 2 = 0) [0;1;2;3;4;5]
here both a function an a list of values are hard coded into the List.filter parameter's.
Bind the value to a name and then reference the name.
let divisibleBy2 = fun x -> x % 2 = 0
let values = [0;1;2;3;4;5]
let result02 = List.filter divisibleBy2 values
here the function parameter for list.filter is bound to divisibleBy2 and the list parameter for list.filter is bound to values.
Create a nameless data structure and pipe it into the function.
let result03 =
[0;1;2;3;4;5]
|> List.filter divisibleBy2
here the list parameter for list.filter is forward piped into the list.filter function.
Pass the result of a function into the function
let result04 =
[ for i in 1 .. 5 -> i]
|> List.filter divisibleBy2
Now that we have all of the data on the stack, how do we process the data using only the stack?
One of the patterns often used with functional programming is to put data into a structure and then process the items one at a time using a recursive function. The structure can be a list, tree, graph, etc. and is usually defined using a discriminated union. Data structures that have one or more self references are typically used with recursive functions.
So here is an example where we take a list and multiply all the values by 2 and put the result back onto the stack as we progress. The variable on the stack holding the new values is accumulator.
let mult2 values =
let rec mult2withAccumulator values accumulator =
match values with
| headValue::tailValues ->
let newValue = headValue * 2
let accumulator = newValue :: accumulator
mult2withAccumulator tailValues accumulator
| [] ->
List.rev accumulator
mult2withAccumulator values []
We use an accumulator for this which being a parameter to a function and not defined mutable is stored on the stack. Also this method is using pattern matching and the list discriminated union. The accumulator holds the new values as we process the items in the input list and then when there are not more items in the list ([]) we just reverse the list to get the new list in the correct order because the new items are concatenated to the head of the accumulator.
To understand the data structure (discriminated union) for a list you need to see it, so here it is
type list =
| Item of 'a * List
| Empty
Notice how the end of the definition of an item is List referring back to itself, and that a list can ben an empty list, which is when used with pattern match is [].
A quick example of how list are built is
empty list - []
list with one int value - 1::[]
list with two int values - 1::2::[]
list with three int values - 1::2::3::[]
Here is the same function with all of the types defined.
let mult2 (values : int list) =
let rec mult2withAccumulator (values : int list) (accumulator : int list) =
match (values : int list) with
| (headValue : int)::(tailValues : int list) ->
let (newValue : int) = headValue * 2
let (accumulator : int list) =
(((newValue : int) :: (accumulator : int list)) : int list)
mult2withAccumulator tailValues accumulator
| [] ->
((List.rev accumulator) : int list)
mult2withAccumulator values []
So putting values onto the stack and using self referencing discriminated unions with pattern matching will help to solve a lot of problems with functional programming.