Comments on: Project Euler in F# – Problem 5 (alternative solution)

By: Joel

Joel — Tue, 08 Dec 2009 09:51:45 +0000

For what it’s worth, this version returns the correct result in about 3.5 ms on my machine, compared to 28 ms for the code in your post. (According to System.Diagnostics.Stopwatch)

let problem5 =
    // least common factor
    let rec lcf = function
        | hd :: tl -> hd :: lcf (List.map (fun x -> if x % hd = 0 then x / hd else x) tl)
        | [] -> []
    [1..20] |> lcf |> List.fold (*) 1

I based the code on the algorithm described at the link below, but in F# I only needed 4 lines of code, vs 30+ lines of C#.

http://srtsolutions.com/blogs/billwagner/archive/2008/04/08/euler-problem-5.aspx

By: claudio

claudio — Mon, 03 Mar 2008 18:21:08 +0000

To measure the execution time I store the value of DateTime.Now at the beginning and the end of the script and subtract them.

It is quite a “naive” solution but I simply wanted to get an idea of the execution time.

By: Michiel Borkent

Michiel Borkent — Mon, 03 Mar 2008 17:47:39 +0000

Btw, how are you measuring the execution time? Using #time;;?

By: Michiel Borkent

Michiel Borkent — Mon, 03 Mar 2008 17:37:39 +0000

… which can of course be written as [11I .. 20I]

By: Michiel Borkent

Michiel Borkent — Mon, 03 Mar 2008 17:36:36 +0000

The list you feed to fold to generate the solution can be simplified itself, by observing that:

lcm(a,b) = b when b can be written as a*n, with n being a natural number.

Examples: lcm(5,20) = 20, lcm(2,100) = 100 etc etc

So, instead of using the whole [1I .. 20I] use [11I; 12I; 13I; 14I; 15I; 16I; 17I; 18I; 19I; 20I]