[2008.07.06] Project Euler: Problem 1

My goal in this series is to use Functional Programming, namely F# to solve the problems given by Project Euler. As previously stated, I am in the process of learning F# and I thought this would be a good way to illustrate my progress.

Project Euler: Problem 1

Add all the natural numbers below one thousand that are multiples of 3 or 5.

Example

If we list all the natural numbers below 10 that are multiples of 3 or 5, we get 3, 5, 6 and 9. The sum of these multiples is 23.

Solution 1

This is the long version where everything is nicely laid out and annotated:
NOTE: I will leave running and getting the results to you.

    1 #light

    2 // Name: Sparky Dasrath

    3 // Date: 2008.07.06

    4 // Code: Project Euler: Problem 1 - Solution 1

    5

    6 // create a function sumNumbers that take on argument, n,

    7 //  which represents the upper limit on the numbers we

    8 //  want to sum

    9 //  n = 999 for this problem

   10 let sumNumbers n =

   11     // generate a list of numbers using

   12     //  Sequence Expression,in the

   13     //  input, n, that are multiples of

   14     //  three, five and fifteen 

   15     let multOfThree   = {for x in 1 .. n when x%3=0 -> x}

   16     let multOfFive    = {for x in 1 .. n when x%5=0 -> x}

   17     let multOfFifteen = {for x in 1 .. n when x%15=0 -> x}

   18

   19     // combine the sequences of multiples of 3 and 5

   20     let firstGroup = Seq.append multOfThree multOfFive

   21

   22     // convert the sequence to an array to take advantage

   23     //  of some of array's built in functionality

   24     let arr1 = Seq.to_array firstGroup

   25

   26     // now convert the multiples of 15 to an array

   27     let arr2 = Seq.to_array multOfFifteen

   28

   29     // use the Sum function of array to add

   30     //  members of the array

   31     let sumOfThreeFive = Array.sumByInt (fun a -> a) arr1

   32     let sumOfFifteen   = Array.sumByInt (fun a -> a) arr2

   33

   34     // return the difference between the two array sums

   35     (sumOfThreeFive-sumOfFifteen)

Solution 2

Here is a much simplified version of the solution.

    1 #light

    2 // Name: Sparky Dasrath

    3 // Date: 2008.07.06

    4 // Code: Project Euler: Problem 1 - Solution 2

    5 // this is a much shorter version using

    6 //  lists instead of sequences

    7 // here using List Expressions two lists

    8 //  are created and appended, then the

    9 //  new list is converted to an array

   10 //  and the items summed

   11 // another list with multiples of 15

   12 //  is subtracted from the previous

   13 //  list

   14 (Array.sumByInt

   15     (fun x -> x)

   16     (List.to_array

   17         ([0 .. 3 .. 999]@[0 .. 5 .. 999]))) -

   18 (Array.sumByInt (fun x-> x) [|0 .. 15 .. 999|])

Solution 3

Even more simplified!

    1 #light

    2 // Name: Sparky Dasrath

    3 // Date: 2008.07.06

    4 // Code: Project Euler: Problem 1 - Solution 3

    5 

    6 // this version avoids using the second

    7 //  list of multiples of 15

    8 // the Set operator actually strips out

    9 //  any duplicate values after the lists

   10 //  are combined and the pipeline

   11 //  operator, |> lets you feed values

   12 //  to a function

   13 let a = [0..3..999]@[0..5..999]

   14 Set.of_list a |> Set.to_array |> Array.sumByInt (fun x->x)