Project Euler Problem 10 in F#

This problem is very similar to several of the other prime generation problems, but being the dedicated blogger that I am here is problem 10:

The sum of the primes below 10 is 2 + 3 + 5 + 7 = 17.

Find the sum of all the primes below two million.

1. Generate primes using Sieve of Eratosthenes.
2. Stop before reaching 2 million.
3. Find the sum of this set of primes.

type public PrimesGenerator() =

member this.getPrimesMax max =

let primes = new BitArray(max+1, true)

let result = new ResizeArray(max/10)

for n = 2 to max do

if primes.[n] then

let start = (int64 n * int64 n)

if start < int64 max then

let i = ref (int start)

while !i <= max do primes.[!i] <- false; i := !i + n

result.Add n

result

member this.Problem9() =

let primesGen = new PrimesGenerator()

primesGen.getPrimesMax 2000000 |> Seq.map (fun prime -> (int64)prime) |> Seq.sum

Download the source code with unit tests.

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Project Euler Problem 9 in F#

In the continuing series of solving Project Euler problems with F#, here is problem 9 with my solution:

A Pythagorean triplet is a set of three natural numbers, a b c, for which,

a² + b² = c²
For example, 3² + 4² = 9 + 16 = 25 = 5².

There exists exactly one Pythagorean triplet for which a + b + c = 1000.
Find the product abc.

1. Enumerate from 1 to 1000.  Create tuples of all sets (m, n) where m > n.
2. Create Pythagorean triplets using Euclid’s formula from the tuples.
3. Find the set where a + b + c = 1000.
4. Find the product of this set.

let problem9 =
    let pythagorean_triplets(m:int, n:int) =
        let a = m*m-n*n
        let b = 2*m*n
        let c = m*m+n*n
        [a;b;c]
    let tops = 1000
    [for m in [1..tops] do
        for n in [1..m-1] do yield (m, n)] |> Seq.map (fun t -> pythagorean_triplets((fst t, snd t)))
            |> Seq.filter (fun x -> x |> Seq.sum = tops) |> Seq.head |> Seq.fold (fun acc elem -> acc * elem) 1

Download the source code with unit tests.

Project Euler Problem 7 in F#

In the continuing series of solving Project Euler problems with F#, here is problem 7:

By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13. What is the 10,001st prime number?

1. Generate primes using Sieve of Eratosthenes.
2. Skip the first ten thousand primes and take the next one.

type public PrimesGenerator() =

member this.PrimesInfinite () =

let rec nextPrime n p primes =

if primes |> Map.containsKey n then

nextPrime (n + p) p primes

else

primes.Add(n, p)

let rec prime n primes =

seq {

if primes |> Map.containsKey n then

let p = primes.Item n

yield! prime (n + 1) (nextPrime (n + p) p (primes.Remove n))

else

yield n

yield! prime (n + 1) (primes.Add(n * n, n))

}

prime 2 Map.empty

member this.Problem7() =

let primes = new PrimesGenerator()

primes.PrimesInfinite() |> Seq.skip 10000 |> Seq.take 1 |> Seq.head

Download the source code with unit tests.

Project Euler Problem 8 in F#

In the continuing series of solving Project Euler problems with F#, here is problem 8 and my solution:

Find the greatest product of five consecutive digits in the 1000-digit number.

73167176531330624919225119674426574742355349194934
96983520312774506326239578318016984801869478851843
85861560789112949495459501737958331952853208805511
12540698747158523863050715693290963295227443043557
66896648950445244523161731856403098711121722383113
62229893423380308135336276614282806444486645238749
30358907296290491560440772390713810515859307960866
70172427121883998797908792274921901699720888093776
65727333001053367881220235421809751254540594752243
52584907711670556013604839586446706324415722155397
53697817977846174064955149290862569321978468622482
83972241375657056057490261407972968652414535100474
82166370484403199890008895243450658541227588666881
16427171479924442928230863465674813919123162824586
17866458359124566529476545682848912883142607690042
24219022671055626321111109370544217506941658960408
07198403850962455444362981230987879927244284909188
84580156166097919133875499200524063689912560717606
05886116467109405077541002256983155200055935729725
71636269561882670428252483600823257530420752963450

1. Parse the input string to a list of integers
2. Create lists of each 5 consecutive digits.
3. Find the product of each list.
4. Find the max product.

open System

let Problem8 =
    let multiply lst = lst |> Seq.fold (fun acc elem -> acc * elem) 1
    let s = "7316717653133062491922511967442657474235534919493496983520312774506326239578318016984801869478851843858615607891129494954595017379583319528532088055111254069874715852386305071569329096329522744304355766896648950445244523161731856403098711121722383113622298934233803081353362766142828064444866452387493035890729629049156044077239071381051585930796086670172427121883998797908792274921901699720888093776657273330010533678812202354218097512545405947522435258490771167055601360483958644670632441572215539753697817977846174064955149290862569321978468622482839722413756570560574902614079729686524145351004748216637048440319989000889524345065854122758866688116427171479924442928230863465674813919123162824586178664583591245665294765456828489128831426076900422421902267105562632111110937054421750694165896040807198403850962455444362981230987879927244284909188845801561660979191338754992005240636899125607176060588611646710940507754100225698315520005593572972571636269561882670428252483600823257530420752963450"
    let values = s.ToCharArray() |> Seq.map (fun x -> Int32.Parse(x.ToString())) |> Seq.toList
    [0..(values.Length - 5)]
        |> Seq.map (fun h -> [0..4] |> Seq.map (fun a -> values.[(h + a)]) |> multiply )
        |> Seq.max

Download the source code with unit tests.

Project Euler Problem 6 in F#

In the continuing series of solving Project Euler problems with F#, here is problem 6 and my solution:

The sum of the squares of the first ten natural numbers is,

1² + 2² + … + 10² = 385
The square of the sum of the first ten natural numbers is,

(1 + 2 + … + 10)² = 55² = 3025
Hence the difference between the sum of the squares of the first ten natural numbers and the square of the sum is 3025 – 385 = 2640.

Find the difference between the sum of the squares of the first one hundred natural numbers and the square of the sum.

1. Enumerate 1-100 and  square each value.  Add the results together
2. Enumerate 1-100 and add each value together.  Square the resulting sum.
3. Find the difference between the two.

let problem6 = 
    let values = [1..100]
    let sumOfSquares = values |> Seq.map (fun x -> pown x 2) |> Seq.sum
    let squareOfSums = values |> Seq.sum |> (fun x -> pown x 2)
    squareOfSums - sumOfSquares

Download the source code with unit tests.

Project Euler Problem 4 in F#

In the continuing series of solving Project Euler problems with F#, here is problem 4 and my solution:

A palindromic number reads the same both ways. The largest palindrome made from the product of two 2-digit numbers is 9009 = 91 99.

Find the largest palindrome made from the product of two 3-digit numbers.

1. Create the set of all products of two 3-digit numbers.
2. Filter the values that are palindromes.
3. Find the max of the palindrome values.

open System

let problem4 =
    let reverseNumber value = Convert.ToInt32(new string(Array.rev (value.ToString().ToCharArray())))
    [for i in [100..999] do
        for j in [100..999] do yield i * j]
            |> Seq.filter (fun x -> x = reverseNumber(x)) |> Seq.max

Download the source code with unit tests.

Project Euler in F#, Problems 1 and 2

I recently stumbled across the Project Euler website. From wikipedia: “Project Euler (named after Leonhard Euler) is a website dedicated to a series of computational problems intended to be solved with computer programs”.

I thought this would be a good opportunity to continue working with F# and have some material for writing more blog articles.

Problem 1 is very straight forward.

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.

Find the sum of all the multiples of 3 or 5 below 1000.

I went with the most logical algorithm I could think of.

1. Enumerate though each of the numbers 1-999.
2. Filter the values that are multiples of either 3 or 5.
3. Add up the resulting values.

The resulting F# code is quite elegant:

let problem1 = 
    [1..999] |> Seq.filter (fun i -> i % 3 = 0 || i % 5 = 0) |> Seq.sum

After looking at other algorithms people had used, there are more efficient methods using summations. But not bad for my first solution. On to problem 2…

Each new term in the Fibonacci sequence is generated by adding the previous two terms. By starting with 1 and 2, the first 10 terms will be:

1, 2, 3, 5, 8, 13, 21, 34, 55, 89, …

By considering the terms in the Fibonacci sequence whose values do not exceed four million, find the sum of the even-valued terms.

Again, there is a pretty obvious solution here.

1. Generate the Fibonacci sequence up to 4 million.
2. Filter the values that are even.
3. Add up the resulting values.

The resulting F# code:

let Problem2 =
    Seq.unfold (fun state ->
        if (snd state > 4000000) then None
        else Some(fst state + snd state, (snd state, fst state + snd state))) (1,1)
            |> Seq.filter (fun x -> x % 2 = 0) |> Seq.sum

Again, a more efficient solution can be found but this is at least semantically easy to follow.

I’m planning on continuing this series of blog articles with a problem or 2 for each entry, and seeing how many of the problems I can get through.  I’ll also provide the source code.   If you’re interested in more in this series stay tuned!