Project Euler – Problem 58 Solution

Problem

Starting with 1 and spiralling anticlockwise in the following way, a square spiral with side length 7 is formed.

image

It is interesting to note that the odd squares lie along the bottom right diagonal, but what is more interesting is that 8 out of the 13 numbers lying along both diagonals are prime; that is, a ratio of 8/13 ? 62%.

If one complete new layer is wrapped around the spiral above, a square spiral with side length 9 will be formed. If this process is continued, what is the side length of the square spiral for which the ratio of primes along both diagonals first falls below 10%?

Solution

let hasDivisor(n) =
let upperBound = int(sqrt(double(n)))
[2..upperBound] |> Seq.exists (fun x -> n % x = 0)

let isPrime(n) = if n = 1 then false else not(hasDivisor(n))

// define function that returns the number on the corners of a spiral of length n
let getCornerNumbers n =
match n with
| 1 -> [1]
| _ when n % 2 = 0 -> []
| _ -> [3..-1..0] |> List.map (fun n’ -> n*n – n’*(n-1))

let answer =
let mutable cornerNumbers, primeNumbers, size = 0, 0, 1
let mutable continueLoop = true

while continueLoop do
// get the numbers that appear at the corners of a spiral of the given size
let newNumbers = getCornerNumbers size

// increment the totals
cornerNumbers <- cornerNumbers + newNumbers.Length primeNumbers <- primeNumbers + (newNumbers |> List.filter isPrime |> List.length)

let ratio = double(primeNumbers) / double(cornerNumbers)
if ratio < 0.1 && size > 1 then continueLoop <- false else size <- size + 2 size [/code] UPDATE: Having stumbled upon some very good algorithms for generating prime number sequence here, I decided to revisit my solution here and using the PGSimple3 algorithm the new code now runs in seconds!

// generate all prime numbers under <= this max let max = 100000 // initialise the list with 2 which is the only even number in the sequence let mutable primeNumbers = [2] // only check the prime numbers which are <= the square root of the number n let hasDivisor n = primeNumbers |> Seq.takeWhile (fun n’ -> n’ <= int(sqrt(double(n)))) |> Seq.exists (fun n’ -> n % n’ = 0)

// only check odd numbers <= max let potentialPrimes = Seq.unfold (fun n -> if n > max then None else Some(n, n+2)) 3

// populate the prime numbers list
for n in potentialPrimes do
if not(hasDivisor n) then primeNumbers <- primeNumbers @ [n] // use the same hasDivisor function instead of the prime numbers list as it offers // far greater coverage as the number n is square rooted so this function can // provide a valid test up to max*max let isPrime n = if n = 1 then false else not(hasDivisor(n)) // define function that returns the number on the corners of a spiral of length n let getCornerNumbers n = match n with | 1 -> [1]
| _ when n % 2 = 0 -> []
| _ -> [3..-1..0] |> List.map (fun n’ -> n*n – n’*(n-1))

let answer =
let mutable cornerNumbers, primeNumbers, size = 0, 0, 1
let mutable continueLoop = true

while continueLoop do
// get the numbers that appear at the corners of a spiral of the given size
let newNumbers = getCornerNumbers size

// increment the totals
cornerNumbers <- cornerNumbers + newNumbers.Length primeNumbers <- primeNumbers + (newNumbers |> List.filter isPrime |> List.length)

let ratio = double(primeNumbers) / double(cornerNumbers)
if ratio < 0.1 && size > 1 then continueLoop <- false else size <- size + 2 size [/code]

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