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+------
+-- Spiral primes
+-- Problem 58
+--
+-- Starting with 1 and spiralling anticlockwise in the following way,
+-- a square spiral with side length 7 is formed.
+--
+-- 37 36 35 34 33 32 31
+-- 38 17 16 15 14 13 30
+-- 39 18  5  4  3 12 29
+-- 40 19  6  1  2 11 28
+-- 41 20  7  8  9 10 27
+-- 42 21 22 23 24 25 26
+-- 43 44 45 46 47 48 49
+--
+-- 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%?
+------
+
+
+main = do
+    print $ firstDropBellow 0.1 3 0
+
+firstDropBellow :: Double -> Int -> Int -> (Int, Int, Double)
+firstDropBellow percent counter n
+    | primesRatio < percent = (n, 2 * n + 1, primesRatio)
+    | otherwise = firstDropBellow percent (counter + countPrimes) (n + 1)
+    where primesRatio =  (fromIntegral countPrimes + fromIntegral counter)
+                         / fromIntegral (4 * n + 1)
+          countPrimes = length $ filter id (map isPrime corners)
+          corners = [dur, ddl, dul]
+          dur = 4 * n ^ 2 - 2 * n + 1
+          ddl = 4 * n ^ 2 + 2 * n + 1
+          dul = 4 * n ^ 2 + 1
+
+isPrime :: Int -> Bool
+isPrime 2 = True
+isPrime 3 = True
+isPrime x
+    | x < 2 = False
+    | x `mod` 2 == 0 || x `mod` 3 == 0 = False
+    | otherwise = divCheck 5
+    where divCheck d
+           | d * d > x = True
+           | x `mod` d == 0 || x `mod` (d + 2) == 0 = False
+           | otherwise = divCheck (d + 6)
+
+-- diagUpRight = [4 * x ^ 2 - 2 * x + 1 | x <- [1..]]
+-- diagdownLeft = [4 * x ^ 2 + 2 * x + 1 | x <- [1..]]
+-- diagUpLeft = [4 * x ^ 2 + 1 | x <- [1..]]
+-- diags = [diagUpRight, diagdownLeft, diagUpLeft]