5 files changed, 134 insertions, 19 deletions
diff --git a/haskell/044-pentagonal_numbers.hs b/haskell/044-pentagonal_numbers.hs
new file mode 100644
index 0000000..db479bc
--- /dev/null
+++ b/haskell/044-pentagonal_numbers.hs
@@ -0,0 +1,37 @@
+-- Pentagon numbers
+-- Problem 44
+--
+-- Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2.
+-- The first ten pentagonal numbers are:
+--
+-- 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
+--
+-- It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their difference,
+-- 70 − 22 = 48, is not pentagonal.
+--
+-- Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference
+-- are pentagonal and D = |Pk − Pj| is minimised; what is the value of D?
+
+
+import Data.Maybe(isJust, fromJust)
+
+main = do
+    print(fromJust $ last $ takeUntil isJust [diffPentagonal p | p <- pentagonals])
+
+pentagonals = [(3 * n ^ 2 - n) `div` 2 | n <- [1..]]
+
+diffPentagonal :: Int -> Maybe Int
+diffPentagonal n
+    | length testBellow == 0 = Nothing
+    | otherwise = Just $ head testBellow
+    where testBellow = [k - (n - k) | k <- takeWhile (< n) pentagonals,
+                        isPentagonal (n - k), isPentagonal (k - (n - k))]
+
+isPentagonal :: Int -> Bool
+isPentagonal x = fromInteger (round n) == n
+    where n = (sqrt (fromIntegral (24 * x + 1)) + 1) / 6
+
+takeUntil :: (a -> Bool) -> [a] -> [a]
+takeUntil f (x : xs)
+    | f x = [x]
+    | otherwise = x : takeUntil f xs
diff --git a/haskell/wip/026-reciprocal_cycles.hs b/haskell/wip/026-reciprocal_cycles.hs
new file mode 100644
index 0000000..f60920f
--- /dev/null
+++ b/haskell/wip/026-reciprocal_cycles.hs
@@ -0,0 +1,24 @@
+-- Reciprocal cycles
+--
+-- Problem 26
+-- A unit fraction contains 1 in the numerator. The decimal representation of the unit
+-- fractions with denominators 2 to 10 are given:
+--
+-- 1/2   =   0.5
+-- 1/3   =   0.(3)
+-- 1/4   =   0.25
+-- 1/5   =   0.2
+-- 1/6   =   0.1(6)
+-- 1/7   =   0.(142857)
+-- 1/8   =   0.125
+-- 1/9   =   0.(1)
+-- 1/10  =   0.1
+-- Where 0.1(6) means 0.166666..., and has a 1-digit recurring cycle. It can be seen
+-- that 1/7 has a 6-digit recurring cycle.
+--
+-- Find the value of d < 1000 for which 1/d contains the longest recurring cycle in
+-- its decimal fraction part.
+
+
+main = do
+    print ()
diff --git a/haskell/wip/044-pentagonal_numbers.hs b/haskell/wip/044-pentagonal_numbers.hs
deleted file mode 100644
index 966cc13..0000000
--- a/haskell/wip/044-pentagonal_numbers.hs
+++ /dev/null
@@ -1,19 +0,0 @@
--- Pentagonal numbers are generated by the formula, Pn=n(3n−1)/2.
--- The first ten pentagonal numbers are:
---
--- 1, 5, 12, 22, 35, 51, 70, 92, 117, 145, ...
---
--- It can be seen that P4 + P7 = 22 + 70 = 92 = P8. However, their difference,
--- 70 − 22 = 48, is not pentagonal.
---
--- Find the pair of pentagonal numbers, Pj and Pk, for which their sum and difference
--- are pentagonal and D = |Pk − Pj| is minimised; what is the value of D?
-
-
-main = do
-    -- print (take 10 pentagonals)
-    let isPentagonal n = n `elem` takeWhile (<=n) pentagonals
-        pentagonals = [(3 * n ^ 2 - n) `div` 2 | n <- [1..]]
-    print (head [j + k| k <- pentagonals, j <- pentagonals,
-            isPentagonal (j + k)]) --, isPentagonal (abs (j - k))])
-
diff --git a/haskell/wip/081-path_sum_two_ways.hs b/haskell/wip/081-path_sum_two_ways.hs
new file mode 100644
index 0000000..ec513c4
--- /dev/null
+++ b/haskell/wip/081-path_sum_two_ways.hs
@@ -0,0 +1,38 @@
+-- Path sum: two ways
+--
+-- Problem 81
+-- In the 5 by 5 matrix below, the minimal path sum from the top left to the bottom right,
+-- by only moving to the right and down, is indicated in bold red and is equal to 2427.
+--
+-- 131 673 234 103 18
+-- 201 96  342 965 150
+-- 630 803 746 422 111
+-- 537 699 497 121 956
+-- 805 732 524 37  331
+--
+-- Find the minimal path sum, in matrix.txt (right click and "Save Link/Target As..."),
+-- a 31K text file containing a 80 by 80 matrix, from the top left to the
+-- bottom right by only moving right and down.
+
+main = do
+    content <- readFile "../data/081_matrix.txt"
+    -- print (content)
+    let rhombus = parseMatrix content
+    print (rhombus)
+
+
+rectToRhombus :: [[Int]] -> [[Int]]
+
+
+parseMatrix :: String -> [[Int]]
+parseMatrix str = map (map (\x -> read x :: Int)) strList
+    where strList = map (split ',') (lines str)
+
+-- https://gist.github.com/yamanobori-old/0f74739f2a97d31ebe05
+split :: Eq a => a -> [a] -> [[a]]
+split _ [] = [[]]
+split delim str =
+    let (before, remainder) = span (/= delim) str
+    in  before : case remainder of [] -> []
+                                   x  -> split delim $ tail x
+
diff --git a/haskell/wip/205-dice_game.hs b/haskell/wip/205-dice_game.hs
new file mode 100644
index 0000000..d21623c
--- /dev/null
+++ b/haskell/wip/205-dice_game.hs
@@ -0,0 +1,35 @@
+-- Dice Game
+--
+-- Problem 205
+-- Peter has nine four-sided (pyramidal) dice, each with faces numbered 1, 2, 3, 4.
+-- Colin has six six-sided (cubic) dice, each with faces numbered 1, 2, 3, 4, 5, 6.
+--
+-- Peter and Colin roll their dice and compare totals: the highest total wins. The
+-- result is a draw if the totals are equal.
+--
+-- What is the probability that Pyramidal Pete beats Cubic Colin? Give your answer
+-- rounded to seven decimal places in the form 0.abcdefg
+
+
+import System.Random
+import Data.List
+
+main = do
+    gen <- getStdGen
+    let n = 20000
+        peter = map sum $ splitIndex 9 $ randN (1, 4) gen (9 * n)
+        colin = map sum $ splitIndex 6 $ randN (1, 6) gen (6 * n)
+    let score = length $ filter (\(p, c) -> p > c) (zip peter colin)
+    print ((fromIntegral score) / (fromIntegral n))
+
+splitIndex :: Int -> [Int] -> [[Int]]
+splitIndex n xs
+    | length xs < n = []
+    | otherwise = h : splitIndex n t
+    where (h, t) = splitAt n xs
+
+randN :: (Int, Int) -> StdGen -> Int -> [Int]
+randN _ _ 0 = []
+randN r gen n = fst next : randN r (snd next) (n - 1)
+    where next = randomR r gen :: (Int, StdGen)
+