From 02d4adc3aa84086240cd2c2a927888035282f312 Mon Sep 17 00:00:00 2001
From: Daniele Fucini <dfucini@gmail.com>
Date: Wed, 20 Nov 2024 19:39:00 +0100
Subject: [PATCH] Add Haskell solutions for Problems 26 and 27

---
 Haskell/p026.hs | 38 ++++++++++++++++++++++++++++++++++++++
 Haskell/p027.hs | 37 +++++++++++++++++++++++++++++++++++++
 2 files changed, 75 insertions(+)
 create mode 100644 Haskell/p026.hs
 create mode 100644 Haskell/p027.hs

diff --git a/Haskell/p026.hs b/Haskell/p026.hs
new file mode 100644
index 0000000..4e34674
--- /dev/null
+++ b/Haskell/p026.hs
@@ -0,0 +1,38 @@
+-- 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.
+
+removeFactor :: (Integral a) => a -> a -> a
+removeFactor f n
+    | n `mod` f /= 0 = n
+    | otherwise = removeFactor f (n `div` f)
+
+findCycleLengthRecursive :: (Integral a) => a -> a -> a -> a
+findCycleLengthRecursive n j k
+    | n == 1 = 0
+    | k `mod` n == 0 = j
+    | otherwise = findCycleLengthRecursive n (j+1) (k * 10 + 9)
+
+findCycleLength :: (Integral a) =>  a -> a
+findCycleLength n = findCycleLengthRecursive ((removeFactor 2 . removeFactor 5) n) 1 9
+
+maxRepeatingCycle :: (Integral a) => a -> a -> a 
+maxRepeatingCycle a b = snd . maximum $ zip xs [1..]
+    where xs = map findCycleLength [a..b]
+
+main = do
+    let result = maxRepeatingCycle 1 999
+    putStrLn $ "Project Euler, Problem 26\n"
+        ++ "Answer: " ++ show result
diff --git a/Haskell/p027.hs b/Haskell/p027.hs
new file mode 100644
index 0000000..b6ce20c
--- /dev/null
+++ b/Haskell/p027.hs
@@ -0,0 +1,37 @@
+-- Euler discovered the remarkable quadratic formula:
+--
+-- n^2+n+41
+--
+-- It turns out that the formula will produce 40 primes for the consecutive integer values 0≤n≤39. However, when n=40,402+40+41=40(40+1)+41 is
+-- divisible by 41, and certainly when n=41,412+41+41 is clearly divisible by 41.
+--
+-- The incredible formula n^2−79n+1601 was discovered, which produces 80 primes for the consecutive values 0≤n≤79.
+-- The product of the coefficients, −79 and 1601, is −126479.
+--
+-- Considering quadratics of the form:
+--
+-- n^2+an+b, where |a|<1000 and |b|≤1000
+--
+-- where |n| is the modulus/absolute value of n
+-- e.g. |11|=11 and |−4|=4
+--
+-- Find the product of the coefficients, a and b, for the quadratic expression that produces the maximum number of primes for consecutive values of n,
+-- starting with n=0.
+
+import Data.List (maximumBy)
+import Data.Function (on)
+import ProjectEuler (isPrime)
+
+findLengthPrimeSequence :: Int -> Int -> Int
+findLengthPrimeSequence a b = length $ takeWhile isPrime [ n*n + a*n + b | n <- [0..] ]
+
+findCoefficients = let as = [-999..999]
+                       bs = filter isPrime [2..1000]
+                       cs = [ (a, b) | a <- as, b <- bs ]
+                    in fst $ maximumBy (compare `on` snd) [ (c, l) | c <- cs, let l = uncurry findLengthPrimeSequence c ]
+
+main = do
+    let coefficients = findCoefficients
+        result = uncurry (*) coefficients
+    putStrLn $ "Project Euler, Problem 27\n"
+        ++ "Answer: " ++ show result