Project Euler Problem 64 Solution
All square roots are periodic when written as continued fractions and can be written in the form:
For example, let us consider :
If we continue we would get the following expansion:
The sequence is repeating. For conciseness, we use the notation , to indicate that the block (1,3,1,8) repeats indefinitely.
The first ten continued fraction representations of (irrational) square roots are:
Exactly four continued fractions, for , have an odd period.
How many continued fractions for have an odd period?
cycles :: Eq a => [a] -> [[a]] cycles xs = map fst $ dropWhile (\(a, b) -> a ++ a /= b) $ zip cs (tail cs) where cs = [left | n <- [1..], let (left, right) = splitAt n xs, left == take n right] isSquare :: Int -> Bool isSquare n = root == fromIntegral (round root) where root = sqrt (fromIntegral n) isPrime :: Int -> Bool isPrime n | n < 1 = False | otherwise = not $ or [n `rem` x == 0 | x <- [2..floor $ sqrt $ fromIntegral n]] expansion :: Int -> [Int] expansion s | isSquare s =  | isSquare (s-1) = [fromIntegral $ 2 * (floor $ sqrt $ fromIntegral (s-1))] | otherwise = head $ dropWhile (all (== first)) cs where cs = cycles $ map a [1..] first = (head . head) cs m = (map m' [0..] !!) m' 0 = 0 m' n = (d (n-1))*(a (n-1)) - (m (n-1)) d = (map d' [0..] !!) d' 0 = 1 d' n = (s - (m n)^2) `quot` (d (n-1)) a = (map a' [0..] !!) a' 0 = floor $ sqrt $ fromIntegral s a' n = floor $ (fromIntegral ((a 0) + (m n))) / (fromIntegral (d n)) main :: IO () main = print $ length $ filter (\x -> odd $ length $ expansion x) [2..10000]
$ ghc -O2 -o continued-fractions continued-fractions.hs $ time ./continued-fractions real 0m4.136s user 0m4.084s sys 0m0.008s
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