Solution 64

064.hs

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+-- All square roots are periodic when written as continued fractions and can be written in the form:
+-- √N = a0 + 	
+--              1
+--   	a1 + 	
+--                1
+--   	  	a2 + 	
+--                  1
+--   	  	  	a3 + ...
+-- 
+-- For example, let us consider √23:
+-- √23 = 4 + √23 — 4 = 4 +  	
+--          1
+-- 	 = 4 +  	
+--          1
+--   	
+--          1
+-- √23—4
+-- 	  	1 +  	
+-- √23 – 3
+-- 7
+-- 
+-- If we continue we would get the following expansion:
+-- √23 = 4 + 	
+-- 1
+--   	1 + 	
+-- 1
+--   	  	3 + 	
+-- 1
+--   	  	  	1 + 	
+-- 1
+--   	  	  	  	8 + ...
+-- 
+-- The process can be summarised as follows:
+-- a0 = 4, 	  	
+-- 1
+-- √23—4
+-- 	 =  	
+-- √23+4
+-- 7
+-- 	 = 1 +  	
+-- √23—3
+-- 7
+-- a1 = 1, 	  	
+-- 7
+-- √23—3
+-- 	 =  	
+-- 7(√23+3)
+-- 14
+-- 	 = 3 +  	
+-- √23—3
+-- 2
+-- a2 = 3, 	  	
+-- 2
+-- √23—3
+-- 	 =  	
+-- 2(√23+3)
+-- 14
+-- 	 = 1 +  	
+-- √23—4
+-- 7
+-- a3 = 1, 	  	
+-- 7
+-- √23—4
+-- 	 =  	
+-- 7(√23+4)
+-- 7
+-- 	 = 8 +  	√23—4
+-- a4 = 8, 	  	
+-- 1
+-- √23—4
+-- 	 =  	
+-- √23+4
+-- 7
+-- 	 = 1 +  	
+-- √23—3
+-- 7
+-- a5 = 1, 	  	
+-- 7
+-- √23—3
+-- 	 =  	
+-- 7(√23+3)
+-- 14
+-- 	 = 3 +  	
+-- √23—3
+-- 2
+-- a6 = 3, 	  	
+-- 2
+-- √23—3
+-- 	 =  	
+-- 2(√23+3)
+-- 14
+-- 	 = 1 +  	
+-- √23—4
+-- 7
+-- a7 = 1, 	  	
+-- 7
+-- √23—4
+-- 	 =  	
+-- 7(√23+4)
+-- 7
+-- 	 = 8 +  	√23—4
+-- 
+-- It can be seen that the sequence is repeating. For conciseness, we use the notation √23 = [4;(1,3,1,8)], to indicate that the block (1,3,1,8) repeats indefinitely.
+-- 
+-- The first ten continued fraction representations of (irrational) square roots are:
+-- 
+-- √2=[1;(2)], period=1
+-- √3=[1;(1,2)], period=2
+-- √5=[2;(4)], period=1
+-- √6=[2;(2,4)], period=2
+-- √7=[2;(1,1,1,4)], period=4
+-- √8=[2;(1,4)], period=2
+-- √10=[3;(6)], period=1
+-- √11=[3;(3,6)], period=2
+-- √12= [3;(2,6)], period=2
+-- √13=[3;(1,1,1,1,6)], period=5
+-- 
+-- Exactly four continued fractions, for N ≤ 13, have an odd period.
+-- 
+-- How many continued fractions for N ≤ 10000 have an odd period?
+
+import Control.Monad
+import Data.List
+import Data.Maybe
+
+f :: Integer -> Integer -> Integer -> [(Integer,(Integer,Integer,Integer))]
+f a b c =
+    let m = a - (b^2)
+        fa = fromInteger a :: Float
+        fb = fromInteger b :: Float
+        fc = fromInteger c :: Float
+        fm = fromInteger m :: Float
+        x = floor $ (fc * ((sqrt fa) - fb))/ (fm)
+        h = c
+        i = -1 *  ( c*b + x*m )
+        j = m
+    in 
+        if ( x^2 == a ) 
+           then (0,(0,0,0)):f a b c
+           else if (m `rem` h /= 0) 
+                then error "Bad division" 
+                else (x,(a,b,c)):f a (i `div` h) (j `div` h)
+ 
+lengthPeriod x =
+    let s = f x 0 x 
+        sig = head $ tail s
+        index = fromJust $ elemIndex sig $( tail $ tail s )
+    in if head s == (0,(0,0,0))
+          then 0
+          else index + 1
+
+main = do
+    print $ length $ filter (\x -> ((lengthPeriod x) `rem` 2) == 1) [2..10000]
+    -- forM_ [2..13] $ \n -> do 
+    --     print $ lengthPeriod n
+    --     -- print $ tail $  take 10 $ map fst $ f n 0 n