Correct, if it can be stated as a Rational then it terminates. I was messing around last night with the first function on a sequence that approximates sqrt(2)
cf (take 25 (1: [2,2..]))
forgetting that while the sequence is infinite, I only grabbed the first 25 elements.
Your replacement code works fine except for a missing
import Data.Ratio
It's quite removed from what I was trying so I'm going to have to read ahead a bit to understand it.
Thanks,
Michael
--- On Sun, 3/29/09, Daniel Fischer
Hi,
Thanks again for the help last night.
The second function cf2 is an attempt to reverse the process of the first function, i.e., given a rational number it returns a list of integers, possibly infinite,
Not for rational numbers.
but you shouldn't get into trouble if you use 98%67 as input (output should be [1,2,6,5]). The interpreter is complaining about the '=' following the 'in' keyword.
That should be '=='.
Is there a better way to state this?
Michael
import Data.Ratio cf :: [Int] -> Rational cf (x:[]) = toRational x cf (x:xs) = toRational x + 1 / cf xs
cf2 :: Rational -> [Int] cf2 a = let ai = toRational (floor ((numerator a) / (denominator a))) in if a = ai then [a] else ai : cf2 ((toRational 1) / (subtract ai a))
import Data.List (unfoldr) cf3 :: Rational -> [Integer] -- Int may overflow cf3 0 = [0] cf3 x = a0:unfoldr f r where a0 = floor x r = x - fromInteger a0 f 0 = Nothing f y = Just (properFraction $ recip y)