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 wrote: From: Daniel Fischer Subject: Re: [Haskell-cafe] Rational and % operator remix To: haskell-cafe@haskell.org Cc: "michael rice" Date: Sunday, March 29, 2009, 2:35 PM -----Inline Attachment Follows----- Am Sonntag 29 März 2009 19:40:19 schrieb michael rice:

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)