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 <daniel.is.fischer@web.de> wrote:
From: Daniel Fischer <daniel.is.fischer@web.de> Subject: Re: [Haskell-cafe] Rational and % operator remix To: haskell-cafe@haskell.org Cc:
"michael rice" <nowgate@yahoo.com> Date: Sunday, March 29, 2009, 2:35 PM
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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)
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