Re: [Haskell-beginners] Is there an "unscan" function?

Hi Jeffrey My version actually contains an error, it should be: unscan :: (a -> a -> b) -> [a] -> [b] unscan f (a:b:bs) = f a b : unscan f (b:bs) unscan _ _ = [] A slightly less concise, but optimized version avoids putting the second element back in a list: unscan f (a:b:bs) = f a b : go b bs where go _ [] = [] go x (z:zs) = f x z : go z zs unscan _ _ = [] The first mention of "speed" for traversing lists at different rates I've seen was in the paper "There and Back Again" by Olivier Danvy and Mayer Goldberg. http://www.brics.dk/RS/02/12/BRICS-RS-02-12.pdf Best wishes Stephen