“Any function on finite lists that is defined by pairing the desired result with the argument list can always be redefined in terms of fold”

Question

I was reading through the paper A tutorial on the universality and expressiveness of fold, and am stuck on the section about generating tuples. After showing of how the normal d

Answer 1

Given the remark that you can / may need to use the original function inside, the claim as stated in your question seems trivial to me:

rewriteAsFold :: ([a] -> (b, [a])) -> [a] -> (b, [a])
rewriteAsFold g = foldr f v where
    f x ~(ys,xs) = (fst (g (x:xs)), x:xs)
    v            = (fst (g []), [])

EDIT: Added the ~, after which it seems to work for infinite lists as well.