Thanks for your reply. I tried a few ways but none worked.

One is like:

shorter as bs = f id id as bs where
    f ca cb [] _ = ca []
    f ca cb _ [] = cb []
    f ca cb (a:as) (b:bs) = f (ca.(a:)) (cb.(b:)) as bs

However this will result in a non-terminating loop for shorter [1..] [2..],
since the first two patterns of f shall never match.

Another way, I could guarantee that the evaluation of
shorter [1..5] (shorter [1..] [2..])
terminate but I lose the information to figure out which list was the shortest one.
Using zips:

shorter = zipWith (\a b -> undefined)
-- this returns the length, but not the content of the shorter list

(\a b -> undefined) could be replaced with something that encode the contents of
the two lists, but it makes no difference since I won't know which one is the answer.

The difficulty is that I cannot have these both:
A. if one list is finite, figure out the shorter one
B. if both are infinite, returning an infinite list could work

BTW, there IS an way to implement this functionality for a finite list of (possibly infinite) lists:

shortest = measureWith [] where
    measureWith ruler as = f matches
        where ruler' = undefined : ruler
              matches = filter p as
              p a = length (zip ruler' a) == length (zip ruler a)
              f [] = measureWith ruler' as
              f matches = matches

which somehow makes it unnecessary to find the function "shorter",

but the original simple problem is interesting itself.

Thanks.

On 10/10/06, Neil Mitchell <ndmitchell@gmail.com> wrote:

Hi,

The trick is not call "length", since length demands the whole of a
list, and won't terminate on an infinite list. You will want to
recurse down the lists.

Is this a homework problem? It's best to declare if it is, and show
what you've managed to do so far.

Thanks

Neil