[Haskell-cafe] beginner's problem about lists
lemming at henning-thielemann.de
Tue Oct 10 09:38:51 EDT 2006
On Tue, 10 Oct 2006, Henning Thielemann wrote:
> On Tue, 10 Oct 2006 falseep at gmail.com wrote:
> > Hi all,
> > I'm trying to implement a function that returns the shorter one of two given
> > lists,
> > something like
> > shorter :: [a] -> [a] -> [a]
> > such that shorter [1..10] [1..5] returns [1..5],
> > and it's okay for shorter [1..5] [2..6] to return either.
> > Simple, right?
> > However, it becomes difficult when dealing with infinite lists, for example,
> > shorter [1..5] (shorter [2..] [3..])
> > Could this evaluate to [1..5]? I haven't found a proper implementation.
> > Again it's ok for shorter [2..] [3..] to return whatever that can solve
> > the above problem correctly.
> > An infinite list could work, I guess, but I don't know how.
> With PeanoNumbers,
> I would first attach a lazy length information to each list before any
> call to 'shorter', then I would remove this information after the last
> call to 'shorter'.
> Untested code follows:
> attachLength :: [a] -> (PeanoNumber.T, [a])
> attachLength xs = (genericLength xs, xs)
> detachLength :: (PeanoNumber.T, [a]) -> [a]
> detachLength = snd
> shorter :: (PeanoNumber.T, [a]) -> (PeanoNumber.T, [a]) -> (PeanoNumber.T, [a])
> shorter (xl,xs) (yl,ys) = (min xl yl, if xl < yl then xs else ys)
Ah, here is a simpler solution:
compareLength :: [a] -> [b] -> Ordering
compareLength (_:xs) (_:ys) = compareLength xs ys
compareLength   = EQ
compareLength (_:_)  = GT
compareLength  (_:_) = LT
shorter :: [a] -> [a] -> [a]
shorter xs ys =
let lt = compareLength xs ys == LT
in zipWith (\x y -> if lt then x else y) xs ys
zipWith returns the length of the shorter list lazily and the elements are
chosen after the shortest list is determined.
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