Possible, yes. Efficient, not really.
inTwain = foldr (\x (ls, rs) -> if length ls == length rs then (x:ls, rs) else (x:(init ls), (last ls):rs)) ([], [])
I have a hunch that everything that reduces a list to a fixed-size
data structure can be expressed as a fold, simply by carrying around
as much intermediate state as necessary. But I'm too lazy and
inexperienced to prove this.
Thomas
On Thu, Jun 4, 2009 at 16:22, Martijn van
Steenbergen
Bonjour café,
A small puzzle:
Consider the function inTwain that splits a list of even length evenly into two sublists:
inTwain "Hello world!" ("Hello ","world!")
Is it possible to implement inTwain such that the recursion is done by one of the standard list folds?
Is there a general way to tell if a problem can be expressed as a fold?
Thank you,
Martijn. _______________________________________________ Haskell-Cafe mailing list Haskell-Cafe@haskell.org http://www.haskell.org/mailman/listinfo/haskell-cafe