This is a classic greedy algorithm, much like the text-wrapping problem. My main suggestion would be that you're not making use of some standard list functions that would simplify things. For example, your runningSum is just scanl1 (+) . Similarly, splitAll should use unfoldr. Another thing is that I would reverse the order of arguments of splitFirst and splitAll, since curried applications are probably more useful that way: splitAll :: (Real a) => a -> [a] -> [[a]] splitAll = unfoldr . split where split _ [] = Nothing split n xs = let (ys,zs) = break ((> n) . snd) (zip xs (scanl1 (+) xs)) in Just (map fst ys, map fst zs) Now, if you're concerned about all that zipping and projecting, you can instead define split via a straightforward recursion, or you could use a different kind of unfold that preserves the terminating value: unfoldrG :: (b -> Either (a,b) b) -> b -> ([a],b) unfoldrG f = unfold where unfold x = case f x of Right y -> ([],y) Left (a,y) -> let (bs,z) = unfold y in (a:bs,z) Here, you will define split by unfolding a pair consisting of a running sum and remaining list. Cheers, --Joe On 2004.04.21 07:42, Steve Schafer wrote:
I have a list of integers, e.g.:
[1,5,3,17,8,9]
I want to split it into a pair of lists, with the criterion being that the sum of the elements in the first list is as large as possible, but not exceeding a threshold value. For example, if the threshold is 10, the result should be:
([1,5,3],[17,8,9])
and then I want to recursively apply this process to the remainder of the list, with the end result being a list of lists of integers. Using the same list along with a threshold of 18, I would get:
[[1,5,3],[17],[8,9]]
I have devised a means of doing this:
1) Create an auxiliary list of integers, where the n'th element is equal to the sum of the first n elements of the original list.
2) Zip the auxiliary list with the original list.
3) Use span to break the list in two according to the threshold.
4) Unzip the two resulting lists and discard the auxiliary portions.
5) Repeat from step 1, operating on the tail of the list, until there's nothing left.
Here's the code that implements this:
runningSum :: (Ord a, Num a) => [a] -> [a] runningSum [] = [] runningSum (i:[]) = i : [] runningSum (i:j:js) = i : runningSum (i+j : js)
zipWithSum :: (Ord a, Num a) => [a] -> [(a,a)] zipWithSum xs = zip (runningSum xs) xs
threshold :: (Ord a, Num a) => [a] -> a -> ([(a,a)],[(a,a)]) threshold xs t = let test x = (t >= (fst x)) in span test (zipWithSum xs)
splitFirst :: (Ord a, Num a) => [a] -> a -> ([a],[a]) splitFirst xs t = let (ys,zs) = threshold xs t in (snd (unzip ys), snd (unzip zs))
splitAll :: (Ord a, Num a) => [a] -> a -> [[a]] splitAll [] _ = [] splitAll xs t = let (ys, zs) = splitFirst xs t in ys : (splitAll zs t)
(One thing that's missing from this code is a check to verify that no single element in the list is greater than the threshold, which should raise an error, rather than get stuck in an infinite loop.)
The algorithm as implemented works fine, but it seems overly complicated and not very elegant. I get the feeling that I'm missing some obvious simplification, but I can't find it. Any ideas?
Thanks,
-Steve Schafer
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