Trying to understand this better I also came across http://groups.google.de/group/fa.haskell/browse_thread/thread/1345c49faff859... where apfulmus gives an implementation of mergesort, which he claims "runs in O(n) time instead of the expected O(n log n)" Does that mean you can have the k minima in O(n) time, where n is length of list, which would seem to be an improvement? mergesort [] = [] mergesort xs = foldtree1 merge $ map return xs foldtree1 f [x] = x foldtree1 f xs = foldtree1 f $ pairs xs where pairs [] = [] pairs [x] = [x] pairs (x:x':xs) = f x x' : pairs xs merge [] ys = ys merge xs [] = xs merge (x:xs) (y:ys) = if x <= y then x:merge xs (y:ys) else y:merge (x:xs) ys 2007/4/13, Thomas Hartman <tphyahoo@gmail.com>:
And for reference, here is again stefan's "stable" quicksort from his earlier post.
" sort [] = [] sort l@(x:_) = filter (<x) l ++ filter (==x) l ++ filter (>x) l
(A stable quicksort, btw) "
This is the code whose legitimacy I am requesting confirmation of.
2007/4/13, Thomas Hartman <tphyahoo@gmail.com>:
You may be missing a few recursive calls there :-)
Indeed.
I'm confused.
Is this a legitimate stable quicksort, or not? (My guess is, it is indeed legit as written.)
This was also the first I have heard of stability as a sort property.
http://perldoc.perl.org/sort.html may shed some light on this...
"A stable sort means that for records that compare equal, the original input ordering is preserved. Mergesort is stable, quicksort is not. "
Is this description a fair characterization of stability for the current discussion?
I'm a bit confused but if I understand correctly sort from the prelude is non stable quicksort, which has O(k n^2) as the worst case, whereas stable quicksort has O( k* log n + n).
non-stable quicksort is just sort from the prelude:
qsort [] = [] qsort (x:xs) = qsort (filter (< x) xs) ++ [x] ++ qsort (filter (>= x) xs)
If any in the above was incorrect, please holler.
2007/4/12, Stefan O'Rear <stefanor@cox.net>:
On Wed, Apr 11, 2007 at 09:20:12PM -0700, Tim Chevalier wrote:
On 4/11/07, Stefan O'Rear <stefanor@cox.net> wrote:
If you want to be really explicit about it, here is a sort that will work:
sort [] = [] sort l@(x:_) = filter (<x) l ++ filter (==x) l ++ filter (>x) l
(A stable quicksort, btw)
You may be missing a few recursive calls there :-)
Indeed.
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