Subject: Re: Quicksearch vs. lazyness Newsgroups: gmane.comp.lang.haskell.general Date: Saturday 14th April 2007 10:05:52 UTC (over 11 years ago) apfelmus wrote: > Steffen Mazanek wrote: >> From my understanding for small k's lazy >> evaluation already does the trick for the naive quicksort >> algorithm (quicksort smaller ++ [x] ++ quicksort greater), >> doesn't it? Is there a search algorithm that makes better >> use of lazy evaluation out of the box? > > No, as far as I thought about quicksort, it needs O(n log n) to produce > the first element. But even the mergesort implementation has to be > chosen carefully as I've seen many that still need O(n log n) just to > return the smallest element. Apparently, I did not think enough about quicksort :) as it is well capable of delivering the minimum element in O(n) time. In fact, given proper pivot choice, take k . qsort is the optimal O(n + k log k) algorithm for returning the first k smallest elements in sorted order! See also http://en.wikipedia.org/wiki/Selection_algorithm Let's assume that the quicksort in qsort [] = [] qsort (x:xs) = qsort ls ++ [x] ++ qsort rs where ls = filter (< x) xs rs = filter (>= x) xs always uses a good pivot x, i.e. that ls and rs have around n/2 elements each. As long as this n/2 is greater than k, taking the first k elements does not evaluate the second recursive call to quicksort. In other words, it takes O(n) -- for filtering xs during the initial call to qsort + O(n/2) -- same as above but with the first half of the list + O(n/4) -- filtering the half of the half + ... + O(k) ________ < O(n + n/2 + n/4 + n/8 + ...) = O(n) time until ls has fewer than k elements. When this becomes the case, the argument list to the recursive call to qsort has a size of at most 2*k and it takes at most O(k log k) time finish sorting it completely and taking the first k elements of this. This gives a total of O(n + k log k). Regards, apfelmus |
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