[Haskell-cafe] excercise - a completely lazy sorting algorithm
Mads Lindstrøm
mads_lindstroem at yahoo.dk
Mon Jul 6 17:50:27 EDT 2009
Hi Petr,
Maybe this will give inspiration
http://en.wikipedia.org/wiki/Selection_algorithm
It seems to me, that you just need a selection algorithm which works in
O(n * k) time for k arbitrary elements. If you combine O(n*k) selection
algorithm with any O(n * lg n) sort, you furfil your time constrain.
Regards,
Mads
> Hi all,
>
> about a month ago, we were discussing sorting in Haskell with a friend. We
> realized a nice property of lazy merge-sort (which is AFAIK the implementation
> of Data.List.sort): Getting the first element of the sorted list actually
> requires O(n) time, not O(n * log n) as in imperative languages. And
> immediately we asked: Would it be possible to create a lazy selection/sorting
> algorithm so that getting any element of the sorted list/array by its index
> would require just O(n) time, and getting all the elements would still be in
> O(n * log n)?
>
> More precisely: The result of sorting an n-element list or array should be a
> structure that allows to ask i-th element of the result (for example, a lazy
> array). Asking k arbitrary elements should take no more than
> O(min(n * log n, k * n))
>
> I believe that this could be done, but so far I wasn't able to implement and
> show it myself. I think the solution would be somewhat modified lazy quicksort
> (or "Median of Medians algorithm" if we want to be sure that we get good
> pivots).
>
> Perhaps somebody else would also like to give it a try? Or perhaps explain me
> why it's not possible?
>
> Best regards,
> Petr
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