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in reply to RE: RE: RE: Shift, Pop, Unshift and Push with Impunity!
in thread Shift, Pop, Unshift and Push with Impunity!

It depends on how qsort is implemented by the C stdlib library with which perl was compiled. My guess is no, but it really depends on how your C stdlib was implemented. It is easy to add a "quicksort worst-case avoider" by not using a "use the first element as the pivot" and instead doing something like: Since not everyone knows the internals of quicksort, there is a worst case performance of O(n^2) with quicksort if the worst pivot is picked for each iteration (if you don't know what a pivot is don;t worry.... if you want to know I can explain it. This worst-case performance can happen if the list is already in is in sorted order and the pivot is picked by choosing the first element of the list as the pivot. However, there are techniques for easily avoiding this pitfall.

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Re: Shift, Pop, Unshift and Push with Impunity!
by jonadab (Parson) on Feb 04, 2004 at 05:56 UTC
    This worst-case performance can happen if the list is already in is in sorted order and the pivot is picked by choosing the first element of the list as the pivot.

    In my data structures and algorithm analysis class, they told us to select one of the elements at random to use as the pivot. This adds a small amount of overhead (the amount of time needed to pick a random number each iteration) to the average-case scenerio, but it basically eliminates the worst-case scenerio, effectively transforming it into an average-case scenerio. "random number" here can be anything that can pass as random. If your system clock has good enough precision, you can just grab that. The key thing is that you won't be picking the same element every iteration -- sometimes an early element, sometimes a late one, sometimes a middle one. So it makes no real difference how the list is sorted initially.

    This is of course all moot now; these days we just use Perl's built-in sort.


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