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RE: RE: RE: Shift, Pop, Unshift and Push with Impunity!

by Eugene (Scribe)
 on Jun 15, 2000 at 09:22 UTC ( #18249=note: print w/replies, xml ) Need Help??

Does that mean that pre-sorted lists are sorted slowly, or is it being randomized before the actual sort?
• Comment on RE: RE: RE: Shift, Pop, Unshift and Push with Impunity!

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RE: RE: RE: RE: Shift, Pop, Unshift and Push with Impunity!
by lhoward (Vicar) on Jun 15, 2000 at 15:13 UTC
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:
• adding a "sorted list detector"
• Picking the pivot randomly (instead of as the first element)
• Shuffling the list before sorting
• Using another pivot picking technique
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.
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.

```\$;=sub{\$/};@;=map{my(\$a,\$b)=(\$_,\$;);\$;=sub{\$a.\$b->()}}
split//,".rekcah lreP rehtona tsuJ";\$\=\$ ;->();print\$/

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