Matthew Flaschen escribió:
Simetrical wrote:
Not as important as intersections, no. But while I've been bitten for spouting off about stuff I'm ignorant of before, surely calculating a union would be cheaper than calculating an intersection anyway?
I don't think so. Basically, the way you do a union is make a list including all the articles in both categories (including duplicates). Then, you remove...the intersection.
Matthew Flaschen
You don't probably need the intersection. If you're giving out the results sorted (as categories are doing), the merge algorithm would remove duplicates 'on the way'. If you have 2 categories of 1000 articles, it's much _easier_ to 'throw' elements of any of them to fill 200 articles, than searching for the (little) ones which may be on both.