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Nov 11, 2011 at 18:07 comment added Suresh Venkat Not sure yet. thinking out loud. But it's an excellent question.
Nov 11, 2011 at 18:05 comment added jkff In a sense yes, but that's far from being the complete answer — eg it doesn't give bisection for the 1d or 2d cases :) what do you suggest to do with the roots?
Nov 11, 2011 at 16:20 comment added Suresh Venkat I see. So actually in your case you'd like a decomposition into the ideals and then examine the roots
Nov 11, 2011 at 15:08 comment added jkff Well - I'm talking about the complexity in terms of the number of evaluations of predicate P, not the comparison predicate.
Nov 11, 2011 at 14:29 comment added Suresh Venkat But that's the point. Without data structures you can't get log n even for s totally ordered set, because all you can do is scan. It's actually a really nice question to try and find a BST equivalent.
Nov 11, 2011 at 5:46 comment added jkff Heh, sounds not too inspiring compared to log(n) :) but thanks anyways!
Nov 11, 2011 at 4:58 history answered Suresh Venkat CC BY-SA 3.0