为nlogn算法提供下限

Question

The original problem was discussed in here: Algorithm to find special point k in O(n log n) time

Simply we have an algorithm that finds whether a set of points in the plane has a center of symmetry or not.

I wonder is there a way to prove a lower bound (nlogn) to this algorithm? I guess we need to use this algorithm to solve a simplier problem, such as sorting, element uniqueness, or set uniqueness, therefore we can conclude that if we can solve eg element uniqueness by using this algorithm, it can be at least nlogn.

It seems like the solution is something to do with element uniqueness, but i couldn't figure out a way to manipulate this into center of symmetry algorithm.

Answer 1

Check this paper

The idea is if we can reduce problem A to problem B, then B is no harder than A.

That said, if problem B has lower bound Ω（nlogn）, then problem A is guaranteed the same lower bound.

In the paper, the author picked the following relatively approachable problem to be B: given two sets of n real numbers, we wish to decide whether or not they are identical.

It's obvious that this introduced problem has lower bound Ω（nlogn）. Here's how the author reduced our problem at hand to the introduced problem (A, B denote the two real sets in the following context):

Answer 2

First observe that that your magical point k must be in the center.

• build a lookup data structure indexed by vector position (O(nlog n))
• calculate the centroid of the set of points (O(n))
• for each point, calculate the vector position of its opposite and check for its existence in the lookup structure (O(log n) * n)

Appropriate lookup data structures can include basically anything that allows you to look something up efficiently by content, including balanced trees, oct-trees, hash tables, etc.