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Re: Question: Centroid given a distance metric
Posted:
Feb 12, 2013 4:39 PM


On Tuesday, February 12, 2013 10:26:24 AM UTC8, Nicolas Bonneel wrote: > On 2/11/2013 12:17 PM, Andrey Savov wrote: > > > Was wondering if you guys can point me in the right direction. > > > > > > Are there any known/studied methods to calculate a centroid (geometric center) of finite set of points in ndimensional real Euclidean space by only knowing a distance metric f(x,y): R^n x R^n > R ? > > > > > > > Have you tried posing it as an optimization problem: > > F(x) = \argmin \sum_i d(x, x_i)^2 > > and running any optimization method ? > > > > There won't likely be a close form solution for an arbitrary distance > > d(x,y), but if it's smooth and the dimension not too large, you can > > manage to find a global optimum. It will not necessarily be unique > > though, but should exist if d is not a strange function (like d=\infty > > everywhere etc.). > > > > > >  > > Nicolas Bonneel
I, as is often the case the distance function is convex, the sum of squares of it is also convex, so a local min will be a global min. However, the problem arises that sometimes the F(x) function is NOT smooth: the minimum mayand in practical problems, often doeslie right on top of one of the points x_i, making F nondifferentiable at the optimal solution. This does not always happen, but it does happen often enough that locationanalysis folks have to devise special algorithms to handle the problem.
I would ask: why do you want to minimize the sum of squares? For Euclidean distance, that F(x) has some physical and statistical meaning, and furthermore leads to a simple solution. However, for other norms such as d(x,y) = x+y or d(x,y) = max(x,y), or for a pnorm with 1 < p < 2, what significance can one attach to the sum of squares? Certainly it makes _some_ problems much harder instead of easier (for example, when d(x,y) = x + y).
Ray Vickson



