I have a set S of n points in dimension d for which I can calculate all pairwise distances if need be. I need to select k points in this set so that t
How about this greedy algorithm:
Lets call the greatest distance between any 2 point D, for each point we add to the solution, we add at least D due to the triangle inequality. so the solution will be at least (k-1)*D, while any solution will have (k-1)^2 distances, none of them exceeds D, so at the worse case you get a solution k times the optimum.
I'm not sure this is the tightest bound that can be proved for this heuristic.