In a paper I am reviewing, the authors define the following problem and construct an algorithm. They give no further references and I suspect it has appeared somewhere in the literature before.

Let $P_1,\ldots,P_n$ be sets of points in a space with a metric $d$. Define the distance to a set as $$\operatorname{dist}(x, P_i) = \min\{ d(x,p) \,|\, x \in P_i\}.$$ The problem is to compute $$\min_x \sum_{i=1}^n \operatorname{dist}(x, P_i).$$ What is the name of this problem? Has it been discussed in the literature? What if $d(x,P_i) = ||x - P_i||_2?$

  • $\begingroup$ I don't know of any references to that specific formula, but wouldn't this be covered under optimization/minimization problems? $\endgroup$
    – Ronny
    Commented Jan 2, 2013 at 17:23
  • 3
    $\begingroup$ It seems like a generalized 1-median problem. $\endgroup$ Commented Jan 2, 2013 at 17:42
  • $\begingroup$ The 1-median problem would be to minimize sum_i sum_{p in P_i} d(x, p), right? Has the generalized version appeared anywhere? $\endgroup$
    – Ben
    Commented Jan 2, 2013 at 18:57
  • $\begingroup$ Not that I'm aware of. Of course it inherits all of the hardness of the 1-median since all the sets can be singletons. It should also admit the 2-approximation that comes from choosing one of the input points as the center. $\endgroup$ Commented Jan 6, 2013 at 6:53

1 Answer 1


It is called p-Median and is known to be NP-hard. Refer to the following paper for more info:

O. Kariv and S. Hakimi, “An Algorithmic Approach to Network Location Problems. II: The p-MEDIANS,” SIAM Journal on Applied Mathematics, vol. 37, no. 3, pp. 539–560, December 1979.

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    $\begingroup$ That seems to be about the special case where the sets $P_i$ consist of every one of the $\binom{n}{p}$ subsets of $p$ points of the input, and the NP-hardness result in that paper requires $p$ to be non-constant. So this doesn't appear to say much about the hardness of the problem asked here in which the $P_i$ are specified, because expanding out all of the $p$-subsets into an explicit list of sets is not a polynomial-time reduction. $\endgroup$ Commented Jan 6, 2013 at 2:02

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