Let $\mathcal{P} = \{P_1, \cdots,P_n\}$ be a family of finite point sets in $\mathbb{R}^d$, each having at most $m$ points. Consider the following objective function

\begin{align} cost(\mathcal{P},c) = \sum_{i\in[n]}\max_{p\in P_i}\Vert p-c\Vert^2 \end{align}

Let $c^\star$ be the point which minimizes the above objective function and the optimal cost is $opt(\mathcal{P}) = cost(\mathcal{P},c^\star)$.

For a point $c\in\mathbb{R}^d$ and $i \in [n]$, let $p_i^{(c)}$ be the point of $P_i$ farthest from $c$, i.,e. \begin{align*} \Vert p_i^{(c)} -c\Vert^2 = \max_{p\in P_i}\Vert p-c\Vert^2 \end{align*} Define $c$-mean to be the mean of these farthest points \begin{align*} \mu^{(c)} = \frac{1}{n}\sum_{i=1}^n p_i^{(c)} \end{align*}

So we have \begin{align*} cost_2(\mathcal{P},c) = \sum_{i=1}^n \Vert p_i^{(c)} -c\Vert^2 \end{align*} For ease of notation, let $p_i^\star = p_i^{(c^\star)}$ and $\mu^\star = \mu^{(c^\star)}$. So \begin{align*} opt_2(\mathcal{P}) = \sum_{i=1}^n \Vert p_i^\star - c^\star\Vert^2 \end{align*}

Note that if every set in $\mathcal{P}$ is singleton, then this is the well-known $k$-means objective and $c^\star = \mu^\star$.

It is easy to show the following, for any $c\in \mathbb{R}^d$ \begin{align} cost(\mathcal{P},c) \leq opt(\mathcal{P}) + n\Vert c - \mu^{(c)}\Vert^2 \end{align}

It follows that if there is a point $c$ such that $c = \mu^{(c)}$, then $c$ is an optimum center. This already helps to find the optimal solution in some simple scenarios, such as $\mathcal{P} = \{P_1,P_2\}$, where $P_1=\{(1,1),(1,-1)\}$ and $P_2=\{(-1,1),(-1-1)\}$. However it seems to me that the reverse direction may not hold: meaning for any optimal center $c^\star$, it may not be the case that $c^\star = \mu^{(c^\star)}$. So far I have not been able to come up with a counter example.

Does anyone have some idea as to finding a counter example.

P.s. Is this objective function already known?

  • 1
    $\begingroup$ Yes, this objective function is known. Check Definition 1.1. of this paper. Yours is a special case for $k = 1$, $p = \infty$, and $q = 1$. $\endgroup$ Commented Dec 1, 2021 at 17:16
  • $\begingroup$ @InuyashaYagami thanks for the reference. I have seen this paper but did not read it. $\endgroup$ Commented Dec 2, 2021 at 3:33

1 Answer 1


I have not verified the forward direction. However, here is a counterexample for reverse direction: $\mathcal{P} = \{ P_1, P_2,P_3\}$, where $P_{1} = \{(-1,0)\}$, $P_{2} = \{(1,0)\}$, and $P_{3} = \{(0,1),(0,-1)\}$. Here $c^{*} = (0,0)$. However, $p_{1}^{*} = (-1,0)$, $p_{2}^{*} = (1,0)$, and $p_{3}^{*} = (0,1)$ or $(0,-1)$ that have mean either $(0,1/3)$ or $(0,-1/3)$

  • $\begingroup$ Why is origin is the optimal center? $\endgroup$ Commented Dec 2, 2021 at 15:17
  • $\begingroup$ @SudiptaRoy Let $(a,b)$ be the center. Then, the cost is $(a-1)^2 + (a+1)^2 + 2b^2 + \max\{a^2+(b-1)^2, a^2+(b+1)^2\}$. Solve it. The function is minimized at the origin. $\endgroup$ Commented Dec 2, 2021 at 15:28
  • $\begingroup$ After my reply, I was thinking about this approach. Thank you $\endgroup$ Commented Dec 2, 2021 at 15:31

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