Let $S$ be a set of $n$ points in the plane. We present several different algorithms for finding a pair of points in $S$ such that any disk that contains that pair must contain at least $cn$ points of $S$, for some constant $c>0$. The first is a randomized algorithm that finds a pair in $O(n\log n)$ expected time for points in general position, and $c = 1/2-\sqrt{(1+2α)/12}$, for any $0<α<1$. The second algorithm, also for points in general position, takes quadratic time, but the constant $c$ is improved to $1/2-1/{\sqrt{12}} \approx 1/4.7$. The second algorithm can also be used as a subroutine to find the pair that maximizes the number of points inside any disk that contains the pair, in $O(n^2\log n)$ time. We also consider variants of the problem. When the set $S$ is in convex position, we present an algorithm that finds in linear time a pair of points such that any disk through them contains at least $n/3$ points of $ S $. For the variant where we are only interested in finding a pair such that the diametral disk of that pair contains many points, we also have a linear-time algorithm that finds a disk with at least $n/3$ points of $S$. Finally, we present a generalization of the first two algorithms to the case where the set $S$ of points is coloured using two colours. We also consider adapting these algorithms to solve the same problems when $S$ is a set of points inside of a simple polygon $P$, with the notion of a disk replaced by that of a geodesic disk.

翻译：设 $S$ 为平面上的 $n$ 个点集。我们提出了几种不同的算法，用于在 $S$ 中寻找一对点，使得任何包含该点对的圆盘必定包含至少 $cn$ 个 $S$ 中的点，其中 $c>0$ 为常数。第一种是随机化算法，对于一般位置的点集，在 $O(n\log n)$ 的期望时间内找到这样一对点，且 $c = 1/2-\sqrt{(1+2α)/12}$，其中 $0<α<1$。第二种算法同样针对一般位置的点集，需要二次时间，但常数 $c$ 改进为 $1/2-1/{\sqrt{12}} \approx 1/4.7$。第二种算法也可用作子程序，在 $O(n^2\log n)$ 时间内找到使得包含该点对的任意圆盘内点数最大化的点对。我们还考虑了该问题的若干变体。当点集 $S$ 处于凸位置时，我们提出了一种线性时间算法，用于找到一对点，使得通过它们的任意圆盘至少包含 $n/3$ 个 $S$ 中的点。对于仅关注寻找一对点，使得该点对的直径圆盘包含多个点的变体，我们同样给出了一个线性时间算法，能够找到一个至少包含 $n/3$ 个 $S$ 中点的圆盘。最后，我们将前两种算法推广到点集 $S$ 被两种颜色着色的情况。我们还考虑了如何调整这些算法，以解决当 $S$ 是简单多边形 $P$ 内部的点集，且圆盘概念被测地圆盘替代时的相同问题。