A disk graph is an intersection graph of disks in the Euclidean plane, where the disks correspond to the vertices of the graph and a pair of vertices are adjacent if and only if their corresponding disks intersect. The problem of determining the time complexity of computing a maximum clique in a disk graph is a long-standing open question. The problem is known to be open even when the radii of all the disks are in the interval $[1,(1+\varepsilon)]$, where $\varepsilon>0$. However, the maximum clique problem is known to be APX-hard for the intersection graphs of many other convex objects such as intersection graphs of ellipses, triangles, and a combination of unit disks and axis-parallel rectangles. Furthermore, there exists an $O(n^3\log n)$-time algorithm to compute a maximum clique for unit disks. Here we obtain the following results. - We give an algorithm to compute a maximum clique in a unit disk graph in $O(n^{2.5}\log n)$-time, which improves the previously best known running time of $O(n^3\log n)$ [Eppstein '09]. - We extend a widely used `co-2-subdivision approach' to prove that computing a maximum clique in a combination of unit disks and axis-parallel rectangles is NP-hard to approximate within $4448/4449 \approx 0.9997 $. The use of a `co-2-subdivision approach' was previously thought to be unlikely in this setting [Bonnet et al. '20]. Our result improves the previously known inapproximability factor of $7633010347/7633010348\approx 0.9999$. - We show that the parameter minimum lens width of the disk arrangement may be used to make progress in the case when disk radii are in $[1,(1+\varepsilon)]$. For example, if the minimum lens width is at least $0.265$ and $ \varepsilon\le 0.0001$, which still allows for non-Helly triples in the arrangement, then one can find a maximum clique in polynomial time.

翻译：圆盘图是欧几里得平面中圆盘的相交图，其中圆盘对应图的顶点，当且仅当对应圆盘相交时，一对顶点相邻。计算圆盘图中最大团的时间复杂度问题是一个长期未解决的公开问题。即使所有圆盘的半径均处于区间 $[1,(1+\varepsilon)]$（其中 $\varepsilon>0$）时，该问题仍被视为开放。然而，对于许多其他凸对象的相交图（如椭圆相交图、三角形相交图、以及单位圆盘与轴平行矩形的组合相交图），最大团问题已知为APX-难。此外，存在一个 $O(n^3\log n)$ 时间复杂度的算法用于计算单位圆盘的最大团。本文取得以下结果： - 我们提出一个 $O(n^{2.5}\log n)$ 时间复杂度的算法，用于计算单位圆盘图中的最大团，改进了此前最佳的 $O(n^3\log n)$ 运行时间 [Eppstein '09]。 - 我们扩展了广泛使用的“co-2-subdivision方法”，证明在单位圆盘与轴平行矩形的组合中计算最大团是NP-难的，且近似比在 $4448/4449 \approx 0.9997$ 以内不可逼近。此前认为该设置中“co-2-subdivision方法”不太可能适用 [Bonnet et al. '20]。该结果改进了此前已知的不可逼近因子 $7633010347/7633010348\approx 0.9999$。 - 我们证明圆盘排列的最小透镜宽度参数可用来推动半径区间为 $[1,(1+\varepsilon)]$ 时的研究进展。例如，若最小透镜宽度至少为 $0.265$ 且 $\varepsilon\le 0.0001$（此时排列中仍允许非Helly三元组存在），则可在多项式时间内找到最大团。