For $p,q\ge2$ the $\{p,q\}$-tiling graph is the (finite or infinite) planar graph $T_{p,q}$ where all faces are cycles of length $p$ and all vertices have degree $q$. We give algorithms for the problem of recognizing (induced) subgraphs of these graphs, as follows. - For $1/p+1/q>1/2$, these graphs correspond to regular tilings of the sphere. These graphs are finite, thus recognizing their (induced) subgraphs can be done in constant time. - For $1/p+1/q=1/2$, these graphs correspond to regular tilings of the Euclidean plane. For the Euclidean square grid $T_{4,4}$ Bhatt and Cosmadakis (IPL'87) showed that recognizing subgraphs is NP-hard, even if the input graph is a tree. We show that a simple divide-and conquer algorithm achieves a subexponential running time in all Euclidean tilings, and we observe that there is an almost matching lower bound in $T_{4,4}$ under the Exponential Time Hypothesis via known reductions. - For $1/p+1/q<1/2$, these graphs correspond to regular tilings of the hyperbolic plane. As our main contribution, we show that deciding if an $n$-vertex graph is isomorphic to a subgraph of the tiling $T_{p,q}$ can be done in quasi-polynomial ($n^{O(\log n)}$) time for any fixed $q$. Our results for the hyperbolic case show that it has significantly lower complexity than the Euclidean variant, and it is unlikely to be NP-hard. The Euclidean results also suggest that the problem can be maximally hard even if the graph in question is a tree. Consequently, the known treewidth bounds for subgraphs of hyperbolic tilings do not lead to an efficient algorithm by themselves. Instead, we use convex hulls within the tiling graph, which have several desirable properties in hyperbolic tilings. Our key technical insight is that planar subgraph isomorphism can be computed via a dynamic program that builds a sphere cut decomposition of a solution subgraph's convex hull.

翻译：对于 $p,q\ge2$，$\{p,q\}$-铺砌图是指所有面均为长度为 $p$ 的环且所有顶点度数为 $q$ 的（有限或无限）平面图 $T_{p,q}$。我们针对识别这些图的（诱导）子图问题给出以下算法。 - 当 $1/p+1/q>1/2$ 时，这些图对应于球面的规则铺砌。这些图是有限的，因此识别其（诱导）子图可在常数时间内完成。 - 当 $1/p+1/q=1/2$ 时，这些图对应于欧几里得平面的规则铺砌。对于欧几里得方形网格 $T_{4,4}$，Bhatt 和 Cosmadakis (IPL'87) 证明了识别子图是 NP 难的，即使输入图是一棵树。我们证明了一个简单的分治算法在所有欧几里得铺砌中可实现亚指数运行时间，并且我们观察到，在指数时间假说下，通过已知规约，在 $T_{4,4}$ 中存在一个几乎匹配的下界。 - 当 $1/p+1/q<1/2$ 时，这些图对应于双曲平面的规则铺砌。作为我们的主要贡献，我们证明了对于任意固定的 $q$，判定一个 $n$ 顶点图是否同构于铺砌 $T_{p,q}$ 的一个子图可以在拟多项式（$n^{O(\log n)}$）时间内完成。我们对双曲情形的结果表明，其复杂度显著低于欧几里得变体，并且不太可能是 NP 难的。欧几里得结果也表明，即使所讨论的图是一棵树，该问题也可能是最大难度的。因此，已知的双曲铺砌子图的树宽界本身并不能导出一个高效算法。相反，我们利用铺砌图内的凸包，其在双曲铺砌中具有若干理想性质。我们的关键技术见解是，平面子图同构可以通过一个动态规划来计算，该规划构建了解决方案子图凸包的球面割分解。