Recognizing Subgraphs of Regular Tilings

Abstract

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.

Figures (4)

• Figure 1: A plane graph $G$ and a sphere cut separator $S$. The vertex separator corresponding to $S$ has size $1$, however the geometric curve $S$ is much more complex.
• Figure 2: (i) The hyperbolic $\{5,4\}$-tiling $\mathcal{T}_{5,4}$ in the Poincaré disk model. All angles are equal and all edges are (straight) line segments of the same hyperbolic length. The Poincaré disk model preserves angles, but distorts distances. (ii) Three nooses in a sphere cut decomposition of a graph $H$. The noose $\nu_L$ and $\nu_R$ are child nooses of the noose $\nu_P$. The region $\mathrm{Reg}(\nu_P)$ contains $8$ vertices of $H$, and the regions $\mathrm{Reg}(\nu_L)$ and $\mathrm{Reg}(\nu_R)$ contain $3$ and $6$ vertices of $H$, respectively.
• Figure 3: Each figure depicts a convex plane graph $H$ embedded in $T_{5,4}$ (in black) as well as: (i) a normalized noose $\nu_1$ of $H$, such that $\nu_1=\nu_1^-$, (ii) a normalized noose $\nu_2$ of $H$ with pre-rays from two intersecting rays (the continuations of the rays are dashed), (iii) a normalized noose $\nu_3$ with two ideal vertices $i_1$ and $i_2$, (iv) two child nooses $\nu_L,\nu_R$ of $\nu_3$.
• Figure 4: The two candidate nooses depicted here are equivalent as one can be obtained from the other via an orientation-preserving isometry.

Theorems & Definitions (23)

• Theorem 1
• Theorem 2
• Lemma 3: Lemma 3.1 of KisfaludiBak25
• Lemma 4: Lemma 3.7 of KisfaludiBak25
• Lemma 5: Lemma 4.3 of KisfaludiBak25
• Lemma 5: Corollary of Lemma 5.5 of KisfaludiBak25, $\star$
• Lemma 5: $\star$
• Lemma 5
• Lemma 6: Theorem 1.3 of KisfaludiBak25
• Lemma 6