Abstract
We give a distributed algorithm in the CONGEST model for property testing of planarity with onesided error in general (unboundeddegree) graphs. Following CensorHillel et al. (Proceedings of the 30th International Symposium on Distributed Computing, pp. 43–56, 2016), who recently initiated the study of property testing in the distributed setting, our algorithm gives the following guarantee: For a graph \(G = (V,E)\) and a distance parameter \(\epsilon \), if G is planar, then every node outputs accept, and if G is \(\epsilon \)far from being planar (i.e., more than \(\epsilon \cdot E\) edges need to be removed in order to make G planar), then with probability \(11/\mathrm{poly}(n)\) at least one node outputs reject. The algorithm runs in \(O(\log V\cdot \mathrm{poly}(1/\epsilon ))\) rounds, and we show that this result is tight in terms of the dependence on V. Our algorithm combines several techniques of graph partitioning and local verification of planar embeddings. Furthermore, we show how a main subroutine in our algorithm can be applied to derive additional results for property testing of cyclefreeness and bipartiteness, as well as the construction of spanners, in minorfree (unweighted) graphs.
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Notes
We note that Ghaffari and Haeupler [23] consider a different, but related question of finding a planar embedding of a planar graph. They give a distributed algorithm for this problem using \(O(D\cdot \min \{\log n,D\})\) rounds (where D is the diameter and n is the number of nodes).
Observe that if the algorithm has onesided error and a constant error probability, then its error probability can be reduced to \(\delta \), for a given parameter \(\delta \), at a multiplicative cost of \(\log (1/\delta )\) in the number of rounds. We are able to obtain error probability \(1/\mathrm{poly}(n)\) without this extra cost.
Recall that a graph H is a minor of a graph G if H is isomorphic to a graph that can be obtained by zero or more edge contractions on a subgraph of G. We say that a graph G is Hminor free (or excludes H as a minor) if H is not a minor of G. We say that G is “minorfree” if it is Hminor free for some fixed H of constant size.
A spanner of a graph G is a (sparse) subgraph of G that maintains distances up to a multiplicative factor, s, which is called the stretch factor, and the spanner is referred to as an sspanner.
It was communicated to us by one of the authors of [23] that their algorithm can be modified so as to detect if the underlying graph is not planar [28]. For the sake of a selfcontained presentation, we rely on the version of the algorithm as provided in [23] (which works under the promise that the graph is planar), and check that the ordering computed by [23] is consistent with a planar embedding.
We note that a previous version of the definition of this condition, given in the conference version of this paper [39], contained an error, which is fixed in the current version.
Note that for any subgraph H of \(G^j\), the ordering \({\tilde{\tau }}\) defines a combinatorial embedding of H as well.
The reason that we use the counterclockwise order rather than the clockwise order is so that the edges in the tree will be in the standard, lefttoright order.
That is, for two (different) nodes u and v, let \(\ell (u) = \sigma _1,\dots ,\sigma _p\) and \(\ell (v) = \sigma '_1,\dots ,\sigma '_q\), where without loss of generality, \(p \le q\) (and for \(u = r^j\) we have \(p=0\)). Let i be the maximum index such that \(\sigma _1,\dots ,\sigma _i = \sigma '_1,\dots ,\sigma '_i\), where if no such index exists, then \(i = 0\). If \(i=p\), then \(\ell (u) < \ell (v)\). Otherwise, \(\ell (u) < \ell (v)\) if \(\sigma _{i+1}<\sigma '_{i+1}\), and \(\ell (u) > \ell (v)\) otherwise.
To be precise, the graphs in the lowerbound construction of CensorHillel have a constant degree, while the graphs in our lower bound construction do not necessarily have a constant degree. However, we can easily modify the construction so that the graphs have a constant degree, in the same manner as in [6] [Theorem 7.3].
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Acknowledgements
We would like to thank Mohsen Ghaffari and Merav Parter for helpful information. We would also like to thank the anonymous PODC reviewers and this journal’s reviewers for their helpful comments.
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This article extends work presented at PODC 2018 [39]. Reut Levi is partially supported by the European Research Council (ERC) under the European Union’s Horizon 2020 research and innovation programme (Grant agreement No. 819702). Reut Levi was also supported by ERC advanced grant 834735. Moti Medina was partially supported by the Israel Science Foundation Grant No. 867/19. Dana Ron was partially supported by the Israel Science Foundation Grants No. 671/13 and 1146/18 and the Kadar family prize.
Proof of Lemma 7
Proof of Lemma 7
In order to prove Lemma 7, we introduce several notions. These notions are very similar to those defined in [Chap. 7] [15], except that that there it was assumed that the graphs in question have vertex connectivity at least two. Here we do not make this assumption.
Let H be a connected graph and let C be a simple cycle in H. Consider the connected components of the graph resulting from removing the nodes in C. For each such connected component D, let A(D) denote the subset of nodes on C that neighbor nodes in D. We refer to A(D) as the attachment nodes of D on C. Let B(D) denote the subgraph induced by nodes of D and A(D), not including edges of C. If \(A(D) \ge 2\), then we refer to B(D) as a bridge, and if \(A(D) = 1\), then it is a halfbridge. We also refer to edges between pairs of nodes on C as bridges. Two bridges B and \(B'\) are said to interlace if one of the following holds:

1.
There are two attachments of B, x and y, and two attachments of \(B'\), w and z, such that all four are distinct, and appear on C in the order (say, clockwise) x, w, y, z.

2.
There are three attachments common to B and \(B'\). That is, \(A(B) \cap A(B') \ge 3\).
The next lemma is a slight modification of Lemma 7.2 in [15] (where here the graph H in question is not assumed to have vertex connectivity at least two).
Lemma 15
Let \(B_1,\dots ,B_s\) be the set of bridges and halfbridges of a graph H with respect to a simple cycle C. Suppose that \(C+B_i\) is planar for every \(1 \le i \le s\) and that no two bridges in the set interlace. Then \(C+B_1+\dots +B_s\) can be embedded in the plane so that all the bridges and halfbridges are inside C.
Proof
We prove the claim by induction on the number of nodes in \(C+B_1+\dots +B_s\). The base case is three nodes (there is just a cycle C and no bridges). For the induction step, as in the proof of Lemma 7.2 in [15], since no two bridges interlace, there must be at least one bridge, \(B_i\) for which the following holds. If we consider the attachments of \(B_i\) in clockwise order, \(a_1,\dots ,a_k\) (for \(k\ge 2\)), then there is no other bridge \(B_j\) with attachments (strictly) after \(a_1\) and before \(a_k\) on C. \(\square \)
As a corollary of Lemma 15 we obtain:
Corollary 5
Let \(B_1,\dots ,B_s\) be the set of bridges and halfbridges of a graph H with respect to a simple cycle C. Suppose that \(C+B_i\) is planar for every \(1 \le i \le s\) and that the set of bridges can be partitioned into two subsets, such that within each subset no two bridges interlace. Then H is planar.
Building on Corollary 5 we are now ready to prove Lemma 7.
Proof of Lemma 7
Consider the bridges of H with respect to C(u, v). They can be partitioned into two pairwise noninterlacing subsets: one corresponding the bridges inside C(u, v) and one corresponding to bridges outside C(u, v) (recall that if the bridges interlace then we cannot embed them in one side of C(u, v)). Therefore H is planar. It remains to show that there exists a planar embedding of H which is consistent with \({\tilde{\tau }}\). For each vertex x in C(u, v), we have the circular order of its edges in the planar embedding of \(S_H^I(u,v)\) and \(S_H^O(u,v)\), respectively. We can merge this pair of orders into a single order that is consistent with \({\tilde{\tau }}\) simply by concatenating them (the only edges that these orders have in common are the edges on C(u, v), which are the first and last edges in both orders). Other vertices of H are either in \(S_H^I(u,v)\) or \(S_H^O(u,v)\) (but not in both), so the ordering remains consistent.
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Levi, R., Medina, M. & Ron, D. Property testing of planarity in the CONGEST model. Distrib. Comput. 34, 15–32 (2021). https://doi.org/10.1007/s00446020003823
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DOI: https://doi.org/10.1007/s00446020003823