Linear Time Algorithm for Isomorphism of Planar Graphs (Preliminary Report)

Abstract

The isomorphism problem for graphs G<subscrpt>1</subscrpt> and G<subscrpt>2</subscrpt> is to determine if there exists a one-to-one mapping of the vertices of G<subscrpt>1</subscrpt> onto the vertices of G<subscrpt>2</subscrpt> such that two vertices of G<subscrpt>1</subscrpt> are adjacent if and only if their images in G<subscrpt>2</subscrpt> are adjacent. In addition to determining the existence of such an isomorphism, it is useful to be able to produce an isomorphism-inducing mapping in the case where one exists. The isomorphism problem for triconnected planar graphs is particularly simple since a triconnected planar graph has a unique embedding on a sphere [6]. Weinberg [5] exploited this fact in developing an algorithm for testing isomorphism of triconnected planar graphs in O(|V|<supscrpt>2</supscrpt>) time where V is the set consisting of the vertices of both graphs. The result has been extended to arbitrary planar graphs and improved to O(|V|log|V|) steps by Hopcroft and Tarjan [2,3]. In this paper, the time bound for planar graph isomorphism is improved to O(|V|). In addition to determining the isomorphism of two planar graphs, the algorithm can be easily extended to partition a set of planar graphs into equivalence classes of isomorphic graphs in time linear in the total number of vertices in all graphs in the set. A random access model of computation (see Cook [1]) is assumed. Although the proposed algorithm has a linear asymptotic growth rate, at the present stage of development it appears to be inefficient on account of a rather large constant. This paper is intended only to establish the existence of a linear algorithm which subsequent work might make truly efficient.

DOI: 10.1145/800119.803896
02040'76'79'83'87'91'95'99'03'07'11'15
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@inproceedings{Hopcroft1974LinearTA, title={Linear Time Algorithm for Isomorphism of Planar Graphs (Preliminary Report)}, author={John E. Hopcroft and J. K. Wong}, booktitle={STOC}, year={1974} }