# Time complexity of counting triangles in planar graphs

Counting triangles in general graphs can be done trivially in $O(n^3)$ time and I think that doing much faster is hard (references welcome). What about planar graphs? The following straightforward procedure shows that it can be done in $O(n\log{n})$ time. My question is two-fold:

• What is a reference for this procedure?
• Can the time be made linear?

From the algorithmic proof of Lipton-Tarjan's planar separator theorem we can, in time linear in the size of the graph, find a partition of vertices of the graph into three sets $A,B,S$ such that there are no edges with one endpoint in $A$ and the other in $B$, $S$ has size bounded by $O(\sqrt{n})$ and both $A,B$ have sizes upper bounded by $\frac{2}{3}$ of the number of vertices. Notice that any triangle in the graph either lies entirely inside $A$ or entirely inside $B$ or uses at least one vertex of $S$ with the other two vertices from $A \cup S$ or both from $B \cup S$. Thus it suffices to count the number of triangles in the graph on $S$ and the neighbours of $S$ in $A$ (and similarly for $B$). Notice that $S$ and its $A$-neighbours induce a $k$-outer planar graph (the said graph is a subgraph of a planar graph of diameter $4$). Thus counting the number of triangles in such a graph can be done directly by dynamic programming or by an application of Courcelle's theorem (I know for sure that such a counting version exists in the Logspace world by Elberfeld et al and am guessing that it also exists in the linear time world) since forming an undirected triangle is an $\mathsf{MSO}_1$ property and since a bounded width tree decomposition is easy to obtain from an embedded $k$-outer planar graph.

Thus we have reduced the problem to a pair of problems which are each a constant fraction smaller at the expense of a linear time procedure.

Notice that the procedure can be extended to find the count of the number of instances of any fixed connected graph inside an input graph in $O(n\log{n})$ time.

• You can count triangles in general graphs by taking the adjacency matrix $A$ and computing $tr(A^3)/6$. This takes $n^{\omega}$ time, where $\omega < 2.373$ is the matrix multiplication exponent. – Ryan Williams Mar 16 '15 at 16:03
• @RyanWilliams You are correct, of course! I will update the question. – SamiD Mar 16 '15 at 16:40