Optimal Offline Dynamic 2, 3-Edge/Vertex Connectivity

Abstract

We give offline algorithms for processing a sequence of 2- and 3-edge and vertex connectivity queries in a fully-dynamic undirected graph. While the current best fully-dynamic online data structures for 3-edge and 3-vertex connectivity require $O(n^{2/3})$ and O(n) time per update, respectively, our per-operation cost is only $O(\log n)$, optimal due to the dynamic connectivity lower bound of Patrascu and Demaine. Our approach utilizes a divide and conquer scheme that transforms a graph into smaller equivalents that preserve connectivity information. This construction of equivalents is closely-related to the development of vertex sparsifiers, and shares important connections to several upcoming results in dynamic graph data structures, including online models.

A Omitted Proofs

Proof

(Proof of Lemma 10). We prove by contradiction. Let G be a graph with no nontrivial 3-edge connected component. Suppose there exists two simple cycles a and b in G with more than one vertex, and thus at least one edge, in common.

Call the vertices in the first simple cycle $a_1, \ldots , a_n$ and the second simple cycle $b_1, \ldots , b_m$, in order along the cycle.

Since these cycles are not the same, there must be some vertex not common to both cycles. Without loss of generality, assume (by flipping a and b) that b is not a subset of a, and (by shifting b cyclically) that $b_1$ is only in b and not a.

Now let $b_{first}$ be the first vertex after $b_1$ in b that is common to both cycles, so

(1)

and let $b_{last}$ be the last vertex in b common to both cycles

(2)

The assumption that these two cycles have more than 1 vertex in common means that

$$\begin{aligned} first < last. \end{aligned}$$

(3)

We claim $b_{first}$ and $b_{last}$ are 3-edge connected.

We show this by constructing three edge-disjoint paths connecting $b_{first}$ and $b_{last}$. Since both $b_{first}$ and $b_{last}$ occur in a, we may take the two paths formed by cycle a connecting $b_{first}$ and $b_{last}$, which are clearly edge-disjoint.

By construction, vertices

$$\begin{aligned} b_{last+1}, \ldots , b_m, b_1, \ldots , b_{first-1} \end{aligned}$$

(4)

are not shared with a. Thus they form a third edge-disjoint path connecting $b_{first}$ and $b_{last}$, and so the claim follows. Therefore, a graph with no 3-edge connected vertices, and thus no nontrivial 3-edge connected component has the property that two simple cycles have at most one vertex in common.