Let G= (V, E) be an undirected graph. An edge e The converse is not true (articulation points do not imply bridges)
However, I claim <x,y> = e is a bridge iff e is on no cycle of G. (1) Consider the validity of this statement (2) consider the algorithm
An Euler partition of a graph G = (V, E) is a partition of E into a set of edge disjoint paths and circuits such that every node of odd degree is at the endpoint of exactly one path. Write an O(|V| + |E|) algorithm to compute the Euler partition.
Example: Observations: <x,y> = e is a bridge => x OR y are articulation point(s) except for base case (only one edge)


Assuming the problem is solvable (i.e., that there exists an Euler Partition), then all paths must start with an odd node - or become a loop. Capitalize on this...
Uses: I found links to these when I did google on Euler Partitions
Using euler partitions to edge color bipartite multigraphs and
here too
Resources say similar idea used to do:
Avoiding memory contention on tightly coupled multiprocessors
Cycles in Wavelength Routed Optical Networks - Lightwave