Unformatted text preview: n vertices has n - 1 edges, such a tree has n - 1 bridges. 5. [BB] No, it cannot. Every edge of an Eulerian graph is part of a circuit; the removal of an edge of a circuit certainly does not disconnect a graph. 6. [BB; (----+)] (----+) If e is a bridge, we can let u and v be its ends. If there were a path from u to v which did not require e, together with e, we would have a circuit containing e. Since the deletion of an edge which is part of a circuit cannot disconnect a graph, we have a contradiction. ( f--) Conversely, if e is not a bridge, then e is part of a circuit, so there is a path between its ends which avoids e. This means that any path between vertices which requires e can be replaced with a path which omits e, so vertices u and v as specified in the question could not exist....
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- Summer '10
- Graph Theory, Glossary of graph theory, Graph theory objects, Eulerian path, Hamiltonian path, Petersen, Cubic graph