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....
View
Full Document
 Summer '10
 any
 Graph Theory, Glossary of graph theory, Graph theory objects, Eulerian path, Hamiltonian path, Petersen, Cubic graph

Click to edit the document details