Discrete Mathematics with Graph Theory (3rd Edition) 355

Discrete Mathematics with Graph Theory (3rd Edition) 355 -...

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Section 12.6 353 2. (b) Strongly connected. A strongly connected orientation is shown at the right above. (c) Not strongly connected; edge 68 is a bridge. 2 5 9 3<JOu<1 4 7 10 (d) Since the graph is not connected, a complete orientation cannot be obtained based on a depth-first search. Also, no orientation will be strongly connected. Tetrahedron [BB] Cube Octahedron Dodecahedron Icosahedron 3. (a) [BB] A strongly connected orientation for the Petersen graph is shown. (b) A Hamiltonian graph always has a strongly connected orientation. Orient the edges of a Hamilto- nian cycle in the direction of the cycle and other edges arbitrarily. (c) The converse of (b) is false. The Petersen graph is not Hamiltonian. 4. Since the removal of any edge from a tree leaves the tree disconnected and since a tree with
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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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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.

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