# 8 15 pts a if g is a nontrivial graph how is κ g the

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_________________________________________________________________ 8. (15 pts.) (a) If G is a nontrivial graph, how is κ (G), the vertex connectivity of G, defined? If G is a complete graph of order n, then κ (G) = n - 1. Otherwise, G has a vertex-cut. In this case, κ (G) = k where k is the cardinality of a minimum vertex-cut. (b) If G is a nontrivial graph, it is not true generally that if v is an arbitrary vertex of G, then either κ (G - v) = κ (G) - 1 or κ (G - v) = κ (G). Give a simple example of a connected graph G illustrating this. [A carefully labelled drawing with a brief explanation will provide an appropriate answer.] Evidently, G K 1 + (K 1 K 2 ). Since G is connected and a is a cut-vertex of G, κ (G) = 1. Note, however, G - b K 3 , and thus κ (G - b) = 2, not 0 or 1. (c) Despite the example above, if G is a nontrivial graph and v is a vertex of G, κ (G - v) κ (G) - 1. Provide the simple proof for this. Proof: Let v be an arbitrary vertex of G. If G is a complete graph of order k, then G - v is complete of order k - 1. Thus, the conclusion follows from definition of κ . So suppose G is not a complete graph. Then G - v is not a complete graph, too. Then either κ (G - v) κ (G) - 1 or κ (G - v) < κ (G) - 1. If κ (G - v) < κ (G) - 1, then there is a subset U 0 of the vertex set of G - v with U 0 = κ (G - v) and (G - v) - U 0 disconnected. But then U = U 0 {v} is a vertex-cut of G for which U = U 0 + 1 < κ (G), an impossibility. Hence we must have κ (G - v) κ (G) - 1.//
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• Summer '12
• Rittered
• Graph Theory, Vertex, Planar graph, κ

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