_________________________________________________________________
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 vertexcut. In this case,
κ
(G) = k where k is
the cardinality of a minimum vertexcut.
(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 cutvertex 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 vertexcut 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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