Fall Semester 2008 CS300 Algorithms
Solution #8
1.
(a)
i.
n

1
ii. min
{
m, n
}
(b)
i. True; By definition, a graph with connectivity 4 is 1connected, 2connected,
3connected, and 4connected, but not 5connected.
ii. False; Think about
K
5
, a complete graph with 5 nodes.
This graph is 1
connected, 2connected, 3connected, and 4connected, but not 5connected.
Therefore, its connectivity is 4.
2.
(a) Suppose that the condensation graph
G
0
contains a cycle (
v
1
, v
2
,
· · ·
, v
m
, v
1
),
m
≥
2.
Let
u
1
be a vertex in the component
S
1
and
u
m
be a vertex in
S
m
. Since
G
0
contains
both a
v
1

v
m
path and a
v
m

v
1
path,
G
has both a
u
1

u
m
path and a
u
m

u
1
path. Thus,
u
1
and
u
m
belong to the same strong component of
G
. This produces a
contradiction. So,
G
0
has no cycle.
(b)
•
Case 1: A newly added edge connects two vertices in the same component. Then
G
*
0
, the condensation graph of
G
*
, has no cycle. Thus, the number of strongly
connected components of
G
and that of
G
*
are same.
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 Spring '08
 Unkown
 Algorithms, Planar graph, complete graph, Vertextransitive graph, Graph connectivity, strongly connected components

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