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290
Solutions to Review Exercises
(b) The answer is 1 since there is an edge from
Vs
to
WI.
(c) The answer is 8 since there are 8 walks oflength two from
W2
to
W4
(via any of VI,
... ,
Vs).
(d) The answer is 12 for the same reason as (c).
(e) The answer is 0 because there is no walk of length two from
WI2 to
V6.
(f) The answer is (12)(8)(12)
=
1152.
(g) The answer is 0 because there is no walk of odd length from
V5
to
V7.
18. Yes it is. By Theorem 10.1.4, a connected graph is Eulerian if and only if the degree of every vertex is
even and this can easily be determined from the adjacency matrix. Remember that the degree of vertex
i is the number of l's in row (or column) i.
19. There are at least four l's in each row. So the degree of each vertex is
d
~
4
~ ~.
The graph has 7
vertices, so the result follows from Dirac's Theorem.
20. Let the vertices of
g
be VI, .
.• ,
Vn.
The number
b
ii
is the number of walks of length 2 from
Vi
to
Vi,
that is, the number of edges incident with
Vi,
equivalently, deg
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 Summer '10
 any
 Graph Theory

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