TEST3/MAD2104
Page 3 of 4
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5. (15 pts.)
(a) Draw a directed graph G
1
whose adjacency
matrix is given on the left below.//
V
1
= { a, b, c, d }.
0 1 0 0
0 0 1 1
0 0 0 1
0 0 0 0
.
(b) Now draw the underlying undirected graph G
2
for the directed
graph G
1
of part (a) of this problem.// V
2
= { a, b, c, d }.
(c) Is G
2
= (V
2
,E
2
), above, isomorphic to the simple graph
G
3
= (V
3
,E
3
) given below?
Either display an isomorphism
f:V
2
→
V
3
or very briefly explain why there is no such function
by revealing an invariant that one graph has that the other
doesn’t.
Yes, define f:V
2
→
V
3
by f(a) = b, f(b) = c,
f(c) = a, and f(d) = d.
Then
{ a ,b }
ε
E
2
↔
{ f(a), f(b) } = { b, c }
ε
E
3
,
{ b ,c }
ε
E
2
↔
{ f(b), f(c) } = { c, a }
ε
E
3
,
{ b ,d }
ε
E
2
↔
{ f(b), f(d) } = { c, d }
ε
E
3
,
{ c ,d }
ε
E
2
↔
{ f(c), f(d) } = { a, d }
ε
E
3
.
[You could also use f(a) = b, f(b) = c,
f(c) = d, and f(d) = a.]
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6. (10 pts.)
Recall that the composition of two relations R
and S on a set A is given by
S R = { (a,c)
ε
A × A
(
∃
b)(b
ε
A and (a,b)
ε
R and (b,c)
ε
S)}.
Also, recall that the n
th
composition power of a relation on a
set is defined recursively by R
1
= R, and for each n
ε
, if
n
≥
1, then R
n+1
= R
n
R.