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Section 8.2
225
16. (a) No. By its very definition,
<
is not reflexive.
(b) Yes,
(A,
j)
is a partially ordered set.
Reflexivity: Since
f
=
f,
we have
f j f·
Antisymmetric:
If
f
j
9
but
f
#
g,
then
f
<
g.
This means
f
=
O(g)
but
9
#
O(f),
so
9
f,
f·
It
follows that if
f
j 9
and
9 j
f,
then
f
=
g.
Transitivity: See the solution to Exercise 10.
17. [BB] Reflexivity: For all
n
~
1,
If(n)1
::;
If(n)l,
so with e
=
1,
no
=
1 in Definition 8.2.1 we see
that
f
is
O(f)
and hence
f
::=::
f·
Symmetry: If
f
::=::
g,
then
f
=
O(g)
and
9
=
O(f).
Hence,
9
=
O(f)
and
f
=
O(g),
so
9
::=::
f·
Transitivity:
If
f
::=::
9
and
9
::=::
h,
then
f
=
O(g)
and
9
=
O(h).
Therefore,
f
=
O(h)
by
Exercise lO(a). Since
h
=
O(g)
and
9
=
O(f),
an argument identical to the one just given yields that
h
=
O(f)
as well. Thus,
f::=:: h.
18. [BB] Since 10g2
n
=
O(n)
by Proposition 8.2.9 and
n
=
O(n),
we have nlog2
n
=
0(n
2
)
by
Proposition 8.2.3. Thus, it remains only to show that
n
2
#
O(n
10g2
n).
Assume to the contrary that
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This note was uploaded on 11/08/2010 for the course MATH discrete m taught by Professor Any during the Summer '10 term at FSU.
 Summer '10
 any
 Graph Theory

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