226
Solutions
to
Exercises
some constant c and all sufficiently large
n.
Now
clog
n
=
log
n
C
and log is an increasing function,
so log
(An
+
B)
~
clogn
~
An
+
B
~
nCo
But
An
+
B
~
(A
+
B)n
~
n
2
for
n
>
A
+
B
so
c
=
2, works.
23. Since
n
=
ab
with 1
<
a, b
<
n
implies either
a
~
..;n
or
b
~
..;n,
n
is prime
if
does
not
have
a proper factor less than
or
equal to
..;n.
So, for each integer
k,
1
<
k
<
r
Vnl,
use the Euclidean
algorithm to determine
if
gcd
(k, n)
=
1. As shown in Problem 19, this algorithm requires
0
(In
k)
=
O(ln..;n)
=
O(ln
n)
divisions. The integer
n
is prime
if
and only
if
gcd(k,
n)
=
1 for all such
k.
The
entire procedure is
0
(
..;n
In
n).
24. [BB] Since
n!
=
n(n

1)(n

2)
..
·3·2.1
<
nn,
logn!
~
n
log
n.
With
c
=
1,
no
=
1 in
Definition 8.2.1,
we
see that
logn!
=
O(n
log
n)
as required.
25. (a)
If
a
=
(akakl
...
a2alaoh,
then the base 2 digits
ao,
al,
...
,
ak
are the remainders given
by
the
following process
of
successive divisions by
2:
a
2qo
+
ao
qo
2ql
+
al
ql
2q2
+
a2
The process terminates when
qk
=
O.
Assuming that each line here is a single operation, the
number
of
operations required is
k
+
1.
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 Summer '10
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 Graph Theory, Division, Natural number, Euclidean algorithm, al + ao

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