CS 1050 Section B, Spring 2001
Homework 2 { due Thursday,February 1
Part I
Problem 1
Exercise 16, page 54.
Problem 2
Exercise 35, page 55.
Problem 3
Exercise 30, page 96. Is the claim true in case
f
is
not
1-to-1?
Problem 4
Exercise 32, page 79.
Problem 5
Exercise 40, page 80.
Part II
Problem 6
Exercise 18, page 91.
Problem 7
What is wrong with the following argument? \Since
n
=
O
(
n
) and 2
n
=
O
(
n
) and 3
n
=
O
(
n
) and
so on, wehave
P
n
k
=1
kn
=
P
n
k
=1
O
(
n
)=
nO
(
n
)=
O
(
n
2
)
:
"
Problem 8
Exercise 26, page 91.
Problem 9
Exercise 28, page 91.
Problem 10
Exercise 36, page 91. If the answer is `yes,' prove the claim. If the answer is `no,' givetwo
functions
f
and
g
such that
f
(
x
)=
O
(
g
(
x
)) but 2
f
(
x
)
6
=
O
(2
g
(
x
)
).