Problem 1
a) 100n + log n =
Q
(n+(log n)^2)
The n term dominates here, and at a large enough n the log n terms are not relevant.
b) log n =
Q
(log(n
2
))
log(n
2
) = 2 log n by the properties of logs
So log n <= 2 log n and vice versa, so this relation is theta.
c) n
2
/logn =
W
(n (log n)
2
)
n
2
/log n >= c(n (log n)
2
)
n
2
>= c(n (log n)
3
)
n*n>=cn(log n)
3
Clearly, (log n)
3
grows more slowly than n.
However, n
2
/logn
∫
Q
(n (log n)
2
)
d)
(
log n)
log n
=
W
(n/log n)
(log n)
log n
>c 2
log n
for any n>4
2
log n
= n
log n
log n
> 2
log n
= n>n/log n
However, (log n)
log n
∫
Q
(n/log n)
e) n
1/2
=
W
((log n)
5
)
n
1/2
>= c (log n)
5
//square both sides
n >= c(log n)
10
This is true for sufficiently large n.
However, n
1/2
∫
Q
((log n)
5
)
f) n 2
n
=
O
(3
n
)
n 2
n
<= c 3
n
n <= c 3
n
/ 2
n
n<=c(1.5)
n
This is true as exponential functions increase faster than polynomials.
However, n 2
n
∫
Q
(3
n
)
CSE 413 Homework 2 Solution Guide

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