Practice Questions
1. Which of the following are true?
(a) 3n = O(n2)
(c) nlgn = Θ(20000nlgn)
(b) 3n2 = O(nlgn)
(d) 2n = Ω(n1000)
Answers: (a), (c), (d)
(a) n2 clearly grows larger than 3n as n gets large.
(c) 20000 is a constant in front of nlgn
(d) All exponential functions grow faster than all polynomial
functions.
(b) is false because 3n2 grows faster than nlgn.
2. What is the run-time of the following segment of code in
terms of n? Give an upper bound and justify it. Only a tight
upper bound will be accepted as a correct answer.
int i=1;
while (i <= n) {
int j = i;
while (j > 0)
j = j/2;
i++;
}
The outer loop runs n times. The inner loop will run log i times
at most since there is repeated halving. Since i never exceeds n,
it is safe to say the inner loop runs at most log n times. Hence
an upper bound on the run time is O(nlgn).
n 1
3. Determine the following sum:  3
i
i 0
This is a geometric sum with a first term of 1, with a common
ratio of 3, with n terms. Here's the sum:
1  3n
3n  1
3 


1 3
2 .
i 0
n 1
i
n
4. Determine the following sum:  log
i 1
n
n
n
i 1
i 1
2
4i
 log 2 4   i(log 2 4)   2i  2(
i
i 1
n( n  1)
)  n( n  1)
2
5. Order the following functions from smallest to largest in
terms of big-theta notation.
n3 2
1
, n n , ,17, lg( n20 ), lg 2 n
lg n
n
1
n3
20
2
2
Answer: n ,17, lg( n ), lg n, n n , lg n
As n grows large, 1/n tends to 0, so this is the smallest function.
17 is a constant, so it comes next. lg n20 = 20lg n which is Θ(lg
n). The following function is slightly bigger than that. Finally
to decide between the last two, note that n is a larger function
than lg n. Thus, when we divide by lg n, we are dividing by
something less than n . (The key here is that n 2 n can be
rewritten as
n3
n
.)
6. Order the following functions from smallest to largest in
terms of big-theta notation.
2n ,4 n ,23n ,7n , n15 , nlg n
Answer:
n15 , nlg n ,4 n ,2n ,7n ,23n
All polynomial functions grow more slowly than exponential
functions, so n15 comes first. We can rewrite the second
function as (2 lg n ) lg n . This makes it comparable to other functions
with an base of 2. The third function can be written as 2 2 n .
Comparing the exponents of these two functions proves their
order. Then the last three functions are each exponential
functions raised to the n power with different bases. It's clear
that these should be ordered by increasing value of the base.
7. Given the following experimental run-times, make the best
guess for the theta-bound for the run-time of the algorithm
described below:
Input Size (n)
10
20
40
80
160
Run-Time (in ms)
5
21
78
325
1277
As the input size doubles, the run time increases by a factor of
4 each time, so the run-time is most likely Θ(n2).
8. Draw the result of inserting 31 into the AVL tree below:
20
/
\
10 40
/
/ \
5
30
50
/ \
25 35
20
/
\
10 35
/
/ \
5
30
40
/ \
\
25 31
50
20
/
---
10
/
5
\
40
/ \
30 50
/ \
25 35
/
31
9. What is the result of inserting 19 into the following 2-4
Tree:
10, 20, 30
/
/
\
5 13, 15, 17
27
\
39
-
10, 20, 30
/
/
\
5 13, 15, 17,19 27
\
39
-
10, 17, 20, 30
/
/
\
5 13, 15,19
27
\
39
-
20
/
10, 17
/
| \
5 13, 15 19
\
30
/ \
27 39
10. Draw the Heap created from running the Make Heap
algorithm (shown in class on the following set of values –
assume a minimum heap):
37
/
21
/
10
/
12
\
58
\
33
\
9
/
19
\
24
37
/
21
/
\
33
9
/
12
\
58
/
19
\ -
24
\
10
37
/
21
/
\
33
9
/
12
\
19
/
58
\ -
24
\
10
37
/
\
19
9
/
10
/
12
\
33
/
58
\ -
24
\
21
9
/
10
/
12
/
37
\
19
\
33
\
21
/
58
\
24
11. Consider a hash table of size 113. Imagine implementing
quadratic probing on such a hash table. If our initial hash
value to insert an item was index 95, list the first five indexes in
which we would look AFTER index 95 in case of collisions.
(Note: Your first answer should be 96.)
The indexes are: 96, 99, 104, 111, and 7, since 120%113 = 7.
12. Here is an array representation of a Disjoint Set of the
elements 1, 2, 3, …, 10. Draw the corresponding tree
representation.
1
8
3
| \
8 10
|
1
2
3
9
4
3
9
|
2
|
4
2
5
5
5
7
|
6
6
7
7
7
8
3
9
9
10
3