This document discusses asymptotic analysis and big-O notation for analyzing the time complexity of algorithms. It begins by defining key concepts like growth rate, asymptotic notations such as O(n), Ω(n) and Θ(n). It then provides examples of analyzing the time efficiency of different algorithms like finding the maximum element in an array and computing prefix averages. The document explains how to determine the asymptotic complexity by counting the total number of operations and expressing it using big-O notation. It also discusses properties of big-O notation like rules for dropping constant factors and lower order terms.

1. Design and Analysis of Algorithms
GROWTH RATE AND ASYMPTOTIC NOTATIONS

2. Objectives
• Understanding Growth Rate
• Functions of Growth Rate
• Asymptotic Notations
• Analyzing Time Efficiency of algorithms and representing in appropriate asymptotic notation
• Examples

3. Analysis: Performance
DESIGN AND ANALYSIS OF ALGORITHMS 3
o We can determine the maximum number of
primitive/basic operations executed by an algorithm, as a
function of the input size.
Algorithm arrayMax(A, n)
# operations
currentMax  A[0] 2
for (i =1; i<n; i++) 2n
(i=1 once, i<n n times, i++ (n-1) times:post increment)
if A[i]  currentMax then 2(n  1)
currentMax  A[i] 2(n  1)
return currentMax 1
Total 6n 1

4. Analysis: Performance
Algorithm arrayMax executes 6n  1 primitive operations in the
worst case.
Define:
a = Time taken by the fastest primitive operation
b = Time taken by the slowest primitive operation
Let T(n) be worst-case time of arrayMax. Then
a (6n  1)  T(n)  b(6n  1 )
Hence, the running time T(n) is bounded by two linear functions
DESIGN AND ANALYSIS OF ALGORITHMS 4

5. Growth Rate of Running Time
DESIGN AND ANALYSIS OF ALGORITHMS 5
• Affects T(n) by a constant factor, but
• Does not alter the growth rate of T(n)
Changing the hardware/ software environment
• The linear growth rate of the running time T(n) is an intrinsic property of algorithm arrayMax
Intrinsic Property
• Growth rate simplifies the Time Efficiency analysis and mechanism to compare algorithms
Analysis and Comparison

6. Seven
Important
Functions
Constant  1
Logarithmic  log n
Linear  n
N-Log-N  n log n
Quadratic  n2
Cubic  n3
Exponential  2n
These seven functions that often appear in algorithm
analysis.
In a chart, the slope of the line corresponds to the growth
rate of the function
DESIGN AND ANALYSIS OF ALGORITHMS 6
1E-1
1E+1
1E+3
1E+5
1E+7
1E+9
1E+11
1E+13
1E+15
1E+17
1E+19
1E+21
1E+23
1E+25
1E+27
1E+29
1E-1 1E+2 1E+5 1E+8
T(n)
n
Cubic
Quadratic
Linear

7. n lgn nlgn n2
n3
2n
0 0 0 1
1 0 0 1 1 2
2 1 2 4 8 4
4 2 8 16 64 16
8 3 24 64 512 256
16 4 64 256 4096 65536
32 5 160 1024 32768 4294967296
64 6 384 4096 262144 1.84467E+19
128 7 896 16384 2097152 3.40282E+38
256 8 2048 65536 16777216 1.15792E+77
512 9 4608 262144 134217728 1.3408E+154
1024 10 10240 1048576 1073741824
2048 11 22528 4194304 8589934592

8. Execution Times
(one time constant is 1 nano seconds)
n f(n) = lgn f(n) = n f(n) = nlgn f(n) = n2
f(n) = n3
f(n) = 2n
1 0 0.003 micro sec 0.01 micro sec 0.033 micro sec 0.1 micro sec 1 micro sec 1 micro sec
2 0 0.004 micro sec 0.02 micro sec 0.086 micro sec 0.4micro sec 8 micro sec 1 milli sec
3 0 0.005 micro sec 0.03 micro sec 0.147 micro sec 0.9 micro sec 27micro sec 1 sec
4 0 0.005 micro sec 0.04 micro sec 0.213 micro sec 1.6 micro sec 64 micro sec 18.3 min
5 0 0.006 micro sec 0.05 micro sec 0.282 micro sec 2.5 micro sec 125 micro sec 13 days
1 02
0.007 micro sec 0.10 micro sec 0.664 micro sec 10 micro sec 1 milli sec 4 exp 13 years
1 03
0.010 micro sec 1.00 micro sec 9.966 micro sec 1 milli sec 1 sec
1 04
0.013 micro sec 10 micro sec 130 micro sec 100 milli sec 16.7 min
1 05
0.017 micro sec 0.10 milli sec 1.67 milli sec 10 s 11.6 days
1 06
0.020 micro sec 1 milli sec 19.93 milli sec 16.7 min 31.7 years
1 07
0.023 micro sec 0.01sec 0.23 sec 1.16 days 31709 years
1 08
0.027 micro sec 0.10 sec 2.66 sec 115.7 days 3.17 exp 7 years
1 09
0.030 micro sec 1 sec 29.90 sec 31.7 years

9. Constant Factors
DESIGN AND ANALYSIS OF ALGORITHMS 9
The growth rate is not affected by
Constant factors or lower-order terms
Select Only higher order term
Examples
102n + 105 is a linear function
105n2 + 108n is a quadratic function

10. Asymptotic Notations
• In Mathematics, The term asymptotic means approaching a value or curve arbitrarily
closely (i.e., as some sort of limit is taken).
• Asymptotic analysis, is a method of describing limiting behavior. As an illustration,
suppose that we are interested in the properties of a function f as n becomes very
large. i.e., growth rate.
• Big Oh (O)
• Omega ()
• Theta ()
DESIGN AND ANALYSIS OF ALGORITHMS 10

11. Big-Oh Notation)
Given functions f(n) and g(n),
f(n) = O(g(n)) iff there exist two positive
constants c and n0 such that
f(n) <= cg(n) for all n >= n0
g(n) is an asymptotic upper bound on f(n).
Example: 2n + 10 is O(n)
◦ 2n + 10  cn
◦ (c  2) n  10
◦ n  10/(c  2)
◦ Pick c = 3 and n0 = 10
DESIGN AND ANALYSIS OF ALGORITHMS 11
Tt
1 n
g(n)
f(n)
n0

12. Big-Oh Example
Example: the function n2 is
not O(n)
◦ n2  cn
◦ n  c
◦ The above inequality cannot
be satisfied since c must be a
constant
DESIGN AND ANALYSIS OF ALGORITHMS 12
1
10
100
1,000
10,000
100,000
1,000,000
1 10 100 1,000
n
n^2
100n
10n
n

13. DESIGN AND ANALYSIS OF ALGORITHMS 13
More Big-Oh Examples
 7n-2
7n-2 is O(n)
need c > 0 and n0  1 such that 7n-2  c•n for n  n0
this is true for c = 7 and n0 = 1
 3n3 + 20n2 + 5
3n3 + 20n2 + 5 is O(n3)
need c > 0 and n0  1 such that 3n3 + 20n2 + 5  c•n3 for n  n0
this is true for c = 4 and n0 = 21
 3 log n + 5
3 log n + 5 is O(log n)
need c > 0 and n0  1 such that 3 log n + 5  c•log n for n  n0
this is true for c = 8 and n0 = 2

14. Big-Oh and
Growth Rate
The big-Oh notation gives an upper
bound on the growth rate of a
function
The statement “f(n) is O(g(n))” means
that the growth rate of f(n) is no more
than the growth rate of g(n)
We can use the big-Oh notation to
rank functions according to their
growth rate
DESIGN AND ANALYSIS OF ALGORITHMS 14
f(n) is
O(g(n))
g(n) is
O(f(n))
g(n) grows
more
Yes No
f(n) grows
more
No Yes
Same
growth
Yes Yes

15. Big-Oh Rules
DESIGN AND ANALYSIS OF ALGORITHMS 15
Use the simplest expression of the class
Say “3n + 5 is O(n)” instead of “3n + 5 is O(3n)”
Use the smallest possible class of functions
Say “2n is O(n)” instead of “2n is O(n2)”
If is f(n) a polynomial of degree d, then f(n) is O(nd), i.e.,
Drop
•Lower-order terms
•Constant factors
Examples:
T(n) = O(2n2) -------- wrong
T(n) = O(n2 + n) ----- wrong

16. Big-Oh Rules
DESIGN AND ANALYSIS OF ALGORITHMS 16
If T(x) is a polynomial of degree n, then
T(x) = (xn)
T1(n) * T2(n) = O( f(n) * g(n) )
If T1(n) = O(f(n)) and T2(n) = O(g(n))
Then T1(n) + T2(n) = Max(O(f(n)), O(g(n)))

17. Asymptotic
Algorithm
Analysis
The asymptotic analysis of an algorithm determines the running time
in big-Oh notation
To perform the asymptotic analysis
◦ We find the worst-case number of primitive operations executed as
a function of the input size (Total Running Time)
◦ We express this function with big-Oh notation
Example:
◦ We determine that algorithm arrayMax executes at most 8n  2
primitive operations
◦ We say that algorithm arrayMax “runs in O(n) time”
Since constant factors and lower-order terms are eventually dropped
anyhow, we can disregard them when counting primitive operations
DESIGN AND ANALYSIS OF ALGORITHMS 17

18. Example: Computing Prefix Averages
We further illustrate asymptotic
analysis with two algorithms for
prefix averages
The i-th prefix average of an array
X is average of the first (i + 1)
elements of X:
A[i] = (X[0] + X[1] + … + X[i])/(i+1)
Computing the array A of prefix
averages of another array X has
applications to financial analysis
DESIGN AND ANALYSIS OF ALGORITHMS 18
0
5
10
15
20
25
30
35
1 2 3 4 5 6 7
X
A

19. DESIGN AND ANALYSIS OF ALGORITHMS 19
Prefix Averages (Quadratic)
The following algorithm computes prefix averages in quadratic time
Algorithm prefixAverages1(X, n)
Input array X of n integers
Output array A of prefix averages of X #operations
A  new array of n integers n
for i  0 to n  1 do n
s  X[0] n
for j  1 to i do 1 + 2 + …+ (n  1)
s  s + X[j] 1 + 2 + …+ (n  1)
A[i]  s / (i + 1) n
return A 1

20. Arithmetic Progression
The running time of
prefixAverages1 is
O(1 + 2 + …+ n)
The sum of the first n integers
is n(n + 1) / 2
◦ There is a simple visual proof of
this fact
Thus, algorithm
prefixAverages1 runs in O(n2)
time
DESIGN AND ANALYSIS OF ALGORITHMS 20
0
1
2
3
4
5
6
7
1 2 3 4 5 6

21. DESIGN AND ANALYSIS OF ALGORITHMS 21
Prefix Averages (Linear)
The following algorithm computes prefix averages in
linear time by keeping a running sum
Algorithm prefixAverages2(X, n)
Input array X of n integers
Output array A of prefix averages of X #operations
A  new array of n integers n
s  0 1
for i  0 to n  1 do n
s  s + X[i] n
A[i]  s / (i + 1) n
return A 1
Algorithm prefixAverages2 runs in O(n) time

22. DESIGN AND ANALYSIS OF ALGORITHMS 22
Relatives of Big-Oh
big-Omega
 f(n) is (g(n)) iff there is a constant c > 0
and an integer constant n0  1 such that
f(n)  c•g(n) for n  n0
big-Theta
 f(n) is (g(n)) iff there are constants
c’ > 0 and c’’ > 0 and an integer constant
n0  1 such that
c’•g(n)  f(n)  c’’•g(n) for n  n0
n
n0
f(n)
g(n)
n
n0
c2g(n)
c1g(n)
f(n)

23. Intuition for Asymptotic
Notation
DESIGN AND ANALYSIS OF ALGORITHMS 23
Big-Oh
 f(n) is O(g(n)) if f(n) is asymptotically
less than or equal to g(n)
big-Omega
 f(n) is (g(n)) if f(n) is asymptotically
greater than or equal to g(n)
big-Theta
 f(n) is (g(n)) if f(n) is asymptotically
equal to g(n)

24. DESIGN AND ANALYSIS OF ALGORITHMS 24
Example Uses of the
Relatives of Big-Oh
f(n) is (g(n)) if it is (n2) and O(n2). We have already seen the former,
for the latter recall that f(n) is O(g(n)) if there is a constant c > 0 and an
integer constant n0  1 such that f(n) < c•g(n) for n  n0
Let c = 5 and n0 = 1
 5n2 is (n2)
f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0  1
such that f(n)  c•g(n) for n  n0
let c = 1 and n0 = 1
 5n2 is (n)
f(n) is (g(n)) if there is a constant c > 0 and an integer constant n0  1
such that f(n)  c•g(n) for n  n0
let c = 5 and n0 = 1
 5n2 is (n2)

25. A. Given f(𝑛) = 3𝑛2 + 2𝑛 + 10, Prove that f(𝑛) = Θ(𝑛2)
Solution:
f(n) = 3𝑛2 + 2𝑛 + 10 ----------- (1)
g(n) = n2 ------------ (2)
We know that,
If c1.g(n) <= 𝑓(𝑛) <= 𝑐2.𝑔(𝑛) ------------(3)
∀ 𝑛 ≥ 𝑛0 ∧ c1,c2 > 0
Then, 𝑓(𝑛) = Θ(𝑔(𝑛))
R.H.S:
𝑓(𝑛) <= 𝑐.𝑔(𝑛) ------------(4)
∀ 𝑛 ≥ 𝑛0 ∧ c > 0
(4)=> 3𝑛2 + 2𝑛 + 10 <= c*n ----------------(5)
Let c=4, n0 = 5
(4)=> 95 <= 100
=> f(n) < c.g(n) ---------------- (6)
Hence Proved R.H.S.
L.H.S
𝑓(𝑛) >= 𝑐.𝑔(𝑛) ------------(7)
∀ 𝑛 ≥ 𝑛0 ∧ c > 0
(7)=> 3𝑛2 + 2𝑛 + 10 >= c*n ----------------(8)
Let c=3, n0 = 1
(8)=> 15 >= 3
=> f(n) > c.g(n) -------------- (9)
Hence Proved L.H.S.
Hence Proved, f(n) = Θ(g(n))