### Complexity Analysis : Asymptotic Analysis

```Nattee Niparnan
Recall
 What is the measurement of algorithm?
 How to compare two algorithms?
 Definition of Asymptotic Notation
Today Topic
 Finding the asymptotic
bound of the algorithm
Interesting Topics of Upper Bound
 Rule of thumb!
 We neglect
 Lower order terms from addition

E.g.
n3+n2 = O(n3)
 Constant
 E.g.
3n3 = O(n3)
Remember that we use =
 From the definition
 We can use any constant
 E.g. 3n = O(n)
 Because

When we let c >= 3, the condition is satisfied
 Consider
 f(n) = n3+n2
 g(n) = n3
 If f(n) = O(g(n))
 Then, for some c and n0


c * g(n)-f(n) > 0
Definitely, just use any c >1
 Try c = 1.1
 Does
1.1 * g(n)-f(n) = 0.1n3-n2
0.1n3-n2 > 0
 It is when
 0.1n > 1
 E.g., n > 10
?
0.1n3-n2 > 0
0.1n3
> n2
0.1n3/n2 > 1
0.1n
> 1
Lower Order only?
 In fact,
 It’s only the dominant term that count
 Which one is dominating term?
 The one that grow faster
The nondominant term
 Why?
 Eventually, it is g*(n)/f*(n)

If g(n) grows faster,
 g(n)/f*(n) > some constant
 E.g, lim g(n)/f*(n)  infinity
The dominant
term
What dominating what?
Left side
dominates
na
n log n
nb
(a > b)
n
n2 log n
n log2 n
cn
Log n
nc
1
n
log n
Putting into Practice
 What is the asymptotic class of
 0.5n3+N4-5(n-3)(n-5)+n3log8n+25+n1.5
 (n-5)(n2+3)+log(n20)
 20n5+58n4+15n3.2*3n2
Putting into Practice
 What is the asymptotic class of
 0.5n3+N4-5(n-3)(n-5)+n3log8n+25+n1.5
O(n4)
 (n-5)(n2+3)+log(n20)
O(n3)
 20n5+58n4+15n3.2*3n2
O(n5.4)
Asymptotic Notation from Program
Flow
 Sequence
 Conditions
 Loops
 Recursive Call
Sequence
Block A
f (n)
f(n) + g(n) =
Block B
g (n)
O(max (f(n),g(n))
Example
Block A
O(n)
O(n2)
Block B
O(n2)
Example
Block A
Θ(n)
O(n2)
Block B
O(n2)
Example
Block A
Θ(n)
Θ(n2)
Block B
Θ(n2)
Example
Block A
O(n2)
Θ(n2)
Block B
Θ(n2)
Condition
Block A
f (n)
Block B
g (n)
O(max (f(n),g(n))
Loops
for (i = 1;i <= n;i++) {
P(i)
}
Let P(i)
takes time ti
n
t
i 1
i
Example
for (i = 1;i <= n;i++) {
sum += i;
}
sum += i Θ(1)
n
 1   ( n )
i 1
Why don’t we use max(ti)?
 Because the number of terms is not constant
for (i = 1;i <= n;i++) {
sum += i;
}
for (i = 1;i <= 100000;i++) {
sum += i;
}
Θ(n)
Θ(1)
With big
constant
Example
for (j = 1;j <= n;j++) {
for (i = 1;i <= n;i++) {
sum += i;
}
}
sum += i Θ(1)
n
n
   (1)
n

j 1 i 1
  (n)
j 1

 ( n )   ( n )  ...   ( n )

 (n )
2
Example
n
for (j = 1;j <= n;j++) {
for (i = 1;i <= ;i++) {
sum += i;
}
}
sum += i Θ(1)
j
   (1)
n

j 1 i 1
  ( j)
j 1
n

 cj
j 1
n

c j
j 1


 n ( n  1) 
c

2


2
 (n )
Example : Another way
n
for (j = 1;j <= n;j++) {
for (i = 1;i <= ;i++) {
sum += i;
}
}
sum += i Θ(1)
j
   (1)
n

j 1 i 1
  ( j)
j 1
n

 O (n)
j 1

2
O (n )
Example
n 1
n 1
j
   (1)

j2 i3
for (j = ;j <=
for (i = ;i <=
sum += i;
}
}
;j++) {
;i++) {
j2




sum += i Θ(1)
 ( j  2)
 n 1 
  j 


 j2 
 ( m  1) m


 1
2


2
m
m
(

 1)
2
2
2
 (m )
Example : While loops
While (n > 0) {
n = n - 1;
}
Θ(n)
Example : While loops
While (n > 0) {
n = n - 10;
}
Θ(n/10)
= Θ(n)
Example : While loops
While (n > 0) {
n = n / 2;
}
Θ(log n)
Example : Euclid’s GCD
function gcd(a, b) {
while (b > 0) {
tmp = b
b
= a mod b
a
= tmp
}
return a
}
Example : Euclid’s GCD
Until the
modding one is
zero
function gcd(a, b) {
while (b > 0) {
tmp = b
b
= a mod b
a
= tmp
}
return a
}
How many
iteration?
Compute mod
and swap
Example : Euclid’s GCD
a
b
Example : Euclid’s GCD
a
b
a mod b
If a > b
a mod b < a / 2
Case 1: b > a / 2
a
b
a mod b
Case 1: b ≤ a / 2
a
b
We can
always put
another b
a mod b
Example : Euclid’s GCD
function gcd(a, b) {
while (b > 0) {
tmp = b
b
= a mod b
a
= tmp
}
return a
}
O( log n)
B always reduces
at least half
Theorem
 If Σai <1 then
 T(n)= ΣT(ain) + O(N)
 T(n) = O(n)
T(n) = T(0.7n) + T(0.2n) + T(0.01)n + 3n
= O(n)
Recursion
try( n ){
if ( n <= 0 ) return 0;
for ( j = 1; j <= n ; j++)
sum += j;
try (n * 0.7)
try (n * 0.2)
}
Recursion
try( n ){
if ( n <=
0 ) return 0;
terminating
Θ(1)
for ( j = 1; j <= n ; j++)
process
Θ(n)
sum += j;
try (n * 0.7)
recursion
try (n * 0.2)
T(0.7n) + T(0.2n)
}
T(n) = T(0.7n) + T(0.2n) + O(n)
T(n)
Guessing and proof by induction
 T(n) = T(0.7n) + T(0.2n) + O(n)
 Guess: T(n) = O(n), T(n) ≤ cn
 Proof:
 Basis: obvious
 Induction:
 Assume T(i < n) = O(i)
 T(n) ≤ 0.7cn + 0.2cn + O(n)

= 0.9cn + O(n)

= O(n)
<<< dominating rule
Using Recursion Tree
 T(n) = 2 T(n/2) + n
n
Lg n
n
n/2
n/4
n/2
n/4
n/4
2 n/2
n/4
4 n/4
T(n) = O(n lg n)
Master Method
 T(n) = aT(n/b) + f (n) a ≥ 1, b > 1
 Let c = logb(a)
 f (n) = Ο( nc - ε )
 f (n) = Θ( nc )
 f (n) = Ω( nc + ε )
→
→
→
T(n) = Θ( nc )
T(n) = Θ( nc log n )
T(n) = Θ( f (n) )
Master Method : Example
 T(n) = 9T(n/3) + n
 a = 9, b = 3, c = log 3 9 = 2, nc = n2
 f (n) = n = Ο( n2 - 0.1 )
 T(n) = Θ(nc) = Θ(n2)
Master Method : Example
 T(n) = T(n/3) + 1
 a = 1, b = 3, c = log 3 1 = 0, nc = 1
 f (n) = 1 = Θ( nc ) = Θ( 1 )
 T(n) = Θ(nc log n) = Θ( log n)
Master Method : Example
 T(n) = 3T(n/4) + n log n
 a = 3, b = 4, c = log 4 3 < 0.793, nc < n0.793
 f (n) = n log n = Ω( n0.793 )
 a f (n/b) = 3 ((n/4) log (n/4) ) ≤ (3/4) n log n = d f (n)
 T(n) = Θ( f (n) ) = Θ( n log n)
Conclusion
 Asymptotic Bound is, in fact, very simple
 Use the rule of thumbs

