1. Advanced Analysis of Algorithms
Lecture # 4
MS(Computer Science)
Semester 1
Spring 2017

2. Asymptotic Notation
After understanding Complexity rules & procedures, we can easily understand that the Time Complexity of any algorithm may be any one or multiple of the followings
1 Constant
log n Logarithmic
n Linear
n2 Quadratic
n3 Cubic
2n Exponential

3. Asymptotic Notation
Let’s Assume the increasing order as
1 < log n < n < n log n < n2 < n3 ……. < 2n < 3n < nn
Upper Bound
Suppose
f(n) = 2n2 + 3n + 2
Higher Degree is n2
All these time complexities greater than n2 becomes upper bound of function f(n). It also includes n2
So We can say f(n) = O(n2)
f(n) = O(n3)
f(n) = O(2n)

4. Asymptotic Notation
Now Coming to Lower Bound Ω (Omega)
1 < log n < n < n log n < n2 < n3 ……. < 2n < 3n < nn
Lower Bound
Suppose
f(n) = 2n2 + 3n + 2
Higher Degree is n2
All these time complexities smaller than n2 becomes lower bound of function f(n). It also includes n2
So We can say f(n) = Ω(n2)
f(n) = Ω(n)
f(n) = Ω(log n)

5. Asymptotic Notation f(n) = Φ(n2)
Now Coming to Tight Bound Φ (Theta)
1 < log n < n < n log n < n2 < n3 ……. < 2n < 3n < nn
Tight Bound
if f(n) = O(n2) and f(n) = Ω(n2)
Then
We can write
f(n) = Φ(n2)
Which is known as
Tight Bound

6. Asymptotic notations f(n) = O(g(n))
If there exist positive constants c and n0
Such that
f(n) < c x g(n)
for all n >= n0
g(n) is an asymptotic upper bound for f(n)

7. Upper Bound (Big – Oh) f(n) = O(g(n)) We can write as
Let us Suppose a function f(n)
f(n) = 2n2 + 3n + 1
Let’s Assume
2n2 + 3n2 + n2 = 6n2
We can write as
2n2 + 3n + 1 = 6n2
f(n) = c x g(n)
Here c is 6 and g(n) = n2
So f(n) = O(n2 )
To Proof the Asymptotic Notation of Big – Oh which is
f(n) = O(g(n))
If there exist positive constants c and n0
Such that
f(n) < c x g(n) for all n >= n0
Consider n = 2 and solve the function f(n) = c g(n)
2n2 + 3n + 1 = 6n2
2x22 + 3x = 6 x 22
= 24
15 = 24
So f(n) < c x g(n) for all n > n0

8. Upper Bound (Big – Oh) f(n) = O(g(n)) We can write as
Let us Suppose a function f(n)
f(n) = 2n2 + 3n + 1
Let’s Assume
22n + 32n + 2n = 62n
We can write as
2n2 + 3n + 1 = 62n
f(n) = c x g(n)
Here c is 6 and g(n) = 2n
So f(n) = O(2n )
To Proof the Asymptotic Notation of Big – Oh which is
f(n) = O(g(n))
If there exist positive constants c and n0
Such that
f(n) < c x g(n) for all n >= n0
Consider n = 2 and solve the function f(n) = c g(n)
2n2 + 3n + 1 = 62n
2x22 + 3x = 6 x 22
= 24
15 = 24
So f(n) < c x g(n) for all n > n0

9. Asymptotic notations f(n) = O(g(n))
If there exist positive constants c and n0
Such that
f(n) > c x g(n)
for all n >= n0
g(n) is an asymptotic Lower bound for f(n)

10. Upper Bound (Big – Omega)
Let us Suppose a function f(n)
f(n) = 2n2 + 3n + 1
Let’s Assume n2
We can write as
2n2 + 3n + 1 = 1 x n2
f(n) = c x g(n)
Here c is 1 and g(n) = n2
So f(n) = Ω (n2 )
To Proof the Asymptotic Notation of Big – Omega which is
f(n) = Ω(g(n))
If there exist positive constants c and n0
Such that
f(n) > c x g(n) for all n >= n0
Consider n = 2 and solve the function f(n) = c g(n)
2n2 + 3n + 1 = 1 x n2
2x22 + 3x = 1 x 22
= 4
15 = 4
So f(n) > c x g(n) for all n > n0

11. Asymptotic notations f(n) = O(g(n))
If there exist positive constants c1,c2 and n0
Such that
c1g(n) < f(n) < c2 x g(n)
for all n >= n0
g(n) is an asymptotic Tight bound for f(n)

12. Upper Bound (Big – Theta)
Let us Suppose a function f(n)
f(n) = 2n2 + 3n + 1
f(n) = O(n2 )
and
f(n) = Ω(n2 )
So according to Asymptotic Notation
Φ (n2 )
Which is Known as Tight Bound