Introduction to analysis algorithm in computer Science 2

1
AsymptoticNotationsand Analysis

Asymptotic Notations Properties
• Categorize algorithms based on asymptotic growth rate
• e.g. linear, quadratic, exponential
• Ignore small constant and small inputs
• Estimate upper bound and lower bound on growth rate of time
complexity function
• Describe running time of algorithm as n grows to .
Limitations
• not always useful for analysis on fixed-size inputs.
• All results are for sufficiently large inputs.
2

Asymptotic Notations
Asymptotic Notations , O, , o, 
 We use  to mean “order exactly”, (Tight Bound)
 O to mean “order at most”, (Tight Upper Bound)
  to mean “order at least”, (Tight Lower Bound)
 o to mean “upper bound”,
•  to mean “lower bound”,
Define a set of functions: which is in practice used to compare
two function sizes.
3

   
 
 
   

    
   
   
 .
for
bound
upper
ally
asymptotic
an
is
function
means
all
for
,
0
such that
and
constants
positive
exist
there
:
functions,
of
set
the
by
denoted
,
0
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given
a
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o
o









Big-Oh Notation (O)
Intuitively:
Set of all functions whose rate of growth is the same as or lower than
that of g(n). f(n) is bounded above by g(n) for all sufficiently large n
We may write f(n) = O(g(n)) OR f(n)  O(g(n))
If f, g: N  R+, then we can define Big-Oh as
4

g(n) is an asymptotic upper bound for f(n).
 c > 0,  n0 > 0 and n  n0, 0  f(n)  c.g(n)
f(n)  O(g(n))
Big-Oh Notation (O)
5

Big-Oh Notation (O)
The idea behind the big-O notation is to establish an upper boundary
for the growth of a function f(n) for large n.
This boundary is specified by a function g(n) that is usually much
simpler than f(n).
We accept the constant C in the requirement
f(n)  Cg(n) whenever n > n0,
We are only interested in large n, so it is OK if
f(n) > Cg(n) for n  n0.
The relationship between f and g can be expressed by stating either
that g(n) is an upper bound on the value of f(n) or that in the long run ,
f grows at most as fast as g.
6

• As a simple illustrative example, we show that the function
2n2 + 5n + 6 is O(n2).
• For all n ≥ 1, it is the case that
2n2 + 5n + 6 ≤ 2n2 + 5n2 + 6n2 = 13n2
• Hence, we can take с = 13 and no =1, and the definition is
satisfied.
7
Example

Prove that 2n2 = O(n3)
Proof:
Assume that f(n) = 2n2, and g(n) = n3
f(n) = O(g(n)) ?
Now we have to find the existence of c and n0
f(n) ≤ c.g(n)  2n2 ≤ c.n3  2 ≤ c.n
if we take, c = 1 and n0= 2 OR
c = 2 and n0= 1 then
2n2 ≤ c.n3
Hence f(n) = O(g(n)), c = 1 and n0= 2
8
Example

Prove that n2 = O(n2)
Proof:
Assume that f(n) = n2, and g(n) = n2
f(n) = O(g(n)) ?
f(n) ≤ c.g(n)  n2 ≤ c.n2  1 ≤ c
if we take, c = 1, n0= 1
Then
n2 ≤ c.n2 for c = 1 and n  1
Hence, n2 = O(n2), where c = 1 and n0= 1 9
Example

Prove that 1000.n2 + 1000.n = O(n2)
Proof:
Assume that f(n) = 1000.n2 + 1000.n, and g(n) = n2
We have to find existence of c and n0 such that
0 ≤ f(n) ≤ c.g(n) ꓯ n  n0
1000.n2 + 1000.n ≤ c.n2
for c = 1001, 1000.n2 + 1000.n ≤ 1001.n2
 1000.n ≤ n2  n2 - 1000.n  0  n (n-1000)  0,
this true for n  1000
Hence f(n) = O(g(n)) for c = 1001 and n0 = 1000
10
Example

Prove that n3 = O(n2)
Proof:
On contrary we assume that there exist some positive constants
c and n0 such that
0 ≤ n3 ≤ c.n2 ꓯ n  n0
n ≤ c
Since c is any fixed number and n is any arbitrary constant,
therefore n ≤ c is not possible in general.
Hence our supposition is wrong and n3 ≤ c.n2, for n  n0 is not
true for any combination of c and n0.
Hence, n3 = O(n2) does not hold
11
Example

Prove that 2n + 10 = O(n)
Proof:
Assume that f(n) = 2n + 10, and g(n) = n
f(n) = O(g(n)) ?
f(n) ≤ c.g(n)  2n + 10 ≤ c.n  (c  2) n  10  n  10/(c  2)
c > 2 for n >0, we pick, c = 3, then n0= 10
Then
2n + 10 ≤ c.n for c = 3 and n  10
Hence, 2n + 10 = O(n), where c = 3 and n0= 10 12
Example

Prove that : 3n3 + 20n2 + 5 = O(n3)
Proof:
Need c > 0 and n0  1 such that
3n3 + 20n2 + 5 ≤ c.n3 for n  n0
This is true for c = 4, n0= 21
OR c = 28, n0= 1
Hence, 3n3 + 20n2 + 5 = O(n), where c = 4 and n0= 21
13
Example

Prove that : 10n + 500 = O(n)
Proof:
Function n will never be larger than the function 500 + 10 n, no
matter how large n gets.
However, there are constants c0 and n0 such that
500 + 10n ≤ c.n when n ≥ n0.
One choice for these constants is c = 20 and n0 = 50.
Other choices for c0 and n0:
• For example, any value of c > 20 will work for n0 = 50.
Therefore, 500 + 10n = O(n). 14
Example

Prove which of the following function is larger by order of growth?
(1/3)n or 17?
• Let’s check if
(1/3)n = O( 17)
(1/3)n ≤ c.17, which is true for c=1,n0 = 1
• Let’s check if
17 = O((1/3)n )
17 ≤ c. (1/3)n , which is true for c > 17. 3n
• And hence can’t be bounded for large n.
• That’s why (1/3)n is less in growth rate then 17.
15
Example

Prove or disprove 22n = O (2n )?
• o prove above argument we have to show
• 22n ≤ C . 2n
• 2n 2n ≤ C .2n
• This inequality holds only when
• C ≥ 2n ,
• which makes C to be non-constant.
• Hence we can’t bound 22n by O(2n)
16
Example

Prove that : 8n2 + 2n - 3  O(n2)
Proof:
Need c > 0 and n0  1 such that
8n2 + 2n - 3 ≤ c.n2 for n  n0
Consider the reasoning:
f(n) = 8n2 + 2n - 3 ≤ 8n2 + 2n ≤ 8n2 + 2n2 = 10n2
Hence, 8n2 + 2n - 3  O(n2), where c = 10 and n0= 1
17
Example

Can you bound 3n = O (2n ) ?
To prove above argument we have to show
3n ≤ C. 2n
3n ≤ C. 2n
3n ≤ (3/2)n 2n
This inequality holds only when C ≥ (3/2)n, which makes C to be non-
constant.
Hence we can’t bound 3n by O(2n)
18
Example

Which of the following function is larger by order of growth? N log N
or N1.5?
Note that g(N) = N 1.5 = N•N 0.5
Hence, between f(N) and g(N), we only need to compare growth
rate of log(N) and N 0.5
Equivalently, we can compare growth rate of log2N with N
Now, we can refer to the previously state result to figure out whether
f(N) or g(N) grows faster!
19
Example

   
 
 
   

    
   
   
 .
for
bound
lower
ally
asymptotic
an
is
function
that
means
,
all
for
0
such that
and
constants
positive
exist
there
:
functions,
of
set
the
by
denote
function
given
a
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o
o








Big-Omega Notation ()
Intuitively:
Set of all functions whose rate of growth is the same as or higher than
that of g(n).
We may write f(n) = (g(n)) OR f(n)  (g(n))
If f, g: N  R+, then we can define Big-Omega as
20

21
g(n) is an asymptotically lower bound for f(n).
Note the duality rule: t(n)(f(n))  f(n)O(t(n))
 c > 0,  n0 > 0 , n  n0, f(n)  c.g(n)
f(n)  (g(n))
Big-Omega Notation ()

Prove that 3n + 2  (n)
Proof:
Assume that f(n) = 3n + 2 , and g(n) = n
f(n)  (g(n)) ?
We have to find the existence of c and n0 such that
c.g(n) ≤ f(n) for all n  n0
c.n ≤ 3n + 2
At R.H.S a positive term is being added to 3n, which will make L.H.S ≤
R.H.S for all values of n, when c = 3.
Hence f(n)  (g(n)), for c = 3 and n0= 1
22
Example

Prove that 5.n2  (n)
Proof:
Assume that f(n) = 5.n2 , and g(n) = n
f(n)  (g(n)) ?
c.n ≤ 5.n2  c ≤ 5.n
if we take, c = 5 and n0= 1 then
c.n ≤ 5.n2 for all n  n0
Hence f(n)  (g(n)), for c = 5 and n0= 1
23
Example

Prove that 5n2 + 2n - 3  (n2)
Proof:
Assume that f(n) = 5n2 + 2n - 3, and g(n) = n2
f(n)  (g(n)) ?
c.g(n) ≤ f(n) ꓯ n  n0
c.n2 ≤ 5.n2 + 2n - 3
We can take c = 5, given that 2n-3 is always positive.
2n-3 is always positive for n  2. Therefore n0 = 2.
And hence f(n)  (g(n)), for c = 5 and n0= 2
24
Example

Prove that 100.n + 5  (n2)
Proof:
Let f(n) = 100.n + 5, and g(n) = n2
Assume that f(n)  (g(n)) ?
Now if f(n)  (g(n)) then there exist c and n0 such that
c.n2 ≤ 100.n + 5
For the above inequality to hold meaning f(n) grows
faster than g(n).
But which means g(n) is growing faster than f(n)
And hence f(n)  (g(n))
25
Example
 
 



 n
g
n
f
n
lim






0
5
100
lim 2
n n
n

26
Theta Notation ()
   
 
 
   

      
   
     
   .
for
bound
ally tight
asymptotic
an
is
and
factor,
constant
a
within
to
to
equal
is
function
means
all
for
0
such that
and
,
constants
positive
exist
there
:
functions,
of
set
the
by
denoted
function
given
a
For
2
1
2
1
n
f
n
g
n
g
n
f
n
g
n
f
n
n
n
g
c
n
f
n
g
c
n
c
c
n
f
n
g
n
g
n
g
o
o









Intuitively: Set of all functions that have same rate of growth as g(n).
When a problem is (n), this represents both an upper and lower
bound i.e. it is O(n) and (n) (no algorithmic gap)
We may write f(n) = (g(n)) OR f(n)  (g(n))
If f, g: N  R+, then we can define Big-Theta as

27
We say that g(n) is an asymptotically tight bound for f(n).
f(n)  (g(n))
 c1> 0, c2> 0,  n0 > 0,  n  n0, c2.g(n)  f(n)  c1.g(n)
Theta Notation ()

Prove that ½.n2 – ½.n = (n2)
Proof
Assume that f(n) = ½.n2 – ½.n, and g(n) = n2
f(n)  (g(n))?
We have to find the existence of c1, c2 and n0 such that
c1.g(n) ≤ f(n) ≤ c2.g(n) for all n  n0
Since, ½ n2 - ½ n ≤ ½ n2 n ≥ 0 if c2= ½ and
½ n2 - ½ n ≥ ½ n2 - ½ n . ½ n ( n ≥ 2 ) = ¼ n2, c1= ¼
Hence ½ n2 - ½ n ≤ ½ n2 ≤ ½ n2 - ½ n
c1.g(n) ≤ f(n) ≤ c2.g(n) n ≥ 2, c1= ¼, c2 = ½
Hence f(n)  (g(n))  ½.n2 – ½.n = (n2)
28
Example

Prove that 2.n2 + 3.n + 6  (n3)
Proof: Let f(n) = 2.n2 + 3.n + 6, and g(n) = n3
we have to show that f(n)  (g(n))
On contrary assume that f(n)  (g(n)) i.e. there exist some
positive constants c1, c2 and n0 such that:
c1.g(n) ≤ f(n) ≤ c2.g(n)
Solve for c2:
f(n) ≤ c2.g(n)  2n2 + 3n + 6 ≤ 2n2 + 3n2 + 6n2 ≤ c2n3  c = 11 and n0= 1
Solve for c1:
c1.g(n) ≤ f(n)  c1n3 ≤ 2n2 + 3n + 6  c1n3 ≤ 2n2 ≤ 2n2 + 3n + 6
c1.n ≤ 2, for large n this is not possible
Hence f(n)  (g(n))  2.n2 + 3.n + 6  (n3)
29
Example

Prove that 3.n + 2 = (n)
Proof: Let f(n) = 3.n + 2, and g(n) = n
Assume that f(n)  (g(n)) i.e. there exist some positive
constants c1, c2 and n0 such that:
c1.g(n) ≤ f(n) ≤ c2.g(n)
c1.n ≤ 3.n + 2 ≤ c2. n 
Take c1 = 3, as 3n ≤ 3n + 2 with n0 = 1
3n + 2 ≤ 3n + 2n ≤ c2. n  5n ≤ c2. n  5 ≤ c2
c1 = 3, c2 = 5, n0 = 1
Hence f(n)  (g(n))  3.n + 2 = (n)
30
Example

Prove that ½.n2 – 3.n = (n2)
Proof
Let f(n) = ½.n2 – 3.n, and g(n) = n2 f(n)  (g(n))?
We have to find the existence of c1, c2 and n0 such that
c1.g(n) ≤ f(n) ≤ c2.g(n) ꓯ n  n0
c1. n2 ≤ ½.n2 – 3.n ≤ c2. n2
Since, ½ n2 - 3 n ≤ ½ n2 n ≥ 1 if c2= ½ and
½ n2 - 3 n ≥ 1/4 n2 ( n ≥ 6 ), c1= ¼
c1.g(n) ≤ f(n) ≤ c2.g(n) n ≥ 6, c1= ¼, c2 = ½
Hence f(n)  (g(n))  ½.n2 – 3.n = (n2)
31
Example from the book

 
 
0
lim
n


 n
g
n
f
   
 
 
 
 
    
all
for
0
such that
constant
a
exists
there
,
constants
positive
any
for
:
functions,
of
set
the
by
denoted
,
0
function
given
a
For





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
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o
o
n
n
n
cg
n
f
n
c
n
f
n
g
o
n
g
o
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g
Little-Oh Notation
   ..
2
but
n
2
e.g., 2
2
2
n
o
n
n
o 

o-notation is used to denote a upper bound that is not
asymptotically tight.
f(n) becomes insignificant relative to g(n) as n approaches
infinity. g(n) is an upper bound for f(n), not asymptotically tight
32

Prove that 2n2  o(n3)
Proof:
Assume that f(n) = 2n2 , and g(n) = n3
f(n)  o(g(n)) ?
Now we have to find the existence n0 for any c
f(n) < c.g(n) this is true
 2n2 < c.n3  2 < c.n
This is true for any c, because for any arbitrary c we can
choose n0 such that the above inequality holds.
Hence f(n)  o(g(n))
33
Example

Prove that n2  o(n2)
Proof:
Assume that f(n) = n2 , and g(n) = n2
Now we have to show that f(n)  o(g(n))
Since
f(n) < c.g(n)  n2 < c.n2  1 ≤ c,
In our definition of small o, it was required to prove for any c
but here there is a constraint over c .
Hence, n2  o(n2), where c = 1 and n0= 1 34
Example

   
 
 
   

    
o
o
n
n
n
f
n
cg
n
c
n
f
n
g
n
g
n
g




all
for
0
such that
constant
a
exists
there
,
constants
positive
any
for
:
functions.
all
of
set
the
by
denote
,
function
given
a
For


 
 



 n
g
n
f
n
lim
Little-Omega Notation
   ..
2
but
2
n
e.g., 2
2
2
n
n
n 
 

Little- notation is used to denote a lower bound that is not
asymptotically tight.
f(n) becomes arbitrarily large relative to g(n) as n
approaches infinity
35

Prove that 5.n2  (n)
Proof:
Assume that f(n) = 5.n2 , and g(n) = n
f(n)  (g(n)) ?
We have to prove that for any c there exists n0 such that
c.g(n) < f(n) ꓯ n  n0
c.n < 5.n2  c < 5.n
This is true for any c, because for any arbitrary c e.g. c =
1000000, we can choose n0 = 1000000/5 = 200000 and the
above inequality does hold.
And hence f(n)  (g(n))
36
Example

Prove that 5.n + 10  (n)
Proof:
Assume that f(n) = 5.n + 10, and g(n) = n
f(n)  (g(n)) ?
We have to find the existence n0 for any c, such that
c.g(n) < f(n) ꓯ n  n0
c.n < 5.n + 10, if we take c = 16 then
16.n < 5.n + 10  11.n < 10 is not true for any positive integer.
Hence f(n)  (g(n))
37
Example

Prove that 100.n  (n2)
Proof:
Let f(n) = 100.n, and g(n) = n2
Assume that f(n)  (g(n))
Now if f(n)  (g(n)) then there n0 for any c such that
c.g(n) < f(n) ꓯ n  n0
 c.n2 < 100.n  c.n < 100
If we take c = 100, n < 1, not possible
Hence f(n)  (g(n)) i.e. 100.n  (n2)
38
Example

If f(n) = Θ(g(n)) we say that f(n) and g(n) grow at the same rate,
asymptotically
If f(n) = O(g(n)) and f(n) ≠ Ω(g(n)), then we say that f(n) is asymptotically
slower growing than g(n).
If f(n) = Ω(g(n)) and f(n) ≠ O(g(n)), then we say that f(n) is asymptotically
faster growing than g(n).
39
Asymptotic Functions Summary

It is not always possible to determine behaviour of an algorithm
using Θ-notation.
For example, given a problem with n inputs, we may have an
algorithm to solve it in a.n2 time when n is even and c.n time when
n is odd. OR
We may prove that an algorithm never uses more than e.n2 time
and never less than f.n time.
In either case we can neither claim (n) nor (n2) to be the order of
the time usage of the algorithm.
Big O and  notation will allow us to give at least partial information
Usefulness of Notations
40

To express the efficiency of our algorithms which of the three
notations should we use?
As computer scientist we generally like to express our algorithms as
big O since we would like to know the upper bounds of our
algorithms.
Why?
If we know the worse case then we can aim to improve it and/or
avoid it.
41

Even though it is correct to say “7n - 3 is O(n3)”, a better statement is
“7n - 3 is O(n)”, that is, one should make the approximation as tight
as possible
Simple Rule:
Drop lower order terms and constant factors
7n-3 is O(n)
8n2log n + 5n2 + n is O(n2log n)
Strictly speaking this use of the equals sign is incorrect
• the relationship is a set inclusion, not an equality
• f(n)  O(g(n)) is better
42

Big Oh Does Not Tell the Whole Story
Question?
• If two algorithms A and B have the same asymptotic complexity, say
O(n2), will the execution time of the two algorithms always be
same?
• How to select between the two algorithms having the same
asymptotic performance?
Answer:
• They may not be the same. There is this small matter of the
constant of proportionality.
• Suppose that A does ten operations for each data item, but
algorithm B only does three.
• It is reasonable to expect B to be faster than A even though both
have the same asymptotic performance. The reason is that
asymptotic analysis ignores constants of proportionality.
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Algorithm_A {
set up the algorithm; /*taking 50 time units*/
read in n elements into array A; /* 3 units per element */
for (i = 0; i < n; i++) {
do operation1 on A[i]; /* takes 10 units */
}
}
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TA(n) = 50 + 3n + (10 + 5 + 15)*n
= 50 + 33*n
Algorithm_B {
set up the algorithm; /*taking 200 time units*/
read in n elements into array A; /* 3 units per element */
for (i = 0; i < n; i++) {
}
}
TB(n) = 200 + 3n + (10 + 5)*n
= 200 + 18*n

Both algorithms have time complexity O(n).
Algorithm A sets up faster than B, but does more operations on the data
Algorithm A is the better choice for small values of n.
For values of n > 10, algorithm B is the better choice
45

A Misconception
A common misconception is that worst case running time is somehow
defined by big-Oh, and that best case is defined by big-Omega.
There is no formal relationship like this.
However, worst case and big-Oh are commonly used together,
because they are both techniques for finding an upper bound on
running time.
46

Relations over AsymptoticNotations
Maximum rule: O(f(n)+g(n)) = O( max(f(n),g(n)) )
Additive and Multiplicative Property:
If e(n) = O(g(n)) & f(n) = O(h(n)) then e(n) + f(n) = O(g(n) + h(n))
If e(n) = O(g(n)) & f(n) = O(h(n)) then e(n) • f(n) = O(g(n) • h(n))
Dichotomy Property: If f(n) = O(g(n)) & g(n) = O(f(n)) then f(n) = (g(n))
If f(n) = (g(n)) and g(n) = (f(n)) then f(n) = (g(n))
Reflexive Property: If f(n) = O(f(n)) and f(n) = (f(n)) and f(n) = (f(n))
f(n)  o(f(n)) and f(n)  (f(n))
Symmetry over : f(n) = (g(n))  g(n) = (f(n))
Transitivity Property:
f(n) = (g(n)) & g(n) = (h(n))  f(n) = (h(n))
f(n) = O(g(n)) & g(n) = O(h(n))  f(n) = O(h(n))
f(n) = (g(n)) & g(n) = (h(n))  f(n) = (h(n))
f(n) = o (g(n)) & g(n) = o (h(n))  f(n) = o (h(n))
f(n) = (g(n)) & g(n) = (h(n))  f(n) = (h(n))
47

Introduction to analysis algorithm in computer Science 2

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