Complexity Analysis : Asymptotic Analysis
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Transcript 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 =
instead of (more correctly)
Why Discard Constant?
From the definition
We can use any constant
E.g. 3n = O(n)
Because
When we let c >= 3, the condition is satisfied
Why Discard Lower Order Term?
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
Why Discard Lower Order Term?
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)
j2 i3
for (j = ;j <=
for (i = ;i <=
sum += i;
}
}
;j++) {
;i++) {
j2
sum += i Θ(1)
( j 2)
n 1
j
j2
( 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
Discard non dominant term
Discard constant
For recursive
Make recurrent relation
Use master method
Guessing and proof
Recursion Tree