Design and analysis of algorithm ppt ppt

• x
UNIT-1
Algorithmic Complexity Measures Methods for expressing and
comparing complexity of algorithms: worst and average cases,
lower bounds, and asymptotic analysis. Abstract Data Type
(ADT) Specification and Design techniques
Elementary ADTs: Lists, Trees, Stacks, Queues, and Dynamic
Sets. Sorting in Linear Time Lower bounds for sorting, Radix sort
and Bucket sort
ADVANCED DATA STRUCTURES AND ALGORITHMS
(Theory and Practice)
Course
Code:
22MCE12TL CIE Marks: 100
Hrs/week L:T:P 3:0:1
42L + 28P
SEE Marks: 100
Credits: 4 SEE: 3 Hrs

Asymptotic Analysis
Asymptotic analysis is the process of calculating the
running time of an algorithm in mathematical units to
find the program's limitations, or “run-time
performance.”
The goal is to determine the best case, worst case and
average case time required to execute a given task.
Asymptotic Notations allow us to analyze an algorithm's
running time by identifying its behavior as the input
size for the algorithm increases

Upper bound & Lower bound
• Upper bound: The maximum time a program can take
to produce outputs, expressed in terms of the size of the
inputs (worst-case scenario).
• Upper bound - the maximum guaranteed time
• Lower bound: The minimum time a program will take
to produce outputs, expressed in terms of the size of the
inputs (best-case scenario).
• Lower bound - the minimum guaranteed time.

Asymptotic notation
Precisely calculating the actual steps is tedious and not
generally useful
Different operations take different amounts of time.
Even from run to run, things such as caching, etc. cause
variations
We want to identify categories of algorithmic runtimes

For example…
f1(n) takes n2 steps
f2(n) takes 2n + 100 steps
f3(n) takes 3n+1 steps
Which algorithm is better?
Is the difference between f2 and f3
important/significant?

Big O: Upper bound
O(g(n)) is the set of functions:
O(g(n)) = f (n):
there exists positive constants c and n0 such that
0 £ f (n) £ cg(n) for all n ³ n0
ì
í
ï
î
ï
ü
ý
ï
þ
ï

Big O: Upper bound
O(g(n)) = f (n):
ì
í
ï
î
ï
ü
ý
ï
þ
ï
We can bound the function
f(n) above by some constant
factor of g(n)

Big O: Upper bound
O(g(n)) = f (n):
ì
í
ï
î
ï
ü
ý
ï
þ
ï
f(n) above by some constant
multiplied by g(n)
For some increasing
range

Big O: Upper bound
O(g(n)) = f (n):
ì
í
ï
î
ï
ü
ý
ï
þ
ï
Generally, we’re most interested in big O
notation since it is an upper bound on the
running time

Omega: Lower bound
Ω(g(n)) is the set of functions:
W(g(n)) = f (n):
0 £ cg(n) £ f (n) for all n ³ n0
ì
í
ï
î
ï
ü
ý
ï
þ
ï

Omega: Lower bound
Ω(g(n)) is the set of functions:
W(g(n)) = f (n):
0 £ cg(n) £ f (n) for all n ³ n0
ì
í
ï
î
ï
ü
ý
ï
þ
ï
f(n) below by some
constant factor of g(n)

Theta: Upper and lower bound
Θ(g(n)) is the set of functions:
Q(g(n)) = f (n):
there exists positive constants c1,c2 and n0 such that
0 £ c1g(n) £ f (n) £ c2g(n) for all n ³ n0
ì
í
ï
î
ï
ü
ý
ï
þ
ï

Q(g(n)) = f (n):
ì
í
ï
î
ï
ü
ý
ï
þ
ï
f(n) above and below by
some constant factor of g(n)
(though different constants)

Q(g(n)) = f (n):
ì
í
ï
î
ï
ü
ý
ï
þ
ï
Note: A function is theta bounded iff it is big O
bounded and Omega bounded

Q(g(n)) = f (n):
ì
í
ï
î
ï
ü
ý
ï
þ
ï
n
n
n
x
f
n
n
x
f
n
x
f
n
x
f
n
log
3
)
(
40
5
)
(
100
2
/
1
)
(
3
)
(
)
(
2
4
2
3
2
2
2
1
2











Worst-case vs. Best-case vs. Average-case
worst-case: what is the worst the running time of the algorithm
can be?
best-case: what is the best the running time of the algorithm can
be?
average-case: given random data, what is the running time of the
algorithm?
Don’t confuse this with O, Ω and Θ. The cases above are
situations, asymptotic notation is about bounding particular
situations

Proving bounds: find constants that satisfy
inequalities
Show that 5n2 – 15n + 100 is Θ(n2)
Step 1: Prove O(n2) – Find constants c and n0 such that
5n2 – 15n + 100 ≤ cn2 for all n > n0
100
15
5 2
2


 n
n
cn
2
/
100
/
15
5 n
n
c 


Let n0 =1 and c = 5 + 100 = 105.
100/n2 only get smaller as n increases and we ignore -15/n
since it only varies between -15 and 0

Proving bounds
Step 2: Prove Ω(n2) – Find constants c and n0 such that
5n2 – 15n + 100 ≥ cn2 for all n > n0
100
15
5 2
2


 n
n
cn
2
/
100
/
15
5 n
n
c 


Let n0 =4 and c = 5 – 15/4 = 1.25 (or anything less
than 1.25). 15/n is always decreasing and we
ignore 100/n2 since it is always between 0 and 100.

Bounds
Is 5n2
O(n)? No
How would we prove it?
O(g(n)) = f (n):
ì
í
ï
î
ï
ü
ý
ï
þ
ï

Disproving bounds
Is 5n2
O(n)?
O(g(n)) = f (n):
ì
í
ï
î
ï
ü
ý
ï
þ
ï
Assume it’s true.
That means there exists some c and n0 such that
5n2
£ cn for n > n0
5n £ c contradiction!

Some rules of thumb
Multiplicative constants can be omitted
 14n2 becomes n2
 7 log n become log n
Lower order functions can be omitted
 n + 5 becomes n
 n2 + n becomes n2
na dominates nb if a > b
 n2 dominates n, so n2+n becomes n2
 n1.5 dominates n1.4

Some rules of thumb
an dominates bn if a > b
 3n dominates 2n
Any exponential dominates any polynomial
 3n dominates n5
 2n dominates nc
Any polynomial dominates any logarithm
 n dominates log n or log log n
 n2 dominates n log n
 n1/2 dominates log n
Do not omit lower order terms of different variables (n2 + m) does not become n2

Big O
n2 + n log n + 50
2n -15n2 + n3 log n
nlog n + n2 + 15n3
n5 + n! + nn

Some examples
• O(1) – constant. Fixed amount of work, regardless
of the input size
 add two 32 bit numbers
 determine if a number is even or odd
 sum the first 20 elements of an array
 delete an element from a doubly linked list
• O(log n) – logarithmic. At each iteration, discards
some portion of the input (i.e. half)
 binary search

Some examples
• O(n) – linear. Do a constant amount of work on each
element of the input
 find an item in a linked list
 determine the largest element in an array
• O(n log n) log-linear. Divide and conquer algorithms
with a linear amount of work to recombine
 Sort a list of number with MergeSort
 FFT

Some examples
• O(n2) – quadratic. Double nested loops that
iterate over the data
 Insertion sort
• O(2n) – exponential
 Enumerate all possible subsets
 Traveling salesman using dynamic programming
• O(n!)
 Enumerate all permutations
 determinant of a matrix with expansion by
minors

A Common Misunderstanding
Confusing worst case with upper bound.
Upper bound refers to a growth rate.
Worst case refers to the worst input
from among the choices for possible
inputs of a given size.

Practical Complexity
0
250
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
f(n) = n
f(n) = log(n)
f(n) = n log(n)
f(n) = n^2
f(n) = n^3
f(n) = 2^n

0
500
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20
f(n) = n
f(n) = log(n)
f(n) = n log(n)
f(n) = n^2
f(n) = n^3
f(n) = 2^n

0
1000
1 3 5 7 9 11 13 15 17 19
f(n) = n
f(n) = log(n)
f(n) = n log(n)
f(n) = n^2
f(n) = n^3
f(n) = 2^n

0
1000
2000
3000
4000
5000
1 3 5 7 9 11 13 15 17 19
f(n) = n
f(n) = log(n)
f(n) = n log(n)
f(n) = n^2
f(n) = n^3
f(n) = 2^n

1
10
100
1000
10000
100000
1000000
10000000
1 4 16 64 256 1024 4096 16384 65536

Comparison of Functions
f  g  a  b
f (n) = O(g(n))  a  b
f (n) = W(g(n))  a  b
f (n) = (g(n))  a = b
f (n) = o(g(n))  a < b
f (n) = w (g(n))  a > b

Review on Summations
• Constant Series: For integers a and b, a  b,
• Linear Series (Arithmetic Series): For n  0,
• Quadratic Series: For n  0,





b
a
i
a
b 1
1
2
)
1
(
2
1
1








n
n
n
i
n
i










n
i
n
n
n
n
i
1
2
2
2
2
6
)
1
2
)(
1
(
2
1 

• Cubic Series: For n  0,
• Geometric Series: For real x  1,
For |x| < 1,








n
i
n
n
n
i
1
2
2
3
3
3
3
4
)
1
(
2
1 











n
k
n
n
k
x
x
x
x
x
x
0
1
2
1
1
1 


 

0 1
1
k
k
x
x

• Linear-Geometric Series: For n  0, real c  1,
• Harmonic Series: nth harmonic number, nI+,














n
i
n
n
n
i
c
c
nc
c
n
nc
c
c
ic
1
2
2
1
2
)
1
(
)
1
(
2 
n
Hn
1
3
1
2
1
1 



 





n
k
O
n
k
1
)
1
(
)
ln(
1

• Telescoping Series:
• Differentiating Series: For |x| < 1,


 


n
k
n
k
k a
a
a
a
1
0
1
 


 

0
2
1
k
k
x
x
kx

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