How to calculate complexity in Data Structure

PROGRAM EFFICIENCY &
COMPLEXITY ANALYSIS
Lecture 03-04
By: Dr. Zahoor Jan
1

ALGORITHM DEFINITION
A finite set of statements that guarantees an optimal
solution in finite interval of time
2

GOOD ALGORITHMS?
 Run in less time
 Consume less memory
But computational resources (time complexity) is usually
more important
3

MEASURING EFFICIENCY
 The efficiency of an algorithm is a measure of the amount of
resources consumed in solving a problem of size n.
 The resource we are most interested in is time
 We can use the same techniques to analyze the consumption of other
resources, such as memory space.
 It would seem that the most obvious way to measure the
efficiency of an algorithm is to run it and measure how much
processor time is needed
 Is it correct ?
4

FACTORS
 Hardware
 Operating System
 Compiler
 Size of input
 Nature of Input
 Algorithm
Which should be improved?
5

RUNNING TIME OF AN
ALGORITHM
 Depends upon
 Input Size
 Nature of Input
 Generally time grows with size of input, so running time of an
algorithm is usually measured as function of input size.
 Running time is measured in terms of number of
steps/primitive operations performed
 Independent from machine, OS
6

FINDING RUNNING TIME OF AN
ALGORITHM / ANALYZING AN
ALGORITHM
 Running time is measured by number of steps/primitive
operations performed
 Steps means elementary operation like
 ,+, *,<, =, A[i] etc
 We will measure number of steps taken in term of size of
input
7

SIMPLE EXAMPLE (1)
// Input: int A[N], array of N integers
// Output: Sum of all numbers in array A
int Sum(int A[], int N)
{
int s=0;
for (int i=0; i< N; i++)
s = s + A[i];
return s;
}
How should we analyse this?
8

SIMPLE EXAMPLE (2)
9
// Input: int A[N], array of N integers
// Output: Sum of all numbers in array A
int Sum(int A[], int N){
int s=0;
for (int i=0; i< N; i++)
s = s + A[i];
return s;
}
1
2 3 4
5 6 7
8
1,2,8: Once
3,4,5,6,7: Once per each iteration
of for loop, N iteration
Total: 5N + 3
The complexity function of the
algorithm is : f(N) = 5N +3

SIMPLE EXAMPLE (3) GROWTH
OF 5N+3
Estimated running time for different values of N:
N = 10 => 53 steps
N = 100 => 503 steps
N = 1,000 => 5003 steps
N = 1,000,000 => 5,000,003 steps
As N grows, the number of steps grow in linear proportion to N for
this function “Sum”
10

WHAT DOMINATES IN PREVIOUS
EXAMPLE?
What about the +3 and 5 in 5N+3?
 As N gets large, the +3 becomes insignificant
 5 is inaccurate, as different operations require varying amounts of time and
also does not have any significant importance
What is fundamental is that the time is linear in N.
Asymptotic Complexity: As N gets large, concentrate on the
highest order term:
 Drop lower order terms such as +3
 Drop the constant coefficient of the highest order term i.e. N
11

ASYMPTOTIC COMPLEXITY
 The 5N+3 time bound is said to "grow asymptotically"
like N
 This gives us an approximation of the complexity of the
algorithm
 Ignores lots of (machine dependent) details, concentrate
on the bigger picture
12

COMPARING FUNCTIONS:
ASYMPTOTIC NOTATION
 Big Oh Notation: Upper bound
 Omega Notation: Lower bound
 Theta Notation: Tighter bound
13

BIG OH NOTATION [1]
If f(N) and g(N) are two complexity functions, we say
f(N) = O(g(N))
(read "f(N) is order g(N)", or "f(N) is big-O of g(N)")
if there are constants c and N0 such that for N > N0,
f(N) ≤ c * g(N)
for all sufficiently large N.
14

EXAMPLE (2): COMPARING
FUNCTIONS
 Which function is better?
10 n2
Vs n3
0
500
1000
1500
2000
2500
3000
3500
4000
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
10 n^2
n^3
17

COMPARING FUNCTIONS
 As inputs get larger, any algorithm of a smaller order
will be more efficient than an algorithm of a larger order
18
Time
(steps)
Input (size)
3N = O(N)
0.05 N2
= O(N2
)
N = 60

BIG-OH NOTATION
 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)
8n2
log n + 5n2
+ n is O(n2
log n)
19

BIG OMEGA NOTATION
 If we wanted to say “running time is at least…” we use Ω
 Big Omega notation, Ω, is used to express the lower bounds on a
function.
 If f(n) and g(n) are two complexity functions then we can say:
f(n) is Ω(g(n)) if there exist positive numbers c and n0 such that 0<=f(n)>=cΩ (n) for all n>=n0
20

BIG THETA NOTATION
 If we wish to express tight bounds we use the theta notation, Θ
 f(n) = Θ(g(n)) means that f(n) = O(g(n)) and f(n) = Ω(g(n))
21

WHAT DOES THIS ALL MEAN?
 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).
22

WHICH NOTATION DO WE USE?
 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.
23

PERFORMANCE CLASSIFICATION
f(n) Classification
1 Constant: run time is fixed, and does not depend upon n. Most instructions are executed once, or
only a few times, regardless of the amount of information being processed
log n Logarithmic: when n increases, so does run time, but much slower. Common in programs which solve
large problems by transforming them into smaller problems. Exp : binary Search
n Linear: run time varies directly with n. Typically, a small amount of processing is done on each
element. Exp: Linear Search
n log n When n doubles, run time slightly more than doubles. Common in programs which break a problem
down into smaller sub-problems, solves them independently, then combines solutions. Exp: Merge
n2 Quadratic: when n doubles, runtime increases fourfold. Practical only for small problems; typically
the program processes all pairs of input (e.g. in a double nested loop). Exp: Insertion Search
n3 Cubic: when n doubles, runtime increases eightfold. Exp: Matrix
2n Exponential: when n doubles, run time squares. This is often the result of a natural, “brute force”
solution. Exp: Brute Force.
Note: logn, n, nlogn, n2
>> less Input>>Polynomial
n3,
2n
>>high input>> non polynomial
24

SIZE DOES MATTER[1]
25
What happens if we double the input size N?
N log2N 5N N log2N N2
2N
8 3 40 24 64 256
16 4 80 64 256 65536
32 5 160 160 1024 ~109
64 6 320 384 4096 ~1019
128 7 640 896 16384 ~1038
256 8 1280 2048 65536 ~1076

COMPLEXITY CLASSES
Time
(steps)
26

SIZE DOES MATTER[2]
 Suppose a program has run time O(n!) and the run time for
n = 10 is 1 second
For n = 12, the run time is 2 minutes
For n = 14, the run time is 6 hours
For n = 16, the run time is 2 months
For n = 18, the run time is 50 years
For n = 20, the run time is 200 centuries
27

STANDARD ANALYSIS
TECHNIQUES
 Constant time statements
 Analyzing Loops
 Analyzing Nested Loops
 Analyzing Sequence of Statements
 Analyzing Conditional Statements
28

CONSTANT TIME STATEMENTS
 Simplest case: O(1) time statements
 Assignment statements of simple data types
int x = y;
 Arithmetic operations:
x = 5 * y + 4 - z;
 Array referencing:
A[j] = 5;
 Array assignment:
 j, A[j] = 5;
 Most conditional tests:
if (x < 12) ...
29

ANALYZING LOOPS[1]
 Any loop has two parts:
 How many iterations are performed?
 How many steps per iteration?
int sum = 0,j;
for (j=0; j < N; j++)
sum = sum +j;
 Loop executes N times (0..N-1)
 4 = O(1) steps per iteration
 Total time is N * O(1) = O(N*1) = O(N) 30

ANALYZING LOOPS[2]
 What about this for loop?
int sum =0, j;
for (j=0; j < 100; j++)
sum = sum +j;
 Loop executes 100 times
 4 = O(1) steps per iteration
 Total time is 100 * O(1) = O(100 * 1) = O(100) = O(1)
31

ANALYZING LOOPS – LINEAR
LOOPS
 Example (have a look at this code segment):
 Efficiency is proportional to the number of iterations.
 Efficiency time function is :
f(n) = 1 + (n-1) + c*(n-1) +( n-1)
= (c+2)*(n-1) + 1
= (c+2)n – (c+2) +1
 Asymptotically, efficiency is : O(n)
32

ANALYZING NESTED LOOPS[1]
 Treat just like a single loop and evaluate each level of nesting as
needed:
int j,k;
for (j=0; j<N; j++)
for (k=N; k>0; k--)
sum += k+j;
 Start with outer loop:
 How many iterations? N
 How much time per iteration? Need to evaluate inner loop
 Inner loop uses O(N) time
 Total time is N * O(N) = O(N*N) = O(N2
) 33

ANALYZING NESTED LOOPS[2]
 What if the number of iterations of one loop depends on the
counter of the other?
int j,k;
for (j=0; j < N; j++)
for (k=0; k < j; k++)
sum += k+j;
 Analyze inner and outer loop together:
 Number of iterations of the inner loop is:
 0 + 1 + 2 + ... + (N-1) = O(N2
)
34

HOW DID WE GET THIS
ANSWER?
 When doing Big-O analysis, we sometimes have to compute a
series like: 1 + 2 + 3 + ... + (n-1) + n
 i.e. Sum of first n numbers. What is the complexity of this?
 Gauss figured out that the sum of the first n numbers is always:
35

SEQUENCE OF STATEMENTS
 For a sequence of statements, compute their complexity
functions individually and add them up
 Total cost is O(n2
) + O(n) +O(1) = O(n2
)
36

CONDITIONAL STATEMENTS
 What about conditional statements such as
if (condition)
statement1;
else
statement2;
 where statement1 runs in O(n) time and statement2 runs in O(n2
)
time?
 We use "worst case" complexity: among all inputs of size n, what is
the maximum running time?
 The analysis for the example above is O(n2
)
37

DERIVING A RECURRENCE
EQUATION
 So far, all algorithms that we have been analyzing have been non
recursive
 Example : Recursive power method
 If N = 1, then running time T(N) is 2
 However if N ≥ 2, then running time T(N) is the cost of each step taken plus time
required to compute power(x,n-1). (i.e. T(N) = 2+T(N-1) for N ≥ 2)
 How do we solve this? One way is to use the iteration method.
38

ITERATION METHOD
 This is sometimes known as “Back Substituting”.
 Involves expanding the recurrence in order to see a pattern.
 Solving formula from previous example using the iteration method
:
 Solution : Expand and apply to itself :
Let T(1) = n0 = 2
T(N) = 2 + T(N-1)
= 2 + 2 + T(N-2)
= 2 + 2 + 2 + T(N-3)
= 2 + 2 + 2 + ……+ 2 + T(1)
= 2N + 2 remember that T(1) = n0 = 2 for N = 1
 So T(N) = 2N+2 is O(N) for last example.
39

SUMMARY
 Algorithms can be classified according to their
complexity => O-Notation
 only relevant for large input sizes
 "Measurements" are machine independent
 worst-, average-, best-case analysis
40

REFERENCES
Introduction to Algorithms by Thomas H. Cormen
Chapter 3 (Growth of Functions)
41

How to calculate complexity in Data Structure

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