Transcript Complexity Analysis

Introduction to Complexity Analysis
• Motivation
• Average, Best, and Worst Case Complexity Analysis
• Asymptotic Analysis
Motivations for Complexity Analysis
• There are often many different algorithms which can be used to
solve the same problem.
– For example, assume that we want to search for a key in a sorted array.
• Thus, it makes sense to develop techniques that allow us to:
o compare different algorithms with respect to their “efficiency”
o choose the most efficient algorithm for the problem
• The efficiency of any algorithmic solution to a problem can be
measured according to the:
o Time efficiency: the time it takes to execute.
o Space efficiency: the space (primary or secondary memory) it uses.
• We will focus on an algorithm’s efficiency with respect to time.
How do we Measure Efficiency?
• Running time in [micro/milli] seconds
– Advantages
– Disadvantages
• Instead,
o we count the number of basic operations the algorithm performs.
o we calculate how this number depends on the size of the input.
• A basic operation is an operation which takes a constant
amount of time to execute.
– E.g., additions, subtractions, multiplications, comparisons, …etc.
• Hence, the efficiency of an algorithm is determined in
terms of the number of basic operations it performs. This
number is most useful when expressed as a function of
the input size n.
Example of Basic Operations:
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Arithmetic operations: *, /, %, +, Assignment statements of simple data types.
Reading of primitive types
writing of a primitive types
Simple conditional tests:
if (x < 12) ...
method call (Note: the execution time of the method itself may
depend on the value of parameter and it may not be constant)
a method's return statement
Memory Access
We consider an operation such as ++ , += , and *= as consisting of
two basic operations.
Note: To simplify complexity analysis we will not consider memory
access (fetch or store) operations.
Simple Complexity Analysis
public class TestAbstractClass {
public static void main(String[] args) {
Employee[] list = new Employee[3];
list[0] = new Executive("Jarallah Al-Ghamdi", 50000);
list[1] = new HourlyEmployee("Azmat Ansari", 120);
list[2] = new MonthlyEmployee("Sahalu Junaidu", 9000);
((Executive)list[0]).awardBonus(11000);
for(int i = 0; i < list.length; i++)
if(list[i] instanceof HourlyEmployee)
((HourlyEmployee)list[i]).addHours(60);
for(int i = 0; i < list.length; i++) {
list[i].print();
System.out.println("Paid: " + list[i].pay());
System.out.println("*************************");
}
}
}
• # of basic operations:
• Assignment statements:
• Additions:
• Print statements:
• Method calls:
Simple Complexity Analysis
• Counting the number of basic operations
is cumbersome
– It is not important to count the number of all
basic operations
• Instead, we count/find the “number of
times” of the statement that gets executed
the most
– E.g. find the number of element comparisons
Simple Complexity Analysis: Simple Loops
• How to find the cost of single unnested
loops
– Find the cost of the body of the loop
– In the case below, consider the number of multiplications
– Find the number of iterations for the for loop
– Multiply the two numbers to get the total
– E.g.
double x, y;
x = 2.5 ; y = 3.0;
for(int i = 0; i < n; i++){
a[i] = x * y;
x = 2.5 * x;
y = y + a[i];
}
Simple Complexity Analysis: Complex Loops
•
Represent the cost of the for loop in summation
form.
–
–
The main idea is to make sure that we find an
iterator that increase/decrease its values by 1.
For example, consider finding the number of times
statements 1, 2 and 3 get executed below:
n 1
for (int i = 1; i < n; i++)
statement1;
for (int i = 1; i <= n; i++)
for (int j = 1; j <= n; j++)
statement2;
}
for (int i = 1; i <= n; i++)
for (int j = 1; j <= i; j++)
statement3;
}
1  n  1
i 1
n
n
n
n
i 1
i 1
2
1

n

n
1

n
 

i 1 j 1
n
i
n
n  n  1
i 1
2
1  i 
i 1 j 1
Useful Summation Formulas
 n 
c  c   1  c .  n  m  1

i m
 i m 
n
n
n  n  1
i 1
2
i 
n
i
2
i 1

n  n  1 2n  1
6
 n  n  1 
i 


2
i 1


n
2
3
n 1
a
1
i
a

,a  1

a 1
i 0
n
Simple Complexity Analysis: Complex Loops
• Represent the cost of the for loop in
summation form.
– The problem in the example below is that
the value of i does not increase by 1
for (int i = k; i <= n; i = i + m)
statement4;
•
i: k , k + m , k + 2m , …, k + rm
–
–
r
Here, we can assume without loss of generality that
k + rm = n, i.e. r = (n – k)/m
i.e., an iterator s from 0, 1, …,r can be used
n k
m
1  
s 0
s 0
n k
n k
1
 0 1 
1
m
m
Simple Complexity Analysis : Loops (with <=)
• In the following for-loop:
for (int i = k; i <= n; i = i + m){
statement1;
statement2;
}
• The number of iterations is: (n – k) / m +1
• The initialization statement, i = k, is executed one time.
• The condition, i <= n, is executed (n – k) / m +1 + 1 times.
• The update statement, i = i + m, is executed (n – k) / m +1 times.
• Each of statement1 and statement2 is executed (n – k) / m +1
times.
Simple Complexity Analysis: Loops (with <)
• In the following for-loop:
for (int i = k; i < n; i = i + m){
statement1;
statement2;
}
The number of iterations is: (n – k ) / m
• The initialization statement, i = k, is executed one time.
• The condition, i < n, is executed (n – k ) / m + 1 times.
• The update statement, i = i + m, is executed (n – k ) / m times.
• Each of statement1 and statement2 is executed (n – k ) / m times.
Simple Complexity Analysis: Complex Loops
• Suppose n is a power of 2. Determine the number of basic
operations performed by of the method myMethod():
static int myMethod(int n){
int sum = 0;
for(int i = 1; i <= n; i = i * 2)
sum = sum + i + helper(i);
return sum;
}
static int helper(int n){
int sum = 0;
for(int i = 1; i <= n; i++)
sum = sum + i; //statement5
return sum;
}
• Solution:
• The variables i and n in myMethod are different from the ones in the
helper method.
– In fact, n of “helper” is being called by variable i in “myMethod”.
– Hence, we need to change the name of variable i in helper because it is
independent from i in myMethod (let us call it k).
• We will count the number of times statement5 gets executed.
• (in myMethod) i : 1 , 2 , 22 , 23 ,…, 2r = n (r = log2 n)
Hence, we can use j where i = 2j j : 0 , 1 , 2 , 3, …, r = log2 n
r
r
i
r
r
 cost(Helper(i ))  1   i   2
j 0
j 0 k 1
j 0
j 0
j
2
r 1
 1  2n  1
Useful Logarithmic Formulas
Best, Average, and Worst case complexities
• What is the best case complexity analysis?
• The smallest number of basic operations carried out by the
algorithm for a given input.
• What is the worst case complexity analysis?
• The largest number of basic operations carried out by the algorithm
for a given input.
• What is the average case complexity analysis?
• The number of basic operations carried out by the algorithm on average.

(Probability of input i * Cost of input i)
for each input i
• We are usually interested in the worst case complexity
• Easier to compute
• Represents an upper bound on the actual running time for all
inputs
• Crucial to real-time systems (e.g. air-traffic control)
Best, Average, and Worst case complexities:
Example
• For linear search algorithm, searching for a key in an array of n
elements, determine the situation and the number of comparisons in
each of the following cases
– Best Case
– Worst Case
– Average Case
Asymptotic Growth
• Since counting the exact number of operations is
cumbersome, sometimes impossible, we can
always focus our attention to asymptotic
analysis, where constants and lower-order terms
are ignored.
– E.g. n3, 1000n3, and 10n3+10000n2+5n-1 are all “the
same”
– The reason we can do this is that we are always
interested in comparing different algorithms for
arbitrary large number of inputs.
Asymptotic Growth (1)
Asymptotic Growth (2)
Running Times for Different Sizes
of Inputs of Different Functions
Asymptotic Complexity
• Finding the exact complexity, f(n) = number of basic
operations, of an algorithm is difficult.
• We approximate f(n) by a function g(n) in a way that
does not substantially change the magnitude of f(n). -the function g(n) is sufficiently close to f(n) for large
values of the input size n.
• This "approximate" measure of efficiency is called
asymptotic complexity.
• Thus the asymptotic complexity measure does not
give the exact number of operations of an algorithm, but
it shows how that number grows with the size of the
input.
• This gives us a measure that will work for different
operating systems, compilers and CPUs.
Big-O (asymptotic) Notation
• The most commonly used notation for specifying asymptotic
complexity is the big-O notation.
• The Big-O notation, O(g(n)), is used to give an upper bound on a
positive runtime function f(n) where n is the input size.
Definition of Big-O:
• Consider a function f(n) that is non-negative  n  0. We say that
“f(n) is Big-O of g(n)” i.e., f(n) = O(g(n)), if  n0  0 and a constant
c > 0 such that f(n)  cg(n),  n  n0
Big-O (asymptotic) Notation
Implication of the definition:
• For all sufficiently large n, c *g(n) is an upper bound of
f(n)
Note: By the definition of Big-O:
f(n) = 3n + 4 is O(n)
it is also O(n2),
it is also O(n3),
...
it is also O(nn)
• However when Big-O notation is used, the function g in
the relationship f(n) is O(g(n)) is CHOSEN TO BE AS
SMALL AS POSSIBLE.
– We call such a function g a tight asymptotic bound of f(n)
Big-O (asymptotic) Notation
Some Big-O complexity classes in order of magnitude from smallest to highest:
O(1)
Constant
O(log(n))
Logarithmic
O(n)
Linear
O(n log(n))
n log n
O(nx)
{e.g., O(n2), O(n3), etc}
Polynomial
O(an)
{e.g., O(1.6n), O(2n), etc}
Exponential
O(n!)
O(nn)
Factorial
Examples of Algorithms and their big-O complexity
Method
Best Case
Worst Case
Average Case
Selection sort
O(n2)
O(n2)
O(n2)
Insertion sort
O(n)
O(n2)
O(n2)
Merge sort
O(n log n)
O(n log n)
O(n log n)
Quick sort
O(n log n)
O(n2)
O(n log n)
Warnings about O-Notation
•
•
•
Big-O notation cannot compare
algorithms in the same complexity class.
Big-O notation only gives sensible
comparisons of algorithms in different
complexity classes when n is large .
Consider two algorithms for same task:
Linear: f(n) = 1000 n
Quadratic: f'(n) = n2/1000
The quadratic one is faster for n < 1000000.
Rules for using big-O
• For large values of input n , the constants and terms with lower
degree of n are ignored.
1. Multiplicative Constants Rule: Ignoring constant factors.
O(c f(n)) = O(f(n)), where c is a constant;
Example:
O(20 n3) = O(n3)
2. Addition Rule: Ignoring smaller terms.
If O(f(n)) < O(h(n)) then O(f(n) + h(n)) = O(h(n)).
Example:
O(n2 log n + n3) = O(n3)
O(2000 n3 + 2n ! + n800 + 10n + 27n log n + 5) = O(n !)
3. Multiplication Rule: O(f(n) * h(n)) = O(f(n)) * O(h(n))
Example:
O((n3 + 2n 2 + 3n log n + 7)(8n 2 + 5n + 2)) = O(n5)
How to determine complexity of code structures
Loops: for, while, and do-while:
Complexity is determined by the number of iterations in
the loop times the complexity of the body of the loop.
Examples:
for (int i = 0; i < n; i++)
sum = sum - i;
O(n)
for (int i = 0; i < n * n; i++)
sum = sum + i;
O(n2)
i=1;
while (i < n) {
sum = sum + i;
i = i*2
}
O(log n)
How to determine complexity of code structures
Nested Loops:
Examples:
sum = 0
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
sum += i * j ;
i = 1;
while(i <= n) {
j = 1;
while(j <= n){
statements of constant complexity
j = j*2;
}
i = i+1;
}
O(n2)
O(n log n)
How to determine complexity of code structures
Sequence of statements: Use Addition rule
O(s1; s2; s3; … sk) = O(s1) + O(s2) + O(s3) + … + O(sk)
= O(max(s1, s2, s3, . . . , sk))
Example:
for (int j = 0; j < n * n; j++)
sum = sum + j;
for (int k = 0; k < n; k++)
sum = sum - l;
System.out.print("sum is now ” + sum);
Complexity is O(n2) + O(n) +O(1) = O(n2)
How to determine complexity of code structures
Switch: Take the complexity of the most expensive case
char key;
int[] X = new int[n];
int[][] Y = new int[n][n];
........
switch(key) {
case 'a':
for(int i = 0; i < X.length; i++)
sum += X[i];
break;
case 'b':
for(int i = 0; i < Y.length; j++)
for(int j = 0; j < Y[0].length; j++)
sum += Y[i][j];
break;
} // End of switch block
Overall Complexity: O(n2)
o(n)
o(n2)
How to determine complexity of code structures
If Statement: Take the complexity of the most expensive case :
char key;
int[][] A = new int[n][n];
int[][] B = new int[n][n];
int[][] C = new int[n][n];
........
if(key == '+') {
for(int i = 0; i < n; i++)
for(int j = 0; j < n; j++)
C[i][j] = A[i][j] + B[i][j];
} // End of if block
else if(key == 'x')
C = matrixMult(A, B);
O(n2)
Overall
complexity
O(n3)
O(n3)
else
System.out.println("Error! Enter '+' or 'x'!");
O(1)
How to determine complexity of code structures
•
Sometimes if-else statements must carefully be checked:
O(if-else) = O(Condition)+ Max[O(if), O(else)]
int[] integers = new int[n];
........
if(hasPrimes(integers) == true)
integers[0] = 20;
O(1)
else
O(1)
integers[0] = -20;
public boolean hasPrimes(int[] arr) {
for(int i = 0; i < arr.length; i++)
..........
O(n)
..........
} // End of hasPrimes()
O(if-else) = O(Condition) =
O(n)