analysis of algorithm. reasons to analyze algorithms performance prediction compare the...

Analysis of Algorithm

Reasons to analyze algorithms

Performance prediction

Compare the performance of different algorithm for the same task and provide some guarantees on how well they perform

Understanding some theoretical bases for how algorithms perform This helps us to avoid performance bugs Clients get poor results because programmer did not

understand performance characteristics of the algorithm

CS2336: Computer Science II

Running time


How many time some operation has to be performed in order to get the computation done

Suppose two algorithms perform the same task. Which one is better?

First approach: implement them in Java and run the programs to get execution time Two problems for this approach:

First, The execution time of a particular program is dependent on the system load.

Second, the execution time is dependent on specific input.

Consider linear search and binary search for example. If an element to be searched happens to be the first in the list, linear search will find the element quicker than binary search.

Running time: Growth Rate

Second approach: Growth Rate Analyze algorithms independent of computers and

specific input Approximates the effect of a change on the size of

the input We can see how fast the execution time increases as

the input size increasesyou can compare two algorithms by examining their

growth rates.


Linear Search

The linear search approach compares the key element, key, sequentially with each element in the array list. The method continues to do so until the key matches an element in the list or the list is exhausted without a match being found. If a match is made, the linear search returns the index of the element in the array that matches the key. If no match is found, the search returns -1.

Linear Search Animation

6 4 1 9 7 3 2 8

6 4 1 9 7 3 2 8

6 4 1 9 7 3 2 8

6 4 1 9 7 3 2 8

6 4 1 9 7 3 2 8

6 4 1 9 7 3 2 8

3

3

3

3

3

3

animation

Key List

Big O Notation

“Linear Search Algorithm”

for (int i = 0; i < n; i++) {

if (key == a[i]) {

return i; // Found key, return index.

}

}


If the key is not in the array, it requires n comparisons for an array of size n

If the key is in the array, it requires n/2 comparisons on average

The algorithm’s execution time is proportional to the size of the array

Big O Notation

If you double the size of the array, you will expect the number of comparisons to double.

The algorithm grows at a linear rate.

The growth rate has an order of magnitude of n. Big O notation to abbreviate for “order of magnitude.”

The complexity of the linear search algorithm is O(n), pronounced as “order of n.”


Best, Worst, and Average Cases

For the same input size, an algorithm’s execution time may vary, depending on the input

An input that results in the shortest execution time is called the best-case input

An input that results in the longest execution time is called the worst-case input worst-case analysis is very useful. You can show that the algorithm will never be slower

than the worst-case. Worst-case analysis is easier to obtain and is thus

common.


Ignoring Multiplicative Constants

The linear search algorithm requires: n comparisons in the worst-case n/2 comparisons in the average-case

both cases require O(n) time The multiplicative constant (1/2) can be omitted Algorithm analysis is focused on growth rate and

multiplicative constants have no impact on growth rates

The growth rate for n/2 or 100n is the same as n i.e., O(n) = O(n/2) = O(100n).


Ignoring Non-Dominating Terms


“Find max number”

int max = a[0];

for (int i = 1; i < n; i++)

if (a[i] > max) max = a[i];

Return max;

It takes n-1 times of comparisons to find maximum number in a list of n elements

The complexity of this algorithm is O(n) The Big O notation allows you to ignore the non-dominating part

Useful Mathematic Summations


12

1222....2222

1

1....

2

)1()1(....321

1)1(3210

1)1(3210

nnn

nnn

a

aaaaaaa

nnnn

Examples: Determining Big-O

Repetition

Sequence

Selection

Logarithm


Repetition: Simple Loops

T(n) = (a constant c) * n = cn = O(n)

for (i = 1; i <= n; i++) {

k = k + 5;

} constant time

executedn times

Ignore multiplicative constants (e.g., “c”).

Time Complexity

Repetition: Nested Loops

T(n) = (a constant c) * n * n = cn2 = O(n2)

for (i = 1; i <= n; i++) {

for (j = 1; j <= n; j++) {

k = k + i + j;

}

} constant time

executedn times

Ignore multiplicative constants (e.g., “c”).

Time Complexity

inner loopexecutedn times


T(n) = c + 2c + 3c + 4c + … + nc = cn(n+1)/2 = (c/2)n2 + (c/2)n = O(n2)

for (i = 1; i <= n; i++) {

for (j = 1; j <= i; j++) {

k = k + i + j;

}

} constant time

executedn times

Ignore non-dominating terms

Time Complexity

inner loopexecutedi times

Ignore multiplicative constants


T(n) = 20 * c * n = O(n)

for (i = 1; i <= n; i++) {

for (j = 1; j <= 20; j++) {

k = k + i + j;

}

}constant time

executedn times

Time Complexity

inner loopexecuted20 times

Ignore multiplicative constants (e.g., 20*c)

Sequence

T(n) = c *10 + 20 * c * n = O(n)

for (i = 1; i <= n; i++) {

for (j = 1; j <= 20; j++) {

k = k + i + j;

}

}

executedn times

Time Complexity

inner loopexecuted20 times

for (j = 1; j <= 10; j++) { k = k + 4;

}

executed10 times

Selection

T(n) = test time + worst-case (if, else) = O(n) + O(n)

= O(n)

if (list.contains(e)) {

System.out.println(e);

}

else

for (Object t: list) {

System.out.println(t);

}

Time Complexity

Let n be list.size().Executedn times.

O(n)

Constant Time

The Big O notation estimates the execution time of an algorithm in relation to the input size. If the time is not related to the input size, the algorithm is said to take constant time with the notation O(1). For example, a method that retrieves an element at a given index in an array takes constant time, because it does not grow as the size of the array increases.

Computation of an

result = 1;

for (i = 1; i <= n; i++) {

result *= a ;

}


O(n)

i result

1 a

2 a2

3 a3

… …

k ak

…. …

n an

Computation of an

result = a;

for (i = 1; i <= k; i++) {

result = result * result ;

}


i result

1 a2î a2

2 a2î a4

3 a2î a8

… …

k a2^k an

n=2k => lg n = k

O(lg n)

If you square the input size, you only double the

time for the algorithm

Binary Search

For binary search to work, the elements in the array must already be ordered. Without loss of generality, assume that the array is in ascending order.

e.g., 2 4 7 10 11 45 50 59 60 66 69 70 79

The binary search first compares the key with the element in the middle of the array.

Binary Search, cont.

If the key is less than the middle element, you only need to search the key in the first half of the array.

If the key is equal to the middle element, the search ends with a match.

If the key is greater than the middle element, you only need to search the key in the second half of the array.

Consider the following three cases:

Binary Search

1 2 3 4 6 7 8 9

1 2 3 4 6 7 8 9

1 2 3 4 6 7 8 9

8

8

8

Key List

animation

Binary Search, cont.

[0] [1] [2] [3] [4] [5] [6] [7] [8] [9] [10] [11] [12]

2 4 7 10 11 45 50 59 60 66 69 70 79

key is 11

key < 50

list

mid

[0] [1] [2] [3] [4] [5] key > 7

key == 11

high low

mid high low

list

[3] [4] [5]

mid high low

list

2 4 7 10 11 45

10 11 45

From Idea to Soluton

/** Use binary search to find the key in the list */

public static int binarySearch(int[] list, int key) {

int low = 0;

int high = list.length - 1;

while (high >= low) {

int mid = (low + high) / 2;

if (key < list[mid])

high = mid - 1;

else if (key == list[mid])

return mid;

else

low = mid + 1;

}

return -1 - low;

}

Logarithm: Analyzing Binary Search

Each iteration in the algorithm contains a fixed number

of operations, denoted by c. Let T(n) denote the time

complexity for a binary search on a list of n elements.

Without loss of generality, assume n is a power of 2

and k=logn. Since binary search eliminates half of the

input after two comparisons

)(lognO

kn 2

T(n) = c + T(n/2) = c + c + T(n/4) = c + c + c + T(n/23) = 4c + T(n/24) = ck + T(n/2k)

Logarithmic Time

An algorithm with the O(logn) time complexity is

called a logarithmic algorithm. The base of the log

is 2, but the base does not affect a logarithmic

growth rate, so it can be omitted. The logarithmic

algorithm grows slowly as the problem size

increases. If you square the input size, you only

double the time for the algorithm.

Quadratic Time

An algorithm with the O(n2) time complexity is

called a quadratic algorithm. The quadratic

algorithm grows quickly as the problem size

increases. If you double the input size, the time

for the algorithm is quadrupled. Algorithms with

a nested loop are often quadratic.

Insertion Sort

2 9 5 4 8 1 62 9 5 4 8 1 6

2 5 9 4 8 1 6

2 4 5 8 9 1 61 2 4 5 8 9 6

2 4 5 9 8 1 6

1 2 4 5 6 8 9

int[] myList = {2, 9, 5, 4, 8, 1, 6}; // Unsorted

animation

The insertion sort algorithm sorts a list of values by repeatedly inserting an unsorted element into

a sorted sublist until the whole list is sorted.

How to Insert?

The insertion sort algorithm sorts a list of values by repeatedly inserting an unsorted element into a sorted sublist until the

whole list is sorted.

[0] [1] [2] [3] [4] [5] [6]

2 5 9 4 list Step 1: Save 4 to a temporary variable currentElement

[0] [1] [2] [3] [4] [5] [6]

2 5 9 list Step 2: Move list[2] to list[3]

[0] [1] [2] [3] [4] [5] [6]

2 5 9 list Step 3: Move list[1] to list[2]

[0] [1] [2] [3] [4] [5] [6]

2 4 5 9 list Step 4: Assign currentElement to list[1]

InsertionSort

public static void insertionSort(int a[]) {

n = a.length;

for (int i = 1; i < n; i++) {

int temp = a[i];

int j;

for(j = i - 1; j >= 0 && temp < a[j]; j--) {

a[j+1] = a[j];

}

a[j+1] = temp;

}

}


Analyzing Insertion Sort

At the kth iteration, to insert an element to a array of size k, it may take k comparisons to find the insertion position, and k moves to insert the element. Let T(n) denote the complexity for insertion sort and c denote the total number of other operations such as assignments and additional comparisons in each iteration. So,

Ignoring constants and smaller terms, the complexity of the

insertion sort algorithm is O(n2).

Towers of Hanoi

There are n disks labeled 1, 2, 3, . . ., n, and three towers labeled A, B, and C.

No disk can be on top of a smaller disk at any time.

All the disks are initially placed on tower A.

Only one disk can be moved at a time, and it must be the top disk on the tower.

Towers of Hanoi, cont.

A B

Original position

C A B

Step 4: Move disk 3 from A to B

C

A B

Step 5: Move disk 1 from C to A

C A B

Step 1: Move disk 1 from A to B

C

A C B

Step 2: Move disk 2 from A to C

A B

Step 3: Move disk 1 from B to C

C A B

Step 7: Mve disk 1 from A to B

C

A B

Step 6: Move disk 2 from C to B

C

Solution to Towers of Hanoi

A B

Original position

C

.

.

.

A B

Step 1: Move the first n-1 disks from A to C recursively

C

.

.

.

A B

Step2: Move disk n from A to C

C

.

.

.

A B

Step3: Move n-1 disks from C to B recursively

C

.

.

.

n-1 disks

n-1 disks

n-1 disks

n-1 disks

The Towers of Hanoi problem can be decomposed into three subproblems.

Solution to Towers of Hanoi

Move the first n - 1 disks from A to C with the assistance of tower B.

Move disk n from A to B. Move n - 1 disks from C to B with the assistance of

tower A.

Analyzing Towers of HanoiLet T(n) denote the complexity for the algorithm that moves disks and c denote the constant time to move one disk, i.e., T(1) is c. So,

)2()12(2...22

2...2)1(2)))2((2(2

)1(2)1()1()(

21

21

nnnn

nn

Occccc

cccTccnT

cnTnTcnTnT

2kT(n-k)

T(1) = C : n – k = 1 => k = n - 1

Exponential algorithms are not practical. If the disk move at rate of 1 disk per second then it would take 232 = 136 years to move

32 disks

Comparing Common Growth Functions

)2()()()log()()(log)1( 32 nOnOnOnnOnOnOO

)1(O Constant time)(lognO Logarithmic

time )(nO Linear time )log( nnO Log-linear time

)( 2nO Quadratic time

)( 3nO Cubic time )2( nO Exponential

time


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