1. Analysis of Algorithm

2. 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
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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.
First, there are many tasks running concurrently on a computer
Second,
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4. 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 increases
you can compare two algorithms by examining their growth rates.
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5. 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.

6. Linear Search Animation
Key
List 3 6 4 1 9 7 3 2 8 3 6 4 1 9 7 3 2 8 3 6 4 1 9 7 3 2 8 3 6 4 1 9 7 3 2 8 3 6 4 1 9 7 3 2 8 3 6 4 1 9 7 3 2 8

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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
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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.”
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9. 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.
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10. 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).
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11. 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
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12. Useful Mathematic Summations
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13. Examples: Determining Big-O
Repetition
Sequence
Selection
Logarithm
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14. Repetition: Simple Loops
for (i = 1; i <= n; i++) {
k = k + 5; }
executed
n times
constant time
Time Complexity
T(n) = (a constant c) * n = cn = O(n)
Ignore multiplicative constants (e.g., “c”).

15. Repetition: Nested Loops
for (i = 1; i <= n; i++) {
for (j = 1; j <= n; j++) {
k = k + i + j; }
executed
n times
inner loop
executed
n times
constant time
Time Complexity
T(n) = (a constant c) * n * n = cn2 = O(n2)
Ignore multiplicative constants (e.g., “c”).

16. Repetition: Nested Loops
for (i = 1; i <= n; i++) {
for (j = 1; j <= i; j++) {
k = k + i + j; }
executed
n times
inner loop
executed
i times
constant time
Time Complexity
T(n) = c + 2c + 3c + 4c + … + nc = cn(n+1)/2 = (c/2)n2 + (c/2)n = O(n2)
Ignore non-dominating terms
Ignore multiplicative constants

17. Repetition: Nested Loops
for (i = 1; i <= n; i++) {
for (j = 1; j <= 20; j++) {
k = k + i + j; }
executed
n times
inner loop
executed
20 times
constant time
Time Complexity
T(n) = 20 * c * n = O(n)
Ignore multiplicative constants (e.g., 20*c)

18. Sequence for (j = 1; j <= 10; j++) { k = k + 4; executed } 10 times
for (i = 1; i <= n; i++) {
for (j = 1; j <= 20; j++) {
k = k + i + j; }
executed
n times
inner loop
executed
20 times
Time Complexity
T(n) = c * * c * n = O(n)

19. T(n) = test time + worst-case (if, else)
Selection
O(n)
if (list.contains(e)) {
System.out.println(e); }
else
for (Object t: list) {
System.out.println(t);
Let n be
list.size().
Executed
n times.
Time Complexity
T(n) = test time + worst-case (if, else)
= O(n) + O(n)
= O(n)

20. 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.

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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
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Computation of an
result = a;
for (i = 1; i <= k; i++) {
result = result * result ; }
O(lg n) i
result 1
a2^i a2 2 a4 3 a8 … k
a2^k an
n=2k => lg n = k
If you square the input size, you only double the time for the algorithm
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23. 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.,
The binary search first compares the key with the element in the middle of the array.

24. Consider the following three cases:
Binary Search, cont.
Consider the following three cases:
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.

25. animation
Binary Search
Key
List 8 1 2 3 4 6 7 8 9 8 1 2 3 4 6 7 8 9 8 1 2 3 4 6 7 8 9

26. Binary Search, cont.

27. 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;

28. 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
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)

29. 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.

30. 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.

31. animation
Insertion Sort
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.
int[] myList = {2, 9, 5, 4, 8, 1, 6}; // Unsorted 2 9 5 4 8 1 6 2 9 5 4 8 1 6 2 5 9 4 8 1 6 2 4 5 9 8 1 6 2 4 5 8 9 1 6 1 2 4 5 8 9 6 1 2 4 5 6 8 9

32. 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.

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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;
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34. 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).

35. 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.

36. Towers of Hanoi, cont.

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

38. 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.

39. Analyzing Towers of Hanoi
Let 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,
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

40. Comparing Common Growth Functions
Constant time
Logarithmic time
Linear time
Log-linear time
Quadratic time
Cubic time
Exponential time

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Questions?
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