Insersion & Bubble Sort in Algoritm

Algorithm Analysis
and Design
White Hat

Insertion Sort
 while some elements unsorted:
 Using linear search, find the location in the sorted portion
where the 1st
element of the unsorted portion should be
inserted
 Move all the elements after the insertion location up one
position to make space for the new element
13 2145 79 47 2238 74 3666 94 2957 8160 16
45
666045
the fourth iteration of this loop is shown here

An insertion sort partitions the array into two regions
Insertion Sort

4
One step of insertion sort
3 4 7 12 14 14 20 21 33 38 10 55 9 23 28 16
sorted next to be inserted
3 4 7 55 9 23 28 16
10
temp
3833212014141210
sorted
less than
10

An insertion sort of an array of five integers
Insertion Sort

Insertion Sort Algorithm
public void insertionSort(Comparable[] arr) {
for (int i = 1; i < arr.length; ++i) {
Comparable temp = arr[i];
int pos = i;
// Shuffle up all sorted items > arr[i]
while (pos > 0 &&
arr[pos-1].compareTo(temp) > 0) {
arr[pos] = arr[pos–1];
pos--;
} // end while
// Insert the current item
arr[pos] = temp;
}
}

public void insertionSort(Comparable[] arr) {
for (int i = 1; i < arr.length; ++i) {
Comparable temp = arr[i];
int pos = i;
// Shuffle up all sorted items > arr[i]
while (pos > 0 &&
arr[pos-1].compareTo(temp) > 0) {
arr[pos] = arr[pos–1];
pos--;
} // end while
// Insert the current item
arr[pos] = temp;
}
}
Insertion Sort Analysis
outer loop
outer times
inner loop
inner times

Insertion Sort: Number of
Comparisons
# of Sorted
Elements
Best case Worst case
0 0 0
1 1 1
2 1 2
… … …
n-1 1 n-1
n-1 n(n-1)/2
Remark: we only count comparisons of elements in the array.

9
Bubble sort
9
© 2006 Pearson Addison-Wesley. All rights reserved 10 A-9
• Compare adjacent elements and exchange
them if they are out of order.
– Comparing the first two elements, the second and
third elements, and so on, will move the largest
elements to the end of the array
– Repeating this process will eventually sort the array
into ascending order

10
Example of bubble sort
7 2 8 5 4
2 7 8 5 4
2 7 8 5 4
2 7 5 8 4
2 7 5 4 8
2 7 5 4 8
2 5 7 4 8
2 5 4 7 8
2 7 5 4 8
2 5 4 7 8
2 4 5 7 8
2 5 4 7 8
2 4 5 7 8
2 4 5 7 8
(done)

Bubble Sort
public void bubbleSort (Comparable[] arr) {
boolean isSorted = false;
while (!isSorted) {
isSorted = true;
for (i = 0; i<arr.length-1; i++)
if (arr[i].compareTo(arr[i+1]) > 0) {
Comparable tmp = arr[i];
arr[i] = arr[i+1];
arr[i+1] = tmp;
isSorted = false;
}
}
}

Bubble Sort: analysis
 After the first traversal (iteration of the main
loop) – the maximum element is moved to its
place (the end of array)
 After the i-th traversal – largest i elements are
in their places

O-notation Introduction
 Exact counting of operations is often difficult (and
tedious), even for simple algorithms
 Often, exact counts are not useful due to other
factors, e.g. the language/machine used, or the
implementation of the algorithm
 O-notation is a mathematical language for
evaluating the running-time (and memory usage) of
algorithms

Growth Rate of an Algorithm
 We often want to compare the performance of
algorithms
 When doing so we generally want to know how they
perform when the problem size (n) is large
 Since cost functions are complex, and may be
difficult to compute, we approximate them using O
notation

Example of a Cost Function
 Cost Function: tA(n) = n2
+ 20n + 100
 Which term dominates?
 It depends on the size of n
 n = 2, tA(n) = 4 + 40 + 100
 The constant, 100, is the dominating term
 n = 10, tA(n) = 100 + 200 + 100
 20n is the dominating term
 n = 100, tA(n) = 10,000 + 2,000 + 100
 n2
is the dominating term
 n = 1000, tA(n) = 1,000,000 + 20,000 + 100
 n2
is the dominating term

Algorithm Growth Rates
Time requirements as a function of the problem size n

Order-of-Magnitude Analysis
and Big O Notation

Big O Notation
 O notation approximates the cost function of an
algorithm
 The approximation is usually good enough, especially
when considering the efficiency of algorithm as n gets very
large
 Allows us to estimate rate of function growth
 Instead of computing the entire cost function we only
need to count the number of times that an algorithm
executes its barometer instruction(s)
 The instruction that is executed the most number of times
in an algorithm (the highest order term)

In English…
 The cost function of an algorithm A, tA(n), can be approximated
by another, simpler, function g(n) which is also a function with
only 1 variable, the data size n.
 The function g(n) is selected such that it represents an upper
bound on the efficiency of the algorithm A (i.e. an upper bound
on the value of tA(n)).
 This is expressed using the big-O notation: O(g(n)).
 For example, if we consider the time efficiency of algorithm A
then “tA(n) is O(g(n))” would mean that
 A cannot take more “time” than O(g(n)) to execute or that
(more than c.g(n) for some constant c)
 the cost function tA(n) grows at most as fast as g(n)

The general idea is …
 when using Big-O notation, rather than giving a precise
figure of the cost function using a specific data size n
 express the behaviour of the algorithm as its data size n
grows very large
 so ignore
 lower order terms and
 constants

O Notation Examples
 All these expressions are O(n):
 n, 3n, 61n + 5, 22n – 5, …
 All these expressions are O(n2
):
 n2
, 9 n2
, 18 n2
+ 4n – 53, …
 All these expressions are O(n log n):
 n(log n), 5n(log 99n), 18 + (4n – 2)(log (5n + 3)), …

Running time depends on not only the size of the array
but also the contents of the array.
Best-case:  O(n)
Array is already sorted in ascending order.
Inner loop will not be executed.
The number of moves: 2*(n-1)  O(n)
The number of key comparisons: (n-1)  O(n)
Worst-case:  O(n2
)
Array is in reverse order:
Inner loop is executed i-1 times, for i = 2,3, …, n
The number of moves: 2*(n-1)+(1+2+...+n-1)= 2*(n-1)+ n*(n-1)/2  O(n2
)
The number of key comparisons: (1+2+...+n-1)= n*(n-1)/2  O(n2
)
Average-case:  O(n2
)
We have to look at all possible initial data organizations.
So, Insertion Sort is O(n2
)
Insertion Sort – Analysis

Bubble Sort – Analysis
Best-case:  O(n)
Array is already sorted in ascending order.
The number of moves: 0  O(1)
The number of key comparisons: (n-1)  O(n)
Worst-case:  O(n2
)
Array is in reverse order:
Outer loop is executed n-1 times,
The number of moves: 3*(1+2+...+n-1) = 3 * n*(n-1)/2  O(n2
)
The number of key comparisons: (1+2+...+n-1)= n*(n-1)/2  O(n2
)
Average-case:  O(n2
)
We have to look at all possible initial data organizations.
So, Bubble Sort is O(n2
)

Insersion & Bubble Sort in Algoritm

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