insertion sort insertion sort animation ana simple

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 21
45 79 47 22
38 74 36
66 94 29
57 81
60 16
45
66
60
45
the fourth iteration of this loop is shown here

An insertion sort partitions the array into two regions

An insertion sort of an array of five integers

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.

Insertion Sort: Cost Function
 1 operation to initialize the outer loop
 The outer loop is evaluated n-1 times
 5 instructions (including outer loop comparison and increment)
 Total cost of the outer loop: 5(n-1)
 How many times the inner loop is evaluated is affected by the
state of the array to be sorted
 Best case: the array is already completely sorted so no “shifting”
of array elements is required.
 We only test the condition of the inner loop once (2 operations = 1
comparison + 1 element comparison), and the body is never
executed
 Requires 2(n-1) operations.

Insertion Sort: Cost Function
 Worst case: the array is sorted in reverse order (so each item
has to be moved to the front of the array)
 In the i-th iteration of the outer loop, the inner loop will perform 4i+1
operations
 Therefore, the total cost of the inner loop will be 2n(n-1)+n-1
 Time cost:
 Best case: 7(n-1)
 Worst case: 5(n-1)+2n(n-1)+n-1
 What about the number of moves?
 Best case: 2(n-1) moves
 Worst case: 2(n-1)+n(n-1)/2

Insertion Sort: Average Case
 Is it closer to the best case (n comparisons)?
 The worst case (n * (n-1) / 2) comparisons?
 It turns out that when random data is sorted, insertion sort is usually
closer to the worst case
 Around n * (n-1) / 4 comparisons
 Calculating the average number of comparisons more exactly would
require us to state assumptions about what the “average” input data set
looked like
 This would, for example, necessitate discussion of how items were
distributed over the array
 Exact calculation of the number of operations required to perform
even simple algorithms can be challenging
(for instance, assume that each initial order of elements has the
same probability to occur)

Bubble Sort
 Simplest sorting algorithm
 Idea:
 1. Set flag = false
 2. Traverse the array and compare pairs of two
consecutive elements
 1.1 If E1  E2 -> OK (do nothing)
 1.2 If E1 > E2 then Swap(E1, E2) and set flag = true
 3. repeat 1. and 2. while flag=true.

Bubble Sort
1 1 23 2 56 9 8 10 100
2 1 2 23 56 9 8 10 100
3 1 2 23 9 56 8 10 100
4 1 2 23 9 8 56 10 100
5 1 2 23 9 8 10 56 100
---- finish the first traversal ----
1 1 2 23 9 8 10 56 100
2 1 2 9 23 8 10 56 100
3 1 2 9 8 23 10 56 100
4 1 2 9 8 10 23 56 100
---- finish the second traversal ----
…

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
 Time cost, number of comparisons, number
of moves -> Assignment 4

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 (different types of
operations do not take the same time anyway)
 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

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)

Big O Notation
 Given functions tA(n) and g(n), we can say that the
efficiency of an algorithm is of order g(n) if there are
positive constants c and m such that
 tA(n) · c.g(n) for all n ¸ m
 we write
 tA(n) is O(g(n)) and we say that
 tA(n) is of order g(n)
 e.g. if an algorithm’s running time is 3n + 12 then the
algorithm is O(n). If c is 3 and m is 12 then:
 4 * n  3n + 12 for all n  12

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)), …

insertion sort insertion sort animation ana simple

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