Algorithms - "Chapter 2 getting started"

Chapter 2: Getting Started
Mutah University
Faculty of IT, Department of Software Engineering
Dr. Ra’Fat A. AL-msie’deen
Algorithms

This material is based on chapter 2 of “Introduction to Algorithms” by
Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest, Clifford Stein.

Overview
• Aims to familiarize us with framework used
throughout text.
• Examines alternate solutions to the sorting problem
presented in Ch.1.
• Specify algorithms to solve problem.
• Argue for their correctness.
• Analyze running time, introducing notation for
asymptotic behavior.
• Introduce divide-and-conquer algorithm technique.
4

Outlines: Getting Started
• Sequential Approach:
o Insertion Sort.
• Divide-and-Conquer Approach:
o Merge Sort.
• pseudo-code.
• Flowchart.
5

The Sorting Problem
 Input: A sequence of n numbers [a1, a2, … , an].
 Output: A permutation or reordering [a'1, a'2, … , a'n ] of the input
sequence such that a'1  a'2  …  a'n .
 An instance of the Sorting Problem:
o Input: A sequence of 6 number [31, 41, 59, 26, 41, 58].
 Expected output for given instance:
o Expected output: The permutation of the input [26, 31, 41, 41, 58 , 59].
6

Insertion Sort
• The main idea …
7
Figure 2.1 Sorting a hand of cards using insertion sort.

Insertion Sort Execution Example
8
9 7 6 1015 16 5 11
9 6 1015 16 5 11
7 9 1015 16 5 11
6 7 9 1015 16 11
5 6 7 9 15 16 11
5 6 7 169 10 15
5 6 7 159 10 11 16
7
6
5
10
11
6 7 1015 16 5 119
6 7 1015 16 5 119

Insertion sort (Cont.)
• Figure 2.2 The operation of INSERTION-SORT on the array A={5; 2; 4;
6; 1; 3}.
• Array indices appear above the rectangles, and values stored in the
array positions appear within the rectangles.
9

Insertion sort - The algorithm …
10
INSERTION-SORT (A)
1. for i = 2 to A.length
2. key = A[i]
3. // Insert A[i] into the sorted sequence A[1 .. i - 1].
4. j = i - 1
5. while j > 0 and A[j] > key
6. A[j + 1] = A[j]
7. j = j - 1
8. A[j + 1] = key

Insertion sort - The algorithm (Cont.)
• (a)–(e) The iterations of the for loop of lines 1–8. In each iteration,
the black rectangle holds the key taken from A[i], which is
compared with the values in shaded rectangles to its left in the test
of line 5.
• Shaded arrows show array values moved one position to the right
in line 6, and black arrows indicate where the key moves to in line
8.
• (f) The ﬁnal sorted array.
11

Loop Invariant
• Property of A[1 .. i  1]
– At the start of each iteration of the for loop of lines 1 8, the
subarray A[1 .. i  1] consists of the elements originally in A[1 ..
i  1] but in sorted order.
• Need to establish the following re invariant:
– Initialization: true prior to first iteration.
– Maintenance: if true before iteration, remains true after
iteration.
– Termination: at loop termination, invariant implies correctness
of algorithm.
12

InsertionSort(A, n) {
for i = 2 to n {
key = A[i] // next key
j = i - 1; // go left
while (j > 0) and (A[j] > key) { // find place for key
A[j + 1] = A[j] // shift sorted right
j = j – 1 // go left
}
A[j+1] = key // put key in place
}
} 13
8 4 9 3 62

• Assignment:
o X := Expression; (or x ← expression;).
― Example: (key ← A[i]; see algorithm 1).
14

Insertion Sort – Java Program
15

An Example: Insertion Sort
int array[] = { 8, 2, 4, 9, 3, 6 };
for (int i = 1; i < array.length; i++) {
int key = array[i];
int j = i - 1;
while ((j > -1) && (array[j] > key)) {
array[j + 1] = array[j];
j = j - 1;
}
array[j + 1] = key;
}
16

An Example: Insertion Sort (Cont.)
int array[] = { 8, 2, 4, 9, 3, 6 };
int key = array[i];
int j = i - 1;
j = j - 1;
}
array[j + 1] = key;
}
17

int array[] = { 8, 2, 4, 9, 3, 6 };
int key = array[i];
int j = i - 1;
j = j - 1;
}
array[j + 1] = key;
}
18

int array[] = { 8, 2, 4, 9, 3, 6 };
int key = array[i];
int j = i - 1;
while ((j > -1) &&
(array[j] > key)) {
j = j - 1;
}
array[j + 1] = key;
}
19
Done!

An Example: Insertion Sort (2)
20
int array[] = { 30, 10, 40, 20 };
int key = array[i];
int j = i - 1;
j = j - 1;
}
array[j + 1] = key;
}

An Example: Insertion Sort (2) (Cont.)
21
int array[] = { 30, 10, 40, 20 };
int key = array[i];
int j = i - 1;
j = j - 1;
}
array[j + 1] = key;
}

An Example: Insertion Sort (2) (Cont.)
22
int array[] = { 30, 10, 40, 20 };
int key = array[i];
int j = i - 1;
j = j - 1;
}
array[j + 1] = key;
}
Done!

Analysis of Insertion Sort
• Time resource requirement depends on input
size.
• Input size depends on problem being studied;
frequently, this is the number of items in the
input.
• Running time: number of primitive operations
or “steps” executed for an input.
• Assume constant amount of time for each line
of pseudocode.
23

Analysis of Insertion Sort (Cont.)
InsertionSort(A, n) {
for i = 2 to n {
key = A[i]
j = i - 1;
while (j > 0) and (A[j] > key) {
A[j+1] = A[j]
j = j - 1
}
A[j+1] = key
}
}
24
How many times
will this while
loop execute?

• Time efficiency analysis …
• For each j = 2, 3, …. , n, where n = A.length, we let tj denote the number of
times the while loop test in line 5 is executed for that value of j.
25

• We assume that comments are not executable statements,
and so they take no time.
• The running time of the algorithm is the sum of running times
for each statement executed; a statement that takes ci steps
to execute and executes n times will contribute cin to the total
running time.
• To compute T(n), the running time of INSERTION-SORT on an
input of n values, we sum the products of the cost and times
columns, obtaining:
26
 

n
j
j
n
j
j
n
j
j nctctct
2
8
22
421 )1()1(7)1(6c51)-(nc1)-(ncncT(n)

• What can C be?
o Best case -- inner loop body never executed (array is
sorted)
 tj= 1 → T(n) is a linear function = a.n + b.
o Worst case -- inner loop body executed for all previous
elements (array sorted in reverse order).
 tj= j → T(n) is a quadratic function = a.n2 + b.n + c.
27
 

n
j
j
n
j
j
n
j
j nctctct
2
8
22
421 )1()1(7)1(6c51)-(nc1)-(ncncT(n)

Best Case Analysis
• Least amount of (time) resource ever needed by algorithm.
• Achieved when incoming list is already sorted in increasing
order.
• Inner loop is never iterated.
• Cost is given by:
T(n) = c1n+c2 (n1)+c4 (n1)+c5(n1)+c8(n1)
= (c1+c2+c4+c5+c8)n  (c2+c4+c5+c8)
= an + b
• Linear function of n.
28

Worst Case Analysis
• Greatest amount of (time) resource ever needed by algorithm.
• Achieved when incoming list is in reverse order.
• Inner loop is iterated the maximum number of times, i.e., tj = j.
• Therefore, the cost will be:
T(n) = c1n + c2 (n1)+c4 (n1) + c5((n(n+1)/2) 1) + c6(n(n1)/2)
+ c7(n(n1)/2) + c8(n1)
= ( c5 /2 + c6 /2 + c7/2 ) n2 + (c1+c2+c4+c5 /2  c6 /2  c7 /2 +c8 ) n
 ( c2 + c4 + c5 + c8 )
= an2 + bn + c
• Quadratic function of n.
29

• Best Case:
 If A is sorted: O(n) comparisons.
• Worst Case:
 If A is reversed sorted: O(n2) comparisons.
• Average Case:
 If A is randomly sorted: O(n2) comparisons.
30

Analyzing Algorithms …
• Has come to mean predicting the resources that the
algorithm requires (resources such as memory).
• Usually computational time is resource of primary
importance.
• Aims to identify best choice among several alternate
algorithms.
• Requires an agreed-upon “model” of computation.
• Shall use a generic, one-processor, random-access
machine (RAM) model of computation.
31

Random-Access Machine
• Instructions are executed one after another (no
concurrency).
• Admits commonly found instructions in “real”
computers, data movement (such as: load, store, copy)
operations (such as: add, subtract, multiply), control
mechanism (such as: subroutine call and return).
• Uses common data types (integer and float).
• Other properties discussed as needed.
• Care must be taken since model of computation has
great implications on resulting analysis.
32

Future Analyses
• For the most part, subsequent analyses will
focus on:
– Worst-case running time:
• Upper bound on running time for any input.
– Average-case analysis:
• Expected running time over all inputs.
• Often, worst-case and average-case have the
same “order of growth”
33

Order of Growth
• Simplifying abstraction: interested in rate of growth
or order of growth of the running time of the
algorithm.
• Allows us to compare algorithms without worrying
about implementation performance.
• Usually only highest order term without constant
coefficient is taken.
• Uses “theta” notation
– Best case of insertion sort is (n).
– Worst case of insertion sort is (n2). (pronounced “theta of
n-squared”).
34

Designing Algorithms
• Several techniques/patterns for designing
algorithms exist.
• Incremental approach: builds the solution one
component at a time.
• Divide-and-conquer approach: breaks original problem
into several smaller instances of the same problem.
– Results in recursive algorithms.
– Easy to analyze complexity using proven techniques.
35

Flowchart
• What is a Flowchart?
o A flowchart is a visual representation of the
sequence of steps and decisions needed to
perform a process. OR
o An organized combination of shapes, lines, and
text that graphically illustrates a process.
• Flowchart - Graphically depicts the logical steps to
carry out a task and shows how the steps relate to
each other.
36

Basic Flowchart Shapes and Definitions
37
Start / End
The start or end of a
workflow.
Project / Task
Process or action.
Split
or
Merge
Upright indicates a process split,
inverted indicates a merge of processes.
Off Page
Connector
Connector used to connect one
page of a flowchart to another.
Connector
Used to connect one part of a
flowchart to another.
Decision
Decision point in a
process or workflow.
Input /
Output
Data: Inputs to, and outputs
from, a process.
Document
Document or
report.
Manual
Input
Prompt for information,
manually entered into a system.
Flowline.

C++ Program
38
Table. 2.1 | Counter-controlled repetition with the for statement.
1. // Counter-controlled repetition with the for statement.
2. #include <iostream>
3. using namespace std;
4. int main()
5. {
6. // for statement header includes initialization,
7. // loop-continuation condition and increment.
8. for ( unsigned int counter = 1; counter <= 10; ++counter )
9. cout << counter << " ";
10. cout << endl; // output a newline
11. } // end main
1 2 3 4 5 6 7 8 9 10

Flowchart Example
39End
Initialize control variable
Increment the
control variable
Display the
counter value
[counter <= 10]
[counter > 10]
Start
unsigned int counter = 1
++countercout << counter << " ";
Determine
whether looping
should continue.

Outlines: Getting Started
• Sequential Approach:
o Insertion Sort.
• Divide-and-Conquer Approach:
o Merge Sort.
• pseudo-code.
• Flowchart.
40

Divide-and-Conquer
• Recursive in structure:
o Divide the problem into several smaller sub-
problems that are similar to the original but
smaller in size.
o Conquer the sub-problems by solving them
recursively. If they are small enough, just solve
them in a straightforward manner.
o Combine the solutions to create a solution to the
original problem.
• A recursive function is a function that calls itself,
either directly, or indirectly (through another
function). 41

Divide-and-Conquer (Cont.)
• Technique (or paradigm) involves:
– “Divide” stage: Express problem in terms of
several smaller subproblems.
– “Conquer” stage: Solve the smaller subproblems
by applying solution recursively – smallest
subproblems may be solved directly.
– “Combine” stage: Construct the solution to
original problem from solutions of smaller
subproblem.
42

An Example: Merge Sort
• Divide: Divide the n-element sequence to be
sorted into two sub-sequences of n/2
elements each.
• Conquer: Sort the two sub-sequences
recursively using merge sort.
• Combine: Merge the two sorted
subsequences to produce the sorted answer.
43

Merge Sort Strategy
• Divide stage: Split the n-
element sequence into two
subsequences of n/2
elements each.
• Conquer stage: Recursively
sort the two subsequences.
• Combine stage: Merge the
two sorted subsequences
into one sorted sequence
(the solution).
44
n
(sorted)
MERGE
n
(unsorted)
n/2
(unsorted)
n/2
(unsorted)
MERGE SORT MERGE SORT
n/2
(sorted)
n/2
(sorted)

Merging Sorted Sequences
1 2 2 3 4 5 6 7
45
Sorted sequence
2 4 5 7 1 2 3 6
2 5 4 7 1 3 2 6
2 5 4 7 1 3 2 6
merge merge
merge
Initial sequence
mergemergemergemerge
Example: Bottom-up view for n = 8

Merging Sorted Sequences (Cont.)
• Example:
Bottom-up
view for n = 8.
46

Merging Sorted Sequences (Cont.)
• Example:
Bottom-up view
for n = 11.
47

Merge Sort
MergeSort(A, p, r) {
if (p < r) { // Check for base case
q = floor((p + r) / 2); // Divide
MergeSort(A, p, q); // Conquer
MergeSort(A, q+1, r); // Conquer
Merge(A, p, q, r); // Combine
}
}
// Merge() takes two sorted subarrays of A and
// merges them into a single sorted subarray of A.
// It requires (n) time
// floor( x ) rounds x to the largest integer not greater than x
// If p ≥ r, the subarray has at most one element and is therefore
already sorted.
48
A
p q r
1 2 3 4 5
x xx xxx xxxx xxxxx

Merge Sort (Cont.)
}
}
49
The key operation of the merge
sort algorithm is the merging of
two sorted sequences in the
“combine” step.

Merge Sort (Cont.)
}
}
50
We merge the two sub-arrays
by calling an auxiliary
procedure Merge(A, p, q, r),
where A is an array and p, q,
and r are indices into the array
such that p ≤ q<r.

Merge Sort (Cont.)
• Input: Array A and indices p, q, r such that:
o p ≤ q < r.
o Subarray A[p . . q] is sorted and subarray A[q + 1 . . r] is
sorted. By the restrictions on p, q, r, neither subarray is
empty.
• Output: The two subarrays are merged into a single sorted
subarray in A[p . . r].
• We implement it so that it takes (n) time, where n = r − p + 1
= the number of elements being merged.
51

Recursive function - Example
Table. 2.2 | Recursive function factorial.
//Recursive function factorial.
#include <iostream>
#include <iomanip>
using namespace std;
unsigned long factorial( unsigned long ); // function prototype
int main() {
//calculate the factorials of 0 through 10
for ( unsigned int counter = 0; counter <= 10; ++counter )
cout << setw( 2 ) << counter << "! = " <<factorial (counter)<< endl;
} // end main
unsigned long factorial( unsigned long number ){
if ( number <= 1 ) // test for base case
return 1; // base cases: 0! = 1 and 1! = 1
else // recursion step
return number * factorial( number - 1 );
} // end function factorial 52
Outputs
0! = 1
1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
8! = 40320
9! = 362880
10! = 3628800

Merge Sort (Cont.)
53
[8, 3, 13, 6, 2, 14, 5, 9, 10, 1, 7, 12, 4]
[8, 3, 13, 6, 2, 14, 5] [9, 10, 1, 7, 12, 4]
[8, 3, 13, 6] [2, 14, 5] [9, 10, 1] [7, 12, 4]
[8, 3] [13, 6] [2, 14] [5] [9, 10] [1] [7, 12] [ 4]
[8 ] [3] [13 ] [6] [2 ] [14] [5] [9 ] [10] [1] [7] [12] [ 4]Divide
Array A and indices left (p), mid (q), right (r).

Merge Sort (Cont.)
54
[1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13,14]
[2, 3, 5, 6, 8, 13, 14] [1, 4, 7, 9, 10,12]
[3, 6, 8, 13] [2, 5, 14] [1, 9, 10] [4, 7, 12]
[3, 8] [6, 13] [2, 14] [5] [9, 10] [1] [7, 12] [ 4]
[8 ] [3] [13 ] [6] [2 ] [14] [5] [9 ] [10] [1] [7] [12] [ 4]
Combine

Merge Sort Revisited
• To sort n numbers:
o if n = 1 done!.
o recursively sort 2 lists of numbers [n/2] and [n/2] elements.
o merge 2 sorted lists in O(n) time.
• Strategy
o break problem into similar (smaller) subproblems.
o recursively solve subproblems.
o combine solutions to answer.
55

Merge Sort Execution Example
56
Input
1 5 2 4 6 3 2 6
1 5 2 4 6 3 2 6
1 5 2 4 6 3 2 6
1 5 2 4 6 3 2 6
1 5 2 4 3 6 2 6
1 2 4 5 2 3 6 6
1 2 2 3 4 5 6 6
Output
Split
Merge

Merging Sorted Sequences – Algorithm - Pseudocode
57
•Combines the sorted
subarrays A[p .. q] and
A[q+1..r] into one sorted
array A[p .. r].
•Makes use of two
working arrays L and R
which initially hold
copies of the two
subarrays.
•Makes use of sentinel
value () as last element
to simplify logic.

Example: A call of MERGE (9, 12, 16)
58

Example: A call of MERGE (9, 12, 16) (Cont.)
59

Analysis of Merge Sort
60
}
}
(1)
(n)
(n)
(1)
(1)
T(n/2)
T(n/2)
(n)
T(n) = 2T(n/2) + (n)

Analysis of Merge Sort (Cont.)
• Divide: The divide step just computes the middle of the
subarray, which takes constant time. Thus, D(n)= (1).
• Conquer: We recursively solve two subproblems, each of size
n/2, which contributes 2T(n/2) to the running time.
• Combine: We have already noted that the MERGE procedure
on an n-element subarray takes time (n), and so C(n)= (n).
61

Statement Effort
}
}
T(n)
(1)
(1)
T(n/2)
T(n/2)
(n)
Let T(n) = running time on a problem of size n.
So T(n) = Θ(1) when n = 1, and 2T(n/2) + Θ(n) + Θ(1) when n > 1.
Solving this recurrence (how?) gives T(n) = n lg n.
This expression is a recurrence.
[Reminder: lg n stands for log2 n].
62

Analysis of Merge Sort (Cont.) - Review
• Divide: computing the middle takes (1).
• Conquer: solving 2 sub-problem takes 2T(n/2).
• Combine: merging n-element takes (n).
• Total:
o T(n) = Θ(1) if n = 1.
o T(n) = 2T(n/2) + Θ(n) + Θ(1) if n > 1.
 T(n) = Θ(n lg n).
• Solving this recurrence (how?) gives T(n) = Θ(n lg n).
• This expression is a recurrence.
• T(n) denote the number of operations required by an algorithm
to solve a given class of problems.
63

• Analysis of recursive calls …
64

65
o Total: cn lg n + cn.
o T(n) = cn (lg n + 1).
= cn lg n + cn.
o Ignore low-order term of
cn and constant coefficient
c ⇒ (n lg n).
o T(n) is (n lg n).

T(n) = 2T(n/2) + Θ(n) + Θ(1) ⇒ if n > 1.
We'll write n instead of Θ(n) in the line below because it makes the algebra
much simpler. We know that T(1) = 1.
 T(n) = 2 T(n/2) + n
T(n) = 2 [2 T(n/4) + n/2] + n
T(n) = 4 T(n/4) + 2n
T(n) = 4 [2 T(n/8) + n/4] + 2n
T(n) = 8 T(n/8) + 3n
o ----------------------------------------
T(n) = 16 T(n/16) + 4n
T(n) = 2k T(n/2k) + k n
 Assume n/2k = 1 OR n = 2k OR log2 n = k
T(n) = 2k T(n/2k) + k n
T(n) = 2log2n T(1) + (log2n) n
T(n) = n + n log2 n [remember that T(1) = 1]
T(n) = O(n log2 n)
 T(n) = Θ(n log2 n) Worst case.
66

Merge Sort – Java Program
67
package Algorithms;
import java.util.Arrays;
public class MergeSort {
public static void main(String args[]) {
int Array[] = { 1, 10, 3, 5, 3, 1, 7, 4, 8, 9, 2, 0, 4 };
mergeSort(Array, 0, Array.length - 1);
System.out.println(Arrays.toString(Array));
}

Merge Sort – Java Program (Cont.)
68
protected static void mergeSort(int array[], int p, int r) {
int q;
if (p < r) {
q = (p + r) / 2;
mergeSort(array, p, q);
mergeSort(array, q + 1, r);
merge(array, p, q, r);
}
}

Merge Sort – Java Program (Cont.)
69
protected static void merge(int array[], int p, int q, int r) {
int n1 = q - p + 1;
int n2 = r - q;
int L[] = new int[n1 + 1];
int R[] = new int[n2 + 1];
for (int i = 0; i < n1; i++) {
L[i] = array[p + i];}
for (int j = 0; j < n2; j++) {
R[j] = array[q + j + 1];}
L[n1] = Integer.MAX_VALUE;
R[n2] = Integer.MAX_VALUE;
int ii = 0;
int jj = 0;
for (int k = p; k <= r; k++) {
if (L[ii] <= R[jj]) {
array[k] = L[ii];
ii = ii + 1;
} else {
array[k] = R[jj];
jj = jj + 1;}}
}
}
Outputs
[0, 1, 1, 2, 3, 3, 4, 4, 5, 7, 8, 9, 10]

Summary
• Sequential Approach
o Insertion Sort:
 Example.
 Time Complexity (Worst, Best, & Average Cases).
 Space Complexity.
 Advantages and Disadvantages.
• Divide-and-Conquer Approach
o Merge Sort:
 Example.
 Time Complexity (Worst Case).
 Space Complexity.
 Advantages and Disadvantages.
70

Algorithms - "Chapter 2 getting started"

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