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1. Module 3
Greedy technique

2. Define greedy technique
• The greedy method is one of the strategies like Divide and
conquer used to solve the problems. This method is used for
solving optimization problems. An optimization problem is a
problem that demands either maximum or minimum results.
• The Greedy method is the simplest and straightforward
approach. It is not an algorithm, but it is a technique. The main
function of this approach is that the decision is taken on the
basis of the currently available information. Whatever the
current information is present, the decision is made without
worrying about the effect of the current decision in future.

3. continued
• This technique is basically used to determine the feasible
solution that may or may not be optimal. The feasible solution is
a subset that satisfies the given criteria. The optimal solution is
the solution which is the best and the most favorable solution in
the subset. In the case of feasible, if more than one solution
satisfies the given criteria then those solutions will be
considered as the feasible, whereas the optimal solution is the
best solution among all the solutions.

4. Applications of Greedy Algorithm
• It is used in finding the shortest path.
• It is used to find the minimum spanning tree using the prim's
algorithm or the Kruskal's algorithm.
• It is used in a job sequencing with a deadline.
• This algorithm is also used to solve the fractional knapsack
problem.

5. Example

6. The Knapsack problem
• Given a Knapsack of a
maximum capacity of W and
N items each with its own
value and weight, throw in
items inside the Knapsack
such that the final contents
has the maximum value.

7. The selection of some things, each with profit and weight
values, to be packed into one or more knapsacks with capacity
is the fundamental idea behind all families of knapsack
problems. The knapsack problem had two versions that are as
follows:
 Fractional Knapsack Problem
 0 /1 Knapsack Problem

8. What is Knapsack Problem Using Greedy Method?
In this method, the Knapsack's filling is done so that the maximum capacity of the knapsack is utilized so that
maximum profit can be earned from it. The knapsack problem using the Greedy Method is referred to as:
Given a list of n objects, say {I1, I2,……, In) and a knapsack (or bag).
The capacity of the knapsack is M.
Each object Ij has a weight wj and a profit of pj
If a fraction xj (where x ∈ {0...., 1)) of an object Ij is placed into a knapsack, then a profit of pjxj is earned.
The problem (or Objective) is to fill the knapsack (up to its maximum capacity M), maximizing the total profit
earned.
Mathematically:

9. Knapsack Problem Algorithm Using Greedy Method
Knapsack Problem Using Greedy Method Pseudocode
A pseudo-code for solving knapsack problems using the greedy method is;
greedy fractional-knapsack (P[1...n], W[1...n], X[1..n]. M)
/*P[1...n] and W[1...n] contain the profit and weight of the n-objects ordered
such that X[1...n] is a solution set and M is the capacity of knapsack*/
{
For j ← 1 to n do
X[j]← 0
profit ← 0 // Total profit of item filled in the knapsack
weight ← 0 // Total weight of items packed in knapsacks
j ← 1
While (Weight < M) // M is the knapsack capacity

10. Fractional Knapsack Problem Using Greedy Method-
Fractional knapsack problem is solved using greedy method in the following steps-
Step-01:
For each item, compute its value / weight ratio.
Step-02:
Arrange all the items in decreasing order of their value / weight ratio
Step-03:
Start putting the items into the knapsack beginning from the item with the highest ratio.
Put as many items as you can into the knapsack.

11. EXAMPLE???

12. Knapsack Problem - 1
Obtain the optimal solution for the knapsack
problem using greedy method given the
following:
M = 15
n = 7
p1,p2,p3,p4,p5,p6,p7 = 10,5,15,7,6,18,3
w1,w2,w3,w4,w5,w6,w7= 2,3,5,7,1,4,1

14. Solution Vector = (1, 0,1, 4/7,0,1,0)= (1, 0,1, 0.57,0,1,0)
Optimal solution using this method is (x1, x2, x3,x4,x5,x6,x7) =
(1, 0,1, 0.57,0,1,0)
with profit = 47

15. Solution Vector = (1, 1,4/5,0,1,1,1)= (1, 1,0.8,0,1,1,1)
Optimal solution using this method is (x1, x2, x3,x4,x5,x6,x7) = (1, 1,0.8,0,1,1,1) with profit = 54
Optimal solution is not guaranteed using method 1 and 2

16. Solution Vector = (1, 2/3,1,0, 1,1,1)= (1, 0.67,1,0, 1,1,1)
Optimal solution is (x1, x2, x3,x4,x5,x6,x7) = (1, 0.67,1,0, 1,1,1) with profit [1*10+0.67*5+1*15+0*7+1*6+1*18+1*3]=
55.34 Weight=[1*2+0.67*3+1*5+0*7+1*1+1*4+1*1]=15
This greedy approach always results optimal solution

17. Problem Statement
In job sequencing problem, the objective is to find a sequence of jobs,
which is completed within their deadlines and gives maximum profit.
Solution
Let us consider, a set of n given jobs which are associated with deadlines and profit is
earned, if a job is completed by its deadline. These jobs need to be ordered in such a way
that there is maximum profit.
It may happen that all of the given jobs may not be completed within their deadlines.
Assume, deadline of ith job Ji is di and the profit received from this job is pi. Hence, the
optimal solution of this algorithm is a feasible solution with maximum profit.
Thus, D(i)>0
 for 1⩽i⩽n
Initially, these jobs are ordered according to profit, i.e. p1⩾p2⩾p3⩾...⩾pn

18. Algorithm: Job-Sequencing-With-Deadline (D,
J, n, k)
D(0) := J(0) := 0
k := 1
J(1) := 1 // means first job is selected
for i = 2 … n do
r := k
while D(J(r)) > D(i) and D(J(r)) ≠ r do
r := r – 1
if D(J(r)) ≤ D(i) and D(i) > r then
for l = k … r + 1 by -1 do
J(l + 1) := J(l)
J(r + 1) := i
k := k + 1

19. Analysis
In this algorithm, we are using two loops, one is within another. Hence, the
complexity of this algorithm is O(n2)
.

20. Examples:
Input:

22. Total profit – 20 + 25 + 35 + 30 = 110

23. We are given the jobs, their deadlines and associated profits as shown-
Jobs J1 J2 J3 J4 J5 J6
Deadlines 5 3 3 2 4 2
Profits 201 181 191 301 121 101

24. Step-01:
Firstly, we need to sort all the given jobs in decreasing order of their profit as follows.
Jobs J4 J1 J3 J2 J5 J6
Deadlines 2 5 3 3 4 2
Profits 300 200 190 180 120 100
Step-02:
For each step, we calculate the value of the maximum deadline.
Here, the value of the maximum deadline is 5.
So, we draw a Gantt chart as follows and assign it with a maximum time on the Gantt chart with 5 units as
shown below.

25. Now,
We will be considering each job one by one in the same order as they appear in the Step-01.
We are then supposed to place the jobs on the Gantt chart as far as possible from 0.
Step-03:
We now consider job4.
Since the deadline for job4 is 2, we will be placing it in the first empty cell before deadline 2
as follows.

26. Step-04:
Now, we go with job1.
Since the deadline for job1 is 5, we will be placing it in the first empty cell before deadline 5 as
shown below.
Step-05:
We now consider job3.
Since the deadline for job3 is 3, we will be placing it in the first empty cell before
deadline 3 as shown in the following figure.

27. Step-07:
Now, we consider job5.
Since the deadline for job5 is 4, we will be placing it in the first empty cell before
deadline 4 as shown in the following figure.

28. Now,
We can observe that the only job left is job6 whose deadline is 2.
Since all the slots before deadline 2 are already occupied, job6 cannot be completed.
Now, the questions given above can be answered as follows:
Part-01:
The optimal schedule is-
Job2, Job4, Job3, Job5, Job1
In order to obtain the maximum profit this is the required order in which the jobs must be
completed.

29. Part-02:
As we can observe, all jobs are not completed on the optimal schedule.
This is because job6 was not completed within the given deadline.
Part-03:
Maximum earned profit = Sum of the profit of all the jobs from the optimal
schedule
= Profit of job2 + Profit of job4 + Profit of job3 + Profit of job5 + Profit of job1
= 181 + 301 + 191 + 121 + 201
= 995 units

30. Minimum Spanning Tree

31. What is a Spanning Tree?
• Given an undirected and connected graph G=(V,E), a spanning tree of
the graph G is a tree that spans G (that is, it includes every vertex of G
) and is a subgraph of G (every edge in the tree belongs to G )

32. What is a Minimum Spanning Tree?
• The cost of the spanning tree is the sum of the weights of all the edges in the
tree. There can be many spanning trees. Minimum spanning tree is the spanning
tree where the cost is minimum among all the spanning trees. There also can be
many minimum spanning trees.
• Minimum spanning tree has direct application in the design of networks. It is
used in algorithms approximating the travelling salesman problem, multi-terminal
minimum cut problem and minimum-cost weighted perfect matching. Other
practical applications are:
Cluster Analysis
Handwriting recognition
Image segmentation

34. Contains all the original graph’s vertices.
Reaches out to (spans) all vertices.
Is acyclic. In other words, the graph doesn’t have any nodes which loop back to itself.

36. Adjacency Matrix
In graph theory, an adjacency matrix is nothing but a square matrix
utilised to describe a finite graph. The components of the matrix
express whether the pairs of a finite set of vertices (also called nodes)
are adjacent in the graph or not. In graph representation, the networks
are expressed with the help of nodes and edges, where nodes are the
vertices and edges are the finite set of ordered pairs.

37. Adjacency Matrix Definition
The adjacency matrix, also called the connection matrix, is a matrix
containing rows and columns which is used to represent a simple labelled
graph, with 0 or 1 in the position of (Vi , Vj) according to the condition
whether Vi and Vj are adjacent or not. It is a compact way to represent the
finite graph containing n vertices of a m x m matrix M. Sometimes
adjacency matrix is also called as vertex matrix and it is defined in the
general form as

41. Prim’s algorithm
• It is a greedy algorithm that is used to find the minimum
spanning tree from a graph. Prim's algorithm finds the subset of
edges that includes every vertex of the graph such that the sum
of the weights of the edges can be minimized.
• Prim's algorithm starts with the single node and explores all the
adjacent nodes with all the connecting edges at every step. The
edges with the minimal weights causing no cycles in the graph
got selected.

42. How does the prim's algorithm work?
Prim's algorithm is a greedy algorithm that starts from one vertex and
continue to add the edges with the smallest weight until the goal is
reached. The steps to implement the prim's algorithm are given as
follows -
 First, we have to initialize an MST with the randomly chosen
vertex.
 Now, we have to find all the edges that connect the tree in the
above step with the new vertices. From the edges found, select the
minimum edge and add it to the tree.
 Repeat step 2 until the minimum spanning tree is formed.

43. The applications of prim's algorithm are -
Prim's algorithm can be used in network designing.
It can be used to make network cycles.
It can also be used to lay down electrical wiring
cables.

44. Example of prim's algorithm
Now, let's see the working of prim's algorithm using an example. It will be
easier to understand the prim's algorithm using an example.
Suppose, a weighted graph is -

45. Step 1 - First, we have to choose a vertex from the above graph. Let's choose B.
Step 2 - Now, we have to choose and add the shortest edge from vertex B. There are two edges from
vertex B that are B to C with weight 10 and edge B to D with weight 4. Among the edges, the edge BD
has the minimum weight. So, add it to the MST.

46. Step 3 - Now, again, choose the edge with the minimum weight among all the other edges. In this case, the
edges DE and CD are such edges. Add them to MST and explore the adjacent of C, i.e., E and A. So, select the
edge DE and add it to the MST.

47. Step 4 - Now, select the edge CD, and add it to the MST.

48. Step 5 - Now, choose the edge CA. Here, we cannot select the edge CE as it
would create a cycle to the graph. So, choose the edge CA and add it to the
MST.

49. So, the graph produced in step 5 is the minimum spanning tree of the
given graph. The cost of the MST is given below -
Cost of MST = 4 + 2 + 1 + 3 = 10 units.

50. Algorithm
Step 1: Select a starting vertex
Step 2: Repeat Steps 3 and 4 until there are fringe vertices
Step 3: Select an edge 'e' connecting the tree vertex and fringe vertex
that has minimum weight
Step 4: Add the selected edge and the vertex to the minimum
spanning tree T
[END OF LOOP]
Step 5: EXIT

51. Data structure used for the minimum edge weight Time Complexity
Adjacency matrix, linear searching O(|V|2)
Adjacency list and binary heap O(|E| log |V|)
Adjacency list and Fibonacci heap O(|E|+ |V| log |V|)

52. Kruskal's Algorithm
Kruskal's Algorithm is used to find the minimum spanning tree
for a connected weighted graph. The main target of the algorithm
is to find the subset of edges by using which we can traverse
every vertex of the graph. It follows the greedy approach that
finds an optimum solution at every stage instead of focusing on a
global optimum.

53. How does Kruskal's algorithm work?
In Kruskal's algorithm, we start from edges with the lowest weight
and keep adding the edges until the goal is reached. The steps to
implement Kruskal's algorithm are listed as follows -
 First, sort all the edges from low weight to high.
 Now, take the edge with the lowest weight and add it to the
spanning tree. If the edge to be added creates a cycle, then reject
the edge.
 Continue to add the edges until we reach all vertices, and a
minimum spanning tree is created.

54. The applications of Kruskal's algorithm are -
Kruskal's algorithm can be used to layout electrical wiring among cities.
It can be used to lay down LAN connections.

55. Example of Kruskal's algorithm
Now, let's see the working of Kruskal's algorithm using an example. It will be easier to
understand Kruskal's algorithm using an example.
Suppose a weighted graph is -

56. The weight of the edges of the above graph is given in the below table -
Edge AB AC AD AE BC CD DE
Weight 1 7 10 5 3 4 2
Now, sort the edges given above in the ascending order of their weights.
Edge AB DE BC CD AE AC AD
Weight 1 2 3 4 5 7 10

57. Step 1 - First, add the edge AB with weight 1 to the MST.
Step 2 - Add the edge DE with weight 2 to the MST as it is not creating
the cycle.

58. Step 3 - Add the edge BC with weight 3 to the MST, as it is not creating any cycle or loop.

59. Step 4 - Now, pick the edge CD with weight 4 to the MST, as it is not
forming the cycle.

60. Step 5 - After that, pick the edge AE with weight 5. Including this edge will
create the cycle, so discard it.
Step 6 - Pick the edge AC with weight 7. Including this edge will create the
cycle, so discard it.
Step 7 - Pick the edge AD with weight 10. Including this edge will also create
the cycle, so discard it.

61. So, the final minimum spanning tree obtained from the given weighted graph
by using Kruskal's algorithm is -
The cost of the MST is = AB + DE + BC + CD = 1 + 2 + 3 + 4 = 10.
Now, the number of edges in the above tree equals the number of vertices minus
1. So, the algorithm stops here.

62. Algorithm
Step 1: Create a forest F in such a way that every vertex of the
graph is a separate tree.
Step 2: Create a set E that contains all the edges of the graph.
Step 3: Repeat Steps 4 and 5 while E is NOT EMPTY and F is not
spanning
Step 4: Remove an edge from E with minimum weight
Step 5: IF the edge obtained in Step 4 connects two different trees,
then add it to the forest F
(for combining two trees into one tree).
ELSE
Discard the edge
Step 6: END

63. Complexity of Kruskal's algorithm
Now, let's see the time complexity of Kruskal's algorithm.
Time Complexity
The time complexity of Kruskal's algorithm is O(E logE) or O(V
logV), where E is the no. of edges, and V is the no. of vertices.

66. Shortest paths – Dijkstra’salgorithm
• The Dijkstra’s algorithm finds the shortest path from a given vertex to
all the remaining vertices in a diagraph.
• We have to find out the shortest path from a given source vertex ‘S’
to each of the destinations (other vertices ) in the graph.

67. Dijkstra’s( s)
// Finds shortest path from source vertex to
all other vertices
//Input: Weighted connected graph G=<V,E>
with nonnegative weights and its vertices s
//Output: The length of distance of a shortest
path from s to v
{
1. for i = 1 to n do // Intialize
S[i] = 0;
d[i] = a[s][i];
2. S[s] = 1; //Assume 1 as the source
vertex
d[s] = 1;

68. 3. for i = 1 to n do
{
Choose a vertex u in v-s such that
d[u] is minimum
S = s ꓴ u
for each vertex v in v-s do
d[v] = min{ d[u], d[u]+c[u,v]}
}
}

69. Lets see an example to understand Dijkstra’s Algorithm
Below is a directed weighted graph. We will find shortest path between all the vertices
using Dijkstra’a Algorithm.
Dijkstra's Algorithm

74. Dijkstra’s Algorithm Applications
• To find the shortest path between source and destination
• In social networking applications to map the connections and
information
• In networking to route the data
• To find the locations on the map

75. Disadvantage of Dijkstra’s Algorithm
• It follows the blind search, so wastes a lot of time to give the
desired output.
• It Negative edges value cannot be handled by it.
• We need to keep track of vertices that have been visited.

76. Optimal Tree : Huffman coding
• Huffman coding provides codes to characters such that the length of
the code depends on the relative frequency or weight of the
corresponding character. Huffman codes are of variable-length, and
without any prefix (that means no code is a prefix of any other). Any
prefix-free binary code can be displayed or visualized as a binary tree
with the encoded characters stored at the leaves.
• Huffman tree or Huffman coding tree defines as a full binary tree in
which each leaf of the tree corresponds to a letter in the given
alphabet.
• The Huffman tree is treated as the binary tree associated with
minimum external path weight that means, the one associated with
the minimum sum of weighted path lengths for the given set of
leaves. So the goal is to construct a tree with the minimum external
path weight

77. Letter frequency table
Letter z k m c u d l e
Frequency 2 7 24 32 37 42 42 120
Huffman code
Letter Freq Code Bits
e 120 0 1
d 42 101 3

78. Letter Freq Code Bits
e 120 0 1
d 42 101 3
l 42 110 3
u 37 100 3
c 32 1110 4
m 24 11111 5
k 7 111101 6
z 2 111100 6
Huffman code

79. The Huffman tree (for the above example) is given below -

80. END OF MODULE 3