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2. Unit 3 (Part-I): Greedy Algorithms

3. Expected Outcomes of topic (CO)
Understand problem and its formulation to design an algorithm
Demonstrate the knowledge of basic data structures and their implementation and decide upon use of particular data structure best suited for implementing solution.
Write efficient algorithms using Greedy, Divide & Conquer to solve the real life problems.
Trace logic formulation, execution path of particular algorithm and data generated during execution of algorithm.

4. Basic Characteristics – Greedy algorithms

5. Greedy Algorithms Characteristics
Greedy Algorithms are short – sighted.
Greedy Algorithms are most efficient if it works.
For every instance of input Greedy Algorithms makes a decision and continues to process further set of input.
The other input values at the instance of decision are lost and not used in further processing. A A 5 8 5 8 S D S D y t 9 2 B x

6. Greedy Algorithm: Knapsack Problem
Principle: Knapsack is a bag of fixed capacity.
In this problem it is assumed that “n” objects with profit value and weight are given.
The main objective is to select the objects and place them in knapsack such that the object placed in the bag will generate maximum profit.
Two types of Knapsack problems:
Fractional Knapsack problem
0/1 Knapsack problem
The Fractional Knapsack problem can be solved using Greedy approach, where as 0/1 Knapsack problem does not have greedy solution.

7. Representation
Capacity = 20
10/2
9/4
15/3
6/6
4/1
11/5

8. Greedy Algorithm: Knapsack Problem
Approaches:
There are three basic approaches to solve knapsack problem:
Select the maximum profit objects
Select the minimum weight objects
Select the objects based on Profit/Weight ratio

9. Significance TIME PROFIT
Process for execution
Generate maximum Profit in available time by proper selection of processes

10. Greedy Algorithm: Knapsack Problem
Example 1: Capacity = 15, number of objects = 7
Profit 10 5 15 7 6 18 3
Weight 2 1 4
Index
Approach: Maximum Profit
Object
Weight
Profit
Capacity Remaining
Partial/Complete O6 4 18
15-4=11 C O3 5 15
11-5=6 O1 2 10
6-2=4
O4*
4-4=0 P
TOTAL 47

11. Greedy Algorithm: Knapsack Problem
Example 1: Capacity = 15, number of objects = 7
Profit 10 5 15 7 6 18 3
Weight 2 1 4
Index
Approach: Minimum Weight
Object
Weight
Profit
Capacity Remaining
Partial/Complete O5 1 6
15-1=14 C O7 3
14-1=13 O1 2 10
13-2=11 O2 5
11-3=8 O6 4 18
8-4=4
O3*
5(4) 12
4-4=0 P
TOTAL 15 54

12. Greedy Algorithm: Knapsack Problem
Example 1: Capacity = 15, number of objects = 7
Profit 10 5 15 7 6 18 3
Weight 2 1 4
Ratio
5.0
1.6
3.0
1.0
6.0
4.5
Index
Approach: Maximum Pi/Wi ratio
Object
Weight
Profit
Capacity Remaining
Partial/Complete O5 1 6
15-1=14 C O1 2 10
14-2=12 O6 4 18
12-4=8 O3 5 15
8-5=3 O7 3
3-1=2
O2*
3(2)
10/3=3.33
2-2=0 P
TOTAL
55.3

13. Greedy Algorithm: Knapsack Problem
Example 2: Capacity = 18, number of objects = 7 Pi 9 15 12 4 6 16 8 Wi 2 3 5
Ratio
4.5
5.0
2.4
1.0
2.0
2.6
Index 1 7
Maximum (Pi/Wi) Ratio
Object
Weight
Profit
Capacity Remaining
Partial/Complete

14. Knapsack Problem: Algorithm
Data Structures:
Weight array: W[n]  To store weight values
Profit array: P[n]  To store profit values
Capacity  Given and will decrease after each step
X[n]  To store 0, 1 or fraction depending upon object placement
Weight = 0 (initially) and will increase towards capacity
Profit = 0 and will be calculated by X[i] and P[i]
Process: To select the object with best Pi/Wi Ratio. (Sorting)

15. Knapsack Algorithm Initialization steps: Decision step
Profit
Weight
x[1..n]  to hold data about object placement [partly, completely, rejected]
Decision step
While ( condition )
Select object //Call function for selecting best suited object//
Place in bag {Completely / Partially }
Update x [i]
Final step: Find profit
Multiply x[i] and associated profit

17. Greedy Algorithms on Graph
Prim’s Algorithm
Kruskal Algorithm
Single Source shortest Path Algorithm (Dijkstra’s Algorithm)

18. Minimum Cost Spanning Tree
Given a graph G =(V,E) V=Number of vertices and E=Number of edges, then Spanning tree is tree generated from graph, with characteristics
All the vertices present in the tree
There is no cycle in the tree i.e., e=v-1
Subset of edges, that forms a tree, where total cost of edges is minimum.
Minimum cost spanning tree is a spanning tree with “edges of minimum cost”.
Applications:
To transmit the message without broadcasting.
To generate travel plan in minimum cost.
Designing network, home electric wiring etc.
Approaches: Prims’ Method and Kruskals’ Method.

19. Minimum Cost Spanning Tree
Principle:
Select an edge of minimum cost. The edge will derive two vertices.
Continue the process from one of the vertex by selecting the next edge of minimum cost.
Once the vertex is visited, then it is marked.
The process will attempt to visit unmarked vertices and terminates when all the vertices are visited.
Process guarantees no cycle in the tree.
“What if the process of tree generation starts from any arbitrary vertex”.
Motivation: To join points as cheaply as possible: Applications in clustering and networking
Acyclic graph to connect all nodes in minimum cost.
One of the term in U.S. legal code (AT&T)

20. Graphs for Examples
Steiner Spanning Tree

21. Spanning tree: Free tree
A free tree has following properties
Exactly n-1 edges for “n” vertices
There exists a unique path between two vertices.
By adding an edge, a cycle will be created in free tree. Breaking any edge on the cycle restores the free tree.
Greedy: Minimization problem.
Repeated selection: of minimum cost edges with certain test cases.
Once decision of adding edge is finalized, it cannot be revoked.
Basic idea: To generate subset of “E” connecting all “V”

22. Minimum Cost Spanning Tree: Prim’s method
Example:
Consider the following graph:
Select an edge of minimum cost.
From selected vertex continue selecting
edge of minimum cost 1 2 3 4 5 6 99 10 30 45 50 40 25 35 15 20 55 10 2 1 50 45 40 3 30 4 25 5 35 2 6 1 55 20 15 3 4 5 6

23. Minimum Cost Spanning Tree: Prim’s method
Example:
Consider the following graph: 1 2 3 4 5 6 99 10 30 45 50 40 25 35 15 20 55 10 2 1 50
The next vertex should be the vertex reachable in minimum cost either from vertex 1 or vertex 2 45 40 3 30 4 25 5 35 T C F R 1 - 2 3
< 4 5 6 6 55 20 15 1 10 2 1 2 3 4 5 6 -
Out of Four possible options: cost[near[j],j] = minimum
Select the index satisfying above condition: 6

24. Minimum Cost Spanning Tree: Prim’s method
The vertex selected is “6” due to which there may be changes in closeness of vertices.
That the vertices which were close to vertex 1 or vertex 2 may modified due to selection of vertex 6.
Example:
Consider the following graph: 1 2 3 4 5 6 99 10 30 45 50 40 25 35 15 20 55 10 2 1 50 45 40 3 30 4 25 5 35 T C F R 1 - 2 3 6
< 4 5 6 55 20 15 1 10 2 1 2 3 4 5 6 - 25 6 1 2 3 4 5 6 -
Out of Four possible options: cost[near[j],j] = minimum
Select the index satisfying above condition: 3

25. Minimum Cost Spanning Tree: Prim’s method
The vertex selected is “3” due to which there may be changes in closeness of vertices.
That the vertices which were close to vertex 1 or vertex 2 or vertex 6 may modified due to selection of vertex 3.
Example:
Consider the following graph: 1 2 3 4 5 6 99 10 30 45 50 40 25 35 15 20 55 10 2 1 50 45 40 3 30 4 25 5 35 1 2 3 4 5 6 - T C F R 1 - 2 3
< 4 6 5 6 55 20 15 1 10 2 1 2 3 4 5 6 - 25 6 3 1 2 3 4 5 6 -
Out of Four possible options: cost[near[j],j] = minimum
Select the index satisfying above condition: 4 15

26. Minimum Cost Spanning Tree: Prim’s method
The vertex selected is “4” due to which there may be changes in closeness of vertices.
That the vertices which were close to vertex 1,2,6,3 may modified due to selection of vertex 4.
Example:
Consider the following graph: 1 2 3 4 5 6 99 10 30 45 50 40 25 35 15 20 55 10 2 1 50 45 40 3 1 2 3 4 5 6 - 30 4 25 5 35 1 2 3 4 5 6 - T C F R 1 - 2 3
< 4 5 6 6 55 20 15 1 10 2 1 2 3 4 5 6 - 25 6 4 3 1 2 3 4 5 6 - 20 15

27. Prim’s method: Output Example: Consider the following graph: 1 2 3 4 56 99 10 30 45 50 40 25 35 15 20 55 10 2 1 50 45 40 3 30 4 25 5 35
Vertex 1
Vertex 2
Cost 1 2 10 6 25 3 15 4 20 5 35
TOTAL
105 6 55 20 15

28. Algorithm: Prim’s Algorithm
Data Structures: Input
cost[1..n, 1..n]  To store the cost information
Intermediate:
near[1..n]  To store near information for selected vertex
Output:
mincost=0
t[1..n-1, 1..3]  To store the spanning tree

29. Algorithm Algorithm Prim(cost,n: mincost,t) {
Step 1: Select an edge of minimum cost from the cost matrix and let it be cost[k,l]
mincost = cost[k,l];
t[1,1] = k; t[1,2] = l;
t[1,3] = cost[k,l]
for 1 = 1 to n do
if cost[i,l] < cost[i,k] then
near[i] = l;
else
near[i] = k;
} //end of for
for i = 2 to n-1 do { let j be an index such that (near[j] ≠ 0 and cost[near[j], j] = minimum) t[i,1] = j; t[i.2] = near[j] t[i,3] = cost[near[j],j] near[j] = 0; for k = 1 to n do if((near[k] ≠ 0) and cost[k,near[k] > cost[k,j])) near[k] = j; } //End of for } //End of algorithm

30. Minimum Cost Spanning Tree: Prim’s Algorithm
Designed by Robert C. Prim in 1957 and was modified by Dijkstra so also called as DJP algorithm.

31. Proof of Correctness: Prims’ Algorithm
The proof of prims’ algorithms is described using CUT property of a graph.
Let G=(V,E) be the graph, then CUT property will divide the graph into two sets of vertices (X and V-X), such that there exists three types of edges.
Edges with both vertices in “X”
Edges with both vertices in “V-X”
Edges with one vertex in “X” and one vertex in “V-X”  Crossing edge
If there are “n” vertices in graph, then number of cuts will be 2n
Empty cut lemma:
A graph is said to be not connected, if cut has no crossing edges. This is denoted as empty cut.
The correctness is proved using straight forward induction method.

32. Minimum Cost Spanning Tree: Kruskals’ Algorithm
Principle: Select an edge of minimum cost. At each step add a new edge with next minimum cost to graph, such that it does not generate cycle.
Kruskals’ algorithm, grows like forest and then generates a tree.
Integration plays major role.
Implementation:
From the set of edges, select an edge of minimum cost
Again continue the process with an edge of next minimum cost, add edge if not generating cycle.

33. Kruskals’ Algorithm Process-1: To find an edge of minimum cost
Sort the edges in ascending order of weight: O(e * log e)
Process-2: To check by adding an edge cycle is generated.
Union-Find Data structure is used to check if cycle is generated.
The algorithm creates set for each vertex: [1] [2] [3] [4] [5] [6] etc.
When an edge [u,v] is selected Find[u] will return the set in which vertex “u” is present and Find[v] will return the set in which vertex “v” is present.
If returned set are distinct, then edge can be added and sets can merged.

34. Kruskals’ Algorithm: Example

35. Kruskals’ Algorithm
Example:

36. Kruskals’ Algorithm: Implementation
Table for generating tree:
Edge
Cost
Action
Component
[1] [2] [3] [4] [5] [6] [7]
(1,2) 1
Accept
[1,2] [3] [4] [5] [6] [7]
(2,3) 2
[1,2,3] [4] [5] [6] [7]
(4,5) 3
[1,2,3] [4,5] [6] [7]
(6,7)
[1,2,3] [4,5] [6,7]
(1,4) 4
[1,2,3,4,5] [6,7]
(2,5)
Reject
(4,7)
[1,2,3,4,5,6,7]

37. Algorithm

38. Complexity of Kruskals algorithm
Let “E” be number of edges in the graph and “n” be number of vertices in the graph.
To Sort “E” edges, the time requirement will be O(E log E) which can be further reduced to O(E log n) since n-1 < E < n(n-1)/2
For creating the vertex set time required will be O(V)
For finding the vertex set of V1 and V2 and merging if the set is distinct, time required will be O(E log n)
Thus the complexity of Kruskals algorithm will be O(E log n)

39. Practice examples

40. Single Source Shortest Path
An approach to find shortest path from given source vertex to all possible destinations.
Minimization problem: To find the path of minimum cost
Path may be direct or indirect
Various algorithms:
Dijkstra’s algorithm can be used if no negative edges are present.
Bellman Ford algorithm can be used if negative edges are present.
Floyd’s and Warshall’s algorithm is extension and called as all pair shortest path algorithm, if no negative edges are given.

41. Example: Dijkstra’s Algorithm 55 25 45

42. Example: Dijkstra’s AlgorithmB C D E F G H A 20
Fix(B) 80 90 -
10=>30
Fix(F)
30=>50
50=>10
Fix (D)
50=>20
Not Reachable A
Source B
20: A-B F
30: A-B-F C
80: A-B-F-C D
90: A-B-F-C-D H
100:A-B-F-C-H G
110:A--G

43. Example: Single source shortest path1 2 3 4 5 1 2 3 4 5

48. Algorithm: Data Structures
Input: Cost matrix [nxn] and source vertex
Intermediate:
s[1..n] to store information about reached vertices
Parent[1..n] is used to hold information about the parent node from which the desired node is reachable.
Output: dist[1..n] and parent[1..n]
Process:
Initially the vertex at minimum distance from the source is selected.
The reachable vertices from source vertex are also initially selected.
When new vertex is added in the reachable list, the existing paths and paths generated from reached vertex are compared to make the next decision.

49. Single Source Shortest Path: Algorithm

50. Job Scheduling: Without Deadline
Principle:
This application is related with execution of jobs on the system in such a way that –
Average time spent by the jobs with machine is minimum
To maximize the profitability by completing the job before deadline.
In banks, where large number of transactions is taking place, it is required to complete the transaction in stipulate time period. If transaction are more and there is only one processor, then the transactions are buffered, but the execution should be so managed that average time spent by transaction in buffer is minimized.

51. Job Scheduling: Without Deadline
Implementation: Shift the shorter jobs ahead of longer jobs
P1 = 4, P2 = 6, P3 = 3, P4 = 8 and P5 = 2 Find the minimal time required to service the customers
Solution:
Rearrange in the ascending order to time requirement:
P5 = 2, P3 = 3, P1 = 4, P2 = 6 and P4 = 8
Formulation:
N * t1 + (N-1) * t2 + (N-2) * t3 + (N-3) * t4 + (N – (N-1)) * tn

52. Job Scheduling without Deadline
Principal: Given set of “n” jobs.
To decide the schedule, so as to complete the job processing in minimum time.
To prove the approach is correct.
Method: SJF: Shortest Job First
Select the job with minimum time requirement and process. Delay the jobs with higher time requirements.
Process required: Sorting of Jobs.
Note: All the jobs present in the set will be executed.

53. Example Total number of jobs=6
Total processing time: (3x6) + (4x5) + (6x4) + (9x3) + (8x2) + (12x1) =
Job 1 2 3 4 5 6
Time 12 8 9
Job 1 2 3 4 5 6
Time 9 8 12
Index

54. Job Scheduling with Deadline
Principal: Given set of “n” jobs.
To decide the schedule, so as to complete the job processing in minimum time.
Each job requires only one cycle for execution.
With each job deadline and profit is associated.
The profit is valid till the deadline.
Process required: Sorting of Jobs.
Note: Does not guarantee the execution of all the jobs present in the set.

55. Example Find the maximum value of deadline
This will provide details about maximum cycles available for execution
For example: Maximum value of deadline = 7
Hence there can be SEVEN slots for execution each of size ONE UNIT
Step 1: Sort the jobs on maximum value of profit
Step 2: Allocation of slot on maximum deadline basis
Step 3: Job rejection process

56. Example .. After sorting Slot Task / Profit 1* T2/20 2 T7/23 3 T9/25 46*
T1/16 7
T8/16
Total Profit

57. Unit 3: Part II: Divide and Conquer
Principle:
Input is divided into small groups before processing using logical division.
Solution space for all the divided groups is same and unique.
The CPU utilization is on higher side whereas the throughput is on lower side.
The division of input reduces the complexity of the process and most of DAC based algorithms have logarithmic complexity.
DAC principle is mostly applied for input of large size.

58. Binary Search Principle: Implemented on the SORTED array
Input is divided into small groups before processing using formula:
𝑚𝑖𝑑= 𝑙𝑜𝑤+ℎ𝑖𝑔ℎ 2 (Consider lower integer value)
Where: “low” represents start of array and “high” represents end of array.
If element is found at position “mid” search is successful, otherwise compare the element at “mid” and element to be searched.
If a[mid] < elem  low = mid + 1
If a[mid] > elem  high = mid -1
Continue process till element is found (a[mid] = elem)
Or otherwise if (low > high)  Search is unsuccessful.

59. Consider the following array of size: 14 (1..14  low=1 and high=14)
Binary Search: Numerical on average number of comparisons for successful and unsuccessful search
Consider the following array of size: 14 (1..14  low=1 and high=14)
Total number of comparisons for successful search: 43
Average number of comparisons = 43/14 = 3.07
For finding total number of comparisons for unsuccessful search:
Draw Binary search tree for the given array
Complete the BST and find the sum of level of BST
Compute average by dividing the sum/n+1 (where n = number of elements) I 1 2 3 4 5 6 7 8 9 10 11 12 13 14 A
-10 -5 21 34 45 56 65 77 89 98
104
110 C

60. Binary Search: Numerical on average number of comparisons for successful and unsuccessful search1 2 3 4 5 6 7 8 9 10 11 12 13 14 A
-10 -5 21 34 45 56 65 77 89 98
104
110 C

61. Binary Search: Algorithm and complexity equation
Algorithm BinarySearch(a,i,l,x) //x is the element to be searched {
If(i==l) then
If(x=a[i])
return(i);
else
return(0); }
Else
Mid=[i+l]/2
If(x==a[mid]) then
Return(mid)
If(x<a[mid])then
Return (bsearch(a,i,mid-1,x)
Return (bsearch(a,mid+1,l,x);

62. Ternary Search

63. Sorting algorithms: Merge Sort
Basis:
Split phase
Divide the array into smaller units: of size 1.
Merge phase
Compare the two adjacent elements, sort elements and merge to generate array at next level. Continue merge process till all units are merged into single unit.
There are various methods to perform split operation.
Divide the array at each element (Cutting the long piece of cake), will require n- 1 operations.
Otherwise apply logarithmic division: divide by 2 option. This will require log n operations.

64. Merge Sort: Example
SPLIT PHASE L1 L2 L3
MERGE PHASE

65. Merge Sort Algorithm and Complexity Equation
There are two components in merge sort: split and merge
Algorithm Merge_Sort(a,low,high) {
if(low<high)
mid= low+high/2
Merge_Sort(a,low,mid)
Merge_Sort(a,mid+1,high)
Merge(a,low,high) }
Complexity Equation:
T(n) = 2T(n/2) + n
For division by factor 2, and two recursive calls
Time required = 2T(n/2)
For merging “n” elements time required = n

66. Merge Sort Algorithm: Characteristics
Merge sort requires auxiliary array for storing the sorted elements after every merge phase.
Discuss time complexity.

67. Min-Max Algorithm
Principle: Method to find minimum and maximum element from an array.
General method requires 2*(n-1) comparisons.
Algorithm min_max(a,n,min,max) {
min = max = a[1]
for i = 2 to n do
if (a[i] < min) then min = a[i]
if (a[i] > max) then max = a[i] }

68. Min-Max: DAC based Example1 2 3 4 5 6 7 8 9 10 22 13 -5 -8 15 60 17 31 47 45 1 2 3 4 5 6 7 8 9 10 1 2 3 4 5 6 7 8 9 10 1 2 13 22 3 -5 6 7 17 60 8 31

69. Min-Max: DAC Algorithm
Algorithm min_max(a,i,j,min,max) {
mid = [i+j] / 2 (Take lower integer)
If (i==j) then min=max=a[i]
min_max(a,i,mid,min,max)
Else
min_max(a,mid+1, j, min1, max1)
If(i=j-1) then
if (max < max1) then max = max1
if (min > min1) then min = min 1
if(a[i] < a[j]) then
min = a[i]; max=a[j] }
Complexity equation:
else
T(n) = T(n/2) + T(n/2) + 2 for n > 2
T(n) = 1 if n=2
min=a[j]; max=a[i]
T(n) = 0 if n=1

70. Quick Sort Analysis
Algorithm QuickSort(a, start, end) { If (start < end) C1 X = partition(a, start, end) Check partition time Quicksort(a, start, x-1) T(n/2) Quicksort(a, x+1, end) T(n/2) } Total time including partition = 2T(n/2) + an + b + c1  2T(n/2) + an Which can be written as: 2T(n/2) + cn, where c = constant. If there is only one element, then T(1) = c, because “if” loop will not be executed.

71. Quick Sort Analysis
Algorithm partition(a, start, end) { pivot = a[end]; 1 unit pindex = start; 1 unit for i = start to end-1 do if (a[i] <= pivot) swap(a[i], a[pindex]) “a” time unit for “n” times pindex++; } swap(a[pindex], pivot) 1 unit return(pindex) 1 unit Total time = an+b

72. Mathematical solving
T(n) = 2T(n/2 ) + cn In the next recursion, the value of n = n/4, i.e., n replaced by n/2. T(n) = 2 2 𝑇 𝑛 4 +𝑐 𝑛/2 +𝑐𝑛 T(n) = 4T(n/4) + 2 cn T(n) = 8T(n/8) + 3cn T(n) = 2 𝑘 𝑇 𝑛 2 𝑘 +𝑘 𝑐𝑛 𝑛 2 𝑘 = 1 hence k = log n T(n) = 2 log 𝑛 ∗ 1+𝑙𝑜𝑔𝑛 𝑐 𝑛=⇒𝑛 log 𝑛

73. Worst Case
In worst case the given array is already sorted and it is required to sort it in reverse order. (ascending to descending).
Next time the array will be partitioned into two un-balanced partitioned.
And in all steps, the resultant arrays will be un-balanced.
In such case: the second Quicksort function will not be executed and time complexity will be controlled by only first quicksort function. This function will be executed for T(n-1) times.
T(n) = T(n-1) + cn
T(n) = T(n-2) + c(n-1) + cn = T(n-2) + 2cn – c
T(n) = T(n-3) + c(n-1) + 2cn – c 
T(n-3) + c(n-2) + 2cn – c = T(n-3) + 3cn – 3c
T(n-4) + 4cn – 6c
T(n-k) + kcn – (k(k-1)/2)
Smallest unit = n-k= 1, hence k = n ignoring the constants.
T(n) = T(1) + ncn – constant term  nxn = n2 1 2 3 4 5 6 7 8