1. Data Structures and Algorithms Ver. 1.0 Session 17 Objectives In this session, you will learn to: Implement a graph Apply graphs to solve programming problems

2. Data Structures and Algorithms Ver. 1.0 Session 17 To implement a graph, you need to first represent the given information in the form of a graph. The two most commonly used ways of representing a graph are as follows: Adjacency Matrix Adjacency List Representing a Graph

3. Data Structures and Algorithms Ver. 1.0 Session 17 Consider the following graph: Adjacency Matrix Adjacency Matrix Representation v1 v2 v3 v4 v10 1 0 0 v20 0 1 0 v30 0 0 0 v41 0 1 0

4. Data Structures and Algorithms Ver. 1.0 Session 17 Consider the following graph: Adjacency List Adjacency List Representation

5. Data Structures and Algorithms Ver. 1.0 Session 17 Traversing a graph means visiting all the vertices in a graph. You can traverse a graph with the help of the following two methods: Depth First Search (DFS) Breadth First Search (BFS) Traversing a Graph

6. Data Structures and Algorithms Ver. 1.0 Session 17 Algorithm: DFS(v) 1.Push the starting vertex, v into the stack. 2.Repeat until the stack becomes empty: DFS a.Pop a vertex from the stack. b.Visit the popped vertex. c.Push all the unvisited vertices adjacent to the popped vertex into the stack.

7. Data Structures and Algorithms Ver. 1.0 Session 17 DFS (Contd.) v1 Push the starting vertex, v1 into the stack

8. Data Structures and Algorithms Ver. 1.0 Session 17 DFS (Contd.) v1 Pop a vertex, v1 from the stack Visit v1 Push all unvisited vertices adjacent to v1 into the stack v1 Visited:

9. Data Structures and Algorithms Ver. 1.0 Session 17 DFS (Contd.) v4 Pop a vertex, v1 from the stack Visit v1 Push all unvisited vertices adjacent to v1 into the stack v1 Visited: v2

10. Data Structures and Algorithms Ver. 1.0 Session 17 DFS (Contd.) Visited: Pop a vertex, v2 from the stack Visit v2 Push all unvisited vertices adjacent to v2 into the stack v1v2 v4 v2

11. Data Structures and Algorithms Ver. 1.0 Session 17 DFS (Contd.) Visited: Pop a vertex, v2 from the stack Visit v2 Push all unvisited vertices adjacent to v2 into the stack v1v2 v3 v6 v4

12. Data Structures and Algorithms Ver. 1.0 Session 17 DFS (Contd.) Visited: Pop a vertex, v6 from the stack Visit v6 Push all unvisited vertices adjacent to v6 into the stack v1v2v6 There are no unvisited vertices adjacent to v6 v3 v6 v4

13. Data Structures and Algorithms Ver. 1.0 Session 17 DFS (Contd.) Visited: Pop a vertex, v3 from the stack Visit v3 Push all unvisited vertices adjacent to v3 into the stack v1v2v6v3 v4

14. Data Structures and Algorithms Ver. 1.0 Session 17 DFS (Contd.) Visited: Pop a vertex, v3 from the stack Visit v3 Push all unvisited vertices adjacent to v3 into the stack v1v2 v5 v6v3 v4

15. Data Structures and Algorithms Ver. 1.0 Session 17 DFS (Contd.) Visited: Pop a vertex, v5 from the stack Visit v5 Push all unvisited vertices adjacent to v5 into the stack v1v2v6v3v5 There are no unvisited vertices adjacent to v5 v5 v4

16. Data Structures and Algorithms Ver. 1.0 Session 17 DFS (Contd.) Visited: Pop a vertex, v4 from the stack Visit v4 Push all unvisited vertices adjacent to v4 into the stack v1v2v6v3v5 There are no unvisited vertices adjacent to v4 v4

17. Data Structures and Algorithms Ver. 1.0 Session 17 DFS (Contd.) Visited: The stack is now empty Therefore, traversal is complete v1v2v6v3v5v4

18. Data Structures and Algorithms Ver. 1.0 Session 17 DFS (Contd.) Although the preceding algorithm provides a simple and convenient method to traverse a graph, the algorithm will not work correctly if the graph is not connected. In such a case, you will not be able to traverse all the vertices from one single starting vertex.

19. Data Structures and Algorithms Ver. 1.0 Session 17 DFS (Contd.) To solve this problem, you need to execute the preceding algorithm repeatedly for all unvisited vertices in the graph. 1.Repeat step 2 for each vertex, v in the graph 2.If v is not visited: a. Call DFS(v)

20. Data Structures and Algorithms Ver. 1.0 Session 17 Algorithm: BFS(v) 1.Visit the starting vertex, v and insert it into a queue. 2.Repeat step 3 until the queue becomes empty. 3.Delete the front vertex from the queue, visit all its unvisited adjacent vertices, and insert them into the queue. BFS

21. Data Structures and Algorithms Ver. 1.0 Session 17 BFS (Contd.) Visit v1 Insert v1 into the queue v1

22. Data Structures and Algorithms Ver. 1.0 Session 17 BFS (Contd.) Remove a vertex v1 from the queue Visit all unvisited vertices adjacent to v1 and insert them in the queue v1 Visited:

23. Data Structures and Algorithms Ver. 1.0 Session 17 BFS (Contd.) Remove a vertex v1 from the queue Visit all unvisited vertices adjacent to v1 and insert them in the queue v2 v1v2 v4

24. Data Structures and Algorithms Ver. 1.0 Session 17 BFS (Contd.) Remove a vertex v2 from the queue Visit all unvisited vertices adjacent to v2 and insert them in the queue v2 v1v2 v4

25. Data Structures and Algorithms Ver. 1.0 Session 17 BFS (Contd.) Remove a vertex v2 from the queue Visit all unvisited vertices adjacent to v2 and insert them in the queue v1v2 v4 v3 v6

26. Data Structures and Algorithms Ver. 1.0 Session 17 BFS (Contd.) Remove a vertex v4 from the queue Visit all unvisited vertices adjacent to v4 and insert them in the queue v1v2 v4 v3 v6 v5

27. Data Structures and Algorithms Ver. 1.0 Session 17 BFS (Contd.) Remove a vertex v3 from the queue Visit all unvisited vertices adjacent to v3 and insert them in the queue v1v2 v4 v3 v6 v5 v3 does not have any unvisited adjacent vertices

28. Data Structures and Algorithms Ver. 1.0 Session 17 BFS (Contd.) Remove a vertex v6 from the queue Visit all unvisited vertices adjacent to v6 and insert them in the queue v1v2 v4 v3v6 v5 v3 does not have any unvisited adjacent vertices

29. Data Structures and Algorithms Ver. 1.0 Session 17 BFS (Contd.) Remove a vertex v6 from the queue Visit all unvisited vertices adjacent to v6 and insert them in the queue v1v2 v4 v3v6v5 v6 does not have any unvisited adjacent vertices

30. Data Structures and Algorithms Ver. 1.0 Session 17 BFS (Contd.) Remove a vertex v5 from the queue Visit all unvisited vertices adjacent to v5 and insert them in the queue v1v2 v4 v3v6v5 v6 does not have any unvisited adjacent vertices

31. Data Structures and Algorithms Ver. 1.0 Session 17 BFS (Contd.) Remove a vertex v5 from the queue Visit all unvisited vertices adjacent to v5 and insert them in the queue v1v2 v4 v3v6v5 v5 does not have any unvisited adjacent vertices

32. Data Structures and Algorithms Ver. 1.0 Session 17 BFS (Contd.) The queue is now empty Therefore, traversal is complete v1v2 v4 v3v6v5 v5 does not have any unvisited adjacent vertices

33. Data Structures and Algorithms Ver. 1.0 Session 17 BFS (Contd.) Although the preceding algorithm provides a simple and convenient method to traverse a graph, the algorithm will not work correctly if the graph is not connected. In such a case, you will not be able to traverse all the vertices from one single starting vertex.

34. Data Structures and Algorithms Ver. 1.0 Session 17 BFS (Contd.) To solve this problem, you need to execute the preceding algorithm repeatedly for all unvisited vertices in the graph. 1.Repeat step 2 for each vertex, v in the graph 2.If v is not visited: a. Call BFS(v)

35. Data Structures and Algorithms Ver. 1.0 Session 17 Problem Statement: You have to represent a set of cities and the distances between them in the form of a graph. Write a program to represent the graph in the form of an adjacency matrix. Activity: Implementing a Graph by Using Adjacency Matrix Representation

36. Data Structures and Algorithms Ver. 1.0 Session 17 Many problems can be easily solved by reducing them in the form of a graph Graph theory has been instrumental in analyzing and solving problems in areas as diverse as computer network design, urban planning, finding shortest paths and molecular biology. Applications of Graphs

37. Data Structures and Algorithms Ver. 1.0 Session 17 Solving the Shortest Path Problem The shortest path problem can be solved by applying the Dijkstra’s algorithm on a graph The Dijkstra’s algorithm is based on the greedy approach The steps in the Dijkstra’s algorithm are as follows: 1. Choose vertex v corresponding to the smallest distance recorded in the DISTANCE array such that v is not already in FINAL. 2. Add v to FINAL. 3. Repeat for each vertex w in the graph that is not in FINAL: a. If the path from v1 to w via v is shorter than the previously recorded distance from v1 to w (If ((DISTANCE[v] + weight of edge(v,w)) < DISTANCE[w])): i. Set DISTANCE[w]=DISTANCE[v] + weight of edge(v,w). 4. If FINAL does not contain all the vertices, go to step 1.

38. Data Structures and Algorithms Ver. 1.0 Session 17 Solving the Shortest Path Problem (Contd.) 5 2 346 3 36 DISTANCE v1v2v3v4v5v6 FINAL 0 ∞ 3∞∞ 5 v1 Suppose you need to find the shortest distance of all the vertices from vertex v1. Add v1 to the FINAL array.

39. Data Structures and Algorithms Ver. 1.0 Session 17 5 2 346 3 36 v1v2v3v4v5v6 0 ∞ 3∞∞ 5 v1 In the DISTANCE array, vertex v4 has the shortest distance from vertex v1. Therefore, v4 is added to the FINAL array. v4 Solving the Shortest Path Problem (Contd.) DISTANCE FINAL

40. Data Structures and Algorithms Ver. 1.0 Session 17 5 2 346 3 36 v1v2v3v4v5v6 0 ∞ 3∞∞ 5 v1v4 v1 → v2 = 5 v1 → v4 → v2 = 3 + ∞ = ∞ ∞ > 5 Therefore, no change is made. Solving the Shortest Path Problem (Contd.) DISTANCE FINAL

41. Data Structures and Algorithms Ver. 1.0 Session 17 5 5 2 346 3 36 v1v2v3v4v5v6 0 ∞ 3∞∞ 5 v1v4 Therefore, the entry corresponding to v3 in the DISTANCE array is changed to 5. v1 → v3 = ∞ v1 → v4 → v3 = 3 + 2 = 5 5 < ∞ Solving the Shortest Path Problem (Contd.) DISTANCE FINAL

42. Data Structures and Algorithms Ver. 1.0 Session 17 ∞ 9 5 2 346 3 36 v1v2v3v4v5v6 03∞ 5 v1v4 5 v1 → v5 = ∞ v1 → v4 → v5 = 3 + 6 = 9 9 < ∞ Therefore, the entry corresponding to v5 in the DISTANCE array is changed to 9. Solving the Shortest Path Problem (Contd.) DISTANCE FINAL

43. Data Structures and Algorithms Ver. 1.0 Session 17 5 2 346 3 36 v1v2v3v4v5v6 03∞ 5 v1v4 5 9 Both the values are equal. Therefore, no change is made. v1 → v6 = ∞ v1 → v4 → v6 = 3 + ∞ = ∞ PASS 1 complete Solving the Shortest Path Problem (Contd.) DISTANCE FINAL

44. Data Structures and Algorithms Ver. 1.0 Session 17 5 2 346 3 36 v1v2v3v4v5v6 03∞ 5 v1v4 5 9 From the DISTANCE array, select the vertex with the shortest distance from v1, such that the selected vertex is not in the FINAL array. v2 and v3 have the shortest and the same distance from v1. Let us select v2 and add it to the FINAL array. v2 Solving the Shortest Path Problem (Contd.) DISTANCE FINAL

45. Data Structures and Algorithms Ver. 1.0 Session 17 5 2 346 3 36 v1v2v3v4v5v6 03∞ 5 v1v4 5 9 v2 v1 → v3 = 5 v1 → v2 → v3 = 5 + 4 = 9 9 > 5 Therefore, no change is made. Solving the Shortest Path Problem (Contd.) DISTANCE FINAL

46. Data Structures and Algorithms Ver. 1.0 Session 17 5 2 346 3 36 v1v2v3v4v5v6 03∞ 5 v1v4 5 9 v2 v1 → v5 = 9 v1 → v2 → v5 = 5 + ∞ = ∞ ∞ > 9 Therefore, no change is made. Solving the Shortest Path Problem (Contd.) DISTANCE FINAL

47. Data Structures and Algorithms Ver. 1.0 Session 17 ∞ 11 5 2 346 3 36 v1v2v3v4v5v6 03 5 v1v4 5 9 v2 v1 → v6 = ∞ v1 → v2 → v6 = 5 + 6 = 11 11 < ∞ Therefore, the entry corresponding to v6 in the DISTANCE array is changed to 11. Pass 2 complete Solving the Shortest Path Problem (Contd.) DISTANCE FINAL

48. Data Structures and Algorithms Ver. 1.0 Session 17 5 2 346 3 36 v1v2v3v4v5v6 03 5 v1v4 5 9 v2 11 v3 From the DISTANCE array, select the vertex with the shortest distance from v1, such that the selected vertex is not in the FINAL array. Let us select v3 and add it to the FINAL array. Solving the Shortest Path Problem (Contd.) DISTANCE FINAL

49. Data Structures and Algorithms Ver. 1.0 Session 17 5 2 346 3 36 v1v2v3v4v5v6 03 5 v1v4 5 9 v2 11 v3 8 v1 → v5 = 9 v1 → v3 → v5 = 5 + 3 = 8 8 < 9 Therefore, the entry corresponding to v5 in the DISTANCE array is changed to 8. Solving the Shortest Path Problem (Contd.) DISTANCE FINAL

50. Data Structures and Algorithms Ver. 1.0 Session 17 8 5 2 346 3 36 v1v2v3v4v5v6 03 5 v1v4 5 v2v3 8 v1 → v6 = 11 v1 → v3 → v6 = 5 + 3 = 8 8 < 11 Therefore, the entry corresponding to v6 in the DISTANCE array is changed to 8. Pass 3 complete Solving the Shortest Path Problem (Contd.) DISTANCE FINAL 11

51. Data Structures and Algorithms Ver. 1.0 Session 17 5 2 346 3 36 v1v2v3v4v5v6 03 5 v1v4 5 v2v3 8 8 v5 From the DISTANCE array, select the vertex with the shortest distance from v1, such that the selected vertex is not in the FINAL array. Let us select v5 and add it to the FINAL array. Solving the Shortest Path Problem (Contd.) DISTANCE FINAL

52. Data Structures and Algorithms Ver. 1.0 Session 17 5 2 346 3 36 v1v2v3v4v5v6 03 5 v1v4 5 v2v3 8 8 v5 v1 → v6 = 8 v1 → v5 → v6 = 8 + ∞ = ∞ ∞ > 8 Therefore, no change is made. Pass 4 complete Solving the Shortest Path Problem (Contd.) DISTANCE FINAL

53. Data Structures and Algorithms Ver. 1.0 Session 17 5 2 346 3 36 v1v2v3v4v5v6 03 5 v1v4 5 v2v3 8 8 v5 Now add the only remaining vertex, v6 to the FINAL array. v6 All vertices have been added to the FINAL array. This means that the DISTANCE array now contains the shortest distances from vertex v1 to all other vertices. Solving the Shortest Path Problem (Contd.) DISTANCE FINAL

54. Data Structures and Algorithms Ver. 1.0 Session 17 Problem Statement: In the previous activity, you created a program to represent a set of cities and the distances between them in the form of a graph. Extend the program to include the functionality for finding the shortest path from a given city to all the other cities. Activity: Solving the Shortest Path Problem

55. Data Structures and Algorithms Ver. 1.0 Session 17 In this session, you learned that: The two most commonly used ways of representing a graph are as follows: Adjacency matrix Adjacency list Traversing a graph means visiting all the vertices in the graph. In a graph, there is no special vertex designated as the starting vertex. Therefore, traversal of the graph may start from any vertex. You can traverse a graph with the help of the following two methods: DFS BFS Summary

56. Data Structures and Algorithms Ver. 1.0 Session 17 Graph theory has been instrumental in analyzing and solving problems in areas as diverse as computer network design, urban planning, finding shortest paths and molecular biology. Summary (Contd.)