Unit - 5

Graphs

Q1) Explain Basic Terminologies of graph.

A1)

A graph can be defined as group of vertices and edges that are used to connect these vertices. A graph can be seen as a cyclic tree, where the vertices (Nodes) maintain any complex relationship among them instead of having parent child relationship.

Definition

A graph G can be defined as an ordered set G (V, E) where V(G) represents the set of vertices and E(G) represents the set of edges which are used to connect these vertices.

Example

The following is a graph with 5 vertices and 6 edges.

This graph G can be defined as G = (V, E)

Where V = {A, B, C, D, E} and E = {(A, B), (A, C) (A, D), (B, D), (C, D), (B, E), (E, D)}.

Directed and Undirected Graph

A graph can be directed or undirected. However, in an undirected graph, edges are not associated with the directions with them. An undirected graph is shown in the above figure since its edges are not attached with any of the directions. If an edge exists between vertex A and B then the vertices can be traversed from B to A as well as A to B.

In a directed graph, edges form an ordered pair. Edges represent a specific path from some vertex A to another vertex B. Node A is called the initial node while node B is called terminal node.

A directed graph is shown in the following figure.

Fig– Directed graph

Graph Terminology

We use the following terms in graph data structure.

Vertex

Individual data element of a graph is called as Vertex. Vertex is also known as node. In above example graph, A, B, C, D & E are known as vertices.

Edge

An edge is a connecting link between two vertices. Edge is also known as Arc. An edge is represented as (startingVertex, endingVertex). For example, in above graph the link between vertices A and B is represented as (A, B). In above example graph, there are 7 edges (i.e., (A, B), (A, C), (A, D), (B, D), (B, E), (C, D), (D, E)).

Edges are three types.

1. Undirected Edge - An undirected edge is a bidirectional edge. If there is undirected edge between vertices A and B then edge (A, B) is equal to edge (B, A).
2. Directed Edge - A directed edge is a unidirectional edge. If there is directed edge between vertices A and B then edge (A, B) is not equal to edge (B, A).
3. Weighted Edge - A weighted edge is an edge with value (cost) on it.

Undirected Graph

A graph with only undirected edges is said to be undirected graph.

Directed Graph

A graph with only directed edges is said to be directed graph.

Mixed Graph

A graph with both undirected and directed edges is said to be mixed graph.

End vertices or Endpoints

The two vertices joined by edge are called end vertices (or endpoints) of that edge.

Origin

If an edge is directed, its first endpoint is said to be the origin of it.

Destination

If an edge is directed, its first endpoint is said to be the origin of it and the other endpoint is said to be the destination of that edge.

Adjacent

If there is an edge between vertices A and B then both A and B are said to be adjacent. In other words, vertices A and B are said to be adjacent if there is an edge between them.

Incident

Edge is said to be incident on a vertex if the vertex is one of the endpoints of that edge.

Outgoing Edge

A directed edge is said to be outgoing edge on its origin vertex.

Incoming Edge

A directed edge is said to be incoming edge on its destination vertex.

Degree

Total number of edges connected to a vertex is said to be degree of that vertex.

Indegree

Total number of incoming edges connected to a vertex is said to be indegree of that vertex.

Outdegree

Total number of outgoing edges connected to a vertex is said to be outdegree of that vertex.

Parallel edges or Multiple edges

If there are two undirected edges with same end vertices and two directed edges with same origin and destination, such edges are called parallel edges or multiple edges.

Self-loop

Edge (undirected or directed) is a self-loop if its two endpoints coincide with each other.

Simple Graph

A graph is said to be simple if there are no parallel and self-loop edges.

Path

A path is a sequence of alternate vertices and edges that starts at a vertex and ends at other vertex such that each edge is incident to its predecessor and successor vertex.

Q2) Explain Representation of graph.

A2)

Graph Representation

By Graph representation, we simply mean the technique which is to be used in order to store some graph into the computer's memory.

There are two ways to store Graph into the computer's memory.

1. Sequential Representation

In sequential representation, we use an adjacency matrix to store the mapping represented by vertices and edges. In adjacency matrix, the rows and columns are represented by the graph vertices. A graph having n vertices, will have a dimension n x n.

An entry Mij in the adjacency matrix representation of an undirected graph G will be 1 if there exists an edge between Vi and Vj.

An undirected graph and its adjacency matrix representation is shown in the following figure.

Fig: Undirected graph and its adjacency matrix

In the above figure, we can see the mapping among the vertices (A, B, C, D, E) is represented by using the adjacency matrix which is also shown in the figure.

There exists different adjacency matrices for the directed and undirected graph. In directed graph, an entry Aij will be 1 only when there is an edge directed from Vi to Vj.

A directed graph and its adjacency matrix representation is shown in the following figure.

Fig- Directed graph and its adjacency matrix

Representation of weighted directed graph is different. Instead of filling the entry by 1, the Non-zero entries of the adjacency matrix are represented by the weight of respective edges.

The weighted directed graph along with the adjacency matrix representation is shown in the following figure.

Fig - Weighted directed graph

2. Linked Representation

In the linked representation, an adjacency list is used to store the Graph into the computer's memory.

Consider the undirected graph shown in the following figure and check the adjacency list representation.

An adjacency list is maintained for each node present in the graph which stores the node value and a pointer to the next adjacent node to the respective node. If all the adjacent nodes are traversed then store the NULL in the pointer field of last node of the list. The sum of the lengths of adjacency lists is equal to the twice of the number of edges present in an undirected graph.

Consider the directed graph shown in the following figure and check the adjacency list representation of the graph.

In a directed graph, the sum of lengths of all the adjacency lists is equal to the number of edges present in the graph.

In the case of weighted directed graph, each node contains an extra field that is called the weight of the node. The adjacency list representation of a directed graph is shown in the following figure.

Q3) What is depth first search and write its algorithm?

A3)

Depth First Search is an algorithm for traversing or searching a graph.

It starts from some node as the root node and explores each and every branch of that node before backtracking to the parent node.

DFS is an uninformed search that progresses by expanding the first child node of the graph that appears and thus going deeper and deeper until a goal node is found, or until it hits a node that has no children.

Then the search backtracks, returning to the most recent node it hadn't finished exploring. In a non-recursive implementation, all freshly expanded nodes are added to a LIFO stack for expansion.

Steps for implementing Depth first search

Step 1: - Define an array A or Vertex that store Boolean values, its size should be greater or equal to the number of vertices in the graph G.

Step 2: - Initialize the array A to false

Step 3: - For all vertices v in G

If A[v] = false

Process (v)

Step 4: - Exit

Algorithm for DFS

DFS(G)

{

For each v in V, //for loop V+1 times

{

Color[v]=white; // V times

     p[v]=NULL; // V times

}

Time=0; // constant time O (1)

For each u in V, //for loop V+1 times

If (color[u]==white) // V times

  DFSVISIT(u) // call to DFSVISIT(v), at most V times O(V)

   }

DFSVISIT(u)

   {

Color[u]=gray; // constant time

t[u] = ++time;

For each v in Adj(u) // for loop

If (color[v] == white)

{

p[v] = u;

DFSVISIT(v); // call to DFSVISIT(v)

}

Color[u] = black; // constant time

f[u]=++time; // constant time

}

DFS algorithm used to solve following problems:

●      Testing whether graph is connected.

●      Computing a spanning forest of graph.

●      Computing a path between two vertices of graph or equivalently reporting that no such path exists.

●      Computing a cycle in graph or equivalently reporting that no such cycle exists.

Q4) Consider the graph G along with its adjacency list, given in the figure below. Calculate the order to print all the nodes of the graph starting from node H, by using depth first search (DFS) algorithm.

A4)

Push H onto the stack

• STACK: H   

POP the top element of the stack i.e. H, print it and push all the neighbours of H onto the stack that are is ready state.

• Print H   
• STACK: A   

Pop the top element of the stack i.e. A, print it and push all the neighbours of A onto the stack that are in ready state.

• Print A  
• Stack: B, D  

Pop the top element of the stack i.e. D, print it and push all the neighbours of D onto the stack that are in ready state.

• Print D   
• Stack: B, F   

Pop the top element of the stack i.e. F, print it and push all the neighbours of F onto the stack that are in ready state.

• Print F  
• Stack: B  

Pop the top of the stack i.e. B and push all the neighbours

• Print B   
• Stack: C   

Pop the top of the stack i.e. C and push all the neighbours.

• Print C   
• Stack: E, G   

Pop the top of the stack i.e. G and push all its neighbours.

• Print G  
• Stack: E  

Pop the top of the stack i.e. E and push all its neighbours.

• Print E  
• Stack:  

Hence, the stack now becomes empty and all the nodes of the graph have been traversed.

The printing sequence of the graph will be:

• H → A → D → F → B → C → G → E  

Q5) Write complexity analysis of depth first search.

A5)

Analysis of DFS

In the above algorithm, there is only one DFSVISIT(u) call for each vertex u in the vertex set V. Initialization complexity in DFS(G) for loop is O(V). In second for loop of DFS(G), complexity is O(V) if we leave the call of DFSVISIT(u).

Now, let us find the complexity of function DFSVISIT(u)

The complexity of for loop will be O(deg(u)+1) if we do not consider the recursive call to DFSVISIT(v). For recursive call to DFSVISIT(v), (complexity will be O(E) as

Recursive call to DFSVISIT(v) will be at most the sum of degree of adjacency for all vertex v in the vertex set V. It can be written as ∑ |Adj(v)|=O(E) vεV

The running time of DSF is (V + E).

Q6) What is Breadth first search and write its algorithm?

A6)

Breadth first search

It is an uninformed search methodology which first expand and examine all nodes of graph systematically in search of the solution in the sequence and then go at the deeper level. In other words, it exhaustively searches the entire graph without considering the goal until it finds it.

From the standpoint of the algorithm, all child nodes obtained by expanding a node are added to a FIFO queue. In typical implementations, nodes that have not yet been examined for their neighbors are placed in some container (such as a queue or linked list) called "open" and then once examined are placed in the container "closed".

Steps for implementing Breadth first search

Step 1: - Initialize all the vertices by setting Flag = 1

Step 2: - Put the starting vertex A in Q and change its status to the waiting state by setting Flag = 0

Step 3: - Repeat through step 5 while Q is not NULL

Step 4: - Remove the front vertex v of Q. Process v and set the status of v to the processed status by setting Flag = -1

Step 5: - Add to the rear of Q all the neighbour of v that are in the ready state by setting Flag = 1 and change their status to the waiting state by setting flag = 0

Step 6: - Exit

Algorithm for BFS

BFS (G, s)

{

For each v in V - {s} // for loop {

   color[v]=white;

      d[v]= INFINITY;

    p[v]=NULL;

}

  color[s] = gray;

  d[s]=0;

  p[s]=NULL;

  Q = ø; // Initialize queue is empty

Enqueue(Q, s); /* Insert start vertex s in Queue Q */

   while Q is nonempty // while loop

{

u = Dequeue[Q]; /* Remove an element from Queue Q*/

   for each v in Adj[u] // for loop

{  if (color[v]  == white) /*if v is unvisted*/

{

Color[v] = gray; /* v is visted */

d[v] = d[u] + 1; /*Set distance of v to no. Of edges from s to u*/

    p[v] = u;  /*Set parent of v*/

Enqueue(Q,v); /*Insert  v in Queue Q*/

}

     }

   color[u] = black; /*finally visted or explored vertex u*/

}

}

Breadth First Search algorithm used in

●      Prim's MST algorithm.

●      Dijkstra's single source shortest path algorithm.

●      Testing whether graph is connected.

●      Computing a cycle in graph or reporting that no such cycle exists.

Q7) Consider the graph G shown in the following image, calculate the minimum path p from node A to node E. Given that each edge has a length of 1.

A7)

Minimum Path P can be found by applying breadth first search algorithm that will begin at node A and will end at E. The algorithm uses two queues, namely QUEUE1 and QUEUE2. QUEUE1 holds all the nodes that are to be processed while QUEUE2 holds all the nodes that are processed and deleted from QUEUE1.

Let’s start examining the graph from Node A.

1. Add A to QUEUE1 and NULL to QUEUE2.

  QUEUE1 = {A}  

  QUEUE2 = {NULL}  

2. Delete the Node A from QUEUE1 and insert all its neighbours. Insert Node A into QUEUE2

  QUEUE1 = {B, D}  

  QUEUE2 = {A}  

3. Delete the node B from QUEUE1 and insert all its neighbours. Insert node B into QUEUE2.

  QUEUE1 = {D, C, F}   

  QUEUE2 = {A, B}  

4. Delete the node D from QUEUE1 and insert all its neighbours. Since F is the only neighbour of it which has been inserted, we will not insert it again. Insert node D into QUEUE2.

  QUEUE1 = {C, F}  

  QUEUE2 = { A, B, D}  

5. Delete the node C from QUEUE1 and insert all its neighbours. Add node C to QUEUE2.

  QUEUE1 = {F, E, G}  

  QUEUE2 = {A, B, D, C}  

6. Remove F from QUEUE1 and add all its neighbours. Since all of its neighbours has already been added, we will not add them again. Add node F to QUEUE2.

  QUEUE1 = {E, G}  

  QUEUE2 = {A, B, D, C, F}  

7. Remove E from QUEUE1, all of E's neighbours has already been added to QUEUE1 therefore we will not add them again. All the nodes are visited and the target node i.e. E is encountered into QUEUE2.

  QUEUE1 = {G}  

  QUEUE2 = {A, B, D, C, F,  E}  

Now, backtrack from E to A, using the nodes available in QUEUE2.

The minimum path will be A → B → C → E.

Q8) Write complexity analysis of breadth first search.

A8)

Analysis of BFS

In this algorithm first for loop executes at most O(V) times.

While loop executes at most O(V) times as every vertex v in V is enqueued only once in the Queue Q. Every vertex is enqueued once and dequeued once so queuing will take at most O(V) time.

Inside while loop, there is for loop which will execute at most O(E) times as it will be at most the sum of degree of adjacency for all vertex v in the vertex set V.

Which can be written as

∑ |Adj(v)|=O(E)

vεV

Total running time of BFS is O(V + E).

Like depth first search, BFS traverse a connected component of a given graph and defines a spanning tree.

Space complexity of DFS is much lower than BFS (breadth-first search). It also lends itself much better to heuristic methods of choosing a likely-looking branch. Time complexity of both algorithms are proportional to the number of vertices plus the number of edges in the graphs they traverse.

Q9) What is the difference between DFS and BFS

A8) Difference between DFS and BFS

Q10) Explain Connected component.

A9)

Given an undirected graph, print all connected components line by line. For example consider the following graph.

We have discussed algorithms for finding strongly connected components in directed graphs in the following posts.

Kosaraju’s algorithm for strongly connected components.

Tarjan’s Algorithm to find Strongly Connected Components

Finding connected components for an undirected graph is an easier task. We simple need to do either BFS or DFS starting from every unvisited vertex, and we get all strongly connected components. Below are steps based on DFS.

1) Initialize all vertices as not visited.

2) Do the following for every vertex 'v'.

(a) If 'v' is not visited before, call DFSUtil(v)

(b) Print new line character

DFSUtil(v)

1) Mark 'v' as visited.

2) Print 'v'

3) Do the following for every adjacent 'u' of 'v'.

If 'u' is not visited, then recursively call DFSUtil(u)

Below is the implementation of the above algorithm.

// C++ program to print connected components in

// an undirected graph

#include <iostream>

#include <list>

Using namespace std;

// Graph class represents a undirected graph

// using adjacency list representation

Class Graph {

Int V; // No. Of vertices

// Pointer to an array containing adjacency lists

List<int>* adj;

// A function used by DFS

Void DFSUtil(int v, bool visited[]);

Public:

Graph(int V); // Constructor

~Graph();

Void addEdge(int v, int w);

Void connectedComponents();

};

// Method to print connected components in an

// undirected graph

Void Graph::connectedComponents()

{

// Mark all the vertices as not visited

Bool* visited = new bool[V];

For (int v = 0; v < V; v++)

Visited[v] = false;

For (int v = 0; v < V; v++) {

If (visited[v] == false) {

// print all reachable vertices

// from v

DFSUtil(v, visited);

Cout << "\n";

}

}

Delete[] visited;

}

Void Graph::DFSUtil(int v, bool visited[])

{

// Mark the current node as visited and print it

Visited[v] = true;

Cout << v << " ";

// Recur for all the vertices

// adjacent to this vertex

List<int>::iterator i;

For (i = adj[v].begin(); i != adj[v].end(); ++i)

If (!visited[*i])

DFSUtil(*i, visited);

}

Graph::Graph(int V)

{

This->V = V;

Adj = new list<int>[V];

}

Graph::~Graph() { delete[] adj; }

// method to add an undirected edge

Void Graph::addEdge(int v, int w)

{

Adj[v].push_back(w);

Adj[w].push_back(v);

}

// Driver code

Int main()

{

// Create a graph given in the above diagram

Graph g(5); // 5 vertices numbered from 0 to 4

g.addEdge(1, 0);

g.addEdge(2, 3);

g.addEdge(3, 4);

Cout << "Following are connected components \n";

g.connectedComponents();

Return 0;

}

Output:

0 1

2 3 4

Time complexity of the above solution is O (V + E) as it does simple DFS for a given graph.

Q11) What is Spanning tree?

A10)

A spanning tree is a subset of Graph G that covers all of the vertices with the fewest number of edges feasible. As a result, a spanning tree has no cycles and cannot be disconnected.

We can deduce from this definition that every linked and undirected Graph G contains at least one spanning tree. Because a disconnected graph cannot be spanned to all of its vertices, it lacks a spanning tree.

Fig: Spanning tree

From a single full graph, we discovered three spanning trees. The greatest number of spanning trees in a full undirected graph is nn-2, where n is the number of nodes. Because n is 3 in the example above, 33-2 = 3 spanning trees are conceivable.

Q12) What are the properties and applications of spanning tree?

A11)

Properties of spanning tree

We now know that a graph can contain many spanning trees. A few features of the spanning tree related to graph G are listed below.

●      There can be more than one spanning tree in a linked graph G.

●      The number of edges and vertices in all conceivable spanning trees of graph G is the same.

●      There is no cycle in the spanning tree (loops).

●      When one edge is removed from the spanning tree, the graph becomes unconnected, i.e. the spanning tree becomes minimally connected.

●      When one edge is added to the spanning tree, a circuit or loop is created, indicating that the spanning tree is maximally acyclic.

Mathematical Properties of Spanning Tree

• Spanning tree has n-1 edges, where n is the number of nodes (vertices).
• From a complete graph, by removing maximum e - n + 1 edges, we can construct a spanning tree.
• A complete graph can have maximum nn-2 number of spanning trees.

Application of spanning tree

The spanning tree is used to discover the shortest path between all nodes in a graph. The use of spanning trees is common in a variety of situations.

●      ​​Civil Network Planning

●      Computer Network Routing Protocol

●      Cluster Analysis

Let's look at an example to help us comprehend. Consider the city network as a large graph, and the current objective is to deploy telephone lines in such a way that we can link to all city nodes with the bare minimum of lines. The spanning tree enters the scene at this point.

Q13) Explain Single source shortest path problem in detail.

A12)

Introduction:

In a shortest- paths problem, we are given a weighted, directed graphs G = (V, E), with weight function w: E → R mapping edges to real-valued weights. The weight of path p = (v0,v1,..... vk) is the total of the weights of its constituent edges:

We define the shortest - path weight from u to v by δ(u,v) = min (w (p): u→v), if there is a path from u to v, and δ(u,v)= ∞, otherwise.

The shortest path from vertex s to vertex t is then defined as any path p with weight w (p) = δ(s,t).

The breadth-first- search algorithm is the shortest path algorithm that works on unweighted graphs, that is, graphs in which each edge can be considered to have unit weight.

In a Single Source Shortest Paths Problem, we are given a Graph G = (V, E), we want to find the shortest path from a given source vertex s ∈ V to every vertex v ∈ V.

Variants:

There are some variants of the shortest path problem.

• Single- destination shortest - paths problem: Find the shortest path to a given destination vertex t from every vertex v. By shift the direction of each edge in the graph, we can shorten this problem to a single - source problem.
• Single - pair shortest - path problem: Find the shortest path from u to v for given vertices u and v. If we determine the single - source problem with source vertex u, we clarify this problem also. Furthermore, no algorithms for this problem are known that run asymptotically faster than the best single - source algorithms in the worst case.
• All - pairs shortest - paths problem: Find the shortest path from u to v for every pair of vertices u and v. Running a single - source algorithm once from each vertex can clarify this problem; but it can generally be solved faster, and its structure is of interest in the own right.

Shortest Path: Existence:

If some path from s to v contains a negative cost cycle then, there does not exist the shortest path. Otherwise, there exists a shortest s - v that is simple.

Fig - Shortest Path: Existence

The single source shortest path algorithm (for arbitrary weight positive or negative) is also known Bellman-Ford algorithm is used to find minimum distance from source vertex to any other vertex. The main difference between this algorithm with Dijkstra’s algorithm is, in Dijkstra’s algorithm we cannot handle the negative weight, but here we can handle it easily.

Fig - The single source shortest path algorithm

Bellman-Ford algorithm finds the distance in bottom up manner. At first it finds those distances which have only one edge in the path. After that increase the path length to find all possible solutions.

Input − The cost matrix of the graph:

0 6 ∞ 7 ∞

∞ 0 5 8 -4

∞ -2 0 ∞ ∞

∞ ∞ -3 0 9

2 ∞ 7 ∞ 0

Output− Source Vertex: 2

Vert: 0 1 2 3 4

Dist: -4 -2 0 3 -6

Pred: 4 2 -1 0 1

The graph has no negative edge cycle

Algorithm

BellmanFord(dist, pred, source)

Input− Distance list, predecessor list and the source vertex.

Output− True, when a negative cycle is found.

Begin

ICount := 1

MaxEdge := n * (n - 1) / 2 //n is number of vertices

For all vertices v of the graph, do

Dist[v] := ∞

Pred[v] := ϕ

Done

Dist[source] := 0

ECount := number of edges present in the graph

Create edge list named edgeList

While iCount < n, do

For i := 0 to eCount, do

If dist[edgeList[i].v] > dist[edgeList[i].u] + (cost[u,v] for edge i)

Dist[edgeList[i].v] > dist[edgeList[i].u] + (cost[u,v] for edge i)

Pred[edgeList[i].v] := edgeList[i].u

Done

Done

ICount := iCount + 1

For all vertices i in the graph, do

If dist[edgeList[i].v] > dist[edgeList[i].u] + (cost[u,v] for edge i), then

Return true

Done

Return false

End

Example(C++)

#include<iostream>

#include<iomanip>

#define V 5

#define INF 999

Using namespace std;

//Cost matrix of the graph (directed) vertex 5

Int costMat[V][V] = {

{0, 6, INF, 7, INF},

{INF, 0, 5, 8, -4},

{INF, -2, 0, INF, INF},

{INF, INF, -3, 0, 9},

{2, INF, 7, INF, 0}

};

Typedef struct{

Int u, v, cost;

}edge;

Int isDiagraph(){

//check the graph is directed graph or not

Int i, j;

For(i = 0; i<V; i++){

For(j = 0; j<V; j++){

If(costMat[i][j] != costMat[j][i]){

Return 1;//graph is directed

}

}

}

Return 0;//graph is undirected

}

Int makeEdgeList(edge *eList){

//create edgelist from the edges of graph

Int count = -1;

If(isDiagraph()){

For(int i = 0; i<V; i++){

For(int j = 0; j<V; j++){

If(costMat[i][j] != 0 && costMat[i][j] != INF){

Count++;//edge find when graph is directed

EList[count].u = i; eList[count].v = j;

EList[count].cost = costMat[i][j];

}

}

}

}else{

For(int i = 0; i<V; i++){

For(int j = 0; j<i; j++){

If(costMat[i][j] != INF){

Count++;//edge find when graph is undirected

EList[count].u = i; eList[count].v = j;

EList[count].cost = costMat[i][j];

}

}

}

}

Return count+1;

}

Int bellmanFord(int *dist, int *pred,int src){

Int icount = 1, ecount, max = V*(V-1)/2;

Edge edgeList[max];

For(int i = 0; i<V; i++){

Dist[i] = INF;//initialize with infinity

Pred[i] = -1;//no predecessor found.

}

Dist[src] = 0;//for starting vertex, distance is 0

Ecount = makeEdgeList(edgeList); //edgeList formation

While(icount < V){ //number of iteration is (Vertex - 1)

For(int i = 0; i<ecount; i++){

If(dist[edgeList[i].v] > dist[edgeList[i].u] + costMat[edgeList[i].u][edgeList[i].v]){

//relax edge and set predecessor

Dist[edgeList[i].v] = dist[edgeList[i].u] + costMat[edgeList[i].u][edgeList[i].v];

Pred[edgeList[i].v] = edgeList[i].u;

}

}

Icount++;

}

//test for negative cycle

For(int i = 0; i<ecount; i++){

If(dist[edgeList[i].v] > dist[edgeList[i].u] + costMat[edgeList[i].u][edgeList[i].v]){

Return 1;//indicates the graph has negative cycle

}

}

Return 0;//no negative cycle

}

Void display(int *dist, int *pred){

Cout << "Vert: ";

For(int i = 0; i<V; i++)

Cout <<setw(3) << i << " ";

Cout << endl;

Cout << "Dist: ";

For(int i = 0; i<V; i++)

Cout <<setw(3) << dist[i] << " ";

Cout << endl;

Cout << "Pred: ";

For(int i = 0; i<V; i++)

Cout <<setw(3) << pred[i] << " ";

Cout << endl;

}

Int main(){

Int dist[V], pred[V], source, report;

Source = 2;

Report = bellmanFord(dist, pred, source);

Cout << "Source Vertex: " << source<<endl;

Display(dist, pred);

If(report)

Cout << "The graph has a negative edge cycle" << endl;

Else

Cout << "The graph has no negative edge cycle" << endl;

}

Output

Source Vertex: 2

Vert: 0 1 2 3 4

Dist: -4 -2 0 3 -6

Pred: 4 2 -1 0 1

The graph has no negative edge cycle

Q14) Explain Floydd’s Algorithm in detail with example.

A13)

Floydd’s Algorithm for All-Pairs Shortest Paths Problem

Let the vertices of G be V = {1, 2........n} and consider a subset {1, 2........k} of vertices for some k. For any pair of vertices i, j ∈ V, considered all paths from i to j whose intermediate vertices are all drawn from {1, 2.......k}, and let p be a minimum weight path from amongst them. The Floyd-Warshall algorithm exploits a link between path p and shortest paths from i to j with all intermediate vertices in the set {1, 2.......k-1}. The link depends on whether or not k is an intermediate vertex of path p.

If k is not an intermediate vertex of path p, then all intermediate vertices of path p are in the set {1, 2........k-1}. Thus, the shortest path from vertex i to vertex j with all intermediate vertices in the set {1, 2.......k-1} is also the shortest path i to j with all intermediate vertices in the set {1, 2.......k}.

If k is an intermediate vertex of path p, then we break p down into i → k → j.

Let dij(k) be the weight of the shortest path from vertex i to vertex j with all intermediate vertices in the set {1, 2.......k}.

A recursive definition is given by

dij(k) = { wij     if  k = 0 min(dijk-1 ,   dikk-1 + dkjk-1)     if k ≥ 1}

FLOYD - WARSHALL (W)

1. n ← rows [W].

2. D0← W

3. For k ← 1 to n

4. Do for i ← 1 to n    

5. Do for j ← 1 to n    

6. Do dij(k)← min (dij(k-1),dik(k-1)+dkj(k-1) )

7. Return D(n)     

The strategy adopted by the Floyd-Warshall algorithm is Dynamic Programming. The running time of the Floyd-Warshall algorithm is determined by the triply nested for loops of lines 3-6. Each execution of line 6 takes O (1) time. The algorithm thus runs in time θ(n3 ).

Example: Apply Floyd-Warshall algorithm for constructing the shortest path. Show that matrices D(k) and π(k) computed by the Floyd-Warshall algorithm for the graph.

Solution:

dij(k) = min (dij(k-1), dij(k-1) + dkj(k - 1))

Step (i) When k = 0

Step (ii) When k =1

Step (iii) When k = 2

Step (iv) When k = 3

Step (v) When k = 4

Step (vi) When k = 5

TRANSITIVE- CLOSURE (G)

1. n ← |V[G]|

2. For i ← 1 to n

3. Do for j ← 1 to n

4. Do if i = j or (i, j) ∈ E [G]

5. The ← 1

6. Else ← 0

7. For k ← 1 to n

8. Do for i ← 1 to n

9. Do for j ← 1 to n

10. Dod ij(k)← 

11. Return T(n).

Q15) What is Topological sort? Explain in detail.

A14)

The topological sorting for a directed acyclic graph is the linear ordering of vertices. For every edge U-V of a directed graph, the vertex u will come before vertex v in the ordering.

As we know that the source vertex will come after the destination vertex, so we need to use a stack to store previous elements. After completing all nodes, we can simply display them from the stack.

Input and Output

Input:

0 0 0 0 0 0

0 0 0 0 0 0

0 0 0 1 0 0

0 1 0 0 0 0

1 1 0 0 0 0

1 0 1 0 0 0

Output:

Nodes after topological sorted order: 5 4 2 3 1 0

Algorithm

TopoSort(u, visited, stack)

Input − The start vertex u, An array to keep track of which node is visited or not. A stack to store nodes.

Output − Sorting the vertices in topological sequence in the stack.

Begin

Mark u as visited

For all vertices v which is adjacent with u, do

If v is not visited, then

TopoSort(c, visited, stack)

Done

Push u into a stack

End

Perform Topological Sorting(Graph)

Input − The given directed acyclic graph.

Output − Sequence of nodes.

Begin

Initially mark all nodes as unvisited

For all nodes v of the graph, do

If v is not visited, then

TopoSort(i, visited, stack)

Done

Pop and print all elements from the stack

End.

Example

#include<iostream>

#include<stack>

#define NODE 6

Using namespace std;

Int graph[NODE][NODE] = {

{0, 0, 0, 0, 0, 0},

{0, 0, 0, 0, 0, 0},

{0, 0, 0, 1, 0, 0},

{0, 1, 0, 0, 0, 0},

{1, 1, 0, 0, 0, 0},

{1, 0, 1, 0, 0, 0}

};

Void topoSort(int u, bool visited[], stack<int>&stk) {

Visited[u] = true;                //set as the node v is visited

For(int v = 0; v<NODE; v++) {

If(graph[u][v]) {             //for allvertices v adjacent to u

If(!visited[v])

TopoSort(v, visited, stk);

}

}

Stk.push(u);     //push starting vertex into the stack

}

Void performTopologicalSort() {

Stack<int> stk;

Bool vis[NODE];

For(int i = 0; i<NODE; i++)

Vis[i] = false;     //initially all nodes are unvisited

For(int i = 0; i<NODE; i++)

If(!vis[i])     //when node is not visited

TopoSort(i, vis, stk);

While(!stk.empty()) {

Cout <<stk.top() << " ";

Stk.pop();

}

}

Main() {

Cout << "Nodes after topological sorted order: ";

PerformTopologicalSort();

}

Output

Nodes after topological sorted order: 5 4 2 3 1 0

Q16) What is Hamiltonian path? Explain with example.

A15)

We must use the Backtracking technique to identify the Hamiltonian Circuit given a graph G = (V, E). We begin our search at any random vertex, such as 'a.' The root of our implicit tree is this vertex 'a.' The first intermediate vertex of the Hamiltonian Cycle to be constructed is the first element of our partial solution. By alphabetical order, the next nearby vertex is chosen.

A dead end is reached when any arbitrary vertices forms a cycle with any vertex other than vertex 'a' at any point. In this situation, we backtrack one step and restart the search by selecting a new vertex and backtracking the partial element; the answer must be removed. If a Hamiltonian Cycle is found, the backtracking search is successful.

Example - Consider the graph G = (V, E) in fig. We must use the Backtracking method to find a Hamiltonian circuit.

To begin, we'll start with vertex 'a,' which will serve as the root of our implicit tree.

Following that, we select the vertex 'b' adjacent to 'a' because it is the first in lexicographical order (b, c, d).

Then, next to 'b,' we select 'c.'

Next, we choose the letter 'd,' which is adjacent to the letter 'c.'

Then we select the letter 'e,' which is adjacent to the letter 'd.'

Then, adjacent to 'e,' we select vertex 'f.' D and e are the vertex next to 'f,' but they've already been there. As a result, we reach a dead end and go back a step to delete the vertex 'f' from the partial solution.

The vertices adjacent to 'e' is b, c, d, and f, from which vertex 'f' has already been verified, and b, c, and d have previously been visited, according to backtracking. As a result, we go back one step. The vertex adjacent to d is now e, f, from which e has already been checked, and the vertex adjacent to 'f' is now d and e. We get a dead state if we revisit the 'e' vertex. As a result, we go back one step.