Check if the given graph represents a Bus Topology

Given a graph G, check if it represents a Bus Topology.

A Bus Topology is the one shown in the image below:

Examples:



Input: 

Output:  YES

Input: 

Output:  NO

A graph of V vertices represents a bus topology if it satisfies the following two conditions:

  1. Each node except the stating end ending ones have degree 2 while the starting and ending have degree 1.
  2. No of edges = No of Vertices – 1.

The idea is to traverse the graph and check if it satisfies the above two conditions. If yes, then it represents a Bus Topology.

Below is the implementation of the above approach:

CPP

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// CPP program to check if the given graph
// represents a bus topology
#include <bits/stdc++.h>
using namespace std;
  
// A utility function to add an edge in an
// undirected graph.
void addEdge(vector<int> adj[], int u, int v)
{
    adj[u].push_back(v);
    adj[v].push_back(u);
}
  
// A utility function to print the adjacency list
// representation of graph
void printGraph(vector<int> adj[], int V)
{
    for (int v = 0; v < V; ++v) {
        cout << "\n Adjacency list of vertex "
             << v << "\n head ";
        for (auto x : adj[v])
            cout << "-> " << x;
        printf("\n");
    }
}
  
/* Function to return true if the graph represented
   by the adjacency list represents a bus topology
   else return false */
bool checkBusTopologyUtil(vector<int> adj[], int V, int E)
{
  
    // Number of edges should be equal
    // to (Number of vertices - 1)
    if (E != (V - 1))
        return false;
  
    // a single node is termed as a bus topology
    if (V == 1)
        return true;
  
    int* vertexDegree = new int[V + 1];
    memset(vertexDegree, 0, sizeof vertexDegree);
  
    // calculate the degree of each vertex
    for (int i = 1; i <= V; i++) {
        for (auto v : adj[i]) {
            vertexDegree[v]++;
        }
    }
  
    // countDegree2 - number of vertices with degree 2
    // countDegree1 - number of vertices with degree 1
    int countDegree2 = 0, countDegree1 = 0;
    for (int i = 1; i <= V; i++) {
        if (vertexDegree[i] == 2) {
            countDegree2++;
        }
        else if (vertexDegree[i] == 1) {
            countDegree1++;
        }
        else {
            // if any node has degree other
            // than 1 or 2, it is
            // NOT a bus topology
            return false;
        }
    }
  
    // if both necessary conditions as discussed,
    // satisfy return true
    if (countDegree1 == 2 && countDegree2 == (V - 2)) {
        return true;
    }
    return false;
}
  
// Function to check if the graph represents a bus topology
void checkBusTopology(vector<int> adj[], int V, int E)
{
    bool isBus = checkBusTopologyUtil(adj, V, E);
    if (isBus) {
        cout << "YES" << endl;
    }
    else {
        cout << "NO" << endl;
    }
}
  
// Driver code
int main()
{
    // Graph 1
    int V = 5, E = 4;
    vector<int> adj1[V + 1];
    addEdge(adj1, 1, 2);
    addEdge(adj1, 1, 3);
    addEdge(adj1, 3, 4);
    addEdge(adj1, 4, 5);
    checkBusTopology(adj1, V, E);
  
    // Graph 2
    V = 4, E = 4;
    vector<int> adj2[V + 1];
    addEdge(adj2, 1, 2);
    addEdge(adj2, 1, 3);
    addEdge(adj2, 3, 4);
    addEdge(adj2, 4, 2);
    checkBusTopology(adj2, V, E);
  
    return 0;
}

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Python3

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# Python3 program to check if the given graph
# represents a bus topology
  
# A utility function to add an edge in an
# undirected graph.
def addEdge(adj, u, v):
    adj[u].append(v)
    adj[v].append(u)
  
# A utility function to prthe adjacency list
# representation of graph
def printGraph(adj, V):
  
    for v in range(V):
        print("Adjacency list of vertex ",v,"\n head ")
        for x in adj[v]:
            print("-> ",x,end=" ")
        printf()
  
# /* Function to return true if the graph represented
# by the adjacency list represents a bus topology
# else return false */
def checkBusTopologyUtil(adj, V, E):
  
    # Number of edges should be equal
    # to (Number of vertices - 1)
    if (E != (V - 1)):
        return False
  
    # a single node is termed as a bus topology
    if (V == 1):
        return True
  
    vertexDegree = [0]*(V + 1)
  
    # calculate the degree of each vertex
    for i in range(V + 1):
        for v in adj[i]:
            vertexDegree[v] += 1
  
    # countDegree2 - number of vertices with degree 2
    # countDegree1 - number of vertices with degree 1
    countDegree2,countDegree1 = 0,0
    for i in range(1, V + 1):
        if (vertexDegree[i] == 2):
            countDegree2 += 1
  
        elif (vertexDegree[i] == 1):
            countDegree1 += 1
  
        else:
            # if any node has degree other
            # than 1 or 2, it is
            # NOT a bus topology
            return False
  
    # if both necessary conditions as discussed,
    # satisfy return true
    if (countDegree1 == 2 and countDegree2 == (V - 2)):
        return True
  
    return False
  
# Function to check if the graph represents a bus topology
def checkBusTopology(adj, V, E):
  
    isBus = checkBusTopologyUtil(adj, V, E)
    if (isBus):
        print("YES")
  
    else:
        print("NO" )
  
# Driver code
  
# Graph 1
V, E = 5, 4
adj1 = [[] for i in range(V + 1)]
addEdge(adj1, 1, 2)
addEdge(adj1, 1, 3)
addEdge(adj1, 3, 4)
addEdge(adj1, 4, 5)
checkBusTopology(adj1, V, E)
  
# Graph 2
V, E = 4, 4
adj2 = [[] for i in range(V + 1)]
addEdge(adj2, 1, 2)
addEdge(adj2, 1, 3)
addEdge(adj2, 3, 4)
addEdge(adj2, 4, 2)
checkBusTopology(adj2, V, E)
  
# This code is contributed by mohit kumar 29

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Output:

YES
NO

Time Complexity : O(E), where E is the number of Edges in the graph.




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Improved By : mohit kumar 29