C++ Program to Check Whether an Undirected Graph Contains a Eulerian Cycle

C++Server Side ProgrammingProgramming

To know about Euler Circuit, we have the idea about Euler Path. The Euler path is a path; by which we can visit every node exactly once. We can use the same edges for multiple times. The Euler Circuit is a special type of Euler path. When the starting vertex of the Euler path is also connected with the ending vertex of that path.

To detect the circuit, we have to follow these conditions:

  • The graph must be connected.
  • Now when no vertices of an undirected graph have odd degree, then it is a Euler Circuit.

Input

Output

The graph has Euler Circuit.

Algorithm

traverse(u, visited)

Input The start node u and the visited node to mark which node is visited.

Output Traverse all connected vertices.

Begin
   mark u as visited
   for all vertex v, if it is adjacent with u, do
      if v is not visited, then
         traverse(v, visited)
      done
End

isConnected(graph)

Input : The graph.

Output : True if the graph is connected.

Begin
   define visited array
   for all vertices u in the graph, do
      make all nodes unvisited
      traverse(u, visited)
      if any unvisited node is still remaining, then
         return false
      done
   return true
End

hasEulerianCircuit(Graph)

Input The given Graph.

Output Returns 0, when no Eulerian Circuit, and returns 1 when it has Euler circuit..

Begin
   if isConnected() is false, then
   return false
   define list of degree for each node
   oddDegree := 0
   for all vertex i in the graph, do
      for all vertex j which are connected with i, do
         increase degree
      done
      if degree of vertex i is odd, then
         increase oddDegree
      done
      if oddDegree is 0, then
      return 1
   else return 0
End

Example Code

#include<iostream>
#include<vector>
#define NODE 5
using namespace std;
/*int graph[NODE][NODE] = {{0, 1, 1, 1, 0},
   {1, 0, 1, 0, 0},
   {1, 1, 0, 0, 0},
   {1, 0, 0, 0, 1},
   {0, 0, 0, 1, 0}};*/ //No Euler circuit, but euler path is present
int graph[NODE][NODE] = {{0, 1, 1, 1, 1},
   {1, 0, 1, 0, 0},
   {1, 1, 0, 0, 0},
   {1, 0, 0, 0, 1},
   {1, 0, 0, 1, 0}}; //uncomment to check Euler Circuit as well as path
/*int graph[NODE][NODE] = {{0, 1, 1, 1, 0},
   {1, 0, 1, 1, 0},
   {1, 1, 0, 0, 0},
   {1, 1, 0, 0, 1},
   {0, 0, 0, 1, 0}};*/ //Uncomment to check Non Eulerian Graph
void traverse(int u, bool visited[]) {
   visited[u] = true; //mark v as visited
   for(int v = 0; v<NODE; v++) {
      if(graph[u][v]) {
         if(!visited[v]) traverse(v, visited);
      }
   }
}
bool isConnected() {
   bool *vis = new bool[NODE];
   //for all vertex u as start point, check whether all nodes are visible or not
   for(int u; u < NODE; u++) {
      for(int i = 0; i<NODE; i++)
         vis[i] = false; //initialize as no node is visited
         traverse(u, vis);
         for(int i = 0; i<NODE; i++) {
            if(!vis[i]) //if there is a node, not visited by traversal, graph is not connected
            return false;
         }
   }
   return true;
}
int hasEulerianCircuit() {
   if(isConnected() == false) //when graph is not connected
   return 0;
   vector<int> degree(NODE, 0);
   int oddDegree = 0;
   for(int i = 0; i<NODE; i++) {
      for(int j = 0; j<NODE; j++) {
         if(graph[i][j])
            degree[i]++; //increase degree, when connected edge found
      }
      if(degree[i] % 2 != 0) //when degree of vertices are odd
      oddDegree++; //count odd degree vertices
   }
   if(oddDegree == 0) { //when oddDegree is 0, it is Euler circuit
      return 1;
   }
   return 0;
}
int main() {
   if(hasEulerianCircuit()) {
      cout << "The graph has Eulerian Circuit." << endl;
   } else {
      cout << "The graph has No Eulerian Circuit." << endl;
   }
}

Output

The graph has Eulerian Circuit.
raja
Published on 28-May-2019 15:01:02
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