Check if an Undirected Graph Contains a Eulerian Path in C++

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 Euler Path, 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, which is also one Euler path.
• When exactly two vertices have odd degree, it is a Euler Path.

Input

Output

Both of the graphs has Euler paths.

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

isEulerian(Graph)

Input : The given Graph.

Output : Returns 1, when Eulerian circuit or path, and returns 0 when it has no Euler Path.

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 > 0, then
      return 0
   else return 1
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}};
/*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 isEulerian() {
   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 > 2) //when vertices with odd degree greater than 2
      return 0;
   return 1; //when oddDegree is 0, it is Euler circuit, and when 2, it is Euler path
}
int main() {
   if(isEulerian() != 0) {
      cout << "The graph has Eulerian path." << endl;
   } else {
      cout << "The graph has No Eulerian path." << endl;
   }
}

Output

The graph has Eulerian path.