C++ Program to Check Whether a Directed Graph Contains a Eulerian Path

C++Server Side Programming Programming

The Euler path is a path; by which we can visit every edge exactly once. We can use the same vertices for multiple times. One graph which contains Euler Circuit is also considered in this case, as it also has the Euler path.

To check whether a directed graph has Euler path or not, we have to check these conditions -

There must be one single vertex a_n where (in-degree + 1 = out_degree)
There must be one single vertex b_n where (in-degree = out_degree + 1)
Rest all vertices have (in-degree = out_degree) if any of these cases fails, the graph has no Euler path.

Vertex b has (in-degree 1, out-degree 2), vertex c has (in-degree 2, out-degree 1). And for the rest of the vertices a, d has (in-degree 2, out-degree 2), e has (in-degree 1, out-degree 1).

Input

Adjacency matrix of the graph.

0	0	1	1	0
1	0	1	0	0
0	0	0	1	0
0	1	0	0	1
1	0	0	0	0

Output

Euler Path Found.

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

hasEulerPath(Graph)

Input The given Graph.

Output True when one Euler circuit is found.

Begin
   an := 0
   bn := 0
   if isConnected() is false, then
      return false
   define list for inward and outward edge count for each node
   for all vertex i in the graph, do
      sum := 0
      for all vertex j which are connected with i, do
         inward edges for vertex i increased
         increase sum
      done
      number of outward of vertex i is sum
   done
   if inward list and outward list are same, then
      return true
   for all vertex i in the vertex set V, do
      if inward[i] ≠ outward[i], then
         if inward[i] + 1 = outward[i], then
            an := an + 1
         else if inward[i] = outward[i] + 1, then
            bn := bn + 1
      done
      if an and bn both are 1, then
          return true
      otherwise return false
End

Example Code

#include<iostream>
#include<vector>
#define NODE 5
using namespace std;
int graph[NODE][NODE] = {{0, 0, 1, 1, 0},
   {1, 0, 1, 0, 0},
   {0, 0, 0, 1, 0},
   {0, 1, 0, 0, 1},
   {1, 0, 0, 0, 0}};
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;
}
bool hasEulerPath() {
   int an = 0, bn = 0;
   if(isConnected() == false){ //when graph is not connected
      return false;
   }
   vector<int> inward(NODE, 0), outward(NODE, 0);
   for(int i = 0; i<NODE; i++) {
      int sum = 0;
      for(int j = 0; j<NODE; j++) {
         if(graph[i][j]) {
            inward[j]++; //increase inward edge for destination vertex
            sum++; //how many outward edge
         }
      }
      outward[i] = sum;
   }
   //check the condition for Euler paths
   if(inward == outward) //when number inward edges and outward edges for each node is same
      return true; //Euler Circuit, it has Euler path
   for(int i = 0; i<NODE; i++) {
      if(inward[i] != outward[i]) {
         if((inward[i] + 1 == outward[i])) {
            an++;
         } else if((inward[i] == outward[i] + 1)) {
            bn++;
         }
      }
   }
   if(an == 1 && bn == 1) { //if there is only an, and bn, then this has euler path
      return true;
   }
   return false;
}
int main() {
   if(hasEulerPath())
      cout << "Euler Path Found.";
   else
   cout << "There is no Euler Circuit.";
}

Output

Euler Path Found.

Arjun Thakur

Updated on: 30-Jul-2019

346 Views

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