algorithm Detecting a cycle in a directed graph using Depth First Traversal


Example

A cycle in a directed graph exists if there's a back edge discovered during a DFS. A back edge is an edge from a node to itself or one of the ancestors in a DFS tree. For a disconnected graph, we get a DFS forest, so you have to iterate through all vertices in the graph to find disjoint DFS trees.

C++ implementation:

    #include <iostream>
     #include <list>
        
    using namespace std; 
    
    #define NUM_V   4

    bool helper(list<int> *graph, int u, bool* visited, bool* recStack)
    {
        visited[u]=true;
        recStack[u]=true;
        list<int>::iterator i;
        for(i = graph[u].begin();i!=graph[u].end();++i)
        {  
            if(recStack[*i]) //if vertice v is found in recursion stack of this DFS traversal
                return true;
            else if(*i==u) //if there's an edge from the vertex to itself
                return true;
            else if(!visited[*i])
            {   if(helper(graph, *i, visited, recStack))
                   return true;
            }
        }
        recStack[u]=false;
        return false;
    }
    /*
    /The wrapper function calls helper function on each vertices which have not been visited. Helper function returns true if it detects a back edge in the subgraph(tree) or false.
*/
    bool isCyclic(list<int> *graph, int V)
    {
      bool visited[V];  //array to track vertices already visited
      bool recStack[V]; //array to track vertices in recursion stack of the traversal.

      for(int i = 0;i<V;i++)
       visited[i]=false, recStack[i]=false;  //initialize all vertices as not visited and not recursed

      for(int u = 0; u < V; u++) //Iteratively checks if every vertices have been visited
      {   if(visited[u]==false)
          {  if(helper(graph, u, visited, recStack)) //checks if the DFS tree from the vertex contains a cycle
               return true;
          }
      }
       return false;
    }
    /*
    Driver function
    */
    int main()
    {
        list<int>* graph = new list<int>[NUM_V];
        graph[0].push_back(1);
        graph[0].push_back(2);
        graph[1].push_back(2);
        graph[2].push_back(0);
        graph[2].push_back(3);
        graph[3].push_back(3);
        bool res = isCyclic(graph, NUM_V);
        cout<<res<<endl;
    }

Result: As shown below, there are three back edges in the graph. One between vertex 0 and 2; between vertice 0, 1, and 2; and vertex 3. Time complexity of search is O(V+E) where V is the number of vertices and E is the number of edges. enter image description here