Given a directed graph, check whether the graph contains a cycle or not. Your function should return true if the given graph contains at least one cycle, else return false. For example, the following graph contains three cycles 0->2->0, 0->1->2->0 and 3->3, so your function must return true.

**Solution**

Depth First Traversal can be used to detect cycle in a Graph. DFS for a connected graph produces a tree. There is a cycle in a graph only if there is a back edge present in the graph. A back edge is an edge that is from a node to itself (selfloop) or one of its ancestor in the tree produced by DFS. In the following graph, there are 3 back edges, marked with cross sign. We can observe that these 3 back edges indicate 3 cycles present in the graph.

For a disconnected graph, we get the DFS forrest as output. To detect cycle, we can check for cycle in individual trees by checking back edges.

To detect a back edge, we can keep track of vertices currently in recursion stack of function for DFS traversal. If we reach a vertex that is already in the recursion stack, then there is a cycle in the tree. The edge that connects current vertex to the vertex in the recursion stack is back edge. We have used recStack[] array to keep track of vertices in the recursion stack.

// A C++ Program to detect cycle in a graph #include<iostream> #include <list> #include <limits.h> using namespace std; class Graph { int V; // No. of vertices list<int> *adj; // Pointer to an array containing adjacency lists bool isCyclicUtil(int v, bool visited[], bool *rs); // used by isCyclic() public: Graph(int V); // Constructor void addEdge(int v, int w); // to add an edge to graph bool isCyclic(); // returns true if there is a cycle in this graph }; Graph::Graph(int V) { this->V = V; adj = new list<int>[V]; } void Graph::addEdge(int v, int w) { adj[v].push_back(w); // Add w to v’s list. } // This function is a variation of DFSUytil() in http://www.geeksforgeeks.org/archives/18212 bool Graph::isCyclicUtil(int v, bool visited[], bool *recStack) { if(visited[v] == false) { // Mark the current node as visited and part of recursion stack visited[v] = true; recStack[v] = true; // Recur for all the vertices adjacent to this vertex list<int>::iterator i; for(i = adj[v].begin(); i != adj[v].end(); ++i) { if ( !visited[*i] && isCyclicUtil(*i, visited, recStack) ) return true; else if (recStack[*i]) return true; } } recStack[v] = false; // remove the vertex from recursion stack return false; } // Returns true if the graph contains a cycle, else false. // This function is a variation of DFS() in http://www.geeksforgeeks.org/archives/18212 bool Graph::isCyclic() { // Mark all the vertices as not visited and not part of recursion // stack bool *visited = new bool[V]; bool *recStack = new bool[V]; for(int i = 0; i < V; i++) { visited[i] = false; recStack[i] = false; } // Call the recursive helper function to detect cycle in different // DFS trees for(int i = 0; i < V; i++) if (isCyclicUtil(i, visited, recStack)) return true; return false; } int main() { // Create a graph given in the above diagram Graph g(4); g.addEdge(0, 1); g.addEdge(0, 2); g.addEdge(1, 2); g.addEdge(2, 0); g.addEdge(2, 3); g.addEdge(3, 3); if(g.isCyclic()) cout << "Graph contains cycle"; else cout << "Graph doesn't contain cycle"; return 0; }

Output:

Graph contains cycle

Time Complexity of this method is same as time complexity of DFS traversal which is O(V+E).

In the below article, another O(V + E) method is discussed :

Detect Cycle in a direct graph using colors

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