Detected Cycle in directed graph if the vertex is found in recursive stack-why? - algorithm

Question

I have read an article from here about how to detect cycle in a directed graph. The basic concept of this algorithm is if a node is found in recursive stack then there is a cycle, but i don't understand why. what is the logic here?
#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);
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.
}
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;
}
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;
}
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;
}

Answer 1

From a brief look, the code snippet is an implementation of depth-first search, which is a basic search technique for directed graphs; the same approach works for breadth-first search. Note that apparently this implementation works only if there is only one connected component, otherwise the test must be performed for each connected component until a cycle is found.
That being said, the technique works by choosing one node at will and starting a recursive search there. Basically, if the search discovers a node that is in the stack, there must be a cycle, since it has been previously reached.
In the current implementation, recStack is not actually the stack, it just indicates whether a specific node is currently in the stack, no sequence information is stored. The actual cycle is contained implicitly in the call stack. The cycle is the sequence of nodes for which the calls of isCyclicUtil has not yet returned. If the actual cycle has to be extracted, the implementation must be changed.

Answer 2

So essentailly, what this is saying, is if a node leads to itself, there is a cycle. This makes sense if you think about it!
Say we start at node1.
{node1 -> node2}
{node2 -> node3}
{node3 -> node4
node3 -> node1}
{node4 -> end}
{node1 -> node2}
{node2 -> node3}.....
This is a small graph that contains a cycle. As you can see, we traverse the graph, going from each node to the next. In some cases we reach and end, but even if we reach the end, our code wants to go back to the other branch off of node3 so that it can check it's next node. This node then leads back to node1.
This will happen forever if we let it, because the path starting at node1 leads back to itself. We are recursively putting each node we visit on the stack, and if we reach an end, we remove all of the nodes from the stack AFTER the branch. In our case, we would be removing node4 from the stack every time we hit the end, but the rest of the nodes would stay on the stack because of the branch off of node3.
Hope this helps!