Given a directed graph, design an algorithm to find out whether there is a route between two nodes.

Given a Directed Graph and two vertices in it, check whether there is a path from the first given vertex to second. For example, in the following graph, there is a path from vertex 1 to 3. As another example, there is no path from 3 to 0.

We can either use Breadth First Search (BFS) or Depth First Search (DFS) to find path between two vertices. Take the first vertex as source in BFS (or DFS), follow the standard BFS (or DFS). If we see the second vertex in our traversal, then return true. Else return false.

Following is C++ code that uses BFS for finding reachability of second vertex from first vertex.

// Program to check if there is exist a path between two vertices of a graph. #include<iostream> #include <list>   using namespace std;   // This class represents a directed graph using adjacency list representation class Graph {     int V;    // No. of vertices     list<int> *adj;    // Pointer to an array containing adjacency lists public:     Graph(int V);  // Constructor     void addEdge(int v, int w); // function to add an edge to graph     bool isReachable(int s, int d);  // returns true if there is a path from s to d };   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. }   // A BFS based function to check whether d is reachable from s. bool Graph::isReachable(int s, int d) {     // Base case     if (s == d)       return true;       // Mark all the vertices as not visited     bool *visited = new bool[V];     for (int i = 0; i < V; i++)         visited[i] = false;       // Create a queue for BFS     list<int> queue;       // Mark the current node as visited and enqueue it     visited[s] = true;     queue.push_back(s);       // it will be used to get all adjacent vertices of a vertex     list<int>::iterator i;       while (!queue.empty())     {         // Dequeue a vertex from queue and print it         s = queue.front();         queue.pop_front();           // Get all adjacent vertices of the dequeued vertex s         // If a adjacent has not been visited, then mark it visited         // and enqueue it         for (i = adj[s].begin(); i != adj[s].end(); ++i)         {             // If this adjacent node is the destination node, then return true             if (*i == d)                 return true;               // Else, continue to do BFS             if (!visited[*i])             {                 visited[*i] = true;                 queue.push_back(*i);             }         }     }       return false; }   // Driver program to test methods of graph class 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);       int u = 1, v = 3;     if(g.isReachable(u, v))         cout<< "\n There is a path from " << u << " to " << v;     else         cout<< "\n There is no path from " << u << " to " << v;       u = 3, v = 1;     if(g.isReachable(u, v))         cout<< "\n There is a path from " << u << " to " << v;     else         cout<< "\n There is no path from " << u << " to " << v;       return 0; }

Output:

 There is a path from 1 to 3
 There is no path from 3 to 1

As an exercise, try an extended version of the problem where the complete path between two vertices is also needed.

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