Find if there is a path between two vertices in a directed graph

2.1

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 are C++,Java and Python codes that use BFS for finding reachability of second vertex from first vertex.

C++

// C++ 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);  
};

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);
            }
        }
    }
    
    // If BFS is complete without visiting d
    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;
}

Java

// Java program to check if there is exist a path between two vertices
// of a graph.
import java.io.*;
import java.util.*;
import java.util.LinkedList;

// This class represents a directed graph using adjacency list
// representation
class Graph
{
    private int V;   // No. of vertices
    private LinkedList<Integer> adj[]; //Adjacency List

    //Constructor
    Graph(int v)
    {
        V = v;
        adj = new LinkedList[v];
        for (int i=0; i<v; ++i)
            adj[i] = new LinkedList();
    }

    //Function to add an edge into the graph
    void addEdge(int v,int w)  {   adj[v].add(w);   }

    //prints BFS traversal from a given source s
    Boolean isReachable(int s, int d)
    {
        LinkedList<Integer>temp;

        // Mark all the vertices as not visited(By default set
        // as false)
        boolean visited[] = new boolean[V];

        // Create a queue for BFS
        LinkedList<Integer> queue = new LinkedList<Integer>();

        // Mark the current node as visited and enqueue it
        visited[s]=true;
        queue.add(s);

        // 'i' will be used to get all adjacent vertices of a vertex
        Iterator<Integer> i;
        while (queue.size()!=0)
        {
            // Dequeue a vertex from queue and print it
            s = queue.poll();

            int n;
            i = adj[s].listIterator();

            // Get all adjacent vertices of the dequeued vertex s
            // If a adjacent has not been visited, then mark it
            // visited and enqueue it
            while (i.hasNext())
            {
                n = i.next();

                // If this adjacent node is the destination node,
                // then return true
                if (n==d)
                    return true;

                // Else, continue to do BFS
                if (!visited[n])
                {
                    visited[n] = true;
                    queue.add(n);
                }
            }
        }

        // If BFS is complete without visited d
        return false;
    }

    // Driver method
    public static void main(String args[])
    {
        // Create a graph given in the above diagram
        Graph g = new Graph(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;
        int v = 3;
        if (g.isReachable(u, v))
            System.out.println("There is a path from " + u +" to " + v);
        else
            System.out.println("There is no path from " + u +" to " + v);;

        u = 3;
        v = 1;
        if (g.isReachable(u, v))
            System.out.println("There is a path from " + u +" to " + v);
        else
            System.out.println("There is no path from " + u +" to " + v);;
    }
}
// This code is contributed by Aakash Hasija

Python

# program to check if there is exist a path between two vertices
# of a graph

from collections import defaultdict
 
#This class represents a directed graph using adjacency list representation
class Graph:
 
	def __init__(self,vertices):
		self.V= vertices #No. of vertices
		self.graph = defaultdict(list) # default dictionary to store graph
 
	# function to add an edge to graph
	def addEdge(self,u,v):
		self.graph[u].append(v)
 	
 	# Use BFS to check path between s and d
	def isReachable(self, s, d):
		# Mark all the vertices as not visited
		visited =[False]*(self.V)
 
		# Create a queue for BFS
		queue=[]
 
		# Mark the source node as visited and enqueue it
		queue.append(s)
		visited[s] = True
 
		while queue:

			#Dequeue a vertex from queue 
			n = queue.pop(0)
			
			# If this adjacent node is the destination node,
            # then return true
 			if n == d:
 				return True

			#  Else, continue to do BFS
			for i in self.graph[n]:
				if visited[i] == False:
					queue.append(i)
					visited[i] = True
 		# If BFS is complete without visited d
 		return False
 
# Create a graph given in the above diagram
g = Graph(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)

u =1; v = 3

if g.isReachable(u, v):
    print("There is a path from %d to %d" % (u,v))
else :
    print("There is no path from %d to %d" % (u,v))

u = 3; v = 1
if g.isReachable(u, v) :
    print("There is a path from %d to %d" % (u,v))
else :
    print("There is no path from %d to %d" % (u,v))

#This code is contributed by Neelam Yadav


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.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

GATE CS Corner    Company Wise Coding Practice

Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.

Recommended Posts:



2.1 Average Difficulty : 2.1/5.0
Based on 51 vote(s)










Writing code in comment? Please use ide.geeksforgeeks.org, generate link and share the link here.