Count all possible Paths between two Vertices

Last Updated : 28 Sep, 2023

Count the total number of ways or paths that exist between two vertices in a directed graph. These paths don’t contain a cycle, the simple enough reason is that a cycle contains an infinite number of paths and hence they create a problem

Examples:

For the following Graph:

Input: Count paths between A and E
Output: Total paths between A and E are 4
Explanation: The 4 paths between A and E are:

A -> E
A -> B -> E
A -> C -> E
A -> B -> D -> C -> E

Input: Count paths between A and C
Output: Total paths between A and C are 2
Explanation: The 2 paths between A and C are:

A -> C
A -> B -> D -> C

Count paths between two vertices using Backtracking:

To solve the problem follow the below idea:

The problem can be solved using backtracking, which says to take a path and start walking on it and check if it leads us to the destination vertex then count the path and backtrack to take another path. If the path doesn’t lead to the destination vertex, discard the path. This type of graph traversal is called Backtracking.

Backtracking for the above graph can be shown like this:

Note: The red color vertex is the source vertex and the light-blue color vertex is destination, rest are either intermediate or discarded paths.

This give four paths between source(A) and destination(E) vertex

Why this solution will not work for a graph which contains cycles?

The Problem Associated with this is that now if one more edge is added between C and B, it would make a cycle (B -> D -> C -> B). And hence after every cycle through the loop, the length path will increase and that will be considered a different path, and there would be infinitely many paths because of the cycle

Follow the given steps to solve the problem:

Create a recursive function that takes the index of a node of a graph and the destination index. Keep a global or a static variable count to store the count.
Keep a record of the nodes visited using a visited array and while returning mark the current node to be unvisited to discover other paths.
- If the current node is the destination then increase the count.
- Else for all the adjacent nodes, i.e. nodes that are accessible from the current node, call the recursive function with the index of the adjacent node and the destination.
Print the Count as the required answer.

Below is the implementation of the above approach.

C++

/*
 * C++ program to count all paths from a source to a
 * destination.
 * https://www.geeksforgeeks.org/count-possible-paths-two-vertices/
 * Note that the original example has been refactored.
 */
#include <bits/stdc++.h>
using namespace std;
 
/*
 * A directed graph using adjacency list representation;
 * every vertex holds a list of all neighbouring vertices
 * that can be reached from it.
 */
class Graph {
public:
    // Construct the graph given the number of vertices...
    Graph(int vertices);
    // Specify an edge between two vertices
    void add_edge(int src, int dst);
    // Call the recursive helper function to count all the
    // paths
    int count_paths(int src, int dst, int vertices);
 
private:
    int m_vertices;
    list<int>* m_neighbours;
    void path_counter(int src, int dst, int& path_count,
                      vector<bool>& visited);
};
 
Graph::Graph(int vertices)
{
    m_vertices = vertices; // unused!!
    /* An array of linked lists - each element corresponds
    to a vertex and will hold a list of neighbours...*/
    m_neighbours = new list<int>[vertices];
}
 
void Graph::add_edge(int src, int dst)
{
    m_neighbours[src].push_back(dst);
}
 
int Graph::count_paths(int src, int dst, int vertices)
{
    int path_count = 0;
    vector<bool> visited(vertices, false);
    path_counter(src, dst, path_count, visited);
    return path_count;
}
 
/*
 * A recursive function that counts all paths from src to
 * dst. Keep track of the count in the parameter.
 */
void Graph::path_counter(int src, int dst, int& path_count,
                         vector<bool>& visited)
{
    // If we've reached the destination, then increment
    // count...
    visited[src] = true;
    if (src == dst) {
        path_count++;
    }
    // ...otherwise recurse into all neighbours...
    else {
        for (auto neighbour : m_neighbours[src]) {
            if (!visited[neighbour])
                path_counter(neighbour, dst, path_count,
                             visited);
        }
    }
    visited[src] = false;
}
 
// Driver code
int main()
{
    // Create a graph given in the above diagram - see link
    Graph g(5);
    g.add_edge(0, 1);
    g.add_edge(0, 2);
    g.add_edge(0, 4);
    g.add_edge(1, 3);
    g.add_edge(1, 4);
    g.add_edge(2, 3);
    g.add_edge(2, 1);
    g.add_edge(3, 2);
     
      // Function call
    cout << g.count_paths(0, 4, 5);
 
    return 0;
}

Java

// Java program to count all paths from a source
// to a destination
import java.util.Arrays;
import java.util.Iterator;
import java.util.LinkedList;
 
// This class represents a directed graph using
// adjacency list representation
 
class Graph {
 
    // No. of vertices
    private int V;
 
    // Array of lists for
    // Adjacency List
    // Representation
    private LinkedList<Integer> adj[];
 
    @SuppressWarnings("unchecked") Graph(int v)
    {
        V = v;
        adj = new LinkedList[v];
        for (int i = 0; i < v; ++i)
            adj[i] = new LinkedList<>();
    }
 
    // Method to add an edge into the graph
    void addEdge(int v, int w)
    {
 
        // Add w to v's list.
        adj[v].add(w);
    }
 
    // A recursive method to count
    // all paths from 'u' to 'd'.
    int countPathsUtil(int u, int d, int pathCount)
    {
 
        // If current vertex is same as
        // destination, then increment count
        if (u == d) {
            pathCount++;
        }
 
        // Recur for all the vertices
        // adjacent to this vertex
        else {
            Iterator<Integer> i = adj[u].listIterator();
            while (i.hasNext()) {
                int n = i.next();
                pathCount = countPathsUtil(n, d, pathCount);
            }
        }
        return pathCount;
    }
 
    // Returns count of
    // paths from 's' to 'd'
    int countPaths(int s, int d)
    {
 
        // Call the recursive method
        // to count all paths
        int pathCount = 0;
        pathCount = countPathsUtil(s, d, pathCount);
        return pathCount;
    }
 
    // Driver Code
    public static void main(String args[])
    {
        Graph g = new Graph(5);
        g.addEdge(0, 1);
        g.addEdge(0, 2);
        g.addEdge(0, 3);
        g.addEdge(1, 3);
        g.addEdge(2, 3);
        g.addEdge(1, 4);
        g.addEdge(2, 4);
 
        int s = 0, d = 3;
       
          // Function call
        System.out.println(g.countPaths(s, d));
    }
}
 
// This code is contributed by shubhamjd.

Python3

# Python 3 program to count all paths
# from a source to a destination.
 
# A directed graph using adjacency
# list representation
 
 
class Graph:
 
    def __init__(self, V):
        self.V = V
        self.adj = [[] for i in range(V)]
 
    def addEdge(self, u, v):
 
        # Add v to u’s list.
        self.adj[u].append(v)
 
    # Returns count of paths from 's' to 'd'
    def countPaths(self, s, d):
 
        # Mark all the vertices
        # as not visited
        visited = [False] * self.V
 
        # Call the recursive helper
        # function to print all paths
        pathCount = [0]
        self.countPathsUtil(s, d, visited, pathCount)
        return pathCount[0]
 
    # A recursive function to print all paths
    # from 'u' to 'd'. visited[] keeps track
    # of vertices in current path. path[]
    # stores actual vertices and path_index
    # is current index in path[]
    def countPathsUtil(self, u, d,
                       visited, pathCount):
        visited[u] = True
 
        # If current vertex is same as
        # destination, then increment count
        if (u == d):
            pathCount[0] += 1
 
        # If current vertex is not destination
        else:
 
            # Recur for all the vertices
            # adjacent to current vertex
            i = 0
            while i < len(self.adj[u]):
                if (not visited[self.adj[u][i]]):
                    self.countPathsUtil(self.adj[u][i], d,
                                        visited, pathCount)
                i += 1
 
        visited[u] = False
 
 
# Driver Code
if __name__ == '__main__':
 
    # Create a graph given in the
    # above diagram
    g = Graph(4)
    g.addEdge(0, 1)
    g.addEdge(0, 2)
    g.addEdge(0, 3)
    g.addEdge(2, 0)
    g.addEdge(2, 1)
    g.addEdge(1, 3)
 
    s = 2
    d = 3
     
    # Function call
    print(g.countPaths(s, d))
 
# This code is contributed by PranchalK

C#

// C# program to count all paths from a source
// to a destination.
using System;
using System.Collections.Generic;
 
// This class represents a directed graph using
// adjacency list representation
public class Graph {
 
    // Array of lists for
    // Adjacency List
    // Representation
    private List<int>[] adj;
 
    Graph(int v)
    {
        adj = new List<int>[ v ];
        for (int i = 0; i < v; ++i)
            adj[i] = new List<int>();
    }
 
    // Method to add an edge into the graph
    void addEdge(int v, int w)
    {
 
        // Add w to v's list.
        adj[v].Add(w);
    }
 
    // A recursive method to count
    // all paths from 'u' to 'd'.
    int countPathsUtil(int u, int d, int pathCount)
    {
 
        // If current vertex is same as
        // destination, then increment count
        if (u == d) {
            pathCount++;
        }
 
        // Recur for all the vertices
        // adjacent to this vertex
        else {
            foreach(int i in adj[u])
            {
                int n = i;
                pathCount = countPathsUtil(n, d, pathCount);
            }
        }
        return pathCount;
    }
 
    // Returns count of
    // paths from 's' to 'd'
    int countPaths(int s, int d)
    {
 
        // Call the recursive method
        // to count all paths
        int pathCount = 0;
        pathCount = countPathsUtil(s, d, pathCount);
        return pathCount;
    }
 
    // Driver Code
    public static void Main(String[] args)
    {
        Graph g = new Graph(5);
        g.addEdge(0, 1);
        g.addEdge(0, 2);
        g.addEdge(0, 3);
        g.addEdge(1, 3);
        g.addEdge(2, 3);
        g.addEdge(1, 4);
        g.addEdge(2, 4);
 
        int s = 0, d = 3;
       
          // Function call
        Console.WriteLine(g.countPaths(s, d));
    }
}
 
// This code is contributed by Rajput-Ji

Javascript

<script>
 
// JavaScript program to count all paths from a source
// to a destination.
 
// No. of vertices
let  V;
 
// Array of lists for
    // Adjacency List
    // Representation
let adj;
 
function Graph(v)
{
    V = v;
        adj = new Array(v);
        for (let i = 0; i < v; ++i)
            adj[i] = [];
}
 
// Method to add an edge into the graph
function addEdge(v,w)
{
    // Add w to v's list.
        adj[v].push(w);
}
 
// A recursive method to count
    // all paths from 'u' to 'd'.
function countPathsUtil(u,d,pathCount)
{
    // If current vertex is same as
        // destination, then increment count
        if (u == d) {
            pathCount++;
        }
  
        // Recur for all the vertices
        // adjacent to this vertex
        else {
             
            for(let i=0;i<adj[u].length;i++) {
                let n = adj[u][i];
                pathCount = countPathsUtil(n, d, pathCount);
            }
        }
        return pathCount;
}
 
// Returns count of
    // paths from 's' to 'd'
function countPaths(s,d)
{
    // Call the recursive method
        // to count all paths
        let pathCount = 0;
        pathCount = countPathsUtil(s, d,
                                   pathCount);
        return pathCount;
}
 
 // Driver Code
Graph(5);
addEdge(0, 1);
addEdge(0, 2);
addEdge(0, 3);
addEdge(1, 3);
addEdge(2, 3);
addEdge(1, 4);
addEdge(2, 4);
 
let s = 0, d = 3;
document.write(countPaths(s, d));
 
 
// This code is contributed by avanitrachhadiya2155
 
</script>

Output

Time Complexity: O(2^n), where n is the number of vertices in the graph. This is because in the worst case scenario, the program will have to recursively visit all possible paths from the source to the destination, which can be exponential in the number of vertices.
Auxiliary Space: O(N), Auxiliary stack space used by recursion calls

shubham_rana_77

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Count all possible Paths between two Vertices

Count paths between two vertices using Backtracking:

Why this solution will not work for a graph which contains cycles?

C++

Java

Python3

C#

Javascript

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