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
Approach:
The problem can be solved using backtracking, that says 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 above graph can be shown like this:
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 as a different path, and there would be infinitely many paths because of the cycle.
Algorithm:
- Create a recursive function that takes index of node of a graph and the destination index. Keep a global or a static variable count to store the count.
- If the current nodes is the destination 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 adjacent node and the destination.
- Print the Count.
Implementation:
C++
// C++ program to count all paths from a // source to a destination. #include <bits/stdc++.h> using namespace std; // A directed graph using adjacency list // representation class Graph { // No. of vertices in graph int V; list<int>* adj; // A recursive function // used by countPaths() void countPathsUtil(int, int, int&); public: // Constructor Graph(int V); void addEdge(int u, int v); int countPaths(int s, int d); }; Graph::Graph(int V) { this->V = V; adj = new list<int>[V]; } void Graph::addEdge(int u, int v) { // Add v to u’s list. adj[u].push_back(v); } // Returns count of paths from 's' to 'd' int Graph::countPaths(int s, int d) { // Call the recursive helper // function to print all paths int pathCount = 0; countPathsUtil(s, d, pathCount); return pathCount; } // 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[] void Graph::countPathsUtil(int u, int d, int& pathCount) { // If current vertex is same as destination, // then increment count if (u == d) pathCount++; // If current vertex is not destination else { // Recur for all the vertices adjacent to // current vertex list<int>::iterator i; for (i = adj[u].begin(); i != adj[u].end(); ++i) countPathsUtil(*i, d, pathCount); } } // Driver Code int main() { // Create a graph given in the above diagram Graph g(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; cout << g.countPaths(s, d); return 0; } |
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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; System.out.println(g.countPaths(s, d)); } } // This code is contributed by shubhamjd. |
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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 print(g.countPaths(s, d)) # This code is contributed by PranchalK |
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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 { // No. of vertices int V; // Array of lists for // Adjacency List // Representation private List<int>[] adj; Graph(int v) { V = 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; Console.WriteLine(g.countPaths(s, d)); } } // This code is contributed by Rajput-Ji |
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Output:
3
Complexity Analysis:
- Time Complexity: O(N!).
If the graph is complete then there can be around N! recursive calls, so the time Complexity is O(N!) - Space Complexity: O(1).
Since no extra space is required.
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