Find maximum number of edge disjoint paths between two vertices

Last Updated : 31 Jan, 2023

Given a directed graph and two vertices in it, source ‘s’ and destination ‘t’, find out the maximum number of edge disjoint paths from s to t. Two paths are said edge disjoint if they don’t share any edge.

edgedisjoint1

There can be maximum two edge disjoint paths from source 0 to destination 7 in the above graph. Two edge disjoint paths are highlighted below in red and blue colors are 0-2-6-7 and 0-3-6-5-7.

edgedisjoint2

Note that the paths may be different, but the maximum number is same. For example, in the above diagram, another possible set of paths is 0-1-2-6-7 and 0-3-6-5-7 respectively.

This problem can be solved by reducing it to maximum flow problem. Following are steps.

Consider the given source and destination as source and sink in flow network. Assign unit capacity to each edge.

Run Ford-Fulkerson algorithm to find the maximum flow from source to sink.

The maximum flow is equal to the maximum number of edge-disjoint paths.

When we run Ford-Fulkerson, we reduce the capacity by a unit. Therefore, the edge can not be used again. So the maximum flow is equal to the maximum number of edge-disjoint paths.

Following is the implementation of the above algorithm. Most of the code is taken from here.

C++

// C++ program to find maximum number of edge disjoint paths 
#include <iostream> 
#include <limits.h> 
#include <string.h> 
#include <queue> 
using namespace std; 
 
// Number of vertices in given graph 
#define V 8 
 
/* Returns true if there is a path from source 's' to sink 't' in 
residual graph. Also fills parent[] to store the path */
bool bfs(int rGraph[V][V], int s, int t, int parent[]) 
{ 
    // Create a visited array and mark all vertices as not visited 
    bool visited[V]; 
    memset(visited, 0, sizeof(visited)); 
 
    // Create a queue, enqueue source vertex and mark source vertex 
    // as visited 
    queue <int> q; 
    q.push(s); 
    visited[s] = true; 
    parent[s] = -1; 
 
    // Standard BFS Loop 
    while (!q.empty()) 
    { 
        int u = q.front(); 
        q.pop(); 
 
        for (int v=0; v<V; v++) 
        { 
            if (visited[v]==false && rGraph[u][v] > 0) 
            { 
                q.push(v); 
                parent[v] = u; 
                visited[v] = true; 
            } 
        } 
    } 
 
    // If we reached sink in BFS starting from source, then return 
    // true, else false 
    return (visited[t] == true); 
} 
 
// Returns the maximum number of edge-disjoint paths from s to t. 
// This function is copy of forFulkerson() discussed at http://goo.gl/wtQ4Ks 
int findDisjointPaths(int graph[V][V], int s, int t) 
{ 
    int u, v; 
 
    // Create a residual graph and fill the residual graph with 
    // given capacities in the original graph as residual capacities 
    // in residual graph 
    int rGraph[V][V]; // Residual graph where rGraph[i][j] indicates 
                    // residual capacity of edge from i to j (if there 
                    // is an edge. If rGraph[i][j] is 0, then there is not) 
    for (u = 0; u < V; u++) 
        for (v = 0; v < V; v++) 
            rGraph[u][v] = graph[u][v]; 
 
    int parent[V]; // This array is filled by BFS and to store path 
 
    int max_flow = 0; // There is no flow initially 
 
    // Augment the flow while there is path from source to sink 
    while (bfs(rGraph, s, t, parent)) 
    { 
        // Find minimum residual capacity of the edges along the 
        // path filled by BFS. Or we can say find the maximum flow 
        // through the path found. 
        int path_flow = INT_MAX; 
 
        for (v=t; v!=s; v=parent[v]) 
        { 
            u = parent[v]; 
            path_flow = min(path_flow, rGraph[u][v]); 
        } 
 
        // update residual capacities of the edges and reverse edges 
        // along the path 
        for (v=t; v != s; v=parent[v]) 
        { 
            u = parent[v]; 
            rGraph[u][v] -= path_flow; 
            rGraph[v][u] += path_flow; 
        } 
 
        // Add path flow to overall flow 
        max_flow += path_flow; 
    } 
 
    // Return the overall flow (max_flow is equal to maximum 
    // number of edge-disjoint paths) 
    return max_flow; 
} 
 
// Driver program to test above functions 
int main() 
{ 
    // Let us create a graph shown in the above example 
    int graph[V][V] = { {0, 1, 1, 1, 0, 0, 0, 0}, 
                        {0, 0, 1, 0, 0, 0, 0, 0}, 
                        {0, 0, 0, 1, 0, 0, 1, 0}, 
                        {0, 0, 0, 0, 0, 0, 1, 0}, 
                        {0, 0, 1, 0, 0, 0, 0, 1}, 
                        {0, 1, 0, 0, 0, 0, 0, 1}, 
                        {0, 0, 0, 0, 0, 1, 0, 1}, 
                        {0, 0, 0, 0, 0, 0, 0, 0} 
                    }; 
 
    int s = 0; 
    int t = 7; 
    cout << "There can be maximum " << findDisjointPaths(graph, s, t) 
        << " edge-disjoint paths from " << s <<" to "<< t ; 
 
    return 0; 
}

Java

// Java program to find maximum number 
// of edge disjoint paths
import java.util.*;
 
class GFG
{
 
// Number of vertices in given graph
static int V = 8;
 
/* Returns true if there is a path from 
source 's' to sink 't' in residual graph.
Also fills parent[] to store the path */
static boolean bfs(int rGraph[][], int s, 
                   int t, int parent[])
{
    // Create a visited array and
    // mark all vertices as not visited
    boolean []visited = new boolean[V];
 
 
    // Create a queue, enqueue source vertex and 
    // mark source vertex as visited
    Queue <Integer> q = new LinkedList<>();
    q.add(s);
    visited[s] = true;
    parent[s] = -1;
 
    // Standard BFS Loop
    while (!q.isEmpty())
    {
        int u = q.peek();
        q.remove();
 
        for (int v = 0; v < V; v++)
        {
            if (visited[v] == false && 
                rGraph[u][v] > 0)
            {
                q.add(v);
                parent[v] = u;
                visited[v] = true;
            }
        }
    }
 
    // If we reached sink in BFS 
    // starting from source, then 
    // return true, else false
    return (visited[t] == true);
}
 
// Returns the maximum number of edge-disjoint 
// paths from s to t. This function is copy of 
// forFulkerson() discussed at http://goo.gl/wtQ4Ks
static int findDisjointPaths(int graph[][], int s, int t)
{
    int u, v;
 
    // Create a residual graph and fill the 
    // residual graph with given capacities 
    // in the original graph as residual capacities
    // in residual graph
     
    // Residual graph where rGraph[i][j] indicates
    // residual capacity of edge from i to j (if there
    // is an edge. If rGraph[i][j] is 0, then there is not)
    int [][]rGraph = new int[V][V];
    for (u = 0; u < V; u++)
        for (v = 0; v < V; v++)
            rGraph[u][v] = graph[u][v];
     
    // This array is filled by BFS and to store path
    int []parent = new int[V]; 
 
    int max_flow = 0; // There is no flow initially
 
    // Augment the flow while there is path 
    // from source to sink
    while (bfs(rGraph, s, t, parent))
    {
        // Find minimum residual capacity of the edges 
        // along the path filled by BFS. Or we can say 
        // find the maximum flow through the path found.
        int path_flow = Integer.MAX_VALUE;
 
        for (v = t; v != s; v = parent[v])
        {
            u = parent[v];
            path_flow = Math.min(path_flow, rGraph[u][v]);
        }
 
        // update residual capacities of the edges and 
        // reverse edges along the path
        for (v = t; v != s; v = parent[v])
        {
            u = parent[v];
            rGraph[u][v] -= path_flow;
            rGraph[v][u] += path_flow;
        }
 
        // Add path flow to overall flow
        max_flow += path_flow;
    }
 
    // Return the overall flow (max_flow is equal to 
    // maximum number of edge-disjoint paths)
    return max_flow;
}
 
// Driver Code
public static void main(String[] args)
{
    // Let us create a graph shown in the above example
    int graph[][] = {{0, 1, 1, 1, 0, 0, 0, 0},
                     {0, 0, 1, 0, 0, 0, 0, 0},
                     {0, 0, 0, 1, 0, 0, 1, 0},
                     {0, 0, 0, 0, 0, 0, 1, 0},
                     {0, 0, 1, 0, 0, 0, 0, 1},
                     {0, 1, 0, 0, 0, 0, 0, 1},
                      {0, 0, 0, 0, 0, 1, 0, 1},
                     {0, 0, 0, 0, 0, 0, 0, 0}};
 
    int s = 0;
    int t = 7;
    System.out.println("There can be maximum " + 
                findDisjointPaths(graph, s, t) + 
                  " edge-disjoint paths from " + 
                                 s + " to "+ t);
}
} 
 
// This code is contributed by PrinciRaj1992

Python

# Python program to find maximum number of edge disjoint paths
# Complexity : (E*(V^3))
# Total augmenting path = VE 
# and BFS with adj matrix takes :V^2 times
  
from collections import defaultdict
  
#This class represents a directed graph using 
# adjacency matrix representation
class Graph:
  
    def __init__(self,graph):
        self.graph = graph # residual graph
        self. ROW = len(graph)
         
  
    '''Returns true if there is a path from source 's' to sink 't' in
    residual graph. Also fills parent[] to store the path '''
    def BFS(self,s, t, parent):
 
        # Mark all the vertices as not visited
        visited =[False]*(self.ROW)
         
        # Create a queue for BFS
        queue=[]
         
        # Mark the source node as visited and enqueue it
        queue.append(s)
        visited[s] = True
          
         # Standard BFS Loop
        while queue:
 
            #Dequeue a vertex from queue and print it
            u = queue.pop(0)
         
            # Get all adjacent vertices of the dequeued vertex u
            # If a adjacent has not been visited, then mark it
            # visited and enqueue it
            for ind, val in enumerate(self.graph[u]):
                if visited[ind] == False and val > 0 :
                    queue.append(ind)
                    visited[ind] = True
                    parent[ind] = u
 
        # If we reached sink in BFS starting from source, then return
        # true, else false
        return True if visited[t] else False
             
     
    # Returns the maximum number of edge-disjoint paths from 
    #s to t in the given graph
    def findDisjointPaths(self, source, sink):
 
        # This array is filled by BFS and to store path
        parent = [-1]*(self.ROW)
 
        max_flow = 0 # There is no flow initially
 
        # Augment the flow while there is path from source to sink
        while self.BFS(source, sink, parent) :
 
            # Find minimum residual capacity of the edges along the
            # path filled by BFS. Or we can say find the maximum flow
            # through the path found.
            path_flow = float("Inf")
            s = sink
            while(s !=  source):
                path_flow = min (path_flow, self.graph[parent[s]][s])
                s = parent[s]
 
            # Add path flow to overall flow
            max_flow +=  path_flow
 
            # update residual capacities of the edges and reverse edges
            # along the path
            v = sink
            while(v !=  source):
                u = parent[v]
                self.graph[u][v] -= path_flow
                self.graph[v][u] += path_flow
                v = parent[v]
 
        return max_flow
 
  
# Create a graph given in the above diagram
 
graph = [[0, 1, 1, 1, 0, 0, 0, 0],
        [0, 0, 1, 0, 0, 0, 0, 0],
        [0, 0, 0, 1, 0, 0, 1, 0],
        [0, 0, 0, 0, 0, 0, 1, 0],
        [0, 0, 1, 0, 0, 0, 0, 1],
        [0, 1, 0, 0, 0, 0, 0, 1],
        [0, 0, 0, 0, 0, 1, 0, 1],
        [0, 0, 0, 0, 0, 0, 0, 0]]
  
 
g = Graph(graph)
 
source = 0; sink = 7
  
print ("There can be maximum %d edge-disjoint paths from %d to %d" %
            (g.findDisjointPaths(source, sink), source, sink))
 
 
# This code is contributed by Neelam Yadav

C#

// C# program to find maximum number 
// of edge disjoint paths
using System;
using System.Collections.Generic; 
 
class GFG
{
 
// Number of vertices in given graph
static int V = 8;
 
/* Returns true if there is a path from 
source 's' to sink 't' in residual graph.
Also fills parent[] to store the path */
static bool bfs(int [,]rGraph, int s, 
                int t, int []parent)
{
    // Create a visited array and
    // mark all vertices as not visited
    bool []visited = new bool[V];
 
    // Create a queue, enqueue source vertex 
    // and mark source vertex as visited
    Queue <int> q = new Queue <int>();
    q.Enqueue(s);
    visited[s] = true;
    parent[s] = -1;
 
    // Standard BFS Loop
    while (q.Count != 0)
    {
        int u = q.Peek();
        q.Dequeue();
 
        for (int v = 0; v < V; v++)
        {
            if (visited[v] == false && 
                rGraph[u, v] > 0)
            {
                q.Enqueue(v);
                parent[v] = u;
                visited[v] = true;
            }
        }
    }
 
    // If we reached sink in BFS 
    // starting from source, then 
    // return true, else false
    return (visited[t] == true);
}
 
// Returns the maximum number of edge-disjoint 
// paths from s to t. This function is copy of 
// forFulkerson() discussed at http://goo.gl/wtQ4Ks
static int findDisjointPaths(int [,]graph, 
                             int s, int t)
{
    int u, v;
 
    // Create a residual graph and fill the 
    // residual graph with given capacities 
    // in the original graph as residual capacities
    // in residual graph
     
    // Residual graph where rGraph[i,j] indicates
    // residual capacity of edge from i to j (if there
    // is an edge. If rGraph[i,j] is 0, then there is not)
    int [,]rGraph = new int[V, V];
    for (u = 0; u < V; u++)
        for (v = 0; v < V; v++)
            rGraph[u, v] = graph[u, v];
     
    // This array is filled by BFS and 
    // to store path
    int []parent = new int[V]; 
 
    int max_flow = 0; // There is no flow initially
 
    // Augment the flow while there is path 
    // from source to sink
    while (bfs(rGraph, s, t, parent))
    {
        // Find minimum residual capacity of the edges 
        // along the path filled by BFS. Or we can say 
        // find the maximum flow through the path found.
        int path_flow = int.MaxValue;
 
        for (v = t; v != s; v = parent[v])
        {
            u = parent[v];
            path_flow = Math.Min(path_flow, 
                                 rGraph[u, v]);
        }
 
        // update residual capacities of the edges 
        // and reverse edges along the path
        for (v = t; v != s; v = parent[v])
        {
            u = parent[v];
            rGraph[u, v] -= path_flow;
            rGraph[v, u] += path_flow;
        }
 
        // Add path flow to overall flow
        max_flow += path_flow;
    }
 
    // Return the overall flow (max_flow is equal to 
    // maximum number of edge-disjoint paths)
    return max_flow;
}
 
// Driver Code
public static void Main(String[] args)
{
    // Let us create a graph shown 
    // in the above example
    int [,]graph = {{0, 1, 1, 1, 0, 0, 0, 0},
                    {0, 0, 1, 0, 0, 0, 0, 0},
                    {0, 0, 0, 1, 0, 0, 1, 0},
                    {0, 0, 0, 0, 0, 0, 1, 0},
                    {0, 0, 1, 0, 0, 0, 0, 1},
                    {0, 1, 0, 0, 0, 0, 0, 1},
                    {0, 0, 0, 0, 0, 1, 0, 1},
                    {0, 0, 0, 0, 0, 0, 0, 0}};
 
    int s = 0;
    int t = 7;
    Console.WriteLine("There can be maximum " + 
               findDisjointPaths(graph, s, t) + 
                 " edge-disjoint paths from " + 
                                s + " to "+ t);
}
} 
 
// This code is contributed by Rajput-Ji

Javascript

<script>
 
// Javascript program to find maximum number 
// of edge disjoint paths
 
// Number of vertices in given graph
let V = 8;
 
/* Returns true if there is a path from 
source 's' to sink 't' in residual graph.
Also fills parent[] to store the path */
function bfs(rGraph, s, t, parent)
{
     
    // Create a visited array and
    // mark all vertices as not visited
    let visited = new Array(V);
      for(let i = 0; i < V; i++)
        visited[i] = false;
   
    // Create a queue, enqueue source vertex and 
    // mark source vertex as visited
    let q = [];
    q.push(s);
    visited[s] = true;
    parent[s] = -1;
   
    // Standard BFS Loop
    while (q.length != 0)
    {
        let u = q[0];
        q.shift();
   
        for(let v = 0; v < V; v++)
        {
            if (visited[v] == false && 
                rGraph[u][v] > 0)
            {
                q.push(v);
                parent[v] = u;
                visited[v] = true;
            }
        }
    }
   
    // If we reached sink in BFS 
    // starting from source, then 
    // return true, else false
    return (visited[t] == true);
}
 
// Returns the maximum number of edge-disjoint 
// paths from s to t. This function is copy of 
// forFulkerson() discussed at http://goo.gl/wtQ4Ks
function findDisjointPaths(graph, s, t)
{
    let u, v;
   
    // Create a residual graph and fill the 
    // residual graph with given capacities 
    // in the original graph as residual capacities
    // in residual graph
       
    // Residual graph where rGraph[i][j] indicates
    // residual capacity of edge from i to j (if 
    // there is an edge. If rGraph[i][j] is 0, 
    // then there is not)
    let rGraph = new Array(V);
    for(u = 0; u < V; u++)
    {
        rGraph[u] = new Array(V);
        for(v = 0; v < V; v++)
            rGraph[u][v] = graph[u][v];
     } 
      
    // This array is filled by BFS and to store path
    let parent = new Array(V); 
     
    // There is no flow initially
    let max_flow = 0; 
   
    // Augment the flow while there is path 
    // from source to sink
    while (bfs(rGraph, s, t, parent))
    {
         
        // Find minimum residual capacity of the edges 
        // along the path filled by BFS. Or we can say 
        // find the maximum flow through the path found.
        let path_flow = Number.MAX_VALUE;
   
        for(v = t; v != s; v = parent[v])
        {
            u = parent[v];
            path_flow = Math.min(path_flow, 
                                 rGraph[u][v]);
        }
   
        // Update residual capacities of the edges and 
        // reverse edges along the path
        for(v = t; v != s; v = parent[v])
        {
            u = parent[v];
            rGraph[u][v] -= path_flow;
            rGraph[v][u] += path_flow;
        }
   
        // Add path flow to overall flow
        max_flow += path_flow;
    }
   
    // Return the overall flow (max_flow is equal
    // to maximum number of edge-disjoint paths)
    return max_flow;
}
 
// Driver Code
let graph = [ [ 0, 1, 1, 1, 0, 0, 0, 0 ],
              [ 0, 0, 1, 0, 0, 0, 0, 0 ],
              [ 0, 0, 0, 1, 0, 0, 1, 0 ],
              [ 0, 0, 0, 0, 0, 0, 1, 0 ],
              [ 0, 0, 1, 0, 0, 0, 0, 1 ],
              [ 0, 1, 0, 0, 0, 0, 0, 1 ],
              [ 0, 0, 0, 0, 0, 1, 0, 1 ],
              [ 0, 0, 0, 0, 0, 0, 0, 0 ] ];
let s = 0;
let t = 7;
 
document.write("There can be maximum " + 
               findDisjointPaths(graph, s, t) + 
               " edge-disjoint paths from " + 
               s + " to " + t + "<br>");
 
// This code is contributed by avanitrachhadiya2155
 
</script>

Output

There can be maximum 2 edge-disjoint paths from 0 to 7

Time Complexity : O(|V| * E²) ,where E is the number of edges and V is the number of vertices.

Space Complexity :O(V) ,as we created queue.

Time Complexity: Same as time complexity of Edmonds-Karp implementation of Ford-Fulkerson (See time complexity discussed here)

Suggest improvement

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