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.
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.
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.
1) Consider the given source and destination as source and sink in flow network. Assign unit capacity to each edge.
2) Run Ford-Fulkerson algorithm to find the maximum flow from source to sink.
3) 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++
// 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 tne 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 tere 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;
} |
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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 tne 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 |
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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 tne 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 |
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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 tne 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 |
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Output:
There can be maximum 2 edge-disjoint paths from 0 to 7
Time Complexity: Same as time complexity of Edmonds-Karp implementation of Ford-Fulkerson (See time complexity discussed here)
References:
http://www.win.tue.nl/~nikhil/courses/2012/2WO08/max-flow-applications-4up.pdf
Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above
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