Find maximum number of edge disjoint paths between two vertices

3.8

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.
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 C++ 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;
}

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


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

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