Find minimum s-t cut in a flow network

4

In a flow network, an s-t cut is a cut that requires the source ‘s’ and the sink ‘t’ to be in different subsets, and it consists of edges going from the source’s side to the sink’s side. The capacity of an s-t cut is defined by the sum of capacity of each edge in the cut-set. (Source: Wiki)
The problem discussed here is to find minimum capacity s-t cut of the given network. Expected output is all edges of the minimum cut.

For example, in the following flow network, example s-t cuts are {{0 ,1}, {0, 2}}, {{0, 2}, {1, 2}, {1, 3}}, etc. The minimum s-t cut is {{1, 3}, {4, 3}, {4 5}} which has capacity as 12+7+4 = 23.

minCut

We strongly recommend to read the below post first.
Ford-Fulkerson Algorithm for Maximum Flow Problem

Minimum Cut and Maximum Flow
Like Maximum Bipartite Matching, this is another problem which can solved using Ford-Fulkerson Algorithm. This is based on max-flow min-cut theorem.

The max-flow min-cut theorem states that in a flow network, the amount of maximum flow is equal to capacity of the minimum cut. See CLRS book for proof of this theorem.

From Ford-Fulkerson, we get capacity of minimum cut. How to print all edges that form the minimum cut? The idea is to use residual graph.

Following are steps to print all edges of minimum cut.

1) Run Ford-Fulkerson algorithm and consider the final residual graph.

2) Find the set of vertices that are reachable from source in the residual graph.

3) All edges which are from a reachable vertex to non-reachable vertex are minimum cut edges. Print all such edges.

Following is C++ implementation of the above approach.

C++

// C++ program for finding minimum cut using Ford-Fulkerson
#include <iostream>
#include <limits.h>
#include <string.h>
#include <queue>
using namespace std;

// Number of vertices in given graph
#define V 6

/* Returns true if there is a path from source 's' to sink 't' in
  residual graph. Also fills parent[] to store the path */
int 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);
}

// A DFS based function to find all reachable vertices from s.  The function
// marks visited[i] as true if i is reachable from s.  The initial values in
// visited[] must be false. We can also use BFS to find reachable vertices
void dfs(int rGraph[V][V], int s, bool visited[])
{
    visited[s] = true;
    for (int i = 0; i < V; i++)
       if (rGraph[s][i] && !visited[i])
           dfs(rGraph, i, visited);
}

// Prints the minimum s-t cut
void minCut(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]; // rGraph[i][j] indicates residual capacity of edge i-j
    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

    // Augment the flow while tere is path from source to sink
    while (bfs(rGraph, s, t, parent))
    {
        // Find minimum residual capacity of the edhes 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;
        }
    }

    // Flow is maximum now, find vertices reachable from s
    bool visited[V];
    memset(visited, false, sizeof(visited));
    dfs(rGraph, s, visited);

    // Print all edges that are from a reachable vertex to
    // non-reachable vertex in the original graph
    for (int i = 0; i < V; i++)
      for (int j = 0; j < V; j++)
         if (visited[i] && !visited[j] && graph[i][j])
              cout << i << " - " << j << endl;

    return;
}

// Driver program to test above functions
int main()
{
    // Let us create a graph shown in the above example
    int graph[V][V] = { {0, 16, 13, 0, 0, 0},
                        {0, 0, 10, 12, 0, 0},
                        {0, 4, 0, 0, 14, 0},
                        {0, 0, 9, 0, 0, 20},
                        {0, 0, 0, 7, 0, 4},
                        {0, 0, 0, 0, 0, 0}
                      };

    minCut(graph, 0, 5);

    return 0;
}

Python

# Python program for finding min-cut in the given graph
# 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.org_graph = [i[:] for i in graph]
        self. ROW = len(graph)
        self.COL = len(graph[0])


    '''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 min-cut of the given graph
    def minCut(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]

        # print the edges which initially had weights
        # but now have 0 weight
        for i in range(self.ROW):
            for j in range(self.COL):
                if self.graph[i][j] == 0 and self.org_graph[i][j] > 0:
                    print str(i) + " - " + str(j)


# Create a graph given in the above diagram
graph = [[0, 16, 13, 0, 0, 0],
        [0, 0, 10, 12, 0, 0],
        [0, 4, 0, 0, 14, 0],
        [0, 0, 9, 0, 0, 20],
        [0, 0, 0, 7, 0, 4],
        [0, 0, 0, 0, 0, 0]]

g = Graph(graph)

source = 0; sink = 5

g.minCut(source, sink)

# This code is contributed by Neelam Yadav


Output:
1 - 3
4 - 3
4 - 5

References:
http://www.stanford.edu/class/cs97si/08-network-flow-problems.pdf
http://www.cs.princeton.edu/courses/archive/spring06/cos226/lectures/maxflow.pdf

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