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.

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; }

## Java

// Java program for finding min-cut in the given graph import java.util.LinkedList; import java.util.Queue; public class Graph { // Returns true if there is a path // from source 's' to sink 't' in residual // graph. Also fills parent[] to store the path private 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[rGraph.length]; // Create a queue, enqueue source vertex // and mark source vertex as visited Queue<Integer> q = new LinkedList<Integer>(); q.add(s); visited[s] = true; parent[s] = -1; // Standard BFS Loop while (!q.isEmpty()) { int v = q.poll(); for (int i = 0; i < rGraph.length; i++) { if (rGraph[v][i] > 0 && !visited[i]) { q.offer(i); visited[i] = true; parent[i] = v; } } } // 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 private static void dfs(int[][] rGraph, int s, boolean[] visited) { visited[s] = true; for (int i = 0; i < rGraph.length; i++) { if (rGraph[s][i] > 0 && !visited[i]) { dfs(rGraph, i, visited); } } } // Prints the minimum s-t cut private static void minCut(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 // rGraph[i][j] indicates residual capacity of edge i-j int[][] rGraph = new int[graph.length][graph.length]; for (int i = 0; i < graph.length; i++) { for (int j = 0; j < graph.length; j++) { rGraph[i][j] = graph[i][j]; } } // This array is filled by BFS and to store path int[] parent = new int[graph.length]; // 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 pathFlow = Integer.MAX_VALUE; for (v = t; v != s; v = parent[v]) { u = parent[v]; pathFlow = Math.min(pathFlow, 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] = rGraph[u][v] - pathFlow; rGraph[v][u] = rGraph[v][u] + pathFlow; } } // Flow is maximum now, find vertices reachable from s boolean[] isVisited = new boolean[graph.length]; dfs(rGraph, s, isVisited); // Print all edges that are from a reachable vertex to // non-reachable vertex in the original graph for (int i = 0; i < graph.length; i++) { for (int j = 0; j < graph.length; j++) { if (graph[i][j] > 0 && isVisited[i] && !isVisited[j]) { System.out.println(i + " - " + j); } } } } //Driver Program public static void main(String args[]) { // Let us create a graph shown in the above example int 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} }; minCut(graph, 0, 5); } } // This code is contributed by Himanshu Shekhar

## 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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