Ford-Fulkerson Algorithm for Maximum Flow Problem

Given a graph which represents a flow network where every edge has a capacity. Also given two vertices source ‘s’ and sink ‘t’ in the graph, find the maximum possible flow from s to t with following constraints:

a) Flow on an edge doesn’t exceed the given capacity of the edge.

b) Incoming flow is equal to outgoing flow for every vertex except s and t.

For example, consider the following graph from CLRS book.

The maximum possible flow in the above graph is 23.

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

Prerequisite : Max Flow Problem Introduction

Ford-Fulkerson Algorithm 
The following is simple idea of Ford-Fulkerson algorithm:
1) Start with initial flow as 0.
2) While there is a augmenting path from source to sink. 
           Add this path-flow to flow.
3) Return flow.

Time Complexity: Time complexity of the above algorithm is O(max_flow * E). We run a loop while there is an augmenting path. In worst case, we may add 1 unit flow in every iteration. Therefore the time complexity becomes O(max_flow * E).

How to implement the above simple algorithm?
Let us first define the concept of Residual Graph which is needed for understanding the implementation.
Residual Graph of a flow network is a graph which indicates additional possible flow. If there is a path from source to sink in residual graph, then it is possible to add flow. Every edge of a residual graph has a value called residual capacity which is equal to original capacity of the edge minus current flow. Residual capacity is basically the current capacity of the edge.
Let us now talk about implementation details. Residual capacity is 0 if there is no edge between two vertices of residual graph. We can initialize the residual graph as original graph as there is no initial flow and initially residual capacity is equal to original capacity. To find an augmenting path, we can either do a BFS or DFS of the residual graph. We have used BFS in below implementation. Using BFS, we can find out if there is a path from source to sink. BFS also builds parent[] array. Using the parent[] array, we traverse through the found path and find possible flow through this path by finding minimum residual capacity along the path. We later add the found path flow to overall flow.
The important thing is, we need to update residual capacities in the residual graph. We subtract path flow from all edges along the path and we add path flow along the reverse edges We need to add path flow along reverse edges because may later need to send flow in reverse direction (See following link for example).

https://www.geeksforgeeks.org/max-flow-problem-introduction/

Below is the implementation of Ford-Fulkerson algorithm. To keep things simple, graph is represented as a 2D matrix.

C++

// C++ program for implementation of Ford Fulkerson algorithm 
#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 */
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 flow from s to t in the given graph 
int fordFulkerson(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 
    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, 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} 
                      }; 
  
    cout << "The maximum possible flow is " << fordFulkerson(graph, 0, 5); 
  
    return 0; 
} 

Java

// Java program for implementation of Ford Fulkerson algorithm 
import java.util.*; 
import java.lang.*; 
import java.io.*; 
import java.util.LinkedList; 
  
class MaxFlow 
{ 
    static final int V = 6;    //Number of vertices in graph 
  
    /* Returns true if there is a path from source 's' to sink 
      't' in residual graph. Also fills parent[] to store the 
      path */
    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]; 
        for(int i=0; i<V; ++i) 
            visited[i]=false; 
  
        // Create a queue, enqueue source vertex and mark 
        // source vertex as visited 
        LinkedList<Integer> queue = new LinkedList<Integer>(); 
        queue.add(s); 
        visited[s] = true; 
        parent[s]=-1; 
  
        // Standard BFS Loop 
        while (queue.size()!=0) 
        { 
            int u = queue.poll(); 
  
            for (int v=0; v<V; v++) 
            { 
                if (visited[v]==false && rGraph[u][v] > 0) 
                { 
                    queue.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 flow from s to t in the given graph 
    int fordFulkerson(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 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 = 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 
        return max_flow; 
    } 
  
    // Driver program to test above functions 
    public static void main (String[] args) throws java.lang.Exception 
    { 
        // Let us create a graph shown in the above example 
        int graph[][] =new int[][] { {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} 
                                   }; 
        MaxFlow m = new MaxFlow(); 
  
        System.out.println("The maximum possible flow is " + 
                           m.fordFulkerson(graph, 0, 5)); 
  
    } 
} 

Python

# Python program for implementation of Ford Fulkerson algorithm 
   
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) 
        #self.COL = len(gr[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 maximum flow from s to t in the given graph 
    def FordFulkerson(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, 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
   
print ("The maximum possible flow is %d " % g.FordFulkerson(source, sink)) 
  
#This code is contributed by Neelam Yadav 

C#

// C# program for implementation  
// of Ford Fulkerson algorithm 
using System; 
using System.Collections.Generic; 
  
public class MaxFlow 
{ 
    static readonly int V = 6; //Number of vertices in graph 
  
    /* 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, int s, int t, int []parent) 
    { 
        // Create a visited array and mark  
        // all vertices as not visited 
        bool []visited = new bool[V]; 
        for(int i = 0; i < V; ++i) 
            visited[i] = false; 
  
        // Create a queue, enqueue source vertex and mark 
        // source vertex as visited 
        List<int> queue = new List<int>(); 
        queue.Add(s); 
        visited[s] = true; 
        parent[s] = -1; 
  
        // Standard BFS Loop 
        while (queue.Count != 0) 
        { 
            int u = queue[0]; 
                queue.RemoveAt(0); 
  
            for (int v = 0; v < V; v++) 
            { 
                if (visited[v] == false && rGraph[u, v] > 0) 
                { 
                    queue.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 flow 
    // from s to t in the given graph 
    int fordFulkerson(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 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.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 
        return max_flow; 
    } 
  
    // Driver code 
    public static void Main () 
    { 
        // Let us create a graph shown in the above example 
        int [,]graph =new int[,] { {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} 
                                }; 
        MaxFlow m = new MaxFlow(); 
  
        Console.WriteLine("The maximum possible flow is " + 
                        m.fordFulkerson(graph, 0, 5)); 
  
    } 
} 
  
/* This code contributed by PrinciRaj1992 */

Output:

The maximum possible flow is 23

The above implementation of Ford Fulkerson Algorithm is called Edmonds-Karp Algorithm. The idea of Edmonds-Karp is to use BFS in Ford Fulkerson implementation as BFS always picks a path with minimum number of edges. When BFS is used, the worst case time complexity can be reduced to O(VE²). The above implementation uses adjacency matrix representation though where BFS takes O(V²) time, the time complexity of the above implementation is O(EV³) (Refer CLRS book for proof of time complexity)

This is an important problem as it arises in many practical situations. Examples include, maximizing the transportation with given traffic limits, maximizing packet flow in computer networks.

Dinc’s Algorithm for Max-Flow.

Exercise:
Modify the above implementation so that it that runs in O(VE²) time.

References:
http://www.stanford.edu/class/cs97si/08-network-flow-problems.pdf
Introduction to Algorithms 3rd Edition by Clifford Stein, Thomas H. Cormen, Charles E. Leiserson, Ronald L. Rivest

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.

My Personal Notes

Recommended: Please solve it on “PRACTICE” first, before moving on to the solution.

C++

Java

Python

C#

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