Ford–Fulkerson algorithm

Algorithm

Let G ( V , E ) be a graph, and for each edge from u to v , let c ( u , v ) be the capacity and f ( u , v ) be the flow. We want to find the maximum flow from the source s to the sink t . After every step in the algorithm the following is maintained:

This means that the flow through the network is a legal flow after each round in the algorithm. We define the residual network G f ( V , E f ) to be the network with capacity c f ( u , v ) = c ( u , v ) − f ( u , v ) and no flow. Notice that it can happen that a flow from v to u is allowed in the residual network, though disallowed in the original network: if f ( u , v ) > 0 and c ( v , u ) = 0 then c f ( v , u ) = c ( v , u ) − f ( v , u ) = f ( u , v ) > 0 .

Algorithm Ford–Fulkerson

Inputs Given a Network G = ( V , E ) with flow capacity c , a source node s , and a sink node t Output Compute a flow f from s to t of maximum value

1. f ( u , v ) ← 0 for all edges ( u , v )
2. While there is a path p from s to t in G f , such that c f ( u , v ) > 0 for all edges ( u , v ) ∈ p :
   1. Find c f ( p ) = min { c f ( u , v ) : ( u , v ) ∈ p }
   2. For each edge ( u , v ) ∈ p
      1. f ( u , v ) ← f ( u , v ) + c f ( p ) (Send flow along the path)
      2. f ( v , u ) ← f ( v , u ) − c f ( p ) (The flow might be "returned" later)

The path in step 2 can be found with for example a breadth-first search or a depth-first search in G f ( V , E f ) . If you use the former, the algorithm is called Edmonds–Karp.

When no more paths in step 2 can be found, s will not be able to reach t in the residual network. If S is the set of nodes reachable by s in the residual network, then the total capacity in the original network of edges from S to the remainder of V is on the one hand equal to the total flow we found from s to t , and on the other hand serves as an upper bound for all such flows. This proves that the flow we found is maximal. See also Max-flow Min-cut theorem.

If the graph G ( V , E ) has multiple sources and sinks, we act as follows: Suppose that T = { t | t  is a sink } and S = { s | s  is a source } . Add a new source s ∗ with an edge ( s ∗ , s ) from s ∗ to every node s ∈ S , with capacity c ( s ∗ , s ) = d s ( d s = ∑ ( s , u ) ∈ E c ( s , u ) ) . And add a new sink t ∗ with an edge ( t , t ∗ ) from every node t ∈ T to t ∗ , with capacity c ( t , t ∗ ) = d t ( d t = ∑ ( v , t ) ∈ E c ( v , t ) ) . Then apply the Ford–Fulkerson algorithm.

Also, if a node u has capacity constraint d u , we replace this node with two nodes u i n , u o u t , and an edge ( u i n , u o u t ) , with capacity c ( u i n , u o u t ) = d u . Then apply the Ford–Fulkerson algorithm.

Complexity

By adding the flow augmenting path to the flow already established in the graph, the maximum flow will be reached when no more flow augmenting paths can be found in the graph. However, there is no certainty that this situation will ever be reached, so the best that can be guaranteed is that the answer will be correct if the algorithm terminates. In the case that the algorithm runs forever, the flow might not even converge towards the maximum flow. However, this situation only occurs with irrational flow values. When the capacities are integers, the runtime of Ford–Fulkerson is bounded by O ( E f ) (see big O notation), where E is the number of edges in the graph and f is the maximum flow in the graph. This is because each augmenting path can be found in O ( E ) time and increases the flow by an integer amount of at least 1 , with the upper bound f .

A variation of the Ford–Fulkerson algorithm with guaranteed termination and a runtime independent of the maximum flow value is the Edmonds–Karp algorithm, which runs in O ( V E 2 ) time.

Integral example

The following example shows the first steps of Ford–Fulkerson in a flow network with 4 nodes, source A and sink D . This example shows the worst-case behaviour of the algorithm. In each step, only a flow of 1 is sent across the network. If breadth-first-search were used instead, only two steps would be needed.

Notice how flow is "pushed back" from C to B when finding the path A , C , B , D .

Non-terminating example

Consider the flow network shown on the right, with source s , sink t , capacities of edges e 1 , e 2 and e 3 respectively 1 , r = ( 5 − 1 ) / 2 and 1 and the capacity of all other edges some integer M ≥ 2 . The constant r was chosen so, that r 2 = 1 − r . We use augmenting paths according to the following table, where p 1 = { s , v 4 , v 3 , v 2 , v 1 , t } , p 2 = { s , v 2 , v 3 , v 4 , t } and p 3 = { s , v 1 , v 2 , v 3 , t } .

Note that after step 1 as well as after step 5, the residual capacities of edges e 1 , e 2 and e 3 are in the form r n , r n + 1 and 0 , respectively, for some n ∈ N . This means that we can use augmenting paths p 1 , p 2 , p 1 and p 3 infinitely many times and residual capacities of these edges will always be in the same form. Total flow in the network after step 5 is 1 + 2 ( r 1 + r 2 ) . If we continue to use augmenting paths as above, the total flow converges to 1 + 2 ∑ i = 1 ∞ r i = 3 + 2 r , while the maximum flow is 2 M + 1 . In this case, the algorithm never terminates and the flow doesn't even converge to the maximum flow.

1. Python program for implementation of Ford Fulkerson algorithm

from collections import defaultdict

1. 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's 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 the 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