The algorithm is identical to the Ford–Fulkerson algorithm, except that the search order when finding the augmenting path is defined. The path found must be a shortest path that has available capacity. This can be found by a breadth-first search, where we apply a weight of 1 to each edge. The running time of is found by showing that each augmenting path can be found in time, that every time at least one of the edges becomes saturated (an edge which has the maximum possible flow), that the distance from the saturated edge to the source along the augmenting path must be longer than last time it was saturated, and that the length is at most . Another property of this algorithm is that the length of the shortest augmenting path increases monotonically. There is an accessible proof in Introduction to Algorithms.[4]
Pseudocode
algorithm EdmondsKarp isinput:
graph (graph[v] should be the list of edges coming out of vertex v in the original graph and their corresponding constructed reverse edges which are used for push-back flow. Each edge should have a capacity 'cap', flow, source 's' and sink 't' as parameters, as well as a pointer to the reverse edge 'rev'.)
s (Source vertex)
t (Sink vertex)output:
flow (Value of maximum flow)
flow := 0 (Initialize flow to zero)repeat(Run a breadth-first search (bfs) to find the shortest s-t path. We use 'pred' to store the edge taken to get to each vertex, so we can recover the path afterwards)
q := queue()
q.push(s)
pred := array(graph.length)
whilenot empty(q) and pred[t] = null
cur := q.pop()
for Edge e in graph[cur] doif pred[e.t] = nulland e.t ≠ s and e.cap > e.flow then
pred[e.t] := e
q.push(e.t)
ifnot (pred[t] = null) then(We found an augmenting path. See how much flow we can send)
df := ∞for (e := pred[t]; e ≠ null; e := pred[e.s]) do
df := min(df, e.cap - e.flow)
(And update edges by that amount)for (e := pred[t]; e ≠ null; e := pred[e.s]) do
e.flow := e.flow + df
e.rev.flow := e.rev.flow - df
flow := flow + df
until pred[t] = null (i.e., until no augmenting path was found)return flow
Example
Given a network of seven nodes, source A, sink G, and capacities as shown below:
In the pairs written on the edges, is the current flow, and is the capacity. The residual capacity from to is , the total capacity, minus the flow that is already used. If the net flow from to is negative, it contributes to the residual capacity.
Path
Capacity
Resulting network
Notice how the length of the augmenting path found by the algorithm (in red) never decreases. The paths found are the shortest possible. The flow found is equal to the capacity across the minimum cut in the graph separating the source and the sink. There is only one minimal cut in this graph, partitioning the nodes into the sets and , with the capacity
Notes
^Dinic, E. A. (1970). "Algorithm for solution of a problem of maximum flow in a network with power estimation". Soviet Mathematics - Doklady. 11. Doklady: 1277–1280.