Edmonds-Karp algorithm

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In computer science and graph theory, the Edmonds-Karp algorithm is an implementation of the Ford-Fulkerson method for computing the maximum flow in a flow network in $O (V E 2)$ . It is asymptotically slower than the relabel-to-front algorithm, which runs in $O (V 3)$ , but it is often faster in practice for sparse graphs. The algorithm was first published by a Russian scientist, Dinic, in 1970^[1], and independently by Jack Edmonds and Richard Karp in 1972^[2] (discovered earlier). Dinic's algorithm includes additional techniques that reduce the running time to $O (V 2 E)$ .

1 Algorithm
2 Pseudocode
3 Example
4 References

[edit] Algorithm

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 the shortest path which has available capacity. This can be found by a breadth-first search, as we let edges have unit length. The running time of $O (V E 2)$ is found by showing that the length of the augmenting path found never decreases, that for every time one of the $E$ edge becomes saturated the augmenting path must be longer than last time it was saturated, that a path is at most $V$ long, and can be found in $O (E)$ time. There is an accessible proof in ^[3].

[edit] Pseudocode

Wikibooks Algorithm implementation has a page on the topic of

Edmonds-Karp

For a more high level description, see Ford-Fulkerson algorithm.

algorithm EdmondsKarp
    input:
        C[1..n, 1..n] (Capacity matrix)
        E[1..n, 1..?] (Neighbour lists)
        s             (Source)
        t             (Sink)
    output:
        f             (Value of maximum flow)
        F             (A matrix giving a legal flow with the maximum value)
    f := 0 (Initial flow is zero)
    F := array(1..n, 1..n) (Residual capacity from u to v is C[u,v] - F[u,v])
    forever
        m, P := BreadthFirstSearch(C, E, s, t)
        if m = 0
            break
        f := f + m
        (Backtrack search, and write flow)
        v := t
        while v ≠ s
            u := P[v]
            F[u,v] := F[u,v] + m
            F[v,u] := F[v,u] - m
            v := u
    return (f, F)

algorithm BreadthFirstSearch
    input:
        C, E, s, t
    output:
        M[t]          (Capacity of path found)
        P             (Parent table)
    P := array(1..n)
    for u in 1..n
        P[u] := -1
    P[s] := -2 (make sure source is not rediscovered) 
    M := array(1..n) (Capacity of found path to node)
    M[s] := ∞
    Q := queue()
    Q.push(s)
    while Q.size() > 0
        u := Q.pop()
        for v in E[u]
            (If there is available capacity, and v is not seen before in search)
            if C[u,v] - F[u,v] > 0 and P[v] = -1
                P[v] := u
                M[v] := min(M[u], C[u,v] - F[u,v])
                if v ≠ t
                    Q.push(v)
                else
                    return M[t], P
    return 0, P

[edit] Example

Given a network of seven nodes, source A, sink G, and capacities as shown below:

In the pairs $f / c$ written on the edges, $f$ is the current flow, and $c$ is the capacity. The residual capacity from $u$ to $v$ is $c f (u, v) = c (u, v) - f (u, v)$ , the total capacity, minus the flow you have already used. If the net flow from $u$ to $v$ is negative, it contributes to the residual capacity.

Capacity	Path
Capacity	Resulting network
$min(c f (A, D), c f (D, E), c f (E, G)) =$ $min(3 - 0,2 - 0,1 - 0) =$ $min(3,2,1) = 1$	$A, D, E, G$

$min(c f (A, D), c f (D, F), c f (F, G)) =$ $min(3 - 1,6 - 0,9 - 0) =$ $min(2,6,9) = 2$	$A, D, F, G$

$min(c f (A, B), c f (B, C), c f (C, D), c f (D, F), c f (F, G)) =$ $min(3 - 0,4 - 0,1 - 0,6 - 2,9 - 2) =$ $min(3,4,1,4,7) = 1$	$A, B, C, D, F, G$

$min(c f (A, B), c f (B, C), c f (C, E), c f (E, D), c f (D, F), c f (F, G)) =$ $min(3 - 1,4 - 1,2 - 0,0 - - 1,6 - 3,9 - 3) =$ $min(2,3,2,1,3,6) = 1$	$A, B, C, E, D, F, G$

Notice how the length of the augmenting path found by the algorithm never decreases. The paths found are the shortest possible. The flow found is equal to the capacity across the smallest 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 ${A, B, C, E}$ and ${D, F, G}$ , with the capacity $c (A, D) + c (C, D) + c (E, G) = 3 + 1 + 1 = 5$ .