Information About

Edmonds-karp Algorithm





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(VE^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 1.


PSEUDOCODE


:''For a more high level description, see Ford-Fulkerson Algorithm .

algorithm EdmondsKarp
input:
C 1..n ''(Capacity matrix)''
E 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 - 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 {Link without Title}
F := F[u,v + m
F := F[v,u - m
v := u
return (f, F)

algorithm BreadthFirstSearch
input:
C, E, s, t
output:
M {Link without Title} ''(Capacity of path found)''
P ''(Parent table)''
P := array(1..n)
for u '''in''' 1..n
P {Link without Title} := -1
P {Link without Title} := -2 ''(make sure source is not rediscovered)''
M := array(1..n) ''(Capacity of found path to node)''
M {Link without Title} := ∞
Q := queue()
Q.push(s)
while Q.size() > 0
u := Q.pop()
for v '''in''' E {Link without Title}
''(If there is available capacity, and v is not seen before in search)''
if C - F[u,v > 0 '''and''' P[v] = -1
P {Link without Title} := u
M := min(M[u , C - F[u,v )
if v ≠ t
Q.push(v)
else
return M {Link without Title} , P
return 0, P


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.

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.


REFERENCES


# Algorithms and Complexity (see pages 63 - 69). http://www.cis.upenn.edu/~wilf/AlgComp3.html