Chapter 27
Maximum Flow
• Maximum Flow Problem
• The Ford-Fulkerson method
• m bipartite matching
Background:
• Just as we can model a road map as a directed graph in
order to find the shortest path from one point to another,
we can also interpret a directed graph as a “flow network”
and use it to answer questions about material coursing
through a system from a source,where the material is
produced, to a sink, where it is consumed.
Flow networks:
• A flow network G=(V,E) is a directed graph in which each
edge (u,v)E has a nonnegative capacity c(u,v)>=0.If
(u,v)E, we assume that c(u,v)=0.We distinguish two
vertices in a flow network:a source s and a sink t.
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20
16
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9
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t
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Flow:
• Let G=(V,E) be a flow network (with capacity function c).
Let s be the source of the network, and let t be the sink.A
flow in G is a real-valued function f:V*V R that
satisfies the following three properties:
• Capacity constraint: For all u,v V,
we require f(u,v) <= c( u,v).
• Skew symmetry: For all u,v V,we require f(u,v)=-f(v,u).
• Flow conservation: For all u V-{s,t}, we require
 f(u,v) =0
vV
Net flow and value of a flow f:
• The quantity f (u,v), which can be positive or negative, is
called the net flow from vertex u to vertex v.
• The value of a flow is defined as
f   f ( s, v )
vV
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10
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A flow f in G with value f  19 .
v1
10 4
v1
8/10 4
v1
8/10 3/4
v1
5/10 4
v1
10 2/4
v2
v2
v2
v2
v2
(a)
(b)
(c)
(d)
(e)
Cancellation.(a) Vertices v1 and v2,with c(v1,v2)=10 and c(v2,v1)
=4.(b)How we indicate the net flow 8 from v1 to v2.
(c) An additional shipment of 3 is made
from v2 to v1. (d) By cancelling flow going in opposite directions,we
can represent the situation in (c) with positive net flow in one
direction only.(e) Another 7 is shipped from v2 to v1.
Maximum-flow problem:
• Given a flow network G with source s and sink t
• Find a flow of maximum value from s to t.
• How to solve it efficiently?
The Ford-Fulkerson method:
• This section presents the Ford-Fulkerson method for
solving the maximum-flow problem.We call it a “method”
rather than an “algorithm” because it encompasses several
implementations with different running times.The FordFulkerson method depends on three important ideas that
transcend the method and are relevant to many flow
algorithms and problems: residual networks,augmenting
paths,and cuts. These ideas are essential to the important
max-flow min-cut theorem,which characterizes the value
of maximum flow in terms of cuts of the flow network.
Continue:
•
•
•
•
•
FORD-FULKERSON-METHOD(G,s,t)
initialize flow f to 0
while there exists an augmenting path p
do augment flow f along p
return f
Residual networks:
• Intuitively,given a flow network and a flow, the residual
network consists of edges that can admit more net
flow.More formally,suppose that we have a flow network
G=(V,E) with source s and sink t.Let f be a flow in G, and
consider a pair of vertices u,vV.The amount of additional
net flow we can push from u to v before exceeding the
capacity c(u,v) is the residual capacity of (u,v), given by:
cf(u,v)=c(u,v)-f(u,v)
• Given a flow network G=(V,E) and a flow f,the residual
network of G induced by f is Gf=(V,Ef),where
Ef={(u,v)V*V: cf (u,v) > 0}
Continue:
• Lemma 27.2
• Let G=(V,E) be a flow network with source s and sink t,
and let f be a flow in G.Let Gf be the residual network of G
induced by f,and let f’ be a flow in Gf.Then, the flow sum
f+f’ is a flow in G with value f  f '  f  f ' .
Augmenting paths:
• Given a flow network G=(V,E) and a flow f, an
augmenting path is a simple path from s to t in the residual
network Gf.
• We call the maximum amount of net flow that we can ship
along the edges of an augmenting path p the residual
capacity of p, given by cf(p)=min{cf(u,v):(u,v) is on p}.
Example:
• (a)The flow network G and flow f .
• (b)The residual network Gf with augmenting path p shaded;
its residual capacity is cf(p)=c(v2,v3)=4.
• (c) The flow in G that results from augmenting along path
p by its residual capacity 4.
• (d) The residual network induced by the flow in (c).
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(a)
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v1
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5 11
s
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(b)
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7 15
v4
t
4
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1/4 9
v2
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(c)
19/20
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v4
t
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5
s
v1
11
1 11
12
12
3
v2
v3
9
3
11
(d)
1
7 19
v4
4
t
Continue:
• Lemma 27.3
• Let G=(V,E) be a flow network, let f be a flow in G,and let
p be an augmenting path in Gf.Define a function fp:V*V 
R by
 cf ( p )
•
if (u,v) is on p,
• fp(u,v)= 
if (v,u) is on p,
  cf ( p )
•
otherwise.
 0

• Then, fp is a flow in Gf with value fp  cf ( p)  0 .
Continue:
• Corollary 27.4
• Let G=(V,E) be a flow network, let f be a flow in G,and let
p be an augmenting path in Gf. Let fp be defined as in
Lemma 27.3.Define a function f’:V*V  R by f’=f+fp.
Then, f’ is a flow in G’ with value f '  f  fp  f .
Cuts of flow networks:
• The Ford-Fulkerson method repeatedly augments the flow
along augmenting paths until a maximum flow has been
found.The max-flow min-cut theorem,which we shall
prove shortly, tells us that a flow is maximum if and only
if its residual network contains no augmenting path.
• A cut (S,T) of flow network G=(V,E) is a partition of V
into S and T=V-S such that sS and t T.If f is a flow,
then the net flow across the cut (S,T) is defined to be
f(S,T).The capacity of the cut (S,T) is c(S,T).
v1
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S
T
A cut (S,T) in the flow network G,where S={s,v1,v2} and
T={v3,v4,t}.The vertices in S are black,and the vertices in
T are white.The net flow across (S,T) is f(S,T)=19,and the
capacity is c(S,T)=26.
Continue:
• Lemma 27.5
• Let f be a flow in a flow network G with source s and sink
t,and let (S,T) be a cut of G.Then, the net flow across (S,T)
is f(S,T)= f .
Continue:
• Corollary 27.6
• The value of any flow f in a flow network G is bounded
from above by the capacity of any cut of G.
Continue:
• Theorem 27.7(Max-flow min-cut theorem)
• If f is a flow in a flow network G=(V,E) with source s and
sink t,then the following conditions are equivalent:
• f is a maximum flow in G;
• The residual network Gf contains no augmenting paths;
• f =c(S,T) for some cut (S,T) of G.
The basic Ford-Fulkerson
algorithm:
• FORD-FULKERSON(G,s,t)
• for each edge (u,v) E[G]
•
do f[u,v]  0
•
f[v,u]  0
• while there exists a path p from s to t in the residual
network Gf
•
do cf(p) min{cf(u,v): (u,v) is in p}
•
for each edge (u,v) in p
•
do f[u,v] f[u,v]+cf(p)
•
f[v,u]  -f[u,v]
Example:
• The execution of the basic Ford-Fulkerson algorithm.(a)-(d)
Successive iterations of the while loop. The left side of
each part shows the residual network Gf from line 4 with a
shaded augmenting path p.The right side of each part
shows the new flow f that results from adding fp to f.The
residual network in (a) is the input network G.(e) The
residual network at the last while loop test.It has no
augmenting paths,and the flow f shown in (d) is therefore a
maximum flow.
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4 9
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(a)
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(b)
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(c)
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7 15
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(d)
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1 11
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7 19
v4
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(e)
t
Time complexity:
• If each c(e) is an integer, then time complexity is O(|E|f*),
where f* is the maximum flow.
• Reason: each time the flow is increased by at least one.
• This might not be a polynomial time algorithm since f* can
be represented by log (f*) bits. So, the input size might be
log(f*).
The Edmonds-Karp algorithm
• Find the augmenting path using
breadth-first search.
• Breadth-first search gives the shortest
path for graphs with edge length 1.
• Time complexity of Edmonds-Karp
algorithm is O(VE2).
• Lemma: With each flow augmentation, the
•
•
•
•
shortest-path distance between s and u for any u
increases monotonically.
Proof: (Fun part, not required)
Prove it by contradiction.
Let f’ is obtained from f by an augmentation and
d(f’,s,v)<d(f, s, v).
Assume that v is a node such that d(f’, s, v) is the
minimum satisfying d(f’,s,v)<d(f, s, v).
• Consider u with d(f’, s, u)=d(f’, s, v)-1.
• Based on the choice of v, we know that
d(f’, s, y)>=d(f, s, u).
Then (u, v) is not in e(
Maximum bipartite matching:
• Given an undirected graph G=(V,E), a matching is a subset
of edges ME such that for all vertices vV,at most one
edge of M is incident on v.We say that a vertex v V is
matched by matching M if some edge in M is incident on
v;otherwise, v is unmatched. A maximum matching is a
matching of maximum cardinality,that is, a matching M
such that for any matching M’, we have M  M ' .
L
R
(a)
L
R
(b)
A bipartite graph G=(V,E) with vertex partition V=LR.(a)A
matching with cardinality 2.(b) A maximum matching with
cardinality 3.
Finding a maximum bipartite
matching:
• We define the corresponding flow network G’=(V’,E’) for
the bipartite graph G as follows. Let the source s and sink t
be new vertices not in V, and let V’=V{s,t}.If the vertex
partition of G is V=L R, the directed edges of G’ are
given by E’={(s,u):uL} {(u,v):u L,v R,and (u,v) E}
{(v,t):v R}.Finally, we assign unit capacity to each
edge in E’.
• We will show that a matching in G corresponds directly to
a flow in G’s corresponding flow network G’. We say that
a flow f on a flow network G=(V,E) is integer-valued if
f(u,v) is an integer for all (u,v) V*V.
s
L
R
(a)
t
L
R
(b)
(a)The bipartite graph G=(V,E) with vertex partition V=LR. A
maximum matching is shown by shaded edges.(b) The corresponding
flow network shown.Each edge has unit capacity.Shaded edges have
a flow of 1,and all other edges carry no flow.
Continue:
• Lemma 27.10
• Let G=(V,E) be a bipartite graph with vertex partition
V=LR,and let G’=(V’,E’) be its corresponding flow
network.If M is a matching in G, then there is an integervalued flow f in G’ with value f  M .Conversely, if f
is an integer-valued flow in G’,then there is a matching M
in G with cardinality M  f .
The End: