Chapter 26 of CLSR
Maximum Network
Flow
by me Dr. M. Sakalli
Ch26 of CSLR and many other sources.
Marmara Univ.
A Flow Network, terminology
• Modeled as a flow graph which is a directed
graph of G=(V,E).
• Flow – a function f : V  V R: Rate passing
through the network from the source vertex s to
the sinking vertex t. In literature called as s-t
network. Net flow, |f(u,v)|.
• Source is vertex from which all the edges are
leaving ‘s’, and sink is the vertex, all the edges
incident to ‘t’ .
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A Flow Network, Constraints
A flow is an assignment of real numbers xij to edges (i,j) of a given network
that satisfy the following: flow-conservation requirements and capacity
constraints
• Capacity constraint: Each edge has a maximum capacity ‘ce’ allowed
to transmit through such that.. Net flow cannot exceed ce…
–  u, v  V: f(u, v) c(u, v)
• Skew symmetry constraint:  u, v  V: f(u,v) = –f(v,u)
• Flow conservation Law: The sum of the net flow incident to a vertex is
equal to the net flow leaving the same vertex.
–  u, v  V – {s, t}, ΣvVf(u, v)=0. for entire vertices.
– What goes in = what leaves out: ΣvVf(u, V)- ΣvVf(V, v) =0,
– s and t vertices:
ΣvVf(v, s)=0, ΣvVf(t, v)=0,
ΣvVf(s, v) = ΣvVf(v, t)>0 and R+,
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• No leakage and no memory is allowed.
• Total value of flow, |f| = f(s, V) = f(V, t)
s
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t
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b
• Among the areas applied are traffic, freight, airflow, hydrodynamic,
communications, and stocks, companies.
• The Maximum Flow Problem Definition: is to find a feasible flow
through a single-source, single-sink flow network that is maximum.
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• Opposite flows bt two vertices (due to skew symmetry)
will cancel each other.
• Super source and super sink connected with infinite
capacity.
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• Find paths to maximize |f|.
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b
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• f(X, Y) means ΣxXΣyYf(x, y)
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9 1/5
s
• f(u, V – {s, t}) = 0
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Lucky Puck Distribution Network
c
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d
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Max Flow, |f| = 19
Or is it?
Best we can do?
t
• For X, Y, Z  V with X  Y =,
• f(X  Y, Z) = f(X, Z) + f(Y, Z) flow between any x and y
through Z….
• f(Z, X  Y) = f(Z, X) + f(Z, Y)
• The value of a flow is the total net flow into the sink; that
is, |f| = f(V, t), and from the source s
|f| = f(s, V) (by definition)
= f(V-{V-s}, V) = f(V, V) - f(V - s, V)
= f(V, V - s)
|f| = f(V, t) = f(V, t) + f(V, V - s - t)
= f(V, t) (by flow conservation)
Cancellation of flows
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Maximum Flow as LP
.
Minimum Cost Flow
.
Ford-Fulkerson method
•
•
Initialize flow value to 0. Start to iterate (dynamic or greedy!)..
Find augmenting paths
– Find a path p from s to t, such that there is some flow value x > 0, and for each
edge (u, v) in p, add x units of flow to |f|, such that
max{x}  min{cf(u, v): (u, v) is on p}.
– Increment flow, f(u, v) = f(u, v) + x
•
Calculate residual capacities: cf(u,v) = c(u,v) – x
i.e. the actual capacity minus the net flow (x) from u to v
Flow may be negative!! That can be interpreted as increasing residual capacity.
•
Residual network: Gf =(V, Ef), where
•
•
Continue till no any augmenting path (any forward channel) left through
Observation – edges in Ef are either edges in E or their reversals: |Ef|  2|E|
Ef = {(u,v)  V  V : cf(u,v) > 0}
Augmenting Path?
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Ford-Fulkerson method
• Contains several algorithms:
– Residue networks
– Augmenting paths
• Find a path p from s to t (augmenting path), such that there is
some value x > 0, and for each edge (u,v) in p we can add x
units of flow
– f(u,v) + x  c(u,v)
Augmenting Path?
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a
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c
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10
b
9
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Residual Network
• To find augmenting path we can find any path in the
residual network:
– Residual capacities: cf(u,v) = c(u,v) – f(u,v)
• i.e. the actual capacity minus the net flow from u to v
• Net flow may be negative
– Residual network: Gf =(V,Ef), where
Ef = {(u,v)  V  V : cf(u,v) > 0}
– Observation – edges in Ef are either edges in E or their
reversals: |Ef|  2|E|
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Sub-graph
With
c(u,v) and
f(u,v)
a
b
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c
Residual
Sub-Graph
10
a
b
19
1
c
5
Residual Graph
• Compute the residual graph of the graph with the
following flow:
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Residual Capacity and Augmenting
Path
• Finding an Augmenting Path
– Find a path from s to t in the residual graph
– The residual capacity of a path p in Gf:
cf(p) = min{cf(u,v): (u,v) is in p}
• i.e. find the minimum capacity along p
– Doing augmentation: for all (u,v) in p, we just
add this cf(p) to f(u,v) (and subtract it from
f(v,u))
– Resulting flow is a valid flow with a larger
value.
Residual network and augmenting path
f(p1:(s, v1),…) = 11, as far as cf(s,v1)<16, and as far as sum v1=0, and f(p2:(s, v2),…),
calculate final residual graph Gf, and next search augmentation path.
The Ford-Fulkerson method
Ford-Fulkerson(G,s,t)
1 for each edge (u,v) in G.E do
2
f(u,v)  f(v,u)  0
3 while there exists a path p from s to t in residual
network Gf do
4
cf = min{cf(u,v): (u,v) is in p}
5
for each edge (u,v) in p do
6
f(u,v)  f(u,v) + cf
7
f(v,u)  -f(u,v)
8 return f
The algorithms based on this method differ in how they choose p in step 3.
Steps 1, 2, O(E), and 4-7 O(E) times for each unit of |f*|.
If chosen poorly the algorithm might not terminate.
Execution of Ford-Fulkerson (1)
Left Side = Residual Graph
Right Side = Augmented Flow
Execution of Ford-Fulkerson (2)
Left Side = Residual Graph
Right Side = Augmented Flow
Path augmenting scenarios
Fig 27.4 of CLSR: The flow network G with f. 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).
The Ford-Fulkerson algorithm, Residual
Capacity and Augmenting Path
In each iteration find any path augmenting the flow along by the residual
capacity câ(p). Following algo computes the max f in G = (V, E) by updating
the net â[u, v] between connected vertices; u, v. If not connected, implicit
assumption is that â[u, v] = 0. Assumption: the c(u, v) is a constant-time
function, which is 0 if (u, v)  E.
c(u, v) might be derived from fields stored within vertices and their adjacency
lists. The residual capacity: câ(u, v).. Here the câ(p) is a temporary variable
for the câ(u, v) of the path p. () refers a function [] refers a mutable
identifier.
Ford-Fulkerson(G,s,t)
1 for each edge (u,v)  G.E do //Initializing
2
f(u,v)  f(v,u)  0
3 while  path p from s to t in residual network Gf do ***
4
cf = min{cf(u,v): (u,v)  p} // determining augmentation
5
for each edge (u,v)  p do // updating residual capacities, new Gf
6
f(u,v)  f(u,v) + cf
7
f(v,u)  -f(u,v)
8 return f
T(n) depends on how line 3 is implemented and the algorithms based on this differ
in how to determine p. If chosen poorly the algorithm might not terminate.
a
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10
4
s
13
c
12
9
b
20
7
14
d
t
Execution
4
Residual Graphs
Augmented Flow
Execution of Ford-Fulkerson
Residual Graph
Augmented Flow
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7
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t
4
Observe that terminates when there is no a path from s to t whose all edges
are in forward direction.
And observe that the edges whose capacities are depleted form bottleneck,
which is the minimum cut and their total capacity of the maximum achivible.
• Does the method find the maximum flow? Yes, if program can get to the
point where the residual graph has no path from s to t. Which is the
moment when the G(V, E) is partitioned into two disjoint sets, by a definite
split, minimum cut, c(S, V-S) = c(S, T), such that s  S and t  T.
• And the net flow (f(S,T)) through the cut (or the capacity) is the total
capacity of the flows f(u,v), where s  S and t  T which includes negative
flows back from T to S as well. Think about it from the graph…
• Minimum cut – a cut with the smallest capacity of all cuts.
|f|= f(S,T) i.e. the value of a max flow is equal to the capacity of a min cut.
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Cut capacity = 24
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Min Cut capacity = 21
Cut properties..
• A cut (S, T) of flow network G = (V, E) is a partition (or a
disjoint set) of V into S and T = V - S such that s  S and t
 T.
MST cut was for undirected graphs, here a directed graph
is cut such that (tails) s  S and (heads) t  T.) If f is a flow,
then the net flow across the cut (S, T) is defined to be â(S,
T).
The capacity of the cut (S, T) is c(S, T).
• f(S, T) = ΣuSΣvTf(u, v)

ΣuSΣvTc(u, v) = c(S, T).
• The value of any flow â in a flow network G is bounded from
above by the capacity of any cut of G.
• Proof Let (S, T) be any cut of G and let â be any flow. By
|â| = â(S, T) (Lemma 27.5 clrs)
and the capacity constraints,
• = ΣuSΣvTf(u, v)  ΣuSΣvTc(u, v) = c(S, T)
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:
1. f is a maximum flow in G.
2. The residual network Gf contains no augmenting
paths.
3. Flow value of f = c(S, T) for some cut (S,T) of G.
(This cut is called min-cut. Bottleneck. Among all
the cuts of G(S, T), it is the cut that returns the
minimum capacity.)
Proof
1→2 Use contradiction. If it does not hold, then the
residual network Gf must have some augmenting
paths. So the net flow |f| could be further increased
which contradicts to that |f| is a maximum flow in G.
2→3 Suppose that Gf has no augmenting path, that
is, that Gf contains no path from s to t. Define
S={vV: there exists a path from s to v in Gf} and
T=V-S. The partition (S,T) is a cut: we have s  S
trivially and t T because there is no path from s to t
in Gf. For each pair of vertices u and v such that
uS and vT, we have f(u,v)=c(u,v) since otherwise
(u,v)Ef, which would place v in set S. Thus, flow
value of f=c(S,T). (Lemma: flow value f=f(S,T))
The end of Proof
3→1 The value of any flow f in a flow network G is
bounded from above by the capacity of any cut of G.
flow value f(S, T)=ΣuSΣvTf(u, v)  ΣuSΣvTc(u, v) = c(S, T).
so the cut capacity c(S,T) is the upper bound of flow f.
Now the flow value f=c(S,T) thus implies that f is a
maximum flow.
Worst case analysis
• In the worst case, suppose that at
each step Ford-Fulkerson’s
algorithm performs some small
1000000
incremental (integer)
s
augmentations of ∆f, then the
number of iterations to reach to the 1000000
maximum flow, |f*|/∆f, where f* is
maximum flow.
v
1/1 p
u
1
v
• In the given example (clrs)
programs starts to traverse
through path p1, suvt and
alternates to the path p2,
svut. instead of completing
the algorithm in two steps.
1000000
1 p
s
1000000-1/1
2
u
1000000-1/1
s
• The running time of FordFulkerson’s algorithm is O(|f*|(E))
1/1000000-1
1000000
v
1 p
u
t
1000000
1/1000000-1
t
1000000
1000000-1/1
t
1
1/1000000-1
Edmonds-Karp algorithm
•
If searching path p in line 3, is implemented with a BFS in the residual network, and if the
augmentation path is a shortest path from s to t, regardless of the weights of edges,
• Edmonds-Karp algorithm. O(VE2), take shortest augmentation path.
• How do we find such a shortest path? DFS or BFS..
•
The number of augmenting paths needed by the shortest-augmenting-path
algorithm never exceeds nm/2, where n and m are the number of vertices and
edges, respectively
•
Since the time required to find shortest augmenting path by breadth-first search is
in O(n+m)=O(m) for networks represented by their adjacency lists, the time
efficiency of the shortest-augmenting-path algorithm is in O(VE2) or O(nm2) for this
representation.
Running time O(VE2), because the number of augmentations is O(VE)
•
•
•
More efficient algorithms have been found that can run in close to O(nm) time
BFS and AP and Maximum Capacity Path (MCP) - find an A.P. that maximizes incremental
flow
•
•
•
•
How many augmenting paths for each method?
Edmonds-Karp - A.P.s to give overall time.
Critical edge on A.P. No capacity will remain on a critical edge after A.P. is recorded
Observations: Edge may become critical several times. A vertex cannot get closer to source
in later rounds of BFS.
s
a
Dist i
First time critical
b
Dist i+1
s
Later
Second time critical
s
a
Dist > i+2
a
Dist > i+2
• Each augmentation has complexity O(E)
• Numbers of augmentations is O(VE)
– Each edge can be a critical edge at most
|V|/2-1 times
– Once edge is critical, its flow is saturated
– Must appear in path the other way next
– Since previous path is shortest, this reversing
path must be two edges longer
df’(s,v)=df(s,u)+1
df’(s,u)=df’(s,v)+1
≥df(s,v)+1 (paths increase
monotonically)
=df(s,u)+2
t
b
Dist > i+1
b
Dist > i+3
t
t
The number of distances
available for the tail of a
repeated critical edge is
(V-2)/2. The number of
edges that become
critical is < E. Thus,
A.P.s overall.
Flow must be sent in
opposite direction by
another A.P. before an
edge can become
critical again.
Shortest-Augmenting-Path Algorithm
Generate augmenting path with the least number of edges by BFS.
Start at s, perform BFS traversal by marking new (unlabeled) vertices with
two labels and a sign mark:
• 1st label – indicates the amount of additional flow that can be
brought from the previous to the current vertex being labeled.
• 2nd label – indicates the vertex from which the vertex being
labeled was reached, with “+” or “–” in forward or in backward,
respectively.
• Vertex labeling: The start is always labeled with ∞,• All other vertices are labeled as follows: If unlabeled vertex j
is connected to the front vertex i with the edge
– from i to j with unused capacity rij = uij –f(uij, vij) (forward edge), vertex
j is labeled with lj, i+, where lj = min{li, rij}.
– from j to i with positive flow xji (backward edge), vertex j is labeled lj,i-,
where lj = min{li, xji}
• If the sink ends up being labeled, then there is an
augmentation path by the amount min{cp} which indicated
is with the sink’s first label.
• The augmentation path is traced along the path from sink
to source; the current flow quantities are increased on the
forward edges and decreased on the backward edges of
this path.
• If the sink remains unlabeled after the traversal queue
becomes empty, the algorithm returns the current flow as
maximum and terminates.
5
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s
2,2+
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∞,-
2,1+
0/2
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2
2,2+
4
Queue: 1 2 4 3 5 6
↑↑↑↑
Augment the flow by 2 (the sink’s first label)
along the path 1→2→3→6
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4
3,1+
t
2,3+
5
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2
3
2/2
1,2+
5
t
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0/1
4
∞,s
1,32/2 2
0/3
1/3
2/2
1/3
4
Queue: 1 4
↑↑
4
1/4
1/5
2/2
t
0/1
5
2
1,4+
2/5
3
1,5+
Queue: 1 4 3 2 5 6
↑↑↑↑↑
s
0/4
3
2/2
3,1+4
Augment the flow by 1 (the sink’s first label)
along the path 1→4→3←2→5→6
t
5
1/1
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∞,s
2/2
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2
4
1/4
1/5
3
2/2
t
1/1
2,1+
No augmenting path (the sink is unlabeled)
the current flow is maximum
• S : (∞, 1, +)
• B : (S, 8, +)
• E : (S, 28, +)
• F : (S, 15, +)
• A : (B, 8, +)
• C : (B, 8, +)
• D : (B, 6, -)
• T : (A, 8, +)
• f updated by 8.
P = (S,B, A, T)
Multiple Sources or Sinks
• What if you have a problem with more than one source
and more than one sink?
• Modify the graph to create a single supersource and
supersink
a
15
s
10
4
14
4
x
c
e
15
g
b
d
13
11
f
9
t
3
13
h

i
14
j
c
e
15

y
3
4
g
b
d
13
11
f
9
13
k
5
9
11
10
4
14
13
10
4
s
5
a
15
13
5
9
11
10
4
14
13
3
t
13

l
5
h

3
Application – Bipartite Matching
• Example – given a community with n men and m
women
• Assume we have a way to determine which
couples (man/woman) are compatible for
marriage
– E.g. (Joe, Susan) or (Fred, Susan) but not (Frank,
Susan)
• Problem: Maximize the number of marriages
– No polygamy allowed
– solve this problem by creating a flow network out of a
bipartite graph
Bipartite Graph
• A bipartite graph is an undirected graph G=(V,E) in
which V can be partitioned into two sets V1 and V2 such
that (u,v)  E implies either u  V1 and v  V2 or vice
versa.
• That is, all edges go between the two sets V1 and V2 and
not within V1 and V2.
Model for Matching Problem
• Men on leftmost set, women on rightmost
set, edges if they are compatible
A
X
B
Y
C
B
B
Y
Y
C
C
D
X
X
Z
Men
A
A
Z
D
Women
A matching
Z
D
Optimal matching
Solution Using Max Flow
• Add a supersouce, supersink, make each
undirected edge directed with a flow of 1
A
A
X
X
B
Y
C
s
B
Y
C
Z
D
Z
D
Since the input is 1, flow conservation prevents multiple matchings
t