university of cambridge · multiple sources and multiple sinks (figure 26.1) 26.1 flow networks 713...

6.6: Maximum flowFrank Stajano Thomas Sauerwald

Lent 2015

Illustration of the Ford-Fulkerson Method

s

2

3

4

5 t

Graph G = (V , E , c):

0/10

0/10

0/2 0/6

0/4

0/8

0/9

0/10

0/10

⇠⇠

0/10 ��0/8

⇠⇠0/108/1

0

0/10

0/2 0/6

0/4

8/8

0/9

0/10

8/10

⇠⇠

8/10 ��0/2

��0/9⇠⇠8/10

10/10

0/10

2/2 0/6

0/4

8/8

2/9

0/10

10/10⇠⇠0/10��2/9

��0/6⇠⇠0/1010

/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10⇠⇠6/10

��2/2

��0/4

⇠⇠6/1010/1

0

8/10

0/2 6/6

2/4

8/8

8/9

8/10

10/10

⇠⇠8/10 ��8/9

��8/8

��2/4⇠⇠8/1010

/10

9/10

0/2 6/6

3/4

7/8

9/9

9/10

10/10

s

2

3

4

5 t

Residual Graph Gf = (V , Ef , cf ):

10

10

2 6

4

8

910

10

2

8

10

2 6

4

8

910

2

8

10

10

2 6

4

8

7

2

1010

10

4

6

2 6

4

8

1

8

4

610

10

2

8

2 6

2

2

8

1

8

2

810

10

1

9

2 6

1

3

7

1

9

1

910

|f | = 0|f | = 8|f | = 10|f | = 16

|f | = 18

|f | = 19

Is this a max-flow?

6.6: Maximum flow T.S. 9

Outline

Recap

Max-Flow Min-Cut Theorem

Analysis of Ford-Fulkerson


History of the Maximum Flow Problem [Harris, Ross (1955)]

21 -C

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5

0! E

0

70 =

In 4'

oS 0

'ma

0

0 O

f

0

3

(a ~

0

za

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On,

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w 0>a

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,,

a cc

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SECR

ET R

7-

Maximum Flow is 163,000 tons per day!


Multiple Sources and Multiple Sinks (Figure 26.1)26.1 Flow networks 713

10

(a)

12

58

147

112

3

15

6

20

1318

1012

58

147

112

3

15

6

20

1318

∞ ∞

∞

∞

∞

∞

∞

∞

s1 s1

s2 s2

s3s3

s4 s4

s5s5

t1 t1

t2t2

t3 t3

(b)

s t

Figure 26.3 Converting a multiple-source, multiple-sink maximum-flow problem into a problemwith a single source and a single sink. (a) A flow network with five sources S D fs1; s2; s3; s4; s5gand three sinks T D ft1; t2; t3g. (b) An equivalent single-source, single-sink flow network. We adda supersource s and an edge with infinite capacity from s to each of the multiple sources. We alsoadd a supersink t and an edge with infinite capacity from each of the multiple sinks to t .

26.1-2Extend the flow properties and definitions to the multiple-source, multiple-sinkproblem. Show that any flow in a multiple-source, multiple-sink flow networkcorresponds to a flow of identical value in the single-source, single-sink networkobtained by adding a supersource and a supersink, and vice versa.26.1-3Suppose that a flow network G D .V; E/ violates the assumption that the networkcontains a path s ! ! ! t for all vertices ! 2 V . Let u be a vertex for which thereis no path s ! u ! t . Show that there must exist a maximum flow f in G suchthat f .u; !/ D f .!; u/ D 0 for all vertices ! 2 V .


Residual Graph

Edge e = (u, v) ∈ Eflow f (u, v) and capacity c(u, v)

Original Edge

cf (u, v) =

c(u, v)− f (u, v) if (u, v) ∈ E ,f (v , u) if (v , u) ∈ E ,0 otherwise.

Residual Capacity

Gf = (V ,Ef , cf ), Ef := {(u, v) : cf (u, v) > 0}Residual Graph

Graph G:

u v6/17

Residual Gf :

u v

11

6


Residual Graph with anti-parallel edges

Edge e = (u, v) ∈ E (& possibly e′ = (v , u) ∈ E)flow f (u, v) and capacity c(u, v)

Original Edge

For every pair (u, v) ∈ V × V ,

cf (u, v) = c(u, v)− f (u, v).

Residual Capacity

Gf = (V ,Ef , cf ), Ef := {(u, v) : cf (u, v) > 0}Residual Graph

Graph G:

u v

6/17

2/4

Residual Gf :

u v

??

??

17-(6-2)

4-(2-6)

13

8


Example of a Residual Graph (Handout)

Flow network Gs

2

3

4

5

t

1/14

1/6

2/3

4/105/12

5/8

4/7

3/4

1/3

3/7

3/4

1/3

3/7

2/3

0/2

2/6

2/3

0/2

2/6

4/105/12

0/101/12

0/101/12

5/8

4/7

1/8

0/7

1/8

0/7

1/8

1/121/14

1/61/8

1/121/14

1/60/8

0/120/14

0/6

0/8

0/120/14

0/6

Residual Graph Gfs

2

3

4

5

t

1111

131

51

78

13

21

34

1 2

By successively eliminating cycles we can sim-plify and reduce the “transportation” cost of a flow.

Comment on anti-parallel edges:You should know about this possibility

In the following proofs we will disallow anti-parallel edges(for convenience)



Flow network Gs

2

3

4

5

t

1/14

1/6

2/3

4/105/12

5/8

4/7

3/4

1/3

3/7

3/4

1/3

3/7

2/3

0/2

2/6

2/3

0/2

2/6

4/105/12

0/101/12

0/10

1/12

5/8

4/7

1/8

0/7

1/8

0/7

1/8

1/121/14

1/6

1/8

1/121/14

1/60/8

0/120/14

0/6

0/8

0/120/14

0/6

Residual Graph Gfs

2

3

4

5

t

1111

131

51

78

13

21

34

1 2






Flow network Gs

2

3

4

5

t

1/14

1/6

2/3

4/105/12

5/8

4/7

3/4

1/3

3/7

3/4

1/3

3/7

2/3

0/2

2/6

2/3

0/2

2/6

4/105/12

0/101/12

0/10

1/12

5/8

4/7

1/8

0/7

1/8

0/7

1/8

1/121/14

1/6

1/8

1/121/14

1/6

0/8

0/120/14

0/6

0/8

0/120/14

0/6

Residual Graph Gfs

2

3

4

5

t

1111

131

51

78

13

21

34

1 2






Flow network Gs

2

3

4

5

t

1/14

1/6

2/3

4/105/12

5/8

4/7

3/4

1/3

3/7

3/4

1/3

3/7

2/3

0/2

2/6

2/3

0/2

2/6

4/105/12

0/101/12

0/10

1/12

5/8

4/7

1/8

0/7

1/8

0/7

1/8

1/121/14

1/61/8

1/121/14

1/6

0/8

0/120/14

0/6

0/8

0/120/14

0/6

Residual Graph Gfs

2

3

4

5

t

1111

131

51

78

13

21

34

1 2






Flow network Gs

2

3

4

5

t

1/14

1/6

2/3

4/105/12

5/8

4/7

3/4

1/3

3/7

3/4

1/3

3/7

2/3

0/2

2/6

2/3

0/2

2/6

4/105/12

0/101/12

0/10

1/12

5/8

4/7

1/8

0/7

1/8

0/7

1/8

1/121/14

1/61/8

1/121/14

1/60/8

0/120/14

0/6

0/8

0/120/14

0/6

Residual Graph Gfs

2

3

4

5

t

1111

131

51

78

13

21

34

1 2





The Ford-Fulkerson Method (“Enhanced Greedy”)

0: def fordFulkerson(G)1: initialize flow to 0 on all edges2: while an augmenting path in Gf can be found:3: push as much extra flow as possible through it

Augmenting path: Pathfrom source to sink in Gf

Questions:How to find an augmenting path? XDoes this method terminate?If it terminates, how good is the solution?


Outline

Recap




From Flows to Cuts

A cut (S,T ) is a partition of V into S and T = V \ S such that s ∈ Sand t ∈ T .

The capacity of a cut (S,T ) is the sum of capacities of the edgesfrom S to T :

c(S,T ) =∑

u∈S,v∈T

c(u, v) =∑

(u,v)∈E(S,T )

c(u, v)

A mininum cut of a network is a cut whose capacity is minimum overall cuts of the network.

Cut

s

2

3

4

5 t

Graph G = (V ,E , c):

10

10

2 6

4

8

9

10

10

c({s, 3}, {2, 4, 5, t}) =

10 + 9 = 19

|f | = 16

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10

10− 2 + 8 = 168 + 8− 6 + 6 = 16


From Flows to Cuts

A cut (S,T ) is a partition of V into S and T = V \ S such that s ∈ Sand t ∈ T .The capacity of a cut (S,T ) is the sum of capacities of the edgesfrom S to T :

c(S,T ) =∑

u∈S,v∈T

c(u, v) =∑

(u,v)∈E(S,T )

c(u, v)


Cut

s

2

3

4

5 t


10

10

2 6

4

8

9

10

10

c({s, 3}, {2, 4, 5, t}) =

10 + 9 = 19

|f | = 16

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10

10− 2 + 8 = 168 + 8− 6 + 6 = 16


From Flows to Cuts

A cut (S,T ) is a partition of V into S and T = V \ S such that s ∈ Sand t ∈ T .The capacity of a cut (S,T ) is the sum of capacities of the edgesfrom S to T :

c(S,T ) =∑

u∈S,v∈T

c(u, v) =∑

(u,v)∈E(S,T )

c(u, v)


Cut

s

2

3

4

5 t


10

10

2 6

4

8

9

10

10

c({s, 3}, {2, 4, 5, t}) = 10 + 9 = 19

|f | = 16

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10

10− 2 + 8 = 168 + 8− 6 + 6 = 16


From Flows to Cuts

Let f be a flow with source s and sink t , and let (S,T ) be any cut of G.Then the value of the flow is equal to the net flow across the cut, i.e.,

|f | =∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u).

Flow Value Lemma (Lemma 26.4)

s

2

3

4

5 t


10

10

2 6

4

8

9

10

10

c({s, 3}, {2, 4, 5, t}) = 10 + 9 = 19

|f | = 16

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10

10− 2 + 8 = 168 + 8− 6 + 6 = 16


From Flows to Cuts


|f | =∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u).


s

2

3

4

5 t


10

10

2 6

4

8

9

10

10

c({s, 3}, {2, 4, 5, t}) = 10 + 9 = 19

|f | = 16

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10

10− 2 + 8 = 16

8 + 8− 6 + 6 = 16


From Flows to Cuts


|f | =∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u).


s

2

3

4

5 t


10

10

2 6

4

8

9

10

10

c({s, 3}, {2, 4, 5, t}) = 10 + 9 = 19

|f | = 16

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10

10− 2 + 8 = 168 + 8− 6 + 6 = 16


From Flows to Cuts


|f | =∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u).


s

2

3

4

5 t


10

10

2 6

4

8

9

10

10

c({s, 3}, {2, 4, 5, t}) = 10 + 9 = 19

|f | = 16

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10

10− 2 + 8 = 16

8 + 8− 6 + 6 = 166.6: Maximum flow T.S. 11

From Flows to Cuts

|f | =∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u).

s

2

3

4

5 t


10

10

2 6

4

8

9

10

10

c({s, 3}, {2, 4, 5, t}) = 10 + 9 = 19

|f | = 16

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10

10− 2 + 8 = 16

8 + 8− 6 + 6 = 166.6: Maximum flow T.S. 11

From Flows to Cuts

|f | =∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u).

|f | =

∑w∈V

f (s,w) =∑u∈S

( ∑(u,w)∈E

f (u,w)−∑

(w,u)∈E

f (w , u))

=∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

s

2

3

4

5 t


10

10

2 6

4

8

9

10

10

c({s, 3}, {2, 4, 5, t}) = 10 + 9 = 19

|f | = 16

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10

10− 2 + 8 = 16

8 + 8− 6 + 6 = 166.6: Maximum flow T.S. 11

From Flows to Cuts

|f | =∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u).

|f | =∑w∈V

f (s,w)

=∑u∈S

( ∑(u,w)∈E

f (u,w)−∑

(w,u)∈E

f (w , u))

=∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

s

2

3

4

5 t


10

10

2 6

4

8

9

10

10

c({s, 3}, {2, 4, 5, t}) = 10 + 9 = 19

|f | = 16

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10

10− 2 + 8 = 16

8 + 8− 6 + 6 = 166.6: Maximum flow T.S. 11

From Flows to Cuts

|f | =∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u).

|f | =∑w∈V

f (s,w) =∑u∈S

( ∑(u,w)∈E

f (u,w)−∑

(w,u)∈E

f (w , u))

=∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

s

2

3

4

5 t


10

10

2 6

4

8

9

10

10

c({s, 3}, {2, 4, 5, t}) = 10 + 9 = 19

|f | = 16

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10

10− 2 + 8 = 16

8 + 8− 6 + 6 = 166.6: Maximum flow T.S. 11

Weak Duality betwen Flows and Cuts

Let f be any flow and (S,T ) be any cut.

Then the value of f is boundedfrom above by the capacity of the cut (S,T ), i.e.,

|f | ≤ c(S,T ).

Weak Duality (Corollary 26.5)

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

≤∑

(u,v)∈E(S,T )

f (u, v)

≤∑

(u,v)∈E(S,T )

c(u, v) = c(S,T ).

|f | = 10− 2 + 8 = 16

s

2

3

4

5 t

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10c(S,T )=10+9=19

s

2

3

4

5 t

10

102 6

4

8

9

10

10



Let f be any flow and (S,T ) be any cut. Then the value of f is boundedfrom above by the capacity of the cut (S,T ), i.e.,

|f | ≤ c(S,T ).


|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

≤∑

(u,v)∈E(S,T )

f (u, v)

≤∑

(u,v)∈E(S,T )

c(u, v) = c(S,T ).

|f | = 10− 2 + 8 = 16

s

2

3

4

5 t

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10c(S,T )=10+9=19

s

2

3

4

5 t

10

102 6

4

8

9

10

10




|f | ≤ c(S,T ).


|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

≤∑

(u,v)∈E(S,T )

f (u, v)

≤∑

(u,v)∈E(S,T )

c(u, v) = c(S,T ).

|f | = 10− 2 + 8 = 16

s

2

3

4

5 t

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10

c(S,T )=10+9=19

s

2

3

4

5 t

10

102 6

4

8

9

10

10




|f | ≤ c(S,T ).


|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

≤∑

(u,v)∈E(S,T )

f (u, v)

≤∑

(u,v)∈E(S,T )

c(u, v) = c(S,T ).

|f | = 10− 2 + 8 = 16

s

2

3

4

5 t

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10c(S,T )=10+9=19

s

2

3

4

5 t

1010

2 6

4

8

9

10

10




|f | ≤ c(S,T ).


|f | =∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

≤∑

(u,v)∈E(S,T )

f (u, v)

≤∑

(u,v)∈E(S,T )

c(u, v) = c(S,T ).

|f | = 10− 2 + 8 = 16

s

2

3

4

5 t

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10c(S,T )=10+9=19

s

2

3

4

5 t

1010

2 6

4

8

9

10

10




|f | ≤ c(S,T ).


|f | =∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

≤∑

(u,v)∈E(S,T )

f (u, v)

≤∑

(u,v)∈E(S,T )

c(u, v)

= c(S,T ).

|f | = 10− 2 + 8 = 16

s

2

3

4

5 t

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10c(S,T )=10+9=19

s

2

3

4

5 t

1010

2 6

4

8

9

10

10




|f | ≤ c(S,T ).


|f | =∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

≤∑

(u,v)∈E(S,T )

f (u, v)

≤∑

(u,v)∈E(S,T )

c(u, v) = c(S,T ).

|f | = 10− 2 + 8 = 16

s

2

3

4

5 t

10/10

6/10

2/2 6/6

0/4

8/8

8/9

6/10

10/10c(S,T )=10+9=19

s

2

3

4

5 t

1010

2 6

4

8

9

10

10



The value of the max-flow is equal to the capacity of the min-cut, that is

maxf|f | = min

S,T⊆Vc(S,T ).

Theorem


Key Lemma

The following three conditions are all equivalent for any flow f :

1. f is a maximum flow

2. There is no augmenting path in Gf3. There exists a cut (S,T ) such that c(S,T ) = |f |

Key Lemma (Theorem 26.6)

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma





Proof 3⇒ 1:

Suppose that (S,T ) is a cut with c(S,T ) = |f |By Corollary 26.5, for any flow f̃ ,

|̃f | ≤ c(S,T ) = |f |.

Hence f is a maximum flow.

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma





Proof 3⇒ 1:Suppose that (S,T ) is a cut with c(S,T ) = |f |

By Corollary 26.5, for any flow f̃ ,

|̃f | ≤ c(S,T ) = |f |.


G

s

2

3

4

5 t

10/10

9/10

0/2 6/6

3/4

7/8

9/99/10

10/10

Gf

s

2

3

4

5 t

10

10

2 6

4

8

9

10

10

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma





Proof 3⇒ 1:Suppose that (S,T ) is a cut with c(S,T ) = |f |By Corollary 26.5, for any flow f̃ ,

|̃f | ≤ c(S,T ) = |f |.Hence f is a maximum flow.

G

s

2

3

4

5 t

10/10

9/10

0/2 6/6

3/4

7/8

9/99/10

10/10

Gf

s

2

3

4

5 t

10

10

2 6

4

8

9

10

10

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma





Proof 3⇒ 1:Suppose that (S,T ) is a cut with c(S,T ) = |f |By Corollary 26.5, for any flow f̃ , |̃f | ≤ c(S,T ) = |f |.


G

s

2

3

4

5 t

10/10

9/10

0/2 6/6

3/4

7/8

9/99/10

10/10

Gf

s

2

3

4

5 t

10

10

2 6

4

8

9

10

10

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma





Proof 3⇒ 1:Suppose that (S,T ) is a cut with c(S,T ) = |f |By Corollary 26.5, for any flow f̃ , |̃f | ≤ c(S,T ) = |f |.Hence f is a maximum flow.

G

s

2

3

4

5 t

10/10

9/10

0/2 6/6

3/4

7/8

9/99/10

10/10

Gf

s

2

3

4

5 t

10

10

2 6

4

8

9

10

10

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma





Proof 1⇒ 2:

For the sake of contradicion, suppose there is an augmenting path withrespect to f .

Then we can improve f by increasing the flow along this path.

Hence f cannot be a maximum flow.

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma





Proof 1⇒ 2:For the sake of contradicion, suppose there is an augmenting path withrespect to f .

Then we can improve f by increasing the flow along this path.

Hence f cannot be a maximum flow.

G

s

2

3

4

5 t

10/10

6/10

2/2 6/6

0/4

8/88/9

6/10

10/10

Gf

s

2

3

4

5 t

10

46

2 6

4

8

18

4

6

10

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma





Proof 2⇒ 3:

Let f be a flow with no augmenting paths.

Let S be the nodes reachable from s in Gf , T := V \ S

⇒ s ∈ S, t 6∈ S.

(u, v) ∈ E(S,T )

⇒ f (u, v) = c(u, v).

(v , u) ∈ E(T ,S)

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma





Proof 2⇒ 3:Let f be a flow with no augmenting paths.


⇒ s ∈ S, t 6∈ S.

(u, v) ∈ E(S,T )

⇒ f (u, v) = c(u, v).

(v , u) ∈ E(T ,S)

G

s

2

3

4

5 t

10/10

9/10

0/2 6/6

3/4

7/8

9/9

9/10

10/10

��

10/10

9/10 ��0/21/2

Gf

s

2

3

4

5 t

10

19

2 6

13

7

1

9

1

9

10

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma







⇒ s ∈ S, t 6∈ S.(u, v) ∈ E(S,T )

⇒ f (u, v) = c(u, v).

(v , u) ∈ E(T ,S)

G

s

2

3

4

5 t

10/10

9/10

0/2 6/6

3/4

7/8

9/9

9/10

10/10

��

10/10

9/10 ��0/21/2

Gf

s

2

3

4

5 t

10

19

2 6

13

7

1

9

1

9

10

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma






Let S be the nodes reachable from s in Gf , T := V \ S ⇒ s ∈ S, t 6∈ S.

(u, v) ∈ E(S,T )

⇒ f (u, v) = c(u, v).

(v , u) ∈ E(T ,S)

G

s

2

3

4

5 t

10/10

9/10

0/2 6/6

3/4

7/8

9/9

9/10

10/10

��

10/10

9/10 ��0/21/2

Gf

s

2

3

4

5 t

10

19

2 6

13

7

1

9

1

9

10

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma






Let S be the nodes reachable from s in Gf , T := V \ S ⇒ s ∈ S, t 6∈ S.(u, v) ∈ E(S,T )

⇒ f (u, v) = c(u, v).(v , u) ∈ E(T ,S)

G

s

2

3

4

5 t

10/10

9/10

0/2 6/6

3/4

7/8

9/9

9/10

10/10

��

10/10

9/10 ��0/21/2

Gf

s

2

3

4

5 t

10

19

2 6

13

7

1

9

1

9

10

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma






Let S be the nodes reachable from s in Gf , T := V \ S ⇒ s ∈ S, t 6∈ S.(u, v) ∈ E(S,T )⇒ f (u, v) = c(u, v).

(v , u) ∈ E(T ,S)

G

s

2

3

4

5 t

10/10

9/10

0/2 6/6

3/4

7/8

9/9

9/10

10/10

��

10/10

9/10

��0/21/2

Gf

s

2

3

4

5 t

10

19

2 6

13

7

1

9

1

9

10

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma






Let S be the nodes reachable from s in Gf , T := V \ S ⇒ s ∈ S, t 6∈ S.(u, v) ∈ E(S,T )⇒ f (u, v) = c(u, v).

(v , u) ∈ E(T ,S)

G

s

2

3

4

5 t

10/10

9/10

0/2 6/6

3/4

7/8

9/9

9/10

10/10

��

10/10

9/10 ��0/21/2

Gf

s

2

3

4

5 t

10

19

2 6

13

7

1

9

1

9

10

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma






Let S be the nodes reachable from s in Gf , T := V \ S ⇒ s ∈ S, t 6∈ S.(u, v) ∈ E(S,T )⇒ f (u, v) = c(u, v).(v , u) ∈ E(T ,S)

G

s

2

3

4

5 t

10/10

9/10

0/2 6/6

3/4

7/8

9/9

9/10

10/10

��

10/10

9/10 ��0/21/2

Gf

s

2

3

4

5 t

10

19

2 6

13

7

1

9

1

9

10

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma






Let S be the nodes reachable from s in Gf , T := V \ S ⇒ s ∈ S, t 6∈ S.(u, v) ∈ E(S,T )⇒ f (u, v) = c(u, v).(v , u) ∈ E(T ,S)⇒ f (v , u) = 0.

G

s

2

3

4

5 t

10/10

9/10

6/6

3/4

7/8

9/9

9/10

10/10

��

10/10

9/10

��0/21/2

Gf

s

2

3

4

5 t

10

19

2 6

13

7

1

9

1

9

10

|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma







|f | =

∑(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma







|f | =∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Key Lemma







|f | =∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v)

= c(S,T )


Key Lemma







|f | =∑

(u,v)∈E(S,T )

f (u, v)−∑

(v,u)∈E(T ,S)

f (v , u)

=∑

(u,v)∈E(S,T )

c(u, v) = c(S,T )


Proof of the Max-Flow Min-Cut Theorem

The following conditions are equivalent for any flow f :

1. f is a maximum flow.

2. There is no augmenting path in Gf .

3. There exists a cut (S,T ) such that c(S,T ) = |f |.

Key Lemma

Max-Flow Min-Cut Theorem algorithm for computing min-cut:

Run Ford-Fulkerson until it terminates with a flow fmaxDefine S as the set of nodes reachable from s in Gf , and T = V \ S

⇒ (S,T ) is a minimal (s, t)-cut whose capacity is equal to |fmax|







Key Lemma


maxf|f | = min

S,T⊆Vc(S,T ).

Theorem










Key Lemma


maxf|f | = min

S,T⊆Vc(S,T ).

Theorem

Proof of “≤”:

For any flow f and cut (S,T ), |f | ≤ c(S,T ) (Corollary 26.5)⇒

maxf|f | ≤ min

S,T⊆Vc(S,T )










Key Lemma


maxf|f | = min

S,T⊆Vc(S,T ).

Theorem

Proof of “≤”:For any flow f and cut (S,T ), |f | ≤ c(S,T ) (Corollary 26.5)

⇒max

f|f | ≤ min

S,T⊆Vc(S,T )










Key Lemma










Key Lemma

Max-Flow Min-Cut Theorem algorithm for computing min-cut:Run Ford-Fulkerson until it terminates with a flow fmax

Define S as the set of nodes reachable from s in Gf , and T = V \ S⇒ (S,T ) is a minimal (s, t)-cut whose capacity is equal to |fmax|







Key Lemma

Max-Flow Min-Cut Theorem algorithm for computing min-cut:Run Ford-Fulkerson until it terminates with a flow fmaxDefine S as the set of nodes reachable from s in Gf , and T = V \ S



Outline

Recap





0: def FordFulkerson(G)1: initialize flow to 0 on all edges2: while an augmenting path in Gf can be found:3: push as much extra flow as possible through it

If all capacities c(u, v) are integral, then the flow at every iteration ofFord-Fulkerson is integral.

Lemma

Flow before iteration integral& capacities in Gf are integral⇒ Flow after iteration integral

For integral capacities c(u, v), Ford-Fulkerson terminates after V · Citerations, where C := maxu,v c(u, v) and returns the maximum flow.

Theorem

at the time of termination, no augmenting path⇒ Ford-Fulkerson returns maxflow (Key Lemma)


Slow Convergence of Ford-Fulkerson (Figure 26.7)

s

u

v

t

G

0/1000

0/1000

0/1000

0/1000

0/1

1/1000

0/1000

0/1000

1/1000

1/1

1/1000

1/1000

1/1000

1/1000

0/1

2/1000

1/1000

1/1000

2/1000

1/1

2/1000

2/1000

2/1000

2/1000

0/1

3/1000

2/1000

2/1000

3/1000

1/1

3/1000

3/1000

3/1000

3/1000

0/1 s

u

v

t

Gf

1000

1000

1000

1000

1

9991

1000

1000

9991

1

9991

9991

9991

9991

1

9982

9991

9991

9982

1

9982

9982

9982

9982

1

9973

9982

9982

9973

1

9973

9973

9973

9973

1

Number of iterations is at least C := maxu,v c(u, v)!

For irrational capacities, Ford-Fulkersonmay even fail to terminate!



s

u

v

t

G

0/1000

0/1000

0/1000

0/1000

0/1

1/1000

0/1000

0/1000

1/1000

1/1

1/1000

1/1000

1/1000

1/1000

0/1

2/1000

1/1000

1/1000

2/1000

1/1

2/1000

2/1000

2/1000

2/1000

0/1

3/1000

2/1000

2/1000

3/1000

1/1

3/1000

3/1000

3/1000

3/1000

0/1

s

u

v

t

Gf

1000

1000

1000

1000

1

9991

1000

1000

9991

1

9991

9991

9991

9991

1

9982

9991

9991

9982

1

9982

9982

9982

9982

1

9973

9982

9982

9973

1

9973

9973

9973

9973

1

Number of iterations is at least C := maxu,v c(u, v)!

For irrational capacities, Ford-Fulkersonmay even fail to terminate!



s

u

v

t

G

0/1000

0/1000

0/1000

0/1000

0/1

1/1000

0/1000

0/1000

1/1000

1/1

1/1000

1/1000

1/1000

1/1000

0/1

2/1000

1/1000

1/1000

2/1000

1/1

2/1000

2/1000

2/1000

2/1000

0/1

3/1000

2/1000

2/1000

3/1000

1/1

3/1000

3/1000

3/1000

3/1000

0/1

s

u

v

t

Gf

1000

1000

1000

university of cambridge · multiple sources and multiple sinks (figure 26.1) 26.1 flow networks 713...

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