7. N F I - Princeton University Computer Science · Ford–Fulkerson algorithm demo 2 s t 0 / 10 0 / 2 0 / 6 0 / 10 0 / 4 0 / 8 0 / 9 network G and ﬂow f 0 / 10 0 value of flow

Lecture slides by Kevin WayneCopyright © 2005 Pearson-Addison Wesley

http://www.cs.princeton.edu/~wayne/kleinberg-tardos

Last updated on 4/17/18 9:07 AM

7. NETWORK FLOW I

‣ Ford–Fulkerson demo

‣ exponential-time example

‣ pathological example

7. NETWORK FLOW I




SECTION 7.1

Ford–Fulkerson algorithm demo

3

s t

0 / 20 /

10 0 / 6

0 / 10

0 / 4

0 / 8

0 / 9

network G and flow f

0 / 10 0

value of flow

0 / 10

flow capacity

s t

2 6

10

4

9

residual network Gf

10

residual capacity

10 10 8


4

2 6

4

9

residual network Gf

10

10

s t

0 / 20 /

10 0 / 6

0 / 10

0 / 4

0 / 8

0 / 9


0 / 10 0

0 / 10

s t

10

10

8

8—

8—

8—

+ 8 = 8


5

4residual network Gf

10

s t

0 / 28 /

10 0 / 6

8 / 10

0 / 4

8 / 8

0 / 9


0 / 10 8

0 / 10

8

8

8

9s

22

—10 2 —

2— + 2 = 10

10 6

10—

2 t


6

4residual network Gf

s t

2 / 210 /

100 / 6

10 / 10

0 / 4

8 / 8

2 / 9


0 / 10 10

0 / 10

8

2

2

10

10

10 7s

10 6

t

6 —

8—

6— + 6 = 16

6—


7

residual network Gf

s t

2 / 210 /

106 / 6

10 / 10

0 / 4

8 / 8

8 / 9


6 / 10 16

6 / 10

8

8

10

10

1

6

6

6

4

4s

4

t

2

8—

0 —

2—

8—

+ 2 = 18

fixes mistake from second augmenting path


8

residual network Gf

s t

0 / 210 /

106 / 6

10 / 10

2 / 4

8 / 8

8 / 9


8 / 10 18

8 / 10

8

10

10 6

8

2

2

8

1

2

s

2

t2

8

9—

9—

7—

3—

9—

+ 1 = 19


9

residual network Gf

s t

0 / 210 /

106 / 6

10 / 10

3 / 4

7 / 8

9 / 9


9 / 10 19

9 / 10

10

10 6

9

2

3

9

9

1

s

1

t1

7

1

nodes reachable from s

min cutvalue of max flow

capacity = 10 + 9 = 19

7. NETWORK FLOW I




SECTION 7.1

Ford–Fulkerson algorithm: exponential-time example

Bad news. Number of augmenting paths can be exponential in input size.

11

v w

t

s

0

1

00

0capacity

flow

100

100

100

100

0

initialize with 0 flow



12

0

00

1

0

0

1st augmenting path

1

1

1

100

100

100

100

t

s

v w



13

1

1

1

2nd augmenting path

0

0

1

1

1

0

100

100

100

100

t

s

v w



14

1

11

1

1

0

3rd augmenting path

2

2

1

100

100

100

100

t

s

v w



15

2

2

1

4th augmenting path

1

1

1

2

2

0

100

100

100

100

t

s

v w



16

. . .



17

99

1

99

0

199th augmenting path

100

100

1

99

99

100

100

100

100

t

s

v w



18

100

100

1

200th augmenting path

99

99

1

100

100

0

100

100

100

100

t

s

v w



19

0

100

100

1

100

100

100

10010

0

100

t

s

v w

7. NETWORK FLOW I




SECTION 7.1

Ford–Fulkerson algorithm: pathological example

Intuition. Let r > 0 satisfy r2 = 1 – r.

・Initially, some residual capacities are 1 and r.

・After two augmenting paths, some residual capacities are r and r2.

・After two more augmenting paths, some residual capacities are r2 and r3.

・After two more, some residual capacities are r3 and r4.

・By carefully choreographing the augmenting paths,infinitely many residual capacities arise!

21

1 – r

r – r2

r2 – r3

r =

�5 � 1

2=� r2 = 1 � r

r � 0.618 =� r4 < r3 < r2 < r < 1


22

v

u

w

xs 0 / C

0 / C 0 / r

0 / 1

0 / 1

t

0 / C

0 / C

0 / C

0 / Cw

x

v

us

t

C sufficiently large that it won’t ever be a bottleneck (we'll suppress)

flow network G

r =

�5 � 1

2=� r2 = 1 � r


23

v

u

w

xs

0 / r

augmenting path 1: s→w→v→t (bottleneck capacity = 1)

0 / 1

t

w

x

v

us

t

0 / 1—

1

r =

�5 � 1

2=� r2 = 1 � r


24

v

u

w

xs

augmenting path 2: s→u→v→w→x→t (bottleneck capacity = r)

t

w

x

v

us

t

0 / r

0 / 1

1 / 1—

1 – r

—r = 1

– r2

—r

r =

�5 � 1

2=� r2 = 1 � r


25

v

u

w

xs

augmenting path 3: s→w→v→u→t (bottleneck capacity = r)

t

w

x

v

us

t

r / r

1 – r

2 / 1

1 – r / 1

1

—

0

r =

�5 � 1

2=� r2 = 1 � r


26

v

u

w

xs

augmenting path 4: s→u→v→w→x→t (bottleneck capacity = r2)

t

w

x

v

us

t

0 / r

1 / 1—

1 – r2

—r2 =

r – r3

1 – r

2 / 1

1

r =

�5 � 1

2=� r2 = 1 � r


27

v

u

w

xs

augmenting path 5: s→x→w→v→t (bottleneck capacity = r2)

t

w

x

v

us

t

1 / 1

1 – r2 / 1

—1 – r

2

1

r – r3 /

r

r =

�5 � 1

2=� r2 = 1 � r


28

v

u

w

xs


t

w

x

v

us

t

1 – r

2 / 1

1 / 1—

1 – r3

1 – r

2 + r3 =

1 –

r4

r – r3 /

rr

r =

�5 � 1

2=� r2 = 1 � r


29

v

u

w

xs

augmenting path 7: s→w→v→u→t (bottleneck capacity = r3)

t

w

x

v

us

t

r / r

1 – r3 / 1

1

—r –

r3

1 – r

4 / 1

r =

�5 � 1

2=� r2 = 1 � r


30

v

u

w

xs


t

w

x

v

us

t

r – r3 /

r

1 / 1—

1 – r4

1

1 – r

4 / 1

r – r3 +

r4 =

r –

r5

r =

�5 � 1

2=� r2 = 1 � r


31

v

u

w

xs

augmenting path 9: s→x→w→v→t (bottleneck capacity = r4)

t

w

x

v

us

t

1 / 1

—1

– r4

1

1 – r4 / 1

r – r5 /

r

r =

�5 � 1

2=� r2 = 1 � r


32

v

u

w

xs

flow after augmenting path 5: { r - r3, 1, 1 - r2 } (value of flow = 1 + 2r + 2r2)

t

w

x

v

us

t

1 / 1

flow after augmenting path 1: { r - r1, 1, 1 - r0 } (value of flow = 1)

1 – r

4 / 1

r – r5 /

r

flow after augmenting path 9: { r - r5, 1, 1 - r4 } (value of flow = 1 + 2r + 2r2 + 2r3 + 2r4)

r =

�5 � 1

2=� r2 = 1 � r


Theorem. The Ford–Fulkerson algorithm may not terminate; moreover, it

may converge to a value not equal to the value of the maximum flow.

Pf.

・After (1 + 4k) augmenting paths of the form just described,the value of the flow

・Value of maximum flow = 2C + 1. ▪

33

= 1 + 22k�i=1

ri

� 1 + 2��

i=1ri

= 3 + 2r

< 5

= 1 + 22k�i=1

ri

� 1 + 2��

i=1ri

= 3 + 2r

< 5

= 1 + 22k�i=1

ri

� 1 + 2��

i=1ri

= 3 + 2r

< 5

r =

�5 � 1

2=� r2 = 1 � r

= 1 + 22k�i=1

ri

� 1 + 2��

i=1ri

= 1 +2r

1 � r

= 3 + 2r

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Reference

34

ELSEVIER Theoretical Computer Science 148 ( 1995) 165-I 70

Theoretical Computer Science

Note The smallest networks on which the Ford-Fulkerson

maximum flow procedure may fail to terminate

Uri Zwick’ Department of Computer Science. Tel Aviv University, Rnmat Ah. 69978 Tel Aviv, Israel

Received November I993 Communicated by M. Nivat

Abstract

It is widely known that the Ford-Fulkerson procedure for finding the maximum flow in a network need not terminate if some of the capacities of the network are irrational. Ford and Fulkerson gave as an example a network with 10 vertices and 48 edges on which their procedure may fail to halt. We construct much smaller and simpler networks on which the same may happen. Our smallest network has only 6 vertices and 8 edges. We show that it is the smallest example possible.

1. Introduction

The maximal flow problem is one of the most fundamental combinatorial optimiza- tion problems. The Ford-Fulkerson augmenting paths procedure is perhaps the most basic method devised for solving it and many more advanced algorithms are based on it.

Ford and Fulkerson themselves point out that their procedure need not terminate if the network it is applied on has some irrational capacities. In their book [3], they describe a network with 10 vertices and 48 edges on which this may happen. Their network is quite complicated and most textbooks (see, e.g., [ 1,2,4-6,8]) that describe their procedure do not present it. A variant of their example appears in [7], it has 14 vertices and 28 edges. We are not aware of any simpler example that had appeared in the literature.

In this note we describe three much smaller and simpler networks, on which the Ford-Fulkerson procedure may fail to terminate. The first two networks contain only 6 vertices and 9 edges each. The third network is yet smaller containing only 6 vertices and 8 edges. All three networks are acyclic and planar. The first two are planar and

* Email: [email protected].

0304-3975/95/$09.50 @ 1995 -EElsevier Science B V. All rights reserved SSDI 0304-3975(95)00022-4

Documents

7. N F I - Princeton University Computer Science · Ford–Fulkerson algorithm demo 2 s t 0 / 10 0 / 2 0 / 6 0 / 10 0 / 4 0 / 8 0 / 9 network G and ﬂow f 0 / 10 0 value of flow