
2/87

SECTION 7.1

7. NETWORK FLOW I



!






3/87

Abstraction for material flowingthrough the edges.

Digraph G= (V,E)with source s !V and sink t !V.Nonnegative integer capacity c(e) for each e !E.

Flow network

3

s t5

15

1015

16

9

15

6

8 10

154

4 10

10

capacity

no parallel edges

no edge enters s

no edge leaves t


4/87

Def. Ast-cut (cut) is a partition (A,B)of the vertices with s!A andt!B.

Def. Its capacityis the sum of the capacities of the edges fromAtoB.

Minimum cut problem

4

5s

15

10

t

capacity = 10 + 5 + 15 = 30

cap(A,B) = c(e)

eout ofA

"


5/87


6/87

Def. Ast-cut (cut) is a partition (A,B)of the vertices with s!A andt!B.

Def. Its capacityis the sum of the capacities of the edges fromAtoB.

Min-cut problem. Find a cut of minimum capacity.

10

Minimum cut problem

6

s

10

t

capacity = 10 + 8 + 10 = 28

8

cap(A,B) = c(e)

eout ofA

"


7/87

Def. Anst-flow (flow) fis a function that satisfies:

For each e!E: [capacity]For each v!V {s, t}: [flow conservation]

7

Maximum flow problem

0 / 4

0 / 4 0 / 15

10/10

10 / 105 / 5 vs t

0/6

5/1

0

5 / 9

5 / 8

5/15

10/1010

/15

10 / 16

inflow at v = 5 + 5 + 0 = 10

outflow at v = 10 + 0 = 10

flow capacity

0 / 15

f(e)

ein to v

" = f(e)eout of v

"0 " f(e) " c(e)


8/87

8




Def. The valueof a flowf is: val(f) =

0 / 4

10/10

10 / 105 / 5s t

5/1

0

5 / 9

5 / 8

5/15

10/1010

/15

0 / 15

value = 5 + 10 + 10 = 25

0 / 4

0/6

10 / 16

0 / 15

f(e)e out ofs

" .

f(e)

ein to v

" = f(e)eout of v

"0 " f(e) " c(e)


9/87

9




Def. The valueof a flowf is: val(f) =

Max-flow problem. Find a flow of maximum value.

0 / 4

10/10

10 / 105 / 5s

8/1

0

8 / 9

8 / 8

10/1013

/15

0 / 15

value = 8 + 10 + 10 = 28

0 / 4

3/6

13 / 16

0 / 15

t

2/15

f(e)

ein to v

" = f(e)eout of v

"0 " f(e) " c(e)

f(e)e out ofs

" .


10/87

SECTION 7.1

7. NETWORK FLOW I



!






11/87

11

Towards a max-flow algorithm

Greedy algorithm.

Start withf(e) = 0for all edge e!E.Find an stpath Pwhere each edge hasf(e) < c(e).

Augment flow along path P.

Repeat until you get stuck.

s t

0 / 2

0/10 0 / 6

0 / 10

0 / 4

0/8

0 / 9

network G

0 / 10 0

value of flow

0/1

0

flow capacity


12/87


13/87

+ 2 = 10

Greedy algorithm.




13


0 / 6

0 / 4

8/8

network G

0 / 10 8

0/1

0

t

0 / 2

8/10

8 / 100 / 9

10

2

2

2

s


14/87

Greedy algorithm.




14


0 / 4

8/8

network G

10

2 / 2

10/10

10 / 10s

0 / 6

0 / 10

0/1

0

t2 / 9

6

8

6

+ 6 = 16

6


15/87

Greedy algorithm.




15


0 / 4

8/8

network G

16

2 / 2

10/10

10 / 10s

6 / 6

6 / 10

6/1

0

t8 / 9

ending flow value = 16


16/87

Greedy algorithm.




16


3 / 4

7/8

network G

19

0 / 2

10/10

10 / 10s

6 / 6

9 / 10

9/1

0

t9 / 9

but max-flow value = 19


17/87


18/87

Def. An augmenting pathis a simple stpath Pin the residual graph Gf.

Def. The bottleneck capacityof an augmenting Pis the minimum

residual capacity of any edge in P.

Key property. Letf be a flow and let Pbe an augmenting path inGf .

Thenf 'is a flow and val(f ') = val(f) + bottleneck(Gf, P).

18

Augmenting path

AUGMENT (f, c, P)___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___

b !bottleneck capacity of pathP.

FOREACHedge e!P

IF (e!E) f(e) !f(e) + b.

ELSE f(eR) !f(eR) b.

RETURN f.___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___


19/87

Ford-Fulkerson augmenting path algorithm.

Start withf(e) = 0for all edge e!E.Find an augmenting path Pin the residual graph Gf.



19

Ford-Fulkerson algorithm

FORD-FULKERSON (G, s, t, c)_______________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___

FOREACHedge e!E:f(e) !0.

Gf !residual graph.

WHILE(there exists an augmenting pathPin Gf )

f!AUGMENT (f, c,P).

Update Gf.

RETURN f.

}_______________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___


20/87

20

Ford-Fulkerson algorithm demo

s t

0 / 2

0/10 0 / 6

0 / 10

0 / 4

0/8

0 / 9

network G

0 / 10 0

value of flow

0/1

0

flow capacity

s t

2 6

10

4

9

residual graph Gf

10

residual capacity

10

10

8


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21


2 6

4

9

residual graph Gf

10

10

s t

0 / 2

0/10 0 / 6

0 / 10

0 / 4

0/8

0 / 9

network G

0 / 10 0

0/1

0

s t

10

10

8

8

8

8

+ 8 = 8


22/87

22


4

residual graph Gf

10

s t

0 / 2

8/10 0 / 6

8 / 10

0 / 4

8/8

0 / 9

network G

0 / 10 8

0/1

0

8

8

8

9s

2

2

10

2

2

+ 2 = 10

106

2

2 t


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23


4

residual graph Gf

s t

2 / 2

10/10

0 / 6

10 / 10

0 / 4

8/8

2 / 9

network G

0 / 10 10

0/1

0

8

2

2

10

10

10 7s

106

t

6

8

6

+ 6 = 16

6


24/87

24


residual graph Gf

s t

2 / 2

10/10

6 / 6

10 / 10

0 / 4

8/8

8 / 9

network G

6 / 10 16

6/1

0

8

8

10

10

1

6

6

6

4

4s

4

t

2

8

0

2

8

+ 2 = 18


25/87

25


residual graph Gf

s t

0 / 2

10/10

6 / 6

10 / 10

2 / 4

8/8

8 / 9

network G

8 / 10 18

8/1

0

8

10

10

6

8

2

2

8

1

2

s

2

t2

8

9

9

7

3

9

+ 1 = 19


26/87


27/87

SECTION 7.2

7. NETWORK FLOW I



! max-flow min-cut theorem

! capacity-scaling algorithm





28/87

Relationship between flows and cuts

Flow value lemma. Letf be any flow and let (A, B)be any cut. Then, the net

flow across (A, B)equals the value of f.

28

0 / 4

10/10

10 / 105 / 5s t

5/1

0

5 / 9

5 / 8

5/15

10/1010

/15

0 / 15

value of flow = 25

0 / 4

0/6

10 / 16

0 / 15

net flow across cut = 5 + 10 + 10 = 25

f(e)eout of A

" # f(e)ein to A

" = v(f)


29/87




29

0 / 4

10/10

10 / 105 / 5s t

5/1

0

5 / 9

5 / 8

5/15

10/1010

/15

0 / 150 / 4

0/6

10 / 16

0 / 15

net flow across cut = 10 + 5 + 10 = 25

f(e)eout of A

" # f(e)ein to A

" = v(f)

value of flow = 25


30/87




30

0 / 4

10/10

10 / 105 / 5s t

5/1

0

5 / 9

5 / 8

5/15

10/1010

/15

0 / 150 / 4

0/6

10 / 16

0 / 15

net flow across cut = (10 + 10 + 5 + 10 + 0 + 0) (5 + 5 + 0 + 0) = 25

edges from B to A

f(e)eout of A

" # f(e)ein to A

" = v(f)

value of flow = 25


31/87



Pf.

=

v#A

" f(e)eout ofv

" $ f(e)ein to v

"%

&'

(

)*

31


v(f) = f(e)eout ofs"

by flow conservation, all terms

except v = s are 0

f(e)eout of A

" # f(e)ein to A

" = v(f)

= f(e)eout ofA

" $ f(e).ein to A

" !


32/87


Weak duality. Letf be any flow and (A, B)be any cut. Then, v(f) !cap(A,B).

Pf.

32

s t

0 / 4

10/10

9 / 105 / 5

8/1

0

8 / 9

7 / 8

2/15

10/10

12/1

5

0 / 4

2/6

12 / 16

0 / 15

0 / 15

s

15

5

10

t

value of flow = 27 capacity of cut = 30

v(f) = f(e)eout of A

" # f(e)ein to A

"

flow-value

lemma

!

$ f(e)eout of A

"

$ c(e)eout of A

"

= cap(A,B) !


33/87

Max-flow min-cut theorem

Augmenting path theorem. A flowfis a max-flow iff no augmenting paths.

Max-flow min-cut theorem. Value of the max-flow = capacity of min-cut.

Pf. The following three conditions are equivalent for any flow f :

i. There exists a cut (A,B) such that cap(A, B) = val(f).

ii. f is a max-flow.

iii. There is no augmenting path with respect tof.

[ i ii ]

Suppose that (A,B)is a cut such that cap(A, B) = val(f).

Then, for any flowf', val(f') # cap(A, B) = val(f).

Thus, fis a max-flow. !

33

weak duality by assumption


34/87


Augmenting path theorem. A flowfis a max-flow iff no augmenting paths.

Max-flow min-cut theorem. Value of the max-flow = capacity of min-cut.

Pf. The following three conditions are equivalent for any flow f :

i. There exists a cut (A,B) such that cap(A, B) = val(f).

ii. f is a max-flow.

iii. There is no augmenting path with respect tof.

[ ii iii ] We prove contrapositive: ~iii ~ii.

Suppose that there is an augmenting path with respect to f.

Can improve flowf by sending flow along this path.

Thus, f is not a max-flow. !

34


35/87

[ iii i ]

Letfbe a flow with no augmenting paths.LetAbe set of nodes reachable from sin residual graph Gf.

By definition of cutA, s!A.

By definition of flow f, t$A.

35


v(f) = f(e)eout of A

" # f(e)ein to A"

original network G

s

t

A B

flow-value

lemma

edge e = (v, w) with v !A, w !B

must have f(e) = c(e)

edge e = (v, w) with v !B, w !A

must have f(e) = 0

= c(e)eout of A

"

= cap(A,B)


36/87

SECTION 7.3

7. NETWORK FLOW I



! max-flow min-cut theorem

! capacity-scaling algorithm





37/87

37

Running time

Assumption. Capacities are integers between 1and C.

Integrality invariant. Throughout the algorithm, the flow valuesf(e)

and the residual capacities cf(e)are integers.

Theorem. The algorithm terminates in at most val(f*) ! nCiterations.

Pf. Each augmentation increases the value by at least 1. !

Corollary. The running time of Ford-Fulkerson is O(mnC).

Corollary. If C= 1, the running time of Ford-Fulkerson is O(mn).

Integrality theorem. Then exists a max-flow f*for which everyflow valuef*(e)is an integer.

Pf. Since algorithm terminates, theorem follows from invariant. !


38/87

Q. Is generic Ford-Fulkerson algorithm poly-time in input size?

A. No. If max capacity is C, then algorithm can take " Citerations.

s%v%w%t

s%w%v%t

s%v%w%t

s%w%v%t

s%v%w%t

s%w%v%t

Bad case for Ford-Fulkerson

38

s

t

w

v

1

m, n, and log C

each augmenting path

sends only 1 unit of flow

(# augmenting paths = 2C)

C

C

CC


39/87

39

Choosing good augmenting paths

Use care when selecting augmenting paths.

Some choices lead to exponential algorithms.Clever choices lead to polynomial algorithms.

If capacities are irrational, algorithm not guaranteed to terminate!

Goal. Choose augmenting paths so that:Can find augmenting paths efficiently.

Few iterations.


40/87

40

Choosing good augmenting paths

Choose augmenting paths with:

Max bottleneck capacity.Sufficiently large bottleneck capacity.

Fewest number of edges.

Theoretical Improvements in Algorithmic Efficiency

for Network Flow Problems

J A C K E D M O N D S

University of Waterloo, Waterloo, Ontario, Canada

A N D

R I C H A R D M K A R P

University of California, Berkeley, California

A BSTRACT. Th i s p ap e r p r e sen t s n ew a l g o r i t h m s f o r t h e m ax i m u m f l o w p r o b l em , t h e H i t ch co ck

t r a n s p o r t a t i o n p r o b l e m , a n d t h e g e n e r a l m i n i m u m - c o s t f lo w pr o b l e m . U p p e r b o u n d s o n t h e

n u m b e r s o f s t e p s i n t h e s e a l g o r i t h m s a r e d e r i v e d , a n d a r e s h o w n t o c o m p a l e f a v o r a b l y w i t h

u p p e r b o u n d s o n t h e n u m b e r s o f s t e p s r e q u i r e d b y e a r l i e r a l g o r i th m s .

F i r s t , t h e p a p e r s t a t e s t h e m a x i m u m f lo w p ro b l e m , g i v e s t h e F o r d - F u l k e r s o n l a b e l i n g m e th o d

f o r i t s so l u t i o n , an d p o i n t s o u t t h a t a n i m p r o p e r ch o i ce o f fl o w au g m en t i n g p a t h s can l ead t o

sev e r e co m p u t a t i o n a l d i f f i cu l t i e s . Th en r u l e s o f ch o i ce t h a t av o i d t h ese d i f f i cu l t i es a r e g i v en .

W e s h o w t h a t , i f e a c h f lo w a u g m e n t a t i o n i s m a d e a l o n g a n a u g m e n t i n g p a t h h a v i n g a m i n i m u m

n u m b e r o f a r c s , t h e n a m a x i m u m f lo w in a n n - n o d e n e t w o r k w i l l b e o b ta i n e d a f t e r n o m o r e t h a n

~( na - n ) au g m en t a t i o n s ; an d t h en w e sh o w t h a t i f each f l o w ch an g e i s ch o sen t o p r o d u ce a

m ax i m u m i n c r ea se i n t h e fl o w v a l u e t h e n , p r o v i d e d t h e cap ac i t i e s a r e i n t eg r a l , a m ax i m u m f l o w

w i l l b e d e t e r m i n ed w i t h i n a t m o s t 1 + l og M/( M-- 1) if t, S) a u g m e n t a t i o n s , wheref* t, s) is the

v a l u e o f t h e m ax i m u m f l o w an d M is t h e m ax i m u m n u m b er o f a r c s ac r o ss a cu t .

N e x t a n e w a l g o r i t h m i s g i v e n f o r t h e m i n i m u m - c o s t f l o w p r o b l e m , i n w h i c h a l l s h o r t e s t - p a t h

c o m p u t a t i o n s a r e p e r f o r m e d o n n e t w o r k s w i t h a l l w e i g h t s n o n n e g a t i v e . I n p a r t i c u l a r , t h i s

a l g o r i t h m s o l v e s th e n X n a s s i g m n e n t p r o b l e m i n

O n 3)

s t ep s . F o l l o w i n g t h a t w e ex p l o r e a

s c a l i n g t e c h n i q u e f o r s o l v i n g a m i n i m u m - c o s t f lo w p r o b l e m b y t r e a t i n g a s e q u e n c e o f d e r i ve d

p r o b l e m s w i t h s c a l e d d o w n c a p a c i t ie s . I t i s s h o w n t h a t , u s i n g t h i s t e c h n i q u e , t h e s o l u ti o n o f

a I i i t c h c o c k t r a n s p o r t a t i o n p r o b l e m w i t h m s o u r ce s a n d n s i n k s , m ~ n , a n d m a x i m u m f lo w B ,

r eq u i r e s a t m o s t ( n + 2 ) l o g 2

B/n)

f l o w au g m en t a t i o n s . S i m i l a r r e su l t s a r e a l so g i v en f o r t h e

g en e r a l m i n i m u m - co s t fl o w p r o b l em .

A n a b s t r a c t s t a t i n g t h e m a i n r e s u l t s o f t h e p r e s e n t p a p e r w a s p r e s e n t e d a t t h e C a l g a r y

I n t e r n a t i o n a l C o n f e r e n c e o n C o m b i n a t o r i a l S t r u c t u r e s a n d T h e i r A p p l i c a t i o n s, J u n e 1 96 9.

I n a p ap e r b y l ) i n i c ( 1 97 0 ) a r e su l t c l o se l y r e l a t ed t o t h e m a i n r e su l t o f S ec t i o n 1 . 2 is o b t a i n ed .

Edmonds-Karp 1972 (USA) Dinic 1970 (Soviet Union)


41/87

41

Capacity-scaling algorithm

Intuition. Choose augmenting path with highest bottleneck capacity:

it increases flow by max possible amount in given iteration.Don't worry about finding exact highest bottleneck path.

Maintain scaling parameter #.

Let Gf(#)be the subgraph of the residual graph consisting only of

arcs with capacity " #.

Gf

t

s

1

122

102

170

110

Gf ("), "= 100

t

s

122

102

170

110


42/87

42

Capacity-scaling algorithm

CAPACITY-SCALING(G, s, t, c)_________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____

FOREACHedge e!E:f(e) !0.

" !largest power of 2 # C.

WHILE(" $ 1)

Gf (") !"-residual graph.

WHILE(there exists an augmenting pathPin Gf ("))

f!AUGMENT (f, c,P).

Update Gf (").

"!

"

/ 2.

RETURN f._________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ___________________________________________ ____________________________________________ ___________________________________________ ____


43/87

43

Capacity-scaling algorithm: proof of correctness

Assumption. All edge capacities are integers between 1and C.

Integrality invariant. All flow and residual capacity values are integral.

Theorem. If capacity-scaling algorithm terminates, then fis a max-flow.

Pf.

By integrality invariant, when #= 1 & Gf (#) = Gf .Upon termination of #= 1phase, there are no augmenting paths. !


44/87

44

Capacity-scaling algorithm: analysis of running time

Lemma 1. The outer while loop repeats 1 + 'log2 C(times.

Pf. Initially C/2 n1/2.

Level graph has more than n1/2levels.

Let 1 ! h! n1/2be layer with min number of nodes: |Vh | ! n1/2.

86

p y p y

VhV0 Vn1/2V1

level graph LG for flow f

Unit-capacity simple networks: analysis


87/87

LEMMA 2. After at most n1/2phases, |f| " |f*| n1/2.

After n1/2phases, length of shortest augmenting path is > n1/2.

Level graph has more than n1/2levels.

Let 1 ! h! n1/2be layer with min number of nodes: |Vh | ! n1/2.

LetA = {v:(v) < h} "{v:(v) = handvhas!1 outgoing residual edge}.

capf (A,B) ! |Vh | ! n1/2 & |f| " |f*| n1/2. !

residual edgesresidual graph Gf

A

07 network flow i

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