LN10_MaxFlow.pdf

8/12/2019 LN10_MaxFlow.pdf

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Algorithms

Professor Chien-Mo James Li Graduate Insti tute of Electronics Engineering

National Taiwan University

Graphs (4)

Maximum Flow

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Outline

Elementary Graph Algorithms, CH22 Minimum Spanning Trees, CH23 Single Source Shortest Paths, CH24 Al l-pairs Shortest Paths, CH25 Maximum Flow, CH26*

Flow Networks, 26.1

Ford-Fulkerson Method 26.2

Edmond-Karp Algorithm

Maximum Biparti te Matching 26.3

*CH 26 is different from 2nd edition


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Flow Networks

Flow Network G = (V,E) is a directed graph Each edge (u,v) has a capacity c(u,v) 0 If (u,v)E, then c(u,v) = 0. If (u,v) E, then reverse edge (v,u) E Two special vertices: source vertex s, sink vertex t , each vertex lies on a path from source to sink

s ~> v ~> t for all vV Imagine: vertices are junctions; edges are conduit of di fferent sizes

capacity is an upper bound on the flow rate = units/time

V ~ junct ionE ~ condui t

C ~ capacity

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Flow* different from 2nd ed.

Flow f: V x V , must satisfy Capacity constraint: For all u,vV, 0 f(u,v) c(u,v) Flow conservation: For all uV - {s, t}

total flow into u = total flow out of u

Value of flow |f| = net flow out of source

Maximum f low problem Given G, s, t , and c, find a flow whose value is maximum

Example: Fig 26.1 max. value of flow= |f| = 19

4342143421

uofoutflowtotaluintoflowtotal

),(),(

=VvVv

vufuvf

=VvVv

svfvsff ),(),(

Flow/capacity


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FFT

Is max flow unique?

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Outline

Elementary Graph Algorithms, CH22 Minimum Spanning Trees, CH23 Single Source Shortest Paths, CH24 Al l-pairs Shortest Paths, CH25 Maximum Flow, CH26

Flow Networks, 26.1





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Ford-Fulkerson Method

FF method contains three concepts Residual Network

Augment Path

Cut

FORD-FULKERSON-METHOD(G,s,t)1 initialize flowf to 02 while there exists an augmenting pathp in the residual network Gf3 augment flowfalongp

4 returnf

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Residual Capacity

Given a flow fin network G =(V,E). Consider a pair of vertices u,v V Residual capacity = additional flow we can push directly from u to v

sending flow back is equivalent to decreasing the flow

Example: Fig 26.4 cf(v3, v2) = 9-4=5

cf(v2, v3) = 4, why?

No (v1, v3) , why?

.),(

,),(

0

),(

),(),(

),(otherwise

E,uvif

Evuif

uvf

vufvuc

vucf

=

f/c

G Gf

residual

capacity


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Residual Network

Residual network Gf = (V, Ef)

Similar to a flow network,

except that it may contain antiparallel edges (u, v) and (v, u) Every edge (u, v) Ef corresponds to

an edge (u, v) E, or an edge (v, u) E, or both therefore

Example: Fig. 26.4 |E| = 9

|Ef|= 15

}0),(:),{( >= vucVVvuE ff

Ef 2E

G Gf

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Augmentation of Flow, ff Given a flow fin G and a flow f in Gf

(ff)= augmentation of fby f equation 26.4

Cancellation : pushing flow on reverse edge in Gf decreases f low in G

otherwise

),(if,

0

),('),('),(),)('( oncancellatiincrease

Evuuvfvufvufvuff

+

= 4342143421

f

f/c

f f(v2, v3)

cancellation

0/9

GfFig. 26.4


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Value of |ff| (Lemma 26.1) Given a flow network G and a flow f.

Let f be a flow in Gf .

Then f f is a flow in G with value |ff | = |f|+|f | Example: Fig 26.4

|f| = 19 |f| = 4

|ff| = 19+4 =23

f

f/c

f f0/9

Gf

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Proof of Lemma 26.1(1)

prove ff is a flow so it obeys capacity constraint

why?)),,(),('(because

(26.4))equation(by

.0),('

),(),('),(

),('),('),(

),)('(

vufuvf

vuf

vufvufvuf

uvfvufvufvuff

+

+

=

=

)ofn(definitio

)constraint(capacity

e)nonnegativareflows(because

(26.4))equation(by

),(

),(),(),(

),(),(

),('),(

),('),('),(

),)('(

f

f

c

vuc

vufvucvuf

vucvuf

vufvuf

uvfvufvufvuff

+

+

+

+

=

=

=


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prove ff is a flow so it obeys flow conservation

( )

( )

),)('(

),('),('),(

),('),('),(

),('),('),(

),('),('),(

),)('(

+

+

+

+

=

=

=

=

=

Vv

Vv

VvVvVv

VvVvVv

VvVv

uvff

vufuvfuvf

vufuvfuvf

uvfvufvuf

uvfvufvufvuff

u

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prove |ff| = |f|+|f|

=

=

21 ),)('(),)('(

),)('(),)('(

'

VvVv

VvVv

svffvsff

svffvsffff

( ) ( )

+=

++

=

+

+=

++=

212121

1 1 22

21

2 22

1 11

1 2

),('),('),(),(

),('),('),('),('

),(),(

),('),('),(

),('),('),(

),('),('),(),('),('),(

VvVv

Vv Vv VvVv

VvVv

Vv VvVv

Vv VvVv

VVvVVv

Vv Vv

svfvsfsvfvsf

svfsvfvsfvsf

svfvsf

vsfsvfsvf

svfvsfvsf

vsfsvfsvfsvfvsfvsf

s V1

V2


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Contd

exercise 26.2-1

+=212121

),('),('),(),(||VvVv VVvVVv

svfvsfsvfvsff'f

'

),('),('),(),(

ff

svfvsfsvfvsfVvVvVvVv

+

+

=

=

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G Gf

Augmenting Path

Augmenting path p is a simple path from s ~> t in Gf p admits more flow along each edge

a sequence of pipes through which we can push more flow

How much more flow can we push from s to t along p? residual capacity of p

smallest residual capacity of all edges on p

(u, v) is called critical on augmenting path p if cf(u, v) = cf(p)

Example: Fig 26.4 Augmenting Path p =

cf(p) = 4 (v2, v3) is cri tical

}onis),(:),(min{)( pvuvucpcff

=

(v2, v3) is critical


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Lemma 26.2 ,corollary 26.3

(Lemma 26.2)Given G and f, let p be an augmenting path in Gf Then fp is a flow in Gf with value |fp| = cf(p) > 0

(corol lary 26.3) Given G and f, and augmenting path p in Gf ffp is a flow in G with value |ffp| = |f| + |fp| > |f|

=

otherwise

26.8eq.,onis),(if

0

)(),(

pvupcvuf

f

p

fp

f

f fp0/9

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CUT

Cut (S, T) of flow network G =(V,E) is a parti tion of V into S and T T = V-S such that sS and tT

Capacity of cut (S,T) is

Net flow across cut (S, T) is

Minimum cut of G is a cut whose capacity is minimum over all cuts Example: Fig 26.5 cut {s, v1, v2} {v3, v4, t}

capacity of cut = 12+14 = 26; net flow cross cut = 12+11-4=19

what is the minimum cut of G? what is the capacity of cut?

=

Su TvSu Tv uvfvufTSf ),(),(),(

=Su Tv

vucTSc ),(),(


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Lemma 26.4* not in exam For any cut (S, T), the net flow across cut f(S, T) = |f|

Proof

0),(),( = VvVv

uvfvuf

=

+

+=

+=

Vv SuVv Su

Vv sSuVv sSu

Vv Vv sSu sSu VvVv

uvfvuf

uvfsvfvufvsf

uvfvufsvfvsff

),(),(

),(),(),(),(

),(),(),(),(

}{}{

}{ }{

44444 344444 214444 34444 21

0126 =

+=

}s{Su VvVv

..eq

VvVv

)u,v(f)v,u(f)s,v(f)v,s(ff

flow conservation, uV {s,t}

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Lemma 26.4 (2)* not in exam Proof (contd) because V = ST, and ST =

split summation over V into summation over S and T

44444 344444 21

0

=

+=

+=

Sv SuSv SuTv SuTv Su

Tv SuSv SuTv SuSv Su

)u,v(f)v,u(f)u,v(f)v,u(f

)u,v(f)u,v(f)v,u(f)v,u(ff

),(

),(),(

TSf

uvfvuffSu TvSu Tv

=

=


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Corollary 26.5* not in exam The value of any flow capacity of any cut

Proof

Therefore, maximum flow capacity of minimum cut

),(

),(

),(

),(),(),(

TSc

vuc

vuf

uvfvufTSff

Su Tv

Su Tv

Su TvSu Tv

=

==

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Max-flow Min-Cut Theorem (1)

(Theorem 26.6) The following are equivalent:1. fis a maximum f low

2. Gf has no augmenting path

3. |f|= c(S, T) for some cut (S, T)

Proof: 12 contrapositive assume Gf has an augmenting path, and fis a maximum f low

by corol lary 26.3, ffp is a flow in G with value |f|+|fp| > |f| so f is not a maximum flow, conflict!

Proof: 31 (corollary 26.5) |f| c(S, T) so |f| = c(S, T) means fis a max flow


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Max-flow Min-Cut Theorem (2)

Proof 23 Suppose Gf has no augmenting path

Let S = {v V: there exists a path s ~> v in Gf}, T = V-S Must have t T ; otherwise there is an augmenting path

Therefore, (S, T) is a cut

Consider uS and vT If (v,u)E, then cf(u,v)= f(v,u)= 0

otherwise, cf(u,v)=f(v,u)>0 (u,v)Ef v S If (u,v)E, then cf(u,v)=0 f(u,v)= c(u,v)

otherwise, (u,v)Ef vS If both (u,v) (v,u) E, must have f(u,v)= f(v,u)= 0

therefore

),(

0),(

),(),(),(||

TSc

vuc

uvfvufTSff

Tv SuSu Tv

Tv SuSu Tv

=

=

==

S T

us tv

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Basic Ford Fulkerson

Keep augmenting f low along an augmenting path unti l there is no augmenting path

FORD-FULKERSON(G,s,t)1 for each edge (u,v) G.E // initialize2 (u,v).f= 03 while there exists a pathp froms totin the residual network Gf4 cf (p) = min {cf (u,v) : (u,v) is inp}5 for each edge (u,v) inp6 if(u,v) E7 (u,v).f= (u,v).f+ cf (p) // increase8 else (v,u).f= (v,u).fcf (p) // cancellation


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Example

Fig 26.6

initially, |f|=0

Gf

=G

GGf

fp=4

fp=?

fp=?

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Example (contd)

Fig 26.6 (contd) Q1: maximum flow =? Q2: can you find the min CUT? Q3: please debug this f igure

no more AP

fp=?

fp=?


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Time Complexity

line 3: finding Gf using BFS or DFS O(V+E) = O(E)

E = {(u,v): (u,v)E or (v,u)E} line 3~8: while loop

Assume capacities are integers. Assume max f low is f *

each i teration increase flow by at least 1

needs |f*| iterations

Time complexity = O(E |f*|)

FORD-FULKERSON(G,s,t)1 for each edge (u,v) G.E2 (u,v).f= 03 while there exists a pathp froms totin the residual network Gf4 cf (p) = min {cf (u,v) : (u,v) is inp}5 for each edge (u,v) inp6 if(u,v) E7 (u,v).f= (u,v).f+ cf (p)8 else (v,u).f= (v,u).fcf (p)

O(E)

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Problem

FF running time is NOT polynomial in input size It depends on |f*|, which is not a function of V and E

called pseudo polynomial

worst case example: Fig 26.7 need 2,000,000 times augmentations!

can we do better?

Gf GfG=Gf


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Outline

Elementary Graph Algorithms, CH22 Minimum Spanning Trees, CH23 Single Source Shortest Paths, CH24

Al l-pairs Shortest Paths, CH25 Maximum Flow, CH26 Flow Networks, 26.1


Edmond-Karp Algor ithm 26.2


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Edmonds-Karp Algorithm

Do FORD-FULKERSON, but compute augmenting paths by BFS AP are shortest paths s ~> t in Gf , with unit edge weights

time complexity O(VE2) (Theorem 26.8)

push-relabel algorithm is even better O(V2E) 26.5 26.6 *not in exam

Exercise 1 show Edmonds-Karp is better than Ford-Fulkerson


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Exercise 2

Use Edmonds-Karp to find max flow Q1: how many iterations do we need?

Q2: what is the max flow?

13

9

2016

4 7

4

Gf

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Lemma 26.7 (1)* not in exam Let f(u, v)= shortest path distance from u to v in Gf

assume unit edge weights

For all vV- {s, t}, f(u, v) increases monotonically with each flowaugmentation.

Proof by contradiction

Suppose there exists vV- {s, t} such that some flowaugmentation causes shortest path distance s ~>v to decrease

Let f= flow before the first augmentation that causes a shortest-

path distance to decrease. Let f = the flow afterward

Let v be a vertex with minimum f(s, v) whose distance wasdecreased by the augmentation

Let a shortest path s to v in Gf be s~>u v

so (u, v)Ef Because of how we chose v, we know the distance from s to u

does not decrease

(26.12)1),(),( '' = vsus ff

(26.13)),(),(' usus ff

),(),(' vsvs ff


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Lemma 26.7 (2)* not in exam Claim (u, v) Ef why? if (u, v) Ef then

contradict assumption

f(s, v) < f(s, v) How can (u, v) Ef but (u, v) Ef ?

The augmentation must have increased flow v to u

Since Edmonds-Karp augments along shortest paths, the

shortest path s to u in Gf has (v, u) as its last edge

contradict assumption f(s, v) < f(s, v) Therefore no v exist such that f(s, v) decreases

(26.12))equation(by

(26.13))inequality(by1

)inequalitytrianglethe24.10,Lemma(by1

)v,s(

)u,s(

)u,s()v,s(

'f

'f

ff

=

+

+

(26.12))equation(by2),(

(26.13))inequality(by1),(inpathshortestonedgelastis),(1),(),(

'

'

=

=

vs

us

Guvusvs

f

f

fff

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Theorem 26.8 (1)* not in exam Edmonds-Karp performs O(VE) augmentations

Proof:

Let p be an augmenting path and (u, v) is cri tical

it d isappears from residual network after augmenting along p

claim: each of the |E| edges can become crit ical |V|/2 times Consider u,vV such that either (u, v)E or (v, u)E or both when (u, v) becomes crit ical first t ime

Af ter augmentation, (u, v) disappears from residual network

(u, v) wont reappear in Gf until flow from u to v decreases, which

happens only i f (v, u) is on an augmenting path in Gf

Let f be the flow when this occurs

1),(),( += usvs ff

1),(),( '' += vsus ff


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Theorem 26.8 (2)* not in exam by Lemma 26.7

Each time, f(s, u) increases by at least 2 Initially, f(s, u) = 0, eventually, f(s, u) |V|-2

augmenting path cant have s, and t as intermediate vertices

u can become crit ical less than (|V|-2) /2 =|V|/2 - 1 times

totally, u can become crit ical less than |V|/2 times

Since O(E) pairs of vertices have an edge between them in Gf Each AP has at least 1 critical edge

total O(VE) augmentations

Use BFS to find each AP in O(E) time Edmons-Karp is O(VE2)

2),(

1),(

1),(),( ''

+=

+

+=

us

vs

vsus

f

f

ff

36NTUEEAlgori thms

Outline

Elementary Graph Algorithms, CH22 Minimum Spanning Trees, CH23 Single Source Shortest Paths, CH24 Al l-pairs Shortest Paths, CH25 Maximum Flow, CH26

Flow Networks, 26.1





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Bipartite Matching

Undirected G = (V, E) is bipartite if we can partition V = LR such that all edges go between L and R

A matching is a subset of edges ME such that for all vV , at most one edge of M is incident on v cardinality = size of M = |M|

Vertex v is matched if an edge of M is incident on it otherwise unmatched

Example: Fig 26.8

(a) cardinality =2

(b) cardinality =3

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Maximum Bipartite Matching

Maximum bipartite matching: a matching of maximum cardinality M is a maximum matching i f |M| |M| for all matching M

Appl icat ions: machine-task matching L = machines, R = tasks

edge (u, v) means machine u is capable of performing a task v

MBM find maximum number of tasks

Example (b) is MBM


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Corresponding Flow Network

Corresponding flow network G=(V, E) V= V {s, t} c(u,v) = 1 for all (u, v) E |E| = |E|+|V|

Each vertex in V has at least one incident edge, |E| |V|/2 |E| = |E| + |V| 3|E|. therefore, |E|=|E|

Idea: a flow in G correspond to a matching in G solve MBM using Ford-Fulkerson method

Example Fig 26.8c max flow = 3

MBM cardinality =3

44 344 21444 3444 2144 344 21

ts

RvtvEvuvuLuusE

toedgesedgesoriginalfromedges

}:),{(}),(:),{(}:),{(' =

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Lemma 26.9

Assume integer-valued flow : f(u,v) is integer are for all edges (u, v) If M is a matching in G, then there is an fin G with value |f| = |M| Conversely, if fis a flow in G, then there is a matching with |M| = |f|

Proof 1: M corresponds to f

if f(u,v) M, then f(s,u) = f(u,v) = f(v,t) = 1 other edges, f(u,v) = 0

(u,v) M corresponds to one unit of flow in G suvt net flow across cut (L{S}, R {T}) = |M| by Lemma 26.4, the value of the flow is |f|=|M|

Proof 2: fcorresponds to M

Let

for each uL, one unit flow enters u if and only i f one vertexvR such that f(u,v) = 1

for every matched vertex uL, we have f(u,v) = 1 net flow across cut (L{S}, R {T}) = |M| by Lemma 26.4, the value of the flow is |M|=|f|

}0),(and,,:),{( >= vufRvLuvuM


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Contd * not in exam (Theorem 26.10) If the capacity function c takes on only integral

values, then maximum flow f produced by FF method |f| is integer.

Moreover, for all vertices u and v, f(u,v) is integer

exercise: 26.3-2

(corollary 26.11) The cardinality of maximum matching M in abipartite graph G equals the value of a maximum f low fin its

corresponding flow network G

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Conclusion

How to solve MBM? create corresponding flow network G

run FF method, O(|f*| E) time

obtain the MBM

MBM Time complexity |f*| = O(V)

|E| = (E)

totally, O(VE)


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Antiparallel Edges

(v1,v2) (v2, v1) are antiparallel edges violate our assumption

how to model this ?

choose one edge, say (v1, v2) create v

replace (v1, v2) by two new edges (v1, v ) and (v , v2)

Example Fig 26.2

v10

10

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Multiple Sources and Sinks

What if more than one sources and sinks? Add a supersource s, add a supersink t

Capacity from supersource to source are Capacity from sinks to supersink are

Fig 26.3


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Reading

CH 26.1-26.3

Documents

LN10_MaxFlow.pdf