Flow Network,Ford Fulkerson Method Notes

8/14/2019 Flow Network,Ford Fulkerson Method Notes

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% # ' / 3 $ % 3 0 . 3 # ) 2 % 3 . ( & $ L / ' + _ 3 ' \ 2 ( Y

X

= ' / $ ` 1 2 K . 5 $ ( 3 5 6 = 3 # $ ] ^ \ ^ ] ' L \ 2 ( . 7 / $ . ) 3 # , > / % 3 % $ . / & # 3 / $ $ ' L Y

X

/ ' ' $ ) . _ H

a / > / % ' % $ / K . ' ( % # . # $ % $ $ K $ % & . ( ' ( ( & / $ . % $ ' K $ / + $ H

8 8 ] ^ \ ^ ]

X b D

Q \ S c ] ^ \ ^ ]

X

Q \ S d e f g h h i j f k l m j n \ g o p l o k j q j f n X r

! 4 3 3 s " # $ & 5 . 2 + 2 % 3 / 2 K 2 . 5 L ' / \ [ _ = % 2 ( & $ ] ^ \ ^ ]

X

Q _ S [ G L ' / . 5 5 X H O # ' ' % $ . ( . / 7 2 3 / . / 6 K $ / 3 $ ; \ t [ _ = . ( )

5 $ 3 _ u v v v u w u \ 7 $ . % # ' / 3 $ % 3 0 . 3 # L / ' + _ 3 ' \ 2 ( Y

X b D

H B Z L 3 # $ / $ 2 % ( ' % 1 & # 0 . 3 # = 3 # $ ( ] ^ \ ^ ]

X b D

Q \ S [ x =

. ( ) 4 $ N / $ ) ' ( $ H C y $ & . 1 % $ 3 # 2 % 2 % . % # ' / 3 $ % 3 0 . 3 # = 4 $ # . K $ ] ^ \ ^ ]

X b D

Q \ S [ ] ^ \ ^ ]

X b D

Q w S b D = . ( ) 3 # $ 2 ( ) 1 & 3 2 K $

# ' # $ % % + $ % # . ] ^ \ ^ ]

X b D

Q w S c ] ^ \ ^ ]

X

Q w S H

z $ ( ' 4 # . K $ 3 4 ' & . % $ % 3 ' & ' ( % 2 ) $ / H Z L w u \ 2 % . ( $ ) : $ 2 ( Y

X

= 3 # $ ( ] ^ \ ^ ]

X

Q \ S { ] ^ \ ^ ]

X

Q w S b D = 7 $ & . 1 % $

# $ $ K $ % . / $ ) $ > ( $ ) / $ . ) # , > / % / . K $ / % . H

a ( 3 # $ ' 3 # $ / # . ( ) = 2 L w u \ 2 % ( ' 3 . ( $ ) : $ 2 ( Y

X

= 3 # $ ( \ u w + 1 % 3 7 $ . ( $ ) : $ 2 ( 3 # $ X 3 # . 1 : + $ ( 3 2 ( : 0 . 3 # H

" # 1 % = \ u w + 1 % 3 5 2 $ ' ( 3 # $ % # ' / 3 $ % 3 0 . 3 # L / ' + _ 3 ' | 2 ( Y

X

= 4 # 2 & # 2 + 0 5 2 $ % 3 # . 3 ] ^ \ ^ ]

X

Q \ S [ ] ^ \ ^ ]

X

Q w S } D {

] ^ \ ^ ]

X

Q w S b D H

Z ( ' # & . % $ % = $ # . K $ ] ^ \ ^ ]

X b D

Q \ S [ ] ^ \ ^ ]

X b D

Q w S b D c ] ^ \ ^ ]

X

Q w S b D c ] ^ \ ^ ]

X

Q \ S H ~

z # $ ( $ K $ / 4 $ . 1 : + $ ( 3 3 # $ < ' 4 = 3 # $ 7 ' 3 3 5 $ ( $ & 9 $ ) : $ 2 ( 3 # $ . 1 : + $ ( 3 2 ( : 0 . 3 # ) 2 % . 0 0 $ . / % L / ' + 3 # $

/ $ % 2 ) 1 . 5 : / . 0 # = . ( ) % ' + $ ' 3 # $ / $ ) : $ 2 ( 3 # $ ^ \ ^ _ ] ' L 3 # $ . 1 : + $ ( 3 2 ( : 0 . 3 # + . 6 B / $ , C . 0 0 $ . / H a 1 / % $ & ' ( )

' % $ / K . ' ( % # . . ( $ ) : $ & . ( ( ' . $ . / ' / ) % . $ . / ' ' + . ( + $ % H

8 8 f l o q k j j j m k l e o e d k j l o l k n p e o p n g f g h q e f l k g o j p q j w u \ p l n g j g f n d f e

k j f j n l p g h q f g Y

W

g k e n k I k l j n r

! 4 3 3 s J 1 0 0 ' % $ w u \ 2 % 2 ( 3 4 ' / $ % 2 ) 1 . 5 : / . 0 # % Y

X

. ( ) Y

b D

= 7 1 3 ( ' 3 2 ( . ( 6 ' L 3 # $ 2 ( 3 $ / + $ ) 2 . 3 $ / $ % 2 ) 1 . 5

: / . 0 # % Y X b D

= H H H = Y

= L ' / % ' + $ X H " # $ ( w u \ + 1 % 3 7 $ 2 ( 3 # $ X 3 # . 1 : + $ ( 3 2 ( : 0 . 3 # = % ' ] ^ \ ^ ] X

Q \ S [



Maximum Flow Networks

CONTENTS

• Network flows on directed ac clic ra hs

• Ford‐Fulkerson Algorithms

‐Residual networks

• Min cut Max flow theorem

• Max Bipartite matching

Network flows on directed acyclic graphs• DAG is a directed graph with no cycles.

• Flow network: a flow network G = (V,E) is a directed graphin which each edge (u,v) E has a nonnegative capacityc(u,v) >= 0. if (u,v) E , we assume that c(u,v) = 0.

We distinguish two vertices in flow network , source s andsink t.

• The source produces the material at a steady rate .e s n consumes e ma er a a a s ea y ra

Objective : how much of material can be imposed throughnetwork that network can handle

• Flow:A flow in G is a real-valued function f : V × V→ R thatsatisfies three properties:

1. Capacity constraint : For all u,v V, we require f(u,v) ≤ c(u,v).The net flow from one vertex to another must not exceed the givencapacity.

2. Skew symmetry : For all u,v V, we require f(u,v) = -f(v,u).

The net flow from a vertex u to a vertex v is the negative of the netflow in the reverse direction

3. Flow conservation : For all u V - {s,t}, we require∑ f(u,v) = 0.v V

The quantity f(u,v),which can be positive or negative , is called thenet flow from vertex u to v.The value of the flow is defined as | f | = ∑ f(s,v) that is the totat

net Flow out of the source. v V

Residual Networks

• Given a flow network and a flow, residual network consists of edges that can admit more flow.

• before exceeding the capacity c(u,v) is the residual capacity of

(u,v), given by

Cf (u,v) - f(u,v)

• Given a flow network G=(V,E) and a flow f, the residual network of G induced by f is Gf =(V,E) and a flow f, the ResidualNetwork of G

Ef ={(u,v) V×V : cf (u,v)>0}

That is, each edge of residual network , or residual edge, canadmit a flow that is greater than 0.



Augmenting Paths

• Given a flow network and a flow f, an augmenting

network Gf .

• Residual capacity of path p is defined as the

maximum amount by which we can increase the flow

on each edge of p, given by

cf p =m n cf u,v : u,v s on p

• Cut(S,T) of flow network G=(V,E) is a partition of V

- suc .

• If f is a flow, then the net flow across the cut(S,T) is

defined to be f(S,T).

• The capacity of the cut (S,T) is c(S,T)

• A minimum cut of a network is a cut whose capacity

is minimum over all cuts of the network

s t

v1 v311/16 15/20

101/4

4/9

7/7

12/12

v4v28/13 11/14

4/4

•The net flow across (S,T) is f(s,t)=19, and the capacity is c(S,T)=26

• e ne ow across cu s

f(v1,v3) + f(v2,v3) + f(v2,v4) = 12+(-4)+11= 19

•Capacity is

c(v1,v3)+c(v2,v4) = 12 + 14 = 26

Ford-Fulkerson-Method(G,s,t)

1. Initialize flow f to 0

2. W e t ere ex sts an augment ng pat p

3. Do augment flow f along p

4. Return f



Ford Fulkerson algorithm• The Ford Fulkerson method is iterative,

• Starts with f(u,v) for (u,v) V, initial flow of value 0.

• The method is based on the augmenting path which is

defined as a path from s to t along which we can push

more flow and then augment flow along this path.

Procedure Ford_Fulkerson_method(G,s,t):

1. For each edge (u,v) E[G]

. ,

3. f[v,u] 0

4. While there exists path p from s to t in residual network Gf

5. do cf (p)min{ cf (u,v):(u,v) is in p}

6. for each edge (u,v) in p

7. do f u v f u v + c. , ,

8. f[v,u] -f[u,v]

• Running time for this algorithms is O(E * |f*|) where f* is the

maximum flow found by algorithm.

• First three Lines take time θ(E).

• The while loop of last five lines is executed at most |f*| times,

since the flow value increases by at least one unit in each

iteration.

s t

v1 v216

12

20

104

9

7

I/P Residual network

v4v313

14

4

v1 v24/16

4/12

20

s t

v4v3

13 4/14

4/4

104

4/9

7

Flow network





Max flow min cut theorem

• This theorem tell us that a flow is max. if and only if its residual

network contains no augmenting path.

Theorem states that ,

If f is a flow in a flow network, G = (V,E) with the source and sink t

then, the following conditions are equivalent:

1. F is a max. flow in G.

2. The residual network G1 contains no augments paths.

3. |F| = c(S,T) for some cut (S,T) of G.

Maximum Bipartite Matching

• Given an undirected graph G=(V,E), matching is a subset of

edges M is subset of E such that for all vertices v ∈ V, at most

one edge of M is incident on v.

• We say that a vertex v ∈V is matched by matching M if some

edge in M is incident on v; otherwise, v is unmatched.

• A maximum matching is a matching of maximum cardinality,

that is, a matching M such that for any matching M’, we have|M|>= |M’|.

A bipartite graph G=(V,E) with vertex partition V= L U R

A matching with

Cardinality 2

A matching with

Cardinality 3

• We can solve the maximum match finding problemusing flow networks.

• Put source and sink on the graph such that sourceconnects a t e vert ces on e t s e o t e grap

with single edge to each vertex and do the same forthe sink. Now assign weight 1 to all the edges in thegraph.

• Run ford Fulkerson al orithm to find max flow.

• Max. cardinality = Max flow

Maximum Flow and Minimum Cut



Princeton University • COS 226 • Algorithms and Data Structures • Spring 2004 • Kevin Wayne • http://www.Princeton.EDU/~cos226

Max Flow, Min Cut

Minimum cut

Maximum flow

Max-flow min-cut theorem

Ford-Fulkerson augmenting path algorithm

Edmonds-Karp heuristics

Bipartite matching

2

n Network reliability.

n Security of statistical data.

n Distributed computing.

n Egalitarian stable matching.

n Distributed computing.

n

Many many more . . .

Max flow and min cut.

n Two very rich algorithmic problems.

n Cornerstone problems in combinatorial optimization.

n Beautiful mathematical duality.

Nontrivial applications / reductions.

n Network connectivity.

n Bipartite matching.

n Data mining.

n Open-pit mining.

n Airline scheduling.

n

Image processing.n Project selection.

n Baseball elimination.

3

Soviet Rail Network, 1955

Source: On the history of the transportation and maximum flow problems .Alexander Schrijver in Math Programming, 91: 3, 2002.

4

Network: abstraction for material FLOWING through the edges.

n Directed graph.

n Capacities on edges.

n Source node s, sink node t.

Min cut problem. Delete "best" set of edges to disconnect t from s.

Minimum Cut Problem

capacity

s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10154

4

sinksource

Cuts Cuts



5

A cut is a node partition (S, T) such that s is in S and t is in T.

n capacity(S, T) = sum of weights of edges leaving S.

s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10154

4

Capacity = 30

S

6



s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10154

4S

Capacity = 62

7



Min cut problem. Find an s-t cut of minimum capacity.

Minimum Cut Problem

s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10154

4S

Capacity = 28

8

Network: abstraction for material FLOWING through the edges.

n Directed graph.

n Capacities on edges.

n Source node s, sink node t.

Max flow problem. Assign flow to edges so as to:n Equalize inflow and outflow at every intermediate vertex.

n Maximize flow sent from s to t.

Maximum Flow Problem

same input as min cut problem

s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10154

4capacity

sinksource

Flows Flows



9

A flow f is an assignment of weights to edges so that:

n Capacity: 0 £ f(e) £ u(e).

n Flow conservation: flow leaving v = flow entering v.

s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10154

4

4

0

0

0

0 0

0 4 4

0

4 0

00

Value = 4

except at s or t

0

capacityflow

10

A flow f is an assignment of weights to edges so that:

n Capacity: 0 £ f(e) £ u(e).

n Flow conservation: flow leaving v = flow entering v.

s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10154

4

10

6

6

11

1 10

3 8 8

0

4 0

00

Value = 2411

capacityflow

except at s or t

11

Max flow problem: find flow that maximizes net flow into sink.

Maximum Flow Problem

s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10154

4

10

9

9

14

4 10

4 8 9

1

0 0

00

Value = 2814

capacityflow

12

Observation 1. Let f be a flow, and let (S, T) be any s-t cut. Then, thenet flow sent across the cut is equal to the amount reaching t.

Flows and Cuts

s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10154

4

10

6

6

0 10

4 8 8

0

4 0

00S 10

10 Value = 24

Flows and Cuts

Flows and Cuts



13

10

6

6

10 0 10

4 8 8

0

4 0

0


s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10154

4

S

0

10 Value = 24

14

10

6

6

10

0 10

4 8 8

0

4 0

0


s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10154

4 0

S10

Value = 24

15

Observation 2. Let f be a flow, and let (S, T) be any s-t cut. Then thevalue of the flow is at most the capacity of the cut.

Flows and Cuts

s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10154

4S

Cut capacity = 30 Þ Flow value £ 30

16

Max Flow and Min Cut

Observation 3. Let f be a flow, and let (S, T) be an s-t cut whose capacityequals the value of f. Then f is a max flow and (S, T) is a min cut.

Cut capacity = 28 Þ Flow value £ 28

Flow value = 28

s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10154

4

10

9

9

15

4 10

4 8 9

1

0 0

00S 15

Max-Flow Min-Cut Theorem Towards an Algorithm



17

Max-flow min-cut theorem. (Ford-Fulkerson, 1956): In any network,the value of max flow equals capacity of min cut.

n Proof IOU: we find flow and cut such that Observation 3 applies.

Min cut capacity = 28Û

Max flow value = 28

s

2

3

4

5

6

7

t

15

5

30

15

10

8

15

9

6 10

10

10154

4

10

9

9

15

4 10

4 8 9

1

0 0

00S 15

18

Find s-t path where each arc has f(e) < u(e) and "augment" flow along it.

s

4

2

5

3 t

4

0 0

0 0 0

0

40

44

10 13 10

Flow value = 0

flowcapacity

19

Towards an Algorithm


n Greedy algorithm: repeat until you get stuck.

s

4

2

5

3 t

4

0 0

0 0 0

0

40

44

10 13 10

10 10 10X X X

Flow value = 10

Bottleneck capacity of path = 10

flow

capacity

20

Towards an Algorithm


n Greedy algorithm: repeat until you get stuck.

n Fails: need to be able to "backtrack."

s

4

2

5

3 t10 13 10

4

4 4

10 6 10

4

44

44

Flow value = 10

Flow value = 14

s

4

2

5

3 t

4

0 0

0 0 0

0

40

44

10 13 10

10 10 10X X X

flow

capacity

Residual Graph Augmenting Paths



21

Original graph.

n Flow f(e).

n Edge e = v-w

Residual edge.

n Edge e = v-w or w-v.

n "Undo" flow sent.

Residual graph.

n All the edges that havestrictly positive residual capacity.

v w176

capacity = u(e)

v w11

6

residual capacity = u(e) – f(e)

residual capacity = f(e)

flow = f(e)

22

Augmenting path = path in residual graph.

n Increase flow along forward edges.

n Decrease flow along backward edges.

s

4

2

5

3 t10 13 10

4

0 0

10 10 10

0

40

44

s

4

2

5

3 t10 10 10

44

44

3

4 4

6

4

4

X

X

X

X

Xoriginal

residual

23

Augmenting Paths

Observation 4. If augmenting path, then not yet a max flow.

Q. If no augmenting path, is it a max flow?

Flow value = 14

s

4

2

5

3 t10 6 10

44

44

7

residual

s

4

2

5

3 t10 13 10

40 0

10 10 10

0

404

4original4 4

6

4

4

X

XX

X

X

24

Ford-Fulkerson Augmenting Path Algorithm

Ford-Fulkerson algorithm. Generic method for solving max flow.

Questions.

n Does this lead to a maximum flow? yes

n How do we find an augmenting path? s-t path in residual graph

n How many augmenting paths does it take?

n How much effort do we spending finding a path?

while (there exists an augmenting path) {

Find augmenting path P

Compute bottleneck capacity of PAugment flow along P

}

Max-Flow Min-Cut Theorem Proof of Max-Flow Min-Cut Theorem



25

Augmenting path theorem. A flow f is a max flow if and only if thereare no augmenting paths.

We prove both simultaneously by showing the following are equivalent:

(i) f is a max flow.

(ii) There is no augmenting path relative to f.

(iii) There exists a cut whose capacity equals the value of f.

(i) Þ (ii) equivalent to not (ii)Þ not (i), which was Observation 4

(ii) Þ (iii) next slide(iii) Þ (i) this was Observation 3

Max-flow min-cut theorem. The value of the maxflow is equal to the capacity of the min cut.

26

(ii) Þ (iii). If there is no augmenting path relative to f, then thereexists a cut whose capacity equals the value of f.

Proof.

n Let f be a flow with no augmenting paths.

n Let S be set of vertices reachable from s in residual graph.

– S contains s; since no augmenting paths, S does not contain t

– all edges e leaving S in original network have f(e) = u(e)

– all edges e entering S in original network have f(e) = 0

T)(S,

)(

)()(

ofout

toinofout

capacity

e u

e f e f f

S e

S e S e

=

=

-=

å

åå

s

t

residual network

S T

27

Max Flow Network Implementation

Edge in original graph may correspond to 1 or 2 residual edges.

n May need to traverse edge e = v-w in forward or reverse direction.

n Flow = f(e), capacity = u(e).

n Insert two copies of each edge, one in adjacency list of v and one in w.

public class Edge {

private int v , w; // from, to

private int cap; // capacity from v to w

private int flow; // flow from v to w

public Edge(int v , int w, int cap) { ... }

public int cap() { return cap; }

public int flow() { return flow; }

public boolean from (int v ) { return this. v == v ; }

public int other(int v ) { return from ( v ) ? this.w : this. v ; } public int capRto(int v ) { return from ( v ) ? flow : cap - flow; }

public void addflowRto(int v , int d ) { flow += from ( v ) ? -d : d ; }

}

28

Ford-Fulkerson Algorithm: Implementation

Ford-Fulkerson main loop.

// while there exists an augmenting path, use it

while (augpath()) {

// compute bottleneck capacity

int bottle = INFINITY;

for (int v = t; v != s; v = ST( v ))

bottle = Math. min( bottle, pred [ v ].capRto( v ));

// augment flow

for (int v = t; v != s; v = ST( v ))

pred [ v ].addflowRto( v , bottle);

// keep track of total flow sent from s to t

value += bottle;

}

Ford-Fulkerson Algorithm: Analysis Choosing Good Augmenting Paths



29

Assumption: all capacities are integers between 1 and U.

Invariant: every flow value and every residual capacities remain aninteger throughout the algorithm.

Theorem: the algorithm terminates in at most | f * | £ V U iterations.

Corollary: if U = 1, then algorithm runs in £ V iterations.

Integrality theorem: if all arc capacities are integers, then thereexists a max flow f for which every flow value is an integer.

not polynomialin input size!

30

1000

0

1

100

100

0

0

0

Use care when selecting augmenting paths.

100

Original Network

s

4

2

t

31

1000

0s

4

2

t1

100

100

0

0

0

Choosing Good Augmenting Paths


100

1

1

1

X

X

X

Original Network

32

1001

100

1

0



100

100

1

10

Original Network

s

4

2

t

Choosing Good Augmenting Paths Choosing Good Augmenting Paths



33

1001

s

4

2

t1

100

100

1

0


100

1

0

1

X

X

X

10

Original Network

34

s

4

2

t1

100

100 100

100

0

1

1 1

1


200 iterations possible!

Original Network

35



n Some choices lead to exponential algorithms.

n Clever choices lead to polynomial algorithms.

n Optimal choices for real world problems ???

Design goal is to choose augmenting paths so that:

n Can find augmenting paths efficiently.

n Few iterations.

Choose augmenting path with: Edmonds-Karp (1972)

n Fewest number of arcs. (shortest path)

n Max bottleneck capacity. (fattest path)

36

Shortest Augmenting Path

Shortest augmenting path.

n Easy to implement with BFS.

n Finds augmenting path with fewest number of arcs.

while (!q .isEmpty()) {

int v = q .dequeue();

IntIterator i = G.neighbors( v );

while(i.hasNext()) {

Edge e = i.next();

int w = e.other( v );

if (e.capRto(w) > 0) { // is v-w a residual edge?

if (wt[w] > wt[ v ] + 1) {

wt[w] = wt[ v ] + 1;

pred [w] = e; // keep track of shortest path

q .enqueue(w);

}

}}

}

return (wt[t] < INFINITY); // is there an augmenting path?

Shortest Augmenting Path Analysis Fattest Augmenting Path



37

Length of shortest augmenting path increases monotonically.

n Strictly increases after at most E augmentations.

n At most E V total augmenting paths.

n O(E2 V) running time.

38

Fattest augmenting path.

n Finds augmenting path whose bottleneck capacity is maximum.

n Delivers most amount of flow to sink.

n Solve using Dijkstra-style (PFS) algorithm.

Finding a fattest path. O(E log V) per augmentation with binary heap.

Fact. O(E log U) augmentations if capacities are between 1 and U.

v w10residual capacity

12 9X 10

if (wt[w] < Math. min(wt[ v ], e.capRto(w)) {

wt[w] = Math. min(wt[ v ], e.capRto(w));

pred [w] = v ;}

39

Choosing an Augmenting Path

Choosing an augmenting path.

n Any path will do Þ wide latitude in implementing Ford-Fulkerson.

n Generic priority first search.

n Some choices lead to good worst-case performance.

– shortest augmenting path

– fattest augmenting path

– variation on a theme: PFS

n Average case not well understood.

Research challenges.

n Practice: solve max flow problems on real networks in linear time.

n Theory: prove it for worst-case networks.

40

History of Worst-Case Running Times

Dantzig

Discoverer

Simplex

Method Asymptotic Time

E V2 U †1951

Year

Ford, Fulkerson Augmenting path E V U †1955

Edmonds-Karp Shortest path E2 V1970

Dinitz Improved shortest path E V21970

Edmonds-Karp, Dinitz Capacity scaling E2 log U †1972

Dinitz-Gabow Improved capacity scaling E V log U †1973

Karzanov Preflow-push V31974

Sleator-Tarjan Dynamic trees E V log V1983

Goldberg-Tarjan FIFO preflow-push E V log (V2 / E)1986

. . . . . . . . .. . .

Goldberg-Rao Length function E3/2

log (V2

/ E) log U†

EV2/3 log (V2 / E) log U †1997

Edmonds-Karp Max capacity E log U (E + V log V) †1970

† Arc capacities are between 1 and U.

An Application Bipartite Matching



41

Jon placement.

n Companies make job offers.

n Students have job choices.

Can we fill every job?

Can we employ every student?

Alice-Adobe

Bob-Yahoo

Carol-HP

Dave-AppleEliza-IBM

Frank-Sun

42

Bipartite matching.

n Input: undirected and bipartite graph G.

n Set of edges M is a matching if each vertex appears at most once.

n Max matching: find a max cardinality matching.

1

3

5

A

C

E

2

4

B

D

Matching M

1-B, 3-A, 4-E

RL

43

Bipartite Matching

Bipartite matching.

n Input: undirected and bipartite graph G.

n Set of edges M is a matching if each vertex appears at most once.

n Max matching: find a max cardinality matching.

1

3

5

A

C

E

2

4

B

D

Matching M

1-A, 2-B, 3-C, 4-D

RL

44

Reduces to max flow.

n Create a directed graph G'.

n Direct all arcs from L to R, and give infinite (or unit) capacity.

n Add source s, and unit capacity arcs from s to each node in L.

n Add sink t, and unit capacity arcs from each node in R to t.

Bipartite Matching

RL

s

1

3

5

A

C

E

t

2

4

B

D

1 1

¥1

3

5

A

C

E

2

4

B

D

G'G

Bipartite Matching: Proof of Correctness Bipartite Matching: Proof of Correctness



45

Claim. Matching in G of cardinality k induces flow in G' of value k.

n Given matching M = { 1-B, 3-A, 4-E } of cardinality 3.

n Consider flow f that sends 1 unit along each of 3 paths:s-1-B-t s-3-A-t s-4-E-t.

n

f is a flow, and has cardinality 3.

RL

s

1

3

5

A

C

E

t

2

4

B

D

1 1

¥1

3

5

A

C

E

2

4

B

D

G'G46

Claim. Flow f of value k in G' induces matching of cardinality k in G.

n By integrality theorem, there exists 0/1 valued flow f of value k.

n Consider M = set of edges from L to R with f(e) = 1.

– each node in L and R incident to at most one edge in M

– |M| = k

RL

s

1

3

5

A

C

E

t

2

4

B

D

1 1

¥1

3

5

A

C

E

2

4

B

D

G'G

47

Reduction.

n Given an instance of bipartite matching.

n Transform it to a max flow problem.

n Solve max flow problem.

n Transform max flow solution to bipartite matching solution.

Issues.

n How expensive is transformation? O(E + V)

n Is it better to solve problem directly? O(E V1/2) bipartite matching

Bottom line: max flow is an extremely rich problem-solving model.

n Many important practical problems reduce to max flow.

n We know good algorithms for solving max flow problems.

Reduction

Documents

Flow Network,Ford Fulkerson Method Notes