Flow Algorithms

Prepared by

John Reif, Ph.D. Distinguished Professor of Computer Science

Duke University

Analysis of Algorithms Week 9, Lecture 2

Flow Algorithms

a)  Max-flow, min-cut Theorem b)  Augmenting Paths c)  O-1 flow d)  Vertex Connectivity e)  Planar Flow

Readings on Flow Algorithms

•  Reading Selection: –  CLR, Chapter 26

Network Definition

+

digraph G (V,E)

distinguished vertices: source s V sink t V

edge capacities: c: E R

! ="""

∈$" ∈"" →&

Reverse Edges

R

R

E reverse of edges in E (u,v) (v,u) the reverse of edge (u,v)u v u v

=

= =

← →

Definition of Network Flow

R +

R

(v,u) E

flow f: (E E ) R (1) f(e) -f(e ) for all e E (2) f(e) c(e) (3) f(v,u) 0 for all v V-{s,t}

∈

∪ →

= ∈

≤

= ∈∑

Value of Flow f

•  Flow f

•  Value v V

( ) (s,v)

sum of flow from source

f f∈

=

=

∑

s

Min Cut = Max Flow

•  cut

v Xu X

X, X is partition of V

where s X, t X

(X,X) ( , )f f v u∈∈

∈ ∈

=∑

s t

X X _

Proof that Min Cut = Max Flow

•  Lemma

•  Proof

,

(X, X) ( , ) ( , )

( , ) ( ) 0 ( ).

. . .

v X v Xu X w X

v w X

f f u v f v w

f v w value f value f

Q E D

∈ ∈∈ ∈

∈

= =

= = − =

∑ ∑

∑

The flow across any cut X, Xis equal to the value( ).f

Residual Flow and Augmenting paths

•  residual capacity of edge e: res(e) = c(e) – f(e)

•  residual graph R: use modified capacities cʹ ( e)=res(e) for res(e) > 0

•  augmenting path p for flow f is path in R from s to t

•  res(p) = min (res(e)) e ∈ p

Characterization of Residual Flow

•  Lemma: R has max flow value value (f *) – value (f ), where f * is the

max flow of G.

•  Proof: If f ‘ is flow in R, then f + f ‘ is flow of G. Also, f ‘= f * – f is a flow in R.

Q.E.D.

Flow f on a path

•  Flow f on path p = (e1, e2, …, ek)

e1 e2 ek

f ( e1 ) f ( ek ) f ( e2 )

Augmenting Flow

•  residual res(p) = Δ = min (c(e) – f (e))

e ∈ p

•  gives Augmented flow f + res(p)

t s e1 e2

ek

f ( e1 ) +Δ f ( ek ) +Δ f ( e2 +Δ

Flow f and Augmenting Flow in Residual Network

s

c d

b a

t

3

3 3

4 3

2

4

1

f = 0

f = 2

f = 2

f = 2

f = 2

f = 3

f = 1

f = 1

Saturated Edge

Network with Flow

s

c d

b a

t

3

2 3

2 2

2

1

1 3

1

2 1

Residual Network

Augmenting Path (s,b,d,a,c,t)

=>

Saturated Edge

Min Cut = Max Flow

•  min cut: cut of minimum capacity •  max flow: max (value ( f ))

f is flow

X

X

s

t

_ Cut

Example of Min Cut = Max Flow

Max Flow = Min Cut = 6

Edges Labeled (Capacity, Flow)

Ford-Fulkerson Proof of Min Cut = Max Flow

Ford - Fulkersons:Theorem: The max flow f is equal to the

min cut , .Proof: (1) If is max flow, then there can be no augmenting path from

X Xf

s to . Let vertices in , reachable from s in residual graph .

-

( ) ( , ) ( , ) ( , )

(2) Clearly, has value at most

u X u Xv X v X

t X VR

X V X

Value f f u v c u v c X X

f

∈ ∈∈ ∈

=

=

= = =∑ ∑

( , )

for any cut , .. . .

c X X

X XQ E D

Bounding Flow augmentations give Max Flow

•  Lemma: At most |E| flow augmentations are required to construct max flow.

Improving Flow by Augmenting Flow from Residue graph

Proof: Suppose * is max flow in . Let * be subgraph of with pos. flow.

0while path from to do

1find a path from to in *let min *( )

for all do *(i

i

i e p

f GG G

is t

i ip s t Gf e

e p f∈

←

∃

← +

Δ =

∈ i) *( )- if *( ) 0, then delete from * odod

Note: Deletes at least on edge per step!

e f ef e e G

⎛⎜⎜⎜⎜⎜ ← Δ⎜

≤⎜⎜⎜⎜⎝

Blocking Flow

Definitions given flow f •  saturated edge e has f(e)=c(e)

•  blocking flow f : every path from s to t has saturated edge

(so cannot augment flow!) Idea: Re-route flow if it is blocking

Shortest Augmenting paths via level Graph

•  Level Graph L is subgraph of residual graph R

•  Note L gives shortest augmenting paths Construct L in O(|V|+|E|) time by Breadth First Search of R

level (v) length of shortest path from s to v in RL contains only edges (v,u) R s.t. level (u) level (v) 1

=

∈

= +

Dinic’s Flow Algorithm

•  Input: network G=(V,E) s.t. capacities ci: (E ∪ ER) → R+

•  Initialization:

for all assign ( ) 0e f e ←

Dinic’s Flow Algorithm (cont’d)

•  Loop:

[1] Construct level graph for by Breadth First Search[2] By augmentations, find blocking flow ' in from .[3] for all assign ( ) ( ) '( )[4] If is not in level graph,

L f

f L fe f e f e f e

t

← +

then return else go to [1]

f

Proof of Dinic’s Flow Algorithm (cont’d)

•  Theorem Dinic’s Algorithm halts after |V|

blocking steps •  Proof Suppose f is flow with

•  R = residual graph (currently) •  level (v) = min length path from s to v in R •  R' = new residual graph •  level' (v) = min length of path s to v in R'

Proof of Dinic’s Flow Algorithm (cont’d)

•  Claim level' (t) > level (t) •  Proof (by contradiction)

–  If level (t) = level' (t), –  then level (w) = level (v)+1 for every edge –  (v,w) ∈ L. –  This contradicts the fact that at least one

edge is saturated (on the blocking flow) on any path p in L.

Q.E.D. Hence n steps suffice for the algorithm

Example of Dinic’s Flow Algorithm (cont’d)

•  Network

3

s

c d

b a

t

3

3

4 3

2

4

1

Example of Dinic’s Flow Algorithm (cont’d)

•  1st Level Graph with Blocking Flow

s

c d

b a

t

3

3 3

4 3

2

4

1

f = 1

f = 3

f = 2

f = 1

f = 1

f = 3

f = 0

f = 1

level 1

level 2

level 3

level 0

Example of Dinic’s Flow Algorithm (cont’d)

•  2nd Level Graph with Blocking Flow

s

c d

b a

t

3

2

3

2 f = 1

f = 1

f = 1

f = 0

f = 1 level 1

level 2

level 3

level 0

1

level 4

Example of Dinic’s Flow Algorithm (cont’d)

•  3rd Level Graph with Blocking Flow

s

c

d

b a

t

2

2

2

1 f = 1

f = 1

f = 1

f = 1 f = 1

level 0

2

level 5

Example of Dinic’s Flow Algorithm (cont’d)

•  Final Flow s

c d

b a

t

3

3 3

4 3

2

4

f = 1

f = 3

f = 2

f = 3

f = 3

f = 3 f = 3

f = 1

1

Finding a Blocking Flow

•  by Karzanov •  Preflow j:

(1) Satisfies capacities’ constraints (2) May have unbalanced vertices where

u vf(u) f(u,v) 0

∈

Δ = >∑

Finding a Blocking Flow (cont’d)

•  Wave method: –  begin with blocking preflow f

(saturates on edge on every path s to t ) –  balance vertices so Δf(v) = 0 to

get blocking flow

Finding a Blocking Flow (cont’d)

•  To balance blocked vertex v:

Repeat (until Δf (v) = 0) do choose edge (u,v) with f(u,v) > 0 decrease f(u,v) by min (f(uv), Δf(v))

Finding a Blocking Flow (cont’d)

•  To attempt to balance unblocked vertex v:

Repeat (until Δf (v) = 0, or there is not an unsaturated edge (v,w) where w is

unblocked). do choose some such edge (v,w) and

decrease f (v,w) by min (C(v,w)- f (v,w), Δf(v)).

Finding a Blocking Flow (cont’d)

•  Wave Algorithm for Blocking Flow Initialize: with preflow that saturates every edge out of s and otherwise 0.

S t

c(e1)

c(ek)

…

0

0 0

0

Example of Finding a Blocking Flow (cont’d)

Edges Labeled (Capacity, Flow)

Increased Flow to Blocked vertex d

Decreased Flow Balanced at vertex d

But Blocked vertex c

Example of Finding a Blocking Flow (cont’d)

Edges Labeled (Capacity, Flow)

Decreased Flow

All vertices Balanced

Decreased Flow Balanced vertex c

But Blocked vertex a

Finding a Blocking Flow (cont’d)

•  Set s blocked, and set V-{s} all unblocked. Repeat until there are no unbalanced vertices do

Increase flow: Scan all vertices between t,s in topological order, balancing every vertex v that is unbalanced and unblocked. (If balancing fails, make v blocked.)

Decrease flow: Scan vertices in reverse topological order, balancing each vertex that is unbalanced and blocked.

Proof of O(n2) Time for Finding a Blocking Flow •  Theorem: Wave Algorithm computes a blocking

flow in O(n2) time (and hence a max flow in O(n3) time).

•  Proof (use invariants): (1) If v blocked then every path from v to t has saturated edge. (2) The preflows constructed by algorithm are blocking.

Proof of O(n2) Time for Finding a Blocking Flow (cont’d) •  Modify: s blocked, and departing edges

saturated.

•  Inductive Step: (a) Scanning in topological order in increase flow guarantees no unblocked, unbalanced vertices. (b) Scanning in reverse topological order guarantees every blocked vertex gets balanced.

Proof of O(n2) Time for Finding a Blocking Flow (cont’d)

•  Note: Each step blocks at least 1 vertex

•  So at most n steps flow on edge e increases and decreases at most once

•  So total time O(|V|2 + |E|) = O(|V|2) = O(n2)

Improved Flow Algorithms

•  Can use data structures to decrease blocking flow algorithms to O(|E| log|V|) time, giving…

•  Theorem Max flow can be computed in O(|V||E| log|V|) time.

0-1 Flow Algorithms

•  Special Case:

•  Theorem (Evan and Tarjan)

0 1 Flow, if for all e E, c(e) = 1− ∈

2 13 2

2 13 2

0 1 Flow requires min (|V| ,|E| ) blocking steps ofDinic's Algorithm, so total time

O(min(|V| ,|E| )|E|log(V)).

−

Unit Flow

•  Unit Network: All capacities are integers and every \vertex v other than s or t has

•  Claim: If Unit Network G has max flow f,

•  then max level is •  Proof: G can be decomposed into value(f )

vertex-disjoint paths from s to t. so value (f ) • (level-1) ≤ |V|

single entering edge orsingle departing edge.!"#

( )|V| 1( )value f≤ +

Illustration of Unit Flow

s t

level

G is decomposed into value(f ) vertex-disjoint paths from s to t.

Unit Flow Algorithms

s-t Vertex Separator

s-t Vertex Separator S ⊆ V: if all paths from s to t contain v ∈ S.

Menger’s Theorem: The size of the smallest s,t Vertex Separator S is exactly the same as the number of vertex disjoint paths from s to t.

Illustration of s-t Vertex Separator S

Solving Vertex Connectivity via Flow

•  Transform Vertex Connectivity to Unit Network Flow Problem

Time for Solving Vertex Connectivity via Flow

•  Total Time O(|V|�|E| log(E)) to compute s-t Vertex Connectivity N(s,t) (from s to t).

•  N(s,t) = number of disjoint paths from s to t.

Vertex Connectivity

•  N(u,v) = min vertex cut size for (G, u, v)

•  G undirected

u,v V

(u,v) E

Vertex Connectivity: c(G)

n-1 if G is complete graphMin N(u,v) else

c(G)∈

∉

⎧⎪⎪

= ⎨⎪⎪⎩

Bounding Vertex Connectivity

•  Lemma

•  Proof

•  Q.E.D. (also true for edge connectivity)

2|E|c(G) |V|≤

v V

v V

Connectivity min degree v,

but degree (v) 2|E|

2|E|Hence, c(G) |V|

∈

∈

≤

=

≤

∑

Algorithm for Vertex Connectivity

•  Lemma:

•  Proof: G – S has at least 2 - components

( , )

If is ( , ) Vertex Separatorwith | | ( ), then( ) min ( , ) for all -

a b E

S u vS c G

c G N a b a V S∉

=

= ∈

Proof of Algorithm for Vertex Connectivity

•  Let b be any node in a component of G-S which does not have a.

Thus, N(a,b) ≤ |S| = c(G).

Q.E.D.

Idea for Randomized Algorithm for Vertex Connectivity

•  Idea: Choose at random a ∈ V.

Time Cost O(log( 1ε)|V

32 ||E| log|E|)

Randomized Algorithm for Vertex Connectivity

•  (Melhorn & Students)

1

|V|

i

ib V

Input: G (V,E), error bound, , o 1 [0] |V| - 2

log ( ) [1] for i=1,2,... until ilog( )

do select a V at random min ( , min N(a , b))

od [

ε

µ

ε ε

µ

µ µ∈

= < <

←

≥

∈

←

2] output µ

Randomized Algorithm for Vertex Connectivity (cont’d)

•  Theorem: Prob (µ ≠ c(G)) ≤ ε •  Proof: Let S be a Vertex Separator with

( ) ( ) 1

1 2 k

1| |

( )

1 k

kk-log ( ) log c(G)|S|

|V| |V|

| | ( ). If c(G), then a , a ,...,aall belong to S, where

log ( )k log( )Hence, prob ( ( )) prob (a ,..., a S)

2 2 .

Vc G

S c G

c G

ε

ε

ε

µ

µ

ε

= >

≥

> = ∈

= = = = =

Definition of Planar Graph

•  G = (V,E) is a planar graph if G can be embedded on plane so no two edges cross.

Dual of Planar Graph

•  Dual: D(G)=(F, D(E)) F = faces of embedding

D(E) = { {Fi, Fj}| e ∈ E is between Fi, Fj }

Max Flow in Planar Graph

Min Cost Cycle in Dual Graph D(G) separating s, t

Definition of Outerplanar Embedded Graph

•  G is outerplanar embedded if the planar embedding has face F0 incident to all vertices.

Algorithm for Max Flow in Outerplanar Embedded Graph

•  Idea: To reduce to Min Cost Path

Add new edge (s,t) with weight ∞.

Time Cost of Algorithm for Max Flow in Outerplanar Embedded Graph

Developing an Efficient Algorithm for Max Flow in a General Planar Graph

•  Lemma: [Reif] If µ(s,t) is a minimum cost path in D(G) from a face bounding on s to a face bounding on t, then any min cost cycle in D(G) separating s,t must contain an

edge of µ(s,t).

Illustration of µ(s,t) Path Separating s and t in Dual Graph D(G)

Proof of Lemma

•  Proof: Suppose not. Then we can shortcut any cycle of D(G) separating s,t to get a shorter one, using edges of the µ(s,t) path.

Efficient Divide and Conquer Algorithm for Max Flow in a General Planar Graph

•  Theorem: [Reif] The min cost flow in a planar graph can be computed in O(|V|log2|V|) time.

•  Proof: Idea: use µ(s,t) cut in D(G) to guide a recursive divided and conquer algorithm. On each step, divide the µ(s,t) path in

half and solve the problem on each half, separately, using s,t cut as separator.

Illustration of an Efficient Divide and Conquer Algorithm for Max Flow in a General Planar Graph

Flow Algorithms

Prepared by

John Reif, Ph.D. Distinguished Professor of Computer Science

Duke University

Analysis of Algorithms Week 9, Lecture 2