Finding Minimum-Cost Circulation By Successive Approximation

Finding Minimum-Cost Circulation By Successive

Approximation

Goldberg and Tarjan 1990

Advanced Algorithms Seminar

Instructor: Prof. Haim Kaplan

Presented by: Michal Segalov & Lior Litwak

Motivation

What is the minimum cost circulation problem?

What is this problem equivalent to in the real world?

The Min-Cost Circulation Problem

In the flow problems we have the “source” and “sink” special vertices, such that flow passes between them.



The circulation problems take these vertices out of the game.




We push flow through edges in cycles along the graph, seeking minimal cost.




We push flow through edges in cycles along the graph, seeking minimal cost.

Notice – we are not interested in pushing maximal flow anymore.

Max-Flow, Min-Cost & What’s Between

Consider the familiar Max-Flow problem. In that problem we want to achieve maximal flow on the edges.



Now think of giving a cost to flow units going through each edge (positive or negative).



Now think of giving a cost to flow units going through each edge (positive or negative).

We now wish to find, amongst all possible maximal flows, the one with minimal cost.

Min-Cost Max-Flow Equivalenceto Min-Cost Circulation

Given a graph G, construct G’ by adding two vertices s and t, not connected to any of the other vertices. Any min cost flow in G’ is a min-cost circulation in G.

s t

G G’

Min-Cost Max-Flow Equivalenceto Min-Cost Circulation

Given a graph G, construct G’ by connecting t and s with a new edge s.t.

, , 1u t s mU c t s C n

ts

ts

G G’

Previous Work

Some other Algorithms out there:

This paper introduced an efficient general scheme, rather than a specific one, for solving the problem

The BIG Picture

In this lecture we will see a general approach for solving this problem.

The BIG Picture


It works by finding an initial approximate solution and then iteratively refining it until reaching an optimal one.

The BIG Picture


It works by finding an initial approximate solution and then iteratively refining it until reaching an optimal one.

The error parameter, ε, is halved in each iteration until the integrality of the arc costs guarantees an optimal solution.

The BIG Picture

We will see two methods for refinement: The Generic method – Uses a sequence of

transformation on the current circulation to produce a finer one. Achieves a runtime of

Blocking Flow method – a generalization of Dinic used for refinement. Complexity is the same for a simple implementation.

Complexity even better with dynamic trees…

3 min log , logO n nC m n

Definitions – The Basics

A circulation network is a symmetric graph G=(V,E) with a real value capacity u(v,w) function and a cost c(v,w) function for each arc s.t.

c(v,w) = -c(w,v) (cost antisymmetry constraint)


A pseudoflow is a function s.t. for each : 1. f(w,v) ≤ u(w,v) (capacity constraint)2. f(w,v) = –f(v,w) (flow antisymmetry constraint)

The cost of a pseudoflow is

,

1cost( ) , ,2 v w E

f c v w f v w

:f E R ,v w E


For a pseudoflow f and a vertex v, the excess flow into v , is defined by

A vertex v is called active if Observe that

(because G is symmetric)

fe v

,

,fu v E

e v f u v

0fe v 0fv V

e v


Pseudoflow example:

4,1,3

6,3,-2

u=5,f=2,c=-1

4,3,1

Total cost = -2

0,-1,-3

0,-3,2

2,-3,-1

0,-2,1

ef = -1

ef = -4

ef = -1

ef = 6


A circulation is a pseudoflow s.t. for each vertex v, 0fe v


A circulation is a pseudoflow s.t. for each vertex v,

This means that there is no excess flow in v, and that flow is not “created” in v(i.e. no “sources” and “sinks” in G)

0fe v


A circulation is a pseudoflow s.t. for each vertex v,

This means that there is no excess flow in v, and that flow is not “created” in v(i.e. no “sources” and “sinks” in G)

A pseudoflow is a circulation iff there are no active vertices (by definition)

0fe v

Definitions – The Residual System

The residual capacity of and edge (v,w) is

An arc is saturated if The residual graph is defined as

Note that all the residual arcs are non-saturated…

, , ,fu v w u v w f v w , 0fu v w

, , | 0f f f fG V E E e u e

Definitions – Prices & Costs

A price function is a function We use the price function to define the

reduced cost of an arc:

:p V R

, ,pc v w c v w p v p w




Note that the cost of a circulation is the same whether the original costs or the reduced costs with respect to some price function are used!

:p V R


Optimality of Circulations

We say the a circulation is optimal if its cost is minimal



Theorem 2.1 A circulation f is minimum cost iff the

residual graph contains no cycles of negative cost


Proof ( ): If there are negative cycles in Gf we can obtain

a “cheaper” circulation by increasing flow on them.

4,0,3

6,3,-2

u=5,f=3,c=-1

4,3,1

4,3

3,-2

uf=2,c=-1

1,1

cost = -6We have a negative cycle

GfG


Proof ( ): If the circulation f is not optimal, we show there

exists a negative cycle in Gf

Let f* be an optimal circulation.We will look at the decomposition of the circulation f*-f into cycles.

Since it is of negative cost and feasible in Gf, it contains a negative cycle!


Theorem 2.2 A circulation f is minimum cost iff there is a

price function p s.t. for each arc (v,w)

, 0 , ,pc v w f v w u v w


Proof ( ): Given such p, there are no negative reduced

cost cycles in Gf. Hence, no negative cost cycles in Gf. Use the previous theorem.


Proof ( ): Given an optimal circulation, such a function

can be calculated using the shortest paths from an external vertex s in Gf.

can you think how?


4,3

2,-2

uf=1,c=-1

Gf

4,1

4,2

4,-1


3

-2

c=-1

Gaux

1

2

-1

S

0

00

0


3

-2

c=-1

Gaux

1

2

-1

S

0

00

0


3

-2

c=-1

Gaux

1

2

-1

S

0

00


3

-2

c=-1

Gaux

1

2

-1

S

0

0


-2

c=-1

Gaux

1

2

-1

S

0

0


-2

c=-1

Gaux

2

-1

S

0

0


-2

c=-1

Gaux

2

S

0

0


-2

c=-1

T

S

0

0


-2

c=-1

T

S

0

0

d = 0

d = 0

d = -1

d = -3

,c v w d v d w


4,3

2,-2

uf=1,c=-1

Gf

4,1

4,2

4,-1

p = 0

p = 0

p = -1

p = -3

Define p = d:


c=-1, cp=0

Gf

p = 0

p = 0

p = -1

p = -3

c=1, cp=0

c=-2, cp=0

c=2, cp=0

c=-1, cp=2

c=3, cp=3

, 0 ,p fc v w v w G

So What Have We Seen So Far?


A circulation network is a symmetric graph G=(V,E) with a real value capacity u(v,w) function and a cost c(v,w) function for each arc s.t.

c(v,w) = -c(w,v) (cost antisymmetry constraint)


A pseudoflow is a function s.t. for each : 1. f(w,v) ≤ u(w,v) (capacity constraint)2. f(w,v) = –f(v,w) (flow antisymmetry constraint)

The cost of a pseudoflow is

,

1cost( ) , ,2 v w E

f c v w f v w

:f E R ,v w E


For a pseudoflow f and a vertex v, the excess flow into v , is defined by

A vertex v is called active if Observe that

(because G is symmetric)

fe v

,

,fu v E

e v f u v

0fe v 0fv V

e v


A circulation is a pseudoflow s.t. for each vertex v, 0fe v




:p V R





Note that the cost of a circulation is the same whether the original costs or the reduced costs with respect to some price function are used!

:p V R






Theorem 2.1 A circulation f is minimum cost iff the

residual graph contains no cycles of negative cost


Theorem 2.2 A circulation f is minimum cost iff there is a

price function p s.t. for each arc (v,w)

, 0 , ,pc v w f v w u v w

Now let’s relax the definition of circulation optimality

Optimality and Approximate Optimality

A circulation f is ε-optimal with respect to a price function p if

(we relax the optimality constraint) A pseudoflow is ε-optimal if it is ε-optimal with

respect to some price function p. Observe that a pseudoflow is ε-optimal w.r.t. p

iff for every residual arc.

, , ,pc v w f v w u v w

,pc v w

Approximate Optimality Properties

Theorem 2.3 If all costs are integers and

then an ε-optimal solution is optimal1/ n

Approximate Optimality Properties

Proof: Consider a simple cycle in Gf. The ε-optimality of f implies that the reduced

cost of the cycle is at least . The reduced cost of the cycle equals the

original cost which is non-negative (all costs are integral).

Since there are no negative cost cycles, f is minimum cost circulation.

1n

Fitting Price Functions and Tight Error Parameters

The definition of ε-optimality raises two interesting questions:

1. Given a pseudoflow f and a constant ε, find a price function p s.t. f is ε-optimal w.r.t p(or show that f is not ε-optimal).

2. Given a pseudoflow f, find an ε s.t. f is ε-tight (the smallest ε s.t. f is ε-optimal).

Finding a Fitting Price Function

Objective:

Given a pseudoflow f and a constant ε, find a price function p s.t. f is ε-optimal w.r.t p, or show that f is not ε-optimal.


Let’s define Extend the residual graph by adding a

vertex s with arcs to all other vertices –

(Also define )

, ,c u v c u v

,aux fG V s E s V

, 0c s u

fG


Theorem 3.3A pseudoflow is ε-optimal if and only if

contains no cycle of negative cost. If T is a shortest path tree of rooted at s w.r.t , and d is the distance function in the tree then f is ε-optimal w.r.t p = d.

auxG

auxG

c

c


Using the previous theorem the first problem can be solved by constructing a short distance tree in (or finding a negative cost cycle in it).

O(n+m) for constructing O(nm) – Building shortest path

tree/looking for negative cost cycle (Bellman Ford)

auxG

auxG

The Last Introduction Slide!

The Minimum-Cost Framework

Now we get to business…We want to construct a general algorithm

framework to solve the Minimum-Cost Circulation problem.


Now we get to business…We want to construct a general algorithm

framework to solve the Minimum-Cost Circulation problem.

The successive approximation framework assumes all costs are integral.


The basic idea: Find an initial C-optimal circulation f w.r.t. p s.t.

p(v) = 0 (for all v) Refine ε by half and adjust f and p to be

ε/2-optimal Repeat until ε is small enough to guarantee

optimality.


Procedure min-cost(V,E,u,c):[Initialization] ε ← C; ; If there exists a circulation then load it into f,

else return NULL;[Main loop] While ε ≥ 1/n do

(ε,f,p) ← refine(ε,f,p) Return f

End

, 0v V p v


Correctness: Under the assumption that refine is correct, the

correctness of the algorithm is immediate from Theorem 2.3.


Correctness: Under the assumption that refine is correct, the

correctness of the algorithm is immediate from Theorem 2.3.

Complexity: iterations of refine Each refine takes and can be reduced

using dynamic trees to .

logO nC 3O n

2log /O nm n m

The Generic Refinement Method

This method uses ideas similar to those presented in Goldberg & Tarjan’s Max-Flow algorithm.

We will use push and relabel operations here as well, modified to fit the ε-optimal environment.


Procedure refine(ε,f,p):[Initialization] ε ← ε/2; For every (v,w) in E do:

If cp(v,w) < 0 then f(v,w) ← u(v,w);[Loop] While there exists a flow/price update operation

that applies do:Select such an operation and apply it;

Return (ε,f,p)End


Refine starts by halving ε and saturating all edges with negative reduced cost.

This achieves an ε-optimal pseudoflow.Next we apply the flow & price update

operations (push & relabel respectively), to transform the pseudoflow into an ε-optimal circulation.

The Generic Refinement MethodProcedure push(v,w):

Applicability: v is active, uf(v,w) > 0 and cp(v,w) < 0 Action: send δ = min {ef(v),uf(v,w)} units of flow

from v to w

Procedure relabel(v):Applicability: v is active and :

uf(v,w) > 0 cp(v,w) ≥ 0 Action: replace p(v) with

w V

,max ,fv w E p w c v w


The push operation (non-saturating):

The vertex is no longer active.

uf=5,cp=-1Gf

uf=2,cp=1

ef=2uf=3,cp=-1

Gf

uf=2,cp=1

ef=0


The push operation (saturating):

The vertex remains active!

uf=1,cp=-1Gf

uf=2,cp=1

ef=2

uf=0,cp=-1Gf

uf=2,cp=1

ef=1


The relabel operation:

No flow is changed, only the price of the vertex.

c=-1,cp=3Gf

c=2,cp=4

p=0

p=-4

p=-2

c=-1,cp=-εGf

c=2,cp=1-ε

p=-3-ε

p=-4

p=-2

Let’s Analyze aComprehensive Example

x1

x2x3

y1

y2

y3

9,0

10,-2 10,-2

10,1 10,1

10,1

10,-2

Example – The Initial G(Assume G = Gf)

Gf

An Iteration of Refine

ε ← ε/2 = 1

x1

x2x3

y1

y2

y3

9,0

10,-2 10,-2

10,1 10,1

10,1

10,-2

Gf


At first all negative cost arcs are saturated

10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,1 10,1

10,1

10,2

Gf

An Iteration of RefineLet’s see how the excess flow moves from

x3 to x1, applying relabels and pushes:

10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,1 10,1

10,1

10,2

Gf

An Iteration of RefineWe start by applying relabel on x3:

p(x3) = 0

10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,1 10,1

10,1

10,2

Gf

An Iteration of RefineWe start by applying relabel on x3:

Relabel:

p’(x3) = -2

10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,1 10,-1

10,-1

10,0

Gf

An Iteration of RefineNow we can push – we choose (arbitrarily)

to push towards x2:

p = -2

10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,1 10,-1

10,-1

10,0

Gf

An Iteration of RefineNow we can push – we choose (arbitrarily)

to push towards x2:

p = -2

10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,1 10, 1

10,-1

10,0

Gf

An Iteration of RefineNow we relabel x2:

p(x2) = 0p = -2

10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,1 10, 1

10,-1

10,0

Gf

An Iteration of RefineNow we relabel x2:

Relabel:

p’(x2) = -2p = -2

10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,-1 10, -1

10,-1

10,0

Gf

An Iteration of RefineAgain, we choose (arbitrarily) to push

towards x1:

p = -2p = -2

10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,-1 10, -1

10,-1

10,0

Gf

An Iteration of RefineAgain, we choose (arbitrarily) to push

towards x1:

p = -2p = -2

10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,1 10, -1

10,-1

10,0

Gf

An Iteration of RefineNow we can relabel x1:

p = -2p = -2

10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,1 10, -1

10,-1

10,0

Gf

p(x1) = 0


Relabel:

p’(x1) = -1

p = -2p = -2

10

-10

x1

x2x3

y1

y2

y3

9,-1

10,2 10,2

10,0 10, -1

10,0

10,0

Gf

An Iteration of RefineWe push 9 flow units towards y1:

p = -2

10

-10

x1

x2x3

y1

y2

y3

9,-1

10,2 10,2

10,0 10, -1

10,0

10,0

Gf

p = -1

p = -2

An Iteration of RefineWe push 9 flow units towards y1:

p = -2

1

-1

x1

x2x3

y1

y2

y3

9, 1

10,2 10,2

10,0 10, -1

10,0

10,0

Gf

p = -1

p = -2

An Iteration of RefineWe can relabel x1 again:

p = -2

1

-1

x1

x2x3

y1

y2

y3

9, 1

10,2 10,2

10,0 10, -1

10,0

10,0

Gf

p = -1

p = -2


Relabel:

p’(x1) = -2

p = -2

1

-1

x1

x2x3

y1

y2

y3

9, 2

10,2 10,2

10,-1 10, -1

10,1

10,0

Gfp = -2

p = -2

1

-1

x1

x2x3

y1

y2

y3

9, 2

10,2 10,2

10,-1 10, -1

10,1

10,0

Gfp = -2

p = -2

Anybody wants to continue?!

p = -2

1

-1

x1

x2x3

y1

y2

y3

9, 2

10,2 10,2

10,-1 10, -1

10,1

10,0

Gfp = -2

p = -2

Anybody wants to continue?!We will, later on – This example is

important to prove certain bounds…

The Correctness of Refine

We need to show a few things: ε-optimality is preserved throughout the

execution of refine. Refine returns an ε-optimal circulation (rather

than just a pseudoflow). Refine terminates (and efficiently…)


Note that push preserves ε-optimality of any pseudoflow – no arcs with negative reduced cost are added to Gf.

Lemma 5.3:Suppose f is an ε-optimal pseudoflow w.r.t

p and relabel is applied to v, then the price of v decreases by at least ε, and f is ε-optimal w.r.t p’ (the new price function).

ε-Optimality Preservation

Proof:Let f be a ε-optimal pseudoflow w.r.t p.

Before the relabeling of v

After the relabel: , , 0 ,f pv w E c v w p v p w c v w

,' max ,fv w Ep v p w c v w p v


Proof:Let f be a ε-optimal pseudoflow w.r.t p.

Before the relabeling of v

After the relabel:

Now we will show that the new pseudoflow is ε-optimal w.r.t. p’. Show that

, , 0 ,f pv w E c v w p v p w c v w

,' max ,fv w Ep v p w c v w p v

', ,f pv w E c v w


Proof – continuedThe only residual arcs effected by the

change are (v,w) or (w,v).An arc of the form (w,v) has its reduced

cost increased by ε, preserving ε-optimal


Proof – continuedThe only residual arcs effected by the

change are (v,w) or (w,v).An arc of the form (w,v) has its reduced

cost increased by ε, preserving ε-optimalFor an arc (v,w) in Gf

since by definition.

' , , 'pc v w c v w p v p w

' ,p v p w c v w


Theorem 5.4If refine terminates, the pseudoflow f it

returns is an ε-optimal circulation.

We saw the ε-optimality preservation. Refine terminates only when there are no

active vertices, which means we have a circulation (by definition).

The Complexity of Refine

Suppose we implement refine naively, choosing update operations arbitrarily.

How many update operations are performed in every invocation of refine?

Bounding the Price Reduction

To limit the number of relabel operations performed, we need to bound the reduction in price for each vertex.


Lemma 5.7Let f be a pseudoflow and f’ a circulation.

For any v with ef(v) > 0 there exists a w with ef(w) < 0 and a path from v to w, such that and for

0 1, ... lv v v v w 1,i i fv v E 1 ',i i fv v E

0 i l


Proof:Suppose ef(v) > 0.


Proof:Suppose ef(v) > 0.Define G+=(V,E+),

G-=(V,E-), , | ' , ,E x y f x y f x y

, | ' , ,E x y f x y f x y



G-=(V,E-), Notice that:

and that iff

, | ' , ,E x y f x y f x y , | ' , ,E x y f x y f x y

',f fE E E E ,x y E ,y x E



G-=(V,E-), Notice that:

and that iffConclusion – it suffices to show the

existence of such path in G+

, | ' , ,E x y f x y f x y , | ' , ,E x y f x y f x y

',f fE E E E ,x y E ,y x E


Proof – Continued:Let S be the set of all vertices reachable

from v in G+


Proof – Continued:Let S be the set of all vertices reachable

from v in G+ Notice that for every edge ,

otherwiseNow we show a crucial inequality

,x y S S , ' ,f x y f x y

y S

Bounding the Price ReductionProof – Continued:

,

,

,,

0

,

0 ' , ' is a circulation

, by previous argument

, , by antisimetry

, by definition of

by antisimetry

x y S S E

x y S S E

x y S S Ex y S S E

x y S V E

fx S

f x y f

f x y

f x y f x y

f x y S

e x


Proof – Continued:Eventually,

But we know thatso there must exist such thatw S

0fx S

e x

0fe v

0fe w


Now we can bound the reduction in price –

Lemma 5.8The price of any vertex v decreases by at

most 3nε during an execution of refine


Proof:We start with , the refine input. 2 2, , 2f p


Proof:We start with , the refine input.

Let f and p be the pseudoflow and the price function after any relabel of v.

2 2, , 2f p




v must be active before the relabel, and it remains active afterwards.

2 2, , 2f p




v must be active before the relabel, and it remains active afterwards.

Use previous lemma with f, f’=f2ε, to show that there is a path from v to w such that

2 2, , 2f p

0fe w

Bounding the Price ReductionProof – Continued:From the ε-optimality of f:

1 1

1 10 0

, ,l l

p i i i ii i

l c v v p v p w c v v


From the 2ε-optimality of f’:

1 1

1 10 0

, ,l l

p i i i ii i


2

1 1

1 2 2 10 0

2 , ,l l

p i i i ii i

l c v v p w p v c v v


From the 2ε-optimality of f’:

Sum the above inequalities to get:

1 1

1 10 0

, ,l l

p i i i ii i


2

1 1

1 2 2 10 0

2 , ,l l

p i i i ii i

l c v v p w p v c v v

2 23 3p v p v l p v n

The Bounds: RelabelLemma 5.9The number of relabel operations during

an execution of refine is at most 3n(n-1)

Proof:Recall that after each relabel, the price of

a vertex decreases by at least εOverall it decreases by at most 3nε We get at most 3n(n-1) relabel operations!

The Bounds: Saturating PushLemma 5.10The number of saturating pushes during

an execution of refine is at most 3nm

Proof:After a saturating push on an arc (v,w),

there must be a relabel operation on v, before (any) other push can occur.

We get at most 3nm saturating pushes!

The Bounds: Non-saturating Push

Lemma 5.11The number of non-saturating pushes

during an execution of refine is at most 3n2(n+m)


Some definitions:An arc (v,w) is admissible if

A push operation is applicable only to active admissible arcs.

, 0 , 0f pu v w and c v w

The admissible graph is

Its most important feature – it is acyclic(as we shall prove).

Lemma 5.5Immediately after a relabel is applied to v,

no admissible arcs enter v.


, |A A AG V E E e e is admissible

Proof:Let . Before the relabeling,

by ε-optimality. , fw v G

,pc w v



by ε-optimality. After the relabeling p(v) decreases by at

least ε, hence it increases by at least ε.

, fw v G

,pc w v

,pc w v



by ε-optimality. After the relabeling p(v) decreases by at

least ε, hence it increases by at least ε.

Thus, after the relabeling, i.e. (w,v) is not admissible.

, fw v G

,pc w v

,pc w v

, 0pc w v



Corollary 5.6Throughout the running of refine,

the admissible graph remains acyclic.Proof:Initially GA contains no arcs,

hence it is acyclic.Pushes preserve acyclicity. From Lemma 5.5, so do relabels.


Now to prove Lemma 5.11:The number of non-saturating pushes

during an execution of refine is at most 3n2(n+m)


Let’s define a potential function for each vertex, , the number of vertices reachable from v in GA.

v



The total potential will be:

v

| is activev v



The total potential will be:

Let’s amortize the number of non-saturating pushes using this potential.

v

| is activev v


In the beginning of refine, there are no admissible arcs, i.e. is at most n.


A non-saturating push decreases by at least 1. In such a push operation over the arc (v,w), v

becomes inactive, so its potential is subtracted from the total.

If w becomes active as a result of the push, its potential is added to the total.

But since GA is acyclic. Hence the total potential decreases!

w v


A saturating push increases by at most n: Such a push may activate at most one other

vertex. At most n vertices are reachable from the newly

activated vertex.


A relabel increases by at most n: From Lemma 5.5, no admissible vertices enter

v after a relabel on v. Recall that no other arc are affected by relabel. Thus, for any w≠v, can only decrease! For v itself, can increase up to n.

v w


Overall, the number of non-saturating pushes equals: The initial potential value Plus the number of relabels, times n Plus the number of saturating pushes, times n

That is:

23 1 3 3n n n n nm n n n m

In conclusion, we have shown that a naïve implementation of refine performs update operations.

Let’s show that such an implementation exists, and works in the above complexity.

The Complexity of Refine

2O n m

A Straight-Forward Implementation

Procedure push/relabel(v):Applicability: v is activeAction: Let {v,w} be the current edge of v If uf(v,w) > 0 && cp(v,w) < 0 then push(v,w) Else If {v,w} is not last of edge list of v then:

Make current edge of v the next edge on the list Else:

Make the first edge of v the current edgeRelabel(v)

End

This implementation of refine uses the push/relabel procedure repeatedly, until all vertices are non-active.

If a relabel is applied on v, all arcs (v,w) are saturated (v is still active).

In order for flow to be pushed back to v, there must be a relabel on w first, after which

But this means


, 0pc w v , 0pc v w

We get a complexity of for refine.


2O n m

By restricting the application of relabel and push we can obtain better complexity, .

We will apply the operations on a vertex until it is either relabeled or rendered inactive.

Now we should decided in which order to process the vertices.

A Faster Implementation of Refine

3O n

A Faster Implementation of Refine

Procedure discharge(v):Applicability: v is activeAction: apply push/relabel on v until it is

relabeled or becomes inactive

A Faster Implementation of Refine:First Attempt - FIFO

In the Max-Flow algorithm of Goldbreg & Tarjan, we saw that FIFO processing of the vertices improves performance.

Does it work here as well?

A Faster Implementation of Refine:First Attempt - FIFO

Unfortunately, we need passes over the queue in the worst case…

Lemma 6.3During an execution of the FIFO

implementation of refine, there arepasses over the queue.

3O n

3O n

A Faster Implementation of Refine:Second attempt: Topological Order

We use a different order of processing than in the FIFO implementation.

Here, we process the vertices in a topological order, with respect to the current admissible graph GA.

Such an order always exists because recall that GA is acyclic!


Just a reminder:An arc (v,w) is admissible if

A push operation is applicable only to active admissible arcs.

The admissible graph is

, 0 , 0f pu v w and c v w

, |A A AG V E E e e is admissible


We maintain a current vertex list in topological order, L.

L is initialized in any order, since GA has no edges to begin with.

In each pass over L, we scan for the first active vertex, and apply discharge on it.

If a vertex is relabeled, it is moved to the beginning of L, and we begin the scan again.


Procedure first-active:Let L be any vertex listLet v be the first vertex in LWhile there exists an active vertex in L

do:If v is active then:

• discharge(v)• If the discharge relabeled v, move v to the front

of LElse replace v by the next vertex in L

End


Note that push operations do not affect the topological order of L.

Recall that after a relabel operation on v: No admissible arcs enter v Only arcs connected to v are affected

Thus, moving v to the front of L retains the topological order with respect to the current GA.


Why is the topological order important?


Why is the topological order important?After relabel we start from the beginning,

so we never miss an active vertex.



so we never miss an active vertex.The topological order guaranties that we

always push flow towards subsequent vertices in L.





If discharge ends with a push, the vertex becomes inactive.





If discharge ends with a push, the vertex becomes inactive.

It will not be active again until the next relabel operation!


We define that a pass over L begins from the first vertex and ends on the first relabel operation.

Lemma 6.4The first-active method terminates after

passes over L.Proof:We have only relabel operations.

2O n

2O n

Theorem 6.5The first-active implementation of refine

runs in time, giving a total time complexity of the Min-Cost Circulation algorithm of


3O n

3 logO n nC


Proof:We can reduce the bound on the number

of nonsaturating push operations to . In a pass over L, each vertex which is not

relabeled becomes inactive. That is, a nonsaturating push is applied to it. We get nonsaturating pushes per pass.

Other bounds remain the same.Total bound for refine is now !

3O n

3O n

O n

Let’s Test the above Method on the Example Shown Earlier

x1 x2 x3 xn

y1

y2 y3yn

h-1,0

h,-2 h,-2 h,-2

h,1 h,1 h,1

h,1

h,-2

The Example Revisited(Assume G = Gf)

Gf


ε ← ε/2 = 1

x1 x2 x3 xn

y1

y2 y3yn

h-1,0

h,-2 h,-2 h,-2

h,1 h,1 h,1

h,1

h,-2

Gf



h

-h

x1

x2 x3 xn

y1

y2 y3yn

h-1,0

h, 2 h, 2 h,2

h,1 h,1 h,1

h,1

Gf

h,2

An Iteration of RefineLet’s see how the excess flow moves from

xn to x1, applying n relabels and pushes:

h

-h

x1

x2 x3 xn

y1

y2 y3yn

h-1,0

h, 2 h, 2 h,2

h,1 h,1 h,1

h,1

Gf

h,2

An Iteration of RefineWe start by applying relabel on xn:

h

-h

x1

x2 x3 xn

y1

y2 y3yn

h-1,0

h, 2 h, 2 h,2

h,1 h,1 h,1

h,1

Gf

p(xn) = 0

h,2

An Iteration of RefineWe start by applying relabel on xn:

h

-h

x1

x2 x3 xn

y1

y2 y3yn

h-1,0

h, 2 h, 2 h,2

h,1 h,1 h,-1

h,-1

GfRelabel:

p’(xn) = -2

h,0


h

-h

x1

x2 x3 xn

y1

y2 y3yn

h-1,0

h, 2 h, 2 h,2

h,1 h,1 h,-1

h,-1

Gf

h,0

p = -2

Now we can push – we choose (unfavorably) to push towards xn-1:


-h

x1

x2 x3 xn

y1

y2 y3yn

h-1,0

h, 2 h, 2 h,2

h,1 h,1 h,1

h,-1

Gf

h,0

p = -2hh,1

xn-1

Now we can push – we choose (unfavorably) to push towards xn-1:

An Iteration of RefineNow we relabel xn-1:

-h

x1

x2 x3 xn

y1

y2 y3yn

h-1,0

h, 2 h, 2 h,2

h,1 h,1 h,1

h,-1

Gf

h,0

p = -2hh,1

xn-1

p(xn-1) = 0

An Iteration of RefineNow we relabel xn-1:

-h

x1

x2 x3 xn

y1

y2 y3yn

h-1,0

h, 2 h, 2 h,2

h,1 h,1 h,-1

h,-1

Gf

h,0

p = -2hh,-1

xn-1

Relabel:

p’(xn-1) = -2

An Iteration of RefineAgain, we choose (unfavorably) to push

towards xn-2:

-h

x1

x2 x3 xn

y1

y2 y3yn

h-1,0

h, 2 h, 2 h,2

h,1 h,1 h,-1

h,-1

Gf

h,0

p = -2hh,-1

xn-1

p = -2

An Iteration of RefineWe end up in the following state (after the

last push to x1):

h

-h

x1

x2 x3 xn

y1

y2 y3yn

h-1,0

h, 2 h, 2 h,2

h,1 h,-1 h,-1

h,-1

Gf

h,0

p = -2h,-1

xn-1

p = -2p = -2p = -2


h

-h

x1

x2 x3 xn

y1

y2 y3yn

h-1,0

h, 2 h, 2 h,2

h,1 h,-1 h,-1

h,-1

Gf

h,0

p = -2h,-1

xn-1

p = -2p = -2p = -2

p = 0


h

-h

x1

x2 x3 xn

y1

y2 y3yn

h-1,-1h, 2 h, 2 h,2

h,0 h,-1 h,-1

h,0

Gf

h,0

p = -2h,-1

xn-1

p = -2p = -2p = -2

Relabel:

p’(x1) = -1

An Iteration of RefineWe push h-1 flow units towards y1:

h

-h

x1

x2 x3 xn

y1

y2 y3yn

h-1,-1

h, 2 h, 2 h,2

h,0 h,-1 h,-1

h,0

Gf

h,0

p = -2h,-1

xn-1

p = -2p = -2p = -2

p = -1

An Iteration of RefineWe push h-1 flow units towards y1:

1

-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 1h, 2 h, 2 h,2

h,0 h,-1 h,-1

h,0

Gf

h,0

p = -2h,-1

xn-1

p = -2p = -2p = -2

p = -1


1

-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 1

h, 2 h, 2 h,2

h,0 h,-1 h,-1

h,0

Gf

h,0

p = -2h,-1

xn-1

p = -2p = -2p = -2

p = -1


1

-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 2h, 2 h, 2 h,2

h,-1 h,-1 h,-1

h,1

Gf

h,0

p = -2h,-1

xn-1

p = -2p = -2p = -2

Relabel:

p’(x1) = -2


1

-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 2

h, 2 h, 2 h,2

h,-1 h,-1 h,-1

h,1

Gf

h,0

p = -2h,-1

xn-1

p = -2p = -2p = -2

The single flow unit will flow back to xn :

p = -2


1

-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 2

h, 2 h, 2 h,2

h,-1 h,-1 h,-1

h,1

Gf

h,0

p = -2h,-1

xn-1

p = -2p = -2p = -2

Push the flow unit to x2 :

p = -2


1

-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 2

h, 2 h, 2 h,2

h-1,-1

h,-1 h,-1

h,1

Gf

h,0

p = -2h,-1

xn-1

p = -2p = -2p = -2

Push the flow unit to x2 :

1,1p = -2


1

-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 2

h, 2 h, 2 h,2

h-1,-1

h,-1

h,1

Gf

h,0

p = -2h,-1

xn-1

p = -2p = -2p = -2

And then push the flow unit to x3 :

1,1

h-1,-1

1,1p = -2


1

-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 2

h, 2 h, 2 h,2

h-1,-1

h,1

Gf

h,0

p = -2

xn-1

p = -2p = -2p = -2

And so on, until finally it reaches xn :

1,1

h-1,-1

1,1

h-1,-1

1,1

h-1,-1

1,1p = -2


1

-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 2

h, 2 h, 2 h,2

h-1,-1

h,0

Gf

h,-1

p = -3

xn-1

p = -2p = -2p = -2

Relabel xn :

1,1

h-1,-1

1,1

h-1,0

1,0

h-1,-1

1,1p = -2


1-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 2

h, 2 h, 2 h,2

h-1,-1

h,0

Gf

h-1,-1

p = -3

xn-1

p = -2p = -2p = -2

Push towards yn :

1,1

h-1,-1

1,1

h-1,0

1,0

h-1,-1

1,1

1,1

p = 0

p = -2


1-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 2

h, 2 h, 2 h,0

h-1,-1

h,0

Gf

h-1,1

p = -3

xn-1

p = -2p = -2p = -2

Relabel yn :

1,1

h-1,-1

1,1

h-1,0

1,0

h-1,-1

1,1

1,-1

p = -2

p = -2


1

-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 2

h, 2 h, 2 h,0

h-1,-1

h,0

Gf

h,1

p = -3

xn-1

p = -2p = -2p = -2

Push back towards xn :

1,1

h-1,-1

1,1

h-1,0

1,0

h-1,-1

1,1

p = -2

p = -2


1

-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 2

h, 2 h, 2 h,0

h-1,-1

h,-1

Gf

h,0

p = -4

xn-1

p = -2p = -2p = -2

Relabel xn again:

1,1

h-1,-1

1,1

h-1,1

1,-1

h-1,-1

1,1

p = -2

p = -2


1

-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 2

h, 2 h, 2 h,0

h-1,-1

h-1,-1

Gf

h,0

p = -4

xn-1

p = -2p = -2p = -2

Choose (very unfavorably) to push back the flow immediately to x1 :

1,1

h-1,-1

1,1

h-1,1

1,-1

h-1,-1

1,1

p = -2

1,1

p = -2


-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 2

h, 2 h, 2 h,0

h-2,-1

h-1,-1

Gf

h,0

p = -41

xn-1

p = -2p = -2p = -2

Now the flow is pushed towards xn-1 :

2,1

h-2,-1

2,1

h-1,1

1,-1

h-2,-1

2,1

p = -2

1,1

p = -2


-1

x1

x2 x3 xn

y1

y2 y3yn

h-1, 2

h, 2 h, 2 h,0

h-2,-1

h-1,-1

Gf

h,0

p = -41

xn-1

p = -4p = -2p = -2

And xn-1 is relabeled:

2,1

h-2,-1

2,1

h-1,-1

1,1

h-2,1

2,-1

p = -2

1,1

p = -2


1

-1

x1

x2 xn-2 xn

y1

y2 y3yn

h-1, 2

h, 2 h, 2 h,0

h-3,-1

h-2,-1

Gf

h,0

p = -4

xn-1

p = -4p = -2p = -2

Now we can push back to xn-2 , but we choose (unfavorably) to do it through xn :

3,1

h-3,-1

3,1

h-2,-1

2,1

h-2,1

2,-1

p = -2

2,1

p = -2


1

-1

x1

x2 xn-2 xn

y1

y2 y3yn

h-1, 2

h, 2 h, 2 h,0

h-n+1,1

h-n+1,-1

Gf

h,0

p = -4

xn-1

p = -4p = -4p = -4

The process continue until, after n iterations, we return the flow to x1 :

n-1,-1

p = -2

n-1,1

p = -2h-n+1,-1

n-1,1

h-n+1,-1

n-1,1

h-n+1,-1

n-1,1


1

-1

x1

x2 xn-2 xn

y1

y2 y3yn

h-1, 4h, 2 h, 2 h,0

h-n+1,-1

h-n+1,1

Gf

h,0

p = -4

xn-1

p = -4p = -4p = -4

We finish by relabeling x1 :

n-1,1

p = -2

n-1,-1

p = -4h-n+1,-1

n-1,1

h-n+1,-1

n-1,1

h-n+1,-1

n-1,1


1

-1

x1

x2 xn-2 xn

y1

y2 y3yn

h-1, 4

h, 2 h, 2 h,0

h-n+1,-1

h-n+1,1

Gf

h,0

p = -4

xn-1

p = -4p = -4p = -4

Notice that all the x nodes had their price reduced by two in this process:

n-1,1

p = -2

n-1,-1

p = -4h-n+1,-1

n-1,1

h-n+1,-1

n-1,1

h-n+1,-1

n-1,1


1

-1

x1

x2 xn-2 xn

y1

y2 y3yn

h-1, 4

h, 2 h, 2 h,0

h-n+1,-1

h-n+1,1

Gf

h,0

p = -4

xn-1

p = -4p = -4p = -4

Also, n-1 flow units have been returned from x1 to xn and passed through this arc:

n-1,1

p = -2

n-1,-1

p = -4h-n+1,-1

n-1,1

h-n+1,-1

n-1,1

h-n+1,-1

n-1,1


1

-1

x1

x2 xn-2 xn

y1

y2 y3yn

h-1, 4

h, 2 h, 2 h,0

h-n,-1

h-n-1,1

Gf

h,0

p = -4

xn-1

p = -4p = -4p = -4

Again, we can push the unit of flow all the way to xn :

n,1

p = -2

n-1,-1

p = -4h-n,-1

n,1

h-n,-1

n,1

h-n,-1

n,1


1

-1

x1

x2 xn-2 xn

y1

y2 y3yn

h-1, 4

h, 2 h, 2 h,0

h-n,-1

h-n-1,0

Gf

h,-1

p = -5

xn-1

p = -4p = -4p = -4

Now relabel xn :

n,1

p = -2

n-1,0

p = -4h-n,-1

n,1

h-n,-1

n,1

h-n,0

n,0


1-1

x1

x2 xn-2 xn

y1

y2 y3yn

h-1, 4

h, 2 h, 2 h,0

h-n,-1

h-n-1,0

Gf

p = -5

xn-1

p = -4p = -4p = -4

And push down to yn :

n,1

p = -2

n-1,0

p = -4h-n,-1

n,1

h-n,-1

n,1

h-n,0

n,0

h-1,-11,1


1-1

x1

x2 xn-2 xn

y1

y2 y3yn

h-1, 4

h, 2 h, 2 h,-1

h-n,-1

h-n-1,0

Gf

p = -5

xn-1

p = -4p = -4p = -4

Relabel yn :

n,1

p = -3

n-1,0

p = -4h-n,-1

n,1

h-n,-1

n,1

h-n,0

n,0

h-1,01,0


-1

x1

x2 xn-2 xn

y1

y2yn

h-1, 4

h, 2 h, 2 h-1,-1

h-n,-1

h-n-1,0

Gf

p = -5

xn-1

p = -4p = -4p = -4

Now we can, for the first time, push the flow unit towards yn-1 :

n,1

p = -3

n-1,0

p = -4h-n,-1

n,1

h-n,-1

n,1

h-n,0

n,0

h-1,01,0

1

yn-1

h, 2

1,1

p = 0


-1

x1

x2 xn-2 xn

y1

y2yn

h-1, 4

h, 2 h, 2 h-1,1

h-n,-1

h-n-1,0

Gf

p = -5

xn-1

p = -4p = -4p = -4

We then relabel yn-1 :

n,1

p = -3

n-1,0

p = -4h-n,-1

n,1

h-n,-1

n,1

h-n,0

n,0

h-1,01,0

1

yn-1

h, 0

1,-1

p = -2


1-1

x1

x2 xn-2 xn

y1

y2yn

h-1, 4

h, 2 h, 2 h,1

h-n,-1

h-n-1,0

Gf

p = -5

xn-1

p = -4p = -4p = -4

The flow is pushed back to yn :

n,1

p = -3

n-1,0

p = -4h-n,-1

n,1

h-n,-1

n,1

h-n,0

n,0

h-1,01,0

yn-1

h, 0

p = -2


1-1

x1

x2 xn-2 xn

y1

y2yn

h-1, 4

h, 2 h, 2 h,0

h-n,-1

h-n-1,0

Gf

p = -5

xn-1

p = -4p = -4p = -4

Relabel yn again:

n,1

p = -4

n-1,0

p = -4h-n,-1

n,1

h-n,-1

n,1

h-n,0

n,0

h-1,11,-1

yn-1

h, 0

p = -2


1

-1

x1

x2 xn-2 xn

y1

y2yn

h-1, 4

h, 2 h, 2 h,0

h-n,-1

h-n-1,0

Gf

p = -5

xn-1

p = -4p = -4p = -4

The flow returns to xn with push:

n,1

p = -4

n-1,0

p = -4h-n,-1

n,1

h-n,-1

n,1

h-n,0

n,0

h,1

yn-1

h, 0

p = -2


1

-1

x1

x2 xn-2 xn

y1

y2yn

h-1, 4

h, 2 h, 2 h,0

h-n,-1

h-n-1,-1

Gf

p = -6

xn-1

p = -4p = -4p = -4

And xn is relabeled again:

n,1

p = -4

n-1,1

p = -4h-n,-1

n,1

h-n,-1

n,1

h-n,1

n,-1

h,0

yn-1

h, 0

p = -2


1

-1

x1

x2 xn-2 xn

y1

y2yn

h-1, 4

h, 2 h, 2 h,0

h-n,-1

h-n-1,-1

Gf

p = -6

xn-1

p = -4p = -4p = -4

We will stop here, the behavior is clear…

n,1

p = -4

n-1,1

p = -4h-n,-1

n,1

h-n,-1

n,1

h-n,1

n,-1

h,0

yn-1

h, 0

p = -2

Can We Get a Better Complexity?

Yes – using dynamic tree, we geta complexity for refine of 2lognm n m

Can We Get a Better Complexity?

Yes – using dynamic tree, we geta complexity for refine of

Use finger search trees to find the first active vertex

Use dynamic trees for performing non-saturating pushes more efficiently (keep the edge list in a dynamic tree)

2lognm n m

So Much for the Generic Method…

Refine Using Dinic’s Ideas

In the Generic Method, we used ideas given by Goldberg & Tarjan in their Max-Flow Algorithm.

Our Min-Cost Circulation problem can also be solved using Dinic’s ideas for solving the Max-Flow problem, namely Blocking Flows.

We call this the Blocking-Flow Method.

Recall that Dinic solved the Max-Flow problem solving O(n) Blocking-Flow problems.

It appears that refine can also be implemented in a similar manner!

Note – this does not imply that refine can be fully implemented by a Max-Flow algorithm (this is an open problem…)

Refine Using Dinic’s Ideas

The Blocking-Flow Method

Some definitions: Let G=(V,E), with 2 distinct vertices s and t. A pseudoflow is a flow if every vertex except s,t

has zero excess. A flow is blocking if every path from s to t

contains at least one saturates arc. Note that a maximum flow is blocking, but not

conversely!


Some definitions (cont.): Define S to be all the vertices reachable in the

current GA from vertices with positive excess.

| s.t. 0

and is reachable from inf

A

v V u V e uS

v u G

The Blocking-Flow MethodGA



Some definitions (cont.): Let N be the auxiliary network formed from GA

by adding:A source and a sink (s & t)An arc (s,v) of capacity ef(v), for every v with ef(v)>0An arc (v,t) of capacity -ef(v), for every v with ef(v)<0



S t

The Blocking-Flow MethodProcedure refine(ε,f,p):

[Initialization] ε ← ε/2; For every (v,w) in E do:

If cp(v,w) < 0 then f(v,w) ← u(v,w);[Loop] While f is not a circulation do:

Compute S, (price update)Build N, and find a blocking-flow fb on N (flow update)

Return (ε,f,p)End

,v S p v p v

, , , , ,A bv w E f v w f v w f v w

x1

x2x3

y1

y2

y3

9,0

10,-2 10,-2

10,1 10,1

10,1

10,-2

Example – The Initial G(Assume G = Gf)

Gf


ε ← ε/2 = 1

x1

x2x3

y1

y2

y3

9,0

10,-2 10,-2

10,1 10,1

10,1

10,-2

Gf



10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,1 10,1

10,1

10,2

Gf


The admissible graph

10

-10

x1

x2x3

y1

y2

y3

GA


S = {x3}

10

-10

x1

x2x3

y1

y2

y3

GA


Decrease price of vertices in S

10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,1 10,0

10,0

10,1

Gfp=-1


Constructing N

10

-10

x1

x2x3

y1

y2

y3

GA

ts

10

10


A blocking flow of 0 is found

10

-10

x1

x2x3

y1

y2

y3

GA

1010t

s

10

10


S is still {x3}

10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,1 10,0

10,0

10,1

Gfp=-1


Decrease price of vertices in S={x3}

10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,1 10,-1

10,-1

10,0

Gfp=-2


The admissible graph now:

10

-10

x1

x2x3

y1

y2

y3

10,-1

10,-1

GAp=-2


Constructing NA blocking flow of 0 is found

10

-10

x1

x2x3

y1

y2

y3

10,-1

10,-1GAp=-2

1010t

s

10

10


S is now {x1,x2,x3}

10

-10

x1

x2x3

y1

y2

y3

9,0

10,2 10,2

10,1 10,-1

10,-1

10,0

Gfp=-2


Decrease price of vertices in S={x1,x2,x3}

10

-10

x1

x2x3

y1

y2

y3

9,-1

10,2 10,2

10,1 10,-1

10,-1

10,-1

Gfp=-3p=-1

p=-1


The admissible graph now:

10

-10

x1

x2x3

y1

y2

y3

9,-1

10,-1

10,-1

10,-1

GAp=-3

p=-1p=-1

An Iteration of RefineConstructing NA BF is found

10

-10

x1

x2x3

y1

y2

y3

9,-1

10,-1

10,-1

10,-1

Gfp=-3

p=-1p=-1

ts

10

10


10

-10

x1

x2x3

y1

y2

y3

9,-1

10,-1

10,-1

10,-1

Gfp=-3

p=-1p=-1

Constructing NA BF is found (9 flow units)

ts

10

10

An Iteration of RefineAdd the BF found to the pseudo flow:

1

-1

x1

x2x3

y1

y2

y3

9, 1

10,2 10,2

10,1 10,-1

1,-1

10,-1

Gfp=-3

p=-1p=-1

9, 1

An Iteration of RefineThe admissible graph:

1

-1

x1

x2x3

y1

y2

y3

10,-1

1,-1

10,-1

GAp=-3

p=-1p=-1


S={x1,x2,x3,y3}

1

-1

x1

x2x3

y1

y2

y3

10,-1

1,-1

10,-1

GAp=-3

p=-1p=-1

An Iteration of RefineDecrease prices of vertices in S

1

-1

x1

x2x3

y1

y2

y3

9, 2

10,2 10,1

10,1 10,-1

10,-1

Gfp=-4p=-2

p=-2

9, 1

p=-1

1,-1

An Iteration of RefineConstruct NA 0 BF is found

1

-1

x1

x2x3

y1

y2

y3

10,-1

10,-1

Gfp=-4

p=-2p=-2

p=-1

1,-1

ts

1

1


S is still {x1,x2,x3,y3}

1

-1

x1

x2x3

y1

y2

y3

9, 2

10,2 10,1

10,1 10,-1

10,-1

Gfp=-4

p=-2p=-2

9, 1

p=-1

1,-1


1

-1

x1

x2x3

y1

y2

y3

9, 3

10,2 10,0

10,1 10,-1

10,-1

Gfp=-5p=-3

p=-3

9, 1

p=-2

1,-1

An Iteration of RefineConstruct NA 0 BF is found

1

-1

x1

x2x3

y1

y2

y3

10,-1

10,-1

Gf 1,-1p=-5

p=-3p=-3

p=-2

ts

1

1


S is still {x1,x2,x3,y3}

1

-1

x1

x2x3

y1

y2

y3

9, 3

10,2 10,0

10,1 10,-1

10,-1

Gfp=-5

p=-3p=-3

9, 1

p=-2

1,-1


1

-1

x1

x2x3

y1

y2

y3

9, 3

10,2 10,-1

10,1 10,-1

10,-1

Gfp=-6p=-4

p=-4

9, 1

p=-3

1,-1


1

-1

x1

x2x3

y1

y2

y3

10,-1

10,-1

10,-1

GAp=-6

p=-4p=-4

p=-3

1,-1

An Iteration of RefineConstruct N

1

-1

x1

x2x3

y1

y2

y3

10,-1

10,-1

10,-1

p=-6p=-4

p=-4

p=-3

1,-1

ts

1

1


S is now {x1,x2,x3,y2,y3}

1

-1

x1

x2x3

y1

y2

y3

9, 3

10,2 10,-1

10,1 10,-1

10,-1

Gfp=-6

p=-4p=-4

9, 1

p=-3

1,-1


1

-1

x1

x2x3

y1

y2

y3

9, 4

10,1 10,-1

10,1 10,-1

10,-1

Gfp=-7p=-5

p=-5

9, 1

p=-4p=-1

1,-1

An Iteration of RefineBuild N

1

-1

x1

x2x3

y1

y2

y3

10,-1

10,-1

10,-1

p=-7p=-5

p=-5

p=-4p=-1

ts

1

1

1,-1


S is still {x1,x2,x3,y2,y3}

1

-1

x1

x2x3

y1

y2

y3

9, 4

10,1 10,-1

10,1 10,-1

10,-1

Gfp=-7

p=-5p=-5

9, 1

p=-4p=-1

1,-1


1

-1

x1

x2x3

y1

y2

y3

9, 5

10,0 10,-1

10,1 10,-1

10,-1

Gfp=-8p=-6

p=-6

9, 1

p=-5p=-2

1,-1

An Iteration of RefineBuild N

1

-1

x1

x2x3

y1

y2

y3

10,-1

10,-1

10,-1

p=-8p=-6

p=-6

p=-5p=-2

1,-1

ts

1

1


S is still {x1,x2,x3,y2,y3}

1

-1

x1

x2x3

y1

y2

y3

9, 5

10,0 10,-1

10,1 10,-1

10,-1

Gfp=-8

p=-6p=-6

9, 1

p=-5p=-2

1,-1


1

-1

x1

x2x3

y1

y2

y3

9, 6

10,-1 10,-1

10,1 10,-1

10,-1

Gfp=-9p=-7

p=-7

9, 1

p=-7p=-3

Decrease prices of vertices in S

1,-1


1

-1

x1

x2x3

y1

y2

y3

10,-1

10,-1

10,-1

GA 1,-1

10,-1

p=-9p=-7

p=-7

p=-7p=-3

An Iteration of RefineBuild N and find a BF (1 flow unit)

1

-1

x1

x2x3

y1

y2

y3

10,-1 10,-1

10,-1

10,-1

Gfp=-9

p=-7p=-7

p=-7p=-3

1,-1

ts

1

1

An Iteration of RefineAdd the BF to the pseudo flow

x1

x2x3

y1

y2

y3

9, 6

9,-1 9,-1

10,1 10,-1

9,-1

Gfp=-9

p=-7p=-7

9, 1

p=-7p=-3

1,1 1,1

1,1

1,-1

p=0

An Iteration of RefineNo more active vertices – iteration over!

x1

x2x3

y1

y2

y3

9, 6

9,-1 9,-1

10,1 10,-1

9,-1

Gfp=-9

p=-7p=-7

9, 1

p=-7p=-3

1,1 1,1

1,1

1,-1

p=0

Blocking-Flow Method Correctness

The following strong lemma proves the main ideas of the Blocking-Flow Method:


Lemma 8.1a. The set S computed in the inner loop

contains only vertices with ef(v)≥0.b. At the beginning of every iteration of the

inner loop, f is ε-optimal w.r.t. p; Decreasing the prices of vertices in S preserve ε-optimality.

c. GA remains acyclic throughout the algorithm (as before).


Lemma 8.1a – Proof:By induction on the number of iterations of

the inner loop: Base – Initially, there are no arcs in GA, so only

ef(v)>0 vertices are contained in S. Step – We add a blocking flow on N, which, by

definition, disconnects all paths from ef(v)>0 vertices to ef(v)<0 vertices!


We will look at a path from s to t

After the BF is found at least one arc is saturated on every path, hence the green vertex is not reachable from the red (active) vertex.

What happens if the first or last arc is saturated?

s t


Lemma 8.1b – Proof:By the same induction:

Base – At the first iteration, f is ε-optimal. Step – The price decrease of vertices in S

preserves ε-optimality:The decrease does not affect arcs from S to S or from

S to S.Arcs from S to S have their reduced cost increased.Arcs from S to S have their reduced cost decreased

by ε, but recall that beforehand cp(v,w)≥0 !


Ss t

Suppose this is the state before the price reduction in S:

S


Ss t

Arcs within S are not effected

S


Ss t

Same goes for arcs in S

S


Ss t

Arcs going from S to S have their reduced cost increased

S


Ss t

Arcs going from S to S have their reduced cost decreased, but still greater than -ε!

S


Lemma 8.1c – Proof:By the same induction:

Base – Initially, there are no arcs in GA. Step – The price decrease can add arcs from S

to S, but will make all arcs from S to S non- admissible, thus GA remains acyclic.


From 8.1b we conclude that f is ε-optimal at all times during the inner loop.

Therefore, if refine terminates, the returned f is an ε-optimal circulation, as needed!

Thus the correction of the Blocking-Flow Method is proven.

Blocking-Flow Method Complexity

We will bound the number of iterations of the inner loop of refine:

Lemma 8.2The number of inner loops is at most 3n


Proof:Note that the inner loop never changes the

vertex excess from non-negative to negative or from non-positive to positive.




Only non-negative excess vertices have their price changed by iε after i iterations (this results from 8.1a).




Only non-negative excess vertices have their price changed by iε after i iterations.

Recall Lemma 5.8 –The price reduction is at most 3nε, and the bound is obtained!


Lemma 8.3The Blocking-Flow Method takes

time, giving a total algorithm bound of

where B(n,m) is the time to find a blocking flow in an acyclic network (here we make use of 8.1c)

,O nB n m

, logO nB n m nC

And That’s That for Dinic…

ConclusionThe main ideas presented:Successive approximation – finding an

approximated (ε-optimal) solution “good enough” for optimality.

The algorithm framework achieves this concept using a refine method.

Recall that this only works for integer costs, to get log(nC) iterations.

ConclusionThe main ideas presented:We saw two ways for implementing refine:

The Generic Method – push & relabel The Blocking-Flow Method

Both methods can be implemented efficiently to achieve complexity of(or using Dynamic Trees)

3O n

2logO nm n m

Questions???

Documents

Finding Minimum-Cost Circulation By Successive Approximation