A New Approach to the Maximum-Flow Problem

Post on 08-Jan-2016

58 views 1 download

description

A New Approach to the Maximum-Flow Problem. Andrew V. Goldberg, Robert E. Tarjan Presented by Andrew Guillory. Outline. Background Definitions Push-Relabel Algorithm Correctness / Termination Proofs Implementation. Maximum Flow Problem. Classic problem in operations research - PowerPoint PPT Presentation

Transcript of A New Approach to the Maximum-Flow Problem

A New Approach to the Maximum-Flow ProblemAndrew V. Goldberg, Robert E. Tarjan

Presented by Andrew Guillory

Outline

Background Definitions Push-Relabel Algorithm Correctness / Termination Proofs Implementation

Maximum Flow Problem

Classic problem in operations research Many problems reduce to max flow

Maximum cardinality bipartite matching Maximum number of edge disjoint paths Minimum cut (Max-Flow Min-Cut Theorem)

Machine learning applications Structured Prediction, Dual Extragradient and Bregman

Projections (Taskar, Lacoste-Julien, Jordan JMLR 2006) Local Search for Balanced Submodular Clusterings

(Narasimhan, Bilmes, IJCAI 2007)

Relation to Optimization

Special case of submodular function minimization

Special case of linear programming Integer edge capacities permit integer

maximum flows (constructive proof)

History of Algorithms

Augmenting Paths based algorithmsFord-Fulkerson (1962) O(mU)Edmonds-Karp (1969) O(nm3)… O(n3) O(nmlog(n)) O(nmlog(U))

Push-Relabel based algorithmsGoldberg (1985) O(n3)Goldberg and Tarjan (1986) O(nmlog(n2/m))Ahuja and Orlin O(nm + n2log(U))

Outline

Background Definitions Push-Relabel Algorithm Correctness / Termination Proofs Implementation

Definitions

Graph G = (V, E) |V| = n |E| = m

G is a flow network if it hassource s and sink tcapacity c(v,w) for each edge (v,w) in Ec(v,w) = 0 for (v,w) not in E

Definitions (continued)

A flow f on G is a real value function on vertex pairs f(v,w) <= c(v,w) for all (v,w) f(v,w) = -f(w,v)∑uf(u,v) = 0 for all v in V - {s,t}

Value of a flow |f| is ∑vf(v,t) Maximum flow is a flow of maximum value

Definitions (continued again)

A preflow f on G is a real value function on vertex pairs f(v,w) <= c(v,w) for all (v,w) f(v,w) = -f(w,v)∑uf(u,v) >= 0 for all v in V - {s}

Flow excess e(v) = ∑uf(u,v) Intuition: flow into a vertex can exceed flow out

Outline

Background Definitions Push-Relabel Algorithm Correctness / Termination Proofs Implementation

Intuition

Starting with a preflow, push excess flow closer towards sink

If excess flow cannot reach sink, push it backwards to source

Eventually, preflow becomes a flow and in fact the maximum flow

Residual Graph

Residual capacity rf(v, w) of a vertex pair is c(v, w) – f(v, w)

If v has positive excess and (v,w) has residual capacity, can push

δ = min(e(v), rf(v, w)) flow from v to w

Edge (v,w) is saturated if rf(v, w) = 0

Residual graph Gf = (V, Ef) where Ef is the set of residual edges (v,w) with rf(v, w) > 0

Labeling

A valid labeling is a function d from vertices to nonnegative integersd(s) = nd(t) = 0d(v) <= d(w) + 1 for every residual edge

If d(v) < n, d(v) is a lower bound on distance to sink

If d(v) >= n, d(v) - n is a lower bound on distance to source

Push Operation

Push(v,w)Precondition: v is active (e(v) > 0) and

rf(v, w) > 0 and d(v) = d(w) + 1

Action: Push δ = min(e(v), rf(v, w)) from v to w

f(v,w) = f(v,w) + δ; f(w,v) = f(w,v) – δ;e(v) = e(v) - δ; e(w) = e(w) + δ;

Relabel Operation

Relabel(v)

Precondition: v is active (e(v) > 0) and

rf(v, w) > 0 implies d(v) <= d(w)

Action: d(v) = min{d(w) + 1 | (v,w) in Ef}

Generic Push-Relabel Algorithm

Starting from an initial preflow

<<loop>>

While there is an active vertex

Chose an active vertex v

Apply Push(v,w) for some w or Relabel(v)

Example

0/30/1

0/2

Flow Network

S T

Example

4

0 0

0

3/30/1

0/2

S T

Initial preflow / labeling

Example

4

0 0

0

3/30/1

0/2

S T

Select an active vertex

Example

4

1 0

0

3/30/1

0/2

Relabel active vertex

S T

Example

4

1 0

0

3/30/1

0/2

Select an active vertex

S T

Example

4

1 0

0

3/31/1

0/2

Push excess from active vertex

S T

Example

4

1 0

0

3/31/1

0/2

Select an active vertex

S T

Example

4

1 1

0

3/31/1

0/2

Relabel active vertex

S T

Example

4

1 1

0

3/31/1

0/2

Select an active vertex

S T

Example

4

1 1

0

3/31/1

1/2

Push excess from active vertex

S T

Example

4

1 1

0

3/31/1

1/2

Select an active vertex

S T

Example

4

5 1

0

3/31/1

1/2

Relabel active vertex

S T

Example

4

5 1

0

3/31/1

1/2

Select an active vertex

S T

Example

4

5 1

0

1/31/1

1/2

Push excess from vertex

S T

Example

4

5 1

0

1/31/1

1/2

Maximum flow

S T

Outline

Background Definitions Push-Relabel Algorithm Correctness / Termination Proofs Implementation

Correctness

Lemma 2.1 If f is a preflow, d is a valid labeling, and v is active, either push or relabel is applicable to v

Lemma 3.1 The algorithm maintains a valid labeling d

Theorem 3.2 A flow is maximum iff there is no path from s to t in Gf (Ford and Fulkerson [7])

Correctness (continued)

Lemma 3.3 If f is a preflow and d is a valid labeling for f, there is no path from s to t in Gf

Proof by contradictionPath s, v0, v1, …, vl, t implies that

d(s) <= d(v0) + 1 <= d(v1) + 2 <= …

<= d(t) + l < nWhich contradicts d(s) = n

Correctness (continued)

Theorem 3.4 If the algorithm terminates with a valid labeling, the preflow is a maximum flow If the algorithm terminates, all vertices have

zero excess (preflow is a flow)By Lemma 3.3 the sink is not reachable from

the sourceBy Theorem 3.2 the flow is maximum

Termination

Lemma 3.5 If f is a preflow and v is an active vertex then the source is reachable from v in Gf

Lemma 3.6 A vertex’s label never decreases

Termination (continued)

Lemma 3.7 At any time the label of any vertex is at most 2n – 1Only active vertex labels are changedActive vertices can reach sPath v, v0, v1, …, vl, s implies that

d(v) <= d(v0) + 1 <= d(v1) + 2 <= …

<= d(s) + l <= n + n - 1

Termination (continued)

Lemma 3.8 There are at most 2n2 labeling operationsOnly the labels corresponding to V-{s,t} may

be relabeledEach of these n – 2 labels can only increaseAt most (2n – 1) (n – 2) relabelings

Termination (continued)

Lemma 3.9 The number of saturating pushes is at most 2nmFor any pair (v,w) d(w) must increase by 2 between

saturating pushes from v to wSimilarly d(v) must increase by 2 between pushes

from w to vd(v) + d(w) >= 1 on the first saturating pushd(v) + d(w) <= 4n - 3 on the lastAt most 2n - 1 saturating pushes per edge

Termination (continued)

Lemma 3.10 The number of nonsaturating pushes is at most 4n2m Φ = ∑v d(v) where v is active

Each nonsaturating push causes Φ to decrease by at least 1 The total increase in Φ from saturating pushes is

(2n – 1) 2nm The total increase in Φ from relabeling is

(2n – 1)(n – 2) Φ is 0 initially and Φ at termination

Termination

Theorem 3.11 The algorithm terminates in O(n2m)

Total time =

# nonsaturating pushes

+ #saturating pushes

+ #relabeling operations

4n2m + 2nm + 2n2 = O(n2m)

Outline

Background Definitions Push-Relabel Algorithm Correctness / Termination Proofs Implementation

Implementation

At each step select an active vertex and apply either Push or Relabel

Problem: Determining which operation to perform and in the case of Push finding a residual edge

Solution: For each vertex maintain a list of edges which touch that vertex and a current edge

Push/Relabel Operation

Push/Relabel(v)

Precondition: v is active

Action:

If Push(v,w) is applicable to current edge (v,w) then Push(v,w)

Else if (v,w) is not the last edge advance current edge

Else reset the current edge and Relabel(v)

Push/Relabel Operation

Lemma 4.1 The push/relabel operation does a relabeling only when relabeling is applicable

Theorem 4.2 The push/relabel implementation runs in O(nm) time plus O(1) time per nonsaturating push operation

O(n3) bound

We can select vertices in arbitrary order Certain vertex selection strategies give

O(n3) boundsFirst-in, first-out method (proved in paper)Maximum distance method (proved here)Wave method

Maximum distance method

At each step, select the active vertex with maximum distance d(v)

Theorem The maximum distance method performs at most 4n3 nonsaturating pushes

Corollary The maximum distance method runs in time O(n3) using the push/relabel implementation

Proof

Consider D = maxx d(x) where x is active D only increases because of relabeling D increases at most 2n2 times D starts at 0 and ends nonnegative D changes at most 4n2 times There is at most one nonsaturating push

per node per value of D