The Primal-Dual Method for Approximation Algorithmsapproximation algorithm for vertex-cover. Goemans and Williamson formalize this approach in 1992. Result is a general toolkit for

Jochen Könemann, September 25, 2004 Group Strategyproof Mechanisms for Steiner Forests - p. 1/44

The Primal-Dual Method for ApproximationAlgorithms

Jochen KönemannUniversity of Waterloo

http://www.math.uwaterloo.ca/~jochen

● What is the primal-dual method?

Primal-Dual: First Steps

Primal-Dual: Hitting Sets

Primal-Dual: Steiner Trees

Primal Dual: MCF


What is the primal-dual method?

■ Originally proposed by Dantzig, Ford, and Fulkerson in 1956as an alternate method to solve linear programs exactly






Primal Dual: MCF




■ Method did not survive... but: Revised version of it hasbecome immensely popular for solving combinatorialoptimization problems.






Primal Dual: MCF





Examples: Dijkstra’s shortest path algorithm, Ford andFulkerson’s network flow algorithm, Edmond’s non-bipartitematching method, Kuhn’s assignment algorithm, ...






Primal Dual: MCF





Examples: Dijkstra’s shortest path algorithm, Ford andFulkerson’s network flow algorithm, Edmond’s non-bipartitematching method, Kuhn’s assignment algorithm, ...

Main feature: Reduce weighted optimization problems toeasier unweighted ones.






Primal Dual: MCF



■ All of the previous problems are in P. Can we extend thismethod to NP-hard problems?






Primal Dual: MCF




Yes! Bar-Yehuda and Even use this in 1981 in theirapproximation algorithm for vertex-cover.






Primal Dual: MCF




Yes! Bar-Yehuda and Even use this in 1981 in theirapproximation algorithm for vertex-cover.

■ Goemans and Williamson formalize this approach in 1992.Result is a general toolkit for developing approximationalgorithms for NP-hard optimization problems.

■ The last 10 years have seen literally hundreds of papers thatuse the primal-dual framework.




● An example: Vertex Cover

● An ILP for Vertex Cover

● Dual LP

● PD: Main Ideas

● Vertex-cover: A PD-Algorithm

● Petersen graph example

● Approximation algorithms



Primal Dual: MCF








● Dual LP

● PD: Main Ideas






Primal Dual: MCF


An example: Vertex Cover

■ Goal: Find a minimum subset of the vertices C such thate ∩ C 6= ∅ for all edges e.

■ Here’s a vertex cover of size 6.






● Dual LP

● PD: Main Ideas






Primal Dual: MCF


An example: Vertex Cover

■ Goal: Find a minimum subset of the vertices C such thate ∩ C 6= ∅ for all edges e.

■ Here’s a vertex cover of size 6.






● Dual LP

● PD: Main Ideas






Primal Dual: MCF


An ILP for Vertex Cover

Variables:






● Dual LP

● PD: Main Ideas






Primal Dual: MCF



Variables: For node i have variable xi.Want

xi =

{

1 : Node i in vertex cover

0 : Otherwise.

Variables like this are called indicatorvariables.






● Dual LP

● PD: Main Ideas






Primal Dual: MCF




xi =

{


0 : Otherwise.

Variables like this are called indicatorvariables.Constraints: Each edge needs to becovered.






● Dual LP

● PD: Main Ideas






Primal Dual: MCF




xi =

{


0 : Otherwise.

Variables like this are called indicatorvariables.Constraints: Each edge needs to becovered.Example: Edge (0, 5)

x0 + x5 ≥ 1






● Dual LP

● PD: Main Ideas






Primal Dual: MCF




xi =

{


0 : Otherwise.


x0 + x5 ≥ 1

Objective function: Minimize cardi-nality of vertex cover.






● Dual LP

● PD: Main Ideas






Primal Dual: MCF




xi =

{


0 : Otherwise.


x0 + x5 ≥ 1

Objective function: Minimize cardi-nality of vertex cover.

minimize9

∑

j=0

xj






● Dual LP

● PD: Main Ideas






Primal Dual: MCF


Primal ILP

minimize∑

v∈V

xv

s.t. xv + xu ≥ 1 ∀(u, v) ∈ E

xv ∈ {0, 1} ∀v ∈ V






● Dual LP

● PD: Main Ideas






Primal Dual: MCF


Dual LP

Dual LP has a variable ye for each edge e ∈ E.

maximize∑

e∈E

ye

s.t.∑

e∈δ(v)

ye ≤ 1 ∀v ∈ V

ye ≥ 0 ∀e ∈ E

δ(v): Edges incident to vertex v ∈ V






● Dual LP

● PD: Main Ideas






Primal Dual: MCF


Primal-dual technique: Main Ideas

■ Construct integral primal and dual feasible solution at thesame time: x and y

■ Show that∑

j

xj ≤ α ·∑

i

yi

For some α.■ Prove a worst-case upper-bound on α.■ Result: For every instance we compute ...

1. an integral and feasible primal solution x, and2. a proof that its value is within a factor of α of the best

possible solution






● Dual LP

● PD: Main Ideas






Primal Dual: MCF




■ Show that∑

j

xj ≤ α ·∑

i

yi

For some α.

■ Prove a worst-case upper-bound on α.■ Result: For every instance we compute ...


possible solution






● Dual LP

● PD: Main Ideas






Primal Dual: MCF




■ Show that∑

j

xj ≤ α ·∑

i

yi

For some α.■ Prove a worst-case upper-bound on α.

■ Result: For every instance we compute ...1. an integral and feasible primal solution x, and2. a proof that its value is within a factor of α of the best

possible solution






● Dual LP

● PD: Main Ideas






Primal Dual: MCF




■ Show that∑

j

xj ≤ α ·∑

i

yi

For some α.■ Prove a worst-case upper-bound on α.■ Result: For every instance we compute ...


possible solution






● Dual LP

● PD: Main Ideas






Primal Dual: MCF


Vertex-cover: A PD-Algorithm

1: yuv ← 0 for all edges (u, v)2: C ← ∅3: while C is not a vertex cover do4: Choose uncovered edge (u, v)5: Increase yuv until

∑

z

ywz = 1

for z ∈ {u, v}6: C ← C ∪ {z}7: end while






● Dual LP

● PD: Main Ideas






Primal Dual: MCF


Petersen graph example

Primal solution:Dual solution:






● Dual LP

● PD: Main Ideas






Primal Dual: MCF









● Dual LP

● PD: Main Ideas






Primal Dual: MCF









● Dual LP

● PD: Main Ideas






Primal Dual: MCF









● Dual LP

● PD: Main Ideas






Primal Dual: MCF









● Dual LP

● PD: Main Ideas






Primal Dual: MCF









● Dual LP

● PD: Main Ideas






Primal Dual: MCF









● Dual LP

● PD: Main Ideas






Primal Dual: MCF









● Dual LP

● PD: Main Ideas






Primal Dual: MCF



Primal solution:Dual solution:Primal solution: 8Dual solution: 5






● Dual LP

● PD: Main Ideas






Primal Dual: MCF




Q: How bad can the primal/dual ratio be?






● Dual LP

● PD: Main Ideas






Primal Dual: MCF




Q: How bad can the primal/dual ratio be?

A: Not too bad. The primal solution has at most two verticesper dual edge. Hence, we always have

|C| ≤ 2 ·∑

e

ye






● Dual LP

● PD: Main Ideas






Primal Dual: MCF


Approximation algorithms

■ The previous algorithm for vertex-cover is a 2-approximationalgorithm.We also say: Its performance guarantee is 2.

■ Formally: The algorithm return a solution that has size atmost twice the optimum for all instances!






● Dual LP

● PD: Main Ideas






Primal Dual: MCF





■ Q: Given a problem and an LP with LP/IP-gap α > 1.Can there be a primal-dual approximation algorithm usingthis LP with performance guarantee less than α?






● Dual LP

● PD: Main Ideas






Primal Dual: MCF





■ Q: Given a problem and an LP with LP/IP-gap α > 1.Can there be a primal-dual approximation algorithm usingthis LP with performance guarantee less than α?A: No! Cannot find a good dual lower-bound for IP/LP-gapinstance.





● How to hit sets...

● Linear Program

● Primal-Dual Algorithm

● Analysis

● Set Cover


Primal Dual: MCF








● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF


How to hit sets...

Let us define a general problem called Hitting Sets.■ Input: Ground set E and m subsets

Ti ⊆ E

for 1 ≤ i ≤ m. Also have cost ce for all e ∈ E.■ Goal: Find H ⊆ E of minimum total cost, s.t.

H ∩ Ti 6= ∅

for all 1 ≤ i ≤ m.

Q: How can we formulate vertex-cover as a hitting-setproblem?Q: Give a hitting-set formulation for shortest s, t-path inundirected graphs!






● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF


How to hit sets...


Ti ⊆ E


H ∩ Ti 6= ∅


Q: How can we formulate vertex-cover as a hitting-setproblem?

Q: Give a hitting-set formulation for shortest s, t-path inundirected graphs!






● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF


How to hit sets...


Ti ⊆ E


H ∩ Ti 6= ∅


Q: How can we formulate vertex-cover as a hitting-setproblem?A: The ground-set E is the set of all vertices.Each edge (u, v) corresponds to a set {u, v} that needs to behit.







● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF


How to hit sets...


Ti ⊆ E


H ∩ Ti 6= ∅









● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF


How to hit sets...


Ti ⊆ E


H ∩ Ti 6= ∅



Q: Give a hitting-set formulation for shortest s, t-path inundirected graphs!A: Ground set E is set of all edges. Sets to hit are all s, t-cuts.Convince yourself that the shortest s, t is the minimum-costsolution!






● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF


Hitting Set: Primal and Dual

minimize∑

e∈E

ce · xe (P )

s.t.∑

e∈Ti

xe ≥ 1 ∀1 ≤ i ≤ m

xe ≥ 0 ∀e ∈ E

maximizem

∑

i=1

yi (D)

s.t.∑

Ti:e∈Ti

yi ≤ ce ∀e ∈ E

yi ≥ 0 ∀1 ≤ i ≤ m






● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF


Hitting Set: Primal-Dual algorithm

■ Want to use the LP formulation and its dual to develop aprimal-dual algorithm.

Ideas, suggestions, ... anybody?






● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF




Ideas, suggestions, ... anybody?■ Recall from vertex-cover PD-algo:

Keep feasible dual solution y and include vertex v into coveronly if

∑

(u,v)

yuv = 1






● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF




Ideas, suggestions, ... anybody?■ Recall from vertex-cover PD-algo:

Keep feasible dual solution y and include vertex v into coveronly if

∑

(u,v)

yuv = 1

■ Idea: Let H be the current hitting set, y be the correspondingdual, and let Ti be a set that has not been hit.Increase yi until

∑

Tj :e∈Tj

yj = ce

for some e ∈ Ti. Include e into H.






● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF


Analysis

Let H be the final feasible hitting set and y the correspondingfeasible dual solution.

∑

e∈H

ce = ?






● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF


Analysis


∑

e∈H

ce =∑

e∈H

∑

Ti:e∈Ti

yi

Since ce =∑

Ti:e∈Tiyi = ce when e is included into H.






● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF


Analysis


∑

e∈H

ce =∑

e∈H

∑

Ti:e∈Ti

yi

=m

∑

i=1

|Ti ∩H| · yi






● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF


Analysis


∑

e∈H

ce =∑

e∈H

∑

Ti:e∈Ti

yi

=m

∑

i=1

|Ti ∩H| · yi

Vertex cover: Ti’s correspond to edges. Hence: |Ti ∩H| ≤ 2!






● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF


Set Cover

Problem input: Elements U and subsets S1, . . . , Sn of U .Goal: Select smallest number of sets that cover U .






● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF


Set Cover


Theorem [Feige]

There is no (log(n) − ε)-approximation for the set-cover problemand any ε > 0 unless NP = P.






● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF


Set Cover


Theorem [Feige]


Q: But what if each element occurs in at most f sets?






● Linear Program


● Analysis

● Set Cover


Primal Dual: MCF


Set Cover


Theorem [Feige]


Q: But what if each element occurs in at most f sets?

A: Hitting set analysis gives f -approximation since eachelement e ∈ U corresponds to a set in hitting set instance and

|Te ∩H| ≤ f

for all e ∈ U .






● Intro

● Example

● Primal LP: Steiner Cuts

● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF








● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF


Steiner trees: Intro

■ Roots of problem can be traced back to Gauss: He mentionsthe problem in a letter to Schumacher.

■ Input:◆ Undirected graph G = (V, E)◆ Terminals R ⊆ V◆ Steiner nodes V \R◆ Edge costs ce ≥ 0 for all e ∈ E.

■ Goal: Compute min-cost tree T in G that spans nodes in R






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF


Steiner trees: Example

Terminal nodes:Steiner nodes:






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF


Steiner trees: Example

Terminal nodes:Steiner nodes:






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF


Primal LP: Steiner Cuts

■ Primal has variables xe for all e ∈ E.xe = 1 if e is in Steiner tree, 0 otherwise

■ Steiner cuts: Subsets of nodes that separates at least oneterminal pair (s, t) ∈ R

s

t

Any feasible Steiner tree must contain at least one of the rededges!






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF



■ Primal has variables xe for all e ∈ E.xe = 1 if e is in Steiner tree, 0 otherwise

■ Steiner cuts: Subsets of nodes that separates at least oneterminal pair (s, t) ∈ R

s

t

Any feasible Steiner tree must contain at least one of the rededges!






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF



Primal LP has one constraints for each Steiner cut.

minimize∑

e∈E

ce · xe

s.t.∑

e∈δ(U)

xe ≥ 1 ∀ Steiner cuts U

xe ≥ 0 ∀e ∈ E

δ(U): Edges with one endpoint in U






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF


Steiner trees: Dual LP

■ Variables: yU for Steiner cut U

■ Constraints?






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




■ Constraints? One for each edge:∑

U :e∈δ(U)

yU ≤ ce






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




■ Constraints? One for each edge:∑

U :e∈δ(U)

yU ≤ ce

■ Objective function: maximize∑

U∈U yU






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF


Steiner trees: Primal and dual LP’s

minimize∑

e∈E

ce · xe (P )

s.t.∑

e∈δ(U)

xe ≥ 1 ∀ Steiner cuts U

xe ≥ 0 ∀e ∈ E

maximize∑

U

yU (D)

s.t.∑

U :e∈δ(U)

yU ≤ ce ∀e ∈ E

yU ≥ 0 ∀U ∈ U






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF


Dual LP: Pictorial View

■ Can visualize yU as a disk around U with radius yU .Example: Terminal pair (s, t) ∈ R, edge (s, t) with cost 4

s t

ys = yt = 0






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




s t1

ys = yt = 1






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




s t2

ys = yt = 2 Have: ys + yt = 4 = cst. Edge (s, t) is tight.






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF


Steiner trees: Algorithm

■ Algorithm always keeps:◆ Infeasible forest F◆ Feasible dual solution y






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




■ Connected components of F are Steiner cuts






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




■ Connected components of F are Steiner cuts■ Increase duals corresponding to connected components of

F .Stop increasing as soon as

∑

e∈P

∑

U :e∈δ(U)

yU =∑

e∈P

ce

for some path P connecting terminals in different connectedcomponents.






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




■ Connected components of F are Steiner cuts■ Increase duals corresponding to connected components of

F .Stop increasing as soon as

∑

e∈P

∑

U :e∈δ(U)

yU =∑

e∈P

ce

for some path P connecting terminals in different connectedcomponents.

At this point add the edges of P to F .






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF


Algorithm SF: An Example Run

Algorithm constructs primal and dual solution at the same time.

s1

t3 t2, s3

t1

s2






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




s1

t3 t2, s3

t1

s2






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




s1

t3 t2, s3

t1

s2






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




s1

t3 t2, s3

t1

s2






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




s1

t3 t2, s3

t1

s2






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF


Steiner trees: Analysis

■ Will now show that SF is a 2-approximation for Steiner treeproblem.

Main tool: Weak duality.Let T be the final tree and y is the constructed dual. Willshow

c(T ) ≤ 2 ·∑

Steiner cut U

yU

■ Proof is by induction over time in algorithm.

Let M be an active moat at time t in SF. TM is the treecontained in M . Want to show

c(TM ) ≤ 2∑

U⊆M

yU − 2t






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF



■ Will now show that SF is a 2-approximation for Steiner treeproblem.

Main tool: Weak duality.Let T be the final tree and y is the constructed dual. Willshow

c(T ) ≤ 2 ·∑

Steiner cut U

yU

■ Proof is by induction over time in algorithm.

Let M be an active moat at time t in SF. TM is the treecontained in M . Want to show

c(TM ) ≤ 2∑

U⊆M

yU − 2t






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF



■ Want to show IH for all t ≥ 0 inductively:

c(TM ) ≤ 2∑

U⊆M

yU − 2t (IH)

■ [t = 0] Clear since forest is empty.■ [t ≥ 0] Suppose (IH) holds at time t. Let t′ > t and no merger

in time interval [t, t′].■ LHS of (IH) changes by: 0

RHS of (IH) changes by: 2(t′ − t)− 2(t′ − t) = 0






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




c(TM ) ≤ 2∑

U⊆M

yU − 2t (IH)

■ [t = 0] Clear since forest is empty.

■ [t ≥ 0] Suppose (IH) holds at time t. Let t′ > t and no merger

in time interval [t, t′].■ LHS of (IH) changes by: 0

RHS of (IH) changes by: 2(t′ − t)− 2(t′ − t) = 0






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




c(TM ) ≤ 2∑

U⊆M

yU − 2t (IH)


in time interval [t, t′].

t

■ LHS of (IH) changes by: 0RHS of (IH) changes by: 2(t′ − t)− 2(t′ − t) = 0






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




c(TM ) ≤ 2∑

U⊆M

yU − 2t (IH)


in time interval [t, t′].

t′

■ LHS of (IH) changes by: 0RHS of (IH) changes by: 2(t′ − t)− 2(t′ − t) = 0






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




c(TM ) ≤ 2∑

U⊆M

yU − 2t (IH)

■ [t ≥ 0] Suppose (IH) holds at time t for moats M ′ and M ′′.SF merges these two components at time t.

v uP

U′ U′′

■ Main idea: Path P costs at most 2t!






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




c(TM ) ≤ 2∑

U⊆M

yU − 2t (IH)


v uP

U′ U′′







● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




v uP

U′ U′′







● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF




v uP

U′ U′′


c(TU ′∪U ′′) = c(TU ′) + c(TU ′′) + c(P )

≤ 2∑

U⊆U ′∪U

yU − 4t + 2t

= 2∑

U⊆U ′∪U

yU − 2t






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF


Steiner trees: Wrapping Up

■ Suppose the algorithm finishes at time t∗ with tree T ∗

■ Previous proof shows

c(T ∗) ≤ 2 ·∑

U∈U

yU − 2t∗ (*)






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF





c(T ∗) ≤ 2 ·∑

U∈U

yU − 2t∗ (*)

■ At any time 0 ≤ t ≤ t∗ at most |R| moats grow. Hence,∑

U∈U

yU ≤ |R| · t∗






● Intro

● Example


● Dual LP

● Dual Moats

● Algorithm

●SF Example

● Analysis

Primal Dual: MCF





c(T ∗) ≤ 2 ·∑

U∈U

yU − 2t∗ (*)

■ At any time 0 ≤ t ≤ t∗ at most |R| moats grow. Hence,∑

U∈U

yU ≤ |R| · t∗

■ Together with (*) we obtain

c(T ∗) ≤ (2− 2/|R|) ·∑

U∈U

yU .






Primal Dual: MCF

● Multicommodity Flows

● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis


Primal-Dual: Multicommodity Flows inTrees






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis


Multicommodity Flows

■ Input:◆ Tree T = (V, E)◆ k terminal pairs R = {(s1, t1), . . . , (sk, tk)}◆ Edge capacities ce ≥ 0 for all e ∈ E.

■ Use Pi for the unique si, ti-path in T

■ For edge e ∈ E, let Re be commodities that have e on theirpath

■ Feasible flow: fi for each 1 ≤ i ≤ k such that∑

i∈Re

fi ≤ ce

for all e ∈ E

■ Goal: Compute feasible flow of maximum value.






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis







i∈Re

fi ≤ ce

for all e ∈ E







Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis







i∈Re

fi ≤ ce

for all e ∈ E







Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis







i∈Re

fi ≤ ce

for all e ∈ E







Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis







i∈Re

fi ≤ ce

for all e ∈ E







Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis


MCF in Trees: An Example

s2

s1 t2

t1

■ All edges have unit capacity

■ Maximum flow has value 1






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis



s2

s1 t2

t1

■ All edges have unit capacity■ Maximum flow has value 1






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis


Primal LP

■ Have a variable fi for the flow on Pi.






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis


Primal LP

■ Have a variable fi for the flow on Pi.

maximizek

∑

i=1

fi

s.t.∑

i∈Re

fi ≤ ce ∀e ∈ E

fi ≥ 0 ∀1 ≤ i ≤ k






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis


Dual LP

■ One variable for each edge e ∈ E: ye

■ One constraint for each commodity 1 ≤ i ≤ k:∑

e∈Pi

ye ≥ 1

■ This is a fractional multi-cut!

Suppose y is integral. Removal of edges e with ye = 1disconnects si from ti for all 1 ≤ i ≤ k.






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis


Dual LP



e∈Pi

ye ≥ 1








Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis


Dual LP



e∈Pi

ye ≥ 1








Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis


Dual LP



e∈Pi

ye ≥ 1



minimize∑

e∈E

ceye

s.t.∑

e∈Pi

ye ≥ 1 ∀1 ≤ i ≤ k

y≥0 ∀e ∈ E






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis



s2

s1 t2

t1

■ All edges have unit cost.■ Multi-cut has value 1.






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis


Multicut in Trees: PD Algorithm

■ Dual is a hitting set LP! Sets to hit are the paths P1, . . . , Pk.

■ Start with zero dual: ye = 0 for all e ∈ E

■ Root the tree T at an arbitrary vertex r ∈ V .

The depth of a vertex v in T is the number of edges on ther, v-path in T

■ Consider terminal pair (si, ti) ∈ R:

Let li be the vertex of lowest depth on Pi

■ Algorithm removes edges from E one by one and ends witha multi-cut.






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis



■ Dual is a hitting set LP! Sets to hit are the paths P1, . . . , Pk.■ Start with zero dual: ye = 0 for all e ∈ E











Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis














Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis














Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis














Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis



1: fi ← 0 for all 1 ≤ i ≤ k2: C ← ∅3: while C is not a multicut do4: Choose (si, ti) ∈ R with Pi ∩ C = ∅ whose

vertex ui has maximum depth5: Increase fi until

∑

j∈Re

fj = ce

for some e ∈ Pi

6: Add the tight edge e ∈ Pi to C that has lowestdepth in T

7: end while






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis


Multicut in Trees: Reverse Delete

■ Algorithm consider sequence of terminal pairs:

(s1, t1), . . . , (sl, tl)

■ For terminal pair (si, ti), let P si be the si, ui-segment of Pi.

Similar: P ti is the ti, ui-segment of Pi

■ Consider (si, ti) in reverse order of algorithm.If |P v

i ∩ C| ≥ 2 for v ∈ {s, t} then keep only the edge oflowest depth. Delete the other one.






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis




(s1, t1), . . . , (sl, tl)










Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis




(s1, t1), . . . , (sl, tl)










Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis


Multicut in Trees: Analysis

■ Let C be the final cut after reverse delete.Can show: |P v

i ∩C| ≤ 1 for all 1 ≤ i ≤ l and for all v ∈ {si, ti}

■ Assume |P vi ∩ C| ≥ 2 for some i and some v ∈ {si, ti}.

Consider e1 and e2 in P vi ∩C and e1 has lower depth than e2.

■ e1 and e2 cannot help any pair (sj , tj) with j < i

■ If e ∈ {e1, e2} helps (sj , tj) with j > i then uj is apredecessor of ui in tree T .

■ This means that e1 helps (sj , tj). Can delete e2.■ Every path P v

i for 1 ≤ i ≤ l and for v ∈ {s, t} has at most oneedge in C






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis
















Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis
















Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis
















Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis









■ This means that e1 helps (sj , tj). Can delete e2.

■ Every path P vi for 1 ≤ i ≤ l and for v ∈ {s, t} has at most one

edge in C






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis
















Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis



■ As in hitting set analysis, can show that

∑

e∈C

ce ≤l

∑

i=1

|Pi ∩ C|fi ≤ 2 ·l

∑

i=1

fi

■ This shows that the cut C is 2-approximate!■ Also: If edge-costs are integral, the flow f is integral.

Moreover, f is a 2-approximation as well.






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis




∑

e∈C

ce ≤l

∑

i=1

|Pi ∩ C|fi ≤ 2 ·l

∑

i=1

fi

■ This shows that the cut C is 2-approximate!

■ Also: If edge-costs are integral, the flow f is integral.Moreover, f is a 2-approximation as well.






Primal Dual: MCF


● Example

● Primal LP

● Dual LP

● Example

● PD Algorithm

● Reverse Delete

● Analysis




∑

e∈C

ce ≤l

∑

i=1

|Pi ∩ C|fi ≤ 2 ·l

∑

i=1

fi

■ This shows that the cut C is 2-approximate!■ Also: If edge-costs are integral, the flow f is integral.

Moreover, f is a 2-approximation as well.


Documents

The Primal-Dual Method for Approximation Algorithmsapproximation algorithm for vertex-cover. Goemans and Williamson formalize this approach in 1992. Result is a general toolkit for