Approximation Algorithm: Iterative Rounding Lecture 15: March 9.

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Approximation Algorithm:

Iterative Rounding

Lecture 15: March 9

Lower bound and Approximation Algorithm

The key of designing a polytime approximation algorithm is to obtain a good (lower or upper) bound on the optimal solution.

For NP-complete problem, we can’t compute an optimal solution in polytime.

The general strategy (for a minimization problem) is:

lowerbound OPT SOL

SOL ≤ c · lowerbound SOL ≤ c · OPT

lowerbound OPT SOL

To design good approximation algorithm, we need a good lowerbound.

For example, if 100 · lowerbound ≤ OPT for some instance, then if we compare SOL to lowerbound to analyze the performance,we could not achieve anything better than an 100-approximation algorithm.

lowerbound OPT SOL

Goal: to find a lowerbound as close to OPT as possible.

In metric TSP and the minimum Steiner tree problem, we use minimum spanning tree as a lowerbound.

In general, it is often difficult to come up with a good lowerbound.

Linear Programming and Approximation Algorithm

lowerbound OPT SOL

Linear programming: a general method to compute a lowerbound in polytime.

To computer an approximate solution,

we need to return an (integral) solution

close to an optimal LP (fractional) solution.

An Example: Vertex Cover

Vertex cover: find an minimum subset of vertices

which cover all the edges.

A linear programming relaxation of the vertex cover problem.

Clearly, LP is a lowerbound on the optimal vertex cover, because every vertex cover corresponds to a feasible solution of this LP.

How bad is this LP?

0.5 0.5

LP = 2.5

OPT = 4

LP = n/2

OPT = n-1

Integrality gap: =

Optimal integer solution.

Optimal fractional solution.

Over all instances.

In vertex cover, there are instances where this gap is almost 2.

Half-integrality

Theorem: For the vertex cover problem, every vertex (or basic) solution of the LP is half-integral, i.e. x(v) = {0, ½, 1}

There is a 2-approximation algorithm for the vertex cover problem.

The integrality gap of the vertex cover LP is at most 2.

Survivable Network Design

Input•An undirected graph G = (V,E),•A cost c(e) on each edge,•A connectivity requirement r(u,v) for each pair u,v.

Output•A minimum cost subgraph H of G which has r(u,v) edge-disjoint paths between each pair u,v. (That is, H satisfies all the connectivity requirements.)

Survivable Network Design

• Minimum spanning tree: r(u,v) = 1 for all pairs.

• Minimum Steiner tree: r(u,v) = 1 for all pair of required vertices.

• Hamiltonian path: r(u,v) = 2 for all pairs and every edge has cost 1.

• k-edge-connected subgraph: r(u,v) = k for all pairs.

• Minimum cost k-flow: r(s,t) = k for the source s and the sink t.

Survivable Network Design is NP-complete.

Linear Programming Relaxation

u vFor each set S separating u and v, there should be at least r(u,v) edges “crossing” S. Let f(S) = max{ r(u,v) | S separates u and v}.

At least r(u,v) edges crossing S

for each subset S of V

Special Case: Minimum Spanning Tree

How bad is this LP?

0.5 0.5

LP = 2.5

OPT = 4

LP = n/2

OPT = n-1

Cannot even solve minimum spanning tree!

Half-integrality

Does the LP has half-integral optimal solution?

Peterson Graph

Consider the minimum spanning tree problem, i.e. f(S)=1 for all S.

All 1/3 is a feasible solution and has cost 5.

Any half-integral solution having cost 5 must be a Hamitonian cycle.

But Peterson graph does not have an Hamitonian cycle!

So, no half-integral optimal solution!

Separation Oracle

There are exponentially many constraints, but this LP can

still be solved in polynomial time by the ellipsoid method.

The reason is that we can design a polynomial time

separation oracle to determine if x is a feasible solution of

the LP.

Separation Oracle

Remember: f(S) = max{ r(u,v) | S separates u and v}.

At least r(u,v) edges crossing S Max-Flow Min-Cut

Every (u,v)-cut has at least r(u,v) edges if and only if

there are r(u,v) flows from u to v.

Separation oracle: check if each pair u,v has a flow of r(u,v)!

Linear Programming Relaxation

What is the integrality gap of this LP?

Moment of Inspiration

All 1/3 is a feasible solution.

But this is not a vertex solution!

Thick edges have value 1/2;

Thin edges have value 1/4.

This is a vertex solution.

Structural Result of the LP

Kamal Jain

Theorem. Every vertex solution has an edge with value at least 1/2

Corollary. There is a 2-approximation algorithm for survivable network design.

Iterative Rounding

Initialization: H = , f’ = f.

While f’ ≠ 0 do:

o Find a vertex solution, x, of the LP with function f’.

o Add every edge with x(e) ≥ 1/2 into H.

o Update f’: for every set S, set

Output H.

A new vertex solution is computed in each

iteration

Guaranteed to exist

Update the connectivity

requirements.

Analysis

Corollary. There is a 2-approximation algorithm for survivable network design.

Intuitive reason: we only pick an edge when the LP picks at least half.

Proof: Let say we pick an edge e.

Key: LP-c(e)x(e) is a feasible solution for the next iteration.

cost(H) = c(e) + cost(H’) ≤ 2·c(e)x(e) + cost(H’)

≤ 2·c(e)x(e) + 2(LP-c(e)x(e)) ≤ 2LP

≤ 2OPT.

Some Remarks

1. The iterative rounding algorithm performs very well in practice.

2. No combinatorial algorithm has an performance ratio better than O(log n).

Spanning Tree with Degree Constraints

•An undirected graph G = (V,E),

•A degree upper bound k.

Output

•A spanning tree with degree at most k.

NP-complete (Hamiltonian path when k=2).

Spanning Tree with Degree Constraints

Motivation: to find a spanning tree in which there is no “overloaded” vertices.

[Furer and Raghavachari ’92]

Given k, there is a polynomial time algorithm which does the following:

Either the algorithm

(i) Show that there is no spanning tree with maximum degree at most k.

(ii) Find a spanning tree with maximum degree at most k+1.

In other words, there is an +1 algorithm for this problem!

Minimum Spanning Tree with Degree Constraints

Input•An undirected graph G = (V,E),•A cost c(e) on each edge e,•A degree upper bound k.

Output

•A minimum spanning tree with degree at most k.

Question: Is there a +1 algorithm for this problem as well?

That is, a polytime algorithm which returns a

minimum spanning tree with maximum degree at most k+1.

Conjecture

Let OPT be the minimum cost of a spanning tree with maximum degree k.

[Goemans]

Conjecture: Given k, there is a polynomial time algorithm which returns a

spanning tree with cost at most OPT and maximum degree at most k+1.

Note that we do not restrict ourselves to MST.

Previous Work

Reference Cost Guarantee Degree

Furer and Raghavachari ‘92 Unweighted Case k+1

Konemann, Ravi ’01 ’02 O(1) O(k+log n)

CRRT ’05 ’06 O(1) O(k)

Ravi, S. 06 MST k+p (p=#distinct costs)

Goemans ’06 1 k+2

Our Result

Theorem: Given k, there is a polynomial time algorithm which returns a

spanning tree with cost at most OPT and maximum degree at most k+1.

[Singh Lau 07]

Mohit Singh

Technique:

Adaptation of iterative rounding,

but we do not round.

Spanning Tree Polytope

• Formulate a linear programming relaxation.

min e2 E ce xe

s.t. e2 E(V) xe= |V|-1

e2 E(S) xe ≤ |S|-1

xe ≥ 0

E(S): set of edges with both endpoints in S.• Separation oracle [Cunningham ’84] ) Optimization in poly time

Any tree has n-1 edges

Cycle elimination constraints

min e2 E ce xe

s.t. e2 E(V) xe= |V|-1

e2 E(S) xe ≤ |S|-1

xe ≥ 0

Recall: A vertex solution is the unique solution of m

linearly independent tight inequalities, where

m denotes the number of variables.

min e2 E ce xe

s.t. e2 E(V) xe= |V|-1

e2 E(S) xe ≤ |S|-1

xe ≥ 0

If there is an edge of 0, delete it.

If there exists a leaf vertex v,

then include the edge

incident at v in and remove v

from G.

min e2 E ce xe

s.t. e2 E(V) xe= |V|-1

e2 E(S) xe ≤ |S|-

xe ≥ 0

Claim: A vertex solution of the LP must have a leaf vertex.

Theorem: There are at most n-1

linearly independent tight

inequalities of this type, where

n denotes the number of vertices.

If there is no leaf vertex,

then every vertex has degree 2,

and hence there are at least 2n/2=n edges,

a contradiction to the above theorem.

• No 1-edge ) many fractional edges

• Vertex Solution with few constraints ) few fractional edges

• Derive contradiction

So a vertex solution must have an edge of 0 or a leaf vertex,

in either case we can finish by induction.

This proves that the linear program has integer optimal solution.

• Extend spanning tree polyhedron

OPT = min e2 E ce xe

s.t. e2 E(V) xe= |V|-1

e2 E(S) xe · |S|-1

e2 (v) xe · Bv 8 v 2 W

xe ≥ 0

A Simple +2 Algorithm

The degree constraint of each vertex could be different

Goal: Find a spanning tree with cost at most OPT

and degree is violated by at most 2.

Initialize F=

While F is not a spanning tree

1. Solve LP to obtain vertex solution x*.

2. Remove all edges e s.t. x*e=0.

3. If there is a leaf vertex v with edge {u,v}, then include {u,v} in F. Decrease Bu by 1. Delete v from G.

• If algorithm works then we solve the problem optimally!

• Cannot pick an edge with 1 > xe¸ ½ : lose optimality of the cost.

First Try

Initialize F=

1. Solve LP to obtain extreme point x*.

4. If there is a vertex v2 W such that degE(v) ≤ Bv+2, then remove the degree constraint of v.

Removing Degree Constraints

If there is a vertex v2 W such that degE(v) ≤

Bv+2, then remove the degree constraint

This is only done one, and the degree constraint is violated by at most +2!

Lemma: For any vertex solution x, one of the following is true:

1) Either there is a leaf vertex v.

2) Or there is a vertex with degree constraint such that degE(v)·Bv+2

Initialize F=

1. Solve LP to obtain extreme point x*.

4. If there is a vertex v2 W such that degE(v) ≤ Bv+2, then remove the degree constraint of v.

OPT = min e2 E ce xe

s.t. e2 E(V) xe= |V|-1

e2 E(S) xe · |S|-1

e2 (v) xe · Bv 8 v 2 W

xe ≥ 0

Theorem: There are at most n-1

linearly independent tight

inequalities of this type, where

n denotes the number of vertices.

Analysis

Proof of the Lemma: Suppose not.

Every vertex has degree at least 2.

Every vertex in W has degree at least 4.

|E| ≥ ½*(2(n-|W|)+4|W|)= n+|W|

The set of tight constraints :

|E| ≤ n-1+|W|

A contradiction.

A quick summary:

Find a leaf vertex v, add the only edge at v and remove v.

- Don’t lose the cost optimality and the degree

bound.

Find a vertex with at most Bv+2 neighbours and has a

degree

constraint, remove the degree constraint.

- Violate the degree bound by at most +2.

By the Lemma, one of the two possibilities must hold.

Obtaining +1 algorithmObtaining +1 algorithm

Initialize F=Initialize F=

While F is not a spanning treeWhile F is not a spanning tree

1.1. Solve LP to obtain extreme point x*.Solve LP to obtain extreme point x*.

2.2. Remove all edges e s.t. x*Remove all edges e s.t. x*ee=0.=0.

3.3. If there exists a leaf vertex v with edge {u,v}, then If there exists a leaf vertex v with edge {u,v}, then include {u,v} in F. Decrease Binclude {u,v} in F. Decrease Buu by 1. Delete v from G. by 1. Delete v from G.

4.4. If there exists a vertex vIf there exists a vertex v22 W such that deg W such that degEE(v)(v)·· B Bvv+2, +2,

then remove the degree constraint of v.then remove the degree constraint of v.

Replace by Bv+1

v leaf ) x{u,v}=1Why not pick any e s.t. xe=1 ?

Concluding Remarks

1. No combinatorial algorithm has a good performance ratio.

2. Similar techniques can be used for more general problems, e.g. minimum maximum degree k-ec subgraph

3. Can deal with lower and upper bounds.

4. A unifying framework for network design problems.

5. Proofs are based on uncrossing techniques in combinatorial optimiation.

Approximation Algorithm: Iterative Rounding Lecture 15: March 9.

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