Approximating Minimum Bounded Degree Spanning Tree (MBDST) Mohit Singh and Lap Chi Lau...

Approximating Minimum Bounded Degree Spanning Tree (MBDST)

Mohit Singh and Lap Chi Lau “Approximating Minimum Bounded Degree Spanning Tress to within One of Optimal” , Proceedings of 39th ACM  Symposium on Theory of Computing, STOC 2007.

Agenda

Introduction and Motivation Iterative Rounding Minimum Spanning Tree BDMST

MBDST

Input Undirected Graph G=(V,E) Cost for each edge, c(e) Integer k (Degree bound)

Goal A minimum spanning tree of G with

degree at most k Motivation

A spanning tree with no overloaded node

MDBST

The problem is NP-Hard Consider k = 2

Conjecture: [Goemans] Polynomial time algorithm for Polynomial time algorithm for optimal optimal costcost and maximum degree at most k+1. and maximum degree at most k+1.

General Case: Given BGeneral Case: Given Bvv , degree , degree bound over each vertexbound over each vertex

Result

Theorem: There exists a polynomial time algorithm for MBDST problem which returns a tree of optimal cost and maximum degree at most k+2

Optimal cost: minimum cost of a tree with max degree <= k

Main Ingredient

Iterative Rounding [Jain ’01]

Use an adaptation of Iterative Rounding, “iterative relaxation”.

Iterative Rounding

1.1. Formulate a LP relaxation Formulate a LP relaxation

2.2. Solve to get a Basic Feasible solution x*.Solve to get a Basic Feasible solution x*.

3.3. If there exists some variable (xIf there exists some variable (x**ii ≥ ½, say) ≥ ½, say)

then include i in the integral solution. then include i in the integral solution.

4.4. Formulate the residual problem and Formulate the residual problem and iterate.iterate.

o Will give 2-approximation for the problemWill give 2-approximation for the problem

Minimum Spanning Tree

xe decision variable for each edge x(U) = Σxe for a subset of edges E(S) = edges with both endpoints in S

min e \in E ce xe

s.t. e \in E(V)xe= |V|-1

e \in E(S)xe ≤ |S|-1

xe ≥ 0

Any tree has n-1 edges

Cycle elimination constraints

for each subset S of V

Minimum Spanning Tree

A Basic Feasible solution (Extreme Point) is the unique solution of m linearly independent tight inequalities, where m denotes the number of

variables.

min e \in E ce xe

s.t. e \in E(V) xe = |V|-1

e \in E(S) xe ≤ |S|-1

xe ≥ 0

Minimum Spanning Tree

There must be a leaf vertex.

F=F=

While F is not a spanning tree

1. Solve LP to obtain an extreme point x*

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

3. If there exists a leaf vertex v, then include the edge incident at v in F and remove v from G.

Minimum Spanning Tree

If algorithm terminates it returns MST

For the leaf vertex x*e = 1 x* restricted to G-v, is a

MST Residual solution will be a

lower bound on MST G-v

min e \in E ce xe

s.t. e \in E(V) xe = |V|-1

e \in E(S) xe ≤ |S|-1

xe ≥ 0

Minimum Spanning Tree

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

min e \in E ce xe

s.t. e \in E(V) xe = |V|-1

e \in E(S) xe ≤ |S|-1

xe ≥ 0

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.

Minimum Spanning Tree

Let E* be the support of x* i.e. E = {e | x*e > 0 }

Theorem implies |E*| <= n-1

Theorem: There are at most n-1 linearly independent tight

inequalities of this type, where n denotes the number of vertices.

Laminar Family: A family of sets is laminar if no two sets are

“intersecting”.

The rank of the tight constraints in a basic solution is

equal to the size of maximal laminar family of tight

sets L [Cornuejols et al ‘88, Jain ‘01]

[Cornuejols et al ‘88, Jain ‘01]

The rank of the tight constraints in a basic solution is equal to the size of maximal laminar family of tight sets L

Proof (idea) Consider the sets corresponding to tight constraints Any two intersecting sets A and B can be uncrossed Both AB and A+B are tight Hence the resulting system is laminar Repeat for all pairs, and we get the maximal

laminar family that spans all tight sets

Let F be family of tight sets F = {S | x*(E(S)) = |S|-1 }

For a subset F of edges letX (F) be the characteristics vector of F

If S and T are in F then so are ST and S+T and X(E(S))+ X(E(T)) = X(E(S+T)) + X(E(ST))

Proof: |S|-1+|T|-1 = |ST|-1 +|S+T|-1

>= x*(E(ST)) + x*(E(S+T))

>= x*(E(S)) + x*(E(T)) = |S|-1+|T|-1

[Cornuejols et al ‘88, Jain ‘01]

[Cornuejols et al ‘88, Jain ‘01]

Let L be maximal laminar subfamily of F then span(L)=span(F)

Assume X(E(S)) is not in span(L). Let it intersect as few sets of L as possible.

By maximality of L some T in L intersect S ST and S+T are in F and

X(E(S))+ X(E(T)) = X(E(S+T)) + X(E(ST))

Either X(E(S+T)) or X(E(ST)) are not in span(L)

Size of maximal laminar family

No singleton set can be tight

A laminar family on ground set of size n, containing no singleton has size at most n-1 By induction on n

Hence there are at most n-1 tight constraints

Minimum Bounded Degree Spanning Tree

Input Undirected Graph G=(V,E) Cost for each edge, c(e) Integer k (Degree bound)

Goal A minimum spanning tree of G with

degree at most k Motivation

A spanning tree with no overloaded node

MBDST LP Formulation

Define δ(S) to be edges with exactly one endpoint in S. Let Bv be the bound on v

min e \in E ce xe

s.t. e \in E(V) xe = |V|-1

e \in E(S) xe ≤ |S|-1

e \in δ(S) xe ≤ Bv

xe ≥ 0

For W, a subset of V

Spanning tree

Degree bounds

First Try

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

1. include {u,v} in F.

2. Decrease Bu by 1. Delete v from G. Delete v from W

If the algorithm works then we solved the problem optimally

A correct +2 AlgorithmInitialize F=

While F is not a spanning tree

1. Solve LP to obtain extreme point x*.

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

3. If there is a leaf vertex v with edge {u,v}, then

1. Include {u,v} in F.

2. Decrease Bu by 1. Delete v from G. Delete v from W

4. If there is a vertex v \in W such that degE(v) ≤ 3, then remove the degree constraint of v. i.e.Delete v from W

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

• Either there is a leaf vertex v.

• Or there is a vertex with degree constraint such that

degE(v) ≤ 3

OPT = min e2 E ce xe

s.t. e \in E(V) xe= |V|-1

e \in E(S) xe ≤ |S|-1

e \in (v) xe ≤ Bv v \in W

xe ≥ 0

Theorem: There are at most n-1+W

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| ≥ (2(n-|W|) + 4|W| ) /2 = n + |W|

The set of tight constraints :

|E| ≤ n-1+|W|

A contradiction to above theorem.

Proof of the theorem

The number of tight constraints from first two types of constraints is <= n-1 By previous analysis

There can be at most W more, i.e. all could be tight.