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.