Post on 30-Dec-2015
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.