Approximating Minimum Bounded Degree Spanning Tree (MBDST)

1
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.

2
Agenda

• Introduction and Motivation
• Iterative Rounding
• Minimum Spanning Tree
• BDMST

3
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

4
MDBST

• The problem is NP-Hard
• Consider k 2
• Conjecture Goemans
• Polynomial time algorithm for optimal cost and
  maximum degree at most k1.
• General Case Given Bv , degree bound over each
  vertex

5
Result

• Theorem There exists a polynomial time algorithm
  for MBDST problem which returns a tree of optimal
  cost and maximum degree at most k2
• Optimal cost minimum cost of a tree with max
  degree lt k

6
Main Ingredient

• Iterative Rounding Jain 01
• Use an adaptation of Iterative Rounding,
  iterative relaxation.

7
Iterative Rounding

• Formulate a LP relaxation
• Solve to get a Basic Feasible solution x.
• If there exists some variable (xi ½, say)
  then include i in the integral solution.
• Formulate the residual problem and iterate.
• Will give 2-approximation for the problem

8
Minimum Spanning Tree

• xe decision variable for each edge
• x(U) Sxe 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
for each subset S of V
Cycle elimination constraints
9
Minimum Spanning Tree
min ? e \in E ce xe s.t. ? e \in E(V) xe
V-1 ? e \in E(S) xe S-1 xe
0

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

10
Minimum Spanning Tree

• F?.
• While F is not a spanning tree
• Solve LP to obtain an extreme point x
• Remove all edges s.t. xe 0
• If there exists a leaf vertex v, then include the
  edge incident at v in F and remove v from G.

• There must be a leaf vertex.

11
Minimum Spanning Tree
min ? e \in E ce xe s.t. ? e \in E(V) xe
V-1 ? e \in E(S) xe S-1 xe
0

• If algorithm terminates it returns MST
• For the leaf vertex xe 1
• x restricted to G-v, is a MST
• Residual solution will be a lower bound on MST
  G-v

12
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.
13
Minimum Spanning Tree
Theorem There are at most n-1 linearly
independent tight inequalities of this type,
where n denotes the number of vertices.

• Let E be the support of x i.e. E e xe gt 0
  
• Theorem implies E lt n-1

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
14
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 AB are tight
• Hence the resulting system is laminar
• Repeat for all pairs, and we get the maximal
  laminar family that spans all tight sets

15
Cornuejols et al 88, Jain 01

• Let F be family of tight sets
• F S x(E(S)) S-1
• For a subset F of edges let
• X (F) be the characteristics vector of F
• If S and T are in F then so are ST and ST and
  X(E(S)) X(E(T)) X(E(ST)) X(E(ST))
• Proof S-1T-1 ST-1 ST-1
• gt x(E(ST)) x(E(ST))
• gt x(E(S)) x(E(T)) S-1T-1

16
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 ST are in F and
• X(E(S)) X(E(T)) X(E(ST)) X(E(ST))
• Either X(E(ST)) or X(E(ST)) are not in span(L)

17
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

18
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

19
MBDST LP Formulation

• Define d(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
d(S) xe Bv xe 0
Spanning tree
For W, a subset of V
Degree bounds
20
First Try

• Initialize F?.
• While F is not a spanning tree
• Solve LP to obtain vertex solution x.
• Remove all edges e s.t. xe 0.
• 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. Delete v from
  W

If the algorithm works then we solved the problem
optimally
21
A correct 2 Algorithm

• Initialize F?.
• While F is not a spanning tree
• Solve LP to obtain extreme point x.
• Remove all edges e s.t. xe 0.
• 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. Delete v from
  W
• 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

22
Analysis
Theorem There are at most n-1W linearly
independent tight inequalities of this type,
where n denotes the number of vertices.
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
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) 4W ) /2
n W The set of tight constraints E
n-1W A contradiction to above theorem.
23
Proof of the theorem

• The number of tight constraints from first two
  types of constraints is lt n-1
• By previous analysis
• There can be at most W more, i.e. all could be
  tight.