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Agenda. Introduction and MotivationIterative RoundingMinimum Spanning TreeBDMST . MBDST. InputUndirected Graph G=(V,E)Cost for each edge, c(e)Integer k (Degree bound)GoalA minimum spanning tree of G with degree at most kMotivationA spanning tree with no overloaded node. MDBST. The problem
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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 k+1.
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 k+2
Optimal cost: minimum cost of a tree with max degree <= 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 (x*i = , 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
9. 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.
10. Minimum Spanning Tree There must be a leaf vertex.
11. 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
12. Minimum Spanning Tree
13. Minimum Spanning Tree Let E* be the support of x* i.e. E = {e | x*e > 0 }
Theorem implies |E*| <= n-1
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 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
15. 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 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]
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 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)
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
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. x*e= 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
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. x*e = 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
23. 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.