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Approximating Minimum Bounded Degree Spanning Tree MBDST

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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Approximating Minimum Bounded Degree Spanning Tree MBDST

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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.

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