1. Approximating Minimum Bounded Degree Spanning Tree (MBDST) Mohit Singh and Lap Chi Lau “Approximating Minimum Bounded DegreeApproximating Minimum Bounded Degree Spanning Tress to within One of Optimal” 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 B v, 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 1. Formulate a LP relaxation 2. Solve to get a Basic Feasible solution x*. 3. If there exists some variable (x * i ≥ ½, say) then include i in the integral solution. 4. Formulate the residual problem and iterate. o Will give 2-approximation for the problem

8. Minimum Spanning Tree  x e decision variable for each edge  x(U) = Σx e for a subset of edges  E(S) = edges with both endpoints in S min  e \in E c e x e s.t.  e \in E(V)  x e = |V|-1  e \in E(S)  x e ≤ |S|-1 x e ≥ 0 Any tree has n-1 edges Cycle elimination constraints for each subset S of V

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. min  e \in E c e x e s.t.  e \in E(V) x e = |V|-1  e \in E(S) x e ≤ |S|-1 x e ≥ 0

10. Minimum Spanning Tree There must be a leaf vertex. 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.

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 min  e \in E c e x e s.t.  e \in E(V) x e = |V|-1  e \in E(S) x e ≤ |S|-1 x e ≥ 0

12. Minimum Spanning Tree Claim: A basic feasible solution of the LP must have a leaf vertex. min  e \in E c e x e s.t.  e \in E(V) x e = |V|-1  e \in E(S) x e ≤ |S|-1 x e ≥ 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  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]

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.  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 δ(S) to be edges with exactly one endpoint in S. Let B v be the bound on v min  e \in E c e x e s.t.  e \in E(V) x e = |V|-1  e \in E(S) x e ≤ |S|-1  e \in δ(S) x e ≤ B v x e ≥ 0 For W, a subset of V Spanning tree Degree bounds

20. 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 B u 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 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 B u by 1. Delete v from G. Delete v from W 4. If there is a vertex v \in W such that deg E (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 deg E (v) ≤ 3

22. OPT = min  e 2 E c e x e s.t.  e \in E(V) x e = |V|-1  e \in E(S) x e ≤ |S|-1  e \in  (v) x e ≤ B v v \in W x e ≥ 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.

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.