1. 1 Spanning Tree Polytope x1 x2 x3 Lecture 11: Feb 21

2. 2 Big Picture LP-solver Problem LP-formulationVertex solution Solution Polynomial time integral

3. 3 Basic Solution Tight inequalities: inequalities achieved as equalities Basic solution: unique solution of n linearly independent tight inequalities

4. 4 Bipartite Perfect Matching Goal: show that any basic solution is an integral solution. Bipartite perfect matching, 2n vertices.Minimal counterexample.

5. 5 Maximum Bipartite Matchings An edge of 0, delete it. An edge of 1, reduce it. So, each vertex has degree 2, and there are at least 2n edges.

6. 6 Maximum Bipartite Matchings An edge of 0, delete it. An edge of 1, reduce it. So, each vertex has degree 2, and there are at least 2n edges. How many tight inequalities?Exactly 2n How many linearly independent tight inequalities?At most 2n-1

7. 7 Linear Dependency x1 x3 x4 x2 Multiply +1 Multiply -1 Each edge is counted twice, one positive, one negative. Sum up to 0 => linear dependency.

8. 8 Maximum Bipartite Matchings An edge of 0, delete it. An edge of 1, reduce it. So, each vertex has degree 2, and there are at least 2n edges. How many tight inequalities?Exactly 2n How many linearly independent tight inequalities?At most 2n-1 Basic solution: unique solution of 2n linearly independent tight inequalities CONTRA!

9. 9 Minimum Spanning Tree There are exponentially many constraints, but this LP can still be solved in polynomial time by the ellipsoid method. The reason is that we can design a polynomial time separation oracle to determine if x is a feasible solution of the LP.

10. 10 Separation Oracle S uv Each cut has total capacity >= 1 Max-Flow Min-Cut Every cut has at capacity >= 1 if and only if we can send 1 unit of flow for all pair. Separation oracle: check if each pair u,v has a flow of 1.

11. 11 Minimum Spanning Tree 0.5 1 1 1 1 Not a good relaxation.

12. 12 Spanning Tree Polytope E(S): set of edges with both endpoints in S. A spanning tree has n-1 edges Cycle elimination constraints

13. 13 Separation Oracle Cycle elimination constraints |S|-1-x(E(S)) is a submodular function Minimizing submodular function can be solved in polytime. A spanning tree has n-1 edges

14. 14 Basic Solution Tight inequalities: inequalities achieved as equalities Basic solution: unique solution of n linearly independent tight inequalities

15. 15 Separation Oracle Goal: Prove that there are at most |V|-1 linearly independent tight constraints If there is an edge of 0, delete it. Theorem: At most |V|-1 linearly independent tight inequalities of this type.

16. 16 Laminar Family A laminar family is a collection of sets with no intersections. Lemma. A laminar family with no singletons has at most n-1 sets. Forest representation

17. 17 Basic Solution Goal: Prove that there are at most |V|-1 linearly independent tight constraints Each tight constraint defines a set. A basic solution is uniquely defined by a laminar family of tight constraints. This would imply the goal.

18. 18 Uncrossing Technique What about two tight sets are intersecting?

19. 19 Laminar Basis The lemma says that a laminar family formed a basis, this implies that there are at most n-1 linearly independent tight constraints.

20. 20 1.Suppose there is a set S which is not in span(L). 2.Consider a set S with smallest intersecting number with L. 3.Let say S intersect with a set T in L. 4.Consider and 5.Both are tight and have smaller intersecting number with L. 6.So both and are in L 7.On the other hand, since S and T are tight, we have 8.This implies that S is in span(L) as well, a contradiction. Proof Sketch of Lemma 2

21. 21 Intersecting Number S T Consider a set R in L. There are only 3 possibilities. 1.R is contained in T. 2.T is contained in R. 3.R and T are disjoint. 1 2 3

22. 22 Looking Forward Uncrossing technique is very important in combinatorial optimization. We will see it in approximation algorithms as well.