1. Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3

2. x1 x3x2 x1 x2 x3 (0.5,0.5,0.5) Linear Programming

3. Good Relaxation Every vertex could be the unique optimal solution for some objective function. So, we need every vertex to be integral. For every objective function, there is a vertex achieving optimal value. So, it suffices if every vertex is integral. Goal: Every vertex is integral!

4. Black Box LP-solver Problem LP-formulationVertex solution Solution Polynomial time integral

5. Convex Combination A point y in R n is a convex combination of if there exist so that and

6. Vertex Solution Fact: A vertex solution is not a convex combination of some other points. A point y in R n is a convex combination of if y is in the convex hull of

7. Prove: a vertex solution corresponds to an integral solution. Every point in the polytope corresponds to a fractional solution. Maximum Bipartite Matchings

8. Pick a fractional edge and keep walking. Prove: a vertex solution corresponds to an integral solution. Because of degree constraints, every edge in the cycle is fractional. Partition into two matchings because the cycle is even.

9. Maximum Bipartite Matchings Since every edge in the cycle is fractional, we can increase every edge a little bit, or decrease every edge a little bit. Degree constraints are still satisfied in two new matchings. Original matching is the average! Fact: A vertex solution is not a convex combination of some other points. CONTRADICTION!

10. Bipartite Stable Matchings Input: N men, N women, each has a preference list. Goal: Find a matching with no unstable pair. How to formulate into linear program?

11. Bipartite Stable Matchings Write if v prefers f to e. Write iffor some v

12. Bipartite Stable Matchings CLAIM: Proof:

13. Bipartite Stable Matchings Focus on the edges with positive value, call them E +. For each vertex, let e(v) be the maximum element of CLAIM: Let e(v) = v,w e(v) is the minimum element of

14. CLAIM: Let e(v) = v,w e(v) is the minimum element of Bipartite Stable Matchings For each vertex, let e(v) be the maximum element of U W e(v) defines a matching for v in U e(w) defines a matching for w in W

15. Bipartite Stable Matchings U W At bottom, blue is maximum, red is minimum. At top, blue is minimum, red is maximum. U W Prove: convex combination.

16. Bipartite Stable Matchings At bottom, blue is maximum, red is minimum. At top, blue is minimum, red is maximum. U W Degree constraints still satisfied. Bottom decreases, top increases, equal! Prove: convex combination!

17. Bipartite Stable Matchings [Vande Vate][Rothblum]

18. Weighted Stable Matchings Polynomial time algorithm from LP. Can work on incomplete graph. Can determine if certain combination is possible.

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

20. Matching Polytope Matching polytope Convex hull of matchings Linear program Define by points Intersections of hyperplanes Define by inequalities Goal: Prove they are equal PQ

21. Matching Polytope Prove P is smaller than Q, and Q is smaller than P. Easy direction: Check all points of P (i.e. all matchings) satisfy all the inequalities Another direction: Check all points of Q (i.e. all fractional solutions) is inside P. How?By showing that all points are convex combination of vertices of P

22. Maximum Bipartite Matchings Goal: show that any fractional solution is a convex combination of matchings How? By induction! Bipartite perfect matching, 2n vertices.Minimal counterexample.

23. 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?At most 2n How many linearly independent tight inequalities?At most 2n-1 Basic solution: unique solution of 2n linearly independent tight inequalities CONTRA!

24. Valid Inequalities Odd set inequalities That’s enough. [Edmonds 1965]