1. 1 Bipartite Matching Polytope, Stable Matching Polytope x1 x2 x3 Lecture 10: Feb 15

2. 2 Perfect Matching

3. 3 x1 x3x2 x1 x2 x3 (0.5,0.5,0.5) Integrality Gap Example

4. 4 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!

5. 5 Black Box LP-solver Problem LP-formulationVertex solution Solution Polynomial time integral

6. 6 Vertex Solutions An optimal vertex solution can be found in polynomial time.

7. 7 Prove: for a bipartite graph, a vertex solution corresponds to an integral solution. Bipartite Perfect Matching

8. 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. Bipartite Perfect Matching

9. 9 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! Bipartite Perfect Matching

10. 10 Boys Girls 1: CBEAD A : 35214 2 : ABECD B : 52143 3 : DCBAE C : 43512 4 : ACDBE D : 12345 5 : ABDEC E : 23415 Stable Matching The Stable Marriage Problem: There are n boys and n girls. For each boy, there is a preference list of the girls. For each girl, there is a preference list of the boys.

11. 11 Stable Matching Boys Girls 1: CBEAD A : 35214 2 : ABECD B : 52143 3 : DCBAE C : 43512 4 : ACDBE D : 12345 5 : ABDEC E : 23415 What is a stable matching? Consider the following matching. It is unstable, why?

12. 12 Stable Matching Boys Girls 1: CBEAD A : 35214 2 : ABECD B : 52143 3 : DCBAE C : 43512 4 : ACDBE D : 12345 5 : ABDEC E : 23415 Boy 4 prefers girl C more than girl B (his current partner). Girl C prefers boy 4 more than boy 1 (her current partner). So they have the incentive to leave their current partners, and switch to each other, we call such a pair an unstable pair.

13. 13 Stable Matching Boys Girls 1: CBEAD A : 35214 2 : ABECD B : 52143 3 : DCBAE C : 43512 4 : ACDBE D : 12345 5 : ABDEC E : 23415 A stable matching is a matching with no unstable pair, and every one is married. What is a stable matching? Does a stable matching always exists?

14. 14 Boys Girls 1: CBEAD A : 35214 2 : ABECD B : 52143 3 : DCBAE C : 43512 4 : ACDBE D : 12345 5 : ABDEC E : 23415 Day 1 Morning: boy propose to their favourite girl Afternoon: girl rejects all but favourite Evening: rejected boy writes off girl

15. 15 Boys Girls 1: CBEAD A : 35214 2 : ABECD B : 52143 3 : DCBAE C : 43512 4 : ACDBE D : 12345 5 : ABDEC E : 23415 Morning: boy propose to their favourite girl Afternoon: girl rejects all but favourite Evening: rejected boy writes off girl Day 2

16. 16 Boys Girls 1: CBEAD A : 35214 2 : ABECD B : 52143 3 : DCBAE C : 43512 4 : ACDBE D : 12345 5 : ABDEC E : 23415 Morning: boy propose to their favourite girl Afternoon: girl rejects all but favourite Evening: rejected boy writes off girl Day 3

17. 17 Boys Girls 1: CBEAD A : 35214 2 : ABECD B : 52143 3 : DCBAE C : 43512 4 : ACDBE D : 12345 5 : ABDEC E : 23415 Morning: boy propose to their favourite girl Afternoon: girl rejects all but favourite Evening: rejected boy writes off girl A stable matching! Day 4

18. 18 Gale,Shapley [1962]: This procedure always find a stable matching in the stable marriage problem. Proof of Gale-Shapley Theorem 1.The procedure will terminate. 2.Everyone is married. 3.No unstable pairs. The stable matching algorithm is boy-optimal That is, among all possible stable matching, boys get the best possible partners simultaneously.

19. 19 Bipartite Stable Matching Input: N men, N women, each has a preference list. Goal: Find a matching with no unstable pair. How to formulate into linear program?

20. 20 Bipartite Stable Matching Write if v prefers f to e. Write iffor some v

21. 21 Bipartite Stable Matching CLAIM: Proof:

22. 22 Bipartite Stable Matching 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

23. 23 Bipartite Stable Matching 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 CLAIM: Let e(v) = v,w e(v) is the minimum element of

24. 24 Bipartite Stable Matching U W At bottom, blue is maximum, red is minimum. At top, blue is minimum, red is maximum. U W construct convex combination.

25. 25 Bipartite Stable Matching 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! construct convex combination!

26. 26 Weighted Stable Matching Polynomial time algorithm from LP. Can work on incomplete graph. Can determine if certain combination is possible. [Vande Vate][Rothblum]

27. 27 Basic Solution Tight inequalities: inequalities achieved as equalities Basic solution: unique solution of n linearly independent tight inequalities

28. 28 Bipartite Perfect Matching Goal: show that any basic solution is an integral solution. Bipartite perfect matching, 2n vertices.Minimal counterexample.

29. 29 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!