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Bipartite Matching Polytope, Stable Matching Polytope

x1. x3. x2. Bipartite Matching Polytope, Stable Matching Polytope. Lecture 10: Feb 15. Perfect Matching. Integrality Gap Example. x1. x2. x3. x1. (0.5,0.5,0.5). x3. x2. Good Relaxation. Every vertex could be the unique optimal solution for some objective function.

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Bipartite Matching Polytope, Stable Matching Polytope

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

  2. Perfect Matching

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

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

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

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

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

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

  9. Bipartite Perfect Matching 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. 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. Boys Girls 1: CBEAD A : 35214 2 : ABECD B : 52143 3 : DCBAE C : 43512 4 : ACDBE D : 12345 5 : ABDEC E : 23415

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

  12. Stable Matching • 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. Boys Girls 1: CBEAD A : 35214 2 : ABECD B : 52143 3 : DCBAE C : 43512 4 : ACDBE D : 12345 5 : ABDEC E : 23415

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

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

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

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

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

  18. Proof of Gale-Shapley Theorem Gale,Shapley [1962]: This procedure always find a stable matching in the stable marriage problem. • The procedure will terminate. • Everyone is married. • 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. 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. Bipartite Stable Matching Write if v prefers f to e. Write if for some v

  21. Bipartite Stable Matching CLAIM: Proof:

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

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

  25. Bipartite Stable Matching At top, blue is minimum, red is maximum. W At bottom, blue is maximum, red is minimum. U Degree constraints still satisfied. construct convex combination! Bottom decreases, top increases, equal!

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

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

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

  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!

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