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1 Matching Polytope x1 x2 x3 Lecture 12: Feb 22 x1 x2 x3.

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Presentation on theme: "1 Matching Polytope x1 x2 x3 Lecture 12: Feb 22 x1 x2 x3."— Presentation transcript:

1 1 Matching Polytope x1 x2 x3 Lecture 12: Feb 22 x1 x2 x3

2 2 Perfect Matching But not every solution is a matching!!

3 3 x1 x3 x2 x1 x2 x3 Matching Polytope x1 x2 x3

4 4 Valid Inequalities Enough? Inequalities which are satisfied by integer solutions but kill all unwanted fractional vertex solution. everywhere

5 5 Valid Inequalities Enough? Odd set inequalities Yes, that’s enough. [Edmonds 1965]

6 6 Exponentially Many Inequalities Can take care by the ellipsoid method. Just need a separation oracle, which determines whether a solution is feasible. If not, find a violating inequality. How to construct a separation oracle for matching?

7 7 Rewriting the Odd-Set Constraints

8 8 Matching Polytope

9 9 Convex Combination A point y in R n is a convex combination of if there exist so that and

10 10 Vertex Solution Fact: A solution is a vertex solution if and only if x is not a convex combination of other feasible solutions. A point y in R n is a convex combination of if y is in the convex hull of

11 11 Convex Combination Goal: Prove that every fractional solution can be written as a convex combination of matchings. This implies that every vertex solution corresponds to a matching.

12 12 Convex Combination of Matchings

13 13 Goal: Prove that every fractional solution can be written as a convex combination of matchings. Convex Combination An edge of 0, delete it. An edge of 1, reduce it.

14 14 Convex Combination of Matching

15 15 Goal: Prove that every fractional solution can be written as a convex combination of matchings. Convex Combination An edge of 0, delete it. An edge of 1, reduce it. Tight odd-set, contract it.

16 16 Fractional solutions satisfying inequalities By induction, each smaller fractional solution is a convex combination of matchings Check degree Check odd-set Perfect Matching

17 17 Fractional solutions satisfying inequalities By induction, each smaller fractional solution is a convex combination of matchings Check degree Check odd-set Perfect Matching

18 18 Fractional solutions Satisfying inequalities By induction, each smaller fractional solution is a convex combination of matchings Check degree Check odd-set So is the original fractional solution! DONE! Perfect Matching

19 19 At most 2n tight (degree) tight constraints! Since each vertex has degree 2, there are at least 2n edges. An edge of 0, delete it. An edge of 1, reduce it. Tight odd-set, contract it. Perfect Matching

20 20 At most 2n tight (degree) tight constraints! Since each vertex has degree 2, there are at least 2n edges. Basic Solution: 2n tight linearly independent constraints for 2n variables. 1.So, exactly 2n edges. 2.Since each vertex has degree 2, the edges form disjoint union of cycles. 3.No odd cycle because of the odd-set constraints, so only even cycles. 4.Even cycle can be decomposed into matchings. 5.Therefore, a convex combination of matchings. Perfect Matching

21 21 Separation Oracle S uv Each odd cut has total capacity >= 1

22 22 Gomory-Hu Tree A compact representation of all minimum s-t cuts in undirected graphs! To compute s-t cut, look at the unique s-t path in the tree, and the bottleneck capacity is the answer! And furthermore the cut in the tree is the cut of the graph!

23 23 Separation Oracle Given a Gomory-Hu tree T, we say an edge e is odd if T-e consists of two odd components.

24 24 Separation Oracle To check if a fractional solution satisfies all odd set constraints, we want to check if the minimum odd-cut has capacity at least 1. If some odd edge has capacity < 1, then we find a violating odd-cut. Surprisingly, this is all we need to check.

25 25 Minimum Odd Cut Let C be an odd cut. C Case 1: there is an odd edge crossing it u v x = 1.2 Let the value of the odd edge be x. Then any u-v cut has capacity at least x >= 1.

26 26 Minimum Odd Cut Let C be an odd cut. C Case 2: there is no odd edge crossing it This is impossible, because each tree component inside C has an even number of vertices, which implies that C has an even number of vertices.

27 27 Big Picture LP-solver Problem LP-formulationVertex solution Solution Polynomial time integral Separation oracle


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