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

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

Matching Polytope, Matching Polytope, Stable Matching PolytopeStable Matching Polytope

Lecture 8: Feb 2

x1

x2

x3

x1

x2x3

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

x1

x3x2x1

x2

x3

(0.5,0.5,0.5)

Linear Programming

Page 3: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

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!

Page 4: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

Black Box

LP-solver

Problem

LP-formulation Vertex solution

Solution

Polynomial time

integral

Page 5: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

Convex Combination

A point y in Rn is a convex combination of

if there exist

so that

and

Page 6: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

Vertex Solution

Fact: A vertex solution is not a convex

combination of some other points.

A point y in Rn is a convex combination of

if y is in the convex hull of

Page 7: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

Prove: a vertex solution

corresponds to an

integral solution.

Every point in the polytope corresponds to a fractional solution.

Maximum Bipartite Matchings

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

Maximum Bipartite Matchings

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 matchingsbecause the cycle is even.

Page 9: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

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!

Page 10: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

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?

Page 11: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

Bipartite Stable Matchings

Write

if v prefers f to e.

Write

if for some v

Page 12: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

Bipartite Stable Matchings

CLAIM:

Proof:

Page 13: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

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

Page 14: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

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

Page 15: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

Bipartite Stable Matchings

U

W

At bottom,blue is maximum, red is minimum.

At top,blue is minimum, red is maximum.

U

WProve: convex combination.

Page 16: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

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!

Page 17: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

Bipartite Stable Matchings

[Vande Vate] [Rothblum]

Page 18: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

Weighted Stable Matchings

Polynomial time algorithm from LP.

Can work on incomplete graph.

Can determine if certain combination is possible.

Page 19: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

Basic Solution

Tight inequalities: inequalities achieved as equalities

Basic solution:unique solution of n linearly independent tight inequalities

Think of 3D.

Page 20: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

Matching Polytope

Matching polytope

Convex hull of matchings

Linear program

Define by points

Intersections of hyperplanes

Define by inequalities

Goal: Prove they are equal

P Q

Page 21: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

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

Page 22: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

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.

Page 23: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

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!

Page 24: Matching Polytope, Stable Matching Polytope Lecture 8: Feb 2 x1 x2 x3 x1 x2 x3.

Valid Inequalities

Odd set inequalities

That’s enough.

[Edmonds 1965]


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