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Bipartite Matching Polytope, Bipartite Matching Polytope,

Stable Matching PolytopeStable Matching Polytope

x1

x2

x3

Lecture 10: Feb 15

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Perfect Matching

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x1

x3x2x1

x2

x3

(0.5,0.5,0.5)

Integrality Gap Example

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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!

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Black Box

LP-solver

Problem

LP-formulation Vertex solution

Solution

Polynomial time

integral

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Vertex Solutions

An optimal vertex solution can be found in polynomial time.

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Prove:

for a bipartite graph,

a vertex solution

corresponds to an

integral solution.

Bipartite Perfect Matching

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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.

Bipartite Perfect Matching

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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

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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.

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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?

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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.

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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?

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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 girlAfternoon: girl rejects all but favouriteEvening: rejected boy writes off girl

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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 girlAfternoon: girl rejects all but favouriteEvening: rejected boy writes off girl

Day 2

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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 girlAfternoon: girl rejects all but favouriteEvening: rejected boy writes off girl

Day 3

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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 girlAfternoon: girl rejects all but favouriteEvening: rejected boy writes off girl

A stable matching!

Day 4

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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.

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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?

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Bipartite Stable Matching

Write

if v prefers f to e.

Write

if for some v

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Bipartite Stable Matching

CLAIM:

Proof:

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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

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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

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Bipartite Stable Matching

U

W

At bottom,blue is maximum, red is minimum.

At top,blue is minimum, red is maximum.

U

Wconstruct convex combination.

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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!

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Weighted Stable Matching

Polynomial time algorithm from LP.

Can work on incomplete graph.

Can determine if certain combination is possible.

[Vande Vate] [Rothblum]

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Basic Solution

Tight inequalities: inequalities achieved as equalities

Basic solution:unique solution of n linearly independent tight inequalities

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Bipartite Perfect Matching

Goal: show that any basic solution is an integral solution.

Bipartite perfect matching, 2n vertices. Minimal counterexample.

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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!