Bipartite Matching Polytope, Stable Matching Polytope
description
Transcript of Bipartite Matching Polytope, Stable Matching Polytope
1
Bipartite Matching Polytope, Bipartite Matching Polytope,
Stable Matching PolytopeStable Matching Polytope
x1
x2
x3
Lecture 10: Feb 15
2
Perfect Matching
3
x1
x3x2x1
x2
x3
(0.5,0.5,0.5)
Integrality Gap Example
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
Black Box
LP-solver
Problem
LP-formulation Vertex solution
Solution
Polynomial time
integral
6
Vertex Solutions
An optimal vertex solution can be found in polynomial time.
7
Prove:
for a bipartite graph,
a vertex solution
corresponds to an
integral solution.
Bipartite Perfect Matching
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 matchingsbecause the cycle is even.
Bipartite Perfect Matching
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
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
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
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
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
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
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 girlAfternoon: girl rejects all but favouriteEvening: rejected boy writes off girl
Day 2
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 girlAfternoon: girl rejects all but favouriteEvening: rejected boy writes off girl
Day 3
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 girlAfternoon: girl rejects all but favouriteEvening: rejected boy writes off girl
A stable matching!
Day 4
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
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
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
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.
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
Weighted Stable Matching
Polynomial time algorithm from LP.
Can work on incomplete graph.
Can determine if certain combination is possible.
[Vande Vate] [Rothblum]
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!