1 matching polytope x1 x2 x3 lecture 12: feb 22 x1 x2 x3
TRANSCRIPT
1
Matching PolytopeMatching Polytope
x1
x2
x3
Lecture 12: Feb 22
x1
x2x3
2
Perfect Matching
But not every solutionis a matching!!
3
x1 x3
x2
x1
x2
x3
Matching Polytope
x1
x2x3
4
Valid Inequalities
Enough?
Inequalities which are satisfied by
integer solutions but kill all
unwanted fractional vertex solution.
everywhereeverywhere
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Valid Inequalities
Enough?
Odd set inequalities
Yes, that’s enough.
[Edmonds 1965]
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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?
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Rewriting the Odd-Set Constraints
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Matching Polytope
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Convex Combination
A point y in Rn is a convex combination of
if there exist
so that
and
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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 Rn is a convex combination of
if y is in the convex hull of
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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.
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Convex Combination of Matchings
0.25 0.35 0.4
0.65
0.25
0.25
0.35 0.750.4
0.35
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.
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Convex Combination of Matching
0.25 0.35 0.4
0.65
0.25
0.25
0.35 0.750.4
0.351
0.25 0.35 0.4
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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.
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Fractional solutionssatisfying inequalities
By induction, each smaller fractional solution
is a convex combination of matchings
Check degree
Check odd-set
0.3
0.3
0.3
0.7
0.70.7
Perfect Matching
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Fractional solutionssatisfying inequalities
By induction, each smaller fractional solution
is a convex combination of matchings
Check degree
Check odd-set
0.3
0.3
0.3
0.7
0.70.7
0.70.7
0.3 0.3
0.70.7
0.3 0.3
Perfect Matching
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Fractional solutionsSatisfying inequalities
By induction, each smaller fractional solution is a convex combination of matchings
Check degree
Check odd-set
0.3
0.3
0.3
0.7
0.70.7
0.70.7
0.3 0.3
0.70.7
0.3 0.3
So is the original fractional solution! DONE!
Perfect Matching
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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
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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
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Separation Oracle
S
u v
Each odd cut has total capacity >=
1
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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!
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Separation Oracle
Given a Gomory-Hu tree T, we say an edge e is odd
if T-e consists of two odd components.
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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.
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Minimum Odd Cut
Let C be an odd cut.
C
Case 1: there is an odd edge crossing itu
vx = 1.2
Let the value of the odd edge be x.
Then any u-v cut has capacity at least x >= 1.
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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.
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Big Picture
LP-solver
Problem
LP-formulation Vertex solution
Solution
Polynomial time
integral
Separation oracle