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Spanning Tree PolytopeSpanning Tree Polytope

x1

x2

x3

Lecture 11: Feb 21

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Big Picture

LP-solver

Problem

LP-formulation Vertex solution

Solution

Polynomial time

integral

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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.

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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? Exactly 2n

How many linearly independent tight inequalities? At most 2n-1

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Linear Dependency

x1

x3

x4

x2

Multiply +1

Multiply -1

Each edge is counted twice,one positive, one negative.

Sum up to 0 => linear dependency.

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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? Exactly 2n

How many linearly independent tight inequalities? At most 2n-1

Basic solution:unique solution of 2n linearly independent tight inequalities

CONTRA!

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Minimum Spanning Tree

There are exponentially many constraints, but this LP can

still be solved in polynomial time by the ellipsoid method.

The reason is that we can design a polynomial time

separation oracle to determine if x is a feasible solution of

the LP.

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Separation Oracle

S

u v

Each cut has total capacity >= 1

Max-Flow Min-CutEvery cut has at capacity >= 1 if a

nd only ifwe can send 1 unit of flow for all pair.

Separation oracle: check if each pair u,v has a flow of 1.

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Minimum Spanning Tree

0.5

0.5

0.5 0.5

0.5 1

1

1

1

Not a good relaxation.

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Spanning Tree Polytope

E(S): set of edges with both endpoints in S.

A spanning tree has n-1 edges

Cycle elimination constraints

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Separation Oracle

Cycle elimination constraints

|S|-1-x(E(S)) is a submodular function

Minimizing submodular function can be solved in polytime.

A spanning tree has n-1 edges

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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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Separation Oracle

Goal: Prove that there are at most |V|-1 linearly independent tight constraints

If there is an edge of 0, delete it.

Theorem: At most |V|-1

linearly independent tight

inequalities of this type.

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Laminar Family

A laminar family is a collection of sets with no intersections.

Lemma. A laminar family with no singletons has at most n-1 sets.

Forest representation

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Basic Solution

Goal: Prove that there are at most |V|-1 linearly independent tight constraints

Each tight constraint defines a set.

A basic solution is uniquely defined by a laminar family of tight constraints.

This would imply the goal.

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Uncrossing Technique

What about two tight sets are intersecting?

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Laminar Basis

The lemma says that a laminar family formed a basis,

this implies that there are at most n-1 linearly independent tight constraints.

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1. Suppose there is a set S which is not in span(L).

2. Consider a set S with smallest intersecting number with L.

3. Let say S intersect with a set T in L.

4. Consider and

5. Both are tight and have smaller intersecting number with L.

6. So both and are in L

7. On the other hand, since S and T are tight, we have

8. This implies that S is in span(L) as well, a contradiction.

Proof Sketch of Lemma 2

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Intersecting Number

S T

Consider a set R in L. There are only 3 possibilities.

1. R is contained in T.

2. T is contained in R.

3. R and T are disjoint.

1

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Looking Forward

Uncrossing technique is very important in combinatorial optimization.

We will see it in approximation algorithms as well.