Spanning Tree Polytope

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Spanning Tree Polytope
Lecture 11 Feb 21
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Big Picture
Problem
Solution
Polynomial time
LP-formulation
Vertex solution
LP-solver
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
x3
x2
x1
x4
Multiply 1
Each edge is counted twice, one positive, one
negative.
Multiply -1
Sum up to 0 gt 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
Max-Flow Min-Cut Every cut has at capacity gt 1
if and only if we can send 1 unit of
flow for all pair.
Each cut has total capacity gt 1
S
Separation oracle check if each pair u,v has a
flow of 1.
u
v
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Minimum Spanning Tree
Not a good relaxation.
0.5
0.5
1
1
0.5
0.5
1
1
0.5
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Spanning Tree Polytope
A spanning tree has n-1 edges
Cycle elimination constraints
E(S) set of edges with both endpoints in S.
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Separation Oracle
A spanning tree has n-1 edges
Cycle elimination constraints
S-1-x(E(S)) is a submodular function
Minimizing submodular function can be solved in
polytime.
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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
Theorem At most V-1 linearly independent tight
inequalities of this type.
If there is an edge of 0, delete it.
Goal Prove that there are at most V-1 linearly
independent tight constraints
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Laminar Family
Forest representation
A laminar family is a collection of sets with no
intersections.
Lemma. A laminar family with no singletons has
at most n-1 sets.
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Basic Solution
Each tight constraint defines a set.
Goal Prove that there are at most V-1 linearly
independent tight constraints
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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Proof Sketch of Lemma 2

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.

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Intersecting Number
1
2
S
3
T

• Consider a set R in L. There are only 3
  possibilities.
• R is contained in T.
• T is contained in R.
• R and T are disjoint.

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Looking Forward
Uncrossing technique is very important in
combinatorial optimization. We will see it in
approximation algorithms as well.