Network Design

1
Network Design

• Lecture 20 March 27

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Spanning Tree with Degree Constraints

• Input
• An undirected graph G (V,E),
• A degree upper bound k.

• Output
• A spanning tree with maximum degree at most k.

NP-complete (Hamiltonian path when k2).
3
Spanning Tree with Degree Constraints
Motivation to find a spanning tree in which
there is no overloaded vertices.
Furer and Raghavachari 92

• Given k, there is a polynomial time algorithm
  which does the following
• Either the algorithm
• Show that there is no spanning tree with maximum
  degree at most k.
• Find a spanning tree with maximum degree at most
  k1.

In other words, there is an 1 algorithm for this
problem!
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Minimum Spanning Tree with Degree Constraints

• Input
• An undirected graph G (V,E),
• A cost c(e) on each edge e,
• A degree upper bound k.

• Output
• A minimum spanning tree with maximum degree at
  most k.

Question Is there a 1 algorithm for this
problem as well? That is, a
polytime algorithm which returns a
minimum spanning tree with maximum degree at
most k1.
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Conjecture
Let OPT be the minimum cost of a spanning tree
with maximum degree k.
Conjecture Given k, there is a polynomial time
algorithm which returns a spanning tree with cost
at most OPT and maximum degree at most k1.
Note that we do not restrict ourselves to MST.
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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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Basic Solution
Tight inequalities inequalities achieved as
equalities
Basic solution unique solution of n linearly
independent tight inequalities
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Finding a Leaf
Theorem At most V-1 linearly independent tight
inequalities of this type.
If there is an edge of 0, delete it.
If there exists a leaf vertex v, then include the
edge incident at v in and remove v from G.
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A Counting Argument
Theorem At most V-1 linearly independent tight
inequalities of this type.
If there is an edge of 0, delete it.
Claim A basic solution of the LP must have a
leaf vertex.
If there is no leaf vertex, then every vertex has
degree 2, and hence there are at least 2n/2n
edges, a contradiction to the above theorem.
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Intuition

• No 1-edge ) many fractional edges
• Vertex solution with few constraints ) few
  fractional edges
• Derive contradiction

So a vertex solution must have an edge of 0 or a
leaf vertex, in either case we can finish by
induction. This proves that the linear program
has an integer optimal solution.
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Minimum Bounded Degree Spanning Tree
The degree constraint of each vertex could be
different
Goal Find a spanning tree with cost at most LP
2.
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Iterative Relaxation

• Initialize F?.
• While F is not a spanning tree
• (Computing a basic solution)
• Solve LP to obtain a basic optimal solution x.
• (Deleting edges)
• Remove all edges e s.t. xe0.
• (Picking the edge of a leaf vertex)
• If there is a leaf vertex v with edge u,v,
  then include u,v in F. Decrease Bu by 1. Delete
  v from G.
• (Removing a degree constraint)
• If there is a vertex v2 W such that degE(v) 3,
  
• then remove the degree constraint of v.

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Removing Degree Constraints
If there is a vertex v2 W such that degE(v) 3,
then remove the degree constraint of v.
This is only done once, and the degree constraint
is violated by at most 2!
1
1/3
Bv1
Bv1
1
1/3
1
1/3
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Key Lemma

• Lemma For any basic solution x, one of the
  following is true
• Either there is a leaf vertex v.
• Or there is a vertex with degree constraint such
  that degE(v)3

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A Counting Argument
Theorem At most V-1 linearly independent tight
inequalities of this type.
Proof of the Lemma Suppose not. Every vertex
has degree at least 2. Every vertex in W has
degree at least 4. E ½(2(n-W)4W)
nW The set of tight constraints E
n-1W A contradiction.
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Iterative Relaxation

• A quick summary
• Find a leaf vertex v, add the only edge at v and
  remove v.
• - Dont lose the cost optimality and the degree
  bound.
• Find a vertex with at most 3 neighbours and has
  a degree
• constraint, remove the degree constraint.
• - Violate the degree bound by at most 2.

By the key Lemma, one of the two possibilities
must hold.
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Remarks

• No combinatorial algorithm has a good performance
  ratio.
• Obtain 1 algorithm if work harder.

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Survivable Network Design

• Input
• An undirected graph G (V,E),
• A cost c(e) on each edge,
• A connectivity requirement r(u,v) for each pair
  u,v.

• Output
• A minimum cost subgraph H of G which has
• r(u,v) edge-disjoint paths between each pair
  u,v.
• (That is, H satisfies all the connectivity
  requirements.)

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Survivable Network Design

• Minimum spanning tree r(u,v) 1 for all pairs.
• Minimum Steiner tree r(u,v) 1 for all pair of
  required vertices.
• Hamiltonian cycle r(u,v) 2 for all pairs and
  every edge has cost 1.
• k-edge-connected subgraph r(u,v) k for all
  pairs.
• Minimum cost k-flow r(s,t) k for the source s
  and the sink t.

Survivable Network Design is NP-complete.
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Linear Programming Relaxation
At least r(u,v) edges crossing S
S
For each set S separating u and v, there should
be at least r(u,v) edges crossing S. Let f(S)
max r(u,v) S separates u and v.
u
v
for each subset S of V
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Special Case Minimum Spanning Tree
How bad is this LP?
LP n/2
LP 2.5
OPT 4
OPT n-1
for each subset S of V
Cannot even solve minimum spanning tree!
0.5
0.5
1
1
0.5
0.5
1
1
0.5
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Half-integrality
Does the LP has half-integral optimal solution?
Consider the minimum spanning tree problem, i.e.
f(S)1 for all S.
Peterson Graph
All 1/3 is a feasible solution and has cost 5.
Any half-integral solution having cost 5 must be
a Hamitonian cycle.
But Peterson graph does not have an Hamitonian
cycle!
So, no half-integral optimal solution!
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Separation Oracle
for each subset S of V
Remember f(S) max r(u,v) S separates u and
v.
At least r(u,v) edges crossing S
S
Max-Flow Min-Cut Every (u,v)-cut has at least
r(u,v) edges if and only if there are r(u,v)
flows from u to v.
u
v
Separation oracle check if each pair u,v has a
flow of r(u,v)!
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Linear Programming Relaxation
for each subset S of V
What is the integrality gap of this LP?
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Insight
All 1/3 is a feasible solution.
But this is not a vertex solution!
This is a vertex solution.
Thick edges have value 1/2 Thin edges have value
1/4.
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Structural Result of the LP Basic Solutions
Theorem. Every vertex solution has an edge with
value at least 1/2
Corollary. There is a 2-approximation algorithm
for survivable network design.
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Iterative Rounding

• Initialization H , f f.
• While f ? 0 do
• Find a vertex solution, x, of the LP with
  function f.
• Add every edge with x(e) 1/2 into H.
• Update f for every set S, set
• Output H.

A new vertex solution is computed in each
iteration
Update the connectivity requirements.
Guaranteed to exist
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Analysis
Corollary. There is a 2-approximation algorithm
for survivable network design.
Intuitive reason we only pick an edge when the
LP picks at least half.
Proof
Let say we pick an edge e.
Key LP-c(e)x(e) is a feasible solution for the
next iteration.
cost(H) c(e) cost(H) 2c(e)x(e) cost(H)
2c(e)x(e) 2(LP-c(e)x(e)) 2LP
2OPT.
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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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Basic Solution
Theorem. Every basic solution has an edge with
value at least 1/3
Using uncrossing technique, we can assume that
the linearly independent tight constraints come
from a laminar family.
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Laminar Family
Forest representation
A laminar family is a collection of sets with no
intersections.
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A Counting Argument

• There are L constraints.
• There are E variables.

Basic solution unique solution of n linearly
independent tight inequalities
Theorem. Every basic solution has an edge with
value at least 1/3
Assume every edge has value 0 1/3. Prove that E L by a counting argument.
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A Counting Argument
Assume every edge has value 0

• At the beginning we give 2 dollars to each edge,
  1 to each endpoint.
• At the end we redistribute the dollars so that
  each member in the
• laminar family has at least 2 dollars, and
  there are still some dollars left.

Then this would imply E L
Redistribute by induction!
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Intuition
Assume every edge has value 0
1/4
1/4
1/4
1/4






1/4




1/4
1/4
1/4
1/4
This can not happen, because the constraints are
linearly independent.
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Laminar Family
Induction Hypothesis The root has 2 extra
dollars.
2
2
2
2
2
2
2
2
By linear independence
Theorem. Every basic solution has an edge with
value at least 1/3
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Important Missing Detail
The connectivity function
at each step is weakly supermodular.
There is a polynomial time separation oracle at
each iteration.
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Concluding Remarks

• No combinatorial algorithm has a good performance
  ratio.
• A unifying framework for network design problems.
• Proofs are based on uncrossing techniques in
  combinatorial optimiation.
• Can use iterative relaxation to get degree
  bounded Steiner networks