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Approximating the Minimum Degree Spanning Tree to within One from the Optimal Degree. R96922115 陳建霖 R96922055 宋彥朋 B93902023 楊鈞羽 R96922074 郭慶徵 R96922142 林佳慶. Authors. Martin Furer Department of Computer Science and Engineering The Pennsylvania State University

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approximating the minimum degree spanning tree to within one from the optimal degree

Approximating the Minimum Degree Spanning Tree to within One from the Optimal Degree

R96922115 陳建霖

R96922055 宋彥朋

B93902023 楊鈞羽

R96922074 郭慶徵

R96922142 林佳慶

authors
Authors
  • Martin Furer

Department of Computer Science and EngineeringThe Pennsylvania State University

  • Balaji Raghavachari

Computer Science Department, EC 3.1University of Texas at Dallas

application
Application
  • Distribution of mail and news on the Internet
  • Designing power grids
similar approximation properties
Similar approximation properties
  • Edge coloring problem
  • 3-colorability of planer graphs
theorem 1
Theorem 1
  • There is a polynomial time approximation algorithm for the minimum degree spanning tree problem which produces a spanning tree of degree
theorem 2
Theorem 2
  • There is a polynomial time approximation algorithm for the minimum degree Steiner tree problem which produces a Steiner tree of degree
theorem 3
Theorem 3
  • There is a polynomial time approximation algorithm for the directed version of the minimum degree spanning tree problem which produces a directed spanning tree of degree
theorem 4
Theorem 4
  • There is a polynomial time approximation algorithm for the minimum degree spanning tree problem which produces a spanning tree of degree at most
improvement
Improvement
  • :the degree of vertex u in T
  • If

we now introduce an “improvement” in T by adding the edge and deleting one of the edge in C incident to

example
Example

G

T

3

2

1

1

<=2

<=2

locally optimal tree
Locally Optimal Tree
  • Def: A locally optimal tree (LOT) is a tree in which none of the non-tree edges produce any improvements. Its maximal degree denoted by k
  • Def: contains those vertices of T which have degree at least i
theorem
Theorem
  • For any b>1, the maximal degree k of a locally optimal tree T is less than

Hence, k =

proof
Proof
  • Observe the ratio of can be larger than b at most times in a row. To be more precise, there is an i with

k- +1 i k such that

Otherwise, we have

which lead to a contradiction

proof1
Proof
  • Suppose we remove the vertices of from T. This split T into a forest F with t trees. And we can say that
proof2
Proof
  • Now, we can say the average degree of vertices in in any spanning tree of G is at least
algorithm sketch
Algorithm (sketch)
  • We start with an arbitrary spanning tree T of G
  • We used the local optimality condition only on “ high”( ) degree vertices.
  • The algorithm stops when no vertex in

has a local improvement.

  • Each phase of the algorithm can be implemented in polynomial time.
theorem 11
Theorem 1
  • There is a polynomial time approximation algorithm for the minimum degree spanning tree problem which produces a spanning tree of degree
slide20
Idea
  • We just need to show that the number of phases is bounded by a polynomial in n.
proof4
Proof
  • Recall :the degree of vertex u in T
  • Define a potential function such that

(c>2)

  • Define total potential of the tree is the sum of the potentials of all vertices.
  • We can say that where k is the maximal degree of T
proof6
Proof
  • For ,

where

proof7
Proof
  • Observe we need phases,the potential reduces by a constant factor.
  • On the other hand, k cannot decrease more than n times.
  • Hence, bound on the number of phases.
steiner problem
Steiner Problem

Consider a graph G = (V,E) and a distinguished set of vertices D V.

Steiner spanning tree: Find a tree whichspans the vertices set D, paths in the tree may go through vertices of V - D.

D1

D2

D3

D4

W1

D5

D6

W4

W2

W3

D7

D8

steiner problem cont
Steiner Problem (cont.)

Minimum Degree Steiner (spanning) Tree

To find a tree of minimum degree, which spans the set D.

This is a generalization of the [Minimum Degree Spanning Tree] problem.

steiner problem cont1
Steiner Problem (cont.)

We start with an arbitrary tree T which spans D and retain only edges which separate the set D in T.

In other words, there are no useless edges since every edge separates at least two distinguished vertices.

slide29
Let W be the set of vertices spanned by T. (W : all vertices in the tree T)

Non-tree path: a path between two vertices in W which goes entirely through vertices of V-W except for the end point.

steiner problem cont2
Steiner Problem (cont.)

Pseudo optimal Steiner tree (POST):

Every edge in the tree separates at least two distinguished vertices.

None of the non-tree paths produce any improvement for any vertex in Si for i = k- ⌈log n⌉ , where k denotes the maximal degree of the tree.

If this property is true for all i, then we call it locally optimal Steiner tree (LOST).

steiner problem cont3
Steiner Problem (cont.)

Similar to the theorem 2.1, we have theorem 3.1: for any b>1, the maximal degree k of a LOST T is at most bΔ* + ⌈log n⌉ . Hence k = O(Δ* + log n ).

This idea can be converted into a polynomial time algorithm which produces a Steiner tree of degree O(Δ* + log n ).

steiner problem cont4
Steiner Problem (cont.)

Theorem 1.2. There is a polynomial time approximation algorithm for the minimum degree Steiner tree problem which produces a Steiner tree of degree O(Δ* + log n ).

slide35

Input : A directed graph G and anyone

spanning tree together with r

which is the root of the tree.

what is a directed spanning tree
What is a directed spanning tree ?
  • Def : A rooted spanning tree T of G is a

subgraph of G with the following

properties:

        • T does not contain any cycles.
        • The outdegree of every vertex except r is exactly one.
        • There is a path in T from every vertex to the root r.
recall
Recall

3

2

1

1

<=2

<=2

The improvement in the undirected case

new improvement
New “improvement”

Idea : Consider a vertex v of indegree i.

Try to reduce the indegree of the

vertex by attaching one of the i

subtrees of v to another vertex of

smaller degree.

slide41

Step1 : Move the root of the subtree T’ that is

being removed from v to a “convenient”

vertex in that subtree.

Step2 : Attach T’ to another vertex outside

the tree to which the new root has a

connection.

what is a convient vertex
What is a convient vertex ?
  • The set of convenient vertices are those in the

strongly connected part of the root of T’ in

the graph induced by the vertices of T’ with all

non-tree edges removed from vertices of

degree i – 1 or greater.

  • We pick a convenient vertex from convenient set
algorithm in directed version
Algorithm in directed version
  • The algorithm tries to decrease the degrees of

vertices in Si for i =k – using the

improvement step above.

a lemma
A Lemma

Let T be a directed spanning tree of degree k.

Let Si consist of those vertices whose indegree

is at least i. Suppose we remove the vertices of

Si from T, breaking T into a forest F.

Then there are at least t = ISiI× (i – 1) + 1 trees

of F whose vertices do not have descendants of

degree i or greater in T.

slide46

Let Y be “ a tree of F whose vertices

do not have descendants of degree i

or greater in T “ ?

What is a Y ?

slide47

i

descendent

ancestor

why is i i i 1 1
Why ISiI× (i – 1) + 1 ?
  • We use induction on the size of |Si|
    • Basis : if |Si| =1, then there are (i-1) +1 = i subtrees like Y (o.k)
    • Note : Each addition of a vertex to the set Si

can remove at most one tree from the

set of candidate trees, but adds at

least i more.

slide49

suppose that |Si| = k holds, i.e : there are k ×(i-1)+1 subtrees like Y.

    • then if |Si| = k+1, we will have at least

[k ×(i-1)+1]+i-1 = k×i-k+i = (k+1) ×(i-1)+1

  • By math induction, we complete the proof.
theorem 1 3
THEOREM 1.3
  • There is a polynomial time approximation algorithm for the directed version of the minimum degree spanning tree problem which produces a directed spanning tree of degree O(∆* + logn).
illustration
illustration
  • As in the proof of the undirected case, we look

for an i in the range k to k – where the

set Si does not “expand” by more than a

constant factor.

  • We will have at least [ ISiI× (i – 1) + 1 ]+ ISiI−1

edges connecting these components and each

one of these edges is incident to at least one

vertex in Si-1

illustration1
illustration
  • Hence the average degree of vertices in Si-1 in

any spanning tree of G is at least

  • The maximal degree ∆* is at least the average

degree of these vertices.

definition
Definition

If ρ(u) ≧ k – 1, we say that u blocks w from (u, v).

w

T

k-1

k

T

w

k-1

k

k-1

k

u

v

≦k-2

k-1

≦k-2

k-1

u

v

If neither u nor v blocks w, then w benefits from (u, v).

the algorithm works in phases
The algorithm works in phases
  • (Recall Def.) Suppose k were the maximal degree of a spanning tree T, Let Si be the set of vertices of degree at least i in T.
  • In each phase we try to reduce the size of Sk by one. If successful, we move on to the next phase.
  • So we also update k after each phase.
  • Even in a phase the number of vertices of degree at most k-1 increases so many, but it’s ok!
local optimality property

Local Optimality Property

Our solution

We use the local optimality property.

many phases

try to improve

Max degree = k

but |Sk| remains the same…

how many phases
How many phases?
  • As the size of Sk reduces by one in each phase (except the last one), there are at most O (n / k) phases when the maximal degree is k.
  • At most phases

k=1

maximal degreefrom k to k-1

k=2

theorem1
Theorem
  • Let △* be the degree of a MDST.
  • Let S be the set Sk.
  • Let B be an arbitrary subset of vertices of degree k–1 in T.

S U B

F

suppose no such edges exist

then k≦△*+1

try to prove something
Try to prove △* ≧ something

△* ≧ average degree of any subset

we want the vertices of high degree

take Sk

take S U B

S U B

Sk

|F| = t

S U B may be either connected or disconnected

(Case 1) (Case 2)

s u b is connected
S U B is connected

S U B

  • Case 1

S U B is connected (a tree), then F contains exactly |S|k+|B|(k-1)-2(|S|+|B|-1) subtrees.

Exactly|S|k+|B|(k-1)-2(|S|+|B|-1)+(|S|+|B|-1) edges connect to S U B.

The average degree of vertices in S U B is exactly

s u b is not connected
S U B is not connected

S U B

  • Case 2

S U B is not connected (not a tree), then F contains more than |S|k+|B|(k-1)-2(|S|+|B|-1) subtrees.

More than|S|k+|B|(k-1)-2(|S|+|B|-1)+(|S|+|B|-1) edges connect to S U B.

The average degree of vertices in S U B is more than

finally we proved k 1
Finally we proved △* ≧ k-1
  • △* ≧ average degree of S U B =
  • Therefore △* ≧ k-1 k ≦ △*+1
  • Note:what if |B|=0?
observation
Observation

Vk U Vk-1

Case 2b

Case 2a

Case 1

k-1

k

Let B = Vk-1, apply Theorem 5.1, we are done.

k-1

k

k-2

k-1

k-2

k

k-1…

≦k-2

≦k-2

≦k-1

≦k-2

If no any Sk on the cycle, just union and mark good

≦k-2

≦k-1

algorithm
Algorithm
  • Step 1. Given a SPT (the degree of all vertices is known), remove Vk U Vk-1 (marked as bad) from T and mark all the connected components as good.

Vk U Vk-1

bad

F

≦k-2

≦k-2

≦k-2

good

algorithm1
Algorithm
  • Step 2. Choose any edge between 2 components and find the cycle generated, also check all bad vertices on the cycle. If no such edges exist, stop.

≦k-2

≦k-2

Vk U Vk-1

F

u

v

≦k-2

≦k-2

algorithm2
Algorithm
  • Step 3. (Case 1) At least a vertex in Vk on the cycle, reduce the size of Vk by one, go to Step 1.

≦k-2

≦k-2

Vk U Vk-1

k

k-1

F

≦k-2

u

≦k-2

v

algorithm3
Algorithm
  • Step 3. (Case 2) No any vertex in Vk on the cycle, (imply) at least a bad vertex in Vk-1 on the cycle, make a union of all components along with all bad vertices on the cycle, also mark the bad vertices as good, go to Step 2.

≦k-2

≦k-2

Vk U Vk-1

k-1

k-1

k-1

u

F

v

≦k-2

≦k-2

good

illustration of example 1
Illustration of example 1

k-1

k

k-1

k-1

k-2

≦k-2

≦k-1

good

good

illustration of example 2
Illustration of example 2

k or k-1

k-1

≦k-2

Must not be choosed

time analysis
Time analysis
  • Step 1. Given a SPT (the degree of all vertices is known), remove Vk U Vk-1 (marked as bad) from T and mark all the connected components as good.
  • Step 2. choose any edge between 2 components and find the cycle generated , also check all bad vertices on the cycle. If no such edges exist, stop.
  • Step 3. (Case 1) At least a vertex in Vk on the cycle, reduce the size of Vk by one, go to Step 1.
  • Step 3. (Case 2) No any vertex in Vk on the cycle, (imply) at least a bad vertex in Vk-1 on the cycle,make a union of all components along with all bad vertices on the cycle, also mark the bad vertices as good, go to Step 2.