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

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

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

### A Simple Approximation Algorithm from the Optimal Degree

### The Steiner tree version from the Optimal Degree

### Directed spanning trees version in the tree T)

By math induction, we complete the proof.

### The △ in the tree T)*+1 algorithm

### Local Optimality Property in the tree T)

R96922115 陳建霖

R96922055 宋彥朋

B93902023 楊鈞羽

R96922074 郭慶徵

R96922142 林佳慶

Authors from the Optimal Degree

- Martin Furer
Department of Computer Science and EngineeringThe Pennsylvania State University

- Balaji Raghavachari
Computer Science Department, EC 3.1University of Texas at Dallas

Application from the Optimal Degree

- Distribution of mail and news on the Internet
- Designing power grids

Similar approximation properties from the Optimal Degree

- Edge coloring problem
- 3-colorability of planer graphs

Theorem 1 from the Optimal Degree

- There is a polynomial time approximation algorithm for the minimum degree spanning tree problem which produces a spanning tree of degree

Theorem 2 from the Optimal Degree

- There is a polynomial time approximation algorithm for the minimum degree Steiner tree problem which produces a Steiner tree of degree

Theorem 3 from the Optimal Degree

- 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 from the Optimal Degree

- There is a polynomial time approximation algorithm for the minimum degree spanning tree problem which produces a spanning tree of degree at most

Improvement from the Optimal Degree

- :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

Locally Optimal Tree from the Optimal Degree

- 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 from the Optimal Degree

- For any b>1, the maximal degree k of a locally optimal tree T is less than
Hence, k =

Proof from the Optimal Degree

- 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

Proof from the Optimal Degree

- Suppose we remove the vertices of from T. This split T into a forest F with t trees. And we can say that

Proof from the Optimal Degree

- Now, we can say the average degree of vertices in in any spanning tree of G is at least

Proof from the Optimal Degree

Algorithm (sketch) from the Optimal Degree

- 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 1 from the Optimal Degree

- There is a polynomial time approximation algorithm for the minimum degree spanning tree problem which produces a spanning tree of degree

Idea from the Optimal Degree

- We just need to show that the number of phases is bounded by a polynomial in n.

Proof from the Optimal Degree

- 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

Proof from the Optimal Degree

Proof from the Optimal Degree

- For ,
where

Proof from the Optimal Degree

- 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 from the Optimal Degree

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.) from the Optimal Degree

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 (cont.) from the Optimal Degree

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.

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.

Some case of non-tree paths in the tree T)

Steiner Problem (cont.) in the tree T)

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 (cont.) in the tree T)

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 (cont.) in the tree T)

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 ).

- Input : A directed graph G and anyone in the tree T)
spanning tree together with r

which is the root of the tree.

What is a directed spanning tree ? in the tree T)

- 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.

An example in the tree T)

Where is the cycle in directed case ? in the tree T)

New “improvement” in the tree T)

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.

Step1 : Move the root of the subtree T’ that is in the tree T)

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 ? in the tree T)

- 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

T’ in the tree T)

Algorithm in directed version in the tree T)

- The algorithm tries to decrease the degrees of
vertices in Si for i =k – using the

improvement step above.

A Lemma in the tree T)

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.

Let Y be “ in the tree T)a tree of F whose vertices

do not have descendants of degree i

or greater in T “ ?

What is a Y ?

Why in the tree T)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.

- suppose that |S in the tree T)i| = 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

THEOREM 1.3 in the tree T)

- 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 in the tree T)

- 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

illustration in the tree T)

- 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.

illustration in the tree T)

Hence,

Definition in the tree T)

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 in the tree T)

- (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!

Our solution

We use the local optimality property.

many phases

…

try to improve

Max degree = k

but |Sk| remains the same…

How many phases? in the tree T)

- 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

…

Theorem in the tree T)

- 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 in the tree T)△* ≧ 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 in the tree T)

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 in the tree T)

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 in the tree T)

- △* ≧ average degree of S U B =
- Therefore △* ≧ k-1 k ≦ △*+1

- Note：what if |B|=0?

Observation in the tree T)

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 in the tree T)

- 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

Algorithm in the tree T)

- 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

Algorithm in the tree T)

- 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

Algorithm in the tree T)

- 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

Time analysis in the tree T)

- 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.

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