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

• Balaji Raghavachari Computer Science Department, EC 3.1

University of Texas at Dallas

Application

• Distribution of mail and news on the Internet

• Designing power grids

Similar approximation properties

• Edge coloring problem

• 3-colorability of planer graphs

Theorem 1

• There is a polynomial time approximation algorithm for the minimum degree spanning tree problem which produces a spanning tree of degree )log( nO

Theorem 2

• There is a polynomial time approximation algorithm for the minimum degree Steiner tree problem which produces a Steiner tree of degree )log( nO

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 )log( nO

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 1

A Simple Approximation Algorithm

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

)(u

2))(),(max()( vuw

),( vuw

Example

3

<=211

2

<=2

G T

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

iS

Theorem

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

Hence, k = nb blog

)log( nO

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

ii SS /1

)(log nO b

nblog ii SbS 1

log

log 1b

b

nkk nS b S n n

Proof

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

iS

)1(2 ii SSit

Proof

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

1iS

1

1

i

i

S

St

Proof

1

1

i

i

S

St

Sib

Si i 1)1(

2( 1) 1 i i i

i

i S S S

b S

b

i 1

b

nk b 1)1log(

b

nk blog

logbk b n

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.

1log nk bS

1log nk bS

Theorem 1

• There is a polynomial time approximation algorithm for the minimum degree spanning tree problem which produces a spanning tree of degree )log( nO

Idea

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

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

)(u

)()( ucu

)(T

kncT )(

Proof

1 1 1

1 1

1

1 2 2

1

1 2, 2

i i a i b i i a i b

i i a i b i i a i b

i i a i b

i i i

c c c c c c

c c c c c c c

c c c c

c c c c a b

Proof

• For ,

where

2 1 2 2( 2 ) (3 ) ( 2 3 )i i i ic c c c c c nki log

2)2()1( iccc

n

c k

3

)2()1(

c

cc

2

)(

n

T

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.

)( 2nO

)( 3nO

The Steiner tree version

Steiner Problem

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

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

W1

W2W3

W4

D1 D2

D3D4

D5D6

D7D8

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 (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.

• 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

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 (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 (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 ).

Directed spanning trees version

• 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 ?

• 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

Recall

• The improvement in the undirected case

3

<=2

11

2

<=2

Where is the cycle in directed case ?

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.

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 ?

• 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’

Algorithm in directed version

• The algorithm tries to decrease the degrees of

vertices in Si for i =k – using the

improvement step above. nlog

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.

Let Y be “ a tree of F whose vertices

do not have descendants of degree i

or greater in T “ ?

What is a Y ?

i

descendent

ancestor

…

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 S i

can remove at most one tree from the

set of candidate trees, but adds at

least i more.

– 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

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

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

nlog

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.

1

1]1)1(|[|

i

ii

S

SiS

illustration

Hence,

nbkbi

S

iS

S

SiS

i

i

i

ii

log

11)1(

*

1

1

*

The △*+1 algorithm

Definition

k-1k

kk-1

≦k-2

u v

w

≦k-2

k

k-1 k-1

k-1

w

u v

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

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

T

T

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!

• We use the local optimality property.

many phases

…

try to improve

Max degree = k

but |Sk| remains the same…

Local Optimality Property

Our solution

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... ( log )1 3

n n nO n n

k k

k=1

k=2

maximal degree from k to k-1

…

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

suppose no such edges exist then k *+1≦△

F

Try to prove * something△ ≧

1

1

1

| | 1*

| |

| | 2(| | 1) | | 1

| |

( 1) | | 1

| |

k

k

k k k

k

k

k

t S

S

k S S S

S

k S

S

△

△* ≧ average degree of any subset

we want the vertices of high degree

take Sk

Sk

|F| = t

take S U BS U B

S U B may be either connected or disconnected (Case 1) (Case 2)

( 1) | | 1

| |

k

k

k S

b S

S U B is connected

• 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|k+|B|(k-1)-(|S|+|B|-1) |S|(k-1)+|B|(k-1)-(|B|-1) |B|-11

|S|+|B| |S|+|B| |S|+|B|k

S U B

S U B is not connected

• 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.

S U B

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

|S|k+|B|(k-1)-(|S|+|B|-1) |S|(k-1)+|B|(k-1)-(|B|-1) |B|-11

|S|+|B| |S|+|B| |S|+|B|k

Finally we proved * k-1△ ≧

• △* ≧ average degree of S U B =

• Therefore * k-1 △ ≧ k *+1≦ △

| | 11

| | | |

Bk

S B

0 1 1* 1 1

|S| 0 |S|

*

k k

k

△

△

• Note ： what if |B|=0?

Observation

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

Vk U Vk-1

≦k-2

≦k-2

≦k-2

≦k-2

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

kk

k-1k-1k-2

kk-1

Case 1Case 2aCase 2b

k-2k-1…

≦k-1≦k-1

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

F

≦k-2≦k-2 ≦k-2

good

bad

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.

Vk U Vk-1

F

uv≦k-2

≦k-2

≦k-2

≦k-2

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.

Vk U Vk-1

F

uv

≦k-2

≦k-2 ≦k-2

≦k-2

kk-1

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.

Vk U Vk-1

Fuv≦k-2

≦k-2 ≦k-2

≦k-2

k-1k-1 k-1

good

Illustration of example 1

≦k-2

k

k-1k-1

good good

k-2

≦k-1

k-1

Illustration of example 2

k or k-1

k-1 ≦k-2

Must not be choosed

Illustration of example 3

k-1k k

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.

( ( ))O m n

(1)O

( ( ))O n n

)(nO

( )O n