1 the steiner problem with edge length 1 and 2 author: marshall bern and paulplassmann reporter:...

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The Steiner problem with edge length 1 and 2

Author: Marshall Bern and PaulPlassmann

Reporter: Chih-Ying Lin (林知瑩 )

Source: Information Process Letter 32 (1989)171-176

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Outline

1. Introduction

2. Problem Definition

3. Preview

4. Algorithm

5. Example

6. Ratio

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Introduction

• G=(V,E) with edge length 0≧ ， and

a set N V of distinguished vertices.• The Steiner problem asks for a minimum length tree

within G that spans all members of N.• NP-complete problem

G=(9,11)

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• Rayward-Smith’s average distance heuristic (ADH) is a 4/3-approximation algorithm for this problem.

• It is the first proof that a polynomial-time heuristic for an NP-complete Steiner problem achieves an approximation bound better than that given by a minimum spanning tree.

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Problem Definition• Steiner(1, 2) 1. In complete graph

2. All length 1 or 2

The Steiner(1, 2) asks for a minimum length tree within G that spans all members of N.

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Preview

7

C

F

D

B

G

H

E

A

2

2

11

1

2

1

1

2

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Algorithm1.Find a vertex v (optional or terminal) and a set

S of terminals (possibly containing v) that minimize the average distance over all choices of v and S

Su

Svud )1/(),(

v=C S={C, F}

minimize the average distance=[d(C,C)+d(C,F)]/(2-1)=1

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C

F

D

B

G

H

E

A

2

2

11

1

2

1

1

2

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2.Replace S {∪ v} by a single terminal vs and

for each vertex u, let d(vs, u) be the minimum distance from u to a vertex of S {∪ v}.

vs D

B

G

H

E

A

2

11

1

2

12

9

Vs1

G

H

E

A

1

1

2

2

C

F

D

B

G

H

E

A

2

2

11

1

2

1

1

2

2

C

F

D

B

G

H

E

A

2

2

11

1

2

1

1

2

2

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Example

A

G

I

H

D

F

EC

B

I

G=(9,36) and terminal node {C, D, H, I} Bold edge length =1 and unseen edge length=2

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minimize the average distance={d(H, D)+d(I, H))}/(3-1)=2/2=1

A

G

I

H

D

F

EC

B

I

V=H, S={D, H, I}

A

G

VS1

F

C

B

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A

G

VS1

F

C

BV=C, S={C, VS1}

minimize the average distance={d(C, VS1)}/(2-1)=2

A

G

VS2

F

B

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A

G

VS2

F

BA

G

VS1

F

C

B

A

G

VS1

F

C

B A

G

I

H

D

F

EC

B

I

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Instance

• For an instance I of STEINER(1, 2)

• |I|:=the number of vertices (terminal and optional)

• |N|:=n

• ratio(I)= HEU(I)／ OPT(I)

• HEU(I) and OPT(I) mean tree length.

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4/3-approximation algorithm for STEINER (1,2)

G(V,H) complete graphSquare:=terminal nodeCircle:=optionalBold edge length is 1,other unseen is 2.

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ADH=(1+1)/1=2

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ADH=2/1=2

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OPT(I)

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OPT(I)=12+5=17

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HEU(I)

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HEU(I)=22

Ratio(I)=22/17=1.29…

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Ratio= HEU(I)/OPT(I)=[1/2(3n-2)]/(2n-2)=4/3

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4/3-approximation algorithm for STEINER (1,2)

For an instance I of STEINER(1, 2)|I|:=the number of vertices(terminal and optional)n:=|N|HEU(I):=the length of the tree found by ADH assuming a worst possible order of breaking tiesOPT(I):= the length of an optimal Steiner treeratio(I)=HEU(I)/OPT(I)

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Lemma 1

• A worst-case instance I contains no pair of terminals 1 apart.

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Lemma 2

• For instance I, the average distance in each reduction is greater than 1.

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Lemma 3

• If I contains a K-star for K 3,≧ then ratio (I) 4/3. ≦

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Lemma 4

• OPT(I)≧n +|P| +1

• P:=A minimum-cardinality set of vertices (optional or terminal ) that dominates all terminals in instance I.

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Lemma 5

• OPT(I) 3/2≧ n- 1/2q-1

• q:= The number of equivalence classes in this partition that contain three terminals.

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Theorem

• ADH is a 4/3-approximation algorithm for STEINER(1, 2).

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4

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424

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424

121

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)(

)()(

sn

sn

qn

sn

qn

sn

IOPT

IHEUIratio

HEU(I) 2≦ n-s-2OPT(I) 3/2≧ n- 1/2q-1

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Thank you

1 the steiner problem with edge length 1 and 2 author: marshall bern and paulplassmann reporter:...

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