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1 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.

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Presentation on theme: "1 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."— Presentation transcript:

1 1 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

2 2 Outline 1.Introduction 2.Problem Definition 3.Preview 4.Algorithm 5.Example 6.Ratio

3 3 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)

4 4 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.

5 5 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.

6 6 Preview

7 7 Algorithm 1.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 v=C S={C, F} minimize the average distance =[d(C,C)+d(C,F)]/(2-1)=1

8 8 2.Replace S ∪ {v} by a single terminal v s and for each vertex u, let d(v s, u) be the minimum distance from u to a vertex of S ∪ {v}.

9 9

10 10 Example G=(9,36) and terminal node {C, D, H, I} Bold edge length =1 and unseen edge length=2

11 11 minimize the average distance ={d(H, D)+d(I, H))}/(3-1) =2/2=1 V=H, S={D, H, I}

12 12 V=C, S={C, V S1 } minimize the average distance ={d(C, V S1 )}/(2-1) =2

13 13

14 14 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.

15 15 4/3-approximation algorithm for STEINER (1,2) G(V,H) complete graph Square:=terminal node Circle:=optional Bold edge length is 1,other unseen is 2.

16 16 ADH=(1+1)/1=2

17 17 ADH=2/1=2

18 18 OPT(I)

19 19

20 20

21 21 OPT(I)=12+5=17

22 22 HEU(I)

23 23

24 24

25 25 HEU(I)=22 Ratio(I)=22/17=1.29…

26 26 Ratio= HEU(I)/OPT(I)=[1/2(3n-2)]/(2n-2)=4/3

27 27 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 ties OPT(I):= the length of an optimal Steiner tree ratio( I )=HEU( I )/OPT( I )

28 28 Lemma 1 A worst-case instance I contains no pair of terminals 1 apart.

29 29 Lemma 2 For instance I, the average distance in each reduction is greater than 1.

30 30 Lemma 3 If I contains a K-star for K ≧ 3, then ratio (I) ≦ 4/3.

31 31 Lemma 4 OPT(I) ≧ n +|P| +1 P:=A minimum-cardinality set of vertices (optional or terminal ) that dominates all terminals in instance I.

32 32 Lemma 5 OPT(I) ≧ 3/2n- 1/2q-1 q:= The number of equivalence classes in this partition that contain three terminals.

33 33 Theorem ADH is a 4/3-approximation algorithm for STEINER(1, 2). HEU(I) ≦ 2n-s-2 OPT(I) ≧ 3/2n- 1/2q-1

34 34 Thank you


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