Introduction to Steiner Tree and Gilbert-Pollak Conjecture

1. Introduction to Steiner Tree and Gilbert-Pollak Conjecture Cheng-Chung Li Dept. of Computer Science and Information Engineering National Taiwan University 2003/02/09

2. Outline • Introduction to Steiner Tree • Historical Background • Some Basic Notions • Some Basic Properties • Full Steiner Trees • Steiner Hulls • The number of Steiner Topologies • Computational Complexity • Physical Models • Gilbert-Pollak Conjecture • n=4, 5 • Historical Background • Other Resources and References

3. Historical Background • Pierre de Fermat(1601-1665) • Fermat Problem • Find in the plane a point, the sum of whose distances from three given point is minimal

4. Torricelli had proposed a geometric solution to this problem before 1640

5. There are two ways to generalize the Fermat Problem • Find a point such that the sum of n distances from the point to n given points achieves minimal – still called Fermat Problem • Find a shortest network interconnecting n given points on the Euclidean plane – Steiner Tree Problem • In Fact, such a shortest network must be a tree, which is called a Steiner minimum tree(SMT), for the given set of points • Courant and Robbins in their famous 1941 book “What is Mathematics” referred to it as the Steiner Problem

6. Some Basic Notions • Let P be a set of n points in a metric space or a graph • A SMT for a point set P may contain vertices such vertices are called Steiner points while vertices in P are called regular points regular point Steiner point OUTPUT INPUT

7. Some Basic Properties • A tree connecting the point set P and satisfying (1), (2), (3) is called a Steiner tree(ST). Its topology(the graph structure of the network) is called a Steiner topology • (1)all leaves are regular points • (2)any two edges meet at an angle of at least 120 • (3)every Steiner point has degree exactly three, and every angle at a Steiner point equals 120

8. Note : if two edges, say AB and BC, meet at an angle ABC of less than 120, then by the solution of the Fermat problem, we see that AB and BC cannot form an SMT for{A, B, C}. Namely, we can shorten the total length of the network by replacing AB and BC with an SMT for{A, B, C}. B B S C A A C ABC < 120 ABC  120

9. Full Steiner Trees(FST) • Theorem 1 : An ST for P contains at most n-2 Steiner points • Suppose that an ST has k Steiner points. Then it has n+k-1 edges. Since each Steiner point has three edges and each regular point at least one, the number of edges must be at least (3k+n)/2 • It follows that n+k-1  (3k+n)/2 or n-2  k • An ST with the maximum n-2 Steiner points is called a FST • Here are some properties about FST • Each terminal is of degree one in an FST • If in an ST every regular point is a leaf, then the ST is a FST • For an given set P of points and a full Steiner topology F, if exists, is unique

10. 2 Fig 1 : an example of FST and circumferential order 1 3 4 6 5

11. If a regular point is not a leaf, the we can split the ST at this regular point, in this way, the ST can be decomposed in to edge-disjoint full sub-STs • Such full sub-STs are called full components of the ST • The topology of a full ST is also called a full topology

12. B C S1 S2 D E F G A Split at A Fig 2 ST S

13. S1 S2 D E F G A A C B split at C split at B Fig 3 full sub-STs of ST S

14. Consider a full topology t and a FST T of topology t, two regular points are said to be adjacent in t if there is a convex path connecting them in T • For n4, there exist at least two Steiner points in an FST which are each adjacent to two terminals

15. Steiner Hulls • A Steiner Hull (characteristic area) for a given set of points P is defined to be a region which is known to contain an SMT • The smaller a Steiner Hull is the better • Lemma 1(The Lune property) Let uv be any edge of an SMT. Let L(u,v) be the region consisting of all points p satisfying |pu| < |uv| and |pv| < |uv|L(u,v) is the lune-shaped intersection of circles of radius |uv| centered on u and v. No other vertex of the SMT can lie in L(u,v)

16. Proof : if q were such a vertex, the SMT would contain either a path from q to u not containing v, or vice versa. In the former case, for example, the SMT can be shortened by deleting [u,v] and adding [q,v], a contradiction

17. Lemma 2(The Wedge property) Let W be any open wedge-shaped region having angle 120 or more and containing none of the terminals. Then W contains no Steiner Points • w.l.o.g.,let W cover the span of angles from-60 to 60 • Suppose to the contrary that W contains a Steiner point • Let s be the Steiner point in W with largest x-coordinate

18. Of the three edges at s, one leaves s in a direction within 60 of the positive x-axis • The edge cannot leave W and so cannot end at a terminal • Furthermore, its endpoint(s’) has a larger x-coordinate than s, a contradiction W s 60 s’

19. Corollary 1 : The convex hull of P is a Steiner Hull • Each supporting line of the convex hull defines a 180 wedge free of terminals

20. The Number of Steiner Topologies • Let f(n), n2, denote the number of full Steiner topologies with (n-2) Steiner points • Clearly, f(2)=1 • Let F be a full Steiner topology with n+1 terminals, if one removes the terminal pn+1 and also its adjacent Steiner point, one obtains a full Steiner topology with n terminals • This shows that every full Steiner topology with n terminals by adding a Steiner point s in the middle of one of the (2n-3) edges and adding an edge connecting s to pn+1 • Hence f(n+1)=(2n-3)f(n) ; which has the solution f(n)=2-(n-2)(2n-4)!/(n-2)!

21. Let F(n,k) denote the number of Steiner topologies with |P|=n and k Steiner points such that no terminal is of degree three • Then F(n,k) can be obtained from f(k) by first selecting k+2 terminals and a full Steiner topology on it, and then adding the remaining n-k-2 terminals one at a time at interior points of some edges • The first terminal can go to one of 2k+1 edges, the second to one of 2k+2 edges,…,and the (n-k-2)nd to one of the n+k-2 edges • Thus F(n,k)=(n,k+2)f(k)[(n+k-2)!]/(2k)!

22. Now consider Steiner topologies with terminals of degree three. Suppose that there are n3 of them • Such topologies are obtainable from Steiner topologies with n-n3 terminals and k+n3 Steiner points by labeling n3 of the Steiner points as terminals • Let F(n) denote the number of Steiner topologies with |N|=n,Then

23. Even though f(n) is much smaller than F(n), it is still a superexponentional function, i.e., increasing faster than an exponential function

24. Computational Complexity • The optimization problem • GIVEN:A set N of terminals in the Euclidean plan • FIND:A Steiner Tree of shortest length spanning N • Can be recast as a decision problem • GIVEN:A set of terminals in the Euclidean plane and an integer B • DECIDE:Is there a Steiner tree T that spans N such that |T|B

25. Discrete Edclidean Steiner problem (perhaps simpler than above two) • GIVEN:A set N of terminals with integer coordinates in the Euclidean plane and integer B • Decide:Is there a Steiner tree T that spans N, such that all Steiner points have integer coordinates, and the discrete length of T is less than or equal to B, where the discrete length of each edge of T is the smallest integer not less than the length of that edge • Unfortunately, the discrete Euclidean Steiner tree problem has been show to be NP-hard by Gary, Graham and Johnson in 1977

26. Physical Models • Two physical devices have been proposed to model the ESP

28. The main disadvantages of these physical models are • They do not produce SMTs • It could be time-consuming to construct a large model • Mechanical errors can build up in large models

29. Gilbert-Pollak Conjecture • Let Ls(P) denote the length of SMT on set P • Let Lm(P) denote the length of minimum spanning tree on set P • To get some feeling on the Steiner ratio, let us look at three points A, B, C forming an equilateral triangle with unit length. • Clearly, Ls(A,B,C)=√3, and Lm(A,B,C)=2 • So Ls(A,B,C)/Lm(A,B,C)=√3/2 • G-P conjecture : =Ls(P)/Lm(P) √3/2 for infinite n points

30. n=3 • A proof for n=3 • Case 1 Ls(A,B,C)=Lm(A,B,C)>(√3/2)Lm(A,B,C) A B C CAB120

31. A • Case 2 • Ls(A,B,C)=d(B,B’)+d(C,C’)+d(A,S)+d(B’,S)+d(C’,S)=d(B,B’)+d(C,C’)+√3/2(d(A,B’)+d(A,C’))√3/2(d(A,B)+d(A,C))√3/2Lm(A,B,C) S ABC has no inner angle larger than 120 C’ B’ C B

32. n=4 • A proof for n=4 • Lemma 1 Ls(P)=|EF|=|BG| proved by Melzak in 1961 G A,B,C,D are four given points ABE,CDF,AFG are equilateral triangles A E D B F Fig 4 C

33. Lemma 2 If two line segments AB and BC meet at an angle of at least 120, then |AC|√3/2(|AB|+|BC|) • Lemma 3 If three line segments AB, BC, CD meet as show in Fig 4, where ABC and BCD are both of at least 120, then |AD|√3/2(|AB|+|BC|+|CD|) C C A B A D D B Fig 5

34. W.L.O.G.,assume BAD+ADC180, we consider four cases • Case 1,BAD120,ADC120 • AHD=180-DAH-HDA=180-(180-BAD-EAB)-(180-ADC-CDF)120 • Ls(P)=|EF|√3/2(|EH|+|HF|)√3/2(|EA|+|AD|+|DF|)=√3/2(|BA|+|AD|+|DC|)√3/2Lm(P) H ABE and CDF areequilateral triangles A D F E C B

35. In the remaining cases, we assume one of the two angles, BAD and ADC is greater than 120, w.l.o.g., we assume BAD>120 • Case 2,ADC60 G E ABE,CDF,AFG areequilateral triangles A D B F Fig 6 C

36. In Fig 6, EAD=360-EAB-BAD120, and ADF=ADC+CDF120 • So Ls(P)=|EF|√3/2(|EA|+|AD|+|DF|)=√3/2(|BA|+|AD|+|DC|)√3/2Lm(p)

37. Case 3, CAD60,since |CF|=|DF|, |AF|=|GF|, AFC=60-DFA=GFD, we have AFC~GFD; hence |AC|=|DG| and FCA=FDG • GDA=360-FDG-CDF-ADC=300-FCA-ADC=300-FCD-DCA-ADC=240-(DCA+ADC)=60+CAD120 • Furthermore, BAD>120 by assumption, therefore Ls(P)=|BG|√3/2(|BA|+|AD|+|DG|)=√3/2(|BA|+|AD|+|AC|)√3/2Lm(P)

38. Case 4, ADC<60, CAD<60, then EAC=EAB+BAC=60+BAD-CAD>120, and FCA=FCD+DCA=60+(180-ADC-CAD)>120 • Hence Ls(P)=|EF|√3/2(|EA|+|AC|+|CF|)=√3/2(|AB|+|AC|+|CD|)√3/2Lm(P)

39. n=5 • For n=5, we can use similar method to prove it, but much more complicated(the proof has 24 cases) • For general case, the basic idea in the proof of G-P conjecture is try to translate the problem into a minimax problem(by Du and Hwang in 1992)

40. Historical Background n for Gilbert-Pollak Conjecture

41.  for general case

42. The Steiner ratio conjecture of Gilbert-Pollak is true D.Z.Du F.K.Hwang

43. The Steiner tree is an optimization problem with applications in telecommunications, computer networks and VLSI design

44. Other Resources and References • Some website about Steiner Tree • http://ganley.org/steiner/ • http://www.cs.sunysb.edu/~algorith/files/steiner-tree.shtml • http://www.nada.kth.se/~viggo/wwwcompendium/node78.html • http://www.nist.gov/dads/HTML/steinertree.html

45. References • A Short Proof of a Result of Pollak on Steiner Minimal Trees,D.Z.Du,E.N.Yao,and F.K.Hwang, J.Combinatorial Theory, Ser. A 32(1982)396-400 • The Steiner Ratio Conjecture is True for Five Points, D.Z.Du,F.K.Hwang,and E.N.Yao, J.Combinatorial Theory, Ser. A 38(1985)230-240 • The Steiner Ratio Conjecture for six points,J.H.Rubinstein and D.A.Thomas, J.Combinatorial Theory, Ser. A 58(1991)54-77 • A Proof of the Gilbert-Pollak Conjecture on the Steiner Ratio, D.Z.Du,F.K.Hwang, Algorithmica(1992)7:121-135 • The Steiner Tree Problem, F.K.Hwang,D.S. Richards, and P.winter,1992 • Computing in Euclidean Geometry, 2nd, D.Z.Du,F.K.Hwang, 1995 • The Steiner Tree Problem, H.J.Promel, A.Steger, 2001

46. Thanks for your attention ! • And please give me some improving suggestions !