Introduction to Steiner Tree and

Gilbert-Pollak Conjecture

Cheng-Chung Li

Dept. of Computer Science and Information Engineering

National Taiwan University2004/02/11

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

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

Torricelli had proposed a geometric solution to this problem before 1640

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

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

INPUT OUTPUT

regular point

Steiner point

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

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}.

A

B

C

S

A

B

C

ABC < 120 ABC 120

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

Fig 1 : an example of FST and circumferential order

12

3

4

5

6

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

A

B

D E

S1

C

F G

S2

Split at A

Fig 2 ST S

B

D E

S1

split at B

C

F G

S2

split at C

Fig 3 full sub-STs of ST S

A A

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 n4, there exist at least two Steiner points in an FST which are each adjacent to two terminals

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)

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

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

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

s’

60

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

The Number of Steiner Topologies Let f(n), n2, 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)!

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)!

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

2

03

2/2

03,33 !/)!(,

3

n

k

kn

n

knknknnFnnnF

n 2 3 4 5 6 7 8

f(n) 1 1 3 15 105 945 10395

F(n) 1 4 31 360 5625 110880 2643795

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

Computational Complexity The optimization problem

GIVEN:A set N of regular points in the Euclidean plan FIND:A Steiner Tree of shortest length spanning N

Can be recast as a decision problem GIVEN:A set of regular points in the Euclidean plane and

an integer B DECIDE:Is there a Steiner tree T that spans N such that |

T|B

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

Physical Models Two physical devices have been proposed to model

the ESP

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

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

n=3 A proof for n=3 Case 1

Ls(A,B,C)=Lm(A,B,C)>(√3/2)Lm(A,B,C)

A

BC

CAB120

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)

A

BC

B’C’

SABC has no inner angle larger than 120

n=4 A proof for n=4

Lemma 1 Ls(P)=|EF|=|BG| proved by Melzak in 1961

A

B

E

C

D

F

G

A,B,C,D are four given pointsABE,CDF,AFG are equilateral triangles

Fig 4

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|)

A B

C

D

A

B

C

D

Fig 5

W.L.O.G.,assume BAD+ADC180, we consider four cases

Case 1,BAD120,ADC120

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)

A

B

E

HD

F

C

ABE and CDF areequilateral triangles

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,ADC60

A

E

B

C

D

G

F

ABE,CDF,AFG areequilateral triangles

Fig 6

In Fig 6, EAD=360-EAB-BAD120, and ADF=ADC+CDF120

So Ls(P)=|EF|√3/2(|EA|+|AD|+|DF|)=√3/2(|BA|+|AD|+|DC|)√3/2Lm(p)

Case 3, CAD60,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+CAD120

Furthermore, BAD>120 by assumption, therefore Ls(P)=|BG|√3/2(|BA|+|AD|+|DG|)=√3/2(|BA|+|AD|+|AC|)√3/2Lm(P)

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)

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)

Historical Background

n Prover Year

4 PollakDu,Yao,Hwang

19781982

5 Du,Hwang,Yao 1985

6 Rubin and Thomas 1991

General case Du and Hwang 1992

n for Gilbert-Pollak Conjecture

Prover Year

0.5 Gilbert,Pollak 1968

0.57 Graham,Hwang 1976

0.74 Chung,Hwang 1978

0.8 Du,Hwang 1983

0.824 Chung,Graham 1985

√3/2 Du,Hwang 1992

for general case

The Steiner ratio conjecture of Gilbert-Pollak is true

D.Z.Du F.K.Hwang

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

Other Resources and References Some website about Steiner Tree

http://ganley.org/steiner/ http://www.cs.sunysb.edu/~algorith/files/steiner-tree.sht

ml http://www.nada.kth.se/~viggo/wwwcompendium/node7

8.html http://www.nist.gov/dads/HTML/steinertree.html

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

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