A Proof of the Gilbert-Pollak Conjecture on the Steiner Ratio

洪俊竹 D89922010蔡秉穎 D90922007洪浩舜 D90922001梅普華 P92922005

The Paper

Ding-Zhu Du (堵丁柱) and

Frank Kwang-Ming Hwang (黃光明).

A Proof of the Gilbert-Pollak Conjecture on the Steiner Ratio. Algorithmica 7(2–3): pages 121–135, 1992. (received April 20, 1990.)

Euclidean Steiner Problem

Given: a set P of n points in the euclidean plane Output: a shortest tree connecting all given poin

ts in the plane

The tree is called a Steiner minimal tree. Notice that Steiner minimal trees may have extr

a points (Steiner points).

Minimum Spanning Tree (MST)

Regular points ( ): P Steiner points ( ): V( SMT(P) ) - P

Steiner Minimum Tree (SMT)

1

2

2

732.13

732.13

History of SMT

Fermat (1601-1665): Given three points in the plane, find a fourth point such that the sum of its distances to the three given points is minimal.

Torricelli solved this problem before 1640.

History of SMT (cont.)

Torricelli Point (or called (First) Fermat Point):

A

B

C

D

E

F

S

History of SMT (cont.)

Jarník and Kössler (1934) formulated the following problem: Determine the shortest tree which connects given points in the plane.

Courant and Robbins (1941) described this problem in their classical book “What is Mathematics?” and contributed this problem to Jakob Steiner, a mathematician at the University of Berlin in the 19th century.

The Complexity of Computing Steiner Minimum Trees

NP-Hard!

• M.R. Garey, R.L. Graham, and D.S. Johnson. The complexity of computing Steiner minimum trees. SIAM J. Appl. Math., 32(4), pages 835–859, 1977.

Approximation

Steiner Ratio

Ls(P): length of Steiner Minimum Tree on P

Lm(P): length of Minimum Spanning Tree on P

Steiner ratio ρ =

PPL

PL

m

s

)(

)(inf

Gilbert-Pollak Conjecture

Gilbert and Pollak conjectured that for any P,

• E.N. Gilbert and H.O. Pollak. Steiner minimal trees. SIAM J. Appl. Math., Vol. 16, pages 1–29, 1968.

)(2

3)( PLPL ms

866.0

2

3

n = 3: Gilbert and Pollak (1968)

Previous Results for n = 3

A

B

C

S

B’C’

Let S be the Torricelli point of △ABC.Suppose that |AS| = min{|AS|,|BS|,|CS|}.

Ls(A,B,C) = AS + BS + CS

= AS + B’S + C’S + BB’ + CC’

= Ls(A,B’,C’) + BB’ + CC’

= (AB’+AC’) + BB’ + CC’

(AB’+BB’+AC’+CC’)

(AB+AC) = Lm(A,B,C)

2

3

2

3

2

32

3

How to Prove the Steiner Ratio

For any SMT(P),

• Ls(P): length of Steiner Minimum Tree on P

• Lm(P): length of Minimum Spanning Tree on P

)(2

3)( PLPL ms

Previous Results for small n

n = 3: Gilbert and Pollak (1968) n = 4: Pollak (1978) n = 5: Du, Hwang, and Yao (1985) n = 6: Rubinstein and Thomas (1991)

Previous Results forlower bound of ρ

ρ 0.5 : Gilbert and Pollak (1968) ρ 0.577 : Graham and Hwang (1976) ρ 0.743 : Chung and Hwang (1978) ρ 0.8 : Du and Hwang (1983) ρ 0.824 : Chung and Graham (1985)

2 |SMT| = |red path| |green path| |MST|

3

1

3

327322

the unique real root of the polynomial x12-4x11-2x10+40x9-31x8-72x7+116x6+16x5-151x4+80x3+56x2-64x+16

Gilbert-Pollak conjecture is true!

Properties of SMT (1/3)

All leaves are regular points.

( Degree of each Steiner point > 1 )

Steiner point

Properties of SMT (2/3)

Any two edges meet at an angle 120°.

( Degree of each point 3 )

120°

120° 120°

Properties of SMT (3/3)

Every Steiner point has degree exactly three.

Steiner Tree

A tree interconnecting P and satisfying previous three properties is called a Steiner tree.

(5,5)

(0,1)

(4,0) (8.0)

(3.17, 1.23)

(5,5)

(0,1)

(4,0) (8,0)

(5.46, 1.09)

length 12.85 length 12.65

Number of Steiner points

Consider a Steiner tree with n regular points P1, …, Pn and s Steiner points.

Σdeg(vertex) = 2 ( # of edges )

3s + = 2 ( n + s - 1 )

s = 2n - 2 - n - 2

n

iiP

1

)deg(

n

iiP

1

)deg(

n

Full Steiner Tree

A Steiner tree with n regular points is called a full Steiner tree if it has exactly n-2 Steiner points, i.e., every regular point is a leaf node.

Decomposition

Any Steiner tree can be decomposed into an edge-disjoint union of smaller full Steiner trees.

Topology

The topology of a tree is its graph structure.

The topology of a Steiner tree is called a Steiner topology.

v1

v2

v3 v4

v5

v1

v2

v3

v4 v5

Steiner Tree

Steiner tree T is determined by its topology t and at most 2n－ 3 parameters.

Parameters:

(1) all edge lengths.

(2) all angles at regular points of degree 2. Writing all parameters into a vector x. A Steiner Tree(ST) of t at x: t(x).

Definitions on a full ST

Convex path. Adjacent regular points. Characteristic area.

Inner spanning tree

The set of regular points on t(x): P(t ; x). Characteristic area: C(t ; x). Inner spanning tree. The vertices of an inner spanning tree for t at

x all lie on the boundary of C(t ; x). Minimum inner spanning tree.

Objective Theorem

l(T): the length of the tree T. Gilbert-Pollack conjecture is a corollary of

the following theorem. Theorem 1:

For any Steiner topology t and parameter vector x, there is an inner spanning tree N for t at x such that l(t(x))( )l(N).

2

3

Gilbert-Pollak Conjecture Is True

Convert conjecture to function form Discuss properties of the function and its

minimum points We only need to discuss minimum point in a

specific structure Prove conjecture with special structure

Convert conjecture to function form

Xt: the set of parameter vectors x in a Steiner topology t such that l(t(x))=1.

Lt(x): the length of the minimum inner spanning tree for t(x).

Lemma 1: Lt(x) is a continuous function with respect to x.

Convert conjecture to function form

Define ft: Xt R by ft(x) = 1 － ( ) Lt(x)

(= l(t(x)) － ( ) Lt(x))

ft(x) is a continuous function.

F(t): the minimum value of ft(x) over all xXt.

Objective theorem(Theorem 1) holds iff for any Steiner topology t, F(t) 0.

2

3

2

3

Gilbert-Pollak Conjecture Is True

Convert conjecture to function form Discuss properties of the function and its

minimum points We only need to discuss minimum point in a

specific structure Prove conjecture with special structure

Proof of Objective Theorem

Prove by contradiction. n: the smallest number of points such that

objective theorem does not hold. F(t*): the minimum of F(t) over all Steiner

topologies t.

F(t*) < 0. Lemma 4: t* is a full topology.

Proof of Lemma 4

Suppose t* is not a full topology.

t1(x(1)):

t*(x):

t2(x(2)):t3(x(3)):

+ +

Proof of Lemma 4

Let Ti = ti(x(i)).

Every Ti has less than n regular points.

Apply Theorem 1, for each Ti find an inner spanning tree mi such that l(Ti)( )l(mi).

∪iC(ti; x(i)) C(t; x), so the union m of mi is an inner spanning tree for t* at x.

For any xXt* , ft(x) 0. So F(t*) 0. Contradiction.

2

3

).(2

3)(

2

3)())(*( mlmlTlxtl i

ii

Proof of Objective Theorem

Since t* is full, every component of x in Xt* is an edge length of t*(x).

An xXt* is called a minimum point if ft*(x)

= F(t*). Lemma 5: Let x be a minimum point. Then x

> 0, that is, every component of x is positive.

Proof of Lemma 5

Suppose x has zero components. (1)zero edge incident to a regular point

Proof of Lemma 5

(2)zero edges between Steiner points.

Gilbert-Pollak Conjecture Is True

Convert conjecture to function form Discuss properties of the function and its

minimum points We only need to discuss minimum point in a

specific structure Prove conjecture with special structure

definition of polygon:

characteristic area one of minimum inner spanning tree

another of minimum inner spanning tree all of minimum inner spanning tree

this called polygon

Critical Structure

Property of polygon• all polygon is bounded by regular points (two minimum

inner spanning tree never cross)

A

B

C

D

E

Assume AB, CD is 2 edge in 2 minimum spanning tree

EA has smallest length among EA, EB, EC, ED.

A is in the component of C when remove CD...

Then, AD < EA + ED CE + ED = CD

Critical Structure

Property of polygon• every polygon has at least two equal longest edges.

ee’

Assume e is longest in the polygon, m is the spanning tree that contain e

e’ not in the m, c is the cycle union by m and e.

Case 1: e is contain in c. m is not minimum.

Case 2: e is not contain in c. another contradition.

ce

e’

Critical Structure

Definition: if a sets of minimum inner spanning tree partition the

characteristic area into n-2 equilateral triangles, we say this sets have critical structure.

Important lemma: Any minimum point x with the maximum number of minimum

inner spanning trees has a critical structure. We only have to prove Theorem 1 hold for all point

with critical structure now. Minimum spanning tree of a point with critical

structure is easy to count.

Critical Structure

Prove idea:• any point x don’t have critical structure the num

ber of minimum inner spanning tree can be increased.

• 3 cases:• There is an edge not on any polygon

• There is a polygon of more then tree edges

• There is a non-equalateral triangle.

Critical Structure

e e’

Case 1: Assume e is the edge not in any polygon.

length of e’ must longer then e. If we shrink the length of e’ between e’ and e, we must can find a minimum point y inside, and the number of minimum inner spanning tree is increased.

e e’

Critical Structure

Gilbert-Pollak Conjecture Is True

Convert conjecture to function form Discuss properties of the function and its

minimum points We only need to discuss minimum point in a

specific structure Prove conjecture with special structure

Minimum Hexagonal Trees

Minimum tree with each crossing of edges has an angle of 120

SMT & MHT

LEMMA 12. Ls(P) (3/2)Lh(P)

• P is point set, Ls(P) is length of Steiner minimum tree on P, Lh(P) is length of minimum hexagonal tree on P

Ls(P ) (3/2)Lh(P )

For triangle ABC with angle A 120• l(BC) (3/2)(l(AB) + l(AC))

Replace each edge of Steiner minimum tree to 2 edges of hexagonal tree Regular point

Steiner point

Characteristic of Hexagonal Tree

Edge has at most two segments Junctions: points on T not in P incident to at

least three lines Exists a MHT such that each junction has at

most one nonstraight edge

Hexagonal Tree for Lattice Points

LEMMA 13. For any set of n lattice points, there is a minimum hexagonal tree whose junctions are all lattice points.

• Bad set: no junction on a lattice point

• Junction in a smallest point set

• Regular point A&B, adjacent junction J, 3nd vertex C adjacent to J A

B

J C

Hexagonal Tree for Lattice Points

If C is regular point, J must be on lattice• Two straight edges fix J to lattice.

If C is junction• J must be on lattice, or J can be moved to regular point or anot

her junction

1. AJ & JB are straight

2. AJ is straight, JB is nonstraight with a segment parallel to AJ

3. AJ is straight, JB is nonstraight without segment parallel to AJ

Case 1: AJ & JB are straight Directions of AJ & JB

• Different directions:

• Same directions:A BJ

Ce

case 1.1

case 1.2

case 1.3

A

B

J

Case 2

Flip JB to line up AJ

A

B

J

Case 3

J can be moved to lattice easily

A

B

J

Proof of Conjecture

• From LEMMA 13, there is a minimum hexagonal tree with junctions all being lattice points (such MHT does not reduce length)

• From LEMMA 11, Lh(P) (n-1)a = Lm(P)

• Minimum spanning tree is also minimum hexagonal tree

• Ls(P) (3/2)Lh(P) = (3/2)Lm(P)

Steiner Ratio of d-dimensional Euclidean Space

d = 2: ρ= • Du and Hwang. 1990.

d = 3: Du and Smith conjectured that

• Ding-Zhu Du and Warren D. Smith. Disproofs of generalized Gilbert-Pollak conjecture on the Steiner ratio in three or more dimensions. Journal of Combinatorial Theory, Series A, Vol. 74, pages 115–130, 1996.

2

3

78419.0140

221119

700

213

700

283

Steiner Ratio of Lp-plane

In the plane with Lp-norm: p = 1: ρ=

• Frank Kwang-Ming Hwang. On Steiner minimal trees with rectilinear distance. SIAM J. Appl. Math., Vol. 30, pages 104–114, 1976.

p = 2: ρ= • Du and Hwang. 1990.

ppp

pxxx

1

21

3

2

2

3