A Proof of the Gilbert-Pollak Conjecture on the Steiner Ratio 洪俊竹 D 蔡秉穎 D 洪浩舜 D 梅普華 P

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, (received April 20, 1990.) Ding-Zhu Du ( 堵丁柱 ) Frank Kwang-Ming Hwang ( 黃光明 )

Euclidean Steiner Problem Given: a set P of n points in the euclidean plane Output: a shortest tree connecting all given points in the plane The tree is called a Steiner minimal tree. Notice that Steiner minimal trees may have extra points (Steiner points).

Minimum Spanning Tree (MST)

Regular points ( ): P Steiner points ( ): V( SMT(P) ) - P Steiner Minimum Tree (SMT) 1 2 2

History of SMT Fermat ( ): Given three points in the plane, find a fourth point such that the sum of its distances to the three given points is minimal. Fermat Torricelli solved this problem before Torricelli

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. Jakob Steiner

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, AApproximation

Steiner Ratio L s (P): length of Steiner Minimum Tree on P L m (P): length of Minimum Spanning Tree on P Steiner ratio ρ =

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.

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|}. L s (A,B,C) = AS + BS + CS = AS + B’S + C’S + BB’ + CC’ = L s (A,B’,C’) + BB’ + CC’ = (AB’+AC’) + BB’ + CC’  (AB’+BB’+AC’+CC’)  (AB+AC) = L m (A,B,C)

How to Prove the Steiner Ratio For any SMT(P), L s (P): length of Steiner Minimum Tree on P L m (P): length of Minimum Spanning Tree on P

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 for lower bound of ρ ρ  0.5 : Gilbert and Pollak (1968) ρ  : Graham and Hwang (1976) ρ  : Chung and Hwang (1978) ρ  0.8 : Du and Hwang (1983) ρ  : Chung and Graham (1985)Chung 2 |SMT| = |red path|  |green path|  |MST| the unique real root of the polynomial x 12 -4x 11 -2x x x 8 -72x x 6 +16x x 4 +80x 3 +56x 2 -64x+16

Previous Results for lower bound of ρ ρ = 0.5 : Gilbert and Pollak (1968) ρ = : Graham and Hwang (1976) ρ = : Chung and Hwang (1978) ρ = 0.8 : Du and Hwang (1983) ρ = : Chung and Graham (1985)Chung

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°

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  12.85length  12.65

Number of Steiner points Consider a Steiner tree with n regular points P 1, …, P n and s Steiner points. Σdeg(vertex) = 2 ( # of edges ) 3s + = 2 ( n + s - 1 ) s = 2n  n - 2  n 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. v1v1 v2v2 v3v3 v4v4 v5v5 v1v1 v2v2 v3v3 v4v4 v5v5

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).

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 X t : the set of parameter vectors x in a Steiner topology t such that l(t(x))=1. L t (x): the length of the minimum inner spanning tree for t(x). Lemma 1: L t (x) is a continuous function with respect to x.

Convert conjecture to function form Define f t : X t  R by f t (x) = 1 － ( ) L t (x) (= l(t(x)) － ( ) L t (x)) f t (x) is a continuous function. F(t): the minimum value of f t (x) over all x  X t. Objective theorem(Theorem 1) holds iff for any Steiner topology t, F(t)  0.

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. t 1 (x(1)): t*(x): t 2 (x(2)): t 3 (x(3)): ++

Proof of Lemma 4 Let T i = t i (x(i)). Every T i has less than n regular points. Apply Theorem 1, for each T i find an inner spanning tree m i such that l(T i )  ( )l(m i ). ∪ i C(t i ; x(i))  C(t; x), so the union m of m i is an inner spanning tree for t* at x. For any x  X t*, f t (x)  0. So F(t*)  0. Contradiction.

Proof of Objective Theorem Since t* is full, every component of x in X t* is an edge length of t*(x). An x  X t* is called a minimum point if f t* (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 treeall 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. e e’e’ 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. c e e’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 number 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

ee’ 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. ee’ 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 L EMMA 12. L s (P)  (  3/2)L h (P) P is point set, L s (P) is length of Steiner minimum tree on P, L h (P) is length of minimum hexagonal tree on P

L s (P )  (  3/2)L h (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 L EMMA 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 JC

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 another 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: ABJ C e 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 L EMMA 13, there is a minimum hexagonal tree with junctions all being lattice points (such MHT does not reduce length) From L EMMA 11, L h (P)  (n-1)a = L m (P) Minimum spanning tree is also minimum hexagonal tree L s (P)  (  3/2)L h (P) = (  3/2)L m (P)

Steiner Ratio of d-dimensional Euclidean Space d = 2: ρ= Du and Hwang 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.

Steiner Ratio of L p -plane In the plane with L p -norm: p = 1: ρ= Frank Kwang-Ming Hwang. On Steiner minimal trees with rectilinear distance. SIAM J. Appl. Math., Vol. 30, pages 104–114, p = 2: ρ= Du and Hwang