Steiner Ratio A Proof of the GilbertPollak Conjecture

Steiner Ratio A Proof of the Gilbert-Pollak Conjecture on the Steiner Ratio D, -Z. Du and F. K. Hwang Algorithmica 1992 The Steiner Ratio Conjecture of Gilbert-Pollak May Still Be Open N. Innami˙B. H. Kim˙Y. Mashiko˙K. Shiohama Algorithmica 2010

Steiner Ratio r 99944014 r 99944015 r 99944033 r 99944020 b 95701241 b 96902118 b 95902077 r 99944012 r 99944009 網媒一網媒一管五資四網媒一姚甯之林書漾黃詩晏吳宜庭高新綠何柏樟王柏易黃彥翔鄭宇婷 Ning-Chih Yao Shu-Yang Lin Shih-Yen Hwang Yi-Ting Wu Hsin-Liu Kao Bo-Jhang Ho Bo-Yi Wang Yan-Hsiang Huang Yu-Ting Cheng

Steiner ratio ß P – a set of n points on the Euclidean plane ß SMT(P) – Steiner Minimum Tree Shortest network interconnecting P � contain Steiner points and regular points � ß ß MST(P) – Minimum Spanning Tree Steiner ratio : L(SMP)/L(MST)

SMT ß Graph SMT Þ Vertex set and metric is given by a finite graph non_terminal ß Euclidean SMT Þ Þ Þ V is the Euclidean space(threedimensional ) and thus infinite Metric is the Euclidean distance Ex: the distance between (x 1, y 1) and (x 2, y 2) terminal

SMT ß SMT(P) Þ Þ Þ Shortest network interconnecting P contain Steiner points and regular points A SMT( Steiner Minimum Tree) follows : 1. 2. 3. All leaves are regular points. Any two edges meet at an angle of at least 120 Every Steiner point has degree exactly three. P: {A, B, C} Steiner points: S Regular points: A , B, C, P: {A, B, C, D} Steiner points: S 1, S 2 Regular points: A , B, C, D

Steiner topology An ST for n regular points Ø Ø at most n-2 Steiner points full ST full topology B B A A D S C Not full ST S 2 S 1 C D full ST

ST Þ not a full ST decomposed into full sub-trees of T � full sub-topologies � edge-disjoint union of smaller full ST � full sub tree E B D A S 3 S 1 C S 2 F G Not full ST

Steiner Trees Ø t(x) – denote a Steiner Tree T Ø vector x – (2 n-3) parameters 1. All edge lengths of T , L(e)>=0 2. All angles at regular points of degree 2 in T B A D S C vector x : { L(SA), L(SB), L(SC), L(BD), Angle(SBD) }

Inner Spanning Trees ß a convex path Þ If a path P denoted S 1. . . Sk � Only one or two segments � Si. Si+3 does not cross the piece Si Si+1 Si+2 Si+3 P 1: S 1˙S 2˙S 3˙S 4˙S 5 P 2: S 1˙S 2˙S 3˙S 4 S 5 S 1 S 4 S 2 S 3 P 1 is a convex path S 1 S 3 S 2 P 2 is a not convex path S 4

Inner Spanning Trees adjacent points Ø regular points a convex path connecting them Adjacent points for examples : {S 1, S 4} {S 2, S 5} {S 1, S 5} S 5 S 1 S 4 S 2 S 3 P 1: S 1˙S 2˙S 3˙S 4˙S 5

Inner Spanning Trees adjacent points Ø in a Steiner topology t they are adjacent in a full subtopology of t B A S 1 C E D S 2 F G

characteristic areas P 3 P(t; x) regular points on t(x) S 2 P 4 S 1 C(t; x) characteristic area of t(x) S 3 P 9 P 8 S 7 P 1 S 4 S 6 P 7 S 5 P 6 P 5

characteristic areas P(t; x) regular points on t(x) C(t; x) characteristic area of t(x) P 3 S 2 P 2 S 1 P 1 S 3 P 8 P 9 P 4 P 9 S 7 S 4 S 6 P 1 P 7 S 5 P 6 P 5

Inner Spanning Trees In the area of C(t; x) An Inner Spanning Trees of t (x) P 3 Spanning on P(t; x) P 2 P 4 P 8 P 1 P 9 P 5 P 7 P 6

Inner Spanning Trees Spanning on P(t; x) Not an Inner Spanning Trees of t (x) Not In the area of C(t; x) P 3 P 2 P 4 P 8 P 1 P 9 P 5 P 7 P 6

Steiner Ratio ü l(T) the length of the tree ü Theorm 1 For any Steiner topology t and parameter vector x, there is an inner spanning tree N for t at x such that t(x) : a Steiner tree N : an inner spanning tree

Steiner Ratio Lt(x) length of the minimum inner spanning tree of t(x) p x ∈ Xt p Xt : the set of parameter vectors x such that l (t (x) ) = 1 Lt(x) x Lemma 1: Lt(x) is a continuous function with respect to x Lt(x) x

Steiner Ratio Thm 1 l (t(x)) ≥ (√ 3/2) l(N) Lemma 1 Lt(x) is a continuous function with respect to x f t ( x) = l( t( x) ) – ( √ 3 / 2 ) Lt ( x ) ft(x) = L(SMT) – (√ 3/2)L (MST) l (t(x)) -> length of a Steiner tree Lt(x) -> length of an min inner spanning tree

Steiner Ratio S te ine r ra tio : L( S M T ) / L ( M S T ) ft(x) = L(SMT) – (√ 3/2)L (MST) ü ü if ft(x) ≥ 0 then L(SMT) /L (MST) ≥ (√ 3/2) ft(x) = L(SMT) – (√ 3/2)L (MST)

Theorem 1 : for any topology y and parameter x, there is an inner spanning tree N for t at x such that: That is , for any x and any t, there exist inner spanning tree N such that:

Between ft(x) and Theorem 1 ß ß Theorem 1 holds if ft(x)>=0 for any t any x. By Lemma 1: ft(x) is continuous, so it can reach the minimum value in Xt.

Between F(t) , F(t*) and Theorem 1 ß ß Let F(t) = minx ft(x) x Xt Then theorem 1 holds if F(t)>=0 for any t. Let t* = argmint F(t) t: all Steiner topologies Then theorem 1 holds if F(t*)>=0.

Prove Theorem 1 by contradiction ß ß ß P : Theorem 1 (F(t*)>=0) ~P : exist t* such that F(t*)<0 Contradiction : If ~P => P then P is true. Assume F(t*)<0 and n is the smallest number of points such that Theorem 1 fail. Some important properties of t* are given in the following two lemmas.

t* is a full Lemma 4. topology Assume t* is not a full topology => for every x Xt ST t*(x) can be decomposed into edge-disjoint union of several ST Ti’s Ti=ti(x(i)) , ti : topology , x(i) : parameter => Ti has less then n regular points => find an inner spanning tree mi such that

=> m : the union of mi => => F(t*) ≥ 0 , contradicting F(t*) ＜ 0.

Lemma 5. Let x be a minimum point. Every component of x is positive. Definition : Companion of t* : 1. t is full topology 2. if two regular point are adjacent in t they are adjacent in t* Minimum point : ,

Assume that x has zero components 1. regular steiner : contradiction! (similar to lemma 4) point 2. steiner : find a “t” with conditions point and P(t; y)=P(t*; x) 實線: t*(x) with zero component (steiner point重和 ) 虛線: t(y)

: find a “t” with conditions point and P(t; y)=P(t*; x) steiner 1. t is a companion of t* 2. there is a tree T interconnecting n points in P(t*; x) , with full topology t and length less than l(t*(x))

find “t” 1. if the ST of topology t exists: let since and t(hy) is similar to t(y)

Definition: any tree of topology t : t(y, Θ) Lt(y, Θ) : the length of minimum inner spanning tree for t G(t)=minimum value of gt(y, Θ)

2. if the ST of topology t does not exist: y has no zero component : t(y, Θ) must be a full ST → G(t)=F(t) → F(t)<F(t*) contradiction! 2. y has zero components : 1. consider subgraph of t induced (1) if every connected component of subgraph having an edge contains a regular point => by Lemma 4 find a full topology t’, G(t’)<0

2. if the ST of topology t does not exist: (2) if exists such connected component of subgraph having an edge contains a regular point => find a full topology t’, G(t’)<G(t) repeating the above argument, we can find infinitely many full topologies with most n regular points contradicting the finiteness of the number of full topology

Lemma 6~9

Lemma 6 ß Let t be a full topology and s a spanning tree topology. Then l(s(t; x)) is a convex function with respect to x.

Convex Function ß contains concave curves

Convex Function ß ß contains concave curves 2 nd deviation function must be nonnegative everywhere

Convex Function ß ß ß contains concave curves 2 nd deviation function non-negative c = λa + (1 -λ)b, then f(c) <= λf(a) + (1λ)f(b)

Lemma 6 ß ß ß Consider each edge of inner spanning tree … Consider one element of the vector … The sum of convex functions is a convex B function B B A Flash demo: http: //www. csie. ntu. edu. tw/~b 96118/convex. swf

Lemma 7

Lemma 7 ß Suppose that x is a minimum point and y is a point in Xt*, satisfying MI(t*; x) MI(t*; y). Then, y is also a minimum point.

Lemma 8 Γ(t; x) is the union of minimum inner spanning trees

Lemma 8 • • • ß • l(AC) < l(AE) + l(EC) Two minimum inner Without loss of Remove the edge CD A is in the connected l(AE) + l(EC) ≤ l(CD) spanning trees can generality, assume from the tree U, the component containing → l(AC) < l(CD) never cross, i. e. , that EA has a smallest remaining tree has D. • We obtain an inner spanning edges meet only at length among EA, EB, two connected tree with length less than that Use AC to connect the vertices. of U, contradicting with the EC, ED. components 2 components minimality of U. ß Proof by contradiction containing C and D, • Therefore, 2 Minimum Inner respectively Spanning Trees can never cross.

Lemma 9 • • • Another example Γ(t; x) • ßProof by A ploygon of Replace e with Let m be the Every polygon of Γ(t; • Therefore, polygon of Replace e with Γ(t; x) is a cycle another edge in the minimum inner • Let n be the contradiction x) has at least two Γ(t; x) cannot have another edge in the which is a polygon spanning tree minimum spanning equal longest edges. only one longest polygon subgraph of Γ(t; x) containing the tree containing the edge longest edge e • The length of the • new tree is shorter The length of the new tree is shorter than the original!

Triangulation Full topology with n regular points P(t*; x) ß Add n-3 diagonals to obtain n-2 triangles ß Embedded Γ(t*; x) ß

Triangulation ß ß Edge-length representation Edge-length independent

Critical Structure Γ(t*; x) is a triangulation of P(t*; x) such that All triangles are equilateral triangles ß

Critical Structure ß ß Let x ∈Xt be a minimum point such that t*(x) has maximum number of minimum spanning trees. Then Γ(t*; x) is a critical structure

Critical Structure ß Proof: if Γ(t*; x) is not critical, then one of the following may happen: (a). Γ(t*; x) has an edge not on any polygon (b). Γ(t*; x) has a non-triangle polygon (c). Γ(t*; x) has a non-equilateral triangle

Critical Structure ß ß Case (a): Γ(t*; x) has an edge e not on any polygon U ⊆ Γ(t*; x) be a minimum spanning tree contain e

Critical Structure ß ß Let e’ not in Γ(t*; x) be an edge on the same triangle of e such that U{e}+{e’} forms a spanning tree l(e’) > l(e)

Critical Structure ß Shrinking e’ until l(e) = l(e’)

Critical Structure ß ß ß Let P(l) be the new set of regular points with l(e’) = l Let L ⊆ [l(e’), l(e)] such that for l ∈ L, exists minimum point y such that P(l) = P(t*; y) L is nonempty

Critical Structure ß ß Let l* be min{L} and y* be the minimum point such that P(l*) = P(t*; y*) Then l* ≠ l(e), otherwise t*(y*) has more number of minimum spanning trees than t*(x)

Critical Structure ß But if l* ≠ l(e), then we can find l < l * such that P(l) and P(l*) has the same set of minimum spanning trees, contradict to that l* is minimum.

Critical Structure For case (b), Γ(t*; x) has a nontriangle polygon. ß We can shrink an edge not in Γ(t*; x) and obtain a contradiction by similar argument. ß

Critical Structure For case (c), Γ(t*; x) has a nonequilateral triangle ß We can increase all shortest edges in Γ(t*; x) and obtain a contradiction by similar argument. ß

Critical Structure ß ß Hence, Γ(t*; x) is a critical structure. Finally, we want to say that a minimum spanning tree m of Γ(t*; x) is not too larger than t*(x) by this critical property.

Lattice point Let a be the length of an edge in Γ(t*; x). ß We can put Γ(t*; x) onto lattice points. ß Then the length of a minimum spanning tree of Γ(t*; x) is (n-1)a ß

Another tree structure… ß ß A Hexagonal tree of points set P is a tree structure using edges with only 3 directions each two meet at 120 ° Permit adding points not in P

Hexagonal Tree Let Lh(P) denote the minimum Hexagonal tree of P, we first show that LS(P) ≥ √ 3/2 Lh(P) ß And then we will show that the points set P with critical structure Γ, Lh(P) = Lm(P) ß

Hexagonal Tree Triangle Property ß ∠A ≥ 120 ° then ß BC ≥ √ 3/2 (AB + AC) ß

Hexagonal Tree ß Hence we have LS(P) ≥ √ 3/2 Lh(P)

Hexagonal Tree ß ß Minimum Hexagonal Tree Straight and Nonstraight edge Full and Sub-full Hexagonal Tree Junction

Hexagonal Tree ß There is a Minimum Hexagonal Tree such that any junction has at most one nonstraight edge

Hexagonal Tree ß ß There is a Minimum Hexagonal Tree such that any junction is on a lattice point Suppose not, consider bad points set P with minimum number of regular points such that, for any Minimum Hexagonal Tree H of P, there is a junction not on a lattice point.

Hexagonal Tree ß ß ß If a Minimum Hexagonal Tree H has a junction not on a lattice point… Then we can either shorten the tree or decrease the number of junctions H is full and no junction is on a lattice point

Hexagonal Tree ß There is a junction J of H not on a lattice point and adjacent to two regular points A, B

Hexagonal Tree Let C be third point adjacent to J ß C is not a regular point C is a junction ß

Hexagonal Tree ß If JA and JB are both straight

Hexagonal Tree ß If JA is straight and JB non-straight

Hexagonal Tree Finally, there is a Minimum Hexagonal Tree H with all junctions on lattice points ß Suppose there is m junctions on H, then l(H) = (m+n-1)a ≥ (n-1)a = Lm(P) ≥ Lh(P) = Lm(P) LS(P) ≥ √ 3/2 Lh(P) = √ 3/2 Lm(P) ß

Steiner Ratio The Steiner Ratio Conjecture of Gilbert-Pollak May Still Be Open

Abstract ß Lemma 1: Lt(x) is a continuous function with respect to x Þ ß Lt(x) : length of the minimum inner spanning tree for t(x) Disproof of the continuity of Lt(x).

Continuity ß

Proof of Discontinuity ß

Steiner tree P 3 regular point Steiner point S 2 P 4 S 1 S 3 P 9 P 8 S 7 P 1 S 4 S 6 S 5 P 7 t( y) P 6 P 5

Convex path P 3 S 2 P 4 S 1 S 3 P 9 P 8 S 7 P 1 S 4 S 6 S 5 P 7 t( y) P 6 P 5 Path Sa. Sb is a convex path if • Only one or two segments • Si. Si+3 does not cross the piece Si. Si+1 Si+2 Si+3, for all a ≤ i ≤ b-3

Characteristic Area P 3 S 2 P 4 S 1 S 3 P 9 P 8 S 7 P 1 S 4 S 6 S 5 P 7 t( y) P 6 P 5 Path Sa. Sb is a convex path if • Only one or two segments • Si. Si+3 does not cross the piece Si. Si+1 Si+2 Si+3, for all a ≤ i ≤ b-3

Minimum Inner Spanning Tree P 3 S 2 P 4 S 1 S 3 P 9 P 8 S 7 P 1 S 4 S 6 S 5 P 7 t( y) P 6 One of segments is removed to get a minimum inner spanning P 5 tree

Minimum Inner Spanning Tree P 3 S 2 P 4 S 1 S 3 P 9 P 8 S 7 P 1 S 4 S 6 P 7 S 5 P 6 t(y ) P 5

Start to Converge P 3 S 2 P 4 S 1 S 3 P 9 P 8 S 7 P 1 S 4 S 6 S 5 P 7 t( y) P 6 P 5

Converging P 3 S 2 P 4 S 1 S 3 P 8 P 9 S 7 P 1 S 4 S 6 P 7 S 5 P 6 t(y>x) P 5

Minimum Inner Spanning Tree P 3 S 2 P 4 S 1 S 3 P 9 P 8 S 7 P 1 S 4 P 7 t( x) S 5 P 6 P 5

P 3 S 2 P 4 S 1 S 3 P 9 P 8 S 7 P 1 S 4 P 7 t( x) S 5 P 6 The Steiner tree is decomposed into two full Steiner trees T 1 and P 5 T 2 at P 7 = S 6.

Characteristic Area P 3 S 2 P 4 S 1 S 3 P 9 P 8 S 7 P 1 S 4 P 7 t( x) S 5 P 6 P 5

Convex Path P 3 S 2 P 4 S 1 P 1 S 3 S 4 P 7 S 5 P 6 P 5 Path Sa. Sb is a convex path if • Only one or two segments • Si. Si+3 does not cross the piece Si. Si+1 Si+2 Si+3, for all a ≤ i ≤ b-3

Characteristic Area P 3 S 2 P 4 S 1 S 3 P 9 P 8 S 7 P 1 S 4 P 7 t( x) S 5 P 6 P 5

Minimum Inner Spanning Tree P 3 S 2 P 4 S 1 S 3 P 9 P 8 S 7 P 1 S 4 P 7 t( x) S 5 P 6 P 5

Comparison P 3 S 2 P 4 S 1 S 2 P 4 S 3 S 1 P 9 P 8 S 3 P 8 S 7 P 1 P 7 S 4 S 6 S 5 P 6 t(x) P 9 P 5 P 1 S 4 S 6 P 7 S 5 P 6 t(y->x) P 5

Comparison P 3 S 2 P 4 S 1 S 2 P 4 S 3 P 8 S 1 P 9 S 3 P 8 S 7 P 1 t(x) P 7 P 9 S 7 S 4 S 6 S 5 P 6 P 5 P 1 S 4 S 6 P 7 S 5 P 6 t(y->x)

Continuity ß

Assumption ß ß ß Lemma 1: Lt(x) is a continuous function with respect to x We disprove the continuity of Lt(x) Lemma 1 is not true.

Minimum Spanning Tree P 3 S 2 P 4 S 1 S 3 P 8 P 9 S 7 P 1 S 4 S 6 P 7 S 5 P 6 P 5 Ø If the minimum spanning tree is not limited in the characteristic area, the minimum spanning can be continuous Ø The Steiner Ratio problem may still open