A sketch proof of the Gilbert-Pollak Conjecture on the Steiner Ratio D.Z.Du, F.K.Hwang, A proof of the Gilbert-Pollak Conjecture on the Steiner Ratio, Algorithmica(1992) 7:121-135 Cheng-Chung, Li 2004/03/01

The way to prove G-P conjecture Lemma 2 Lemma 3 Lemma 1 ft(x)=1-/(3/2)Lt(x) Lemma 4 Lemma 6 Lemma 5 Lemma 7 Lemma 10 Lemma 11 Lemma 12 Lemma 13 Ls(P)(3/2)Lm(P)

Part 1 Inner Spanning Trees Lemma 2 Lemma 3 Lemma 1 ft(x)=1-/(3/2)Lt(x) Lemma 4 Lemma 6 Lemma 5 Lemma 7 Lemma 10 Lemma 11 Lemma 12 Lemma 13 Ls(P)(3/2)Lm(P)

Some Defines-(1) An ST T can be determined by its topology t and at most 2n-3 parameters, including all edge length of T and all angles at regular points of degree 2 in T When writing all parameters into a vector x, the ST T is denoted by t(x) Given a full topology t and a parameter vector x, connecting every pair of adjacent regular points in t(x), we obtain a polygon, if this polygon is simple, i.e., not self-intersecting, then it bonds an area containing the tree t(x), this area is called the characteristic area of t at point x

Some Defines-(2) For a Steiner topology t not necessary full, the characteristic area C(t;x) of t at t can be defined in a similar way. Clearly, it is the union of characteristic areas of full subtopologies of t Given t and x, the set of regular points on tree t(x) is denoted by P(t;x) A spanning tree on P(t;x) is called an inner spanning tree for t at x if it lies in the area C(t;x) An inner spanning tree is called a minimum inner spanning tree if it has the minimum length over all inner spanning trees for the given topology at the given point

Some Defines-(3) Let l(t) denote the length of the tree T Given a Steiner topology t, let Xt be the set of parameter vectors such that l(t(x))=1 Let Lt(x) denote the length of the minimum inner spanning tree for t(x) For a topology s of spanning tree, denote by s(t;x) the spanning tree on the point set P(t;x) with topology s For each Steiner topology t and each parameter vector x, let I(t;x)(MI(t;x)) be the set of spanning tree topologies s such that the s(t;x) is an (a minimum) inner spanning tree for t at x

G-P conjecture is a corollary of the below 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))(3/2)l(N) This is our final goal in this paper

Lemma 1 Lemma 1: Lt(x) is a continuous function with respect to x Lemma 1 is a consequence of the following two lemmas Lemma 2: If mMI(t;x), then there exits a neighborhood of x such that for any point y in the neighborhood, mI(t;y) Lemma 3: For every x, there is a neighborhood of x such that for any y in the neighborhood, MI(t;y)MI(t;x)

Function ft: XtR Lemma 2 Lemma 3 Lemma 1 ft(x)=1-/(3/2)Lt(x) Lemma 4 Ls(P)(3/2)Lm(P)

Some Defines-(1) Define a function ft: XtR by setting ft(x)=1-(3/2)Lt(x)(=l(t(x))-(3/2)Lt(x)) By Lemma 1, ft(x) is continuous and hence reaches the minimum in Xt Let F(t) denote the minimum value of ft(x) over all xXt Note that every ST is similar to an ST with length one, thus Theorem 1 holds iff for any Steiner topology t, F(t)0

Some Defines-(2) We prove Theorem 1 by contradiction Suppose that Theorem 1 is not true and let F(t*) be the minimum of F(t) over all Steiner topologies t Then, F(t*)<0

Lemma 4 & 5 Lemma 2 Lemma 3 Lemma 1 ft(x)=1-(3/2)Lt(x) Lemma 4 Ls(P)(3/2)Lm(P)

Lemma 4 and some defines Lemma 4: t* is a full topology Since t* is a full topology, every component of the parameter vector x in Xt* is an edge length of the ST t*(x) A full topology t is said to be a companion of t* if two regular points are adjacent in t iff they are adjacent in t* A point xXt* is called a minimum point if ft*(x)=F(t*)

Lemma 5 Lemma 5: Let x be a minimum point. Then, x>0, that is, every component of x is positive By Lemma 4, we know that every length in ST t*(x) is positive

Lemma 6 & 7 Lemma 2 Lemma 3 Lemma 1 ft(x)=1-/(3/2)Lt(x) Lemma 4 Ls(P)(3/2)Lm(P)

Lemma 6 & 7 Lemma 6: 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 By Lemma 4 and Lemma 6, we have Lemma 7 : 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

Some Defines and Lemma 8 Let t be a full topology and x a parameter vector Denote by (t;x) the union of minimum inner spanning trees for t(x) Lemma 8: Two minimum inner spanning trees can never cross, i.e., edges meet only at vertices From Lemma 8, we can see that (t;x) divides the characteristic area C(t;x) into smaller areas each of which is bounded by a polygon with vertices all being regular points Such polygon is called a polygon of (t;x), if it is a subgraph of (t;x)

Some Defines When t is a full topology, the characteristic area of t(x) is bounded by a polygon of n edges Partitioning the area into n-2 triangles by adding n-3 diagonals, we will obtain a network with n vertices and 2n-3 edges This network will be called a triangulation of C(t;x) A (t;x) is said to have a critical structure if (t;x) partitions C(t;x) into exactly n-2 equilateral triangles A (t;x) with a critical structure is also said to be critical

Lemma 10 Lemma 2 Lemma 3 Lemma 1 ft(x)=1-/(3/2)Lt(x) Lemma 4 Lemma 6 Ls(P)(3/2)Lm(P)

Lemma 10 Lemma 10: Any minimum point x with the maximum number of minimum inner spanning trees has a critical (t;x) The proof of Lemma 10 needs the results of Lemma 1, 5, 7 Now, in order to derive a contradiction to F(t*)<0, it suffices to show that for any ST t(x) with critical (t;x), Theorem 1 holds for t(x) Now that a critical (t;x) contains n-2 equilateral triangles which form a framework fixing all regular points Let a be the length of an edge of the equilateral triangles in the (t;x) If we divide the plane into a union of disjoint equilateral triangles with edge length a, then all regular points in a critical (t;x) can be placed on the lattices points

Lemma 11,12,13 Lemma 2 Lemma 3 Lemma 1 ft(x)=1-/(3/2)Lt(x) Lemma 4 Ls(P)(3/2)Lm(P)

Lemma 11 Lemma 11: The minimum spanning tree for n lattice points has length at least (n-1)a. For the point set P(t;x) with critical (t;x), the minimum inner spanning tree has length exactly (n-1)a From Lemma 10, A minimum point x, …, has a critcal (t;x), then any minimum inner spanning tree is a minimum spanning tree Thus, to show that Theorem 1 holds for t(x), it suffices to verify the truth of Gilbert-Pollak conjecture for the point set P(t;x) with critical (t;x)

Some Defines Given three directions, each two of which meet an angle of 120。, a shortest network interconnecting a given set P of points and having edges all parallel to the three directions is called a minimum hexagonal tree on P Let T be a minimum hexagonal tree for a given set P of points. A point on T but not in P is called a junction if the point is incident to at least three lines

Lemma 12 & 13 Lemma 12: Ls(P)(3/2)Lh(P), Lh(P) denote the length of the minimum hexagonal tree for P Lemma 13: For any set of n lattice points, there is a minimum hexagonal tree whose junctions are all lattice points

Concludes Lemma 2 Lemma 3 Lemma 1 ft(x)=1-/(3/2)Lt(x) Lemma 4 Lemma 6 Ls(P)(3/2)Lm(P)

Ls(P)Lm(P) Consider the lattice containing the critical (t;x) Since all regular points are lattice points, by Lemma 13, there is a minimum hexagonal tree, with junctions all being lattice points By Lemma 11, the length of the minimum hexagonal tree is at least (n-1)a, the length of a minimum spanning tree. Note that a minimum spanning tree, in this case, is a hexagonal tree Thus, a minimum spanning tree is also a minimum hexagonal tree. By Lemma 12, we have Ls(P)(3/2)Lh(P)=(3/2)Lm(P) Theorem 1 is proved

Discussions Recently, W.D.Smith showed that in the euclidean space of dimension d, 3d9, the Steiner Ratio is not achieved by the vertex set of regular simplex as conjectured This suggests that determining the Steiner Ratio in higher-dimension space is much more difficult but an interesting topic for further research