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

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

• ??? D89922010
• ??? D90922007
• ??? D90922001
• ??? P92922005

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The Paper

• Ding-Zhu Du (???) and Frank
  Kwang-Ming Hwang (???). A Proof of the
  Gilbert-Pollak Conjecture on the Steiner Ratio.
  Algorithmica 7(23) pages 121135, 1992.
  (received April 20, 1990.)

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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).

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Minimum Spanning Tree (MST)
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Steiner Minimum Tree (SMT)

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

1
2
2
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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.

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History of SMT (cont.)

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

D
F
B
S
A
C
E
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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.

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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 835859, 1977.
• Approximation

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Steiner Ratio

• Ls(P) length of Steiner Minimum Tree on P
• Lm(P) length of Minimum Spanning Tree on P
• Steiner ratio ?

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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 129,
  1968.

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Previous Results for n 3

• n 3 Gilbert and Pollak (1968)

Let S be the Torricelli point of ?ABC. Suppose
that AS minAS,BS,CS. Ls(A,B,C)
AS BS CS AS BS CS BB CC
Ls(A,B,C) BB CC (ABAC) BB
CC ? (ABBBACCC) ? (ABAC)
Lm(A,B,C)
A
S
C
C
B
B
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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

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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)

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Previous Results forlower bound of ?

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

2 SMT red path ? green path ? MST
the unique real root of the polynomial
x12-4x11-2x1040x9-31x8-72x7116x616x5-151x480x3
56x2-64x16
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Previous Results forlower bound of ?

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

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Gilbert-Pollak conjecture is true!
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Properties of SMT (1/3)

• All leaves are regular points.
• ( Degree of each Steiner point gt 1 )

Steiner point
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Properties of SMT (2/3)

• Any two edges meet at an angle ? 120.
• ( Degree of each point ? 3 )

120
120
120
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Properties of SMT (3/3)

• Every Steiner point has degree exactly three.

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Steiner Tree

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

(5,5)
(5,5)
(0,1)
(0,1)
(5.46, 1.09)
(3.17, 1.23)
(4,0)
(8.0)
(4,0)
(8,0)
length ? 12.85
length ? 12.65
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Number of Steiner points

• Consider a Steiner tree with n regular points P1,
  , Pn and s Steiner points.
• Sdeg(vertex) 2 ( of edges )
• 3s 2 ( n s - 1 )
• s 2n - 2 - ? n - 2

? n
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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.

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Decomposition

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

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Topology

• The topology of a tree is its graph structure.
• The topology of a Steiner tree is called a
  Steiner topology.

v1
v2
v1
v5
v2
v3
v3
v4
v4
v5
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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).

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Definitions on a full ST

• Convex path.
• Adjacent regular points.
• Characteristic area.

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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.

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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).

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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

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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.

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

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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

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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) lt 0.
• Lemma 4 t is a full topology.

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Proof of Lemma 4

• Suppose t is not a full topology.

t(x)
t3(x(3))
t1(x(1))
t2(x(2))


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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 x?Xt , ft(x) ? 0. So F(t) ? 0.
  Contradiction.

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Proof of Objective Theorem

• Since t is full, every component of x in Xt is
  an edge length of t(x).
• An x?Xt is called a minimum point if ft(x)
  F(t).
• Lemma 5 Let x be a minimum point. Then x gt 0,
  that is, every component of x is positive.

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Proof of Lemma 5

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

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Proof of Lemma 5

• (2)zero edges between Steiner points.

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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

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Critical Structure
definition of polygon
characteristic area
one of minimum inner spanning tree
this called polygon
another of minimum inner spanning tree
all of minimum inner spanning tree
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Critical Structure

• Property of polygon
• all polygon is bounded by regular points (two
  minimum inner spanning tree never cross)

A
C
E
B
D
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 lt EA ED ? CE ED CD
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Critical Structure

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

c
e
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.
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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.

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Critical Structure

• Prove idea
• any point x dont 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.

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Critical Structure
e
e
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.
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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

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Minimum Hexagonal Trees

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

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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

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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
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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

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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 AB, adjacent junction J, 3nd
  vertex C adjacent to J

A
J
C
B
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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

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Case 1 AJ JB are straight

• Directions of AJ JB
• Different directions
• Same directions

A
J
B
A
B
J
case 1.1
case 1.2
C
e
case 1.3
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Case 2

• Flip JB to line up AJ

A
J
B
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Case 3

• J can be moved to lattice easily

A
B
J
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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)

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Steiner Ratio of d-dimensional Euclidean Space

• d 2 ?
• 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 115130, 1996.

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Steiner Ratio of Lp-plane

• In the plane with Lp-norm
• p 1 ?
• Frank Kwang-Ming Hwang. On Steiner minimal trees
  with rectilinear distance. SIAM J. Appl. Math.,
  Vol. 30, pages 104114, 1976.
• p 2 ?
• Du and Hwang. 1990.