Fermat-Steiner Problem in the Metric Space of Compact Sets endowed with Hausdorff Distance

2 MB·

by  Alexandr Ivanov

Fermat–Steiner Problem in the Metric Space
6
1
0 of Compact Sets endowed with Hausdorff
2

n Distance
a
J

5
1 A. Ivanov, A. Tropin, A. Tuzhilin

]
G
M Abstract
.
h Fermat–Steiner problem consists in finding all points in a metric
at space Y such that the sum of distances from each of them to the
m points from some fixed finite subset of Y is minimal. This problem is
[ investigated for the metric space Y = H(X) of compact subsets of a

1 metricspaceX,endowedwiththeHausdorffdistance. Forthecaseofa
v proper metric space X a description of all compacts K H(X) which
2 ∈
the minimum is attained at is obtained. In particular, the Steiner
5
8 minimal trees for three-element boundaries are described. We also
3 construct an example of a regular triangle in H(R2), such that all its
0
shortest trees have no “natural” symmetry.
.
1
0
6
1 Introduction and Preliminaries
1
:
v
i The initial statement of the problem that is now referred as the Steiner
X
Problem probably belongs to P. Fermat, who asked to find a point in the
r
a
plane, such that the sum of distances from it to three fixed points is minimal.
J. Steiner generalized the problem considering an arbitrary finite subset of
theplaneorofthethree-space. Nowadays theSteiner Problemusuallystands
for the general problem of finding a shortest tree connecting a finite subset
of a metric space, which was stated for the case of the plane by V. Jarnik
and M. Ko¨ssler [1]. By Fermat–Steiner problem we mean the J. Steiner type
generalization of the P. Fermat problem, namely, for a fixed finite subset
= a ,...,a of a metric space (Y,ρ) find allthe points y Y minimizing
1 n
A { } ∈
the value S (y) := ρ(y,a ). The set of all such points we denote by Σ( ).
A i i A
P
1
Put d(y, ) = ρ(y,a ),...,ρ(y,a ) and Ω( ) = d(y, ) : y Σ( ) .
1 n
A A A ∈ A
For each d Ω( )(cid:0)we put Σ ( ) = y (cid:1) Σ( ) : d(y, (cid:8)) = d . Thus, the s(cid:9)et
d
∈ A A ∈ A A
Σ( ) is partitioned into the classes (cid:8)Σ ( ), d Ω( ). Notic(cid:9)e that generally
d
A A ∈ A
the set Σ( ) may be empty, but the following result can be easily obtained
A
from standard compactness and continuity arguments.
Assertion 1.1 (Existence of Steiner Compact in Proper Metric Space). Let
Y be a proper metric space, then the set Σ( ), and hence, the set Ω( ) is
A A
not empty for an arbitrary nonempty finite Y.
A ⊂
In the present paper we consider the Fermat–Steiner problem inthe space
Y = H(X) of nonempty compact subsets of a given metric space X, endowed
with the Hausdorff distance [2], see also [3]. Geometry of the space H(X)
is used in such important applications as patterns recognition and compari-
son, constructing of continuous deformations of one geometrical object into
another, see for example [4] and [5].
In the present paper the following results are obtained.
Let be a finite subset of Y = H(X). Each element from Σ( ) we
A A
call Steiner compact. Notice that in most examples of the boundaries the
Steiner compacts are not uniquely defined even in Σ ( ). In each class
d
A
Σ ( ) minimal and maximal Steiner compacts with respect to the natural
d
A
order generated by inclusion are naturally defined. We prove that if a class
Σ ( ) is not empty, then each Steiner compact K Σ ( ) contains some
d d
A ∈ A
minimal Steiner compact K Σ ( ).
λ d
∈ A
For the case of a proper metric space X we show that the set Σ( ) is
A
not empty for any finite nonempty . Maximal Steiner compact is unique in
A
each Σ ( ) and is equal to the intersection of the corresponding closed balls
d
A
centered at the boundarycompact sets. This maximal compact is denoted by
K ( ). Also in this case we describe the whole class Σ ( ): a compact set
d d
A A
K belongs to Σ ( ), if and only if the inclusions K K K ( ) hold for
d λ d
A ⊂ ⊂ A
some minimal Steiner compact K Σ ( ). Thus, the problem of finding
λ d
∈ A
the Steiner compacts is reduced to the description of maximal and minimal
Steiner compacts in each class Σ ( ). Notice that finding a vector d Ω( )
d
A ∈ A
is a nontrivial problem.
As an example, we consider the boundary set = A ,A ,A H(R2),
1 2 3
A { } ⊂
where each A consists of the following pair of points: a vertex of the regular
i
triangle, and its image under the rotation by the angle π around the center o
3
ofthe triangle. Forthis example we completely describe the set ofthe Steiner
compacts. It turns out that this symmetrical boundary possesses exactly
2
three classes Σ ( ) that are transferred one into another under rotations by
d
A
the angles 2π and 4π around o. The maximal Steiner compact in each class is
3 3
a nonconvex curvilinear 4-gon. Moreover, each class contains unique minimal
Steiner compact, and this Steiner compact consists of two points. The length
of Steiner minimal tree is less than three radii of the circle containing .
A
Let us pass to Hausdorff distance geometry. Let (X,ρ) be a metric space.
Recall that the set BX(A) := x X: ρ(x,A) r , where ρ(x,A) =
r ∈ ≤
inf ρ(x,a), is called a closed n(cid:8)eighborhood of radius r(cid:9)of a set A X in
a A
(X,∈ρ). Notice that if A consists of a single point, then BX(A) is the⊂closed
r
ball of radius r in X centered at this point. Below we omit the reference to
the space and write B (A) instead of BX(A), providing it does not lead to
r r
misunderstandings.
The Hausdorff distance between subsets A and B of a metric space (X,ρ)
is defined as the following value:
d (A,B) = inf r : BX(A) B,BX(B) A .
H r ⊃ r ⊃
(cid:8) (cid:9)
An equivalent definition is given by the expression
d (A,B) = max sup inf ρ(a,b),sup inf ρ(a,b) ,
H
(cid:8)a∈Ab∈B b∈B a∈A (cid:9)
i.e.,
d (A,B) = max supρ(a,B),supρ(b,A) .
H
(cid:8)a∈A b∈B (cid:9)
It is well-known that the Hausdorff distance is a metric on the space H(X)
of compact subsets of X, see for example [3].
Consider an arbitrary metric space (X,ρ). For any points a and b from
X we sometimes write ab instead of ρ(a,b).
| |
We need the following technical results.
Assertion 1.2. Let A and B be compact sets in a metric space (X,ρ). Then
for any point a A there exists a point b B such that ρ(a,b) d (A,B).
H
∈ ∈ ≤
Proof. Indeed,
d (A,B) sup inf ρ(a,b) inf ρ(a,b),
H
≥ a Ab B ≥ b B
∈ ∈ ∈
where the first inequality follows directly from the definition of the Hausdorff
distance, and the second inequality is valid for any point a A because of
∈
the supremum definition. Since B is compact and the function ρ(a,b) is
continuous, then the infimum inf ρ(a,b) is attained at some point b. For
b B
∈
this b the inequality ρ(a,b) d (A,B) holds. Assertion is proved.
H
≤
3
Assertion 1.3. Let A and B be compact sets in a metric space (X,ρ), and
let d (A,B) = r. Then A BX(B) and B BX(A).
H ⊂ r ⊂ r
Proof. It suffices to show that an arbitrary point a A belongs to a closed
∈
neighborhood BX(B). Due to Assertion 1.2, there exists a point b B such
r ∈
that ρ(a,b) d (A,B) = r. But then ρ(a,B) r.
H
≤ ≤
Assertion 1.4. Let compact sets A, B, C, and D be such that A B C
⊂ ⊂
and d (A,D) = d (C,D) = d. Then d (B,D) d.
H H H
≤
Proof. The equality d (A,D) = d and Assertion 1.3 imply that D B (A).
H d
⊂
The inclusion A B imply the inclusion B (A) B (B), and hence, D
d d
⊂ ⊂ ⊂
B (B). Similarly, the equality d (C,D) = d implies the inclusion C
d H
⊂
B (D). But the inclusion B C implies that B B (D). Thus, D B (B)
d d d
⊂ ⊂ ⊂
and B B (D), that implies the inequality desired.
d
⊂
Figure 1: An example to Assertion 1.4
with d (B,D) < d.
H
Remark 1.5. An example of compact sets A, B, C, and D from Asser-
tion 1.4, such that d (B,D) < d is shown in Figure 1. Here A = [a ,a ],
H 1 2
B = [b ,b ], C = [c ,c ], D = [d ,d ] are segments such that A B C,
1 2 1 2 1 2
⊂ ⊂
a b = b c = a b = b c , the polygon b d d b is a rectangle, and
1 1 1 1 2 2 2 2 1 1 2 2
| | | | | | | |
put a d = d. Then d (A,D) = d (C,D) = a d = d, but d (B,D) =
1 1 H H 1 1 H
| | | |
b d < d.
1 1
| |
Assertion 1.6. Let A be a compact subset in a proper space X. Then for
any d > 0 the set B (A) is also a compact set.
d
4
Proof. Since the ambient space X is proper, then it suffices to show that
B (A)is closed and bounded. Since A is bounded, then we have A U (x)
d r
⊂
for some point x from X and some r > 0. Then B (A) U (x), and hence,
d r+d
⊂
B (A) is bounded. Now, let us show that B (A) is closed.
d d
Let x beanadherent point oftheset B (A). Then forany positive integer
d
i each U (x) contains some point b from B (A). Due to definition of B (A),
1/i i d d
for each i there exists a from A such that b belongs to B (a ). Since A is
i i d i
compact, then any sequence a of points from A contains a subsequence
i
{ }
a converging to some point a A. Consider the subsequence b of the
{ ik} ∈ { ik}
sequence b corresponding to a . It converges to x, and hence, ax d,
{ i} { ik} | | ≤
due to continuity of the distance function. Therefore, x B (A), and so,
d
∈
B (A) is closed.
d
Figure 2: To Assertion 1.7.
Assertion 1.7. Let A be a convex compact set in Rn. Then the compact set
B (A) is also convex for any d > 0.
d
Proof. For any two points b , b from B (A) chose points a ,a from A such
1 2 d 1 2
thatb B (a ). Evidently, aclosedd-neighborhoodB [a ,a ] ofasegment
i d i d 1 2
∈
is convex. So, the segment [b ,b ] is contained in B [a(cid:0),a ] . (cid:1)But [a ,a ]
1 2 d 1 2 1 2
⊂
A, therefore B [a ,a ] B (A), and hence, [b ,b (cid:0)] is cont(cid:1)ained in B (A).
d 1 2 d 1 2 d
⊂
(cid:0) (cid:1)
Assertion 1.8. If a metric space X is proper, then the space H(X) of its
compact subsets endowed with a Hausdorff metric is also proper.
5
Proof. Consider an arbitrary bounded closed set W H(X). It lies in
⊂
H(X)
some ball B (K) H(X) centered at some compact set K H(X).
r
⊂ ∈
Therefore, each compact set C from W is contained in a closed neighborhood
B = BX(K) X of the compact set K in the space X. The set B is closed,
r ⊂
bounded and, hence, it is compact due to our assumptions.
It is well-known, see [3], that the space H(Y) is compact, if and only if
the space Y is compact itself. Therefore, C, and hence, W are contained in
a compact set H(B). Thus, W is a closed subset of a compact set, therefore
W is compact. Assertion is proved.
The next result directly follows from Assertions 1.1 and 1.8.
Corollary 1.9. Let X be a proper metric space, and H(X) be a
A ⊂
nonempty finite subset. Then Σ( ) is not empty.
A
2 Structure of Steiner Compacts in Proper Met-
ric Spaces
We start with the case of an arbitrary metric space X.
Assertion 2.1 (Existence of Minimal Steiner Compacts). Let X be a met-
ric space, = A ,...,A H(X), Σ( ) = , and d Ω( ). Then
1 n
A { } ⊂ A 6 ∅ ∈ A
each compact set from the class Σ ( ) contains at least one minimal Steiner
d
A
compact from Σ ( ).
d
A
Proof. The set Σ ( ) is naturally ordered with respect to inclusion. Let us
d
A
show that each chain K : s from Σ ( ) possesses a lower bound, and
s d
{ ∈ S} A
apply Zorn’s Lemma.
Lemma 2.2. The intersection K := K of all the elements of an arbi-
s s
∩ ∈S
trary chain is a nonempty compact set.
e
Proof. Compact subsets of any metric (and hence, Hausdorff) space are
closed, therefore the intersection of an arbitrary family of compact subsets
of a metric space is compact (because it is a closed subset of a compact set).
So, it remains to verify that K is nonempty.
Choose an index s and consider the index subset := s : K
0 ′ s
e S { ∈ S ⊂
Ks0}. In accordance with definition of a chain, K = ∩s∈S′Ks for any s0.
The family K : s consists of closed subsets of the compact set
s ′
{ ∈ S } e
6
K satisfying the finite intersection property, therefore its intersection is
s0
nonempty, see for example [7].
Lemma 2.3. For any ε > 0 there exists an element Ks′′ of the chain, such
that for any element Ks of the chain, that is contained in Ks′′, the inequality
d (K ,K) < ε is valid.
H s
Proof. Aessume the contrary. Then d = inf d (K ,K) > 0. Indeed, if d = 0,
s H s
then for any ε > 0 there exists a compact set Ks′′ suech that dH(Ks′′,K) < ε,
but then the inequality dH(Ks,K) < ε is valid for all Ks ⊂ Ks′′ ealso, a
contradiction.
e
Fix some s and consider = s : K K again. Open sets
0 ∈ S S′ { ∈ S s ⊂ s0}
X K : s together with an arbitrary open neighborhood U (K) of
s ′ ε
{ \ ∈ S }
K, ε > 0, form an open covering of the compact set K . Indeed, if it is not
s0 e
so, then there exists a point x K that does not belong to any set X K ,
e ∈ s0 \ s
s , and also does not lie in U (K). But in this case x belongs to all K ,
′ ε s
∈ S
s ∈ S′, but s 6∈ K = ∩s∈S′Ks, a conteradiction.
Take ε = d/2, and choose a finite subcovering from the corresponding
e
open covering. Let X Ks′ be the greatest element with respect to inclusion
\
among the elements of the form X Ks from this covering. Then X Ks′
\ \
and Ud/2(K) cover the compact set Ks0. But since K ⊂ Ks′, then d ≤
dH(Ks′,K)e= inf{r: Ks′ ⊂ Ur(K)}. So, Ks′ is not coentained in Ud/2(K),
and hencee the sets X \ Ks′ andeUd/2(K) do not cover Ks′, and therefoere,
they do not cover Ks0 ⊃ Ks′. This ceontradiction completes the proof of
Lemma.
For any bounded compact A and any s , the triangle inequality
j
∈ S
implies that d (A ,K) d (A ,K ) d (K ,K), where d (A ,K ) = d ,
H j H j s H s H j s j
| − | ≤
because K Σ ( ). Due to Lemma 2.3, for any ε > 0 there exists an
s ∈ d Ae e
element Ks′ of the chain, such that for any element Ks of the chain that is
contained in Ks′, the inequality dH(Ks,K) < ε is valid. Then dH(Aj,K)
−
d < ε and, due to arbitrariness of ε > 0, the equality d (A ,K(cid:12)) = d holds.
j e H j (cid:12) j e
Th(cid:12)erefore, the compact set K also belongs to Σ ( ). Due to construction, it
(cid:12) d e
A
is a lower bound of the chain under consideration. Therefore, due to Zorn’s
e
Lemma, any compact set K from Σ ( ) majorizes some minimal element.
d
A
In our terms the latter means that any K from Σ ( ) contains at least one
d
A
minimal Steiner compact.
7
Assertion 2.4 (Intermediate Steiner Compacts). Let X be a metric space,
= A ,...,A H(X), Σ( ) = , and d Ω( ). Assume that
1 n
A { } ⊂ A 6 ∅ ∈ A
K , K Σ ( ), such that K K . Then any compact set K such that
1 2 d 1 2
∈ A ⊂
K K K belongs to Σ ( ) too.
1 2 d
⊂ ⊂ A
Proof. Due to Assertion 1.4, for any compact set K such that K K
1
⊂ ⊂
K the inequalities d (A ,K) d (A ,K ) are valid for all i = 1,...,n.
2 H i H i 1
≤
Therefore, S (K) S (K ) (recall that by S (K) we denote the value
1
A ≤ A A
d (K,A )). But the function S attains it least value at the compact set
i H i
A
KP1, therefore S (K) = S (K1), and hence, dH(Ai,K) = dH(Ai,K1) for all
A A
i. So, K Σ ( ).
d
∈ A
Let X be a proper metric space. Let d = (d ,...,d ) be a vector with
1 n
nonnegative components. Put K ( ) := n BX(A ). Due to Assertion 1.6,
d A ∩i=1 di i
K ( ) is a compact set (it may be empty).
d
A
The following Assertion generalizes results from [6] describing compact
sets located between two fixed compact sets in a given distances in the sense
of Hausdorff (so-called compact sets in s-location) to the case of three and
more compact sets.
Assertion 2.5 (Existence and Uniqueness of Maximal Steiner Compact).
Let X be a proper metric space, = A ,...,A H(X). Then Σ( ) and
1 n
A { } ⊂ A
Ω( ) are not empty, and for any d Ω( ) the set Σ ( ) contains unique
d
A ∈ A A
maximal Steiner compact, and this Steiner compact is equal to K ( ).
d
A
Proof. The sets Σ( ) and Ω( ) are not empty due to Corollary 1.9. Let d
A A
be an arbitrary element of Ω( ), and let K be an arbitrary Steiner compact
A
from Σ ( ). Recall that d = (d ,...,d ), where d = d (A ,K). Due to
d 1 n i H i
A
Assertion 1.3, K BX(A ) for all i, and hence, K BX(A ) = K ( ).
⊂ di i ⊂ ∩i di i d A
Putd = d (A ,K ( )). Showthatd d . Todothatitsufficestoverify
′i H i d A ′i ≤ i
that K ( ) BX(A ) and A BX(K ( )). The first inclusion is valid
d A ⊂ di i i ⊂ di d A
in accordance with the definition of K ( ). The second inclusion is valid,
d
A
because K K ( ), and hence, BX(K) BX(K ( )), and A BX(K)
⊂ d A di ⊂ di d A i ⊂ di
in accordance with Assertion 1.3.
Thus, S K ( ) S (K). But the function S attains its least value
d
A A ≤ A A
at K, therefo(cid:0)re S (cid:1)K ( ) = S (K), and hence, d = d for any i, and
A d A A ′i i
so K ( ) Σ ( ).(cid:0) The in(cid:1)clusion K K ( ) proved above implies that
d d d
A ∈ A ⊂ A
K ( ) is the greatest element with respect to inclusion in the class Σ ( ).
d d
A A
Therefore, it is unique maximal element in Σ ( ).
d
A
8
Theorem 1 (Structure of Σ ( ) in a Proper Metric Space). Let X be a
d
A
proper metric space, = A ,...,A H(X). Then Σ( ) and Ω( )
1 n
A { } ⊂ A A
are not empty, and for any d Ω( ) a compact set K belongs to the class
∈ A
Σ ( ), if and only if K K K ( ) for some minimal Steiner compact
d λ d
A ⊂ ⊂ A
K Σ ( ) and the unique maximal Steiner compact K ( ) from Σ ( ).
λ d d d
∈ A A A
Proof. The sets Σ( ) and Ω( ) are not empty due to Corollary 1.9. The
A A
existence, uniqueness andformofamaximalSteiner compact inΣ ( )follow
d
A
from Assertion 2.5. Due to Assertion 2.1, each Steiner compact K contains
some minimal Steiner compact K . Thus, for any Steiner compact K the
λ
inclusions K K K ( ) are valid.
λ d
⊂ ⊂ A
Conversely, due to Assertion 2.4, the inclusions K K K ( ) imply
λ d
⊂ ⊂ A
that K Σ ( ).
d
∈ A
Consider the particular case X = Rm in more details.
Corollary 2.6. Let = A ,...,A H(Rm), and d Ω( ). If the
1 n
A { } ⊂ ∈ A
maximal Steiner compact K ( ) from the class Σ ( ) is convex and does
d d
A A
not coincide with some (and hence, with any) minimal Steiner compact K
λ
∈
Σ ( ), then the cardinality of Σ ( ) is continuum.
d d
A A
Proof. ConsideranarbitrarypointxfromK ( )andsomepointy K ( )
d d
A ∈ A \
K that exists due to our assumptions. Then the segment [x,y] belongs to
λ
K ( ) due to its convexity. Further, since K is a closed subset of K ( ),
d λ d
A A
then there exists an open ball U centered at the point y that does not in-
tersect K . Therefore, the interval [x,y] U does not intersect K , it is
λ λ
∩
contained in K ( ), and its cardinality is continuum. Thus, for any point
d
A
z [x,y] U, the compact set K(z) = K z satisfies the inclusions
λ
∈ ∩ ∪ { }
K K(z) K ( ), and hence, in accordance with Thorem 1, each K(z)
λ d
⊂ ⊂ A
belongs to Σ ( ). Thus, Σ ( ) contains a subset of cardinality continuum.
d d
A A
It remains to notice, that the cardinality of the set of all compact subsets
of Rm is continuum also. Indeed, it is well-known, see for example [8], that
the cardinality of the set of all real sequences is continuum. So, the same is
valid for the family of all the sequences of points in Rm. On the other hand,
each compact set in Rm is a closure of some its at most countable subset that
can be considered as a sequence of points in Rm.
Assertion 2.7 (Convexity of Maximal Steiner compact for a Convex Bound-
ary). Let = A ,...,A H(Rm), and let all A be convex. Then for
1 n i
A { } ⊂
any d Ω( ) the maximal Steiner compact K ( ) is also convex.
d
∈ A A
9
Proof. Due to Assertion 2.5, K ( ) = n B (A ). Since A are convex,
d A ∩i=1 di i i
then in accordance with Assertion 1.7, the sets B (A ) are convex for all d ,
di i i
and so, their intersection K ( ) is also convex.
d
A
3 Example of Symmetric Boundary in H(R2)
with Three Classes of Steiner Compacts
In this Section we consider an example of three-element boundary in H(R2)
(an equilateral triangle). For it we construct explicitly all minimal and max-
imal Steiner compacts (and, hence, all Steiner compacts). In spite of the
symmetry of the boundary with respect to the rotations of the plane by the
angles 2π/3, the corresponding solutions to Fermat–Steiner problem turn
±
out to be not invariant with respect to those rotations.
In what follows we write xA instead of inf xa for any point x R2
a A
and any compact set A R2.| By| E(ab,s) we den∈ot|e th|e ellipse with foc∈i are
⊂
located at the points a, and b and and whose sum of focal radii is equal to s.
Consider a boundary = A ,A ,A H(R2), where A = a ,b ,
1 2 3 i i i
A { } ⊂ { }
i = 1, 2, 3, the points a are located at the vertices of a regular triangle
i
inscribed in the unit circle centered at the origin o, and the points b are
i
obtained from the corresponding points a by the rotation with respect to o
i
by the angle π/3.
Let ω (√3/2,1). By k (ω) and k (ω) we denote the closest to o points
a b
∈
from the sets ∂B (a ) ∂B (a ) and ∂B (b ) ∂B (b ), respectively (see
ω 1 ω 2 ω 1 ω 2
∩ ∩
Figure3). Forthechosenω thepointsk (ω)andk (ω)belongtothesegments
a b
[o,b ] and [o,a ], respectively. Put t = ok (ω) = ok (ω) . It is clear that
1 2 a b
t (0,1/2), and ω = √1+t2 t. In w(cid:12) hat fo(cid:12)llow(cid:12)s it is c(cid:12)onvenient to use
∈ − (cid:12) (cid:12) (cid:12) (cid:12)
the parameter t. Consider the family of two-point compacts T (t) =
ab
T
k (t),k (t) , t (0,1/2), where k (t) = k ω(t) , and k (t) = k ω(t) . To
a b a a b b
∈
b(cid:8)e short, we(cid:9)sometimes omit the explicit pa(cid:0)rame(cid:1)ter and write k (cid:0)inste(cid:1)ad of
a
k (t), etc.
a
Put t = √5 1 √5 7/4, ω = 1+t2 t , and T (t ) = K . The
0 4 − 2q − 0 0 − 0 ab 0 0
value t is found as a root of degree 4palgebraic equation, see Lemma 3.1
0
below. The following result solves completely the Steiner Problem for the
chosen boundary in H(R2).
A
Theorem 2. Under the above notations,
10