#### Curve Reconstruction, the Traveling Salesman Problem, and Menger

Discrete Comput Geom
0 Institut fu ̈r Theoretische Informatik, ETH Zu ̈rich , CH-8092 Zu ̈rich , Switzerland
1 Curve Reconstruction, the Traveling Salesman Problem, and Menger's Theorem on Length
We give necessary and sufficient regularity conditions under which the curve reconstruction problem is solved by a traveling salesman tour or path, respectively. For the proof we have to generalize a theorem of Menger [12], [13] on arc length.
Introduction
In 1930 Menger [
12
] proposed a new definition of arc length:
The length of an arc be defined as the least upper bound of the set of all numbers that
could be obtained by taking each finite set of points of the curve and determining
the length of the shortest polygonal graph joining all the points.
: : :
We call this the messenger problem (because in practice the problem has to be
solved by every postman, and also by many travelers): finding the shortest path
joining all of a finite set of points, whose pairwise distances are known.
This statement is one of the first references to the Traveling Salesman Problem.
Arc length is commonly defined as the least upper bound of the set of numbers
obtained by taking each finite set of points of the curve and determining the length of
the polygonal graph joining all the points in their order along the arc. In [
13
] Menger
proves the equivalence of his definition and the common one (Menger’s theorem).
It seems natural to think that this equivalence holds due to the fact that the shortest
polygonal graph coincides with the polygonal graph joining the points in their order along
¤ This work was supported by grants from the Swiss Federal Office for Education and Science (Projects
ESPRIT IV LTR No. 21957 CGAL and No. 28155 GALIA).
the arc, provided the set of points is sufficiently dense. In other words, for sufficiently
dense point sets a traveling salesman path solves the polygonal reconstruction problem
for arcs. This problem was stated by Amenta et al. [
4
] as follows:
Given a curve ° 2 Rd and a finite set of sample points S ½ ° . A polygonal
reconstruction of ° from S is a graph that connects every pair of samples adjacent
along ° , and no others.
However, Menger’s proof does not show this at all. So we want to study the question
whether the polygonal reconstruction problem is solved by a traveling salesman path,
provided the sample points are sufficiently dense in the curve. Since a traveling salesman
path is always simple, we cannot expect that it solves the reconstruction problem for
curves with intersections. Even worse, the traveling salesman path may not coincide with
the polygonal reconstruction for arbitrarily dense samples of simple curves. Consider
the following example:
Let ° be the simple arc which consists of the unit interval on the x -axis and the graph
of y D x 2 on this interval. That is,
° : [0; 1] ! R2; t 7!
(.1 ¡ 2t; 0/;
.2t ¡ 1; .2t ¡ 1/2/;
t · 21 ;
t > 12 :
For large n the samples
with
Sn D f pn1; pn2; pn3; pn4g [
[n ½µ i
iD2
n
¶ µ i i 2 ¶¾
; 0 ; ;
n n2
1
pn D
3
pn D
µ 1
n
µ 2
¶
; 0 ;
;
n3 n6
4 ¶
;
2
pn D
4
pn D
1 ¶
µ 1
;
n3 n6
µ 1 1 ¶
;
n2 n4
;
become arbitrarily dense in ° . However, the traveling salesman path through Sn is
different from the polygonal reconstruction from Sn (see Fig. 1) because
j pn ¡ pnj C j pn ¡ pnj C j pn ¡ pn4j > j pn ¡ pnj C j pn ¡ pnj C j pn ¡ pn4j:
1 2 2 3 3 1 3 3 2 2
In this example, the arc ° has finite length and finite total absolute curvature. Thus, even
finite curvature, which is a stronger property than finite length, see [
1
], is not sufficient
for the polygonal reconstruction problem to be solved by a traveling salesman path,
provided the points are sampled densely enough. The crucial point is that ° behaves
quite well, but is not regular at .0; 0/ 2 ° . The regularity conditions necessary turn out
to be only slightly stronger.
In this paper we prove:
Suppose that for every point of a simple curve:
1. The left and the right tangents exist and are nonzero.
2. The smaller angle between these tangents is less than ¼ .
Under these conditions there exists a finite sampling density such that the traveling
salesman path solves the polygonal reconstruction problem for all samples with larger
sample density.
Regularity is a local property. In contrast to that, it is a global property for a path to be
a shortest polygonal path through a finite point set. One of the most interesting aspects of
our work is this transition from a local to a global property and the methods used therein.
We first prove a local version of our global theorem by using a projection technique
from integral geometry. We believe that this technique could be useful in many other
contexts, even in the study of higher-dimensional objects than curves. At a first glance
it is not obvious how to derive the global version from the local one. This extension is
achieved by an application of two corollaries of Menger’s theorem. As a by-product we
generalize Menger’s theorem to simple closed curves in Euclidean space.
2. Basic Definitions
In this section we give definitions and cite theorems that we need for our proofs. The
results that we present are only valid for connected simple curves in Euclidean space.
Hence in the following ° : [0; 1] ! Rd is always a connected simple curve. We use the
symbol ° either to denote a mapping or to denote the image of the mapping. It should
always be clear from the context what ° denotes.
We call a curve ° s(emi)-regular if in every point on ° nonzero left and right tangents
exist. This is expressed in the following definition:
Definition 2.1. Let
and
T D f.t1; t2/: t1 < t2; t1; t2 2 [0; 1]g
¿ : T ! Sd¡1; .t1; t2/ 7!
° .t2/ ¡ ° .t1/ :
j° .t2/ ¡ ° .t1/j
The curve ° is called left (right) regular at ° .t0/ with left tangent 1.° .t0// or right tangent
r.° .t0// if for every sequence .»n/ in T which converges from the left (right) to .t0; t0/
in closure.T / the sequence ¿ .»n/ converges to 1.° .t0// or r.° .t0//, respectively. We call
° s-regular if it is left and right regular in all points ° .t /, t 2 [0; 1]. A curve is called
regular if it is s-regular and in every point left and right tangent coincide.
The relationship between s-regularity and two of the most interesting geometric
properties of curves, length and total absolute curvature, was shown by Aleksandrov and
Reshetnyak [
1
].
Theorem 2.1. Every curve ° of finite total absolute curvature C .° / is s-regular and
every s-regular curve has finite length L.° /.
Both length and total absolute curvature are defined via inscribed polygons. For the
definition of inscribed polygons we need the definition of a sample, which we give next.
Definition 2.2. A sample S of ° is a finite sequence
S D f p1; : : : ; pjSjg
of points where pi 2 ° . We assume that the sample points pi are ordered according
to the order of the ° ¡1. pi / 2 [0; 1]. To every sample S its density is defined to be the
inverse of the following number:
".S/ D sup minfj pi ¡ x j: i D 0; : : : ; ng:
x2°
In the following we sometimes make use of the convention pjSjC1 D pjSj if ° is open
and pjSjC1 D p1 if ° closed.
Sometimes, especially for closed curves, it is more appropriate to make use of an ordering
of the sample points that is independent of a specific parameterization.
Definition 2.3. Let S D f p1; : : : ; pjSjg be a sample of ° . If ° is open one writes i G j
if ° ¡1. pi / < ° ¡1. p j /. This notion is independent of an orientation preserving change
of parameterization. If ° is closed and has finite length choose an orientation along °
and write i G j if i 6D j and L.° . pi : p j // · L.° . p j : pi //, where ° . pi : p j / ½ ° is the
arc connecting pi and p j in the orientation along ° . One writes i G j if the possibility
that i D j is included.
Now we are prepared to give the definition of length and total absolute curvature. Let
P.S/ denote the polygon that connects a sample S in the right order.
Definition 2.4.
1. The (Jordan) length L.° / of a curve ° is the following number:
supfL. P.S//: S is a sample of ° g;
where
L. P.S// D XjSj j piC1 ¡ pi j:
iD1
2. The curvature C .° / of a curve ° is defined as the number
where
if ° is open, and
where pjSjC1 D p1, if ° is closed.
iD1
iD1
supfC . P.S//: S is a sample of ° g;
jSj¡2
C . P.S// D X j\. piC1 ¡ pi ; piC2 ¡ piC1/j;
jSj¡1
C . P.S// D X j\. piC1 ¡ pi ; piC2 ¡ piC1/j;
In the following we make use of a theorem of Reshetnyak [
14
], which states how one gets
the length of a curve by integration over the length of the orthogonal projections of the
curve on all one-dimensional subspaces ` 2 RdC1. The collection of all one-dimensional
subspaces of RdC1 forms the d-dimensional projective space Pd . In the following the
elements of Pd are sometimes called lines. From a standard construction in integral
geometry one gets a probability measure ¹d on Pd from the surface area measure on the
d-dimensional sphere Sd :
The continuous and locally homeomorphic mapping
': Sd D fx 2 RdC1: jx j D 1g ! Pd ;
x 7! f¸x : ¸ 2 Rg;
is a double cover. Since ' is continuous, it is also measurable. That is the inverse image
'¡1.B/ of a Borel set B 2 Pd is a Borel set in Sd . On Sd we have the usual surface area
measure ºd . Therefore we get a measure ¹0d on Pd as ¹0d .B/ D ºd .'¡1.B// for all Borel
sets B µ Pd . We have
which we use to normalize ¹0d . That is, we set
for all Borel sets B ½ Pd .
The theorem of Reshetnyak now reads as follows:
Theorem 2.2. For any s-regular curve ° one finds
Z
L.¼`.° // d¹d .`/ D cd L.° /
0..d C 1/=2/
with cd D p¼ 0..d C 2/=2/ :
.1/
3. A Local Analysis
In this section we give several reformulations of our notion of regularity. We end up with
a reformulation which is a local version of the theorem we want to prove in this paper.
The main tool to prove this version is Theorem 2.2. To apply this theorem we have to
study projections of sample points on one-dimensional subspaces.
The first reformulation of regularity in the following lemma is a pure metric
interpretation of our definition of regularity.
Lemma 3.1. Let ° be a simple closed curve, which is left (right) regular in p 2 ° .
Let . pn/; .qn/, and .rn/ be sequences of points from ° , that converge to p from the left
(right), such that pn < qn < rn for all n 2 N in an order locally around p along ° . Let
®n be the angle at qn of the triangle with corner points pn; qn, and rn. Then the sequence
of angles .®n/ converges to ¼ .
Proof. Since ° seen as a mapping from an interval is locally a homeomorphism the
sequences .° ¡1. pn//; .° ¡1.qn//, and .° ¡1.rn// converge to ° ¡1. p/. Thus by our
definition of left (right) regularity asymptotically the three oriented segments
convf pn; qng; convfqn; rng; and convf pn; rng
have to point in the direction of the left (right) tangent l. p/ or r. p/, respectively. That
is, limn!1 ®n D ¼ .
We believe that this metric aspect of regularity is the most important one for our
theorem to hold, but in the following we have to make use of the linear structure of
RdC1. We exploit this linear structure by studying projections of a set of sample points
on one-dimensional subspaces of RdC1.
In the following let ° be s-regular and let p 2 ° be a fixed point. We want to introduce
the following notions. Let ´ > 0. The connected component of
fq 2 ° : j p ¡ qj < ´g
which contains p is denoted by B´. p/. We assume that every one-dimensional subspace
` of RdC1 not perpendicular to l. p/ is oriented according to the orientation induced by
the orthogonal projection ¼`.l. p// of l. p/ on ` and that every one-dimensional subspace
` perpendicular to l. p/ carries an arbitrary orientation.
We want to compare the ordering of a set of sample points close to p on ° with
the ordering of the projections of these points on one-dimensional subspaces of RdC1.
The following reformulation of regularity states to which extent these orderings can be
different:
Lemma 3.2. Assume there exists a sequence .`n/ of one-dimensional subspaces of
RdC1 and sequences . pn/; .qn/; .rn/ 2 B1=n. p/ with pn < qn < rn · p in the order
along ° , but
in the order on `n. Then the limit of every convergent subsequence of .`n/ has to be
perpendicular to l. p/.
Proof. We want to do the proof by contradiction. That is, we can assume that .`n/
converges in the topology of Pd and that
¼
0 · ¯ :D nl!im1 \.l. p/; `n/ < 2
;
where \.l. p/; `n/ denotes the smaller of the two angles defined by the subspace `n and
the subspace spanned by l. p/. For the proof we assume without loss of generality that
¼`n .qn/ < ¼`n . pn/ < ¼`n .rn/. Furthermore, we can assume by moving qn or rn a little
bit on ° that there exists 0 < ¸ < 1 (independent of n) such that
j¼`n . pn/ ¡ ¼`n .rn/j D ¸ cos.¯/j¼`n .qn/ ¡ ¼`n .rn/j:
Why can we do so? By our assumption we have
j¼`n .qn/ ¡ ¼`n .rn/j D j¼`n .qn/ ¡ ¼`n . pn/j C j¼`n . pn/ ¡ ¼`n .rn/j;
and hence
j¼`n . pn/ ¡ ¼`n .rn/j D ¸nj¼`n .qn/ ¡ ¼`n .rn/j
with 0 < ¸n < 1. From the continuity of ° and the continuity of the projection maps ¼`n
we find that moving pn toward qn on ° can make j¼`n .qn/ ¡ ¼`n . pn/j arbitrarily small.
Similarly, moving rn toward qn on ° can make j¼`n . pn/ ¡ ¼`n .rn/j arbitrarily small. Now
move pn toward qn on ° if ¸n < ¸ cos.¯/ or move rn toward qn on ° if ¸n > ¸ cos.¯/
as long as ¸n 6D ¸ cos.¯/.
From the regularity of ° we find that
lim \.convfqn; rng; l. p// D nl!im1 \.convf pn; rng; l. p// D 0
n!1
and together with the triangle inequality for spherical triangles
lim \.convfqn; rng; ¼`n .convfqn; rng//
n!1
· nl!im1 ¡\.convfqn; rng; l. p// C \.l. p/; ¼`n .convfqn; rng//¢
D nl!im1 \.convfqn; rng; l. p// C nl!im1 \.l. p/; ¼`n .convfqn; rng//
D 0 C nl!im1 \.l. p/; `n/
¼
D ¯ < :
2
Analogously we find
Hence
lim \.convf pn; rng; ¼`n .convf pn; rng// · ¯ <
n!1
¼
2
:
0 < cos.¯/ · nl!im1 j¼`n .qn/ ¡ ¼`n .rn/j · 1
jqn ¡ rnj
and
Therefrom we find
0 < cos.¯/ · nl!im1 j¼`n . pn/ ¡ ¼`n .rn/j · 1:
j pn ¡ rnj
and thus
cos.¯/ ¸ nl!im1 j¼`n .qn/ ¡ ¼`n .rn/j
1
jqn ¡ rnj
j pn ¡ rnj
j¼`n . pn/ ¡ ¼`n .rn/j
D nl!im1 jj¼¼``nn..qpnn// ¡¡ ¼¼``nn..rrnn//jj jjqpnn ¡¡ rrnnjj
D nl!im1 jj¼¼``nn..qpnn// ¡¡ ¼¼``nn..rrnn//jj nl!im1 jjqpnn ¡¡ rrnnjj
1 lim j pn ¡ rnj
D ¸ cos.¯/ n!1 jqn ¡ rnj
lim j pn ¡ rnj · ¸ < 1:
n!1 jqn ¡ rnj
Hence there exists N 2 N such that j pn ¡rnj · jqn ¡rnj for all n ¸ N . Now we consider
the triangles with corner points pn; qn, and rn. Let ®n be the angle at qn. From the law
of cosines together with j pn ¡ rnj · jqn ¡ rnj we find
j pn ¡ rnj2 ¡ j pn ¡ qnj2 ¡ jqn ¡ rnj2
2j pn ¡ qnjjqn ¡ rnj
> 0:
Thus ®n has to be smaller than or equal to ¼=2 for all n ¸ N . Since ° is left regular in
p that is a contradiction to Lemma 3.1.
From this lemma we can derive another one that states the measure ¹d of the set of
lines on which the sample points in a small neighborhood left from p are not projected
in the right order turns to zero as the neighborhood shrinks to p.
Lemma 3.3. For all n 2 N let pn; qn; rn 2 B1=n. p/ with pn < qn < rn · p in the
order along ° . Let
Nn D f` 2 Pd : ¼`.qn/ < ¼`. pn/ < ¼`.rn/ or ¼`. pn/ < ¼`.rn/
< ¼`.qn/ in the order on `g;
where ¼` denotes the orthogonal projection on the line `. Then limn!1 ¹d .Nn/ D 0.
Proof. In [
14
] Reshetnyak shows: Let V be a d-dimensional subspace of RdC1 and let
E ½ Pd be the set of all one-dimensional subspaces contained in V . Then ¹d .E / D
0. The proof of the lemma follows directly from this theorem of Reshetnyak and
Lemma 3.2.
Obviously an analogous result for sample points larger than p in the order along °
also holds.
Now we are prepared to prove a local version of our theorem. For a given sample S of a
neighborhood of p we fix the smallest and the largest sample point along ° and consider
paths through S which connect these points. One such path is the polygonal reconstruction
P.S/ which connects the sample points in their order along ° . We distinguish three types
of lines ` 2 Pd :
1. There exists a path through the sample points different from P.S/ which has a
shorter projection on ` than P.S/.
2. Every path through the sample points different from P.S/ has a larger projection
on ` than P.S/.
3. There exist paths through the sample points different from P.S/ which has
projections on ` with the same length as the projection of P.S/, but there is no path
with a shorter projection.
The proof is subdivided into three steps. First, we show that the measure of the first set
of lines tends to zero as the neighborhood of p shrinks to p itself. Second, there exists
a constant larger than zero such that the measure of the second set of lines is larger than
this constant for arbitrary small neighborhoods of p. In the third step we want to apply
Theorem 2.2. We use this theorem and the results of the first two steps to show that for
small neighborhoods of p the polygonal reconstruction P.S/ has to be the shortest path
through the sample points.
Theorem 3.1. Assume
® D supf\.l.q/; r.q//: q 2 ° g < ¼
and let .Sn/ be a sequence of samples of B1=n. p/. Let TSP¤.Sn/ be a path of minimal
length through the sample points Sn with fixed startpoint min Sn and fixed endpoint
max Sn. Here min and max are taken with respect to the order induced on Sn by ° . Then
there exists N 2 N such that
TSP¤.Sn/ D P.Sn/;
for all n ¸ N . Furthermore, TSP¤.Sn/ is unique for all n ¸ N .
Proof. First Step. Let Ln ½ Pd be the set of lines for which the projection ¼`. P.Sn//
is not a shortest path through the projected sample points ¼`.Sn/ and let ¹d be the
probability measure on Pd introduced in (1).
We show that
lim ¹d .Ln/ D 0:
n!1
For the proof we use the following abbreviations (min and max denote the minimum
and the maximum along ° and min` and max` denote the minimum and the maximum
along `):
m1 D ¼`.min Sn/;
m2 D min ¼`.Sn/;
`
m3 D max ¼`.Sn/;
`
m4 D ¼`.max Sn/:
Take ` 2 Pd together with its orientation. A path of minimal length through the points
¼`.Sn/ which connects m1 with m4 (see Fig. 2) consists of
convfm1; m2g [ convfm2; m3g [ convfm3; m4g:
That is, the points of ` between m1 and m2 are covered twice by a path of minimal
length through the points ¼`.Sn/, the points between m1 and m4 are covered once, and
the points between m3 and m4 are covered twice again. If ¼`.P.Sn// is not a path of
minimal length through the points ¼`.Sn/, then there has to exist an interval
I D
·
m`inf¼`. pni/; ¼`. pniC1/g; m`axf¼`. pni/; ¼`. pniC1/g¸
on ` with pni; pniC1 2 Sn, which is covered by ¼`.P.Sn//
1. 2 C 2k times, k ¸ 1, if
or
m2 · m`inf¼`. pni/; ¼`. pniC1/g < m`axf¼`. pni/; ¼`. pniC1/g · m1
m4 · m`inf¼`. pni/; ¼`. pniC1/g < m`axf¼`. pni/; ¼`. pniC1/g · m3;
2. 1 C 2k times, k ¸ 1, if
m1 · m`inf¼`. pni/; ¼`. pniC1/g < m`axf¼`. pni/; ¼`. pniC1/g · m4:
For all pnj 2 Sn; j 2 f2; : : : ; jSnj ¡ 2g, we call the interval
Ij D
·
m`inf¼`. pnj /; ¼`. pnjC1/g; m`axf¼`. pnj /; ¼`. pnjC1/g¸
positive if ¼`. pnj / < ¼`. pnjC1/ in the order on ` and negative otherwise. The intervals
Ij which cover the interval I must have alternating signs. That is, if ¼`.P.Sn// is not a
shortest path through the points ¼`.Sn/ we find, using that P.Sn/ connects the pnj 2 Sn
in their order along ° and using the continuity of ° and of the projection map ¼`, three
points pn; qn; rn 2 B1=n. p/ with pn < qn < rn · p in the order along ° , but
¼`.qn/ < ¼`. pn/ < ¼`.rn/ or ¼`. pn/ < ¼`.rn/ < ¼`.qn/
in the order on `I
or we find three points pn; qn; rn 2 B1=n. p/ with pn > qn > rn ¸ p in the order along
° , but
¼`.qn/ > ¼`. pn/ > ¼`.rn/ or ¼`. pn/ > ¼`.rn/ > ¼`.qn/
in the order on `:
See Fig. 3 for an example.
In general the points pn; qn, and rn need not be sample points. From Lemma 3.3 we
can conclude that
lim ¹d .Ln/ D 0:
n!1
Second Step. Let Mn ½ Pd be the set of lines ` for which we have:
1. The order of Sn along ° and the order of Sn induced by the order of ¼`.Sn/ on `
coincide.
2. For all convf pi ; piC1g ½ P.Sn/ with pi ; piC1 2 Sn we have
1
j¼`. piC1/ ¡ ¼`. pi /j ¸ 2 cos
µ ¼ C ® ¶
4
We show that there exist c > 0 and N 0 2 N such that for all n ¸ N 0 we have
¹d .Mn/ ¸ c:
For the proof we construct a set of lines C ½ Pd with ¹d .C / > 0 and show that there
exists N 0 2 N such that C ½ Mn for all n ¸ N 0. The set C is defined as follows: Let
`0 be the line, `0 ½ spanfl. p/; r. p/g such that `0? ½ spanfl. p/; r. p/g halves the angle
\.l. p/; ¡r. p// between the lines determined by l. p/ and r. p/. Now we define
C D f` 2 Pd : \.`; `0/ · 41 .¼ ¡ ®/g:
By \.`; `0/ for `; `0 2 Pd we denote the value of the minimum of the two angles
determined by ` and `0. The set S`2C fx 2 `g ½ RdC1 is a double cone (see Fig. 4).
By construction ¼`.l. p// and ¼`.r. p// point in the same direction on every ` 2 C .
Since ® < ¼ we have ¹d .C / > 0. It remains to check conditions 1 and 2 to prove that
for sufficiently large n we have C ½ Mn.
1. Assume that for arbitrary large n we find `n 2 C such that the first condition is
violated. Then we can find three points pn; qn; rn 2 B1=n. p/ with pn < qn < rn · p in
the order along ° , but
in the order on `n. By Lemma 3.1 the limit of every convergent subsequence of .`n/ has
to be perpendicular to either l. p/ or r. p/. Since C is compact we find ` 2 C as the limit
of a convergent subsequence of .`n/ which is perpendicular to either l. p/ or r. p/. On
the other hand we find for all ` 2 C using the triangle inequality for spherical triangles,
lim \.`; convf pni; pniC1g/ D \ ³`; lim convf pni; pniC1g´
n!1 n!1
That is, for all sufficiently large n and all ` 2 C we have
1
j¼`. pi / ¡ ¼`. piC1/j ¸ 2 cos
For pni; pniC1 2 Sn with p · pni < pniC1 we get the same result using r. p/ instead of
l. p/ in the triangle inequality for spherical triangles. It remains to consider the case that
pni < p < pniC1. We choose ´ > 0 such that
and show that, for all ` 2 C and all sufficiently large n,
cos
µ ¼ C ®
4
¶
C ´ >
\.`; convf pni; pniC1g/ ·
To do so let ¼ be the projection on spanfl. p/; r. p/g. The s-regularity of ° implies
lim sup.\.`0; ¼.convf pni; pniC1g/// · \.`0; l. p// D \.`0; r. p//:
We define the distance between two two-dimensional subspaces of RdC1 by the smaller
of the two dihedral angles defined by them. With this metric the span seen as a
mapping becomes continuous. From the continuity of \ and the continuity of span we
get
(2)
D nl!im1 \.spanfl. p/; r. p/g; convf pni; pniC1g/:
; convf pni; pniC1g
¶
; lim convf pni; pniC1g
n!1
¶
That is, for all sufficiently large n we have
\.¼.convf pni; pniC1g/; convf pni; pniC1g/ · ´:
.3/
Finally we get, for all sufficiently large n using the inequalities (2), (3) and the triangle
inequality for spherical triangles,
\.`; convf pni; pniC1g/ · \.`; `0/ C \.`0; convf pni; pniC1g/
· \.`; `0/ C \.`0; ¼.convf pni; pniC1g//
C \.¼.convf pni; pniC1g/; convf pni; pniC1g/
· \.`; `0/ C \.`0; l. p// C ´
¼ ¡ ® ®
· 4 C 2 C ´
¼ C ®
D 4 C ´:
Hence for all sufficiently large n, all ` 2 C , and all pni; pniC1 2 Sn we have
1
j¼`. pi / ¡ ¼`. piC1/j ¸ 2 cos
That is, there exists N 0 2 N such that for all n ¸ N 0 we have C ½ Mn. Hence
¹d .Mn/ ¸ ¹d .C / > 0:
Third Step. In the first two steps we considered the measures of the subspaces Ln
and Mn of Pd , the space on which we want to integrate. In this step we compare
integrands.
If we compare the length of the projections ¼`.P.Sn// and ¼`.PQ .Sn// on ` 2 Ln (see
Fig. 5) we find
L.¼`.P.Sn/// ¡ L.¼`.PQ .Sn/// · 2 X j¼`. pni/ ¡ ¼`. pniC1/j:
i2In
On the other hand we find, for the length of projections on ` 2 Mn (see Fig. 5),
L.¼`.PQ .Sn/// ¡ L.¼`.P.Sn/// ¸ 2 X j¼`. pni/ ¡ ¼`. pniC1/j:
i2In
Furthermore, we have, for all pni 2 Sn; i 2 f1; : : : ; jSnj ¡ 1g and all ` 2 Mn,
1
j¼`. piC1/ ¡ ¼`. piC1/j ¸ 2 cos
From this inequality we get another inequality which is valid for all ` 2 Mn and all
`0 2 Ln,
Xi2In j¼`. pni/ ¡ ¼`. pniC1/j ¸ 21 cos µ ¼ C4 ® ¶ Xi2In j¼`0 . pni/ ¡ ¼`0 . pniC1/j
which implies, together with (4) and (5),
L.¼`.PQ .Sn/// ¡ L.¼`.P.Sn/// ¸ 21 cos µ ¼ C4 ® ¶ ³L.¼`0 .P.Sn/// ¡ L.¼`0 .PQ .Sn///´:
Since this inequality is valid for all ` 2 Mn and all `0 2 Ln we get for the increase of
length on Mn
Z
Mn
³L.¼`.PQ .Sn/// ¡ L.¼`.P.Sn///´ d¹d .`/
and for the decrease of length on Ln
Ln
Z ³L.¼`. P.Sn/// ¡ L.¼`. PQ .Sn///´ d¹d .`/
³ ´
· ¹d .Ln/ sup L.¼`. P.Sn/// ¡ L.¼`. PQ .Sn/// :
`2Ln
Because of limn!1 ¹d .Ln/ D 0 there exists N ¸ N 0 such for all n ¸ N we have
¹d .Ln/ < ¹d .C /
cos
Using Theorem 2.2 we find that for all n ¸ N there is no shortcut possible and that
P.Sn/ is the unique path of minimal length through the points Sn with fixed start- and
endpoints, because the polygon connecting the points in the order induced by ° has a
shorter projection on all ` 2 C than every other polygon through the points Sn with fixed
start- and endpoints.
Menger’s Theorem
We need Menger’s theorem and some corollaries to achieve the transition from the local
results of the last section to the global.
Menger’s elementary proof of his theorem is only valid for simple open curves. Here
we give a new proof that also holds for simple closed curves. For this proof, which in
contrast to the original proof of Menger is restricted to curves in Euclidean spaces, we
need the following lemma.
Lemma 4.1. Let ° be a simple curve and let .Sn/ be a sequence of samples of ° with
limn!1 ".Sn/ D 0 and lim sup L.TSP.Sn// < 1. Let
°n: [0; L.TSP.Sn//] ! Rd
be the parameterization by length of TSP.Sn/. Then the sequence .°n/ has a Fre´chet
convergent subsequence.
Proof. By turning to an appropriate subsequence we can assume that
L.TSP.Sn// < 2 lim sup L.TSP.Sn// D: c < 1
for all n 2 N. To prove the lemma it is sufficient to check that the premises of the theorem
of Ascoli [
9
] are fulfilled.
1. The sequence of reduced parameterizations
°n0: [0; 1] ! Rd ; t 7! °n.t L.TSP.Sn///
is equicontinuous, because
j°n0.t 0/ ¡ °n0.t /j · .t 0 ¡ t /L.°n0/
D .t 0 ¡ t /L.TSP.Sn//
Now the theorem of Ascoli states that the sequence .°n0/ has a convergent subsequence
.°m0 / which converges to a continuous function
° 0: [0; 1] ! Rd :
Note that the condition lim sup L.TSP.Sn// < 1 is always fulfilled if ° has finite
length. Now we are prepared to prove Menger’s theorem. In this proof we make use of
the one-dimensional Hausdorff measure H1, see [
8
] for the definition and its relation to
the geometry of curves.
Theorem 4.1. Every simple curve ° satisfies
L.° / D supfL.TSP.S//: S is a sample of ° g;
where TSP.S/ denotes a shortest path through the sample points S if ° is an open curve
and a shortest tour otherwise.
Proof. Assume the contrary. That is, we can assume that
supfL.TSP.S//: S is a sample of ° g < L.° /;
because TSP.S/ · P.S/ · L.° / for every sample S of ° . Take any sequence .Sn/ of
samples from ° with limn!1 ".Sn/ D 0. By our assumption we have
lim sup L.TSP.Sn// < L.° /:
Let °n: [0; 1] ! Rd be the reduced parameterization of TSP.Sn/. From Lemma 4.1
we know that there exists a Fre´chet convergent subsequence .°m / of .°n/. Let the
continuous function ° 0: [0; 1] ! Rd be the limit of .°m /. We have, for the convergent
subsequence,
lim sup L.TSP.Sm // · lim sup L.TSP.Sn// < L.° /:
Next we show that ° ½ ° 0.[0; 1]/: Take p 2 ° . Since limn!1 ".Sn/ D 0, one finds a
sequence . pn/ with pn 2 Sn which converges to p. Of course the subsequence . pm / with
pm 2 °m also converges to p. For all k 2 N there exists m.k/ 2 N and pm0.k/ 2 ° 0.[0; 1]/
such that
and j pm.k/ ¡ pj <
Hence by the triangle inequality we have j pm0.k/ ¡ pj < 1= k. From the compactness of
° 0.[0; 1]/ we get p 2 ° 0.[0; 1]/.
Both ° and ° 0.[0; 1]/ are as compact sets H1 measurable. Using the properties of the
one-dimensional Hausdorff measure H1, see [
8
], and our assumption we find
L.° / D H1.° /
The following corollary states that the length of the traveling salesman path (tour)
through a sequence of sample points converges to the length of the curve when the density
of the sample goes to infinity. To avoid confusion we remark here that in the following
¼ and ¼n always denote permutations and no longer projections.
Corollary 4.1. Let ° be a curve and let .Sn/ be a sequence of samples of ° , with
limn!1 ".Sn/ D 0. Then
L.° / D nl!im1 L.TSP.Sn//;
where TSP.Sn/ denotes a shortest path through the sample points Sn if ° is an open
curve and a shortest tour otherwise.
Proof. Assume L.° / < 1. For a given ´ > 0 consider three sets:
1. From Menger’s theorem we know that there exists a sample
S D fq1; : : : ; qjSjg
´
with L.TSP.S// > L.° / ¡ 2 :
2. Let S0 D f p1; : : : ; pjS0jg 2 fSn: n 2 Ng be such that ".S0/ < ´=4jSj.
3. Let S00 D fr 1; : : : ; r jSjg ½ S0 be a multiset with jr i ¡ qi j · ".S0/ for all i D
1; : : : ; jSj.
In the following we make use of the conventions for jSj C 1 and jS0j C 1 introduced in
Definition 2.2. Let ¼ be a permutation of f1; : : : ; jSjg such that
XjSj jq¼.iC1/ ¡ q¼.i/j D L.TSP.S//;
let ¼ 0 be a permutation of f1; : : : ; jS0jg such that
and let ¼ 00 be a permutation of f1; : : : ; jSjg such that
jS0j
X j p¼0.iC1/ ¡ p¼0.i/j D L.TSP.S0//;
Then it follows that:
1.
2.
by construction.
iD1
by the definition of ¼ .
XjSj jr ¼00.iC1/ ¡ r ¼00.i/j · XjS0j j p¼0.iC1/ ¡ p¼0.i/j
jSj
X jq¼00.iC1/ ¡ q¼00.i/j
iD1
iD1
Since ´ can be arbitrary small, we are done. The proof in the case L.° / D 1 is quite
similar.
The next lemma states that the maximal length of a segment in a traveling salesman
path (tour) of a sequence of sample points from a curve of finite length tends to zero as the
density of the sample goes to infinity. The proof again makes use of the one-dimensional
Hausdorff measure H1.
Lemma 4.2. Assume that the curve ° has finite length and let .Sn/ be a sequence
of samples of ° with limn!1 ".Sn/ D 0. For all n 2 N let ¼n be the permutation of
f1; : : : ; jSnjg induced by TSP.Sn/. Then
lim maxfj p¼n.iC1/ ¡ p¼n.i/j: i D 1; : : : ; jSnjg D 0;
n!1
where ¼n.jSn C 1j/ D ¼n.jSnj/ if ° is an open curve and ¼n.jSn C 1j/ D ¼n.1/ if ° is
closed.
Proof. Assume the contrary. That is, by choosing an appropriate subsequence one can
assume that there exists c > 0 such that for every n 2 N it holds that
maxfj p¼n.iC1/ ¡ p¼n.i/j: i D 1; : : : ; jSnjg > c:
Let pn D p¼n.i/ 2 Sn and qn D p¼n.iC1/ 2 Sn be such that j pn ¡ qnj > c. By the
compactness of ° we can turn to an appropriate subsequence such that . pn/ converges
to p 2 ° and .qn/ converges to q 2 ° . Of course it holds that j p ¡ qj ¸ c. Remove the
interior of convf pn; qng from TSP.Sn/. We consider two cases:
First, if ° is an open curve, then TSP.Sn/ decomposes into two paths Pn1 which
connects p¼n.1/ with p¼n.i/ and Pn2 which connects p¼n.iC1/ with p¼n.jSnj/. Let Pn
denote Pn1 [ Pn2 and let °nj : [0; 1] ! Rd , j D 1; 2, be the reduced parameterization
of Pnj . As in Lemma 4.1 one can use the theorem of Ascoli to show that .°nj / has a
Fre´chet convergent subsequence. That is, we can assume by considering a common
subsequence that .°n1/ and .°n2/ converge to continuous functions ° 1: [0; 1] ! Rd and
° 2: [0; 1] ! Rd . Let ° 0 be ° 1.[0; 1]/ [ ° 2.[0; 1]/. Second, if ° is a closed curve,
then TSP.Sn/ becomes after the removal of the interior of convf pn; qng a path denoted
by Pn which connects p¼n.i/ with p¼n.iC1/. Let °n be the reduced parameterization of
Pn. As in Lemma 4.1 one can show that .°n/ has a Fre´chet convergent subsequence.
That is, we can assume by taking an appropriate subsequence that .°n/ converges to
a continuous function ° 0: [0; 1] ! Rd . We use the symbol ° 0 also to denote the set
° 0.[0; 1]/.
Next we show that ° ½ ° 0: Take r 2 ° . Since limn!1 ".Sn/ D 0, one finds a
sequence .rn/ with rn 2 Sn which converges to r . For all k 2 N there exists n.k/ 2 N
and r n0.k/ 2 ° 0 such that
and jrn.k/ ¡ r j <
Hence by the triangle inequality jr n0.k/ ¡ r j < 1= k. From the compactness of ° 0 we get
r 2 ° 0.
The sets ° ; ° 0; Pn, and convf pn; qng are all as compact sets H1 measurable. Using the
properties of the one-dimensional Hausdorff measure H1, see [
8
], we find
H1.° 0/ ¸ H1.° [ convf pn; qng/
D H1.° / C H1.convf p; qg/
> H1.° /
D L.° /
¸ sup L. Pn/
D sup H1. Pn/
¸ H1.° 0/:
That is a contradiction.
5. From Local to Global
In this section we finally want to prove the promised theorem. That is, here we achieve
the transition from the local version of the theorem to the global. To do so we need
another definition.
Definition 5.1. Given a curve ° and a sample S from ° . We call r 2 S a return point,
if r is connected to p; q 2 S in a traveling salesman path (tour) of S and r G p; q or
r > p; q in the order along ° . In the first case we call the return point positive and in
the second case we call it negative. For open curves we also call the start- and endpoints
of TSP.S/ the return points.
For example in Fig. 1 points 2 and 3 are return points. In the traveling salesman path,
point 2 is connected to points 3 and 4, which are both larger than point 2 in the order
along the curve. Point 3 is connected to points 1 and 2, which are both smaller than
point 3 in the order along the curve. Hence point 2 is a negative return point and point 3
is a positive return point.
First we prove the global result for closed curves. The proof is again subdivided into
three steps.
Theorem 5.1. Let ° be a closed curve. Assume
and let .Sn/ be a sequence of samples of ° with limn!1 ".Sn/ D 0. Then there exists
N 2 N such that TSP.Sn/ D P.Sn/ for all n ¸ N . Here TSP.Sn/ is a shortest tour
through the points Sn. Furthermore, TSP.Sn/ is unique for all n ¸ N .
First Step. We show that for large n there exist at least four return points. First we
show that there exists at least one return point for large n. Assume the contrary. That is,
there does not exist a return point in Sn for arbitrary large n. By turning to a subsequence
one can assume without loss of generality that there does not exist a return point for all
n 2 N. Since TSP.Sn/ 6D P.Sn/ there exists pni 2 Sn which is not connected to pni¡1 in
TSP.Sn/. Cut TSP.Sn/ in two polygonal arcs P1, with startpoint pni and endpoint pni¡1,
and Pn2, with startpoint pni¡1 and endpoint pni.nBy our assumption that there does not
exist a return point in Sn the sample points in both polygonal arcs are connected in their
order along ° .
From Lemma 4.2 and Definition 2.4 of the Jordan length we can conclude that
lim L. Pn1/; lim L. Pn2/ D L.° /:
n!1 n!1
That is, limn!1 L.TSP.Sn// D 2L.° /. Which is a contradiction. Hence there has to
exist a return point for large n.
Observe that the signs, see Definition 5.1, of the return points in Sn always sum to zero
and that return points incident along TSP.Sn/ always have different signs. So one can
conclude that for sufficiently large n there exist at least two return points. Now assume
that we have only two return points pn and qn for arbitrary large n. Cut TSP.Sn/ into two
polygonal arcs Pn1 and Pn2 that connect pn with qn. The points along these arcs, running
from pn to qn, are ordered in the same way as they are ordered along ° . From Lemma 4.2
and Definition 2.4 of the Jordan length we can conclude that
lim L. Pn1/; lim L. Pn2/ D L.° /:
n!1 n!1
That is, limn!1 L.TSP.Sn// D 2L.° /. Which is a contradiction to Corollary 4.1. Hence
for large n there are at least four return points in Sn.
Second Step. We show that for large n there must exist two return points rn1 G rn2
neighborly along TSP.Sn/, i.e., rn1 and rn2 are consequent return points along TSP.Sn/, such
that the other return points rQn1 neighborly to rn1 and rQn2 neighborly to rn2 along TSP.Sn/
are not in between rn1 and rn2. That is, one cannot find the following situation:
1 1 2
rn G rQn G rn
However, it is possible that rQn1 D rn2 and rQn2 D rn1.
Assume that for all neighborly return points one finds situation (6) then all return
points have to accumulate in between two return points (see Fig. 6). That is impossible
since TSP.Sn/ is closed.
Third Step. In this step the transition from the local version of the theorem to the global
one is achieved. In this transition use is made of the return points rn1 and rn2 found in
the second step. Choose the orientation of TSP.Sn/ such that rn2 G rn1along TSP.Sn/. Let
0
rn 2 Sn be the last sample point one encounters running through TSP.Sn/ with
rn0 G rn1 along ° and rn0 G rn2 along TSP.Sn/
and let rn3 2 Sn be the first sample point one encounters running through TSP.Sn/ with
rn2 G rn3 along ° and rn1 G rn3 along TSP.Sn/:
That is, one finds the situation shown in Fig. 7.
By the compactness of ° we can assume by turning to convergent subsequences that
.rn0/; .rn1/; .rn2/, and .rn3/ converge to r 0; r 1; r 2; r 3 2 ° . Let sn 2 Sn be the successor of
rn0 and let pn 2 Sn be the predecessor of rn3 along TSP.Sn/. By construction we have
From Lemma 4.2 we can conclude that
That is, r 0 D r 1 and r 2 D r 3. Now assume r 1 G r 2. Consider three sets of sample points
rn0 G rn1 G sn and pn G rn2 G rn3 along ° :
0 3
lim jrn ¡ snj D nl!im1 j pn ¡ rn j D 0:
n!1
Mn1 D f p 2 Sn: p · rn2 along TSP.Sn/g;
Mn2 D f p 2 Sn: rn2 · p · rn1 along TSP.Sn/g;
Mn3 D f p 2 Sn: p ¸ rn1 along TSP.Sn/g:
Using Corollary 4.1 and Theorem 4.1 one has
lim L.TSP.Sn//
n!1
¸ nl!im1 ¡L.TSP.Mn1// C L.TSP.Mn2// C L.TSP.Mn3//¢
D nl!im1 L.TSP.Mn1// C nl!im1 L.TSP.Mn2// C nl!im1 L.TSP.Mn3//
D L.° j[0;° ¡1.r2/]/ C L.° j[° ¡1.r1/;° ¡1.r2/]/ C L.° j[° ¡1.r1/;1]/
D L.° / C 2L.° j[° ¡1.r1/;° ¡1.r2/]/
> L.° /:
That is a contradiction to Corollary 4.1. Hence we have
r 0 D r 1 D r 2 D r 3 D: r 2 ° :
By turning to an appropriate subsequence of .Sn/ we can assume without loss of
generality that
rn0; rn1; rn2; rn3 2 Sn \ B1=n.r /:
That is a contradiction to Theorem 3.1, which is the local version of this theorem.
Next we want prove the global result for open curves. To do so we start with three
lemmas.
Lemma 5.1. Let S be a sample of an open curve ° , let E .S/ be the set of edges of
P.S/, and let E 0.S/ be the set of edges of P0.S/, where P0.S/ is another polygonal path
on S. Let ¼ be the permutation of f1; : : : ; jSjg induced by P0.S/. Then there exists a
bijection f : E 0.S/ ! E .S/ with
2. f jE.S/\E0.S/ D id.
1. f : convf p¼.i/; p¼.iC1/g 7! convf p j ; p jC1g with ¼.i / · j and j C 1 · ¼.i C 1/.
Proof. The proof is done by induction on the cardinality of S. If jSj D 2 we must have
P.S/ D P0.S/ and set f D id. Assume the lemma is proven for jSj < n. Now assume
that jSj D n. We distinguish two cases, which are depicted graphically in Fig. 8. The lines
in this figure denote edges of Pmin.S/. In both cases map the edge e to convf p1; p2g.
Now remove p1 and consider the induced polygon TSP.S ¡ f p1g/ on the vertex set
f p2; : : : ; png. In the second case add convf pi ; p j g to the induced polygon TSP.S ¡f p1g/.
We are left with the problem of finding a suitable bijection on edge sets with n ¡ 1
elements. That is, we have reduced the problem to finding an appropriate bijection to the
case jSj D n ¡ 1.
The next two lemmas are about regular curves. Nevertheless we can make use of
them, since an open s-regular curve is regular in its endpoints.
Lemma 5.2. Let ° be a regular curve. Then there exists an " > 0 such that
1. j pi ¡ pi¡1j < j pi ¡ pk j if pk G pi¡1,
2. j pi ¡ piC1j < j pi ¡ pk j if pk > piC1
for all samples S D f p1; : : : ; pjSjg with ".S/ < " and all i 2 f1; : : : ; jSjg.
Proof. Assume the contrary. Then there exists a sequence .Sn/ of samples with
limn!1 ".Sn/ D 0 and pni; pnk 2 Sn such that
pk G pi¡1 and j pni ¡ pni¡1j ¸ j pni ¡ pnk j
pk > piC1 and j pni ¡ pniC1j ¸ j pni ¡ pnk j:
or
By choosing a subsequence one can always assume that for all n 2 N one of the above
possibilities holds. Without loss of generality assume that this is the first one. Since ° is
compact, one can also assume by choosing a subsequence that . pni/ converges to p 2 ° .
From limn!1 ".Sn/ D 0 it follows that
lim j pni ¡ pnk j · nl!im1 j pni ¡ pni¡1j D 0:
n!1
aHnedncpek.. pFnir¡o1m/ atnhde.lapwnk/ oalfscoocsoinnvesertgoegteothpe.rNwoiwthlojponik ¡at tphnie¡t1rjia¸ngjlepwni¡ithpvnkejr,tiictefsolploni¡w1s; pfonir,
n
the angle ®n at pni¡1 that
Thus ®n has to be smaller than or equal to ¼=2, but that is a contradiction to Lemma 3.1.
The proof of the third lemma is similar to the proof of the second one, so we omit it
here.
Lemma 5.3. Let ° be a regular curve. Then there exists an " > 0 such that
1. j pi ¡ pm j < j pi ¡ pk j for pk G pm G pi , if pk G pi¡1 and j pi ¡ pk j · j pl ¡ plC1j
for some l 2 f1; : : : ; jSjg,
2. j pi ¡ pm j < j pi ¡ pk j for pi G pm < G pk , if pk > piC1 and j pi ¡ pk j · j pl ¡ plC1j
for some l 2 f1; : : : ; jSjg
for all samples S D f p1; : : : ; pjSjg with ".S/ < " and all i 2 f1; : : : ; jSjg.
Now we are prepared to prove the global result for open curves.
Theorem 5.2. Let ° be an open curve. Assume
and let .Sn/ be a sequence of samples of ° with limn!1 ".Sn/ D 0. Then there exists
N 2 N such that TSP.Sn/ D P.Sn/ for all n ¸ N . Here TSP.Sn/ is a shortest path
through the points Sn. Furthermore, TSP.Sn/ is unique for all n ¸ N .
Proof. The proof is done by contradiction. Assume without loss of generality that
TSP.Sn/ 6D P.Sn/ for all n 2 N.
First Step. We show that for large n there exist at least four return points. Because
TSP.Sn/ 6D P.Sn/ there have to exist at least three return points. If in Sn there is only one
return point besides the start- and endpoints of TSP.Sn/, then this point is either min Sn
or max Sn along ° . Assume that there exists N 2 N such that the number of return points
in Sn is three for all n ¸ N . Without loss of generality we can assume that the only
return point besides the endpoints of TSP.Sn/ is pn1 D min Sn.
In TSP.Sn/ we also have that pn1 is connected to pn2, because otherwise Sn has more
than three return points. Cut TSP.Sn/ into two polygonal arcs Pn1 and Pn2 with endpoint
pn1. Since S contains only three return points one of these arcs has as its second endpoint
pnjSnj. Assume without loss of generality that this arc is always Pn1. From Lemma 4.2
and the definition of Jordan length we can conclude that limn!1 L. Pn1/ D L.° /. This
implies
lim L. Pn2/ D 0:
n!1
Let pni, i 2 f3; : : : ; jSnjg, be the second endpoint of Pn2. We observe two things:
1. limn!1j j pn1 ¡ pnij D 0, because j pn1 ¡ pnij · L. Pn2/ for large n.
2. convf pn ; pnjC1g ½ TSP.Sn/ for all i < j < jSnj and large n. That is, the shortcuts
take place on the first i indices. Using that TSP.Sn/ 6D P.Sn/ and the statement
and notions of Lemma 5.1, with P0.S/ D TSP.S/, we find that there exists
convf pn¼. j/; pn¼. jC1/g 2 E 0.Sn/ and
convf pnk ; pnkC1g 2 E .Sn/
with
j pn¼. j/ ¡ pn¼. jC1/j · j pnk ¡ pnkC1j;
j; k < i;
¼. j / · k;
k C1 · ¼. j C1/;
and
¼. j / 6D k or k C1 6D ¼. j C1/.
From limn!1 ".Sn/ D 0 we can conclude that pn1 converges to ° .0/. Since ° .0/ is a
regular point of ° we can apply here Lemmas 5.2 and 5.3. Now Lemma 5.3 tells us that
there exists n 2 N with:
1. j pnk ¡ pn¼. jC1/j · j pn¼. j/ ¡ pn¼. jC1/j for all n ¸ N . That means
j pn ¡ pn¼. jC1/j < j pn ¡ pnkC1j
k k
for all n ¸ N :
2. j pnkC1 ¡ pn¼. j/j · j pn¼. j/ ¡ pn¼. jC1/j for all n ¸ N . That means
j pnkC1 ¡ pn¼. j/j · j pnk ¡ pnkC1j
for all n ¸ N :
That is a contradiction to Lemma 5.2. Hence for large n there are at least four return
points in Sn.
Second Step. We show that for large n there must exist two return points rn1 Grn2 incident
along TSP.Sn/ such that the other return points rQn1 incident to rn1 and rQn2 incident to rn
2
along TSP.Sn/ are not in between rn1 and rn2. That is, one cannot find the following
situation:
However, it is possible that rQn1 D rn2 and rQn2 D rn1.
The path Pn1 ½ TSP.Sn/ connecting pn1 with pnjSnj either contains two return points,
as we are looking for, or it does not contain any return point besides pn1 and pnjSnj. So
assume that Pn1 does not contain any return point besides pn1 and pnjSnj. Cut TSP.Sn/ into
three paths Pn1, Pn2 with endpoint pn1 and Pn3 with endpoint pnjSnj. At least one of the the
paths Pn2; Pn3 cannot be empty since Sn has at least four return points. From Lemma 4.2
and the definition of Jordan length we can conclude that limn!1 L. Pn1/ D L.° /. This
implies
lim L. Pn2/ D nl!im1 L. Pn3/ D 0:
n!1
Without loss of generality we can assume that Pn2 is not empty for all sufficiently large
n 2 N. The same reasoning as at the end of the first step shows that this leads to a
contradiction. Hence for large n the sets Sn have to contain two return points, as we are
looking for.
Third Step. In this step the transition from the local version of the theorem to the global
one is done. This step is the same as the third step in the proof of Theorem 5.1.
The example in the Introduction shows that the regularity conditions required to prove
this theorem are necessary. That is, this theorem is best possible.
Finally we want to put our work in perspective to related recent work on curve
reconstruction. We showed that there exists a global bound on the sampling density such that
the curve reconstruction problem is solved by a traveling salesman path or tour,
respectively. Obviously this bound is much too demanding for many smooth regions of the
curve. That is, locally a much lower sampling density should be sufficient. Amenta et
al. [
4
] concretize the idea of a locally dense sampling using the concept of feature size.
The medial axis of a plane curve ° is the set of points in the plane which have more
than one closest point on ° . The feature size f . p/ of a point p 2 ° is the distance of
p to the closest point on the medial axis. Amenta et al. define sampling density based
on a parameter " by requiring that each point p 2 ° has a sample point within distance
" f . p/. Several algorithm with this assumption of sampling density have been developed
that provably can reconstruct simple, closed, smooth curves [
4
]–[
6
], [
11
]. There is also
an experimental study by Althaus et al. [
3
] that compares several of these algorithms.
For nonsmooth curves this notion of sampling density breaks down, since the medial
axis passes through the corner points of the curve. Thus one is required to sample the
curve infinitely near the corners to satisfy the sampling condition.
Many curve reconstruction algorithm are based on picking edges from the Delaunay
triangulation. In [
10
] we show that the same regularity assumptions on the curve are
necessary to find the correct reconstruction as a subgraph of the one-skeleton Delaunay
triangulation. Dey and Wenger [
7
] present another algorithm that can reconstruct curves
with sharp corners.
We showed that the traveling salesman path can reconstruct simple open curves and
the traveling salesman tour can reconstruct simple closed curves. However, we do not
give a method to detect only from a sample if a curve is open or closed. Dey et al. [
6
]
give an algorithm that does so in the case of simple smooth curves.
Finally, in general it is NP-hard to compute a traveling salesman path or tour,
respectively. Althaus and Mehlhorn [
2
] show that the traveling salesman path=tour can
be computed in polynomial time for dense samples from plane curves, satisfying the
regularity conditions we specified in this paper.
Acknowledgments References
I want to thank Emo Welzl and Nicola Galli for helpful discussions.
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