Curve Reconstruction, the Traveling Salesman Problem, and Menger’s Theorem on Length
Discrete Comput Geom
Geometr Y
Curve Reconstruction 0
the Traveling Salesman Problem 0
Menger's Theorem on Length" 0
0 Institut fur Theoretische Informatik, ETH Zurich , CH8092 Zurich, Switzerland giesen @inf.eth2.ch
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.

1. 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 graphjoining 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 GALlA).
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 y E ]Rd and a finite set of sample points S C y. A polygonal
reconstruction of y from S is a graph that connects every pair of samples adjacent
along y, 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 atraveling 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 y be the simple arc which consists of the unit interval on the xaxis and the graph
of y = x 2 on this interval. That is,
: 0 1 ~]R2 ,
tH'Y [ , ]
I(
1  2t, 0),
(2t  1, (2t  1)2),
t ~ 21'
s. = {p~, p~, p~, p:l u ~ { (~, 0) ,(~, ::) }
p~ = (~, 0) ,
p~ = (:3' ~ ),
p~ = (:3' :6) ,
p: = (:2' :4 )
become arbitrarily dense in y. However, the traveling salesman path through Sn is dif
ferent from the polygonal reconstruction from S« (see Fig. 1) because
Ip~  p~1 + Ip~  p~1 + Ip~  p~1 > Ip~  p~1 + Ip~  p~1 + Ip~  p~l.
In this example, the arc y 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 y behaves
quite well, but is not regular at (0, 0) E y. The regularity conditions necessary tum 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 x .
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 higherdimensional 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 byproduct we
generalize Menger's theorem to simple closed curves in Euclidean space.
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 y: [
0, 1
] ~ }Rd is always a connected simple curve. We use the
symbol y either to denote a mapping or to denote the image of the mapping. It should
always be clear from the context what y denotes.
We call a curve y s(emi)regular if in every point on y nonzero left and right tangents
exist. This is expressed in the following definition:
Definition 2.1.
Let
and
The curve y is called left (right) regular at y (to) with left tangent 1(y (to)) or right tangent
r(y(to)) if for every sequence (~n) in T which converges from the left (right) to (to, to)
in closure(T) the sequence r (~n) converges to 1(y (to)) or r(y (to)), respectively. We call
y sregular if it is left and right regular in all points y(t), t E [
0, 1
]. A curve is called
regular if it is sregular and in every point left and right tangent coincide.
The relationship between sregularity and two of the most interesting geometric prop
erties of curves, length and total absolute curvature, was shown by Aleksandrov and
Reshetnyak [
1
].
Theorem 2.1. Every curve y offinite total absolute curvature C(y) is sregular and
every sregular curve hasfinite length L(y).
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 y is a finite sequence
S = {pI, ... , plSI}
of points where pi E y. We assume that the sample points pi are ordered according
to the order of the y 1 (pi) E [
0, 1
]. To every sample 8 its density is defined to be the
inverse of the following number:
8(8) = sup minllp'  x]: i = 0, ... , n}.
XEy
In the following we sometimes make use of the convention plSI+l = plSI if y is open
and plSI+l = pI if y 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 8 = {pI, ... , plSI} be a sample of y. If y is open one writes i <J j
if Y1 (pi) < y 1 (pi). This notion is independent of an orientation preserving change
of parameterization. If y is closed and has finite length choose an orientation along y
and write i <J j if i =F j and L(y(pi: pi)) ::: L(y(pi: pi)), where y(pi: pi) C Y is the
arc connecting pi and pi in the orientation along y. One writes i ~ j if the possibility
that i = 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(y) of a curve y is the following number:
sup{L(P(S)): S is a sample of y},
where
lSI
L(P(S)) = L Ipi+l  pi I.
i=1
2. The curvature C(y) of a curve y is defined as the number
where
if y is open, and
where plSI+l = pI, if y is closed.
i=1
i=1
sup{C(P(S)): S is a sample of y},
ISI2
C(P(S)) = L IL(pi+l  pi, pi+2 _ pi+l)l,
ISIl
C(P(S)) = L
IL(pi+l  pi, pi+2 _ pi+l)l,
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 onedimensional subspaces z E JRd+ I . The collection of all onedimensional
subspaces of JRd+l forms the ddimensional projective space JP>d. In the following the
elements of JP>d are sometimes called lines. From a standard construction in integral
geometry one gets a probability measure J,td on JP>d from the surface area measure on the
ddimensional sphere §d:
The continuous and locally homeomorphic mapping
is a double cover. Since q; is continuous, it is also measurable. That is the inverse image
q;1(B) of a Borel set B E JP>d is a Borel set in s. On §d we have the usual surface area
measure vs Therefore we get a measure J,t~ on JP>d as J,t~(B) = Vd(q;I (B)) for all Borel
sets B ~ JP>d. We have
( 21l'(d+l)/2
Jpd d/L~ = vol(§d) = r«d + 0/2) ,
which we use to normalize J,t~. That is, we set
x ~ {Ax: A E JR},
for all Borel sets B C JP>d.
The theorem of Reshetnyak now reads as follows:
Theorem 2.2. For any sregular curve y one finds
. h
wlt
r«d + 1)/2)
Cd = ,J1ir«d + 2)/2) .
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 onedimensional subspaces.
The first reformulation of regularity in the following lemma is a pure metric interpre
tation of our definition of regularity.
Lemma 3.1. Let y be a simple closed curve, which is left (right) regular in p E y.
Let (Pn), (qn), and (rn) be sequences ofpoints from y, that converge to p from the left
(right), such that Pn < qn < rn for all n E N in an order locally around p along y. Let
an be the angle at qn ofthe triangle with corner points Pn, qn, and rn. Then the sequence
of angles (an) converges to Jr.
Proof. Since y seen as a mapping from an interval is locally a homeomorphism the
sequences (yl (Pn)), (yl (qn)), and (yl (rn)) converge to yl (p). Thus by our defi
nition of left (right) regularity asymptotically the three oriented segments
have to point in the direction of the left (right) tangent l(p) or r(p), respectively. That
is, limn~oo an = Jr. 0
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
JRd+ 1• We exploit this linear structure by studying projections of a set of sample points
on onedimensional subspaces of JRd+l.
In the following let y be sregular and let p E Y be a fixed point. We want to introduce
the following notions. Let 17 > O. The connected component of
{q E y: Ipql < 17}
which contains p is denoted by Bl1(p ). We assume that every onedimensional subspace
i of JRd+l not perpendicular to l(p) is oriented according to the orientation induced by
the orthogonal projection Jrt(l(p)) ofl(p) on i and that everyonedimensional subspace
i perpendicular to l(p) carries an arbitrary orientation.
We want to compare the ordering of a set of sample points close to p on y with
the ordering of the projections of these points on onedimensional subspaces of JRd+ 1•
The following reformulation of regularity states to which extent these orderings can be
different:
Lemma 3.2. Assume there exists a sequence (in) of onedimensional subspaces of
JRd+l and sequences (Pn), (qn), (rn) E Bl/n(P) with Pn < qn < rn ~ p in the order
along y, but
in the order on in. Then the limit of every convergent subsequence of (in) has to be
perpendicular to l(p).
Proof We want to do the proof by contradiction. That is, we can assume that (in)
converges in the topology of JP>d and that
• 1f
o ~ fJ := nl~imoo L(l(p), in) < ,
2
where L(l(p), in) denotes the smaller of the two angles defined by the subspace in and
the subspace spanned by l(p). For the proof we assume without loss of generality that
ne, (qn) < 1fln (Pn) < 7rln (rn). Furthermore, we can assume by moving qn or rn a little
bit on y that there exists 0 < A < 1 (independent of n) such that
Why can we do so? By our assumption we have and hence
l1fl n(Pn)  1fl n(rn)1 = Anl1fl n(qn)  1ft n(rn)1
with 0 < An < 1. From the continuity of y and the continuity of the projection maps 1ftn
we find that moving Pn toward qn on y can make l1ft n(qn)  1ftn (Pn)1 arbitrarily small.
Similarly, moving Tn toward qn on y can make l1ft n(Pn) 1ftn (rn)1 arbitrarily small. Now
move Pn toward qn on y if An < Acos(fJ) or move rn toward qn on y if An > Acos(fJ)
as long as An =I= Acos(fJ).
From the regularity of y we find that
lim L(conv{qn, rn}, l(p)) = lim L(conv{Pn, rn}, l(p)) = 0
n~oo n~oo
and together with the triangle inequality for spherical triangles
nl~imoo L(conv{qn, rn}, 1ftn (conv{qn, Tn}))
~ lim (L(conv{qn, rn}, l(p)) + L(l(p), 1ftn (conv{qn, Tn})))
n~oo
= nl~imoo L(conv{qn, rn}, l(p)) + nl~imoo L(l(p), 1ftn (conv{qn, rn}))
= 0 + lim L(l(p), in)
n~oo
n
= fJ < 2"'
Analogously we find
Hence
and
Therefrom we find
and thus
nI+1OmO IIPqnn  rrnn II <"I\. < 1.
·
Hence there exists N E N such that IPn rnl ~ Iqn rnl for all n 2: N. Now we consider
the triangles with comer points Pn, qn, and rn. Let an be the angle at qn. From the law
of cosines together with IPn  rnl ~ Iqn  rnl we find
Thus an has to be smaller than or equal to n /2 for all n 2: N. Since y is left regular in
P that is a contradiction to Lemma 3.1. 0
From this lemma we can derive another one that states the measure ~d 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 E N let Pn, qn, rn E B1/n(p) with Pn < qn < rn ~ P in the
order along y. Let
N; = {l E JPd: Jre(qn) < Jrt(Pn) < Jrt(rn) or Jrt(Pn) < Jrt(rn)
< .7rt(qn) in the order on l},
where .7re denotes the orthogonal projection on the line l. Then limn+ oo ~d(Nn) = o.
Proof In [
14
] Reshetnyak shows: Let V be a ddimensional subspace of JRd+l and let
E C JP>d be the set of all onedimensional subspaces contained in V. Then ~d(E) =
O. The proof of the lemma follows directly from this theorem of Reshetnyak and
Lemma 3.2. 0
Obviously an analogous result for sample points larger than p in the order along y
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 y 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 y. We distinguish three types
of lines l E Jp>d:
1. There exists a path through the sample points different from P(S) which has a
shorter projection on l than P (S).
2. Every path through the sample points different from P(S) has a larger projection
on l than P(S).
3. There exist paths through the sample points different from P(S) which has pro
jections on l 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
ex = sup{L(I(q), r(q»: q E y} < n
and let (Sn) be a sequence of samples of Bl/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 S; by y. Then
there exists N E N such that
TSP*(Sn) = P(Sn),
for all n 2:: N. Furthermore, TSP*(Sn) is unique for all n 2:: N.
Proof First Step. Let L; C Jp>d be the set of lines for which the projection 7r/,(P(Sn»
is not a shortest path through the projected sample points 7r/,(Sn) and let JLd be the
probability measure on Jp>d introduced in (1).
We show that
For the proof we use the following abbreviations (min and max denote the minimum
and the maximum along y and min, and max, denote the minimum and the maximum
along l):
m2
•
mt
m4
Jn3
Take .e E pd together with its orientation. A path of minimal length through the points
Jrl(Sn) which connects mi with m4 (see Fig. 2) consists of
That is, the points of e between m 1 and mz are covered twice by a path of minimal
length through the points Jrl(Sn), the points between ml and m« are covered once, and
the points between m3 and m« are covered twice again. If Jrl(P(Sn)) is not a path of
minimal length through the points Jrl(Sn), then there has to exist an interval
I = [mln{1l't(p~). 1l't(p~+I)}, mF{1l't(p~). 1l't(p~+I)}]
on z with p~, p~+l E Sn, which is covered by Jrl(P(Sn))
1. 2 + 2k times, k ~ 1, if
or
m2 s min{Jrl(p~), Jrl(p~+l)} < max{Jrl(p~), Jrl(p~+l)} s ml
l l
m« ~ ~n{1l't(p~), 1l't(p~+I)} < mF{1l't(p~). 1l't(p~+I)} ~ mv,
2. 1 + 2k times, k ~ 1, if
ml s ~in{1l't(p~). 1l't(p~+I)} < mF{1l't(p~), 1l't(p~+I)} ~ ms.
For all p~ E Sn, j E {2, ... , ISn I  2}, we call the interval
Ij =
[mln{1l't(p~). 1l't(p~+I)}, mF{1l't(p~), 1l't(p~+I)}]
positive if Jrf,(p~) < Jrl(p~+l) in the order on e and negative otherwise. The intervals
I j which cover the interval I must have alternating signs. That is, if Jrf,(P(Sn)) is not a
shortest path through the points Jrl(Sn) we find, using that P(Sn) connects the p~ E S«
in their order along y and using the continuity of y and of the projection map ne. three
points Pn, qn, r n E B1/n(p) with Pn < qn < r« ~ P in the order along y, but
or we find three points Pn, qn, r n E BI/n(P) with Pn > qn > r n ~ P in the order along
y, but
Jrf,(qn) > Jrf,(Pn) > Jrf,(rn)
or
Jrl(Pn) > Jrf,(rn) > Jrf,(qn)
in the order on e.
See Fig. 3 for an example.
in the order on .e;
,
.' ,
"
.5...•.. .•6 .. ......
••4..._ _8..I.I.._ _... _ •   ,
 ,................ 1• •.._ _..2.......3......... __ .. '
~
In general the points Pn, qn, and 'n need not be sample points. From Lemma 3.3 we
can conclude that
Second Step. Let M; C pd be the set of lines i for which we have:
1. The order of S; along y and the order of S; induced by the order of 11't(Sn) on i
coincide.
2. For all convlp', pi+l} C P(Sn) with pi, pi+l E s; we have
111't(p''+1 )  11't(p',)1 1~"2 cos (114'+a) Ip''+1  pi' I.
We show that there exist c > 0 and N' E N such that for all n ~ N' we have
J.Ld(Mn ) ~ c.
For the proof we construct a set of lines C C pd with J.Ld (C) > 0 and show that there
exists N' E N such that C C M; for all n ~ N'. The set C is defined as follows: Let
io be the line, io C span{I(p), r(p)} such that l~ C span{I(p), r(p)} halves the angle
L(l(p), r(p)) between the lines determined by l(p) and r(p). Now we define
C = {l E F: L(i, io) ::: i(11'  a)}.
By L(l, l') for i, i' E pd we denote the value of the minimum of the two angles
determined by land t'. The set UtEC{X E i} C JRd+l is a double cone (see Fig. 4).
By construction 11't(l(p)) and 11't(r(p)) point in the same direction on every i E C.
Since a < 11' we have J.Ld (C) > O. It remains to check conditions 1 and 2 to prove that
for sufficiently large n we have C C Mn •
r(p)
I(p)
Fig. 4. The double cone C.
1. Assume that for arbitrary large n we find in EC such that the first condition is
violated. Then we can find three points Pn, qn, rn E Bl/n(P) with Pn < qn < rn ~ P in
the order along y, but
Ttl n (qn) < Ttl n (Pn) < Ttl n (rn) or Ttl n (Pn) < Ttl n (rn) < Ttl n (qn)
in the order on in; or we find three points Pn, qn, Tn E B1/n(p) with Pn > qn > r n ~ P
in the order along y, but
Ttl n (qn) > Ttl n (Pn) > Ttl n (rn) or Ttl n (Pn) > Ttl n (rn) > Ttl n (qn)
in the order on in. By Lemma 3.1 the limit of every convergent subsequence of (in) has
to be perpendicular to either I(p) or r(p). Since C is compact we find i E C as the limit
of a convergent subsequence of (in) which is perpendicular to either l(p) or r(p). On
the other hand we find for all f E C using the triangle inequality for spherical triangles,
L(i,l(p)) < L(i, io) + L(io, l(p))
< !(Tta) + (~_ Tta)
4 2 2
=
Tt+a n
  <
4 2
n
L.(i, r(p)) < "2.
and analogously
That is, every i E C is neither perpendicular to l(p) nor to r(p). So we have a contra
diction. That means, for all sufficiently large n the first condition cannot be violated.
2. From the sregularity of y and the continuity of L we have, for p~, p~+l E S« with
p~ < p~+l :::: p and alIi E C,
lim L(i, conv{p~, p~+l}) = L. (i, lim conv{p~, p~+l})
n+oo n+OO
= L(i, l(p)) s tt 4+a.
That is, for all sufficiently large n and all z E C we have
For p~, p~+l E Sn with p ~ p~ < p~+l 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
p~ < p < p~+l. We choose 1] > 0 such that
and show that, for all f E C and all sufficiently large n,
cos(Jr;a +ry) > ~cos(Jr;a)
L(i, conv{p~., p~·+1 }) s Tt 4+ a + 1].
To do so let n be the projection on span{l(p), r(p)}. The sregularity of y implies
limsup(L(lo, Tl(conv{p~, p~+I}))) ~ L(io,l(p)) = L(io, r(p)).
We define the distance between two twodimensional subspaces oflRd+1 by the smaller
of the two dihedral angles defined by them. With this metric the span seen as a map
ping becomes continuous. From the continuity of L and the continuity of span we
get
That is, for all sufficiently large n we have
O = n1+imOO
°
L (
span { Ip'~  p Ip  p~'+I1' cony Pn' r;i+l} )
Pni  PI' P  Pni+I } {i
= L ( nl'i+moo span { IPp'~ni  Ppi , IpP  pP~in'++l1I· } , I'im cony {Pin' Pni+l} )
n+OO
= L (span{l(p), r(p)}, nli+moo conv{p~, p~+I})
= lim L(span{l(p), r(p)}, conv{p~, p~+I}).
n+oo
L(Tl(conv{p~, p~+1 }), conv{p~, p~+l}) s TJ.
Finally we get, for all sufficiently large n using the inequalities (2), (3) and the triangle
inequality for spherical triangles,
L(i, conv{p~, p~+I}) < L(i, io) + L(io, conv{p~, p~+I})
(2)
(3)
< L(i, lo) + L(io, Tl(conv{p ~,, p~'+1 }))
+ L(Tl(conv{p~, p~+1 }), conv{p~, p~+l})
< c«, io) + L(io, I(p)) + TJ
n  a a
< 4+2"+TJ
Tl+a
=  4  + TJ.
Hence for all sufficiently large n, all f E C, and all p~, p~+1 E S; we have
ITll(p',)  1rl(p''+1 )1 ~ 21cos
(Tl + a ) ,
 4  Ip'  pi'+1 I·
That is, there exists N' E N such that for all n ?: N' we have C C M n • Hence
Third Step. In the first two steps we considered the measures of the subspaces L n
and M; of JP>d, the space on which we want to integrate. In this step we compare
integrands.
• • 1
2 6 3
•
T
T
•p • • L • 1
2 3 4 5 6 7
• • • • • • •
p
•
Pn _ {l· E {2, ... , ISnl 2}.. rk(P ni' P (Sn)) > rk(Pni+l ,P(Sn))}.
If we compare the length of the projections Tlt,(P(Sn)) and Tlt,(P(Sn)) on i E L; (see
Fig. 5) we find
L(Tlt,(P(Sn)))  L(Tlt,(P(Sn))) s 2 L ITlt,(p~)  Tlt,(p~+I)I.
ieln
On the other hand we find, for the length of projections on i E M; (see Fig. 5),
L(Tlt,(P(Sn)))  L(Tlt(P(Sn))) ~ 2 L ITlt,(p~)  Tlt,(p~+I)I.
ieln
Furthermore, we have, for all P~ E Sn, i E {I, ... , ISnl  I} and all f E Mn ,
11l'l(pl'+1 ) 1l'l(pl'+)11 ~ 21cos (Tl 4+ a) Ip''+1  p'' I·
(4)
(5)
From this inequality we get another inequality which is valid for all i E M; and all
i' E t.;
which implies, together with (4) and (5),
L(Tlt,(P (Sn)))  L(Tlt(P(Sn))) ~ 21cos
(Tl+a)(
 4
 )
L(Tlt,,(P(Sn)))  L(Tlt,,(P(Sn))) .
Since this inequality is valid for all z E M; and all z' E L; we get for the increase of
length on M;
1 (L(1l'l(P(Sn)))  u». (P (Sn))) ) df.Ld(l)
Mn
~ f.Ld(Mn) ! cos (Tl +a) sup (L(Tlt,(P(Sn)))  L(Tlt(P(Sn))))
2 4 t,eL n
and for the decrease of length on L;
( (L(1rt ( P (Sn)))  L(1rt ( P (Sn)))) dILd(l)
JLn
s ILd(Ln) sup (L(1rl ( P (Sn)))  Lin, (P (Sn))) ) .
leLn
Because of limn+ oo ~d(Ln) = 0 there exists N 2:: N' such for all n 2:: N we have
Using Theorem 2.2 we find that for all n 2:: 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 Y has a
shorter projection on all Z E C than every other polygon through the points Sn with fixed
start and endpoints. 0
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 y be a simple curve and let (Sn) be a sequence ofsamples of y with
limn.+ oo e(Sn) = 0 and lim sup L(TSP(Sn)) < 00. Let
be the parameterization by length of TSP(Sn). Then the sequence (Yn) has a Frechet
convergent subsequence.
By turning to an appropriate subsequence we can assume that
L(TSP(Sn)) < 2limsupL(TSP(Sn)) =: c < 00
for all n EN. 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
o
is equicontinuous, because
Iy~(t')  y~(t)1 ~ (t'  t)L(y~)
= (t'  t)L(TSP(Sn))
~ c(t'  t)
for all °~ t < t' ~ 1 and all n E N.
2. We have sUPneN sup{ly~(t)l: t E [0, I]} < 00, because y is compact.
Now the theorem of Ascoli states that the sequence (y~) has a convergent subsequence
(y~) which converges to a continuous function
y': [
0, 1
] ~ jRd.
Note that the condition lim sup L(TSP(Sn)) < 00 is always fulfilled if y has finite
length. Now we are prepared to prove Menger's theorem. In this proof we make use of
the onedimensional Hausdorff measure fi', see [
8
] for the definition and its relation to
the geometry of curves.
Theorem 4.1.
Every simple curve y satisfies
L(y) = sup{L(TSP(S)): S is a sample oiy},
where TSP(S) denotes a shortest path through the sample points S ify is an open curve
and a shortest tour otherwise.
Assume the contrary. That is, we can assume that
sup{L(TSP(S)): S is a sample of y} < L(y),
because TSP(S) ~ peS) ~ L(y) for every sample S of y. Take any sequence (Sn) of
samples from y with limn~oo e(Sn) = 0. By our assumption we have
limsupL(TSP(Sn)) < L(y).
Let Yn: [
0, 1
] ~ jRd be the reduced parameterization of TSP(Sn). From Lemma 4.1
we know that there exists a Frechet convergent subsequence (Ym) of (Yn). Let the con
tinuous function y': [
0, 1
] ~ Rd be the limit of (Ym). We have, for the convergent
subsequence,
limsupL(TSP(Sm)) ~ limsupL(TSP(Sn)) < L(y).
Next we show that y C y' ([
0, 1
]): Take p E y. Since limn~oo e(Sn) = 0, one finds a
sequence (Pn) with Pn E Sn which converges to p. Of course the subsequence (Pm) with
Pm E Ym also converges to p. For all kEN there exists m(k) E N and P~(k) E Y'([
0, 1
])
such that
, 1
IPm(k)  Pm(k) I < 2k
and
1
IPm(k)  pi < u:
Hence by the triangle inequality we have Ip~(k)  pi < II k. From the compactness of
y'([O, 1]) we get p E y'([O, 1]).
Both Y and y'([O, 1]) are as compact sets fjl measurable. Using the properties of the
onedimensional Hausdorff measure fjl, see [
8
], and our assumption we find
L(y) = fjl(y)
::; fj I (y' ([
0, 1
]))
s lim inf jjl (TSP(Sm))
= liminf L(TSP(Sm))
< limsupL(TSP(Sm))
< L(y).
That is a contradiction.
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
1f and st; always denote permutations and no longer projections.
Corollary 4.1. Let y be a curve and let (Sn) be a sequence of samples of y, with
limn~oo 8(Sn) = 0. Then
L(y) = lim L(TSP(Sn)),
n~oo
where TSP(Sn) denotes a shortest path through the sample points S« if y is an open
curve and a shortest tour otherwise.
Assume L(y) < 00. For a given TJ > °consider three sets:
1. From Menger's theorem we know that there exists a sample
with L(TSP(S» > L(y)  i.
2. Let S' = {pI, ... , plS'I} E {Sn: n E N} be such that 8(S') < TJI4ISI.
3. Let S" = {r l , ••• , r lSI} C S' be a multiset with Ir i  qi I ::; 8(S') for all i =
1, ... , lSI.
In the following we make use of the conventions for ISI + 1 and IS'I + 1 introduced in
Definition 2.2. Let n be a permutation of {I, ... , IS I} such that
lSI
L Iq1r(i+I)  q1r(i)I = L(TSP(S)),
;=1
let 11:' be a permutation of {I, ... , IS'I} such that
and let 11:" be a permutation of {I, ... , ISI} such that
Then it follows that:
1.
2.
by construction.
IS'I
L Ip 1f' (i + l )  p1f'(i) I = L(TSP(S')),
i=1
lSI
L Ir1f" (i + l )  r1r" (i ) I is minimal.
i=1
i=1
i=1
lSI IS'I
L Ir1f" (i + l )  r1f" (i ) I s L
Ip 1f' (i + l )  p1f'(i) I
lSI lSI
L Iq 1f(i + l )  q1f(i) I s L Iq 1f" (i + l )  q1f"(i) I
i=1
i=1
by the definition of 11: •
3. From Ir1f" (j )  q1f"(j) I < 8(S') for all j = 1, ... , lSI and the triangle inequality it
follows that
IlrlTlI(i+I) _rlTl(j)I_lqlT"(i+I) _qlTlI(j)11 < 2e(S').
Combining these inequalities leads to
L(TSP(S')) =
~
~
IS'I
L Ip 1f' (i + l )  p1f'(i) I
i=1
lSI
L Ir1f" (i + l )  r1f" (i ) I
i=1
lSI
L Iq 1f" (i + l )  q1f"(i)l_ 2ISI8(S')
i=1
lSI
~ L Iq1f(i + l)  q1f(i) I  2ISI8(S')
i=1
= L(TSP(S))  2ISI8(S')
Since 1] can be arbitrary small, we are done. The proof in the case L(y) = 00 is quite
similar. 0
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 onedimensional
Hausdorff measure ij 1•
Lemma 4.2. Assume that the curve y has finite length and let (Sn) be a sequence
of samples of y with limn.+>oo e(Sn) = 0. For all n E N let 1T:n be the permutation of
{I, ... , ISnl} induced by TSP(Sn). Then
where 1T:n(ISn + 1I) = 1T:n(ISn I) if y is an open curve and 1T:n(ISn + 1I) = 1T:n(1) if y is
closed.
assume that there exists c > °such that for every n E N it holds that
Proof Assume the contrary. That is, by choosing an appropriate subsequence one can
Let Pn = p1rn(i ) E Sn and qn = p1rn(i + l) E S; be such that IPn  qnI > c. By the
compactness of y we can tum to an appropriate subsequence such that (Pn) converges
to p E Y and (qn) converges to q E y. Of course it holds that Ip  ql ~ c. Remove the
interior of conv{Pn, qn} from TSP(Sn). We consider two cases:
First, if y is an open curve, then TSP(Sn) decomposes into two paths P': which
connects p1rn(l ) with p1rIl (i) and P; which connects p1rn(i + l) with p1rIl (ISnD. Let r;
denote P': U P; and let yj: [
0,1
] + jRd, j = 1,2, be the reduced parameterization
of pj. As in Lemma 4.1 one can use the theorem of Ascoli to show that (yj) has a
Frechet convergent subsequence. That is, we can assume by considering a common sub
sequence that (Y':) and (y;) converge to continuous functions y 1: [
0, 1
] + jRd and
y2: [
0, 1
] + jRd. Let y' be y 1([
0, 1
]) U y2([
0, 1
]). Second, if y is a closed curve,
then TSP(Sn) becomes after the removal of the interior of conv{Pn, qn} a path denoted
by Pn which connects p1rn(i) with p1rl (i + l ) . Let Yn be the reduced parameterization of
Pn. As in Lemma 4.1 one can show that (Yn) has a Frechet convergent subsequence.
That is, we can assume by taking an appropriate subsequence that (Yn) converges to
a continuous function y': [
0,1
] + Rd. We use the symbol y' also to denote the set
y' ([
0, 1
]).
Next we show that y C y': Take r E y. Since limn.+>oo e(Sn) = 0, one finds a
sequence (rn) with rn E S« which converges to r. For all kEN there exists n(k) E N
and r~(k) E y' such that
, 1
Irn(k)  rn(k) I < 2k
and
1
Irn(k)  r] < 2k·
Hence by the triangle inequality Ir~(k)  r] < 1/ k. From the compactness of y' we get
r E y'.
S)1(y') ~ S)1(y U conv{Pn, qn})
= S)l (y) + S)l (conv{p, q})
> S)l(y)
That is a contradiction.
5. From Local to Global
o
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 y and a sample S from y. We call reS a return point,
if r is connected to p, q E S in a traveling salesman path (tour) of Sand r <] p, q or
r ~ p, q in the order along y. 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 y be a closed curve. Assume
a = sup{L(I(q), r(q)): q e y} < n
and let (Sn) be a sequence of samples of y with limn.+oo e(Sn) = O. Then there exists
N E N such that TSP(Sn) = 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.
Proof The proof is done by contradiction. Assume without loss of generality that
TSP(Sn) ¥= P(Sn) for all n e 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 E N. Since TSP(Sn) i= P(Sn) there exists p~ E S; which is not connected to p~l in
TSP(Sn). Cut TSP(Sn) in two polygonal arcs P;, with startpoint p~ and endpoint p~l,
and P;, with startpoint p~l and endpoint p~. By 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 y.
From Lemma 4.2 and Definition 2.4 of the Jordan length we can conclude that
lim L(P~), lim L(P;) = L(y).
n~OO n~oo
That is, limn~oo L(TSP(Sn» = 2L(y). 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 P; and P; that connect P» with qn' The points along these arcs, running
from Pn to qn, are ordered in the same way as they are ordered along y. From Lemma 4.2
and Definition 2.4 of the Jordan length we can conclude that
That is, limn~oo L(TSP(Sn» = 2L(y). 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 r~ <l r; neigh
borly along TSP(Sn), i.e., r~ and r; are consequent return points along TSP(Sn), such
that the other return points r~ neighborly to r~ and r; neighborly to r; along TSP(Sn)
are not in between r~ and r;. That is, one cannot find the following situation:
However, it is possible that r~ = r; and r; = r~.
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 r~ and r; found in
the second step. Choose the orientation of TSP(Sn) such that r; <3r~along TSP(Sn). Let
r~ E Sn be the last sample point one encounters running through TSP(Sn) with
r~ <3r~ along y and r~ <3r; along TSP(Sn)
and let r~ E Sn be the first sample point one encounters running through TSP(Sn) with
r; <3r~ along y and r~ <3r~ along TSP(Sn).
That is, one finds the situation shown in Fig. 7.
By the compactness of y we can assume by turning to convergent subsequences that
(r~), (r~), (r;), and (r~) converge to rO, r 1, r2 , r3 E y. Let Sn E Sn be the successor of
r~ and let Pn E Sn be the predecessor of r~ along TSP(Sn). By construction we have
From Lemma 4.2 we can conclude that
r: o«r, l<3Sn dan Pn <3rn2 <3rn3 aIong y.
lim lr~  snl = nlim..oo IPn  r~l = O.
n..oo
M~ = {p E s; P ~ r; along TSP(Sn)},
M; = {p E Sn: r; ~ p ~ r~ along TSP(Sn)},
M~ = {p E s; P !::: r~ along TSP(Sn)}'
That is, rO = r 1 and r2 = r3• Now assume r 1 <3r2• Consider three sets of sample points
Using Corollary 4.1 and Theorem 4.1 one has
lim L(TSP(Sn))
n~oo
~ lim (L(TSP(M~)) + L(TSP(M;)) + L(TSP(M~)))
n..oo
= lim L(TSP(M~)) + lim L(TSP(M;)) + lim L(TSP(M~))
n~oo n..OO n~oo
= L(Ylro,yl(r2)]) + L(YI[yl(r1),yl(r2) ] ) + L(YI[yl(r1),1])
= L(y) + 2L(YI[yl(r1),yl(r2)])
> L(y).
That is a contradiction to Corollary 4.1. Hence we have
rO = r 1 = r2 = r3 =: r E y.
e
~
e
~
....... ~
By turning to an appropriate subsequence of (Sn) we can assume without loss of gener
ality that
That is a contradiction to Theorem 3.1, which is the local version of this theorem.
0
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 y, let E (S) be the set of edges of
P (S), and let E' (S) be the set ofedges of P' (S), where P' (S) is another polygonal path
on S. Let Tl be the permutation of {I, ... , lSI} induced by P'(S). Then there exists a
bijection f: E'(S) + E(S) with
1. f: conv{p1r(i), p1r(i+l)} t1> convjp', pi+ I} with Tl(i) :5 j and j + 1 :5 Tl(i + 1).
2. fIE(s)nE'(s) = ide
Proof The proof is done by induction on the cardinality of S. If lSI = 2 we must have
P(S) = P'(S) and set f = ide Assume the lemma is proven for lSI < n. Now assume
that lSI = 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 convlp ', p2}.
Now remove pI and consider the induced polygon TSP(S  {pI}) on the vertex set
{p2, ... , pn}. In the second case add convlp', pi} to the induced polygon TSP(S _ {pI }).
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 IS I = n  1. 0
The next two lemmas are about regular curves. Nevertheless we can make use of
them, since an open sregular curve is regular in its endpoints.
Lemma 5.2.
Let y be a regular curve. Then there exists an 8 > 0 such that
1. Ipi  piII < Ipi  pkl if pk <J piI,
2. Ipi  pi+II < Ipi  pk I if pk I> pi+I
for all samples S= {pI, ... , plSI} with 8(S) < 8 and all i E {I, ... , ISIl.
or
Proof Assume the contrary. Then there exists a sequence (Sn) of samples with
limn~oo e(Sn) = 0 and p~, p~ E S; such that
pk <J pil and Ip~ _ p~11 :::: Ip~ _ p~1
By choosing a subsequence one can always assume that for all n E N one of the above
possibilities holds. Without loss of generality assume that this is the first one. Since y is
compact, one can also assume by choosing a subsequence that (p~) converges to p E y.
From limn+ oo e(Sn) = 0 it follows that
Hence (p~I) and (p~) also converge to p. Now look at the triangle with vertices p~I, p~,
and p~. From the law of cosines together with Ip~  p~11 :::: Ip~  p~ I, it follows for
the angle an at p~1 that
cos(a ) 
n
Ipi _ pkl2 Ipi _ pi 112 I pil _ pkl2
n n n n n n
21 Pni _ PniIII PniI _ Pnkl
> 0
 .
Thus an has to be smaller than or equal to 1! /2, but that is a contradiction to Lemma 3.1.
o
The proof of the third lemma is similar to the proof of the second one, so we omit it
here.
Lemma 5.3. Let y be a regular curve. Then there exists an e > 0 such that
1. Ipi  pml < Ipi  pklfor pk «o" «r'. if pk <Jpil and Ipi _ pkl ~ Ipl _ pl+ll
for some 1 E {I, ... , IS/},
2. Ipi_pml < Ipipklforpi<Jpm < <Jpk,ifpk I>pi+landlpi_pkl s Ipl_pl+ll
for some 1 E {I, , IS/}
for all samples S = {pI,
, plSI} with e(S) < e and all i E {I, ... ,IS/}.
Now we are prepared to prove the global result for open curves.
Theorem 5.2. Let y be an open curve. Assume
a = sup{L(l(q), r(q)): q E y} < 1!
and let (Sn) be a sequence of samples of y with lim n+ oo e(Sn) = O. Then there exists
N E N such that TSP(Sn) = 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) ¥= P(Sn) for all n E N.
First Step. We show that for large n there exist at least four return points. Because
TSP(Sn) ¥= P (Sn) there have to exist at least three return points. If in S; there is only one
return point besides the start and endpoints of TSP(Sn) , then this point is either min S«
or max S; along y . Assume that there exists N E N such that the number of return points
in S; 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 p~ = min S«.
In TSP(Sn) we also have that p~ is connected to p~, because otherwise S« has more
than three return points. Cut TSP(Sn) into two polygonal arcs Pnl and P; with endpoint
p~. Since S contains only three return points one of these arcs has as its second endpoint
p~Snl. Assume without loss of generality that this arc is always P;. From Lemma 4.2
and the definition of Jordan length we can conclude that limn+>oo L(P;) = L(y). This
implies
lim L(P;) = O.
n+>oo
Let p~, i E {3, ... , ISnl}' be the second endpoint of P;. We observe two things:
1. limn+>oo Ip~  p~ I = 0, because Ip~  p~ I ~ L(P;) for large n.
2. conv{p~, p~+l} C TSP(Sn) for all i < j < ISnl and large n. That is, the shortcuts
take place on the first i indices. Using that TSP(Sn) ¥= P(Sn) and the statement
and notions of Lemma 5.1, with P'(S) = TSP(S), we find that there exists
conv{p:(j) , p:(j+l)} E E' (Sn) and
conv{p~, p~+l} E E(Sn)
with
Ip:(j)  p:(j+l) I ~ Ip~ _ p~+ll,
j, k < i,
7T: (j) ~ k,
k + 1 ~ 7T: (j + 1),
and 7T: (j) ¥= k or k + 1 ¥= 7T: (j + 1).
From limn+>oo e(Sn) = 0 we can conclude that p~ converges to y(O). Since y(O) is a
regular point of y we can apply here Lemmas 5.2 and 5.3. Now Lemma 5.3 tells us that
there exists n E N with:
1. Ip~  p:(j+l) I s Ip:(j)  p:(j+l) I for all n ~ N. That means
for all n ~ N.
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 r~ <Jr; incident
along TSP(Sn) such that the other return points r~ incident to r~ and r; incident to r;
along TSP(Sn) are not in between r~ and r;. That is, one cannot find the following
situation:
(7)
However, it is possible that r~ = r; and r; = r~.
The path P; C TSP(Sn) connecting p~ with p~Snl either contains two return points,
as we are looking for, or it does not contain any return point besides p~ and p~Snl. So
assume that P; does not contain any return point besides p~ and p~Snl. Cut TSP(Sn) into
three paths pn1, P; with endpoint p~ and P~ with endpoint p~Snl. At least one of the the
paths pn2 , P; 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+ oo L(P;) = L(y). This
implies
lim L(P;) = lim L(P;) = O.
n+oo n+oo
Without loss of generality we can assume that P; is not empty for all sufficiently large
n EN. The same reasoning as at the end of the first step shows that this leads to a
contradiction. Hence for large n the sets S; 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. 0
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 recon
struction. 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, respec
tively. 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 y is the set of points in the plane which have more
than one closest point on y. The feature size f (p) of a point p E Y is the distance of
p to the closest point on the medial axis. Amenta et al. define sampling density based
on a parameter e by requiring that each point p E Y has a sample point within distance
ef (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 comer points of the curve. Thus one is required to sample the
curve infinitely near the comers 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 oneskeleton Delaunay
triangulation. Dey and Wenger [
7
] present another algorithm that can reconstruct curves
with sharp comers.
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 NPhard to compute a traveling salesman path or tour, re
spectively. 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
I want to thank Emo Welzl and Nicola Galli for helpful discussions.
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