Curve Reconstruction, the Traveling Salesman Problem, and Menger’s Theorem on Length

Discrete & Computational Geometry, Aug 2000

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.

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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 , CH-8092 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 x-axis 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 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. 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 s-regular if it is left and right regular in all points y(t), t E [ 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 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 s-regular and every s-regular 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 Y-1 (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}, ISI-2 C(P(S)) = L IL(pi+l - pi, pi+2 _ pi+l)l, ISI-l 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 one-dimensional subspaces z E JRd+ I . The collection of all one-dimensional subspaces of JRd+l forms the d-dimensional 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 d-dimensional 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 s-regular 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 one-dimensional 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 (y-l (Pn)), (y-l (qn)), and (y-l (rn)) converge to y-l (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 one-dimensional subspaces of JRd+l. In the following let y be s-regular 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: Ip-ql < 17} which contains p is denoted by Bl1(p ). We assume that every one-dimensional 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 everyone-dimensional 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 one-dimensional 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 one-dimensional 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 d-dimensional subspace of JRd+l and let E C JP>d be the set of all one-dimensional 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 (1-14'+-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)) < !(Tt-a) + (~_ Tt-a) 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 s-regularity 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 s-regularity 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 two-dimensional 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)}, nl-i+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 (T-l 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 one-dimensional 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 one-dimensional 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 one-dimensional 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 = nl-i-m..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,y-l(r2)]) + L(YI[y-l(r1),y-l(r2) ] ) + L(YI[y-l(r1),1]) = L(y) + 2L(YI[y-l(r1),y-l(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)} t-1> 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 s-regular 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 - pi-II < Ipi - pkl if pk <J pi-I, 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 pi-l 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 pi-l _ pkl2 n n n n n n 21 Pni _ Pni-III Pni-I _ 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 <Jpi-l and Ipi _ pkl ~ Ipl _ pl+ll for some 1 E {I, ... , IS/}, 2. Ipi_pml < Ipi-pklforpi<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 one-skeleton 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 NP-hard 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. 1. A.D. Aleksandrov , Yu.G. Reshetnyak, Integral Curvature of a Curve in n-Dimensional Euclidean Space , Siberian Math. J. 29 ( 1 ) ( 1988 ), 1 - 16 . 2. E. Althaus , K. Mehlhorn , TSP-Based Curve Reconstruction in Polynomial Time , Proc. 11th SIAM Symp. on Discrete Algorithms , pp. 686 - 695 ( 2000 ). 3. E. Althaus , K. Mehlhorn , S. Naber , S. Schirra , Experiments on Curve Reconstruction, Proc. ALENEX 2000 ( 2000 ). 4. N. Amenta , M. Bern, D. Eppstein, The Crust and the fj-Skeleton: Combinatorial Curve Reconstruction, Graph . Models Image Process. 60/2 ( 2 ) ( 1998 ), 125 - 135 . 5. T.K. Dey , P. Kumar , A Simple Provable Algorithm for Curve Reconstruction , Proc. 10th SIAM Symp. on Discrete Algorithms , pp. 893 - 894 ( 1999 ). 6. T.K. Dey , K. Mehlhorn , E.A. Ramos , Curve Reconstruction: Connecting Dots with Good Reason , Proc. 15th ACM Symp. on Computational Geometry , pp. 197 - 206 ( 1999 ). 7. T.K. Dey , R. Wenger , Reconstructing Curves with Sharp Corner , Proc. 16thACM Symp. on Computational Geometry , pp. 233 - 241 ( 2000 ). 8. KJ. Falconer, The Geometry of Fractal Sets , Cambridge University Press, Cambridge ( 1985 ). 9. G.B. Folland , Real Analysis , Modern Techniques and Their Applications , 2nd edn., Wiley-Interscience, New York ( 1999 ). 10. U. Adamy, 1 . Giesen, M. John , Surface Reconstruction Using Umbrella Filters, submitted to Comput. Geom. Theory Appl . ( 2000 ). 11. C. Gold, Crust and Anti-Crust : A One-Step Boundary and Skeleton Extraction Algorithm , Proc. 15th ACM Symp. on Computational Geometry , pp. 189 - 196 ( 1999 ). 12. K. Menger (ed.), Ergebnisse eines Mathematischen Kolloquiums 2, Kolloquium 5 . 11 . 1930 , Teubner, Leipzig ( 1932 ). 13. K. Menger , Untersuchungen tiber eine allgemeine Metrik . Vierte Untersuchung. Zur Metrik der Kurven , Math. Ann. 103 ( 1932 ), 467 - 501 . 14. Yu .G. Reshetnyak, Some Applications of Integral Geometry to the Theory of Curves of Finite Rotation , Siberian Math. J. 29 ( 1 ) ( 1988 ), 109 - 116 .


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J. Giesen. Curve Reconstruction, the Traveling Salesman Problem, and Menger’s Theorem on Length, Discrete & Computational Geometry, 2000, 577-603, DOI: 10.1007/s4540010061