2010
DOI: 10.1007/978-3-642-11805-0_3
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Drawing Hamiltonian Cycles with No Large Angles

Abstract: Let n ≥ 4 be even. It is shown that every set S of n points in the plane can be connected by a (possibly self-intersecting) spanning tour (Hamiltonian cycle) consisting of n straight line edges such that the angle between any two consecutive edges is at most 2π/3. For n = 4 and 6, this statement is tight. It is also shown that every even-element point set S can be partitioned into at most two subsets, S 1 and S 2 , each admitting a spanning tour with no angle larger than π/2. Fekete and Woeginger conjectured t… Show more

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Cited by 5 publications
(7 citation statements)
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References 12 publications
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“…The improvement is based on Lemma 2.4 below. This lemma -in fact a stronger version, stating that any two opposite cones defined by the three concurrent lines contain the same number of points-has been proved for even n by Dumitrescu et al [11]. Our proof of Lemma 2.4 is similar to their proof.…”
Section: (I)supporting
confidence: 57%
See 1 more Smart Citation
“…The improvement is based on Lemma 2.4 below. This lemma -in fact a stronger version, stating that any two opposite cones defined by the three concurrent lines contain the same number of points-has been proved for even n by Dumitrescu et al [11]. Our proof of Lemma 2.4 is similar to their proof.…”
Section: (I)supporting
confidence: 57%
“…We give it because we also need it for odd n, and because we will need an understanding of the proof to describe our algorithm for computing the concurrent triple in the lemma. Our algorithm will run in O(n log 2 n) time, a significant improvement over the O(n 4/3 log 1+ε n) running time obtained (for the case of even n) by Dumitrescu et al [11].…”
Section: (I)mentioning
confidence: 71%
“…The next lemma shows that this is indeed the case. The lemma-in fact, a stronger version, stating that any two opposite cones defined by the three concurrent lines contain the same number of pointshas been proved for even n by Dumitrescu et al [11]. Our proof of Lemma 2.5 is similar to their proof.…”
Section: Proof Consider a Votersupporting
confidence: 55%
“…We give it because we also need it for odd n, and because we will need an understanding of the proof to describe our algorithm for computing the concurrent triple in the lemma. Our algorithm will run in O (n log n) time, a significant improvement over the O (n 4/3 log 1+ε n) running time obtained (for the case of even n) by Dumitrescu et al [11]. Lemma 2.5.…”
Section: Proof Consider a Votermentioning
confidence: 80%
“…Note that the ζ-median and the (−ζ)-median of V are the same, so that L V,ζ and L V,−ζ are the same line, but with opposite directions. These considerations lead to the following useful observation, first proved in this precise form in [21, Lemma 3] (see also [22,Lemma 2] and [14,Lemma 2]). Lemma 8.2.…”
Section: Equilateral Triangles In the Planementioning
confidence: 91%