Drawing Hamiltonian Cycles with no Large Angles

Abstract

Let $n \geq 4$ 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 $2\pi/3$. 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 $\pi/2$. Fekete and Woeginger conjectured that for sufficiently large even $n$, every $n$-element set admits such a spanning tour. We confirm this conjecture for point sets in convex position. A much stronger result holds for large point sets randomly and uniformly selected from an open region bounded by finitely many rectifiable curves: for any $\epsilon>0$, these sets almost surely admit a spanning tour with no angle larger than $\epsilon$.

Bounded-Angle Spanning Tree: Modeling Networks with Angular Constraints

“…Fekete and Woeginger also conjectured that for every set of 2k ≥ 8 points there exists a Hamiltonian cycle, such that all its angles are bounded by π/2. Recently, Dumitrescu et al [11] showed how to construct a Hamiltonian cycle whose angles are bounded by 2π/3. As for lower bound, in [9] and, independently, in [11] it is shown that, for any ε > 0, there exists a set of points, for which any Hamiltonian path has an angle greater than π/2 − ε.…”