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Let $T_{\mathrm{MIN}}$ and $T_{\mathrm{MAX}}$ denote the shortest, resp. longest Hamilton cycle through a set of $n=2k+1$ points.

Let further $S_{\mathrm{MIN}}$ and $S_{\mathrm{MAX}}$ be the "antipodal" Hamilton cycles related to $T_{\mathrm{MIN}}$ and $T_{\mathrm{MAX}}$.

If $\left(\pi(1),\,\dots,\,\pi(n=2k+1)\right)$ is the order in which the points are encountered on a Hamilton cycle, then $\left(\pi'(1),\,\dots,\,\pi'(n\right)$ with $\pi'(i)=\pi(i+(i-1)*k)$ is the order in which they are encountered on the related antipodal Hamilton cycle with additions modulo $n$.

Questions:

  • what does $\frac{\left(\left|T_{\mathrm{MIN}}\right|-\left|S_{\mathrm{MIN}}\right|\right)\ +\ \left(\left|T_{\mathrm{MAX}}\right|-\left|S_{\mathrm{MAX}}\right|\right) }{2}$ tell about the point set given that the measure becomes 0 for regular $(2k+1)$-gons?
  • what is the maximal value of the measure, say depending for points from Euclidean spaces, depending on dimension and distance measure?
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  • $\begingroup$ @SamHopkins thank you for pointing me to the error; Ihave edited my question accordingly. $\endgroup$
    – Manfred Weis
    Commented Jul 10, 2021 at 4:05
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    $\begingroup$ @SamHopkins no, you start at the first point on the "original" tour and then move ahead $k$ tour edges to get the next point on the antipodal tour, or put differently $\pi'(i)=\pi(i+(i-1)*k)$ $\endgroup$
    – Manfred Weis
    Commented Jul 10, 2021 at 4:16

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