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Let $C=\{ c_1, c_2, \cdots,c_n \}$ be a set of $n$ alternatives and $T$ be the set of all strict complete orderings on $C$. For any two $t_1$ and $t2$ in $T$, their (Kendal-tau) distance $d(t_1, t_2)$ is defined as the number of pairwise disagreements between $t_1$ and $t_2$.

My Question: How to find $k$ (much smaller than $n!$) different elements from $T$ such that they are "evenly ditributed" in $T$ with respect to this (Kendal-tau) distance $d$?

For example, the k+1 elements $0, 1/k, 2/k, \cdots, (k-1)/k, 1$ are evenly distributed in the interval [0,1].

Joe Zhou
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  • $k=n$ cyclic permutations have mutual distance $n$. -- Whenever $n=a+b$ with $a,b>0$, cyclic permutations of the first $a$ and of the last $b$ elements give us $k=ab$ elements such that each has (several) nearest neighbours at distance $\min{a,b}$. – Hagen von Eitzen Aug 21 '21 at 13:45
  • Thank you so much. So these n cyclic permutations (with mutual distance n) are evenly distributed in the set of all permutations? – Joe Zhou Aug 21 '21 at 15:19
  • Can we simply apply the k-medoids algorithm? – Joe Zhou Aug 22 '21 at 00:50

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