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Suppose the original natural numbers are sorted as 1, 2, ..., N. The distances of two neighbors are 1. Is there any method to reorder the natural number list to maximize the distance of ALL neighbors?

Furthermore, is it possible to find the list to maximize all distances between ith and (i+j)th numbers, where 1<=i<=N, 1<=j<=N, and i+j<=N?

Thanks, Tang Laoya

Tang Laoya
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    What do you mean by maximizing the distance of all neighbours? Do you mean maximize the minimum of the distances, or maximize the sum of the distances, or ??? – Robert Israel Mar 02 '14 at 03:15
  • thanks. i mean maximize ALL the distance between elements i, i+1, elements i, i+2, ..., elements i, i+j, where 1<=i<=N, 1<=j<=N, and i+j<=N. – Tang Laoya Mar 02 '14 at 03:23
  • You can't maximize ALL the distances. To maximize the distance between $i$ and $i+1$, put one of them at the start of the list and the other at the end. But then you're not maximizing the distance between any other pair. – Robert Israel Mar 02 '14 at 07:29
  • thanks. i wish to get an opposite distance distribution compare to the list 1, 2, ..., N. For the list 1, 2, ..., N, the distances between i and i+j are larger when j is larger. Now I wish that the distance between i and i+j are smaller when j is larger. Is it possible to find such a list? – Tang Laoya Mar 02 '14 at 08:33

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