Minimal number of different distances

[franktaw at netscape.net]

For n=12, we have:

 X X X
X X X X
 X X X
  X X

with distances 1, sqrt(3), 2, sqrt(7), and 3.

For n=19, it's:

  X X X
 X X X X
X X X X X
 X X X X
  X X X

with distances 1, sqrt(3), 2, sqrt(7), 3, sqrt(12), sqrt(13), and 4.

All distances for a hexagon can be measured from a corner.  The 
distances are of three types:

* "Orthogonal" distances, which each occur 3 times:
  + X X
 X X X X
O X + X X
 X X X X
  + X X

* "Diagonal" distances, which each occur twice:

  X X +
 X X X X
O X X X X
 X X X X
  X X +

* "Other" distances, which occur 4 times:

  X + X
 X X X +
O X X X X
 X X X +
  X + X

As the size of the hexagon increases, the "other" distances 
predominate, giving the asymptotic upper bound of n/4 + o(n).

Franklin T. Adams-Watters

-----Original Message-----
From: Dan Dima <dimad72 at gmail.com>

   Frank can you share both your distributions for a(19) <= 8 & a(12) <= 
5 (maybe some graphics will help). I can't figure either if there is a 
smaller n such that a(n) < floor(n/2).
  At the first sight it seems to me that hexagonal grids with n points 
are not better than the regular n-gon.
  As I have seen on http://schoolar.google.com Erdos conjectured that 
a(n) = floor(n/2).
Sorry for using non-standard notations, I will follow the rules next 
time.

 Regards,
 Dan


  On 5/3/06, franktaw at netscape.net <franktaw at netscape.net> wrote:     I 
will note that a_2(n) is not always floor(n/2); in particular, 19 
points in a regular hexagonal pattern have only 8 distances and 12 
points in a slightly irregular hexagon have only 5 distances. I think 
hexagonal grids give a bound of n/4 + o(n) on a_2(n). (It may be lower, 
since there are non-trivial solutions to a^2+ab+b^2 = c^2+cd+d^2; I'm 
not immediately able to estimate their density.)

 Franklin T. Adams-Watters







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