Avoiding Colinear Points

[Kennedy]

There is no solution to the n-queens problem for n=2 or n=3.
Besides, superimposing two n-queens solutions does not always yield 2n 
points.
For example, here are two 8-queens solutions and their superposition, which 
only has 13 distinct points.

--*-----
-----*--
-*------
------*-
*-------
---*----
-------*
----*---


--*-----
-----*--
-------*
-*------
---*----
*-------
------*-
----*---

Superimposed (13 points):

--*-----
-----*--
-*-----*
-*----*-
*--*----
*--*----
------**
----*---

----- Original Message ----- 
From: "Jud McCranie" <j.mccranie at adelphia.net>
To: "Leroy Quet" <qq-quet at mindspring.com>
Cc: "Leroy Quet" <qq-quet at mindspring.com>; <seqfan at ext.jussieu.fr>
Sent: Thursday, October 21, 2004 3:54 PM
Subject: Re: Avoiding Colinear Points


> At 04:44 PM 10/21/2004, Leroy Quet wrote:
>
>>I describe a sequence and game based on the following idea.
>>
>>If we have a rectangular grid of n lattice-points by n lattice-points,
>>how many dots can we place at the lattice-points, at most one dot per
>>lattice-point, so that no 3 or more dots are colinear (in any direction)?
>>
>>I get the sequence:
>>1, 4, 6, 8, 10,..
>>And Richard Mathar has extended it:
>>12, 14, (15 or 16),...
>>
>>The upper bound on each term is obviously 2n, since we can have at most 2
>>dots in each row  and column.
>
> There is an easy approach to this for nxn.  If you superimpose any two 
> different n-queens solutions, you get a solution with no three colinear 
> points.  Therefore 2n is always possible (for n > 1).
>
>
>