[seqfan] Re: How many squares can you make from n points in the plane?

[Neil Sloane]

I can answer Benoit's first question, at least.

Given a configuration of M squares on n points, form the M x n incidence
matrix.
There are M rows, each has 4 ones, and any two rows can meet in at most 2
ones.

That is exactly a binary constant weight code of length n containing M
codewords,
and the minimum Hamming distance between rows is at least 4.

The codewords are the rows.



Best regards
Neil

Neil J. A. Sloane, Chairman, OEIS Foundation.
11 South Adelaide Avenue, Highland Park, NJ 08904, USA.
Also Visiting Scientist, Math. Dept., Rutgers University, Piscataway, NJ.
Phone: 732 828 6098; home page: http://NeilSloane.com
Email: njasloane at gmail.com



On Thu, Sep 30, 2021 at 2:18 PM Peter Munn <techsubs at pearceneptune.co.uk>
wrote:

> On Thu, September 30, 2021 6:38 am, Neil Sloane wrote:
> > This is just about a(n), A051602, where the n points are restricted to
> the
> > integer lattice Z x Z.
> [...]
> >
> > Is there a clever argument that says that if n = m^2, the square mXm grid
> > arrangement  is optimal?
>
> On the contrary, it isn't for m > 6 (probably also m = 6, I need to double
> check). De-select a grid point from a corner of the square and select
> instead a point just beyond the middle of an opposite side. For m = 7, I
> reckon this increases the square count by 3, and I think another 2 is
> possible by repeating this for another corner. For m = 9, I suspect the
> optimal arrangement is achieved when this adjustment is enacted for all 4
> corners.
>
> Best regards,
>
> Peter
>
>