[seqfan] Re: Improved lower bound for A250000

[Benoît Jubin]

Dear Rob,
I improved by one the final board on the webpage by using the same idea as
for my asymptotic bound, but you are right that in this specific case, the
result is not centrally symmetric. From the last example on the webpage, I
made marginal changes along the diagonals (in the [0,1]^2 coordinates)
y=x+1/3 (deleted 3 Ws, added 5 Bs) and y=x-1/3 (deleted 4 Bs, added 4Ws):

------------------------

......WWWWWW............

......WWWWWW...........W

......WWWWWW..........WW

......WWWWWW.........WWW

......WWWWWW........WWWW

......WWWWW........WWWWW

.......WWW........WWWWWW

........W.........WWWWWW

..................WWWWWW

..................WWWWW.

..................WWWW..

..................WWW...

....BB..................

...BBB..................

..BBBB..................

.BBBBB..................

BBBBBB..........B.......

BBBBBB.........BBB......

BBBBBB........BBBB......

BBBBB........BBBBB......

BBBB........BBBBBB......

BBB.........BBBBBB......

BB..........BBBBBB......

B...........BBBBBB......

------------------------



On Tue, Feb 24, 2015 at 6:13 PM, Rob Pratt <Rob.Pratt at sas.com> wrote:

> Please share your n =24 solution.  Under the central symmetry constraint,
> I get a maximum of 80, not 84.
>
> -----Original Message-----
> From: SeqFan [mailto:seqfan-bounces at list.seqfan.eu] On Behalf Of Benoît
> Jubin
> Sent: Tuesday, February 24, 2015 11:50 AM
> To: Sequence Fanatics Discussion list
> Subject: [seqfan] Improved lower bound for A250000
>
> Dear seqfans,
>
> Let a(n)=A250000(n) be the maximum size of coexisting armies of queens on
> an (n,n) chessboard.
>
> By modifying the Pratt--Selcoe configuration, I improved the best known
> lower bound from
> a(n) > (9/4)*(n/4)^2
> to
> a(n) > (7/3)*(n/4)^2.
> I have been sloppy with side effects, but to be on the safe side, let's say
> a(n) > (7/3)*(floor(n/4))^2 - (3+8*sqrt(2)/3)*ceiling(n/4), where the
> coefficient 3+8*sqrt(2)/3 is a perimeter that you can compute from the
> following description.
>
> The configuration in the limit n = infinity is as follows: denoting by x,y
> in [0,1] the coordinates on the chessboard, the queens of one color are in
> the two regions x<1/4, y<1/2, x<y<x+1/3 and 1/2<x<3/4, y<x-1/3, y<1-x and
> the queens of the other color are obtained by central symmetry. As you can
> guess, I obtained these coefficients by equalizing the lengths of the
> "opposite" boundaries of the armies (this already improves (by 1) on the
> "Board 4" example of the webpage).
>
> Using an easy upper bound, one has asymptotically
> (2+1/3)*(n/4)^2 < a(n) < 4*(n/4)^2.
> Anyone to help fill the gap?
>
> Best,
> Benoit
>
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