[seqfan] Re: Improved lower bound for A250000

[Bob Selcoe]

Hi Benoit,

Excellent!  This certainly is an improvement.

But I'm having a little difficulty following how you've obtained the upper 
and lower bounds, as well as your description of coordinates.

So for my n=24 a(n)=83 board, wouldn't you also say 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, while the queens of the other color are 
obtained by central symmetry (or at least something approximating central 
symmetry)???

So I don't see how the definitions between your and my configurations 
differ; that is, when you say:

>> 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).

I'm not sure what you "equalized" to gain the improvement.

But still, there is definite improvement.

I probably won't have time to look into a structural pattern for n = 4m for 
several days; my guess is one exists based on your approach, which will 
improve upon mine.  Will you have a chance to see if such a pattern exists?

Best Wishes,
Bob Selcoe

--------------------------------------------------
From: "Benoît Jubin" <benoit.jubin at gmail.com>
Sent: Tuesday, February 24, 2015 11:25 AM
To: "Sequence Fanatics Discussion list" <seqfan at list.seqfan.eu>
Subject: [seqfan] Re: Improved lower bound for A250000

> 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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