Re: [math-fun] [seqfan] Re: Re: Squares packings
Looks good to me. WFL On 10/13/20, Leonardo Costa <leonardocostalesage@gmail.com> wrote:
Part (A) can be settled by using the same argument as in the original 23x23 problem. The idea is to paint the squares of the grid white and black. To be more precise, all squares in odd columns will be black and all even ones will be white. Observe that there are as many black and white squares in any 2x2 square you place and any 3x3 square you place will have 3 more white squares than black ones or vice versa.
Now suppose we can cover a nxn square (n = 6k+1, 6k-1) with 2x2 and 3x3 pieces. Then, the difference between the amount of black and white squares will be n, but it will also be a multiple of 3, because only 3x3 squares cover a different amount of black and white squares. But 3 doesn't divide n, contradiction.
So if I haven't made any mistakes, I think this problem is settled. Good job everyone!
Leonardo Costa
On Tue, 13 Oct 2020 at 03:39, Fred Lunnon <fred.lunnon@gmail.com> wrote:
Doh --- what's worse, JG had just beaten me to making the the exact same remark myself!
WFL
On 10/13/20, Rob Pratt <robert.william.pratt@gmail.com> wrote:
(B) is already settled by Jack Grahl’s argument.
On Oct 12, 2020, at 9:11 PM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
I don't know --- I was hoping that we have some seasoned packer of squares on hand who can tell us what might be known.
In the meantime, just to set up a formal aunt Sally: Conjecture: For natural n mod 6 in {1, 5} : (A) no packing of a n x n box by 1 x 1, 2 x 2, 3 x 3 tiles exists with no 1 x 1 tile; (B) there exists some packing with only a single 1 x 1 tile.
WFL
On 10/13/20, rcs@xmission.com <rcs@xmission.com> wrote: Is it established that large 6k+-1 squares require any 1x1s? --Rich
----- Quoting Fred Lunnon <fred.lunnon@gmail.com>:
<< Sounds good for the upper bound. I have also confirmed the lower bound up through n = 100. >>
Unclear: does your program establish that _NO_ solution exists without some 1x1 tile for edge length n mod 6 in {1, 5} & n < 100 ?
If so, this suggests an interesting conjecture ... WFL
On 10/12/20, Rob Pratt <robert.william.pratt@gmail.com> wrote: > Sounds good for the upper bound. I have also confirmed the lower bound > up > through n = 100. > > On Mon, Oct 12, 2020 at 3:06 AM Jack Grahl <jack.grahl@gmail.com> > wrote: > >> If we have a solution with 1 small square for n=11 and n=13, >> doesn't >> that >> imply a solution with 1 small square for all larger numbers 6k+1 >> and >> 6k-1 >> (ie all for which the solution is greater than zero)? >> >> Simply form an L-shape with width 6, to extend a solution for 6k+1 to >> a >> solution for 6(k+1)+1. The L shape is made up of a 6x6 square, and two >> 6x(6k+1) strips. Since any number >1 can be formed by a sum of 2's and >> 3's, >> we can always make the strips. Same for 6k-1 of course. >> >> Jack Grahl >> >> On Sun, 11 Oct 2020, 20:25 Rob Pratt, < robert.william.pratt@gmail.com> >> wrote: >> >>> Via integer linear programming, I get instead >>> 1 0 0 0 4 0 3 0 0 0 1 0 1 0 0 0 1 0 1 0 0 0 1 >>> >>> Here's an optimal solution for n = 17, with the only 1x1 appearing in >> cell >>> (12,6): >>> 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 >>> 1 2 2 3 3 38 38 38 4 4 5 5 6 6 7 7 8 8 >>> 2 2 2 3 3 38 38 38 4 4 5 5 6 6 7 7 8 8 >>> 3 9 9 10 10 38 38 38 11 11 12 12 13 13 14 14 15 15 >>> 4 9 9 10 10 39 39 39 11 11 12 12 13 13 14 14 15 15 >>> 5 16 16 17 17 39 39 39 18 18 19 19 40 40 40 41 41 41 >>> 6 16 16 17 17 39 39 39 18 18 19 19 40 40 40 41 41 41 >>> 7 42 42 42 20 20 21 21 22 22 23 23 40 40 40 41 41 41 >>> 8 42 42 42 20 20 21 21 22 22 23 23 24 24 25 25 26 26 >>> 9 42 42 42 27 27 43 43 43 44 44 44 24 24 25 25 26 26 >>> 10 45 45 45 27 27 43 43 43 44 44 44 28 28 29 29 30 30 >>> 11 45 45 45 31 31 43 43 43 44 44 44 28 28 29 29 30 30 >>> 12 45 45 45 31 31 1 46 46 46 47 47 47 32 32 48 48 48 >>> 13 33 33 34 34 35 35 46 46 46 47 47 47 32 32 48 48 48 >>> 14 33 33 34 34 35 35 46 46 46 47 47 47 36 36 48 48 48 >>> 15 49 49 49 50 50 50 51 51 51 52 52 52 36 36 53 53 53 >>> 16 49 49 49 50 50 50 51 51 51 52 52 52 37 37 53 53 53 >>> 17 49 49 49 50 50 50 51 51 51 52 52 52 37 37 53 53 53 >>> >>> Clearly, even 2n can be partitioned into 2x2 only, and 3n can be >>> partitioned into 3x3 only. Otherwise, it appears that the minimum is >>> 1 >> for >>> n >= 11. >>> >>> On Sun, Oct 11, 2020 at 3:11 AM <michel.marcus@free.fr> wrote: >>> >>>> >>>> Hello Seqfans, >>>> >>>> >>>> "Images du CNRS" French site 4th problem for September 2020 is: >>>> . >>>> What is the minimum number of 1X1 pieces to fill a 23X23 square >>>> with >> 1X1, >>>> 2X2, and 3X3 pieces. >>>> Problem is at >>>> http://images.math.cnrs.fr/Septembre-2020-4e-defi.html. >>>> Solution is at >>>> http://images.math.cnrs.fr/Octobre-2020-1er-defi.html >>>> (click in Solution du 4e défi de septembre) >>>> >>>> >>>> I wondered what we get for other squares and painstakingly obtained >>>> for >>>> n=1 up to n=23: >>>> 1 0 0 0 4 0 3 0 0 0 1 0 1 0 0 0 4 0 4 0 0 0 1 >>>> >>>> >>>> Do you see some mistakes? Is it possible to extend it? >>>> >>>> >>>> Thanks. Best. >>>> MM >>>> >>>> -- >>>> Seqfan Mailing list - http://list.seqfan.eu/ >>>> >>> >>> -- >>> Seqfan Mailing list - http://list.seqfan.eu/ >>> >> >> -- >> Seqfan Mailing list - http://list.seqfan.eu/ >> > > -- > Seqfan Mailing list - http://list.seqfan.eu/ >
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Fred Lunnon