[seqfan] Re: [math-fun] Re: Squares packings

Rob Pratt robert.william.pratt at gmail.com
Tue Oct 13 03:35:22 CEST 2020


(B) is already settled by Jack Grahl’s argument.

> On Oct 12, 2020, at 9:11 PM, Fred Lunnon <fred.lunnon at 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 at xmission.com <rcs at xmission.com> wrote:
>> Is it established that large 6k+-1 squares require any 1x1s?  --Rich
>> 
>> -----
>> Quoting Fred Lunnon <fred.lunnon at 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 at 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 at 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 at 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 at 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/
>>>>>>> 
>>>>>> 
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