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

Tue Oct 13 09:26:26 CEST 2020

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 at 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 at gmail.com> wrote:
> > (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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> >>>>>>
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> >>>>>
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> >>>>>
> >>>>
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