<< 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
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