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It is known that squares with side 1-24 can't quite pack in a side 70 square. However, use four copies of each and a side 140 square is possible, as shown by Erich Friedman.

squares 1-24, four times

I just noticed that squares 1-24, each twice, might fit into a side 99 square with one empty space. Is that possible?

Here are a few similar problems, usually with 1 empty space. I'd also be interested in solutions for $(n+1)(n-1)$ rectangles.

  1. Squares 1-6 × 8 in a side 27 square - 1.
  2. Squares 1-7 × 6 in a side 29 square - 1.
  3. Squares 1-8 × 5 in a side 32 square - 4. -- Solvable
  4. Squares 1-8 × 6 in a side 35 square - 1.
  5. Squares 1-10 × 3 in a side 34 square - 1.
  6. Squares 1-11 × 4 in a side 45 square - 1.
  7. Squares 1-12 × 4 in a side 51 square - 1.
  8. Squares 1-15 × 3 in a side 61 square - 1.
  9. Squares 1-13 × 5 in a side 64 square - 1.
  10. Squares 1-16 × 3 in a side 67 square - 1.
  11. Squares 1-23 × 2 in a side 93 square - 1.
  12. Squares 1-24 × 2 in a side 99 square - 1. -- opening problem
  13. Squares 1-47 × 1 in a side 189 square - 1.
  14. Squares 1-48 × 1 in a side 195 square - 1.
  15. Squares 1-57 × 3 in a side 436 square - 1.
  16. Squares 1-57 × 7 in a side 666 square - 1.

Do any of these have solutions?

Closely related: Balanced Consecutive Tilings.

Vaguely related -- I've updated solutions for Mrs. Perkins's Quilt up to size 40000.

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    $\begingroup$ if nothing else this got me looking at how to make color in MathJax turns out it's $\style{background-color:red}{\begin{matrix} & &\\& &\end{matrix}}$ etc. $\endgroup$
    – user451844
    Commented Jul 7, 2017 at 19:58

5 Answers 5

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Corrected Bouwkamp code (may or may not agree with pictured) Computer solutions for problems 1-10.
Brute force, greedy backtracking algorithm:

Problem #1: 49 27 27 (6,6,6,6,3)(3)(6,6,6,6,3)(3)(5,5,5,5,5,2)(2)(2)(5,4,4,4,4,1,1,1,1)(5,1)(1)(1)(4,4,3,3,2)(1)(5)(1)(4,4)(3,3)(2,2,2,2)

Problem #2: 43 29 29 (7,7,7,7,1)(1)(1)(1)(1)(1)(1)(7,7,6,6,3)(3)(6,6,3)(6,4,4)(3)(2,6)(5,3,3,4)(4,4)(2,2,2)(4)(5,5,5,5,5)(2,2)

Problem #3: 44 32 32 (8,8,8,8)(8,7,7,7,3)(3)(3)(1,1,1,1,1,1,1,7,5,2)(6,6,3)(5)(3)(5)(6,6,6,2,2)(5)(2,1,1)(7)(2)(5)(4,4,4,4,4)

Problem #4: 49 35 35 (8,8,8,8,3)(3)(3)(8,8,7,7,2)(3)(2)(5)(1,1,1,1,1,1,6,2)(6,4,7,5)(7)(4)(2,6,3)(6)(5,4)(4)(5,5)(1,3,4,2)(7,7,2)(6,6)(5)(4)

Problem #5: 31 34 34 (10,10,9,5)(5)(1,1,1,5,1)(7,7,7,2)(6)(2)(3,4)(2,4)(9,9,6)(6)(4)(3,3)(10)(8,8,8)

Problem #6: 45 45 45 (11,11,11,10,2)(2)(2)(2)(1,1)(1,1)(1,11)(9,9,8,8)(9,7)(10,8)(5,6)(5,7)(6)(7,7,3)(10)(8)(9,4)(10,4)(4)(6,6)(5,5)(4)(3,3,3)

Problem #7: 49 51 51 (12,12,12,12,3)(3)(3)(3)(11,11,11,11,7)(7)(10,10,10,8,6)(7)(6)(8)(9,9,8,4)(7)(6)(4)(2,4,2)(1,6)(10,2,2)(5,4)(9,9)(8)(1,1,1,1)(5,5,5)

Problem #8: 46 61 61 (15,15,15,12,4)(4)(4)(3,13)(14,14,14,6)(6)(13)(2,2,2)(13,10,7,9,9)(7)(12,9,10)(10)(8,5)(7)(3,1,1)(3,6)(1,1)(5,5)(12)(11)(11,11)(8,8)

Problem #9: 66 64 64 (13,13,13,13,12)(12)(13,12,12,12,3)(3)(3)(3)(1,11)(1,1,1,1,1,11,11,9,4)(9,9)(4)(4)(2,7)(8,10)(11)(6,7,11)(4)(6,5)(6)(8)(5,2)(10)(2)(8,8)(5,7,7)(7,4)(8)(3,6)(10,10,10)(2,9,9)(2)(6)(5,5)

Problem #10: 49 67 67 (16,16,16,15,4)(4)(4)(3,1)(1)(1)(1,8,10)(13,14,11,11)(6,2)(12)(10,10,2)(13)(8)(14)(7,9,9,15)(13)(11,8,2)(6,14)(3,6,5)(15)(12,12,3)(5)(9)(7,7)(5)

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  • $\begingroup$ Can you explain your notation? $\endgroup$
    – Joffan
    Commented Jan 23, 2018 at 2:59
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    $\begingroup$ squaring.net/downloads/downloads.html#Bouwkampcode $\endgroup$
    – Brian Trial
    Commented Jan 23, 2018 at 9:20
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    $\begingroup$ My first solution above should be flipped horizontally to match the "ordered by decreasing height" criterion for the Bouwkamp Code I listed. My groups are ordered by increasing height, not decreasing height. $\endgroup$
    – Brian Trial
    Commented Jan 23, 2018 at 9:28
  • $\begingroup$ Can you double check the code you posed and see whether there are any typos? I have tried the code you have for Problem #2 - #5 and I cannot reconstruct squares from anyone of them. The code you posted in another answer is prefectly fine. If there aren't any typos, can you post a picture for say Problem #2 so that I can figure out what I'm missing in my reconstruction. Thanks ahead. $\endgroup$
    – achille hui
    Commented Jan 25, 2018 at 6:34
  • $\begingroup$ oops, the problem is on my end (bug): The groups are correct and represent real solutions, but the are not necessarily in the right order. Try problem 2 as (maybe?) 43 29 29 (7,7,7,7,1)(1)(1)(1)(1)(1)(1)(7,7,6,6,3)(3)(6,6,3)(6,4,4)(3)(2,6)(3,3,4,5)(4,4)(2,2,2)(4)(5,5,5,5)(5)(2,2) I'll post a picture of this one too. $\endgroup$
    – Brian Trial
    Commented Jan 26, 2018 at 0:54
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Answer to problem #1 : 49 27 27 (6,6,6,6,1,1,1)(2,1)(1)(2,1)(1)(3)(6,6,4,4,4)(3)(4,4,4)(3)(6,6)(3,3,3,3,3)(5,5,5)(2,2,4,4)(2,2)(2,5,5,5,5,5)(2)(1,1) Hand solution using Visio

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Answer to Problem #2 (1-7)x6 in 43x43Answer to Problem #2 (1-7)x6 in 43x43

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Problem 6: Squares 1-11 × 4 in a side 45 square - 1. Problem 6: Squares 1-11 × 4 in a side 45 square - 1.

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Problem #9 Solution

Problem #9: 66 64 64 (13,13,13,13,12)(12)(13,12,12,12,3)(3)(3)(3)(1,11)(1,1,1,1,1,11,11,9,4)(9,9)(4)(4)(2,7)(8,10)(11)(6,7,11)(4)(6,5)(6)(8)(5,2)(10)(2)(8,8)(5,7,7)(7,4)(8)(3,6)(10,10,10)(2,9,9)(2)(6)(5,5)

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