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While relocating my wine barrels (which actually are cubic, more or less) I came up with this question : How can $7$ squares be arranged so that the overall perimeter of the whole shape (top-view) is the minimum?

Hopefully, the image below illustrates my question with some examples of guesses:

7_squares

Is it possible to go below $12$ (without squares overlapping of course)?

N. F. Taussig
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Bikay
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    A trivial lower bound by the isoperimetric inequality is $2\sqrt{7\pi}\approx 9.4$. – Benjamin Wang Nov 27 '23 at 12:27
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    Observations: (a) 7 squares fit in a 3x3 square, which has perimeter 4x3 = 12. (b) If you remove squares from a 3x3 squares without removing a full row or column, then you cannot decrease the perimeter, because every removed square exposes at least as many sides as it removes. This proves that your 1st, 3rd, 5th, 8th and 9th attempts couldn't work. Other attempts, which don't fit inside a 3x3 square, are "intuitively worse" because in order to minimise the perimeter your want the shape to be most compact, but I'd need more thought to make that a proof. – Stef Nov 27 '23 at 12:43
  • @BenjaminWang so a perimeter of around 9.4 is possible? Should I code a random-try program to find an arrangement or there's also a clue on how to solve this problem? (I mean what would be the actual arrangement to get that minimum perimeter?) – Bikay Nov 27 '23 at 12:52
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    9.4 is a lower bound, not a minimum. That means we're absolutely sure that we cannot have a perimeter smaller than 9.4. But it's pretty likely that we cannot have a perimeter smaller than 12, either. – Stef Nov 27 '23 at 12:54
  • Oh, got it. Such beautiful details in math. – Bikay Nov 27 '23 at 12:55
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    @Stef I think it is enough to say that in the non-3x3 cases the alignment exposes more sides than it removes, because one square needs as many neighbours as possible. – Dávid Laczkó Nov 27 '23 at 13:27

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