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submitted 1 week ago by [email protected] to c/[email protected]
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[-] [email protected] 16 points 1 week ago

It's important to note that while this seems counterintuitive, it's only the most efficient because the small squares' side length is not a perfect divisor of the large square's.

[-] [email protected] 9 points 1 week ago

What? No. The divisibility of the side lengths have nothing to do with this.

The problem is what's the smallest square that can contain 17 identical squares. If there were 16 squares it would be simply 4x4.

[-] [email protected] 14 points 1 week ago

He's saying the same thing. Because it's not an integer power of 2 you can't have a integer square solution. Thus the densest packing puts some boxes diagonally.

[-] [email protected] 2 points 1 week ago

And the next perfect divisor one that would hold all the ones in the OP pic would be 5x5. 25 > 17, last I checked.

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this post was submitted on 01 Jul 2025
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