• @bitjunkie@lemmy.world
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    177 days 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.

      • dream_weasel
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        66 days ago

        Bro, the people here, like the people everywhere, ARE stupid.

        It’s always better to be explicit. I’m one of the stupid people who learned some things reading the comments here and I’ve got a doctoral degree in aero astro engineering.

    • jeff 👨‍💻
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      147 days 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.

      • @bitjunkie@lemmy.world
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        26 days 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.

      • Natanael
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        147 days 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.

    • @sga@lemmings.world
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      47 days ago

      this is regardless of that. The meme explains it a bit wierdly, but we start with 17 squares, and try to find most efficient packing, and outer square’s size is determined by this packing.