• bitjunkie@lemmy.world
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    3 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.

    • jeff 👨‍💻@programming.dev
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      3 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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        2 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.