• datelmd5sum@lemmy.world
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    10 months ago

    I mean it makes sense when you think about how the circles arrange in an infinte square and e.g. 4r square. There has to be some fuckery between the perfect packing and the small square packing. You can see a triangle of almost perfect packing in the middle of the 49 circle square, surrounded by fault lines in the structure and then some more good packing, and garbage in the bottom.

    slightly related Steve Mould video

    • Maggoty@lemmy.world
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      10 months ago

      Or, they could do 6x8 with one obviously extra at the end. But this is a funny not a rational thing.

    • intensely_human@lemm.ee
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      10 months ago

      Well-put. One perfect pattern at one scale, another perfect pattern at a different scale, and then there has to be a transition between them of optimal steps along the way. I like that.

  • hsdkfr734r@feddit.nl
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    10 months ago

    How?

    Yes, if you push the circles down a bit, it forms a 7 by 7 matrix. But if pushing the circles into a square matrix is not allowed: how?

    Edit: I get it now. It is about (efficient) packing not about counting. I also get the 4th panel now…

  • VeganPizza69 Ⓥ@lemmy.world
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    10 months ago

    This is the kind of stuff the timber mafia needs to know so that they can efficiently pack trees and send them to IKEA.

    • enkers
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      10 months ago

      Because it’s a smaller area than 7x7.

      If you consider the regular packing in an infinite plane, tri/hex packing is the most space efficient (least wasted space), so I’d assume larger packings would tend towards that. But in smaller packings, the efficiency loss from the extra size needed to offset the circles outweighs the efficiency gained by hex packing.

      7x7 is the boundary where those efficiency tradeoffs switch.

      • tquid
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        10 months ago

        Thank you for this explanation!

      • ddh@lemmy.sdf.org
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        10 months ago

        Trying to think how tri/hex is more efficient than any regular tiling, say squares.

        • enkers
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          10 months ago

          Want a hint? Think about a circle bound by an n-sided polygon. What happens to the space between the bounding polygon and the circle as n increases? And when n is infinite?

          So of three possible regular tilings, which will be most and least efficient?

          (Btw, strictly speaking, I shouldn’t have said tri/hex before, as it’s really just hex tiling.)

          You could also use some fancy trig to calculate the efficiency %, but that’s way too much work for me. :)

    • enkers
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      10 months ago

      I think you skipped a row.

      Also, 6*6+7=???

      • boatswain@infosec.pub
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        10 months ago

        I did yeah; deleted my content almost immediately after posting it because I went to double check. Counting is hard!