And the next perfect divisor one that would hold all the ones in the OP pic would be 5x5. 25 > 17, last I checked.
Comment on Cursed
jeff@programming.dev 3 weeks agoWhat? 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 3 weeks ago
Natanael@infosec.pub 3 weeks 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.