We understand that for packing identical circles (indeed packing copies of any centrally symmetric convex shape) in an infinite plane, the best arrangment is a always a lattice arrangement (hexagonal lattice for packing circles).
Does the above lead naturally to the following?
Consider any specified 2d shape as a container for identical circles.
Claim: Some finite critical magnification(scaling) of this shape exists such that
1.the best packing in it of circles is a hexagonal lattice of circles arranged inside and
2.for any further scaling up of this shape the best pack is the hexagonal pack.
This implies: If the container is a square, there is some critical size of the square beyond which the best circle packing of unit circles is ALWAYS the hexagonal lattice.
Some more thoughts on 'automatable' arrangments to follow...
Update (3rd May 2009): The main guess above is invalid. For the square the hex lattice is tbe best packing only in the asymptotic limit, as the square goes to infinity and is the same as the plane itself. No finite threshold and stuff. This is a bit surprising to me but that is how things probably are!
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