A bit more on Kissing Numbers and Spheres
PART 1
The kissing number of spheres in 3D is 12 as is well known; in 2D the kissing number of disks is 6.
Question: Can a unit sphere be kissed by 12 spheres which are all mutually equal but with radius slightly greater than 1? The reason for asking this is that there is quite a bit of room left after 12 unit spheres kiss the central unit sphere; this left over space has been proved to be not enough to squeeze in a 13th unit kissing sphere but what prevents the 12 kissing spheres from being slightly bigger?
If the answer to above question is "Yes", then how much bigger can the 12 kissers be? Further in that case, can one have an infinite (in all 3D directions) arrangement of spheres such that every one of them has a unique and finite radius and every one of them is touched by exactly 12 other spheres?
Note: In 2D, we cannot form an infinite (in every direction in 2D) arrangement of mutually non-identical and finite disks with every sphere being touched by 6 other disks. Indeed, there has to be a finite smallest disk; it cannot be kissed by 6 disks all of which are strictly larger.
And as usual, what about 4D?
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Note: The face centered cubic and hexagonal close packed structures both have a packing factor of 0.74, consist of closely packed planes of atoms (ie in each plane every sphere has 6 others touching it), and have a coordination number of 12. The difference between the fcc and hcp is the stacking sequence. The hcp layers cycle among the two equivalent shifted positions whereas the fcc layers cycle between three positions. So, obviously, in 3D, any arrangement of slightly bigger but mutually identical 12 spheres kissing a central unit sphere cannot be either hcp or fcc. But there may well be other options.
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PART 2
Another direction to extend the question:
In an arrangement of copies of a certain region C, let us define the neighborhood of any single unit, C0 as the set of copies of itself kissing C0 plus the relative positions and orientations of these kissing copies. Let residents of two unit C's in the arrangement, C1 and C2, note their respective neighborhoods. If these neighborhoods are different for such every pair {C1, C2}, we say every unit C in the arrangment has a unique neighborhood.
Question: Given any 2D convex shape C, fill the plane with a rigid arrangement (there could be gaps among units) of identical copies of C such that each unit sees a 'unique neighborhood'. No constraint on the number of neighbors each unit has. C could be a disk or square or... This question goes naturally to 3D.
A constrained version: Form infinite (in all directions)and connected - maybe with rigidity relaxed - layouts of copies of C such that every unit C touches a specified number n of other units and has a 'unique neighborhood'
Special cases: One can try to form a connected and infinite arrangement of disks on a plane such that every disk touches exactly 4 (or 5 or even 3) other disks and has a unique neighborhood. In the 3-disk case, one has an equivalent problem of tiling the plane with hexagons all with all sides of equal length and at every vertex, the 3 angles meeting being a unique set.
Guess: Insisting on rigidity and fixing the number of neighbors might make such 'locally unique' layouts hard to achieve especially for disks where each unit is fully specified by its position alone (no scope for changing orientation).
Note: In the '4 disks touching each disk' case, there is the similar but perhaps not quite equivalent problem of forming a rigid and plane-filling arrangement of rhombuses with all the sides equal to 1 but with the 4 angles meeting at a vertex unique. In both cases, the angles are constrained to be at least 60 degrees and also have an upper bound.
And here is something more basic that I don't know as of now: When arranging identical disks to fill the plane and such that each disk touches exactly n other disks, there is only one arrangement possible when n =6. However if n =3, 4 or 5, will the possible arrangements be infinitely many?
Update (June 2019): Some of the issues raised in this post got resolved in a future post: https://nandacumar.blogspot.com/2019/03/more-on-2d-layouts-with-unique.html
The kissing number of spheres in 3D is 12 as is well known; in 2D the kissing number of disks is 6.
Question: Can a unit sphere be kissed by 12 spheres which are all mutually equal but with radius slightly greater than 1? The reason for asking this is that there is quite a bit of room left after 12 unit spheres kiss the central unit sphere; this left over space has been proved to be not enough to squeeze in a 13th unit kissing sphere but what prevents the 12 kissing spheres from being slightly bigger?
If the answer to above question is "Yes", then how much bigger can the 12 kissers be? Further in that case, can one have an infinite (in all 3D directions) arrangement of spheres such that every one of them has a unique and finite radius and every one of them is touched by exactly 12 other spheres?
Note: In 2D, we cannot form an infinite (in every direction in 2D) arrangement of mutually non-identical and finite disks with every sphere being touched by 6 other disks. Indeed, there has to be a finite smallest disk; it cannot be kissed by 6 disks all of which are strictly larger.
And as usual, what about 4D?
-----
Note: The face centered cubic and hexagonal close packed structures both have a packing factor of 0.74, consist of closely packed planes of atoms (ie in each plane every sphere has 6 others touching it), and have a coordination number of 12. The difference between the fcc and hcp is the stacking sequence. The hcp layers cycle among the two equivalent shifted positions whereas the fcc layers cycle between three positions. So, obviously, in 3D, any arrangement of slightly bigger but mutually identical 12 spheres kissing a central unit sphere cannot be either hcp or fcc. But there may well be other options.
-----
PART 2
Another direction to extend the question:
In an arrangement of copies of a certain region C, let us define the neighborhood of any single unit, C0 as the set of copies of itself kissing C0 plus the relative positions and orientations of these kissing copies. Let residents of two unit C's in the arrangement, C1 and C2, note their respective neighborhoods. If these neighborhoods are different for such every pair {C1, C2}, we say every unit C in the arrangment has a unique neighborhood.
Question: Given any 2D convex shape C, fill the plane with a rigid arrangement (there could be gaps among units) of identical copies of C such that each unit sees a 'unique neighborhood'. No constraint on the number of neighbors each unit has. C could be a disk or square or... This question goes naturally to 3D.
A constrained version: Form infinite (in all directions)and connected - maybe with rigidity relaxed - layouts of copies of C such that every unit C touches a specified number n of other units and has a 'unique neighborhood'
Special cases: One can try to form a connected and infinite arrangement of disks on a plane such that every disk touches exactly 4 (or 5 or even 3) other disks and has a unique neighborhood. In the 3-disk case, one has an equivalent problem of tiling the plane with hexagons all with all sides of equal length and at every vertex, the 3 angles meeting being a unique set.
Guess: Insisting on rigidity and fixing the number of neighbors might make such 'locally unique' layouts hard to achieve especially for disks where each unit is fully specified by its position alone (no scope for changing orientation).
Note: In the '4 disks touching each disk' case, there is the similar but perhaps not quite equivalent problem of forming a rigid and plane-filling arrangement of rhombuses with all the sides equal to 1 but with the 4 angles meeting at a vertex unique. In both cases, the angles are constrained to be at least 60 degrees and also have an upper bound.
And here is something more basic that I don't know as of now: When arranging identical disks to fill the plane and such that each disk touches exactly n other disks, there is only one arrangement possible when n =6. However if n =3, 4 or 5, will the possible arrangements be infinitely many?
Update (June 2019): Some of the issues raised in this post got resolved in a future post: https://nandacumar.blogspot.com/2019/03/more-on-2d-layouts-with-unique.html
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