Here's an upper bound. First, two simple facts that I won't take time to prove. My covering disks have radius 1. (1) A disk of radius 1+2r, with r = 2/sqrt(3) - 1, will always contain a disk of radius r that is completely uncovered. (Consider the maximum size disk that can be squeezed into the hole formed by three mutually tangent covering disks.) (2) A disk of radius r placed anywhere in the plane, where there is a triangular lattice of minimum distance sqrt(3) r, will always contain at least one lattice point. (The triangular Delone cells define the largest disks that "fit" inside the lattice.) So all I need to do is generate a triangular lattice with minimum distance sqrt(3) r and select all the lattice points within 1+2r of the origin. Result: 85 points. Veit On Dec 10, 2010, at 5:55 PM, Stephen B. Gray wrote:
Obvious extension: unit spheres and n points in R3. d>3 dimensions, anyone?
I too have no idea how to go about these. It seems like it would have been good for Erdos, and it deserves a place in the next edition of Discrete and Computational Geometry. Fine problem! I've lost track of where the R2 version came from.
Steve Gray
On 12/10/2010 12:21 PM, Allan Wechsler wrote:
I could quickly prove that 3 points can always be covered. I could not immediately prove the same for 4, though I have no doubt that it's true. My intuition is that an uncoverable set can be constructed with on the order of 15 to 20 points, but I really have no idea how to go about it.
On Fri, Dec 10, 2010 at 9:21 AM, Fred lunnon<fred.lunnon@gmail.com> wrote:
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