Four circles in a hexagonal close-packing completely cover a circle of radius 1, of course. So for this to work, the arbitrary circle needs to have (optimally) a radius somewhat off of 1. If you add the center point, forcing one of the unit circles deeply into the the circle you draw, that might help reduce the point count. On Mon, Dec 13, 2010 at 9:56 AM, Allan Wechsler <acwacw@gmail.com> wrote:
I got the idea, regardless, and while I haven't actually crunched the numbers, it does look like you beat my configuration. Yay, glory!
You see what I was trying to do, though. I suspect that something like this will go through. I want to find the most devilish radius, and that suggests a new problem.
Imagine two players, say "Allan" and "Tom". Allan draws an arbitrary circle on the plane, and then Tom positions a hexagonal close-packing of unit disks on the same plane. Then Tom has to pay Allan X grams of gold, where X is the size in radians of the largest contiguous arc of Allan's circle left uncovered by the disk array. (If the circle is completely covered, Tom pays nothing.) Assuming optimal play, how many grams of gold can Allan win?
Once we know the answer to this problem, we can space points around that circle at slightly less than X radians, and this arrangement will solve the original problem, though it might not be minimal. It doesn't feel to me like the number of points is going to be much more than 7.
On Mon, Dec 13, 2010 at 12:41 PM, Tom Rokicki <rokicki@gmail.com> wrote:
Oops; forget the +1; just use 2/sqrt(3) e^ (2k pi i / 3).
On Mon, Dec 13, 2010 at 9:37 AM, Tom Rokicki <rokicki@gmail.com> wrote:
Unless I'm misunderstanding something, these points are covered by three unit circles centered at
2/sqrt(3) e^ (2k pi i / 3 + 1) for 0 <= k < 3
-tom
On Mon, Dec 13, 2010 at 9:08 AM, Allan Wechsler <acwacw@gmail.com> wrote:
I'm going to up the ante a little. I claim that I have a 7-point solution, but I haven't been able to prove it. The seven points are 1.01 e ^ (2k pi i / 7), for 0 <= k < 7. Gain glory by covering these seven points with a hexagonal close-packing of unit disks.
On Sun, Dec 12, 2010 at 1:14 PM, Veit Elser <ve10@cornell.edu> wrote:
I managed to reduce the number of points, such that one must go uncovered, to 55. It's a simple modification of my earlier construction. An entirely newr approach is needed to go significantly below this.
Veit
On Dec 11, 2010, at 8:29 AM, Veit Elser wrote:
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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