On Fri, Feb 19, 2016 at 4:42 PM, <rcs@xmission.com> wrote:
I think this can be improved by staging the circle sizes, assuming you're allowed to use circles of different sizes.
If by "improved". you mean "fills a higher percentage of the circle" sure. And by having largish, much smaller, and much smaller than that circles, we can have A^3 inside the circles. But we're not trying to maximize the area inside the circles, we're trying to maximize the sum of their radii. If we have n numbers,. and a constraint on the sum of their squares, we maximize their sum by making them all equal. We have a hard constraint that the sum of the squares of the radii must be less than 1, because the areas of the circles must sum to less than the area of the large circle. So we probably want to make the circles close to the same size. If the circles are all the same size, we have a tighter soft constraint that the sum of the radii must be less than sqrt(A), because we can't fill any area bigger than a hexagonal close packing. My intuition says that if we try to use circles of different radii, the gain by needing to satisfy the soft constraint instead of the hard one is less than the loss by not having the radii nearly equal. The difference between the soft and hard constraints isn't that big, because sqrt(A) is already close to 1, and I don't think we get to fill much extra area by varying the circle size until the size of the small circles is small enough that we can fit one into the deltoid made by three mutually tangent larger circles, and I think that this much variation in size will lead to a loss of more than a factor of sqrt(A). Not a proof by any means, just my intuitions. Andy Latto andy.latto@pobox.com
Let N be humungous.
(stage 1) Allocate some minor portion n of N to a close packing of largish circles that comes close to achieving the coverage of hexagonal close-packing. This leaves ~10% of the area uncovered.
(stage 2) Now use the remaining N-n (much smaller) circles to cover that 10%. The coverage of that will approach the value for hex-close-packing, since the second stage circles are tiny, generally tinier than the wastage in the irregularly shaped holes left after stage 1. For sufficiently large N, and a sufficient ratio between the stage 1 and stage 2 circles, the uncovered area approaches 10% x 10% = 1%.
The idea can be used with more stages to make the uncovered area approach 0%.
Unclear if this maximizes the radial sum.
Rich
----------- Quoting Andy Latto <andy.latto@pobox.com>:
My intuitions say that the maximal sum of radii for circles within a circle will approach, not sqrt(n), but sqrt(An), where A = pi sqrt(3)/6 ~ .9069 is the ration of the area inside the circles to the total area in a hexagonal close-packing. To get a radius-sum of sqrt(n), you'd want n circles of radius 1/sqrt(n). But those will have total area pi, the same as the area of the bounding circle, while my intuition says that when you have lots of small circles, they will be roughly the same size in roughly a hexagonal close-packing, so they won't have total area more than A pi. So if there are n circles, each should have area A pi/n, or radius sqrt(A)/sqrt(n), and n of them have total radius sqrt(nA).
Andy Latto andy.latto@pobox.com
On Fri, Feb 19, 2016 at 3:07 PM, Charles Greathouse <charles.greathouse@case.edu> wrote:
Maybe it's obvious, but I'd just like to know if the maximal sum of the radii of n circles in a circle of unit radius approaches its upper bound sqrt(n). (Or have I been caught in the sort of trap that inspired this conversation?)
Charles Greathouse Analyst/Programmer Case Western Reserve University
On Fri, Feb 19, 2016 at 9:08 AM, Veit Elser <ve10@cornell.edu> wrote:
On Feb 19, 2016, at 8:48 AM, Fred Lunnon <fred.lunnon@gmail.com> wrote:
I'm a tad suspicious: are the circle packings on Friedman's page proved to be maximal, or are they merely the best known so far? There is no link to any proofs, or key to diameters ? WFL
I have a conjecture that the maximum-sum-of-diameters packings have triangulated contact graphs. If that?s true, one could enumerate the graphs and calculate the packing for each one to see which is maximal.
-Veit
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