The reason I chose University of Michigan is because Thomas Hales hails from University of Michigan, yet Hales has a fake proof of Kepler Packing. The Kepler Packing has many counterexamples that Hales never even mentions. Below is a detailed discussion of counterexamples.
Alright, the reason Kepler Packing Problem KPP, was never proven in mathematics, is not because it is a difficult proof, but because there are two definitions involved that are very much cloudy, foggy and ill-defined. When you have one definition ill-defined is enough to prevent there ever being a math proof, but in the case of KPP, there are two ill-defined definitions: (1) finite versus infinity with a borderline between them (2) density definition
If we define density to be surface area contact by equal spheres, then the maximum kissing points is 12 and there is a theorem in mathematics that proves 12 is maximum. In this definition of density, the surface area is the "Container involved". So if Kepler had stated his KPP as kissing point density, he probably could have proven his own conjecture.
However, if you do not define density as surface contact intersection, then density is defined as to Space as a container. And since Kepler and all his later day followers never defined "infinity" with precision, that there would never be a proof of KPP, so long as infinity is ill-defined.
If you take the unit Tractrix area and unit circle area and follow them out to Floor-pi*10^603, you discover there is a crossover in amount of area. That the area of circle is always greater except when it reaches floor-pi*10^603 and then the tractrix area becomes momentarily greater than the circle. In that crossover is the Infinity borderline.
So, to prove KPP, with a precision definition of what it means to be finite and infinite, that the 1*10^603 by 1*10^603 by 1*10^603 is a container, and the KPP question becomes, is the Hexagonal Closed Pack the most dense packing of that container (and all the smaller sized cube containers)?
Well, the answer surprises most people, in that a Hybrid form of Hexagonal Closed Pack is more dense than the pure Hexagonal Closed Pack, because near the walls, edges and corners, we can rearrange some of the Hexagonal Closed Pack to fit more unit spheres and make a denser packing.
I wrote a few days ago:
Alright, well, we could never have asked Kepler to define precisely infinity with a borderline, for he would never have had the ability to know the Tractrix and Circle area crossover at Floor-pi*10^603 and thus infinity border is 1*10^603. And then Kepler could have said for his Conjecture, that the hexagonal closed pack is the most dense packing of equal unit spheres in a container that is 1*10^603 by 1*10^603 by 1*10^603. Then, mathematicians would have looked to see if there was a more dense packing. And it turns out, that there is a more dense packing as we manipulate the last plane and last edge and last corners to fit more spheres in an otherwise pure Hexagonal Closed Pack. The most dense packing depends on the size of the container.
Now, if density is defined as how many other spheres can be packed onto a single sphere-- the kissing points, then in that density definition, the proof of Kepler Packing is a one paragraph proof that 12 kissing points is maximum. The surface of a single sphere would be considered "the container".
But Kepler had in mind something different for density as a container. Now if one does not define finite with infinity properly as a borderline, then empty space that is infinite empty space is meaningless, because the Space cannot tell you whether you are in "finite space" or infinite space. In total empty space, means there is only finite space, because no border was reached to say, now we are in infinite space.
So if we define Infinite Space as a borderline that is crossed and that forms a container such as the 1*10^603 by 1*10^603 by 1*10^603 cube.
Now we can get an induction notion of this container of Space by smaller cubes such as 10 by 10 by 10 or the 100 by 100 by 100 cube and ask whether the most dense packing is the same as we go higher in cubes?
It turns out that a Hybrid form of Hexagonal Closed Pack is the most dense packing possible for it allows more unit spheres into a container than the pure Hexagonal Closed Pack.
In the below old post of mine, I illustrate this Hybrid Hexagonal Closed Pack.
I wrote a few days ago:
Alright, in my prior post, the illustration was messed up in format, so here I hope to have cleared up view of the illustration.
Now I think that the maximum hybrid of hexagonal closed pack is going to add spheres in the corners and edges, but I fail to see or recollect in my memory how an entire plane of new spheres can be added to a stock of hexagonal closed pack. If the body of the packing is hexagonal closed pack, I fail to see how an entire plane can be added in a hybrid. I think I came to that conclusion many years back when I first did these hybrid packings, but want to make sure of that memory.
Here is the illustration:
when the headroom of the ceiling face is large enough to scoot ?that upper layer to the rightward side then we can add a new sphere ?shown by the X
And if the upper layer is full then we do the scooting on the second ?to top ?layer as such: OOOOO ? XOOOO
And here is the illustration of Oblong Hex when the square or cube has ?a 0.99 after the integer
O 9 ?_O 8 ?O 7 ?_O 6 ?O 5 ?_O 4
The underline indicates a distance of 0.99 to the wall. Those ?are either spheres in 3D or circles in 2D and they are numbered ?for easy reference. In the Oblong technique we scoot 5 downwards leaving a gap ?large enough to pack a new sphere or circle. We leave 6 undisturbed ?but move 7 downwards leaving a gap which we add a new circle or ?sphere. And we do that entire face of both the length face and width ?face, depending on whether we are in 3D or 2D.
Recently I re-opened the old newsgroup of 1990s and there one can read my recent posts without the hassle of mockers and hatemongers.