Sphere Packing in Distance Geometry
Sphere Packing in Distance Geometry
In distance geometry the only condition we have is that any (n+1)-simplex made from (n+2) points is zero in n dimensions. Let R_nm be the squared distance between point n and point m.The kissing number problem is then stated as:
R0n = 1
Rnm > 1
Then we ask is there a solution in terms of R such that the above holds and also the Cayley-Menger determinants are all zero?
In 2 dimensions with 7 circles this gives us
R0n = 1
Rnm > 1
VolumeOfTetrahedron( R_ab,R_ac,R_ad,R_bc,R_bd,R_cd) = 0
Which gives cubic equalities for the R.
We could set:
Rnm = 1+(Snm)^2
This gives us a set of 6th degree polynomials in S. Then we should be able to use the resultant to see if they all have a simultaneous solution!
But the resultant does not distinguish between real and imaginary solutions. :(
e.g.
(x^2+1)(x-3)=0
(x^2+1)(x-4)=0
Share an imaginary solution! (Only possible if share common factor?)
What we need to solve is when a system of polynomial equations has positive solutions. Equally hard question. No simpler??? :(
x^2+1 = 0 has no solution in reals since det(Q)<0. So it IS possible to ignore imaginary solutions.
The determinants in 1 dimension are are:
D(1,2,3) = det
[ 0 , 1+S12^2 , 1+S13^2 , 1]
[1+S12^2, 0 , 1+S23^2, 1]
[1+S13^2, 1+S23^2, 0 , 1]
[1 , 1 , 1 , 0]
2 dimensions: n-3 equations:
D(1,2,3,n)
Now the sums of the squares of all these must be zero.
D(1,2,3,4)^2+D(1,2,3,5)^2+D(1,2,3,6)^2 +... = 0
So now we have a single polynomial in S (in several variables) which we have to see if it intersects the y=0 plane. (Which it might not for example x^2+y^2+1=0 doesn't.)
Anmo.. Xn Xm Xo..... = 0
or
Anmo... x^n y^m z^o ..... = 0
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