wu :: forums
« wu :: forums - hypersphere packing »

Welcome, Guest. Please Login or Register.
Nov 28th, 2024, 8:40am

RIDDLES SITE WRITE MATH! Home Home Help Help Search Search Members Members Login Login Register Register
   wu :: forums
   riddles
   putnam exam (pure math)
(Moderators: Icarus, Eigenray, towr, SMQ, william wu, Grimbal)
   hypersphere packing
« Previous topic | Next topic »
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print
   Author  Topic: hypersphere packing  (Read 580 times)
JocK
Uberpuzzler
*****






   


Gender: male
Posts: 877
hypersphere packing  
« on: Jul 25th, 2004, 1:30pm »
Quote Quote Modify Modify

In N dimensions, N+1 equally sized hyperspheres can be configurated in a cluster such that each hypersphere touches all other hyperspheres.  
 
For N[to][infty] calculate the radius of the smallest hypersphere that can contain such a cluster (assign a radius of unity to the spheres in the cluster).
 
JCoolCK
IP Logged

solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.

xy - y = x5 - y4 - y3 = 20; x>0, y>0.
Eigenray
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
Posts: 1948
Re: hypersphere packing  
« Reply #1 on: Jul 26th, 2004, 6:59am »
Quote Quote Modify Modify

First let's inductively find (n+1) equally spaced vectors in Sn-1 [subset] [bbr]n.
For n=1 we have of course v0=-1, v1=1.
Suppose v0, ..., vn are equally spaced in Sn-1.  Now let's rotate each of these up into the (n+1)th dimension, giving vectors vi' = (vi*sqrt(1-t2),t) [in] Sn, which gives us room to insert the vector (0, -1).
Let dn = <vi,vj>, i [ne] j.  d1 = -1.
We pick t in such a way to make them equally spaced, i.e.,
dn+1 = <vi', vj'> = <vk',(0,-1)>
dn(1-t2) + t2 = -t
t = -1, dn/(dn-1).
dn+1 = -dn/(dn-1)
Inductively we find dn = -1/n.
The distance rn between two of these vectors is given by
rn2 = <vi-vj, vi-vj> = 1 + 1 - 2(-1/n) = 2 + 2/n.
So, with our equally spaced vectors in place, we scale them such that each pair are a distance 2 apart to make room for the hyperspheres:
wi = vi*2/rn
The radius of the containing hypersphere is |wi|+1 = [sqrt](2n/(n+1)) + 1 [to] 1+[sqrt]2
« Last Edit: Jul 26th, 2004, 9:15am by Eigenray » IP Logged
Eigenray
wu::riddles Moderator
Uberpuzzler
*****






   


Gender: male
Posts: 1948
Re: hypersphere packing  
« Reply #2 on: Jul 27th, 2004, 6:14am »
Quote Quote Modify Modify

Compare to this packing, where the radius of the inscribed hypersphere is unbounded as n[to][infty].
IP Logged
JocK
Uberpuzzler
*****






   


Gender: male
Posts: 877
Re: hypersphere packing  
« Reply #3 on: Jul 31st, 2004, 12:15pm »
Quote Quote Modify Modify

Correct Eigenray!
 
Your reference to the inscribed hypersphere problem is interesting, as it is (losely) related to the follow-up question:

 
 
Things get a bit more complicated as we now investigate unequally sized hyperspheres:
 
In n dimensions, if properly sized, n+2 hyperspheres can be configurated in a cluster such that each hypersphere touches all other hyperspheres. In particular, for n dimensions this holds for properly chosen [alpha][subn]>1 if the hyperspheres have radii R,  [alpha][subn] R,  [alpha][subn]^2 R, .. ,  [alpha][subn]^(n+1) R.  
 
This means that an infinite nesting of hyperspheres can be constructed with each subsequent hypersphere being a factor 1/[alpha] smaller in size, such that within this nesting each group of n+2 subsequently sized hyperspheres consist of hyperspheres that all touch each other. For such a cluster of n+2 subsequent hyperspheres, the smallest hypersphere is the inscribed hypersphere of the n+1 larger spheres.
 
For n=3,  [alpha] < 2. Can you prove this? What is the exact value?  
 
What happens if we increase n beyond 3? What is the value for [alpha][subn] for n [to] [infty]? And how does [alpha][subn]^n (the volume-ratio of subsequent hyperspheres) behave for n [to] [infty]?  
 
JCoolCK  
« Last Edit: Aug 1st, 2004, 12:22pm by JocK » IP Logged

solving abstract problems is like sex: it may occasionally have some practical use, but that is not why we do it.

xy - y = x5 - y4 - y3 = 20; x>0, y>0.
Pages: 1  Reply Reply Notify of replies Notify of replies Send Topic Send Topic Print Print

« Previous topic | Next topic »

Powered by YaBB 1 Gold - SP 1.4!
Forum software copyright © 2000-2004 Yet another Bulletin Board