sphere packing

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3 total messages Started by byron elbows Tue, 26 Jan 1993 23:54

#2640

Author: byron elbows
Date: Tue, 26 Jan 1993 23:54

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I need help on a sphere packing question.  Suppose you have a white ping
pong ball, and 13 red ones, all the same size.  Can you glue all of the red
ping pong balls to the white one?  In other words, is it possible to place
13 unit spheres around one unit sphere, so that each of the 13 is touching
the central sphere, and none of the 13 intersect each other?

It is a simple matter to do it with 12.  Simply place the spheres at the
corners of an icosahedron centered at the center of the white ping pong ball.
It turns out this leaves some extra room for spheres to be moved around, but
is it possible to move them around so that another sphere can be squashed
in there?

In the meantime, I leave you with two simpler questions:

1.  Suppose you pick a number at random from 0 to 1.  Assume uniform
    distribution (ie, pdf = 1 from 0 to 1).  Clearly, that number is
    not going to be greater than 1.  Now you pick another number at
    random from 0 to 1.  The probability that the sum of these two
    numbers is greater than 1 is simple to calculate; it is 1/2.  If
    it is greater than 1, stop; otherwise, continue picking numbers
    at random until the accumulated sum is greater than 1.  Derive the
    average number of picks required.  (What is the answer if the
    distribution is arbitrary?)

2.  Suppose I have a deck of cards numbered from 1 to 100.  You and I
    are dealt two hands of 50 cards each, randomly, from the deck.  We
    both sort the cards in increasing order.  In the first round, we
    both play our first card; the player with the higher card wins
    one dollar, and the two cards are discarded.  This routine is
    repeated for our second card, third card, etc.  What is the
    probability that I win this warped version of War by a score of
    50 to nothing?

byron elbows
(mail to brian@cs.ucla.edu)

Re: sphere packing

#2606

Author: J. Theodore Schuerzinger
Date: Wed, 27 Jan 1993 02:39

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In article <1993Jan26.235433.18005@cs.ucla.edu>
byron elbows writes:

> I need help on a sphere packing question.  Suppose you have a white ping
> pong ball, and 13 red ones, all the same size.  Can you glue all of the red
> ping pong balls to the white one?  In other words, is it possible to place
> 13 unit spheres around one unit sphere, so that each of the 13 is touching
> the central sphere, and none of the 13 intersect each other?
> 
> It is a simple matter to do it with 12.  Simply place the spheres at the
> corners of an icosahedron centered at the center of the white ping pong ball.
> It turns out this leaves some extra room for spheres to be moved around, but
> is it possible to move them around so that another sphere can be squashed
> in there?

You're asking a question about the 'kissing number'; that is, the
greatest number of objects that can be touching another object of the
same size.  For two dimensions (ie. circles), the kissing number is 6,
and for three dimensions (spheres -- the question you ask), it's been
proven that the kissing number is 12.  Therefore, there is no way to
move those twelve spheres to make a 13th one touch the center sphere. 
Sometime within the last year (sorry I don't have the exact date!)
Scientific American ran a piece about the kissing number in its math
column.

--Ted Schuerzinger
email: .zed@Dartmouth.EDU
"I should have realized it would be bad vodka when all the label said
was 'Russian Vodka'."

Re: sphere packing

#2694

Author: David Karr
Date: Thu, 28 Jan 1993 19:25

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In article <1993Jan26.235433.18005@cs.ucla.edu> byron elbows writes:
>I need help on a sphere packing question.  Suppose you have a white ping
>pong ball, and 13 red ones, all the same size.  Can you glue all of the red
>ping pong balls to the white one?  [...]
>
>It is a simple matter to do it with 12.  Simply place the spheres at the
>corners of an icosahedron centered at the center of the white ping pong ball.
>It turns out this leaves some extra room for spheres to be moved around [...]

This remark caused me to wonder what is the diameter of a sphere
circumscribed about an icosahedron of side 1.  (Presumably it's less
than 2, so that the above works.)  It would be nice if there were a
simple, painless answer to this, but I couldn't think of anything that
didn't involve lots of trigonometry.  Does anyone know a very simple
method to compute this?

By the way, you also get 12 spheres around a central one if you use
either hexagonal or cubic close packing, in which 6 of the spheres lie
in one plane (along with the center sphere) with centers at the
vertices of a regular hexagon.  In this configuration all the
surrounding spheres just barely fit in around the central one with no
"wiggling room," but it's easy enough to verify by diagram that the
configuration works.

-- David Karr (karr@cs.cornell.edu)

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