home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: rec.puzzles
- Path: sparky!uunet!charon.amdahl.com!pacbell.com!decwrl!ames!sun-barr!cs.utexas.edu!zaphod.mps.ohio-state.edu!usc!news.service.uci.edu!ucivax!ucla-cs!maui.cs.ucla.edu!brian
- From: byron elbows
- Subject: sphere packing
- Message-ID: <1993Jan26.235433.18005@cs.ucla.edu>
- Summary: not an open question, but who knows the answer?
- Originator: brian@maui.cs.ucla.edu
- Sender: the waste land
- Nntp-Posting-Host: maui.cs.ucla.edu
- Organization: UCLA, Computer Science Department
- Date: Tue, 26 Jan 93 23:54:33 GMT
- Lines: 35
-
- 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)
-