Drexel dragonThe Math ForumDonate to the Math Forum



Search All of the Math Forum:

Views expressed in these public forums are not endorsed by Drexel University or The Math Forum.


Math Forum » Discussions » sci.math.* » sci.math

Topic: Billiards Puzzle
Replies: 8   Last Post: Oct 3, 2004 4:23 AM

Advanced Search

Back to Topic List Back to Topic List Jump to Tree View Jump to Tree View   Messages: [ Previous | Next ]
Michael Mendelsohn

Posts: 43
Registered: 12/13/04
Re: Billiards Puzzle
Posted: Oct 1, 2004 11:18 AM
  Click to see the message monospaced in plain text Plain Text   Click to reply to this topic Reply


Alan Sagan schrieb:
> poopdeville@gmail.com (Acid Pooh) wrote in message
> > Suppose you're racking up 15 billiard balls in one of the standard
> > configurations (I'm not going to try to typeset these, so just picture
> > an equilateral triangle instead of a right one):
> >
> > S
> > T S
> > S E T
> > T S T S
> > S T S T T
> >
> > where S is a "solid," T is a stripe, and E is the eight ball. A
> > configuration is also standard if every S is mapped to a T, or if the
> > triangle is reflected across its verticle axis of symmetry. Anyway,
> > so you're racking up and you dump 15 balls into the rack randomly.
> > Assuming you don't make any mistakes, what's the maximum number of two
> > ball permuations necessary to get to any of the 4 standard
> > configurations?

> I get 4

Proof:

a) You need at most 4 moves.

First place the eight-ball, using up one move.
For each misplaced solid, there's a misplaced stripe now.
Of the 7 solids and stripes each, there can be at most 3 such misplaced
pairs; because if there were more wrong, you'd aim for a solution with
solids and stripes exchanged, and have less wrong.
Thus, you never need more than 4 moves.

b) You may need 4 moves.

Proof by example:

E
T T
S S S
T T T T
S S T S S

One move is needed to move the E inside.
2 moves are needed to rectify either the left or the right edge.
Now the top and ball and the center ball in the bottom row are of
different kinds, but they have to be the same, so a fourth move is
needed to fix this.
These same-same relations (between row endpoints and the top-bottom) are
invariant under remapping T<->S and vertical mirroring, so there is no
way to do it shorter.

Cheers
Michael
--
Still an attentive ear he lent Her speech hath caused this pain
But could not fathom what she meant Easier I count it to explain
She was not deep, nor eloquent. The jargon of the howling main
-- from Lewis Carroll: The Three Usenet Trolls




Point your RSS reader here for a feed of the latest messages in this topic.

[Privacy Policy] [Terms of Use]

© Drexel University 1994-2014. All Rights Reserved.
The Math Forum is a research and educational enterprise of the Drexel University School of Education.