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

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Glenn C. Rhoads

Posts: 31
Registered: 12/13/04
Re: Billiards Puzzle
Posted: Oct 1, 2004 4:47 AM
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poopdeville@gmail.com (Acid Pooh) wrote in message news:<4765002.0409301719.70ddf1ba@posting.google.com>...
> Here's a neat little puzzle I thought of, though I still haven't
> figured out its answer.
>
> 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 don't understand your description of "standard configuration."
What do you mean by "every S is mapped to a T"? There are seven
of each so it is always possible to think of each S being paired
with a T. Also, the two back corners cannot be the same (by the
rules of the 8 ball) and hence, the rack is never symmetric across
the vertical axis.




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