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

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Michael Mendelsohn

Posts: 43
Registered: 12/13/04
Re: Billiards Puzzle
Posted: Oct 1, 2004 6:45 AM
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> Glenn C. Rhoads" schrieb:
> poopdeville@gmail.com (Acid Pooh) wrote in
> > 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"?

I understand this to mean that

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

maps to

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

> 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.

Well, though the rack is not symmetric, the triangle is; which means
that the above setups reflect to those setups below:


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

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

These appear to be the only 4 legal billiard setups. (Although I can't
fathom why the bottom row couldn't be T S S T S, for example.)
The problem then boils down to what the maximum of the minimums of
2-ball permutations from any permutation to any of these 4 permutations
is. ;)

Cheers
Michael
--
Feel the stare of my burning hamster and stop smoking!




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