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Re: Billiards Puzzle
Posted:
Oct 1, 2004 6:45 AM


> 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 2ball permutations from any permutation to any of these 4 permutations is. ;)
Cheers Michael  Feel the stare of my burning hamster and stop smoking!



