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Re: Billiards Puzzle
Posted:
Oct 1, 2004 3:17 PM


Please ignore my last post since it was in error, sorry Puzzle used as spoiler space 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? > > 'cid 'ooh
E SS SSS SSTT TTTTT takes 4 moves
I think it easy too show that 4 is the max. Leaving the eight ball in place take any configuration of S&T. Now compare this to a specific standard configuration, call it ConfigX. Suppose you find N Sballs are out of place where N=07. If you swap S and T in ConfigX than 7N Sballs will now be out of place. So I can always pick a configuration where the maximum number of Sballs that can be out of place is 3 so we have a total of 6 balls out of place. Now swap out the eight ball and you have 8 balls out of place, which is 4 swaps. Done.



