Alan Sagan schrieb: > email@example.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 > E > SS > SSS > STST > TTTTT
Takes only 3, though:
(E) S S S(S)S S T S T T T T T T
S S(S) S E S S T S T T T T T(T)
S S T (S)E S S T S T T T(T)T S
S S T T E S S T S T T T S T S
That is a mirror of the setup Pooh gave.
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