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

 Messages: [ Previous | Next ]
 Alan Sagan Posts: 13 Registered: 12/13/04
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:&lt;4765002.0409301719.70ddf1ba@posting.google.com&gt;...
&gt; Here's a neat little puzzle I thought of, though I still haven't
&gt;
&gt; Suppose you're racking up 15 billiard balls in one of the standard
&gt; configurations (I'm not going to try to typeset these, so just picture
&gt; an equilateral triangle instead of a right one):
&gt;
&gt; S
&gt; T S
&gt; S E T
&gt; T S T S
&gt; S T S T T
&gt;
&gt; where S is a "solid," T is a stripe, and E is the eight ball. A
&gt; configuration is also standard if every S is mapped to a T, or if the
&gt; triangle is reflected across its verticle axis of symmetry. Anyway,
&gt; so you're racking up and you dump 15 balls into the rack randomly.
&gt; Assuming you don't make any mistakes, what's the maximum number of two
&gt; ball permuations necessary to get to any of the 4 standard
&gt; configurations?
&gt;
&gt; '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&amp;T. Now
compare this to a specific standard configuration, call it ConfigX.
Suppose you find N S-balls are out of place where N=0-7. If you swap S
and T in ConfigX than 7-N S-balls will now be out of place. So I can
always pick a configuration where the maximum number of S-balls 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.

Date Subject Author
9/30/04 Acid Pooh
10/1/04 Alan Sagan
10/1/04 Michael Mendelsohn
10/1/04 Michael Mendelsohn