From: ransom@monu6.cc.monash.edu.au (Mr S. Ransom)
Subject: Re: simple puzzle
Organization: Monash University, Melb., Australia.
Date: Thu, 10 Sep 1992 23:18:42 GMT

The 12 coins problem - there may be a simpler solution to this
but this one seems to work.

The non-standard coin can be isolated and it can be determined 
whether it is heavier or lighter than standard in all cases in 
exactly 3 weighings.

First weighing:

Divide the coins into 3 equal piles of 4.  Place one pile to one side
and weigh the remaining 2 piles against each other.  The result is
that either they balance each other or they do not.

If they balance then we know that the coin is in the pile of 4 which
we put aside and that the other 8 are all standard BUT we know nothing
of whether the rogue coin is heavy or light.

If the balance tips then we have reduced the problem to the coin being
in the "down" pile +and+ being heavier than standard or being in the
"up" pile +and+ being lighter than standard.  We also know that the 4
coins put aside are all standard.

Second weighing - first weighing balanced:

   Take the pile which we set aside and, keeping 1 of the 4 coins out,
   place the remaining 3 coins two to one side and one to the other.  
   Use a (now) known standard coin to make up the numbers to 2 each.  
   Either this 2nd weighing results in a balance or it doesn't.  

   Third weighing:

      If it does balance then the single coin kept out is the 
      rogue and weighing it against a standard coin in a 3rd 
      weighing will determine whether it is heavy or light.

      If the weighing does not balance then the rogue is either 
      the singleton (with the standard) and is heavier (lighter) 
      or it is one of the pair and is lighter (heavier).  For 
      the third weighing split the pair and weigh one against 
      one.  If this balances then it is the heavier (lighter) 
      singleton.  If it doesn't then it is the one on the 
      lighter (heavier) side.

Second weighing - first weighing didn't balance:

   We now have a stock of 4 known standard coins.  Take 3 of the 
   +lighter+ 4 (the pile which went up) and place to one side.
   This leaves 4 possibly heavy and 1 possibly light coins on the
   balance.  Leave the possibly light coin where it was and move 3
   of the possibly heavy coins across to join it.  Add 3 standards
   to the scale with the single remaining possibly heavy so that
   there are 4 coins on either side.

   Third weighing:

      If they now balance then the rogue is one of the 3 which we
      removed prior to the second weighing +and+ is lighter than
      standard.  In this case the 3rd weighing matches 2 
      of the 3 against each other.  We know at this point that 
      the rogue is lighter than standard so if the 2 balance 
      then the rogue is the remaining one (and is light) or if 
      they do not balance then it is the one which went up. 

      If the scales tip in the same direction as they did at the 
      1st weighing then we have eliminated the 3 moved coins 
      leaving one possibly heavy and one possibly light coin
      which the 3rd weighing can choose between by weighing either 
      one against a standard coin.

      If the beam changes direction then the rogue is one of the 
      3 moved +and+ is heavier.  As above, weigh one against one.
      We know the rogue is heavier so and if the two balance then
      the rougue is the third, unweighed coin and if they don't
      balance then it is the one which went down.
      
-- 
Stephen Ransom                        | Department of Software Development
Telnet: ransom@monu6.cc.monash.edu.au | Monash University
Phone: (613) 573 2181                 | Australia
Newsgroups: rec.org.mensa