Date:         Fri, 3 Jan 1997 14:46:33 +0000
Reply-To:     R.A.Reese@UCC.HULL.AC.UK
Sender:       "SPSSX(r) Discussion" <SPSSX-L@UGA.CC.UGA.EDU>
From:         "R. Allan Reese" <R.A.Reese@UCC.HULL.AC.UK>
Subject:      12 balls problem
Comments: To: Joel York <jyork@SPSS.COM>
In-Reply-To:  <E0vg9uu-0001MY-00@adelphi.ucc.hull.ac.uk>
Content-Type: TEXT/PLAIN; charset=US-ASCII

On Thu, 2 Jan 1997, Joel York wrote:
 
> I believe this method requires sometimes requries 4 measurements.
>
> Say group A and B are measured in the first round and are equal.
> The odd ball is in Group C (nmeasured, which has 4 balls.
>
> Compare C1 to C2 and they are the same.
>
> It will take two more measures to decide if it is ball C3 or 4 for a
> total of four measurements.
 
I didn't get a complete solution, but if A,B,C are groups of four balls
and weighing 1 gives A=B, take the four balls in C and label them 1,2,3,4.
Weighing 2: 1 vs 2
Weighing 3: 1 vs 3
gives sufficient information to deduce which three balls are equal, and
hence which is different (you don't have to say if it's lighter or
heavier.
 
If weighing 1 gives A <> B, you can discard C.  I then proceeded to take
one ball from each of A and B and to switch one more ball from each set
into the other pan for a second weighing.  Whatever the result of that
weighing then identifies the odd ball as one of two, and all the balls in
the other pan are the same.  So take the odd pair and one normal ball
and one more weighing serves to identify the odd one.
 
I think we got there!  I'll check the logic when I'm not supposed to be
working.
 
R. Allan Reese                        Email: r.a.reese@ucc.hull.ac.uk
Head of Applications                  Direct voice:   +44 1482 465296
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Hull University, Hull HU6 7RX, UK.    Fax:            +44 1482 466441