What a neat approach ... I haven't checked your answer, but the approach looks sound !

I will put it as part of official solution when I get a chance. Should I credit "cnaumann" ?

I have added your solution here.

Hello. I'm a new guy in here. Mr. Cnaumann 's solution is so impressive and systematic. Bravo !
I figured out a solution several months ago, although less nimble but...I tried my best :

Weigh (1st time) 4 balls against other 4 (choose randomly).
IF THE SCALE NOT BALANCED: the last 4 balls (not on the scale) are normal ones.
Name the 4 balls on heavier side A,B,C,D. If the odd ball is in group ABCD, it must be heavier than normal one.
Name the 4 balls on lighter side E,F,G,H. If the odd ball is in group EFGH, it must be lighter than normal one.
Weigh (2nd time) : (N is a normal ball)

A B N ~ C D E

1a/ If  ABN = CDE : the odd ball is in group F,G,H.
Weigh F against G (3rd time):
        If not balanced, the lighter one is the odd ball.
        If       balanced, H is the odd ball (and is lighter).
1b/ If ABN ≠  CDE :
1b1/  If the scale tipped towards A,B,N side (heavier): the odd ball is in group A,B,E.
Weigh A against B (3rd time) :
        If not balanced, the heavier one is the odd ball.
        If       balanced, E is the odd ball and lighter.
1b2/ If the scale tipped towards C,D,E side : the odd ball is in group C,D.
Weigh C against D. The heavier one is the odd ball.

IF THE SCALE BALANCED (at the 1st weighing) : these 8 balls are normal ones.
Name the last 4 balls A,B,C,D.
Weigh (2nd time) :

A N ~ C D

1a/ If not balanced (AN≠ CD) : the odd ball is in group A,C,D.
1a1/ If the scale tipped towards A,N side :
If the odd ball is A, it must be heavier than normal one. If the odd ball is C or D, it must be lighter than normal one.
Weigh C against D (3rd time):
       If not balanced, the lighter one is the odd ball.
       If       balanced, A is the odd ball and heavier.
1a2/ If the scale tipped towards C,D side :
If the odd ball is A, it must be lighter than normal one. If the odd ball is C or D, it must be heavier than normal one.
Weigh C against D (3rd time) :
       If not balanced, the heavier one is the odd ball.
       If       balanced, A is the odd bal and lighter.
1b/ If balanced (AN=CD) : B is the odd ball.
Weigh B against a normal ball to know if it is lighter or heavier than normal one .

Last edited by tt (2005-07-05 09:33:46)

It helps if you have skills with balls. (j/k)

What does tt stand for, tt?

Or isn't it meant to stand for anything?

Aha, I see. Heeheeheeheeheeheehee.......

hello. i'm sorry to say it, but the neat answer isn't as new as you would expect... it has been published in at least five books (however, it isn't very popular in the Internet, if it makes you feel proud ). just take a look at http://www.cut-the-knot.com/blue/OddballProblem1.shtml. however, i'm not accusing anybody of plagiarism--this solution had been found many times independently (quite a good exapmle of convegence, you would say).

the problem with this solution is the rules of inference cannot be observed so clearly as in the ``classical'' solution. it would be quite good if someone (the author?) explained what is the rule of putting the balls on the left and on the right throughout the 3 weightings.

Last edited by rszopa (2005-10-05 08:09:28)

who can juggle 12 balls?

Twelve!? I can't even juggle three properly. Can you juggle, wcy?

Hi koundinya17;

Welcome to the forum! That is a solution!