As per the diagram above, first weighing, you put balls 1, 2, 3, 4 on the left side of the scale, 5, 6, 7, 8 on the right.

Second weighing, you put balls 4, 5, 6, 7 on the left side, 8, 9, 10, 11 on the right.

You have 9 possible outcomes:
/ means left side goes down
means right side goes down
- means the scales are balanced
w1 = first weighing, w2 = second weighing:
[CODE]

case w1 w2 means

==== == == =====

1

 

Cases 2, 3, 4, 5, 7 and 9 is trivial (you solve it like you did the original riddle - you've got three balls, you know one of them is heavier or lighter)

Case 8 is trivial too, you know ball 12 is the odd ball. You weigh it against one of the other balls to find out if it's heavier or lighter.

Case 1: You put balls 4 and 8 on one the left side, and any other two on the right side of the scale.
Left side goes down means ball 4 is the odd ball and it's heavier, left side goes up means ball 8 is odd and lighter.

Case 6: Reverse of case 1, you put balls 4 and 8 on the left side, and any of the other two on the right.
Left side goes down means ball 8 is the odd ball and is heavier,
left side goes up means ball 4 is the odd ball and is lighter.

 

It are true.

 

There are a few routes to the solution; it appears that you found one of them.

Cool!

 

Geez, there's more than one solution? Got to put my thinking cap back on.

 

"What if C-A-T really spelled DOG?" "Woah, that's heavy Ogre." (Revenge of the Nerds II: Nerds in Paradise)

 

I can't think of any other solutions for now. Since you mentioned that I found one route to the solution, I'm guessing you mean the solution can be described in a more general fashion, and there is in fact only one solution?

 

Another way is to weigh balls 1, 2, 3, 4, on one side and 5, 6, 7, 8 on the other. Depending on outcome is what you weigh next. If it comes out even, then the heavy ball is in the last group of 9, 10, 11, or 12. If one side is heavy, then that group has the heavy ball. You then split up the group with the heavy ball with 2 on each side. Again, this will tell you which new group of two has the heavy ball. The third weighing consists of only one ball on each side- that will tell you which one is the heavy ball.

Shoot! I just re-read the question and it's if they are lighter or heavier... This only works if one is only heavier...

 

Here's another approach:

[CODE]Weigh 1, 2, 3 & 4 against 5, 6, 7 & 8.

 

[QUOTE]Originally posted by magician
Here's another approach:
[CODE]

Weigh 1, 2, 3 & 4 against 5, 6, 7 & 8.