[Guest]

In there were 12 balls, one of which weighed a bit different from all the others. Good solutions were given that used 3 weighings on a balance scale in to determine which ball was different and whether it was lighter or heavier.

Now comes the catch. Solve the same problem, but decide ahead of time which balls to use for each weighing before seeing the outcome of any weighing.

Edited March 11, 2010 by Lath

[Guest]

This can still be done in 3. Call the coins 1234567890AB and do the following:

1: 1234 vs 5678

2: 5234 vs 10AB

3: 2560 vs 379A

The 24 possible outcomes are revealed as follows:

1H : ><=

1L : <>=

2H : >>>

2L : <<<

3H : >><

3L : <<>

4H : >>=

4L : <<=

5H : <>>

5L : ><<

6H : <=>

6L : >=<

7H : <=<

7L : >=>

8H : <==

8L : >==

9H : ==<

9L : ==>

0H : =<>

0L : =><

AH : =<<

AL : =>>

BH : =<=

BL : =>=