Puzzle:finding the scale weight of 12 solid balls in 3 tries

weighing the balls
Number the balls A-L.

Begin by weighing four against four. If they balance, then weigh any of the remaining three against any three of the good balls.

If they balance then the odd one is the remaining ball and can be
identified (whether it's lighter or heavier) in the final weighing.

If three against three do not balance, take the three containing the odd
ball and weigh any one against another.

Thus:

If the first weighing of four against four does not produce a balance, then
the second weighing involves three against three with the balls switched
between the two pans and a good ball introduced.

So:

If A+B+C+D > E+F+G+H

We try A+B+E against C+F+J

If A+B+E + C+F+J then either D is heavier or G or H is lighter, so weigh
G against H.

If A+B+E > C+F+J then either F is lighter or A or B is heavier, so we
weigh A against B.

If A+B+E < C+F+J then either E is lighter or C is heavier, so weigh
either against a good ball, eg. K against E.

An alternative second weighing is A+B+E against C+D+F, which follows
similar lines to the above.

This is a familar one !

A major problem is to simplify the answer.

Bumble, you do good job with the proviso that it might be easier to label all balls found to be normal as "n" as you go along. So the 4-4 unbalance leaves you with ABCD EFGH nnnn etc.

The second weighing is of course the crucial one
using ABE vs FCn and then to state the three outcomes in terms of the original direction of the 4-4 unbalance

i.e (1) If unbalance SAME DIRECTION then target is A, B or F (Test A against B last weighing and either take the one the goes in its original direction or if they balance it must be F)
(2) If unbalance OPPOSITE DIRECTION then target is E or C Test either against n last weighing)
(3) If balance, then target is D, G or H. (Test G against H last weighing and either take the one that goes in its original direction or if they balance it must be D)

This question was in my physics exam a couple of years ago. The trouble is, the equal-arm balance wasn't an equal-arm balance but an un-equal arm balance. You had to know torque and other things to first get the mass of the heavier arm, then eliminate the balls. That was a bit of a shock to me

If there was just 9 balls and 2 weighs it would be:

Measure 1 - Weigh six on the scales (three either side) and keep three in your hand

Measure 2 - If the scales balance
Weight two of the ball bearings in your hand
The scales will show up the lightest or
If these balance the last one left is the lightest

Measure 2 " Pick the three lightest ball bearings from the first weighing and weigh two of these three
The scales will show up the lightest or
If these balance the one you’re not weighing from the first three is the lightest

In the original problem, there were 12 balls and the "bad" one could be either too heavy or too light. That's 24 possibilities. With each weighing there are 3 outcomes, so in 3 weighings you could explore up to 3^3 = 27 possibilities.

In your modified problem, if there are 9 balls, you still have 18 possible solutions. In two weighings you could explore 3^2 = 9 possibilities which wouldn't be enough to solve the problem.

If you were told whether the "bad" ball was too heavy or too light, then you could find it in two weighings.