9 billiard balls, which one is heavier/lighter?

I at one stage solved this problem with 12 balls in 3 steps. But I don't think you can do 9 in 2 steps so it must 3 steps too.

My 2C.

Think i can do it in 3 with a slightly changed method:

1. Place any 3 balls on each side of the scales. If balanced goto 2 else 3.

2. Knowledge: One of the 2 balls not used in the first test is rouge; the 6 used in the test are normal.

Action:  The 2 balls that might be rouge can be compared each in turn with a normal ball to find which is rouge and whether it's too heavy or too light. End (2 or 3) 

3. Knowledge: The 2 balls not used are normal. Either one of the 3 on the heavier side is too heavy or one of the 3 on the lighter side is too light.

Action: take the 3 balls form the heavier side a one ball knowned to be normal. Place 2 of these balls on each side of the scales. If balanced goto 4 else goto 5.

4. Knowledge: All 4 balls from the last test are normal therefore one of the 3 balls form the lighter side in step1 is too light and rouge.

Action: Place one of the 3 balls on each side of the scales (one left over) if the scales balance the left over ball is rouge and too light else the lighter one of the two on the scales is rogue. End(3)

5. Knowlege: One of the 2 balls on the heavier side of the scalse is rouge; The rouge ball is too heavy; one of the balls on the scales is already known to be normal (see step 3).

Action: Take 2 balls on the heavier side. if one of them is the ball already known to be normal then the other is rouge and too heavy. if either of the 2 balls could be rouge weight those 2 balls and the heavier of them is rouge and too heavy. End (2 or 3)

Is it possible to do this in less than 3 attempts?

Sam. 

I believe it will take either 2, 3 or 4 attempts depending on how lucky you are!

Here how i'd do it (the ball that has a different mass i refer to as the rouge ball):

1. place 4 balls on each side if scales balance goto 2 else goto 3.

2. The rouge ball is the ball left over: swaping it with another ball and seeing if it make the side it's put on heavier or light will tell you if the rouge ball is heavier or lighter than usual. End (scales used 2 times)

3. Take the 4 balls on the side that was lighter and put them somewhere safe. Take the 4 balls on the side that was heavier and put 2 of them on each side of the scales. If the scales don't balance goto 4 elses goto 5.

4. The rouge ball is one of the 4 on the sacles (easy to understand) and it's heavier  than usual (a bit harder to understand but i still shouldn't have to explain why!). It is therefore one of the 2 balls on the heavier side of scales. Put one of those 2 balls of each side of the scales and the heavier one is the rouge ball. End (scales used 3 times)

5. The rouge ball is one of the 4 that you put somewhere safe and it's lighter than usual (similar logic to how i know in step 4 that the rouge ball must be heavier than usual). Therefore put 2 of the 4 ball in question on each side of the scales. Take the 2 lighter ones and put one on each side of the scales the lighter one is the rouge ball! End (scales used 4 times)

If there's a problem with my logic or a quicker method please post

Sam.

1 - 4?

Nevermind. I did for lighter ball only. I seperated one ball from rest of the 8. If both sides were equal, I picked the right ball. Lucky bastered.

But if weights were not equal, I put 4 on each sides and picked the ligher side. Again, put took 4 balls and put 2 balls on each sides and so on.