12 snooker balls Logic problem.

[Bitter'n'twisted]

For anyone who likes logic problems (and hasn't already done this one).

You have 12 red snooker balls. Identical in every way except one weighs a very slightly different amout to the others.
You also have a pair of balancing (comparison) scales.

What is the least number of comparisons on the scales you can do to find which is the different snooker ball, and to find out whether it is lighter or heavier?
Also, and more importantly, how is it done?

I'm going home now, and expect lots of correct answers by the morning!

[simpo two]

Bitter'n'twisted said:

Also, and more importantly, how is it done?

You post it on PH and let someone else work it out




Well he did, now it's gone...

>> Edited by simpo two on Wednesday 13th July 17:52

[samwilliams]

Is it 3?

1st. Put 6 on each side, the side containing the lighter ball with be lighter, discard the heavier 6
2nd. Put 3 on each side, same as before
3rd. Pick any two, put one on each side. If one is lighter than the other, that's the lighter one, if they're equal, the other one is the lightest.

[bjwoods]

BUT you don't know if it is a lighter or heavier ball

[samwilliams]

bjwoods said:

BUT you don't know if it is a lighter or heavier ball

That'll teach me to not read things through properly.

oops


Is it 4 then?

1st, 6 on each side. take the lighter 6
2nd, 3 on each side. If they're equal, then the odd ball is heavier, and in the other 6. If one is lighter than the other, that contains the odd ball, and it's lighter.
3rd, If the odd ball was lighter, you can do it in 3, putting one on each side, if the odd ball was heavier, do 3 balls on each side with the heavier group of 6, and then follow it up with one on each side.

Does that make any sense? Where have I gone wrong this time?!


>> Edited by samwilliams on Wednesday 13th July 17:59

[bridgdav]

1... Put 4 balls on each side.

If equal the lighter one is in the remaining 4
If unequal the lighter ball is in the side gone up.

From the lighter set of 4

2... Put 2 in each side

The lighter ball is in the pair gone up...

3... Put a ball in each

The lighter ball is the one gone up...

Makes 3 weighs...

[foster3jd]

Spilt balls into 3 group of 4 balls each... A, B and C

1. A v B
2. B v C

example1
A<B, B=C
hence keep group A, which contains one lighter ball
discard groups B and C

example2
A>B, B=C
hence keep group A, which contains one heavier ball
discard groups B and C


split A into 2 groups of 2.... A1 and A2

3. A1 v A2

example1
A1>A2
hence keep lighter group A2, discard A1

example2
A1<A2
hence keep heavier group A2, discard A1


4. compare remaining balls in group A2

example1
keep lightest

example2
keep heaviest

>> Edited by foster3jd on Wednesday 13th July 18:05

[justayellowbadge]

bridgdav said:

1... Put 4 balls on each side. If equal the lighter one is in the remaining 4 If unequal the lighter ball is in the side gone up. From the lighter set of 4 2... Put 2 in each side The lighter ball is in the pair gone up... 3... Put a ball in each The lighter ball is the one gone up... Makes 3 weighs...

Sadly, you don't know the ball is lighter.
That's why I deleted my first idiot post!

[justayellowbadge]

foster3jd said:

3. A1 v A2 example1 A1>A2 hence keep lighter group A2, discard A1 example2 A1<A2 hence keep heavier group A2, discard A1 4. compare remaining balls in group A2 example1 keep lightest example2 keep heaviest >> Edited by foster3jd on Wednesday 13th July 18:05

Not quite. Compare A1 to 2 previously discarded balls. If balanced, select A2 otherwise A1

Compare 1 of the previous balls to one of the selected. If balance, other ball is the odd one out, otherwise selected ball is.
Max 4 weighs.
I think.

[Fluffy]

justayellowbadge said:

Max 4 weighs. I think.

I agree with your logic, seems fine to me, can't fault it.
(Maths teacher)

[2 Smokin Barrels]

..but who nicked the other three reds?

[wizzpig]

It can be done in 3 moves folks

[foster3jd]

justayellowbadge said:

Not quite. Compare A1 to 2 previously discarded balls. If balanced, select A2 otherwise A1 Compare 1 of the previous balls to one of the selected. If balance, other ball is the odd one out, otherwise selected ball is. Max 4 weighs. I think.

Why go back to discarded balls which will weigh the same as those in either A1 or A2?

The whole point is to isolate the lightest/heaviest ball, hence do binary chop on the 4 balls selected from first two compares, which already give you the lighter/heavier information.

What you described will have exactly the same effect as my method.... and with the same number of compares!

Can I take my bow now?

[foster3jd]

wizzpig said:

It can be done in 3 moves folks

This better be good!

[foster3jd]

Fluffy said:

justayellowbadge said: Max 4 weighs. I think. I agree with your logic, seems fine to me, can't fault it. (Maths teacher)

Hang on .... that was my logic!!!!

[2 Smokin Barrels]

The least is three.

If you were lucky, with your first selection of one ball on each scale you may find a disparity.

Remove one ball and place on another from the pile. If it balances the ball you removed is the diparity.

Compare the disparate ball with the third ball (which we know to be standard) to see if it'd heavier or lighter.

The question was what could happen, not a guaranteed solution.

[2 Smokin Barrels]

A bit of a cheat

[wizzpig]

foster3jd said:

wizzpig said: It can be done in 3 moves folks This better be good!

individually identify each ball with letters or numbers. You then need to change the balls depending on the result. Each ball need to be weighed twice No more clues

But i bet ya nervy could tell you.