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12 Ball Bearings Puzzle

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bellenuit

bellenuit

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There are twelve ball bearings, all the same size, shape and colour. All weigh exactly the same, except that one ball bearing is slightly different in weight, but not noticeably so in the hand. Moreover, the odd ball bearing might be lighter or heavier than the others.

Your challenge is to discover the odd ball bearing and whether it is lighter or heavier. You must use a beam balance only, and you are restricted to three weighing operations.

A beam balance is a simple scales that has two trays, one on either side. Whichever tray contains the heavier load will go down, the other tray up. So it only tells you which side is heavier, but doesn't tell you the actual weight of either side.
 
mazzatelli said:
To start off:

Separate into groups of 3 - i.e. 4 balls/group

Weigh 4 against 4 first :)
Click to expand...

If one group is heavier than the other, then you know the rogue bearing is in that group. If they are equal you know the rogue bearing is the unweighed group.

Now you have a group of 4 that you know the bearing is in. Split the group and weigh 2 against 2, split the heavier group and weigh again to find the bearing.
 
mazzatelli said:
To start off:

Separate into groups of 3 - i.e. 4 balls/group

Weigh 4 against 4 first :)
Click to expand...

If we number the balls 1 to 12.

First Weighing

Weigh 1 to 4 against 5 to 8

If they balance, then the odd ball is in 9 to 12.

Assuming they balance

Second Weighing

Weight 9 and 10 against 11 and 1 (1 we know is a good ball)

If 9 and 10 are heavier, then either 9 or 10 is an odd ball and is heavier or 11 is an odd ball and is lighter.

Assume 9 and 10 are heavier than 11 and 1

Third Weighing

Weigh 9 against 10

If they balance, then 11 is the odd ball and is lighter
If they don't balance, then whichever is the heavier is the odd ball and is heavier.

Assume 9 and 10 are lighter than 11 and 1

Third Weighing

Weigh 9 against 10

If they balance, then 11 is the odd ball and is heavier
If they don't balance, then whichever is the lighter is the odd ball and is lighter.

Assume 9 and 10 are the same as 11 and 1

Then 12 must be the odd ball

Third Weighing

Weigh 12 against any other ball. That will tell you whether 12 (the odd ball) is heavier or lighter.

That's only part of the solution......

Assume 1 to 4 weighed against 5 to 8 (first weighing above) don't balance...

I'll need to think a bit more on that outcome
 
I don't know but I suspect it comes down to two weighings of 6 ball bearings.

Weighing 4 (or 8) at a time can result is possibilities where there are 4 ball bearings left over after two weighings.

EDIT:
Same problem from two weighings of 6 balls as the number of balls that can be weighed at any given time must be even to yield any useful information.
 
bellenuit said:
There are twelve ball bearings, all the same size, shape and colour. All weigh exactly the same, except that one ball bearing is slightly different in weight, but not noticeably so in the hand. Moreover, the odd ball bearing might be lighter or heavier than the others.

Your challenge is to discover the odd ball bearing and whether it is lighter or heavier. You must use a beam balance only, and you are restricted to three weighing operations.

A beam balance is a simple scales that has two trays, one on either side. Whichever tray contains the heavier load will go down, the other tray up. So it only tells you which side is heavier, but doesn't tell you the actual weight of either side.
Click to expand...

I advocate 4x4 because 6x6 doesn't tell us what we already do not know - i.e. one side will be heavier than the other

There are various scenarios - I'll go through one

Scenario 1:
1) Weigh 4x4 [Let's call it pile A & B]
If they balance - meaning the odd ball is in the other pile we left out [Pile C]

2) Now leave 3 of the balls that we have already weighed [it is safe to assume they are normal] and weigh it against 3 from Pile C
If they balance - the remaining ball from Pile C must be odd one out

3) Then weigh the odd ball against a normal ball to see if it is heavier or lighter

Now there's also the scenario where 4x4 doesn't balance out straight away or 3x3 in step 2 doesn't balance
 
weigh 6x6

3x3

1x1

if 1x1 is equal the 3rd ball is odd one out

if not equal you have your answer there too


That will be $59.95 thankyou.

I would appreciate no questions or arguments on my calculations as i am a ramper not a scientist :)
 
bellenuit said:
That's only part of the solution......

Assume 1 to 4 weighed against 5 to 8 (first weighing above) don't balance...

I'll need to think a bit more on that outcome
Click to expand...

Carry on from where I left off.

Assume 1 to 4 is heavier than 5 to 8 after the first weighing

Second Weighing

Weight 1 2 and 5 against 4 6 and 9 (9 we know is a good ball bearing)

If 1 2 and 5 are heavier, then either 1 or 2 is the odd ball and is heavier, or 6 is the odd ball and is lighter.

Assume 1 2 and 5 are the heavier

Third Weighing

Just weight 1 against 2. If they are the same, then 6 is the odd ball and is lighter. If they are different, then whichever is the heavier is the odd ball and is heavier than the rest.

Assume 1 2 and 5 are the lighter

Then either 5 is the odd ball and is lighter or 4 is the odd ball and is heavier.

Third Weighing

Weigh 5 against a good ball (9 say). If they are the same, then 4 is the odd ball and is heavier. Otherwise 5 will be the odd ball and is lighter

Assume 1 2 and 5 are the same weight as 4 6 9

Then either 3 is heavier or 7 and 8 are lighter

Third Weighing

Weigh 7 against 8

If 7 is heavier than 8, then 8 is the odd ball and is lighter.
If 7 is lighter than 8, 7 is the odd ball and is lighter
If 7 and 8 are the same, 3 is the odd ball and is heavier

The only option not taken into account is....

Assume 1 to 4 is lighter than 5 to 8 after the first weighing

This is the same as the above, except heavier and lighter are reversed in the logic.
 
drsmith said:
This might help.

http://puzzlingtimes.com/puzzles/12ballbearings/12ballbearings.php

It's easy in 4 weighs.
Click to expand...
Did it in two:cool:

Weighed ball 1 against 2 = 2 heavier
Weighed ball 2 against 3 = 2 heavier

Ball 2 is the odd one and it's heavier;)

Just lucky on the location of the odd ball, if it were ball 12 it would have taken 12 weighs with this method.



Tried it again, seems 4 weighs is the max you need to determine the location and weight by:

Weigh ball 1-6 against 6-12 = uneven scales

Weigh ball 1-3 against 4-6 = either even or uneven depending on whether the odd ball is in this group. This step tells you which group of 6(and if lucky 3, you can knock out one weigh) the odd ball is in and whether it is heavier or lighter.

If scales were even:
Weigh ball 7-9 against 10-12 = uneven scales. Now you know which group of 3 has the odd ball.

Weigh the first two in the group of three. If the scales are even or uneven you know which ball is the odd one, by both the scales and because you know whether it's heavier or lighter

cheers
 
You need to weigh 4 vs 4

Eg

Weighing 1
1-4 vs 5-8 - say 1-4 is heavier

Weighing 2
1-2 vs 3-4 - say 1-2 is heavier. So now we know that the fake bearing is heavier because 1-4 was heavier and now 1 or 2 is heavier but 3 or 4 cannot be lighter.

Weighing 3
2 vs 3 - say 2 is heavier. Now we know that the fake bearing must be 2.

There are different scenarios but this is the gist of it, will try to play more out later.
 
MACCA350 said:
If scales were even:
Weigh ball 7-9 against 10-12 = uneven scales. Now you know which group of 3 has the odd ball.
Click to expand...

No you don't. Because you don't know whether the odd ball is heavier or lighter, if 7-9 are the heavier, it could be because one of these are the odd ball and is heavier, or 10-12 has the odd ball and is lighter
 
mazzatelli said:
I advocate 4x4 because 6x6 doesn't tell us what we already do not know - i.e. one side will be heavier than the other

There are various scenarios - I'll go through one

Scenario 1:
1) Weigh 4x4 [Let's call it pile A & B]
If they balance - meaning the odd ball is in the other pile we left out [Pile C]

2) Now leave 3 of the balls that we have already weighed [it is safe to assume they are normal] and weigh it against 3 from Pile C
If they balance - the remaining ball from Pile C must be odd one out

3) Then weigh the odd ball against a normal ball to see if it is heavier or lighter

Now there's also the scenario where 4x4 doesn't balance out straight away or 3x3 in step 2 doesn't balance
Click to expand...

Scenario 2:
1) Repeat as above and the result is 4x4 balances

2) Repeat as above
BUT in this case Pile C [3x3] does have an odd ball
If Pile C weighs down, it is a heavier ball or vice versa [lighter]

3) From the 3 in Pile C
weigh 1x1
If the result balances - the odd ball is the remaining one
If it does not balance - since we figured out if it is heavier/lighter in step 2 - we can figure out which ball is the odd one.

So there is the scenario where the initial 4x4 does not balance...
 
bellenuit said:
No you don't. Because you don't know whether the odd ball is heavier or lighter, if 7-9 are the heavier, it could be because one of these are the odd ball and is heavier, or 10-12 has the odd ball and is lighter
Click to expand...
Yes you do, you found out the weight in the second weigh as I mentioned.

cheers
 
bellenuit, with that process you seem to end up at the third weighting with 3 possibilities...

If you split the balls into 4 groups of 3;

First Weighting:
Weight 1,2,3 against 4,5,6.

If not equal:
Weight heavier group against 7,8,9. If this turns out to be equal, the ball is lighter, and in group 4,5,6. If 1,2,3 is heavier then 7,8,9 then the ball is heavier and in 1,2,3.
So the ball can easily be worked out, by using the thrid go to do 1,2 or 7,8 to see which one is the heavier or lighter ball.


If they are equal, then the ball is in 7,8,9,10,11,12.
So use the second weight to do 7,8,9 on 1,2,3. If 7,8,9 is heavier, implies that the ball is heavier, and you can weight 2 balls from 7,8,9 to work out which one is the right one. If its lighter, same process.

If 7,8,9 equals 1,2,3 we are screwed, as 10,11,12 has the right ball, and could either be heavier or lighter. We have one weighting to work it out... I'm not sure how this can be done for sure?
 
Seems that the first weighing has to be a 3v3.

4v4 = rogue ball in a group of [4 or 8]
3v3 = rogue ball in group of [6]

(ie.
if 3v3 balance, rogue ball is in other group of 6
if doesnot balance, rogue ball in current group
)

as we are looking for a scenario where it 'always works' so 3v3 > 4v4.
 
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