The Hardest Riddle I Know

[mcfbridge]

This is the hardest riddle or logic puzzle I have ever run across. I know I had a riddle thread, but I wanted to post this one seperately. I'll leave this alone until for at least a week. If after that, no one has posted the correct answer, and people have requested it, I will post the solution.


You have 12 ball bearings (metal marbles) Each of them weighs an identical amount, except for one which is slightly different from the others.

You have a balance scale, the kind with a basket at each end. Using the scale only 3 times, find the odd ball. Remember, you don't know if the odd one is heavier or lighter, only that it is different.

There is no cute trick to this or funny answer, it is a straightforward, but very difficult logic puzzle. Good luck.

PS. Anyone is welcome to post an answer, and I will respond to it as soon as I can. I do ask, that if anyone has actually heard this before and knows the answer, please don't post it for a few days.

 

[Nemasis Enforcer]

You sure there is an answer to this...

Im just racking my brain and hope to have it if there is one in a few hours or maybe the next day or too...

 

[SummerMorning]

You know, you could just use the scale about 12 times and you'd find the odd one faster...

hmm...

1st step would probably be to weigh six and six, to find out where the odd one is... no... that wouldn't work. Hmmm...

1. weigh two ball bearings (there's a 1 in 6 chance you've already hit the odd one, in that case, replace one sphere and weigh again. This way you could have the odd one in 2 steps, but this method, like I said, has only a 1 in 6 probability.

2. If they weigh the same, add 1 more to each side. Now you have a 1 in 5 chance of getting the odd one, but this means there's only a 11 in 30 chance of actually hitting on the right one in 3 steps, which is slightly not good enough...

hmm...

 

[Nemasis Enforcer]

You know, you could just use the scale about 12 times and you'd find the odd one faster...

hmm...

1st step would probably be to weigh six and six, to find out where the odd one is... no... that wouldn't work. Hmmm...

1. weigh two ball bearings (there's a 1 in 6 chance you've already hit the odd one, in that case, replace one sphere and weigh again. This way you could have the odd one in 2 steps, but this method, like I said, has only a 1 in 6 probability.

2. If they weigh the same, add 1 more to each side. Now you have a 1 in 5 chance of getting the odd one, but this means there's only a 11 in 30 chance of actually hitting on the right one in 3 steps, which is slightly not good enough...

hmm...

You only get to use the scales 3 times and thats where my mind goes blank

How to get an odd ball from 12 not knowing if it lighter of heavier then the others in three moves

 

[gauchecritic]

Still working on it but 4 lots of 3 ball will give give you two groups of three in one weighing, one of the groups being different, whether it be those weighed or those not.

If you weighed the group with a difference first time then removing one by one only takes two more chances.

Don't know what happens after that.

Edited to add. it doesn't matter which is weighed first because the next weighing is two balls.

Gauche

 

[Tanuki]

You said all 12 balls have identical weight. Is that wrong? I think you mean one of the 12 is either a little lighter or a little heavier, right?

 

[Lauren Hynde]

Well, I'm down to doing it in 4 measurements, but not in 3 yet (unless I'm lucky with the 2nd)

33% chance of needing to use the scale 3 times, 67% chance of having to use it 4 times...

 

[R. Richard]

I won't post the answer. However, I will post a hint.

The key to solving the problem is to obtain the maximum amount of useful information from each weighing. In order to obtain the maximum amount of useful information, it is necessary to weigh only some of the balls. In that manner, if the balls weighed balance then the remaining fewer balls can be evaluated in the two remaining weighings.

The correct approach is to split the balls into three groups of four and weigh two of the groups, one agaionst the other. The answer is still a bit tedious to work out and requires one 'non-standard' approach.

Happy puzzling!

R. Richard

 

[Tanuki]

Ok here you go, that wasn't so hard = )

group the balls into groups of 3, call them A, B, C, D

1. Measure group A and B.
- if they balance, A and B are normal, go to step 2a
- if they don't balance, then C and D are normal, go to step 2b
(take note of whether the odd group was heavier or lighter)

2a Measure group A and C
- if they balance, then A, B, and C are normal, go to step 3d
- if they don't balance, then A, B and D are normal, go to step 3c
(take note of whether the odd group was heavier or lighter)

2b Measure group A and C
- if they balance, then A, C, and D are normal, go to step 3b
- if they don't balance, then B, C and D are normal, go to step 3a
(take note of whether the odd group was heavier or lighter)

3a Group A has the weird ball
3b Group B has the weird ball
3c Group C has the weird ball
3d Group D has the weird ball
- measure two the 3 balls - if they balance, the 3rd ball is the weird one. If they don't balance, one of the two remaining balls is weird. To know which is the one, go back to step 1 or 2 and see if the weird ball caused the balance to be heavier or lighter, that'll tell you which of the two balls is the odd one.

 

[Tanuki]

I won't post the answer. However, I will post a hint.

The key to solving the problem is to obtain the maximum amount of useful information from each weighing. In order to obtain the maximum amount of useful information, it is necessary to weigh only some of the balls. In that manner, if the balls weighed balance then the remaining fewer balls can be evaluated in the two remaining weighings.

The correct approach is to split the balls into three groups of four and weigh two of the groups, one agaionst the other. The answer is still a bit tedious to work out and requires one 'non-standard' approach.

Happy puzzling!

R. Richard

Ah, you beat me to it by 3 minutes! It took quite a while to type all that stuff out.

 

[Guest]

1 - 6 v 6

2 - 3 v 3

3 - 1 v 1 (if it's equal, then you know the odd one out. if it's not equal, you still know the odd one out.)

1 minute, tops.

 

[Tanuki]

1 - 6 v 6

2 - 3 v 3

3 - 1 v 1 (if it's equal, then you lnow the odd one out. if it's not equal, you still know the odd one out.)

1 minute, tops.

Sorry Chilled, it's not that simple for this reason. you don't know whether the ball is heavier or lighter. So how do you know which of the six has the odd ball?

 

[smutpen]

I won't post the answer. However, I will post a hint.

The key to solving the problem is to obtain the maximum amount of useful information from each weighing. In order to obtain the maximum amount of useful information, it is necessary to weigh only some of the balls. In that manner, if the balls weighed balance then the remaining fewer balls can be evaluated in the two remaining weighings.

The correct approach is to split the balls into three groups of four and weigh two of the groups, one agaionst the other. The answer is still a bit tedious to work out and requires one 'non-standard' approach.

Happy puzzling!

R. Richard

Fascinating puzzle.


I think your hint is not too good, though.

Saluki has it right; the key to the problem is recognizing that you need groups of three, because if you know whether the oddball is heavier or lighter, you can pick out one of three with just one measurement.

 

[neonlyte]

Tanuki, doesn't your solution require a fourth weighing for confirmation?

I think I have a solution it if the first weighing is balanced, I cannot yet see a solution if the first weighing is unbalanced.

 

[mcfbridge]

Ok here you go, that wasn't so hard = )

group the balls into groups of 3, call them A, B, C, D

1. Measure group A and B.
- if they balance, A and B are normal, go to step 2a
- if they don't balance, then C and D are normal, go to step 2b
(take note of whether the odd group was heavier or lighter)

2a Measure group A and C
- if they balance, then A, B, and C are normal, go to step 3d
- if they don't balance, then A, B and D are normal, go to step 3c
(take note of whether the odd group was heavier or lighter)

2b Measure group A and C
- if they balance, then A, C, and D are normal, go to step 3b
- if they don't balance, then B, C and D are normal, go to step 3a
(take note of whether the odd group was heavier or lighter)

3a Group A has the weird ball
3b Group B has the weird ball
3c Group C has the weird ball
3d Group D has the weird ball
- measure two the 3 balls - if they balance, the 3rd ball is the weird one. If they don't balance, one of the two remaining balls is weird. To know which is the one, go back to step 1 or 2 and see if the weird ball caused the balance to be heavier or lighter, that'll tell you which of the two balls is the odd one.

Sorry won't work.

Assume A & B balance.
You now know it's in C or D.

Now measure groups A & C
If they balance ball is in Group D.

Since all steps previously balanced, you still don't know if the ball is heavier or lighter, so your last step won't work.

Nice try, though. Well thought out.

 

[Tanuki]

Sorry won't work.

Assume A & B balance.
You now know it's in C or D.

Now measure groups A & C
If they balance ball is in Group D.

Since all steps previously balanced, you still don't know if the ball is heavier or lighter, so your last step won't work.

Nice try, though. Well thought out.

Ah yes, you're right! In that case it wouldn't work. Interesting! It's a great riddle by the way, love it!

 

[Colleen Thomas]

1. Measure 6 vs 6. while the balls are still on the scal remove 1 from each tray. If they remain in the same balance, discard those 2 & repeat until you remove a pair that changes the scale to balance.

2. You now hold one regular & 1 iregular. Take one of the discarded balls (which have to be regular) and weight it against one of the two in your hand.

3. Weigh the other. In any case You have your oddball.

-Colly

 

[mcfbridge]

1. Measure 6 vs 6. while the balls are still on the scal remove 1 from each tray. If they remain in the same balance, discard those 2 & repeat until you remove a pair that changes the scale to balance.

2. You now hold one regular & 1 iregular. Take one of the discarded balls (which have to be regular) and weight it against one of the two in your hand.

3. Weigh the other. In any case You have your oddball.

-Colly

Cute, but no. The puzzle is straightforward. Changing the number of balls in the basket is considered a use.

 

[Tanuki]

1. Measure 6 vs 6. while the balls are still on the scal remove 1 from each tray. If they remain in the same balance, discard those 2 & repeat until you remove a pair that changes the scale to balance.
-Colly

I assume that counts as multiple weighings.

Does this remind anyone of the movie Die Hard 2?

 

[Nemasis Enforcer]

If you don't know if its heavier or lighter how can you possibly work out which is odd in 3 moves... Im dying to see this answer!!!

 

[Nemasis Enforcer]

I can do it ie find the odd one but things could go wrong and then I can only get it 1 in 2... but what I've got is...

Move 1 - 3 balls on each side if they balance its one of the other 6 so put these to one side... if it doesn't then its one of these so put the others aside...

Move 2 - now your down to 6 put 2 on each side... if they balanced they are all the same, put these to the side with the others and your left with 2...

Move 3 - place one of discarded ones on the scales and one of the 2 left on the other side... if they balance its the other one left thats odd if not its this one.

(But of cause if Move 2 doesn't balance then your left with 1 of 4 being odd and not enough moves to fine out which .)

 

[smutpen]

Sorry won't work.

Assume A & B balance.
You now know it's in C or D.

Now measure groups A & C
If they balance ball is in Group D.

Since all steps previously balanced, you still don't know if the ball is heavier or lighter, so your last step won't work.

Nice try, though. Well thought out.

Damn!

I went at it another way, but I still got caught with a 50-50 choice in the last two if every measurement turned up even.

Either there's a way to ensure that I get the info about weight, or there's a way to make the determination without knowing...

You are evil, Mcfbridge.

 

[neonlyte]

If you don't know if its heavier or lighter how can you possibly work out which is odd in 3 moves... Im dying to see this answer!!!

I agree! I can find an odd ball(s) in three weighings but can't determine if it's heavier on LH side or lighter on RH side

I'm sure the way is to swap sides,
eg:
1. 5 LH / 5 RH if equal then one of 2 unused is odd and can be found in two moves.
If unequal, note direction.
2. Take, and note, 3 LH swap with 3 RH, if stays same problem is with 2 in each tray unmoved, remove the 3 swapped between trays leaving two unmoved in each.
3. Swap one from each tray, if stays same problem is with one of the balls not moved, but which? If direction changes, problem is with one of the balls moved, but which?

You can solve it with another move.

PS If direction changes on second weighing you can solve in one move by switching sides again and removing balls.

 

[Colleen Thomas]

All right. The hard way then.

1.) Break em up into sets of four (A, B, C) then measure A vs B.

If they balance your odball is in set C. (case A)
If not then it is in A or B (Case B)

2-a).If your oddball is in C, then measure 3 C's vs. 3 As. Should they balance your 4th C was the odd. Should they not and the C side falls then one c is heavy, if they rise then 1 C is light.

3-a) You then weight c vs c. If they balance, you have your oddball in hand, if not, since you now know your C is heavy or light, you have your answer by which dise is heavy or light.

On the other hand if A or B is not in balance then you know C is all regular. this leades to case B.

Gets ugly if A & B don't balance. But you know by the scale you either have a heavy A or a light B if the A side falls or a heavy B or light a if the B side falls.

Reset your groups. (A1,A2,A3, C1) (B1,B2,B3,A4) (C2,C3,C4,B4)

2-B) Measure (B1,B2,B3,A4) vs (C2,C3,C4,B4)

If they Balance your oddball is in (A1,A2,A3, C1) (Case B1)
3-B1) then Measure A1 vs A2, if they balance A3 is odd, if they don' you will by the above you know your A was heavy or light, so if it's light the one that rises is your odd ball, if they are heavy the one that falls is your oddball.

If they don't balance (Case B2)

3-B2)You measure A4 vs. any C. If they balance you know you have a light B, if they don't then you are holding the Heavy A.

3-b3) If it's a light B, weight B1 Vs B2. If they balance B3 is your odd ball, if they don't, the one theat rises is your light B.

The hard way.

-Colly

 

[neonlyte]

Brilliant Colly !