You have 12 identical balls. One of them is slightly different in weight than the rest. You have three weighs of a balance scale to determine which ball is different and whether it is heavier or lighter.
This is not easy… try hard to figure this out before looking at the answer.
• 1st weigh: Balls 1-4 vs. Balls 5-8 o There are 3 possible outcomes:
• They Balance. In the case that balls 1-4 balance with balls 5-8, we know the odd ball is in balls 9-12. So we weigh (2nd) balls 9,10, and 11 against three neutral balls (1,2, and 3) • If 9-11 balances with 1-3, the odd ball is number 12 in which case we weigh (3rd) it against another neutral ball to determine whether it is light or heavy • If 9-11 is heavier than 1-3 we weigh (3rd) ball 9 against 10. o If they balance: it is heavy ball 11 o If they do not: the heavier ball is the odd ball
• 1-4 is heavier than 5-8***. In this case we make balls 1-4 potentially heavy balls, balls 5-8 potentially light balls and balls 9-12 neutral balls. So, we weigh (2nd) balls 1 and 2 (potentially heavy) and 5 (potentially light) against balls 3 (potentially heavy), 7 (potentially light) and 12 (neutral) • If 1, 2, and 5 are heavier than 6,7, and 12, we weigh(3rd) the two potentially heavy balls (1 and 2) against each other o If they balance, it’s the third light ball (5) o If they don’t the heavier of the two is the odd ball • If 3, 7, and 12 are heavier, it’s either the heavy ball (3) on this side or the light ball (5) on the other we weigh (3rd) the heavy ball against a neutral one, if it’s heavier, then it’s the odd ball, if it balances, it’s the light ball that wasn’t weighed. • If 1,2,5 balance with 3,7,12 then the odd ball is in 4 (potentially heavy), 6, or 8 (potentially light). We then weigh (3rd) the 2 potentially lights (6 and 8) against each other. o If they balance, it’s the heavy ball (4) o If they don’t it’s the lighter of the two.
***The Third Outcome is 5-8 is heavier than 1-4 in which case the same procedure is followed but using balls 5-8 rather than 1-4
Comments
I found an alternate solution to this one: The first step is the same, but in step 2, when you've got 1-4 and 5-8 as containing the odd one out, weigh (1,2,5) x (3,4,9). If they're equal, then 6,7,or 8 is the light one, so weigh one against another. Otherwise, there are two outcomes: 1) if (1,2,5) is lighter, then either (1,2) is heavy, or 5 is light, so weigh (1) x (2). 2) if (1,2,5) is heavier, then either (1,2) is heavy, so again weigh (1) x (2) to find out which.
I've heard the variant of this where there's one ball that's heavier, but that's muuuch easier.
Yeah I've heard that one as well. Here's another one with 8: http://www.mytechinterviews.com/8-identical-balls-problem
So much more manageable...this one is tricky and takes some time.
There's also the 9 ball version with 2 weighs.
Am I missing something, but isn't the solution much easier?
Divide the twelve balls into 2 groups of six and weigh them Discard the lightest set Divide the remaining balls into 2 groups of three and weigh them Discard the lightest set You now have three balls Take two of the three balls and weigh them If they are different weights, you have your answer. If they are the same weight, the ball you did't weigh is the heaviest one
That would work if you knew that you were searching for a ball that's heavier. However, if the odd one out is lighter, you could just end up with three balls of equal weight in your final three.
Actually, you can find the lighter ball too. We select the first ball and compare it with the second. If they're equal (two balls are always identical from the three), the third ball is unique. Next, we compare the first one with the third one. It'll be either lighter or heavier. If it's lighter the third ball is lighter than the 11 remaining balls, otherwise we've found the heaviest ball.
I think this still assumes that you know whether the odd ball is heavier or lighter in advance. The question doesn't appear to give you that assumption.
This is a great one. I remember when I first saw this puzzle. I saw it in a logic puzzles book. I had a bit of trouble, so I looked in the answer section, and it only gave the solution if the two initial piles were even - which I had already figured out. It forgot to put in the alternative - so I spent a fair bit figuring this out myself, using a pack of cards to simulate the scenarios. Incidentally, my answer is slightly different, but not significantly.
If there's 12 identical balls and one of them is different, well, they're not identical are they...
I agree. I assume it means "identical in appearance".
This seems easy enough. What about something like this
1) Divide balls into 3 equal groups: [1-4] [5-8] [9-12] 2) Weigh each group side by side 3) In case we are looking for lighter ball, two groups of the three will weigh the same one will weigh slightly less. Now we know we are looking for lighter ball. Similar observation of heavier ball. 4) Say [1-4] weighs slightly less than [5-8] and [9-12] 5) Now make [1-4] into 2 groups: [1-2] and [3-4] 6) Pick one that weighs less, then further divide them so that finally you are comparing a ball to another ball side by side, and now you can pick the one that weighs the less.
Similar logic can be followed for heavier ball.
Unless I'm misunderstanding this explanation, your Step 2 will involve three weighings, so you won't be able to proceed with the final weigh.
A simpler solution would be -- Pick any 6 balls. Divide them into two groups of 3 and weigh them.
-- Now pick 3 balls from the GOOD set and weigh them against 3 balls from the NOT OK set. If they balance -- the ODD one is in the remaining 3 If they don't balance -- the ODD one is in the set of 3 balls that you had picked from the NOT OK set. From this we would also come to know if the ODD ball is heavier or lighter than the rest. Mark this set of 3 balls as ODD
-- From the ODD set of 3 balls, weigh any 2. If they balance, ODD one is the one that's left. If they don't, we can find which one is ODD ( based on the knowledge which we gained previously about the ODD ball being heavy or light)
this doesn't work if the first weighing was balanced and the second weighing was also balanced.
you need two weighing to distinguish 3 balls with 1 unknown
I agree with aquaepulse - it's easy to misread the question and think it's a special case.
Though this simplified problem is a good starting point, I think.