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## 12 Identical Balls 28

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.

see explanation

• 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

faizan

amazing......

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city_slick

I've heard the variant of this where there's one ball that's heavier, but that's muuuch easier.

Jay Lone

Yeah I've heard that one as well. Here's another one with 8: http://www.mytechinterviews.com/8-identical-balls-problem

Aakash

So much more manageable...this one is tricky and takes some time.

Ravi

There's also the 9 ball version with 2 weighs.

Alwyn

Weigh 1: Divide into 6 balls each and weigh. Pick the heavier group. Weigh 2 : Divide into 3 balls each from the heavier group in Weigh 1, Pick the heavier group Weigh3 : From 3 balls left, weigh any two if it balances the third ball was the heaviest. else the heavier ball was either of the ball which was heavier.

aquaepulse

This assume the unknown is heavier, it can be lighter.

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nqeron

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.

Reshu Singh

The solution is confusing. Definitely not correct. In case 1-4 is heavier than 5-8, what is your second weigh?1,2,5 vs 3,7,12 or 1,2,5 vs 6,7,12? You seem to be switching between these combinations while explaining the second weigh. Either way, you wont be able to determine the odd ball with your answer.

aquaepulse

Solution is definitely correct, but not explained the best.

1,2,5 v 3,7,12 is correct, the 6 is a typo and if you assume he meant 3 the logic is fine.

I think my explanation is clearer, but OPs is fine.

Shreyas Kanakpur

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Best Answer Yes, this problem can be solved. I had found a solution for this problem soon after it was published many years ago, in the Journal of the American Mathematical Association, if I remember correctly. It requires some detailed explanation. Anyone interested in knowing the steps, please send me an e mail via the YA e mail system, when I would give the solution via the YA e mail system. If you have not registered your e mail with YA in your profile, then I would not be able to send you the solution. I pose this problem for people to think about, when I am in a group of strangers interested in jokes/chat/puzzles, and my experience is that very few persons come back with the right solution. It is not an easy problem, but not too difficult either.

EDIT: On second thoughts, I find that the solution would not fit in the limited space available in the YA e mail box. So I have posted the solution below. I request interested members to go through the solution and point out any errors they find, in the form of an e mail to me through the YA e mail system. Thanks.

Solution for the 12 Balls Problem 1. Place any 4 balls in the left pan, and any 4 balls in the right pan. Mark the remaining balls A, B, C and D lightly so that the markings don’t add to the weight. 2. Case 1: If the scale shows no variation, then the marked balls contain the defective ball. Now, retain only 3 balls in the left pan and remove the other 5 balls from both pans. Take any 3 of the marked balls and place them in the right pan. (a) If the scale is still even, this means the 4th marked ball is defective. To find out if this is heavier or lighter than the remaining 11 balls, weigh the 4th marked ball with any one of the remaining 11 equal balls. (b) If the right pan tilts down during the second weighing, this means that those 3 balls (say A,B,D) contain the defective ball which also would be heavier than the other 11 balls. To isolate the heavier ball, clear the scales of all balls, take any two of the balls A, B and D, place say A in the left pan and B in the right pan. Whichever pan tilts down contains the heavier ball. If however the scale is even during this third weighing, then ball D is the heavier ball. (c) During the second weighing, if the left pan tilts down, then it means that one of the balls A, B and D in the right pan is the defective ball which is also lighter than the other 11 balls. Follow the same procedure in 2(b) to isolate the lighter ball. 3. Case 2: If the scale during the first weighing shows say the right pan tilting down, this means one of these 8 balls is defective and balls A,B, C and D are of equal weight. Now lightly mark the balls in the left pan P, Q, R and S and those in the right pan T,U, V and W. Leave ball P in the left pan and ball T in the right pan, retain the balls U,V and W separately. Transfer balls Q, R and S from the left pan to the right pan (so that the right pan would have balls T, Q, R and S). Place balls A, B and C (which are of equal weight) in the left pan so that the left pan would now have P, A, B and C. So now we get the second weighing P, A, B and C in the left pan and T, Q, R and S in the right pan. (a): If the scale is now even, then it means that one of the balls removed from the right pan, viz. one of U,V and W is not only defective, but also is heavier than the other 11, since during the first weighing the right pan tilted down. The heavier among U,V and W can then be isolated in the 3rd weighing by weighing one of them against another of them in the manner mentioned in step 2 above. (b): If the right pan still tilts down, it means either P is lighter than the rest or T is heavier then the rest. Determination of which between P (lighter) and T (heavier)- only one of these is possible- is done by weighing either P or T against any of the other 10 balls which are of equal weight. (c): If now the left pan tilts down, it would mean that one of three balls Q, R or S transferred from the left pan to the right pan, is the lighter one. The lighter among Q, R and S can then be isolated in the third weighing by weighing two of them separately against each other, one in each pan

cubixguy77

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.

Kara Luke

A better solution could be to divide the balls into three groups of 4.Then you only need 2 weighs to find out which ball is the different one and whether it is Lighter or Heavier....

Kara Luke

Lets say we have the groups A B and C.We randomly weigh the groups A and B , now let us say they are equal so group C is the one with the different ball. The only thing to do then is to weigh group C with one of the equals.If we get unlucky and group A is heavier than B, we weigh group A and C... It's always 2 weighs that way!!!

Nohairdave

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

Jay Lone

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.

ghostrider08

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.

Aakash

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.

tsubasa

A simpler solution would be -- Pick any 6 balls. Divide them into two groups of 3 and weigh them.

``````If they balance -- mark the set of these 6 balls as OK
If they don't balance -- mark the set of these 6 balls as NOT OK.
``````

-- 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)

aquaepulse

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

Aakash

I agree with aquaepulse - it's easy to misread the question and think it's a special case.

Aakash

Though this simplified problem is a good starting point, I think.

MrNutty

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.

ace25

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.

capt_obvious

If there's 12 identical balls and one of them is different, well, they're not identical are they...

Ravi

I agree. I assume it means "identical in appearance".

aquaepulse

Initially we know the set is 11 identical and 1 unknown which is either light or heavy (11i + 1u).

Weight 4v4.

This yields 2 possible states. When the scale balances (CASE1) we know 8 are identical and 4 are unknown either light or heavy (8i + 4u). When the scale is unbalanced (CASE2) we know 4 balls are potentially light and 4 are potentially heavy and 4 are identical, (4i + 4l + 4h).

CASE 1

Weigh iiu v iuu and exclude 5i+u.

If balanced then we mark the balls iii v iii and our state is 11i + u.

If the left was lighter then mark the balls iil v ihh and our state is 9i + l + 2h

If the right was lighter then mark the balls iih v ill and our state is 9i +2l +h.

CASE 2

Weigh lhh v lhh and exclude 4i + 2l

If balanced mark the balls iii v iii and our state is 10i + 2l

If left was light then mark the balls lii v ihh and our state is 9i +l + 2h

If right was light then mark the balls ihh v lii and our state is 9i +l +2h.

After 2 weighings we see there are only four cases: [9i+l+2h, 9i+2l+h, 10i+2l, 11i+u]

CASE 1: weight the 2 potentially heavy balls. If they balance the potentially light ball is the odd ball and it is light. If they do not balance the heavier ball is the odd ball and it is heavy.

CASE 2: Same as CASE1 by symmetry. Simply exchange light for heavy.

CASE 3: Weight the 2 potentially light balls, whichever is lighter is the odd ball and it is light.

CASE 4: Weight the unknown ball against any ball to determine if it is lighter or heavier.

Nigel Parsons

There is a solution to this which allows you to know in advance what three weighings/comparisons you will make, so that the choice of balls for the second and third weighing is not dependant on what has gone before. Label/identify each ball with a single letter: THE KP FORMULA Weigh: TAKE against FOUR PARK against THEM HALF against MORE Record the results of the three weighings (Left pan heavy / left pan light / pans balance) By comparing the results with the pan contents you can identify the odd ball and whether it is heavy or light every time. eg if the left pan always goes down then 'A' is heavy, if it always goes up 'A' is light. each ball has its own set of three unique weighings.

Pea Wormsworth

I found a solution like this. But I really like the way you worded it. After this I tried to find a solution where the rearrangement of the balls could be done consistently. It can. use 4 balls on each side of the scale and 4 balls not on the scale. Number each ball as having a position (1-12). Then after each measure, move say ball in slot #1 to slot #6, and ball in position #2 to position #4, etc. Each ball takes a new position based on this and re-measure. Then repeat the movement and take the 3rd measure. The result is a unique set of measurements that reveal which ball is unique and it's weight (heavy or light). I am saying that not only are the positions of the balls for each measure known, but also the re-arrangement of the balls between measures is also the same. Then I tried to apply a formula for the rearrangement process. So instead of a lookup table, the rearrangement was based on the position number. And I think there is also a solution for that. I don't have the equation any more, but I think it was like: (A*n1+B)%C = n2. Where n1 is the starting position, A, B and C are integers, "%" is modulus and n2 is the new position. And I think it worked. But I lost the formula, so I cannot prove it now. The end result would also be a unique set of measurements that results in a unique set that can find the solution from a lookup table. Finally, I thought that since each weight results in one of 3 outcomes, that the sum of the weightings is a number in base 3. So, the final step in simplification would be to either modify the rearrangement equation or perform a final equation on the resulting number to make the solution be a number which reveals the solution in order of the starting positions of the balls. This would remove the lookup table required on the measurements made to identify the final ball and its mass. But alas, I never found the answer to this last problem before I lost interest and lost all my notes. The reason for doing all this is that I like "golf", in a programming sense. I think taking a puzzle and reducing it down to minimal code to solve it is very interesting. I suspect that if a computer program was written to provide a structure for the problem at hand, that the resulting solution code would be about 2 or 3 lines of recursive code. Which would be far more efficient than all the if/then type conditions and lookup table required for most of the obvious solutions people are so proud of online. I have searched a little to see if anyone has done it, but I haven't seen it yet. I am passing this to you, because I think someone with math knowledge and an understanding of the problem should be able to either find the solution I described or prove that it is not possible. If you ever hear of someone with that solution you described, encourage them to complete the job and simplify their solution down to its most reduced form.

Sai Krishna

I have a doubt if this is right.

There are 12 balls out of which 11 are identical in mass. Take the first one. Keep it on one side of a balance. Then in a series, keep the other balls one by one, on the other plate.

There are 2 possible outcomes. 1) out of the 11 balls, 10 are identical in weight with the 1st ball. Find out which is different and if it's heavier or lighter. 2) the ball you took is the different one. In this case, check whether the other balls are different in weight compared to this one, and note down.

Is it right? Or am I missing something?

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