Another brain teaser – 12 balls puzzle

It’s a sample of using Component Search (Shainin DOE) in Keki Bhote’s book.
BTW
I wonder how it could be solved with e.g. Ishikawa diagram???

 
Lets take 4 balls on both sides in our FIRST Trial
Case I
OOOO = OOOO ( Both sides are equal)
Both Sides are equal.
This means the defective is in the remaining four
Now Lets take three out of remaining 4 on the one side
and THREE PROVEN GOOD BALLS on other side our second trial .
Now again if both of them are equal ie OOO = OOO
the defective is the last one which could be detected in the
LAST trial with a good ball and could be identified whether it is heavy or light.
Now again if both of them are not equal in second trial OOO # OOO then we know that these three have one defective and we know whether it is light or heavy depending upon which side the balance moves now
From these three balls keep one each on both sides in third trial.
If they are equal O = O then defective is the left one. ( Light or Heavy is decided on the second trial itself)
If they are not equal, defective one is identified depending upon inference of the previous trial ( Light or heavy)
Case II
OOOO # OOOO (Both sides not equal)
It means the remaining 4 does not have a defective ball
Second Trial :
Remove one ball from one side (lets say form up side) and two balls from the other side (lets say which is down) and put a good ball on the side where two balls are removed.
Eg if a b c d are in down side and e f g h are in up side and I j k l are the proven good balls then
Make it b c g and I d h ( ie a and e removed and f replaced by a good ball and position of d and g has changed)
Now look at the possible cases of this trial ( Second One )
IF BALANCE DIRECTION CHANGES that means the removed balls are good balls and there is a interchange of defective ball ie either g is light or d is heavy. Which can be easily detected in third trial by putting g or d in the third trial with a good ball.
 
NO CHANGE IN BALANCE DIRECTION : This means removed balls are good and interchanged balls are also good. So either b or c is heavy or h is light. Put these two ( b, c) in the third trial whichever is heavy is the defective and if equal h is defective (light).
BALANCE BECOMES EQUAL: This means removed balls are defective ie a is heavy or e / f is light.
Put e and f in third trial and if they are equal a is heavy otherwise the lighter one is defective.
Nayan
BB

Ypu are right. I wrongly assumed (scarey word) that the out of spec ball would be heavy. Someone else posted a reply earlier that had a more valid solution.
Thanks
Ed

Both explanations are confusing. I have some objections. I wonder if you guys are still checking this message board. If so, I will write you what steps are not clear.
Thanks,
Alex
 

I think dividing 12 balls into 4 groups of 3 is a little bit easier to solve this brain teaser. group1: a , b, c; group 2: d, e, f; group 3: g, h, k; group 4: l, m, n;
put group 1 and group 2 on both sides of scale, if  they are equal, the defective one is in the rest 6 balls of group 3 and group 4. Otherwise, the defective one is in the 6 balls of group 1 and group 2.  Then use the same idea Nayan uses to find the defective. Make clearer, in this way, the most complicated case is to find the defective among 6 balls, which is a little bit easier than to find the defective among 8 balls in Nayan’s solution.
Besides, other cases also seems a little bit easier. Try yourself if you are interested.
Jacqueline 
  

Alex
Sorry I sent you the wrong E-mail address
The correct address is:
[email protected]
Tony
 

You have i too easy.  In three trials you should not only pick out the odd ball but should tell whether it was lighter or heavier than the others. Ed

Place 6 balls in each pan. The heavier of the two contains the odd ball.  Set the other 6 off. 
From the heavier set of 6, split this into two sets of three.  Retain the heavier set and set the other off to the side.
Select two of the remaining 3 and place one in each pan.  If they are of equal weight, then the odd one is the one that’s not in the pan.  Otherwise, the scale will reveal the heavier one.

Let us concentrate in the way of of placing the balls in the pans.I think is not a big assumption that we can place the balls in the pans one by one. If you accept this, then is possible to complete the task using the balance two times. Start placing the balls in the pans in pairs one in each pan at times. At some moment the balance will be out of balance. Then one of the two balls placed in the last moment will be the defective one. Name this balls B1 and B2.  (B1 the heavier, B1 > B2) B2) B2) Take this ball and place it in pan 1. Place any other ball (say B) in pan 2. Case 1:   B1=B   then the defective one is B2 and is light.Case 2:  B1>BBB  then B1 is defective and is heavy.Case 3:  B1

Sure you do!  All of the previous separations have been based on keeping the set that is heavier.  So, when you are down to the last 3 balls, one of them is heavier than the other 2.  If the two on the scale are of equal weight, then the remaining ball has to be the heavy one.

Hello folks,
On the possum farm we keep things simple!  If I had 12 balls and needed to find the light or heavy one I’d put them on a scale one at a time and write the weights down.  Then I would look at the weights I wrote down and would see which one is different and say..”hey, that one’s different!” 
And I bet I’d get the answer quicker …so much for muda!  If you want to discuss this I’ll be out in the possum pen playing catch with a couple of the smart possums.  You can continue to group and messarge your balls until you get an answer that feels good.
Later,
Billybob

Hello Bobby Joe Jim Bob,
Great story….are we related?
Later,
 Billybob

You said:
“B: If one side or the other weighs out then that group of four has the out of weight ball.”
So yopu put 4 and 4 for the first weighting. Then one side goes up and one side goes down. Which is “that group of four has the out of weight ball”? Since you don’t know if the out-of-weight ball is heavier or lighter, it could be a heavy one in the lower side or a light one in the upper sisde.

This is an old problem which has once again been raised in Britain’s “Daily Telegraph” the solution can be found using a mnemonic (this is not my solution)
Label the balls with individual letters “THE KP FORMULA” then make three test weighings using
TAKE against FOUR
PARK against THEM
HALF against MORE
Recording the results of these three weighings will give you the result you require, identifying which ball is ‘non-standard’ and whether it is light or heavy (compared to the standard). For example, ball ‘A’ is in the left pan every time. if it consistently goes up/down, you know ‘A’ is light/heavy.
‘E’ is in the left pan initially, then in the right pan twice. If results for these pans are consistent then ‘E’ is the odd ball. (similarly for ‘R’)
If any of the three ‘weighings’ balance then the 8 balls in that weighing can be eliminated from the record of the other two weighings and a single solution will be found.
 
Nigel

OK, but what do you do if this produces a balanced scale?
The bad ball must then be found in one weighing from 3 UP balls and 2 DOWN balls.
Weighing 2 v 2 (UP +DOWN vs UP + GOOD) doesn’t work.  If UP + DOWN goes down, its either that DOWN or the opposite UP.
What am I missing?

I suggest the following solution:
Part one
1. Weigh 4 and 4. 1.1. They are equal. From the remaining four take one and replace it with one of those determinet as being OK. Weigh them 2 and 2. 1.1.1. They are equal. Ball taken at 1.1. is quilty the third weigh will determine if is lighter or heavier. 1.1.2. It is heavier on the side with one  suspected ball. Take out the suspected ball and replace with a ball from the other side of the balance and replace with an inocent ball. 1.1.2.1. They are equal. Secondball taken out is guilty, and is heavier. 1.1.2.2. The balance changes its inclination. The moved ball is quilty, is lighter. 1.1.2.3. The balance stays the same. The only one ball moved is guilty and is lighter.  
Part two to follow…

first put 6 ball on each pan. see which weighs lighter . share the lighter six on two pan. again you will find one lighter . out of 3 put one of tem on each pan you will find one of them will be lighter or weighing the same. by this you will come to know which ball is weighs less of the 12.

Irfan ur assuming the balls are lighter which is a no- no, the balls cud have been lighter and by the time ur wiser u’d have exhausted a chance. I think Nayan has solved it brilliantly and if you cud look at his answer carefully u’ll get it. Lemme try to put it the way I understand what he said.
3 sets of 4 balls.
1)abcd     2)efgh    3)ijkl
lemme get to the most complicated path which is after weighing 1 and 2 , they don’t balance, meaning 3(i,j,k,l) has all good balls. Lets assume 1 goes up and 2 goes down ( either way its symmetric)
make 3 groups
 4)aeh   5)bdf    6)cgi
now weigh 5 and 6. 3 possibilities
5 goes up:    its either because b and d were lighter or because g was heavier. next weigh b and d, the one thats lighter is the defect ball and if they balance g was heavier.
6 goes up:  its either because f was heavier or c was lighter. Weigh either of them with a good ball and find the defective ball.
they balance: the defective ball is one of a, e or h. Weigh e and h (both of whom were heavier in the second weigh), the heaveier one is the defective ball and if they balance a is lighter and defective.

I just looked more carefully at the previous comments and noticed that Nayan had already solved the puzzle (his and my answers are slightly different, but either one works).  Why keep talking about it? 

And who said that the odd ball was heavier?
Your solution does not work, so your request “let me know if there is an easier, better and quick solution for this” does not apply.

Ok. Let’s start from scratch:
You have 12 balls from which 11 have the same weight and 1 weights different, either more or less.
You put 6 balls in one pan and 6 balls in the other pan. The 2 pans will not balance for sure (you don’t need to waste 1 wighting to learn that), so you will have either 1 too heavy ball in the low pan or 1 too light ball in high pan.
Now please go on and identify the out-of-weight ball in the next two weightings left… (you can’t)

I stand corrected.

Gabriel,
I stand corrected as well. However, I have another solution-
We know 1 ball is odd ball and we have 3 weighings. This ball is lighter or heavier.
Start with 1 ball each in the left and right pan. If the pans balance, we know that these 2 balls are ideal weight. We set them aside. If the pans do not balance, we remove 1 of the balls and replace with one of the ones from the remaining 10. If the pans balance, one we took out is odd if not the one left in the pan we did not take out is the odd one.
If we have not found our ball yet, we then place 5 balls in each pan (left and right). One of the pans is going to be high. This is the 2nd weighing. We carefully remove 1 ball each from the 2 pans. We have 3 balls left in each pan. If the pans balance we have the odd ball in our hands from the removed ones. If we do, we compare the one of the ones with the ideal one we have set aside and find the odd ball (using our third weighing). If we find that the pans are not balanced, we remove 1 each again from each of the pans. Eventually we will have found the odd ball in the pans or in our hand.
The third weighing will be the comparison weighing.
PB
PS – No one says I can not remove balls from the pans. So I resolve it this way.

You must be joking!
I can identify 1 different ball among 1,000,000 with only ONE weighting!
Just put one and one.
a) If they do not balance jsut swithch one of the balls with a not weighted on. If they balance the switched one is the faulty ball. If they do not balance the not removed ball is the faulty one.
b) If they balance add one and one, and so on. Eventually the pans will unbalance. Then remove one of the last balls from one pan and one of the previous (not-last) balls from the other. If the pans balance the removed last ball was the faulty one. If they do not balance the not removed last ball is the faulty one.
PS – No one says I can not add, remove or switch balls from the pans. So I resolve it this way.
PS II: As said, you must be joking!
PS III: Read the messages of the thread. The answer is there.

Hi Gabriel,
 
The best I could come up with is:
1 = cup one 2 = cup 2 0 = no included
 
Ball     Test 1    Test 2    Test 2
1            2            2             0
2            2             1             0
3            2            0             2
4            2             0             1
5            1             2             2
6            1             2            1
7            1             1            1
8            1             0            0
9            0             2             0
10          0             1             2
11          0             1             1
12          0             0             2
 
The results will be evaluated as follows:
 (Number Represents heavy cup (side)
                            
Ball one heavy               2 2 even
Ball one Light                1 1 even
Ball two Heavy              2 1 even
Ball two light                  1 2 even
Ball three heavy             2 even 2
Ball Three Light              1 even 1
Ball Four heavy              2 even 1
Ball Four Light                1 even 2
Ball Five Heavy               1 2 2
Ball Five Light                  2 1 1
Ball six Heavy                 1 2 1
Ball six light                     2 1 2
Ball seven heavy             111
Ball seven light                 2 2 2
Ball eight heavy               1 even even
Ball eight light                   2 even even
Ball nine heavy                 even 2 even
Ball nine light                    even 1 even
Ball ten heavy                   even 1 2
Ball ten light                      even 2 1
Ball eleven heavy              even 1 1
Ball eleven light                 even 2 2
Ball twelve heavy               even even 2
Ball twelve light                  even even 1
 
Is this correct?
 
I think it is
 
Statman

I didn’t know of any mnemonic code for this solution. 
 
I knew I needed 24 unique combinations of results across the three tests (e.g.,1-2-even).  If you enumerate the possible combinations there is 3*3*3 = 27 possible combinations. Then eliminate even-even-even as it won’t tell you anything leaves 26 combinations and half of these are the reciprocal of the other half (1-1-even is the reciprocal of 2-2-even) that will help you determine if the ball is heavy or light.  This leaves 13 possible placements for each ball across the 3 tests.  Then it was just a matter of choosing the 12 combinations that would give you 4 in cup 1, 4 in cup 2 and 4 not included for each of the thee tests.
 
The array for the test is the 12 combinations
 
QED!  I guess those couple of semesters of advanced combinatorial mathematics in grad school has some pay back beyond determining how many ways you can arrange the letters in Mississippi and the odds of various poker hands.
 
Cheers,
 
Statman
 

Weigh six on one side and six on the other side. Which ever side is lightest can be eliminated. Then split the heavy side in half. Weigh three and three. Again, the lighter can be eliminated. Then weigh one and one. If they are balanced, then the ball not weighed is your culprit. If they are unbalanced, the heavier of the two is the heavy ball.

Arun,
That really does not matter. Please read the threads and you will find out.(I would look at postings by Gabriel on this.)
PB

Since now we know the answer, how about try finding the ultimate solution:If number of weighting is given, what is the maximum number of balls that can be dealt with? Is it a NP complete problem??

But the question says that one of ’em is WRONG WEIGHT…does not specify light or heavy…might also turn out that on the first weigh in the ball is in the heavier side…

Dear Jimgggg,
This should suffice. For all those who like solving puzzles.
This is tougher problem than the ones which say if one of the weights is lighter or heavier. The solution is long, I could not find any other alternative solution. So here I go.
Select any eight ball and label them from 1 to 8 and put four of them on each side of the balance, assume we have 1234 on one side and 4567 on other, now there are two possibilities:
(1) One side of the balance comes out lighter (assume 1234 is lighter than the other). In this case, you know that the abnormal ball is one of the balls already on the balance.
(2) The balance is even. in this case, you know that the abnormal ball is one of the balls not on the balance. Label the balls not on the balance as ABCD.
 
Lets start with possibility 1:
Side 1234 came out higher (or lighter) than the side 5678 due to:
One of 1234 is the abnormal ball, and it’s lighter than the other balls.
Or
One of 5678 is the abnormal ball, and it’s heavier than the other balls.
 
Let’s do another weighing, with two of the 1234 balls on either side, and one of the 5678 balls on either side. For simplicity, let’s weigh 125 versus 346. Given our earlier inference following can happen:
if one of 12 is a light ball, the 125/346 weighing will come out with 125 high.
if one of 34 is the light ball, the 125/346 weighing will come out with 125 low.
if 5 is the heavy ball, the 125/346 weighing will come out with 125 low.
if 6 is the heavy ball, the 125/346 weighing will come out with 125 high.
if one of 78 is the heavy ball, the 125/346 weighing will come out even.
 
Now, the last weighing for possibility 1:
(a) if it comes out even, then we know that the abnormal ball is either 7 or 8. for our third weighing, we can weigh 7 against one of the balls we already know to be normal. If it comes out even, then 7 is normal and 8 must be the abnormal ball.
(b) as we can see from our chart above, if the 125/346 weighing comes out with 125 high, then the situation is either: 1 is the light ball, 2 is the light ball, or 6 is the heavy ball. So we can do 1 versus 2 weighing and conclude the problem (trivial).
(c) as we can see from our chart above, if the 125/346 weighing comes out with 125 low, then the situation is either: 3 is the light ball, 4 is the light ball, or 5 is heavy ball. So we can do 3 versus 4 weighing and conclude the problem (trivial).
 
Now possibility 2, where we know that one of ABCD is abnormal and we have two weighing to find the abnormal ball:
 
Pick ABC from the lot and weigh against any three from 12345678, for simplicity lets take 123. Let’s look at the possibilities:

if the weighing is equal, than D is abnormal and can be weighed against any of normal ball to figure out if it light or heavy.
If the weighing says ABC is light. Weigh A against B and conclude by saying which one one of A or B or C is light (trivial, so not writing details). Similarly the case when ABC is heavy can be worked out.
Thnx
Ajit

The problem is, you don’t know if the odd ball is heavier or lighter.

start with weighing 5 balles on eahc side……….
 
the details are then easy to reach

hello,
 
thanks for your reply.  my solution doesnt end with weignhing 5 balls each….next you take 2 from the knwon good ones……..and continue…..
in the second step you will know if the defective ball is heavier or lighter……
thanks again…….
i look forward to another teaser
regards
sunil
 
 
 

“1) You weight 5 balls against 5 more. If one side tips you know one of the five balls is heavier. “
Wrong. It could also be that one of the 5 balls in the opposite side is lighter. Noone said that the odd ball must be heavier. Noone said that it must be lighter either. Whith this point wrong, the rest of the reasoning remains wrong.
Keep trying.

Number the balls 1-12.> >> > At each of the three weighings, there are three outcomes: L> > (the left side weighs more), E (both sides weigh the same),> > and R. Note that there are 3*3*3 = 27 possible outcomes.> > However, we can avoid the outcomes LLL, RRR, and EE.> >> > This leaves 24 outcomes, each of which has a natural> > opposite: to get the opposite outcome, replace L’s with R’s,> > R’s with L’s, and leave the E’s the same. So, the opposite> > of LER is REL. Now we are down to 12 outcomes and their> > opposites.> >> > Now the trick is to do the weighings so that each outcome> > corresponds to a particular ball, and if say RRE corresponds> > to ball 8 being heavier, then the opposite outcome LLE> > corresponds to ball 8 being lighter.> >> > Say we want to make LRR correspond to ball 1 being heavy.> > Then put ball 1 on the left side once and the right side> > twice. Now, to each of the 12 outcomes, pick a ball which> > you want to correspond to that outcome, and do the same> > thing.> >> > That’s the idea. Here’s the answer:> >> > 1 2 7 10 | 3 6 8 9> > 4 5 9 11 | 1 2 3 8> > 2 3 5 12 | 1 4 6 7> >> > Each row is a weighing of 8 of the balls, 4 on each side.> > The 24 possible results, with corresponding ball, are given> > below:> >> > LRR/RLL = 1> > RLR/LRL = 2> > RRL/LLR = 3> > ELR/ERL = 4> > ELL/ERR = 5> > LEL/RER = 6> > LER/REL = 7> > LLE/RRE = 8> > LRE/RLE = 9> > LEE/REE = 10> > ELE/ERE = 11> > EEL/EER = 12> >>

Note: I’m a little lazy in my typing, and did not fully explain the obvious at the end of measurement three.
 
The answer is:                 
 
 
1                                              2                                              3
 
 
1234  5678  =  Balance           123  9,10,11 = Balance           1  12  =     You know. 
        A                                           A                                          A
 
                                                                         _______
                                                 123  9,10,11 = Balance          9  10  =      You know 
                                                       A                                         A
 
   V                  _______
1234  5678  =  Balance           1235  4,9,10,11 = Balance      6  7  =        You know    
        A                                             A                                       A
 
 
                                                   V                       _______ 
                                                1235  4,9,10,11 = Balance      1  2 =         You know
                                                        A                                       A
 
                                               
                                                               V           _______    
                                                1235  4,9,10,11 = Balance      12  45 =     You know
                                                        A                                         A        
 
 
 
This is my favorite challenge and have thought of variations such as two balls, one heavier and one lighter by the same amount.
 

losi mo ayam ka

first put 4-4 ball on each pan. see which weights lighter. if not then take last 4 ball.  share the lighter 2-2 on two pan again. take lighter.  out of 2 put 1-1 of them on each pan you will find one of them will be lighter. by this you will come to know which ball is weighs less of the 12.