# Another brain teaser – 12 balls puzzle

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• #29354

jimgggg
Participant

The six-sigma team must test 12 balls, one of which is the wrong weight.  The only tool they have is a two-pan balance scale, so they can only measure the relative weight of one side vs the other.
Design a test to determine which ball is outside of weight specs.  Use only three  weighings.  The test design should provide for each possible outcome of each weighing.
Jim

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#74977

jimgggg
Participant

And is it too heavy or too light?

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#74990

Ed Van Haute
Participant

It does not matter if the out of spec ball is too heavy or too light. This test assumes the out of spec ball is too heavy, but it works just as well if it is too light
1. conduct a test with 6 balls on each pan. Select the six balls that weigh heavy
2. Conduct the second test (with the six balls that weighed heavy) with three balls on each pan. Select the three that weigh heavy.
3. select any two of the three remaining balls, and place one on each pan. If one weighs heavy, that is the out of spec ball. If they weigh equal, the one that was not selected is the out of spec ball.

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#74994

kony
Participant

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???

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#74995

Tony Reed
Member

Since we do not know if the ‘out of spec ‘ ball is heavy or light, then we can not assume anything.A sytematic method to solve this problem is as follows1. Arrange the 12 balls into 3 groups of 4
Label these G1,G2 and G3.

2. Place the G1 into pan1 of the scales and then
place G2 into pan1.(a) If they balance then the odd ball in G3
(b) If they dont balance and pan1 goes down then the odd is either heavy and in G1 or light and in G2
(c) If they dont balance and pan1 goes up the the odd ball is either light and in G1 or heavy and in G2.lets take a condition, any condition, and apply the next step. Pick condition (b) ie Pan1 goes downUnder this condition we know that all of the balls in G3 in within spec.Remove 1 G1 ball fron pan 1 and replace with 1 G3 ball. Then remove 3 G2 balls from pan 2 and replace with 3 G3 balls.Once again only three conditions will result (d) pan1 will go level.
(e) pan 1 goes up
(f) pan 1 goes downLets pick a condition say (f) pan 1 goes down
If this occurs then either 1 of the three G1 balls is heavy or the G1 ball in pan2 is light.Remove 2 of the G1 balls in pan 1,
the remove 1 of the G1 balls and 1 of the G2 balls in pan2In this case only two conditions will result.(g) pan 1 will go level
(h) pan 1 remains downLets take condition (g) then this will only occur if
One of the 2 G1 balls fron pan 1 was heavy of the G1 ball from pan 2 was lightPlace 1 of the G1 balls that was in pan1 back into pan1 and the other G1 ball that was in pan 1 into pan 2Again only 2 conditions will occur(i) pan 1 will balance
(j) pan 2 goes downIf (i) happens then the G1 ball in pan 2 is light
If (j) happens then the G1 ball in pan 1 is heavy.
This sytem can be used for any nominated combinations you may wish to tryTony

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#74998

Keshri
Participant

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

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#74999

Tony Reed
Member

To correct my previous messageStep 2 should read:2. Pleace the G1 into pan1 of the scales and then place G2 into pan 2.Thanks

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#75002

jimgggg
Participant

This solution works only half the time, when the odd ball happens to be heavy.  If it is light there are not enough weighings left to find which is the light one.
Hint:  6 vs 6 never narrows the possibilities enough to complete all cases in the remaining 2 weighings.

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#75008

Ian S.
Participant

(Good teaser by the way.)

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#75010

jimgggg
Participant

Nayan has got it, and well explained at that.  Tony’s solution may be equivalent just a bit harder for me to follow.  The key concepts are
– 3 groups of 4
-use every bit of knowledge gained in each test
-recognize and use known “standard” balls to sort out the non-standard.
Jim

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#76925

Ajit Mankottil
Participant

Hi Ed,
Suppose the out of spec ball is light. Then if you weigh 6 balls on each side and you take the 6 balls that are heavy and remove the 6 balls wich weighed light, then the out of spec ball will be in the collection of light balls. Right?
Thanks

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#76963

Ed Van Haute
Participant

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

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#78753

Layla
Participant

Your approach works if the ball is heavier.  It doesn’t work however if the ball is lighter.

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#79674

Gutierrez
Participant

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

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#81697

Tony Reed
Member

Alex
I can provide more details directly.
[email protected]
Tony

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#81700

Jacqueline CUI
Participant

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

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#81706

Gabriel
Participant

I am not sure that your method works. Can you clarify it, please? My doubts are regarding the following situation. You put 3 and 3 ant the plates are equal. Then the defective is among the other 6. How do you find, with only 2 weightings left, a defective among 6 knowing that the other 6 are all not defecticve?
Dividing in 3 groups of 4 has an adventage: If the plates are equal, you have to find the defective among only the remaining 4, and you have 8 “good” balls to use as standard. If the paltes are not equal, you have to find the defective among 8 knowing that the remaining 4 are “good” but you already know that if the defective is in the lower plate it is a heavier and if it is in the higher it is lighter. That’s a piece of information you will use later.
Dividing in 4 groups of 3, if the plates are not equal it is better than before because you have the same kind of information and only 6 balls instead of 8. But if they are equal again you have to search among 6 instead of 4, and with no extra information about the weight of the defective.

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#81709

Tony Reed
Member

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

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#81720

Jacqueline CUI
Participant

Gabriel:
You are right. I thought I did not need to weigh the rest 6 balls if the the 3 and 3 on the both sides of scale were equal. So, I guess dividing in 3 groups of 4 is the only way to solve this brain teaser.
Thank you for reminding me.
Jacqueline

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#81747

Ed Thomas
Participant

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

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#81760

Rakesh
Participant

The six-sigma team must test 80 balls, one of which is the wrong weight. The only tool they have is a two-pan balance scale, so they can only measure the relative weight of one side vs the other. and you have only four chances

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#82530

Mark Wright
Participant

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.

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#82531

Anonymous
Participant

But we don’t know if the off-weight ball is heavier or lighter….

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#82533

H. Flores
Participant

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

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#82536

Gabriel
Participant

Yes, you broke a rule: The restricion was “three weightings”, and not “three fillings of plates”.
Anyway, it was a nice example of creativity.

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#82556

Mark Wright
Participant

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.

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#82562

Anonymous
Participant

I reread you post, and I still think your method won’t work.
You assume the off-weight ball is heavy and not light.  If your assumption is wrong, and it is light, then when you compare the 6 heavy balls from the first weighing, the scale will read that the two sets of 3 weigh the same.  By then, you’ve burned two weighings and I don’t think you can discern which of the remining 6 balls is the light one.

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#82566

billybob
Participant

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

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#82572

Bobby Joe Jim Bob
Participant

Billy bob,

I find the amount of Six Sigma practitioners that have had personal experiences with possums to be astonishing.  The last two winters I have fought the evil that is know as the possum.  Sure, the possum looks friendly, but I believe he an agent of the devil.  I live down by the river in a vintage (1972 model) mobile home.  Several of your possum cousins decided to climb into by duct work and live in my trailer.  I have two vehicles in the front yard on blocks that the possums were welcome to stay during the winter, but no they wanted my trailer.   I banged pots and pans together to scare um out the house. Them boys were not going to leave. I decide to use my sophisticated problem solving skills to find an answer to my dilemma.  So I got my firework stash out of the closet from the 4th of July, and I found some bottle rockets.  The ducting under the trailer is in a straight line going the length of the trailer.  I went to the back room, and pulled up the floor vent.  I then taped two bottle rockets together and lit the wick.  I jammed the rockets in duct, and put the vent back into the floor.  The rockets fired up and caught my trailer on fire and it burned to the ground.  Now I do all my DOEs in the back of my station wagon.

Bobby Joe Jim Bob

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#82576

billybob
Participant

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

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#82783

Swaggerty
Participant

I read most answers and I think that out of the 12 balls, I would put 4 on one side and 4 on the other and weigh. (four balls excluded) (1st weighing)A: If they are equal, the out of weight ball is in the excluded four balls.B: If one side or the other weighs out then that group of four has the out of weight ball.If A is true, discard all 8 balls and place the excluded 4 balls on the scale 2 to a side. (second weighing).If B is true take the four balls and place 2 of each on either side of the scale and weigh.(second weighing.)From the second weighing, select take the two balls and place one on either side of the scale and weigh.
(third weighing)You will now know which is the out of weight ball

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#82809

Gabriel
Participant

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.

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#82888

Schmidt
Participant

I don’t thing anyone’s described the full solution yet!  I still haven’t quite got it figured it out.  Here’s the problem:Use Tony / Nayans first weighing (4 vs 4) and assume the pans don’t balance.  Then use their second weighing with 3 of the four “UP” balls and a “GOOD” against 2 “DOWN”s and 2 “GOOD”s.  Now there are only TWO outcomes here:1) Balance or 2) UPs go UP again (The UPs can not go down because you removed a potential light ball and put a good one in and on the other side you removed potentail heavies and put good ones in.  You could not have put a heavy ball into the side that previously was light)1) If Balance:  Then either you took out the light ball from the up side, or you took out a heavy from the other pan, but it is one of two balls.  So now you put one of the possible heavy balls with the possible light ball and weigh against the other possible heavy and a good ball.  Lets call this L H1 vs. H2 GWe know this won’t balance.If LH1 goes up:  L is the ball ball and it’s lightIf L H1 goes down, H1 is the ball ball and it’s heavyIf H2 G goes down, H2 is the ball ball and it’s heavy3 weighings  – problem solved.  That works so far…………OK – now back to UPs go UP as a result of the second weighing.That means either:  Light ball is in the 3 UPs on the UP side, or Heavy ball is in the 2 DOWNS on the Down side.I can’t figure out 1 weighing to identify  the bad ball now !

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#82899

Nigel Parsons
Participant

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

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#82944

Tony Reed
Member

Marcus
In the second weighing you need to weigh 1 “UP” ball and 1 “GOOD” ball and 1 “DOWN” ball with 2 “GOOD” balls and 1 “DOWN” ball
ie weigh three with three not  four with four.
The resulting conditions can then be deduced.
Hope this helps.
Tony.

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#82958

Schmidt
Participant

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?

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#82966

Tony Reed
Member

Your right 3 on 3 doesn’t work.
The second weighing should be
1″GOOD” ball + 1 “UP” ball +2 “DOWN” balls  balanced against
2″GOOD” balls+1″UP” ball + 1 “DOWN”

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#83209

Mares
Participant

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…

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#84083

Mohammed Irfan
Participant

First, split 12 balls into 3 sets(each of 4):
Weighing 1: Take 2 sets each on either side (ie. 4 – 4 balls in the balance). If you find weight is dropping one side, then the weighted ball is situated in that set, else if the weight remain the same on two sides, then the 3rd set of 4 balls contains weighted ball.
Weighing 2:Now take the identified weighted set of 4 balls and split into 2 balls(each set: 2+2=4).Keep each 2 on either side. Now, you will find weight is dropping one side, then the weighted ball is situated in that set(in 2 balls).
Weighing 3: Take those weighing set of 2 balls, split into 1 each and weigh in the balance by keeping each on either side.Now the pan of weighted ball will surely goes down.
with wishes,
Mohd. Irfan

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#84568

aravind
Participant

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.

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#86198

Member

Mohammed,
This solution works on the circumstances particular only as described. The error is in the statement: “If you find weight is dropping one side, then the weighted ball is situated in that set, else ….”
This is error because even if the pan drops on that set, the deviant ball could possibly be lighter and is situated in the other set of the pan that is going up.
The correct thought process to follow is : If  LeftGroupof4 does not balance RightGroupof4, then we must follow the procedure described by Nayan/Tony in previous post of this topic. In fact, it is those very procedures that hallmarks the depth of this problem which results to the acheiving the requirement of ascertaining the deviant within the constraints.
I have not fully internalized that particular logic. But I believe, it is those very logic that solves the problem. Other scenarios have lighter solutions.
truth

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#86799

anon
Participant

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.

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#87936

phaedrus
Participant

here is the correct answer, which works whether the odd ball is lighter or heavier:
Step 1) put 4 balls on each pan, 1,2,3,4 on one side, 5,6,7,8 on the other
case 1i) if they balance, then the odd ball is 9, 10, 11 or 12.  now that it is down to 4 possible balls, the rest is trivial.
case 1ii) 1-4 weigh more than 5-8, then
Step 2) put 4,5,6 on one pan, 7,8,9 on the other pan
case 2i) if they are equal, then odd ball is heavier and it’s 1,2 or 3, weigh 1 vs 2, if it’s heavier, odd ball is 1, if it’s lighter odd ball is 2, if they’re equal, odd ball is 3
case 2ii) 4,5,6 weigh less than 7,8,9, then odd ball is lighter, and it’s 5 or 6, you do the rest
case 2iii) 4,5,6 weigh more than 7,8,9, then either 4 is heavier than the rest or either 7 or 8 is lighter, weigh 7 vs 8.  if they are unequal, then the lighter one is the odd ball.  if they weigh equal, then 4 is the odd ball.
csae 1iii) 1-4 weighs less than 5-8, then redo the above, reversing everything

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#87973

phaedrus
Participant

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?

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#92624

Preeti Handa
Participant

Solution to 12 ball puzzle:
The odd weight ball can be seperated as follows:
Weigh 6 balls on each side of the weighing balance.Say each ball is 1gm.One side will be heavier than the other(6gms<7 gms).Take the balls on the heavier side & put three each on both sides of the balance. Again the odd ball side is heavier.Now out of these three balls take randomly any two balls and put on each side of the weighing balance;if the balls in the balance weigh same, the third one is the odd ball. If the odd ball is in the balance, it will weigh more than the other. Thus we can identify the odd ball using only three weighings.
Please let me know if there is an easier, better and quick solution for this.
Regards,
Preeti

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#92630

Gabriel
Participant

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.

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#92633

DaveG
Participant

Gabriel,
I disagree.  His solution will work for a sufficiently lighter or heavier ball.

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#92634

Gabriel
Participant

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)

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#92638

PB
Participant

You would take the 6 balls in the pan that was high or low and place 3 each in the pans. Again one pan will be high or low. (This is weighting 2)  Then take 1 each and place them in pan. If the pans balance the one remaining in your hand will be the odd ball out. Or you will have a pan high or low depending upon whether the odd ball is higher or lower in weight. (This is weighting 3)
OR
You would take 2 bals and place them in the pan. If the weights balance you would weigh the remaing 2 balls or take from the pan that is high/low depending.
PB

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#92641

DaveG
Participant

I stand corrected.

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#92642

Gabriel
Participant

“You would take the 6 balls in the pan that was high or low and place 3 each in the pans. Again one pan will be high or low”
Not necesarely. You put 6 in each pan. One pan goes up and the other goes down. Either there is a heavy ball in the low pan or a light ball in the high pan. You take the 6 balls of the low pan and put 3 and 3 in each pan. The pans remain leveled. That menas that these 6 balls were Ok and that there is a ligh ball among the other 6. Now you have only one weighting left…

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#92650

PB
Participant

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.

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#92652

Statman
Member

Every time you add or replace a ball it is a new wieghing.
The only solution is to seperate the balls into 3 groups of four and sequentially test using the information from the last comparison.
Statman

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#92655

Gabriel
Participant

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!

0
#92658

Gabriel
Participant

This is “a” solution, but not the only one.
It is possible to design a test that would be fully run without the need to take decisions based on the partial results before going on with the testing (i.e. without the need to “sequentially test using the information from the last comparison”, as you said). In this way, when you finish the test you get all the data and, after analyzing it, you can discover which is the odd ball and wether it was heavier or lighter.
How wold such a test be? Do you dare to solve this?
(If you do, congratulations! I didn’t solve it and I was told the answer)

0
#92666

Statman
Member

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

0
#92677

Gabriel
Participant

Looks Great!
Click here to check the message 23422, which is in this same thread and is the one that told me and the rest of the forum this kind of solution.

0
#92693

Statman
Member

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

0
#97058

Bahero
Participant

I don’t beleive there’s an easier way to solve it other than this…

0
#98044

mccordre
Participant

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.

0
#98045

Arun Nanjakutty
Participant

Hi
I just have one concern regarding the puzzle, one of the 12 balls is said to be out of specified limits , but it doesnt say whether it weighs more or less than the other 11 balls.
Thanks
Arun

0
#98047

PB
Participant

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

0
#98048

PB
Participant

Mccordre,
You made the same assumptions that we all had made that one ball is lighter than the others. You are not told that one ball is lighter or heavier. The trick is to find the odd ball in 3 measures.
PB
PS – Also in your answer, you are eliminating the lighter side of the scale and then weighing the other side and looking for lighter side again.

0
#98679

Chao
Participant

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??

0
#99898

moebius
Participant

with a known ball, you can deal with up to 13 balls with three trials.

0
#102678

aniketdeshpande
Participant

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…

0
#103889

Wil
Member

Arun’s doesn’t work.  It implies that the odd ball msut be lighter.  What if it were heavier?

0
#103969

A.Singh
Participant

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.

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

0
#105648

Participant

YOU  weigh two sets of four, saving four, if one of them is heavier you wigh two and two and then one and one for the heavier of the two.
In case both sets of four have the same weight, you weigh the saved four two against two and then one versus one to identify the heavier.

0
#105658

Gabriel
Participant

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

0
#106314

Kinga
Participant

I would build 4 blocks of 3 balls A,B,C,D
1. weigh:  let us compare the first two blocks, A and B
two cases arise: A=B, AB
a) If A=B we know that the odd block is either C or D
2a) weigh take A&C, if they are equal the odd ball is in D, if not then it is C
3a) weigh take two balls from D/C to determine the odd ball
b) case AB
Take the heavier/lighter whichever you feel like, let us assume A>B
2b) compare A &C or D, we know that A either must be heavier than  C/D/ if the odd ball was heavier/, or equal to C/D /in case the odd ball was lighter/.  If it is equal to C/D then take B if it is heavier then take A
3b) similar to the previous

0
#107486

Member

the details are then easy to reach

0
#107489

PB
Participant

Sunil,
Good thinking. However, I beleive it is unknown whether the odd ball is heavier or lighter than the other 11 balls (all you are told is that ‘one of the balls is of the wrong weight’). Therefore, if you place 5 on each side and one side goes up you are not sure whether that is the side you want because the odd ball (heavier one) is in the other pan.
Within this thread is where Gabriel has posted the solution.
PB

0
#107559

Lala
Member

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

0
#112920

Gaurav
Participant

OK, here it is hope its correct
1) You weight 5 balls against 5 more. If one side tips you know one of the five balls is heavier.
2) Divide the 5 balls out of which one is heavier into a group of 2 and three. Weigh these.
3) Assuming the three side is heavier. You weigh two balls. If they are equal you know the ball that was not weighed was heavier. If one side tips you know that is the ball.
4) Going back to step one, if both the sides are equal (in which case you were lucky chosing the balls). weight the two balls not on the scale and you will which is heavier.

0
#112924

Gabriel
Participant

“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.

0
#114120

Edwards
Participant

Nice puzzle!In coming up with a solution, I realized that it’s actually less complicated to solve it with 13 balls than with 12! This is because with 13 balls, you have an extra ball to use as a control after the first weighing, and this makes a huge difference.Anyway, here’s my solution (as a side note, this process not only determines which is the odd ball, but also whether it’s lighter or heavier than the others):Label the balls A through L. Weigh ABCD against EFGH.Case 1:
They balance. Thus the odd ball must be either I,J,K, or L. Weigh IJK against ABC.
Case 1.1:
They balance. This implies that L is the odd ball. To determine whether it’s lighter or heavier, simply weigh it against A and see.
Case 1.2:
IJK is lighter than ABC, so one of I, J, K is the odd ball, and it’s lighter. Weigh I against J. If they balance, then K is lighter. Otherwise, the odd ball is the lighter of I and J.
Case 1.3:
IJK is heavier than ABC. This case is symmetric to Case 1.2.Case 2:
ABCD is lighter than EFGH. This means that the odd ball is either lighter and one of A,B,C, or D, or it’s heavier and one of E,F,G, or H. It also implies that I,J,K, and L are not the odd ball, and thus can be used as controls. Now weigh ABCEF against DIJKL.
Case 2.1:
They balance. Therefore the odd ball is either G or H, and so it’s heavier. Weigh G against H. Whichever one is heavier is the odd ball.
Case 2.2:
ABCEF is lighter than DIJKL. Note that this means that the odd ball must be either A, B, or C. This is because, as stated earlier, E and F can only be the odd ball if the odd ball is heavier. But since ABCEF is lighter, neither E nor F can be heavier, and so neither can be the odd ball. Also, D can’t be the odd ball because it would have to be lighter, but it’s on the heavier side of this balance. Hence either A, B, or C is the odd ball, and so the odd ball must be lighter. Now weigh A against B. If they balance, then C is the odd ball and it’s lighter. Otherwise, the odd ball is the lighter of A and B.
Case 2.3:
ABCEF is heavier than DIJKL. Following the same lines of reasoning as in Case 2.2, we can now eliminate A,B, and C from being the odd ball. Thus either E or F is the odd ball and it’s heavier, or D is the odd ball and it’s lighter. Now weigh E against F. If they balance, then D is the odd ball and it’s lighter. Otherwise, the odd ball is the heavier of E and F.Case 3:
ABCD is heavier than EFGH. Again, we weigh ABCEF against DIJKL, and the reasoning is perfectly symmetric to that of Case 2 (except that A,B,C,D are now in the heavy group, while E,F,G,H are now in the light group).Voila! In three weighings, we can determine which ball is the odd one, AND whether it’s lighter or heavier than the other balls! Now I’m trying to generalize the problem to N balls, although it seems quite difficult. I think I might be able to write a dynamic programming algorithm to calculate the minimum required number of weighings, since you can break down the larger problem into the sum of smaller subproblems.

0
#128141

sammy boychick
Member

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> >>

0
#129378

Member

Right I have a slightly different solution to the problem. We have 12 balls out of which one weights lighter than the rest. I will randomly divide the 12 into two groups of 6.
First Weighing
Pick any group and divide it into 3 and 3 on each scale. Identify lighter scale
Second Weighing
Could be the lighter ball is in this group you picked. Then proceed to weigh one and one on each side. If scales weigh the same then you have the isolated ball as being the lighter ball.

In this solution at least you have 50% chance of getting it right the second time round.

If you are not right you can still weigh the other group of six into 3 and 3. And in the third weighing you can figure out which the lighter one is.
Kind Regards
Department of Nuclear Physics
Massachusetts Institute of Technology

0
#136673

Bill Henry
Participant

Note: Im a little lazy in my typing, and did not fully explain the obvious at the end of measurement three.

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.

0
#139064

Christopher Wheat
Participant

Assume the out of spec ball is lighter.  Place six balls on each side.  One pan will be lighter.  From the six lighter balls place 2 random balls on each pan.  If the scale is even, the lighter, out of spec ball, is one of the remaining two balls.  If the scale is not even, take the lighter pan balls and place one on each scale.  That will determine the lighter ball.

0
#143363

kulot
Participant

losi mo ayam ka

0
#160623

Tomaalimosh
Member

That is interesting… but you can expand the puzzle: 1. same problem for 36 balls and 4 weighings
2. same problem for 4*(3^(x)) balls and x+2 weighings where x > 1 1 1 Let’s see you solve this one :). And it is solvable too :D

0
#163226

Yuvraj
Member

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.

0
#168374

Dixit
Member

This can be solved in Minimum of 2 Tests or Maximum of 12 Tests.
Keep Ball 1 on one side of the Scale.
Start checking other Balls, one by one on the other scale and compare.

0
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