QUIZ: How heavy are my balls?

[Simon]

Another puzzle team.

12 balls -11 of equal weight and 1 weighing differently. It can be lighter or heavier than the others.

You have a pair of balancing scales.

How can you identify which ball is lighter or heavier with only three weighings allowed!


Good luck!

[asy]

Originally posted by Simon
12 balls -11 of equal weight and 1 weighing differently. It can be lighter or heavier than the others.

You have a pair of balancing scales.

How can you identify which ball is lighter or heavier with only three weighings allowed!




Divide the balls in half (6 and 6), weigh the two groups against each other...

Take the lighter group and divide it in half (3 and 3), weigh the two groups against each other...

Take the lighter group, pick any two balls, weigh them against each other, if they are the same, then the third ball in the group is the odd one out, but if one ball is lighter than the other you have your odd one out!

This probably isn't right, but it's all I can come up with in 5 mins... hehe

asy :D

[Nicholas H]

1. weighing .. 3 each side .. 6 left aside .. if they are equal then remove all 6 .. otherwise it is the other 6 .. you have 6 ones left
2. Weight 3 from the remaining amount, and 3 from the removed amount, remembering which one was removed, if they are equal, then remove these 3, otherwise remove the other 3
3. We now know it is in 1 of 3 .. weigh 2 of the 3 .. if they are equal then you know it is the other one .... hmm.. otherwise it is the one ..

not the correct answer .. it has a 33% chance of getting it in 3 weighing .. I probably could get it up higher by working out if it was heavier or lighter, then you can decided out of the last 2 that way .. but if it has never been in the comparison, then you do not know .. I will see what else I can think of .. I am sure someone else will create a solution

[Simon]

Sorry Asy - we don't know if the odd ball is heavier or lighter. So setting aside the light group will not always work.

Sorry Nick you need to think a bit longer but your logic is on the right track.

I used to swap these with another Mensa buddy - and it took me a few tries to get this one out.

Cheers,

[geoffw]

mensa Where do people get the name for organisations like this? Not from English... it must be a wordin another language.

So let's try the second most common Euopean language spoken worldwide-Spanish.

"Mensa ... silly, stupid (feminine form... masculine form is "menso")

Similar in fact, to "tonto" for those Lone Ranger fans.

[Simon]

mmm geoff I am guessing you don't know the answer.

I will give email you a hint if you like.

Cheers,

[geoffw]

Simon,

I am no good at these puzzles... for whatever reasons. My post was merely a digression.

I went to the bank last week to get a loan. I wanted to speak to the manager. But I got Tonto.

I told them that was good enough. After all, I was looking for the Loan Arranger.

[asy]

Sorry Asy - we don't know if the odd ball is heavier or lighter. So setting aside the light group will not always work.

Yeah, thanks Simon, I realised that after I posted!!

Not a bad first effort, though, I thought??

OK, here is a more measured thought...

Label the balls A, B, C, D, E, F, G, H, I, J, K, L. To start with...

I figure you have to break the balls into equal groups, to start the weighings. There are 4 possibilities for groups...

6 piles of 2 balls
4 piles of 3 balls
3 piles of 4 balls
2 piles of 6 balls

I have already proven why the last permutation doesn't work! hehe :)

Six Piles of 2 balls is too lengthy in that three weighings will not achieve anything substantial.

4 Piles of 3 balls wouldn't work... (tried to figure it out, but was too long...



Three piles of 4 balls works thus:

Take ABCD[set1] and EFGH [set2] (Leave IJKL [set3] aside).

Weigh them against each other.

If they balance the odd one out is in IJKL. Weigh any three of sets 1 and 2 (say, ABC for example) set against I, J and K. If they are equal K is the odd one out, weigh K against any other ball to see if it is lighter or heavier. (3 weighings, with answer)

If I, J, K are heavier or lighter than ABC, then the odd ball is in this set, weigh I against J. If they are equal then K is the odd one, weigh it against any other ball to detemine whether it is heavier or lighter. If they are not equal, pick, say, the heavier of the two (say I) and compare it to A. If it is equal, then the light ball is J. If I is heavier than A, then it is the odd one out.



if the two initial sets don't weigh the same
say set 1 Are lighter, then either the light ball is in this set, or the heavy ball is in set 2.

Since we only have two weighings left, we need to split the sets, and import a known ball. (did the theory, too long to type)
Weigh ABE against CFI.

If they weigh the same, the odd ball is from set 1(light) or set 2(heavy). So either D is light, or G or H are heavy. Weigh G against H and if they are the same, then D is light, otherwise the heavier of G and H is the heavy ball.

If CFI are heavy, the odd ball is either the odd one from this set, F (heavy) or the original balls from the first set, A or B (light), similarly to above, weigh A and B, if they are equal, F is the odd ball, otherwise, whichever of A and B is lighter is the oddity.

If CFI are light, the oddity is either a light C or a heavy E (being the ringin on the other side). weigh either one, say C, against a known ball, say L. From this you will determine the outcome. If they balance, E is the heavy ball, if not C will be lighter than L, and is the light ball of the set.

If, when you weighed ABCD against EFGH you found that EFGH were lighter rather than heavier, you reverse the above, and determine the answer that way.


Can someone please check this... :D

I think it will work, but I may have just spent half an hour rambling and mutterring for nothing...

If it does work, I have to thank a friend who assisted me when I got stuck on the second weighing, and helped work out that I had to use the 'ringin ball'...

asy :D

[asy]

Sorry guys, not sure if it was plainly enough said the first time,

I HAD HELP DOING THIS ONE!!!

Got half way, then got really stuck, so employed a mathematically minded friend.

Seems I am only half way to mensa level (if that!!)

he he...

(didn't want you all to get the wrong idea)

asy :D

[Les]

Jeez, Simon,

I know I am close, but can't quite get the very last option sorted, but, to help others out, I got pretty close this way:-

(Call the 12 balls A - L)

Weigh 1:- Compare ABC with DEF - if equal, then one of GHIJKL is the odd one. (See further down - "A")

If unequal, then one of ABCDEF is odd one out. FIRST, note which way the balance is swinging !!!!!!!!! Prior to second weighing, swap A with D, remove B and E, and leave C and F where they are.

Weigh 2:- If balance is now even, then one of B or E is the odd ball. If balance swings the same way as before, then C or F is the odd ball. If balance swings the OPPOSITE way, then A or D is the odd ball.

Weigh 3:- Depending on answer to 2. check ONE of the two suspects against a known good ball - if unequal, then the ball on the scales is the odd one. If even, then the other of the pair is the odd one.
50% chance of succeeding so far :)


Now, back to "A" (where we go if first weighing was equal where one of GHIJKL is the odd one) We are one "weighing" behind the eight ball in this scenario - and this is where I can't quite make it - but, since we know that one of GHIJKL is odd, compare GHI with ABC

Weigh 2:- comparing ABC with GHI - is it even or uneven? If uneven, we do what we did in previous scenario (swap A with G, remove B and H, and leave C and I where they are). If even, then one of J, K, or L is the odd ball - and this is where I have run out of weighings (I have ONE left, but can't be really sure, depending on the results - see "B" below)

Weigh 3:- Depending on whether balance swings the same way, the opposite way, or stays even, we eliminate down to one of two ..... No, we don't!! We eliminate to one only as we know ONLY G, H, or I is odd (not A, B, or C).
75% chance of success :D


"B" So, my problem is if one of JKL is odd - I have only ONE weighing left - depending on the results, I may not be able to choose between one of two balls. Anyway, here goes:-

Weigh 3: Put J and K on the scales - if even, then L is the odd ball - BUT, if the balance dips, then I DON'T KNOW which of J or K is odd (as we don't know if "odd one" is lighter or heavier than the rest.
So, is 83.33% a pass ???? :(

So, there we are - close, but no cigar !!! But the mechanisms used might give a clue to someone else who can then come up with the complete answer.

Good luck - it's a beauty - and thanks, Simon :)


Addendum:-- And I see Asy has had another crack at it since I was typing - and, Asy, you might have the goods here. Try the "swapping", and "removing" techniques with 4 balls - it might just provide the answer.... Or, maybe you're already there? (I didn't check your answer "blow-by-blow" to be sure...)


Regards,

[Les]

G'day Asy,

Spent a little time looking at your answer - maybe you've struck "the key" by starting with 4 balls, then switching to 3....

As I said in previous post, I know I'm close with what I'm doing, but could not quite "nail" it.

Thanks for the clue - I might give it another 10 mins and see if "4 then 3" provides me with the answer that "just 4" or "just 3" couldn't.....

Regards,

PS Your answer sounded like it is working - but I did get lost in the words somewhat so I can't be sure....

[asy]

It works Les...

Well, it works in my head...

which doesn't necesarilly mean it works in real life.. hehehe

Please try to test it for me...

It would be good to know if it really does work!!!

asy :D

[Les]

Asy,

I thought I'd spotted a "ring-in" weighing in this lot !!!! But, I really just got a bit lost in the words. I think you might have aced this little sucker, Asy - well done !!!

Regards,

[Steve Navra]

Three groups of 4 balls:

A 1,2,3,4
B 5,6,7,8
C 9,10,11,12

Scenario Alpha:
1st Weigh:
A 1,2,3,4 V’s B 5,6,7,8 if equal weight, then C 9,10,11,12 contains a lighter or heavier ball.

2nd Weigh:
A 1,2,3 V’s C 9,10,11 if equal weight then C12 is the lightest or heaviest ball.

3rd Weigh:
C12 V’s any other ball, will confirm if it is lighter or heavier.

Scenario Beta:
1st Weigh:
A 1,2,3,4 V’s B 5,6,7,8 and A1,2,3,4 is heavier: meaning that A contains the heavier ball or B contains the lighter ball. (note this can be visa versa assuming B was heavier, but then the reverse of this sequence applies * see note at the end) The C balls are then not in play.

NOW THE REAL TRICK IS!!!

2nd Weigh:
A 1,2 and B 5 V’s B 6,7 and A 3 (Aha!) resulting in two possibilities: Gamma (equal) and Delta (scale tips)

Gamma:
A 1,2 and B 5 V’s B 6,7 and A 3 are equal (as per 2nd weigh)
3rd Weigh:
A 4 V’s C 12
If A 4 is heavier then this is the heavy ball
And if equal, then B 8 is the light ball. (whew, nearly there)

Delta:
A 1,2 and B 5 V’s B 6,7 and A 3 and A 1,2 and B 5 is heavier (as per 2nd weigh)

Then:
3rd Weigh:
A1 V’s A2 one of these has to be the heaviest because of 2nd weigh.

Theta:
The alternative that B 6,7 and A3 was heavier in the 2nd weigh would automatically mean that A3 was the heavy ball.

· Note:
The reverse principle applies assuming that the B side was heavier from the visa versa note from Beta 1st weigh. (A had to contain the heavy ball and B the lighter, or visa versa)

It is now 3-45 am!

GRRRRRRRRRRRRRRRRR

Steve
:mad:

[Lissy]

Get a life people.... ;)
It's puzzles like these that bring out my lazy streak....

[Simon]

Looks like ASY and Steve have both nailed it. Much quicker than I did too.

Here is my solution

1st Weighing

Weigh two lots of four balls. And leave a third lot to the side.

If the scales tip discard the third lot as being all Standard Weight.

If they balance go to scenario 2 below.

Scenario 1 (scales didn't balance)

Mark the four that went down as potentially heavy and the four that went up as potentially light.

2nd Weighing

Weigh two potentially heavy and one potentially light on each side of the scales ie three balls each end. With two potentially light balls off to the side

If the scales tip then mark the two potentially heavy balls that went up as Standard Weight and the potentially light that went down as Standard Weight.

You are left with two potentially light and a potentially heavy ball. See A below under third weighing.

If they balanced then see B below under third weighing.

3rd Weighing

You are now left with two potentially light and a potentially heavy ball

A. Weigh the two potentially light balls. If the scales tip then the up one is light. If they balance then the potentially heavy ball is the one.

B. If in the second weigh above the scales balanced then you have two potentially light balls to choose from. Weigh one against a known Standard Weight ball.


SCENARIO 2 (scales balanced in first weighing)
This was when the 1st weigh balanced. You now have 8 Standard Weight Balls and four Suspect balls.

Discard the 8 Standard Weight balls

2nd Weighing

Weigh two of the four balls. If they balance then they are Standard Weight. See A below.

If they don't balance go to B below.

3rd weighing A

compare one of the last two balls to a known Standard Weight ball. This will determine which of the last two is suspect and you have identified the culprit.

3rd weighing B

Compare one of the balls from the second weighing with a known Standard Weight ball. This will also identify the culprit.




But a simpler explanation I have seen is





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.