Brain Teaser 2

[Wrahn]

Okay all you rocket scientists out there.

12 balls all identical except you know one is a little heavier or a little lighter than the rest (not sure which)

You have a set of balanced scales and 3 tests to figure out which.

have fun.

EDIT: Extraneous words (that shouldn't have been there)

[Herbert Dypp]

nice one. i have a solution and will post when i figure out how to make it readable "> ">

[Herbert Dypp]

ok. here goes...

divide the 12 coins into 3 sets of 4 and then divide each set of 4 into a group of 3 and a single coin. for ease sake, put each set of three into a bag. (you could keep track of them using labels i guess)

place one 4 coin group in each pan and leave one set alone.

note down the condition of the balance scales.
(this is the first weighing)

Rotate the bags of 3 coins, shifting the one from the right pan to the table, the one from the left pan to right pan, and the one from the table to the left pan.

Observe the condition of the balance. If it changes, it will show you the bag that has the dodgy coin (and its relative weight)
(this is the second weighing)

in this case, you have 3 coins, one of which is known to be either heavier or lighter. you can then use the usual method (similar to one solution posted in the first brainteaser) to determine the odd one out.

If the balance didnt change, the dodgy coin is one of the single coins and can be pinpointed by rotating them as with the bags of three. (this would be the third weighing)

...bit tricky but i think i solved it. good problem ">

[Wrahn]

I believe in essence you have it. The only problem is, you need to apply that logic to the balls and forget about the coins.

[Herbert Dypp]

lol. looking at websites with many of these problems - just got coins into my head

[Wrahn]

The twelve ball question I first saw on EQDiva about a year ago, tortured my brain for hours figuring it out.

[Quillamaya]

Herb did you cut and paste that answer=) /giggle

[papyscott]

This problem was originally posed (at least the first time I saw it) in one of the old Zork trilogy. It was a tough one then too

Here is how it goes as I recall:

split balls into 3 groups of 4.

Weigh two of the groups of 4.

If they are equal
Tthen you know the funny ball is in the remaining group of 4.
Take 3 of the non-dodgy balls and 3 from this group of 4 and weigh them. If they are equal you know the remaining ball is dodgy and you can weigh it against a non-dodgy one to figure out whether it is heavy or light. If the 3 are lighter or heavier then you know the dodgy ball is lighter or heavier. At this point you take two of these 3 balls and compare them. If the dodgy ball is lighter and one of these is lighter then that is the one. If they are the same then the remaining ball is the dodgy one.

If the original two groups of 4 are unequal,
Then you have to label one group of 4 as heavy and one group of 4 as light based on this weighing. Let's call the heavy group 1h, 2h, 3h, 4h and the light group 1L, 2L, 3L, 4L. The other 4 balls you know are ok so lets call them 1K, 2K, 3K, 4K.

In this case, you do this in the second weighing:

1h 2h 1L on the left side say and 3h 4h 2L on the right side.

Suppose the left side is heavier.
Then you know 1h, 2h or 2L is the funny one. Your third and final weigh in this case is 1h vs 2h. If one is heavier that is your funny one. If they are the same then 2L is the funny one (and is light).

Suppose the right side is heavier
Then you know 1L, 3h or 4h is the funny one. Your third and final weigh in this case is 3h vs 4h. If one is heavier that is your funny one. If they are the same then 1L is the funny one (and is light).

If both the left and right side are equal
Then you know the funny one is either 3L or 4L. Put them on the scale together and you know the lighter one is the funny one.

It's late, but I think that works

Papy

[Rocsalt]

Edit : now I revisit the original problem, I see that the person weighing the balls doesn't know if the dodgy ball is heavier or lighter... making this solution false (it's based on knowing if the ball is heavier or lighter in the first place) ... I read it as the question poser not remembering the full question

I have a simpler idea for the solution...

(Assuming the dodgy ball is heavier...)
split the 12 balls into 2 sets of 6, weigh each set. --- TEST 1 ---

The lighter set remove from the scales and discard.

Split the remaining 6 balls into 2 groups of 3, weigh each set --- TEST 2 ---

The lighter set remove from the scales and discard.

Pick one ball and remove it from the scales, holding it in your hand.

Split the remaining 2 balls into 2 groups of 1, weigh each set. ---TEST 3 ---

From that last weighing:

If the weight is equal -- the ball in your hand is the heavier ball.
If the weight is not equal -- the heavier ball on the scale is the heavier ball.

hope I made it readable.

NOTE: if you assume the dodgy ball is lighter instead of heavier then just do the oposite discarding
_________________

No Peeking
65 Monk

[ZigonZagoff]

Everyone is also assuming a 2 trayed scale when if fact if you have every had to work on a ceiling fan you would know that you can have as many points on a "balanced scale" as you design into the scale.

But I was keeping quiet until you bumped this.