balance problem

The problem is simple:

You have 12 marbles, who look all the same. One of them differs slightly in mass from the others. You don't know if it's havier or lighter. How do you find out which one differs, by only using a balance, and only using it trice? You don't have to know if it's havier or lighter (there is a solution though, that gives you a high probability that you do know it.)

If it's too easy for you, then try to figure out out of how many marbles you can find the odd one by only using the balance four times.

Originally posted by Fiathahel
The problem is simple:

You have 12 marbles, who look all the same. One of them differs slightly in mass from the others. You don't know if it's havier or lighter. How do you find out which one differs, by only using a balance, and only using it trice? You don't have to know if it's havier or lighter (there is a solution though, that gives you a high ...[text shortened]... igure out out of how many marbles you can find the odd one by only using the balance four times.

Originally posted by belgianfreak
I may have an answer, but probably not. I can't tell because you've got a typo on the number of times your allowed to use the balance - "trice"? Could be "twice", or "thrice", or if I'm lucky "five times", because then my theory will work 😉

I think this problem has been posed before. I'll bump the older version when I find it.

Ack, can't find it. Here's a reconstruction of the solution:

Assign a 'weighing code' to each marble, indicating for each weighing whether you are going to put it on the left pan (L), the right pan (R), or neither, eg LRR. After each weighing record which pan is heavier with either L, N or R, and there are two possibilities: if the odd marble is heavier, you'll get its weighing code. If it's lighter, you'll get the inverse of its weighing code (ie L instead of R, R instead of L.) This will work provided the weighing codes obey three rules:

1. They must be distinct.
2. No code may be the inverse of another.
3. They must balance, ie equal numbers of marbles on the L and R pans for each weighing.

First consider the problem with 12 marbles and 3 weighings. NNN is self-inverse. Here are the other codes, paired with inverses and arranged in groups. For brevity, some groups will be indicated just by a description.

2 'N's, 1 'L' / 2 'N's, 1 'R' (3 pairs)

1 'N', 2 'L's / 1 'N', 2 'R's (3 pairs)

2 'R's, 1 'L' / 2 'L's, 1 'R' (3 pairs)

NLR NRL
RNL LNR
LRN RLN

LLL RRR

A suitable set of weighing codes would be the left-hand side of each group, except the last group, which is ignored. Incidentally this guarantees that you will know whether the odd marble is heavier or lighter.

With 4 weighings, only NNNN is self-inverse. The remaining groups are:

3 'N's, 1 'L' / 3 'N's, 1 'R' (4 pairs)

2 'N's, 2 'L's / 2 'N's, 2 'R's (6 pairs)

3 'R's, 1 'N' / 2 'L's, 1 'N' (4 pairs)

2 'R's, 1 'L', 1 'N' / 2 'L's, 1 'R', 1 'N' (12 pairs)

3 'L's, 1 'R' / 3 'R's, 1 'L' (4 pairs)

LRRL RLLR
LRLR RLRL
LLRR RRLL

NNRL NNLR
NLNR NRNL
NRLN NLRN
RNNL LNNR
RNLN LNRN
RLNN LRNN

LLLL RRRR

The odd one of 40 marbles can be found by using the left-hand side of every group except the last, and using NNNN as the weighing code for the 40th marble. I don't think you can do it with 41, and certainly not with more marbles.

Originally posted by Acolyte
Ack, can't find it. Here's a reconstruction of the solution:

Assign a 'weighing code' to each marble, indicating for each weighing whether you are going to put it on the left pan (L), the right pan (R), or neither, eg LRR. After each weighing record which pan is heavier with either L, N or R, and there are two possibilities: if the odd marble is hea ...[text shortened]... for the 40th marble. I don't think you can do it with 41, and certainly not with more marbles.

You can determine the marble and defect of the marble with 4 weightings and 12 marbles in total...

Originally posted by TheMaster37
You can determine the marble and defect of the marble with 4 weightings and 12 marbles in total...