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Posers and Puzzles

Posers and Puzzles

  1. Standard member Fiathahel
    Artist in Drawing
    26 Aug '03 14:08
    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.
  2. Donation belgianfreak
    stitching you up
    26 Aug '03 14:58
    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.
    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
  3. Standard member Fiathahel
    Artist in Drawing
    26 Aug '03 16:22
    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 meant thrice. Perhaps I shouldn't use words I don't know how to write
  4. Donation Acolyte
    Now With Added BA
    26 Aug '03 18:11
    I think this problem has been posed before. I'll bump the older version when I find it.
  5. Donation Acolyte
    Now With Added BA
    26 Aug '03 19:34 / 2 edits
    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.
  6. Donation richjohnson
    TANSTAAFL
    26 Aug '03 20:16
    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.
    Check the General forum - I bumped the thread you're looking for.

    -Rich
  7. Standard member TheMaster37
    Kupikupopo!
    27 Aug '03 08:22
    You can determine the marble and defect of the marble with 4 weightings and 12 marbles in total...
  8. Donation Acolyte
    Now With Added BA
    27 Aug '03 15:49
    Originally posted by TheMaster37
    You can determine the marble and defect of the marble with 4 weightings and 12 marbles in total...
    You can do it with 3 weighings