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12 objects and only three experiments

  1. Aug 9, 2005 #1
    YOu have 12 objects of the same apperance. 11 of theme have the same mass and one has a different one (larger or smaller). You allso have a ballance but you can use it only three times (and it's the only way to distinguish betwen masses). Is it posibel to find the object with the different mass (be warned that we dont know if the mass is larger or smaller, and that the ballance only showes three stages: balalanced, left side down and right side down). If there is a posibilty explain how it's done and if there isn't explain why it isn't posible.
  2. jcsd
  3. Aug 9, 2005 #2
    Yes it is possible.

    Separate the 12 objects into 3 groups of 4.
    Weigh 2 groups of them.

    CASE 1:
    If they balance, the 8 objects are good, and the bad object is in the remaining group.
    Use 3 of the good objects to weigh against 3 of the remaining ones.
    If it balances, the last object that has not been weighted is the bad object.
    If it doesn't, one of the 3 suspected objects is either too heavy or too light.
    Just compare any 2 of them on the last weigh.

    CASE 2:
    If it doesn't balance on the 1st weigh, either 1 of the heavier group is too heavy, or 1 of the lighter group is too light. The 4 unused objects are definitely good.
    Label the light group of objects L1, L2, L3, L4.
    Label the heavy group H1, H2, H3, H4.
    The good objects will simply be called G.

    2nd Weigh: H1 H2 H3 L1 vs H4 G G G
    If it balances, we are left with L2 L3 L4, one of which is too light. Simply compare any 2 on last weigh to find out the bad object.
    If left side is heavier, we know for sure that 1 of {H1 H2 H3} is too heavy. Again, compare 2 of them on last weigh.
    If right side is heavier, it means either:
    L1 is too light or H4 is too heavy. Compare any one of them with a G to find out.

    I hope I am clear. In fact, I think the maximum number of objects we can start with is 13 if we are given 3 weighs.
  4. Aug 13, 2005 #3

    Well if they aren't equal you cant tell if one of them is a lighter or heavier ball compared to the Good balls. But you sorta answered it in the next part.

    Except that you used 4 weighs when you "compare any one of them with a G"
  5. Aug 13, 2005 #4
    I think you don't get my answer. The explanation is very long if I go into full details I think.
    Please note that:
    1) when you have 3 balls left, and one of them is too light (or heavy), you can find out which is the one with simply one weigh. Point is, you must already know if the bad ball is too light or too heavy.

    2) My "compare any one of them with a G" is the 3rd weigh. On the 2nd weigh, if the right side is too heavy... then I have either L1 too light or H4 too heavy. Let's say I compare L1 to a G on the 3rd weigh. If it balances, it must mean H4 is too heavy. If it doesn't, then obviously it must be too light.
  6. Aug 17, 2005 #5
    Correction to case 1)
    If you wind up with three untested unknown after two and compare two against each other only if they balance is the 3rd one the odd ball, BUT if the the third test does not balance you only know one might be High or the other could be Low - you'd need one more test!
    Instead after one test if you have 8 G's (good) and 4 U's (untested) name them U1 U2 U3 U4. Then test G G vs U1 U2 balance or not brings it down to U1 U2 or U3 U4. Worst case balance and it's U3 U4.
    Final 3rd test - G vs U3 still balance means it's U4 but you don't know if high or low. But knowing that was not required.

    Your case 2)
    looks OK but also can reach an answer without knowing if the odd ball is high or low. Last test balances to decide between 3 H's or 3 L's.
    Therefore I disagree about 13
    - 12 is the max you can deal with in a three test. 13 balls would be to many to work.
  7. Aug 17, 2005 #6
    I think many of you have problems understanding my solution (case 1). I shall attempt once more...

    First, let's establish the fact that if you are left with 3 untested objects, one of which is too heavy, and you have only 1 weigh left, then it is possible to find out which one is the bad ball.
    This can be done easily just by comparing any 2 of them, say U1 and U2.
    If U1 is heavier than U2, obviously U1 is the bad object, vice versa.
    If U1 = U2, then U3 must be the bad object.

    That said, you MUST know whether the ball is too heavy/light in order for this to work. This is an important condition.

    Separate objects into 3 groups of 4. Weigh 2 of the groups.

    CASE 1:
    The 2 groups balance, thus I am able to identify the "bad group" with the 1st weigh (the untested group).
    Now, I have 8 definitely good objects and 4 untested ones (we don't know if it's too heavy/light yet).

    On the 2nd weigh, I will compare 3 good objects with 3 untested ones
    - if the 3 untested objects are HEAVIER, then surely one of the 3 untested objects is HEAVIER.
    - if the 3 untested objects are LIGHTER, then surely one of the 3 untested objects is LIGHTER.

    We now know whether it is too HEAVY/light. Therefore, we can now apply the method mentioned above on the last weigh.

    (Note: if the 3 untested objects balance the 3 good objects on the 2nd weigh, then they must be good as well and the last untested object must be the bad object. We won't even need the last weigh.)

    CASE 2:
    Read my first post...

    Now, if you think about it using similar methods, 13 is definitely possible. I think that is the maximum you can go.
  8. Aug 18, 2005 #7
    OK I see what I missed
    And since your approch always identifies down to the 12th ball if it is was heavy or light. --
    It means if you find even balance all the way though to the Third test you'd know that the bad ball had been replaced with a good ball and was held out as a 13th ball never included in the testing. You'd know the 13th one was the odd one just not if it was high or low.
  9. Aug 18, 2005 #8


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    Since each weighing is either up,down, or balanced, there are only [itex]3^3=27[/itex] possible measurement results. So, with three weighings, if you need figure out whether the ball is light or heavy, then you have to distinguish between twice the number of balls in possibilities, but fourteen balls requires distinguishing between 28 possibilities - that's one to many.

    It might be possible to include the possibility that all 13 balls are the same weight.
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