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Sat 21 Jun, 2008 07:40 pm
We have 8 marbles: 6 are same weight - say m , 1 is lighter - say (m - m0) , 1 is heavier - say (m + m0)
In how many times we use a balance scale (like old version) to figure which is which?
I can do it in four weighings. I doubt it can be done in three.
There are 8*7=56 possibilities for the heavy/light combination. Each weighing has three possible outcomes (left heavier, right heavier, equal). Three weighings has 3^3=27 possible outcomes, Four weighings has 3^4=81 possible outcomes. That doesn't prove that three is impossible, but it can't be done with three predetermined weighings - decisions would have to be made after each weighing.
Here are my first three weighings (these remain the same regardless of the outcomes):
Left | Right | Excluded
--------------------------
ABC | DEF | GH
BFG | CDH | AE
AEH | BCD | FG
The fourth weighing depends on the outcomes of the first three. No combination of outcomes occurs more than three times for the 56 cases; so it is always possible to differentiate them on the fourth weighing.
I'd prefer to have four predetermined weighings with the solution strictly determined by the four outcomes, but I'll need to write a program to either do that or prove that it can't be done.
Yes, 4times I think the best approach.
Quote:I'd prefer to have four predetermined weighings with the solution strictly determined by the four outcomes, but I'll need to write a program to either do that or prove that it can't be done.
Good luck friend.
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