- 22 May '05 01:01

I have already posed this problem before with coins vice marbles. When I did, no one came forth with a general solution.*Originally posted by Hegemon***There are 12 marbles all weighing the same except one. You have an old fashioned counterbalanced scale. By placing the marbles on the scales you can determine which marble is different from the rest. You can only use the scale three times. How?**

It's a very good problem that took me a while to solve, so I don't want to spoil it by posting the general method. - 22 May '05 03:57

There are two situations here. An easy one, and a hard one. I'll start with the easy situation. We will label the marbles 1 to 12. We will then weigh marbles 1 to 4 against marbles 5 to 8. The easy situation is that these two sets of marbles weigh the same. So, we can use marbles 1 to 4 as our control group. We know our test group is marbles 9 through 12. So, we weigh marbles 1 through 3 against marbles 9 through 11. If the weight matches, then marble 12 is different. If you need to know whether it is lighter or heavier, you can use the third weighing to weigh marbles 1 against marble 12. Situation 1 part B is that marbles 1 through 3 are heavier(or lighter, doesn't matter) than marbles 9 through 11. Then the different marble is lighter, and it is either 9, 10, or 11. Weigh marble 9 against 10, if they weigh the same, then marble 11 id lighter, if 9 is lighter than 10, then 9 is the different marble. If 9 is heavier, then marble 10 is lighter. That takes care of the easy situation. The hard situation is when marbles 1 through 4 are heavier (Or lighter) than marbles 5 through 8. We have marbles 9 through 12, which we know all weigh the same as the majority. So, we remove marbles 1, 5, and 6 from the scale, move marbles 2 and 3 to the lighter side, and place marbles 7, and 9 in the heavy side. Now, we weigh marbles 4, 7, and 9 against marbles 2, 3, and 8. If the two sides balance, then either 1 is heavy, or 5, or 6 is light. So, weigh 5 and 6, and the lighter of the two is different, unless they balance, in which case 1 is heavy. Again, that would be too easy. So, let's say that the side with marbles 4, 7 and 9 is heavier. Then either 4 is heavy, or 8 is light, so weigh 4 against 1, if they balance, 8 is light, otherwise, 4 better weigh more than 1, meaning 4 is heavy. If the side with 2, 3, and 8 is heavy, then either 2, or 3 is heavy, or 7 is light, so weigh 2 against 3 If they balance, then 7 is light, otherwise, the heavier of the two marbles is different. Whew!! That takes care of all the possibilities.*Originally posted by Hegemon***There are 12 marbles all weighing the same except one. You have an old fashioned counterbalanced scale. By placing the marbles on the scales you can determine which marble is different from the rest. You can only use the scale three times. How?**

Worst case scenerio weighings:

1: 1, 2, 3, 4 against 5, 6, 7, 8.

Side 1 is heavier.

2: 4, 7, 9 against 2, 3, 8

side 2 is heavier

3: 2 against 3. ~~skeeter~~515 + 30 days22 May '05 10:52What a load of bull dock. Solution is;

Divide the marbles into two groups of six and balance. Heaviest set is divided into two groups of three and balanced. From the heaviest set take any two and balance for the last time. If they balance then the remaining marble is the heaviest and of course if they don't then you know which is the heaviest marble.

skeeter- 22 May '05 10:57

My solution was to divide the marbles into 3 groups of 4 and balance two against each other. If they are equal, the one not balanced has the heavier one. If they are different the heavier set has the heavier marble.*Originally posted by skeeter***What a load of bull dock. Solution is;**

Divide the marbles into two groups of six and balance. Heaviest set is divided into two groups of three and balanced. From the heaviest set take any two and balance for the last time. If they balance then the remaining marble is the heaviest and of course if they don't then you know which is the heaviest marble.

skeeter

From this group of 4 you can figure out which two has the heavier marble in one weighing, and the last weighing lets you figure out which one it is.

The general idea is that for one of the weighings you divide the marbles you have left into three groups and measure two against each other. You can do this in the first step (my solution), the last step (skeeter's solution), or the second step. In the last case you'd measure 6 vs 6, then 2 v 2 (one group of 2 left out), then 1 v 1. - 22 May '05 10:59 / 1 edit

You don't know wether the odd marble is lighter or heavier. This way, you will not find the guilty one if it is a lighter marble.*Originally posted by skeeter***What a load of bull dock. Solution is;**

Divide the marbles into two groups of six and balance. Heaviest set is divided into two groups of three and balanced. From the heaviest set take any two and balance for the last time. If they bala ...[text shortened]... y don't then you know which is the heaviest marble.

skeeter

edit. perhaps you need to smell again at that bull dock.... - 22 May '05 11:03 / 1 edit

As Mephisto already pointed out, the one odd marble may be lighter or heavier, and you don't know which it is. So your post is the 'load of bull dock', and your solution is no solution at all.*Originally posted by skeeter***What a load of bull dock. Solution is;**

Divide the marbles into two groups of six and balance. Heaviest set is divided into two groups of three and balanced. From the heaviest set take any two and balance for the last time. If they bala ...[text shortened]... y don't then you know which is the heaviest marble.

skeeter ~~skeeter~~515 + 30 days22 May '05 11:08

You must assume that the "marble" is heavier in differance as there is no solution if this is not stated. Perhaps you need to come round to the front of the Bull.*Originally posted by Mephisto2***You don't know wether the odd marble is lighter or heavier. This way, you will not find the guilty one if it is a lighter marble.**

edit. perhaps you need to smell again at that bull dock....

skeeter- 22 May '05 11:15 / 2 edits

You have to assume no such thing. You are making yourself look more and more foolish with each posting.*Originally posted by skeeter***You must assume that the "marble" is heavier in differance as there is no solution if this is not stated. Perhaps you need to come round to the front of the Bull.**

skeeter - 22 May '05 11:17

Of course there is a solution in the case you don't know wether "the marble" is heavier or lighter. Of course, not all mammals would find the solution, and bulls are not amongst them. And another of course, "the marble" could have been made out of bull load (more likely if it is a lighter one, though)..*Originally posted by skeeter***You must assume that the "marble" is heavier in differance as there is no solution if this is not stated. Perhaps you need to come round to the front of the Bull.**

skeeter - 22 May '05 12:32

And how do you know the different marble is heavier?*Originally posted by skeeter***What a load of bull dock. Solution is;**

Divide the marbles into two groups of six and balance. Heaviest set is divided into two groups of three and balanced. From the heaviest set take any two and balance for the last time. If they balance then the remaining marble is the heaviest and of course if they don't then you know which is the heaviest marble.

skeeter - 22 May '05 12:38

It does not matter once you know which it is, but you don't at the beginning. This piece of uncertainty makes the problem much more difficult. If you know beforehand that the special marble is heavier, then the solution is much easier to get.*Originally posted by Duck Duck Goose***call this dumb (not really, it's hypothetical, seriously...), but does it matter if it is lighter or not?**

P.S. i lost my bull