Logic Problem

[morgs]

how about 6 on each side, then 3 on each, then 2, then 1.

I'm not too sure which one to choose first to split up, but it seems like that would be the best way.

 

[Dan]

Right, sorry about that. I'll clean up the terminology once I figure out 2.12. I was sure I had it, but now the way in which I'm choosing the balls in the last step doesen't make sense anymore.

Edit: How about I break 2.12 into two cases.

First I have that A' is above B' on the scale. Now that means that the element from A' that was originally in B is lighter then the rest OR that ONE of the two elements in B' that were originally in B is heavier. Having the element in A' be lighter is the only one of these two that makes sense since my arbitrary assumption that A be below B would be contradicted if the one of the 2 elements of B' was the odd one.

The second case is that A' is below B'. So this breaks down into the same thing above except I pick one of the two in B' since otherwise the conclusion would contradict my arbitrary decision. Now I can weigh these two and choose the heavier one.

You are kaffufling yourself again.

I will try to clear it up for you using your terminology:

So you say that A is below (heavier than) B. Relative to each other, A is heavier than B. This could mean that one of the balls in B is light, or one of the balls in A is heavy.

Next you make the groups A' and B'. Where A' has 2 balls from A and 1 from B, and B' has 2 balls from B and 1 from A.

Now you weigh them, and there are three possibilities: A' and B' are equal, A' is above (lighter than) B', A' is below (heavier than) B'. If they are equal it is easy to solve. If A' is above (lighter than) B', it is also easy to solve. But if A' is below (heavier than) B', it is impossible to solve. If A' is below (heavier than) B', either one of the two balls from A in A' is heavy, or one of the two balls from B in B' is light. You can't solve that in one measurement.

If you want to run a test for a particular case, write down on paper the 12 balls and label them a through j, or 1 through 12. Then pick one as either heavy or light and go through the steps that gets you to your particular case. If it is completely determinate then your solution is correct, (or you did the test wrong).

 

[ericms]

Right, I understand it now. I just skimmed over the last case not even thinking about the other 2 balls that were originally in A.

I'm not that great at logic apparently. I've pigeonholed myself here with what is probably a more complicated argument than needed.

Edit: Ugh.. this begins to become more of an ugly problem every second. I'm trying one more time and if I don't get it then I'll have you show me your solution. This time let A' be as it was, but have B' have 1 element from A, 1 from B, and 1 from C (which I know is of normal weight). So if I'm thinking about this clearly I can have group A be heavier then group B just as before. That part seems fine to me. I weigh A' and B' for my second weighing and can have three cases just as before.

1.) A' and B' are equal which means that either one of the two balls I excluded from B' that were originally in B is light or that the one I excluded from A' which was originally part of A is heavy. There are 3 balls so we're done.

2.) A' is above B' which means one of the elements I swapped is the special ball. I swapped 2 of them so the one that came from A that is now in B' is heavy or the other way around. There are 2 balls so we're done.

3.) This is the case where A' is below B'. So the intersection of A and A' gives 2 elements of which 1 can be heavy and similarly the intersection of B and B' gives 1 element which can be light. We have 3 balls so we're done.

 

[Dan]

I don't get why you can't just say heavier or lighter instead of above an below, but yeah that solution works.

Here is what I did:
http://www.mediafire.com/?513fcbxfxx9

 

[ericms]

You really broke it down into cases didn't you?

Anyway, there is no reason for me saying above/below. I didn't think it would bother anyone.

By the way, you should try the problem in the other thread I started or post another one. These things are fun.