need proof

exponents 12 into two scales first try, then 6 into two scales second try and then last 2 stones on one scale and the other stone on other scale. it's either a hit or miss situation, but if you're lucky this riddle is solvable.

the problem is i need a solution that would work no matter what. I need a solution that at the end, wont involve guessing. Like i said, in 3 steps i got 1 to 1, but when i get to that point, i still dont know if the oddball is lighter or heavier than the rest of them

Place six stones on each side of the scale. Since only one stone is different in weight, take the side that is lighter. Split them into two groups of three. Compare these. Again, one will be lighter. Take that group.

Now, pick any of the two stones. If they are equal in weight, the other stone is lighter. If they aren't equal, the lighter stone being weighed is the lighter stone.

Three weighings, not impossible.

zex... your second weighing could be equal... you don't know if the one stone is lighter or heavier.

I can't remember the exact details, and I haven't time right now to puzzle it out, but it involves splitting the group into thirds...

Whoops. Should have read the instructions. I saw "the stone is lighter".

I was able to determine that starting at 8, every 4 stones that are added, another step is needed. I know that my riddle is possible with 8 stones and 3 steps. so twelve stones, you need 4 steps. This is just a hunch. But I still can't solve it

not impossible...just tedious
forgive me but to explain this will take a while


first number the stones to make it easy to keep track of them (1-12). then you split them into three groups of four.
weigh 1 2 3 4 vs 5 6 7 8

if the two are equal you know it has to be 9 10 11 or 12 so for your second measure you would weigh 9 10 vs 11 1 (you use 1 because you know it is good.) if they weigh equal you know 12 is bad and all you have to do for a third measurement is weigh it against any other stone to find if it is light or heavy. if the second measure is uneven you measure 9 vs 10 as your third measure. if 9 & 10 are equal it is 11 and is is light or heavy dependant on how it weighed in the second measure. if 9 & 10 are uneven it is the one that made the side light or heavy...again depending on what you got in the second measurement.

now if the first measurement is uneven you measure
1 2 5 9 vs 6 3 10 11 as your second measurement. if they are even then weigh 7 vs 8 if they are even then it was 4 and it was light or heavy dependant on the first measurement. if the third measure is uneven it was the one that was heavier or lighter depending on the weight of the first measure. now, if the second measurement is uneven you measure 1 vs 2 if they are even 6 was light or heavy depending on it's previous measures. if the third is uneven, again it was the one that was heavy or light depending on the previous measures.

that should cover the possibilities if i missed anything or you have trouble understanding something please let me know

golazydude

This is a well known brainteaser with an equally well known solution in 3 weighings.

http://www.able2know.com/forums/viewtopic.php?t=13054&highlight=

Lookhowifeel

Interesting! I did it a bit different from you - not entirely differently though - and mine works too.

The major difference is in the second weighing if the first does not balance. My way: take three from the 'light' pan and put them to one side. Take three from the heavy pan and put them into the light pan. Put three 'known to be good' into the heavy pan.

Either these balance, or the side that was light remains light, or the side that was light is now heavy.

If they now balance one of the three removed is light.
If the same side that was light remains light then it is one of the two that did not change that is wrong but it is unknown which, only that either the one in the heavy pan is heavy or the one in the light pan is light.
If the side that was light is now heavy, one of the three moved from one pan to the other is heavy.

After that it is a simple matter of weighing one of a three against another of the three, or weighing one of the two against one known to be good.

But Golaz asked for a proof.

Each stone has three propositions; that it is light, that it is heavy, or that it is good. Two of these propositions must be false for any given stone, so with twelve stones there are two times twelve false propositions to be eliminated.

Each weighing has three outcomes for any given stone; that it ends up in the light pan (which eliminates the proposition that it is heavy) in the heavy pan (which eliminates the proposition that it is light) or balanced which eliminates two propositions for each of the stones in the pans but leaves the unweighed stones still with two false propositions.

With four stones in each pan the outcomes are either - if unbalanced -that one false proposition is removed from each stone in the pans and the unweighed stones have two removed, or - if balanced - removes two from each of the stones in the pans and none from the unweighed stones. Sixteen removed in either case. This leaves eight which is less than the number of possible outcomes to the power of the remaining weighings, so a sloution is still available.

So with three weighings, three outcomes, and two false propositions per stone, it follows that the number of stones times the number of false propositions must be less than the number of possible weighing outcomes to the power of the number of weighings for a solution to be possible.

This means - or should do if I am right - that the same problem could be set with thirteen stones instead of twelve and still be possible of being solved - two times thirteen being less than three to the power of three! - but would not be solvable, as set, with fourteen stones.

And sure enough I have found a solution for thirteen stones. I found it tricky but you might not.

Soreeeee!
<< Hangs head in shame>>
I made a foolish mistake when I thought I had found a solution for thirteen stones. It can only be done if you have a fourteenth stone known to be good, and that is not the same thing, is it?