Earlier, this question was answered incorrectly - A Puzzle from the Heart of Mathematics?

Question

Here's the question:
"There are 12 stones. The diamond is embedded in one of the stones. Eleven of the stones weigh the same, but the stone containing the jewel weighs either slightly more or slightly less than the others. Find the right stone and find whether it's heavier or lighter. You could use each of the three balance scales exactly once.

Can anyone help me out? My boyfriend and me are working on this one and haven't solved it yet."

I know the answer. Now let's see if you can get it.

Not my boyfriend but the original quesioner.

Answer 1

The complication is that you don't know if the embedded diamond causes the stone to be heavier or lighter. That's what makes this a tricky problem.

Here's an elegant way to find the diamond in exactly 3 weighings and to know if it is heavier or lighter.

Number the stones 1 though 12:

First Weighing:
(1 2 3 4) against (5 6 7 8)
Record the heavier side (L=left, R=right, or B=balanced)

Second Weighing:
(1 2 5 9) against (3 4 10 11)
Record the heavier side (L=left, R=right or B=balanced)

Third Weighing:
(3 7 9 10) against (1 4 6 12)
Record the heavier side (L=left, R=right or B=balanced)

There are 27 (3^3) possible combination of scale readings. A complete sorted list of the scale reading appears below. Impossible combinations (BBB, LLL, RRR) are eliminated. Notice how it results in 24 combinations (12 possible stones, 2 possible weights of heavier or lighter).

BBL - stone #12 is light (12L)
BBR - stone #12 is heavy (12H)
BLB - stone #11 is light (11L)
BLL - stone #9 is heavy (9H)
BLR - stone #10 is light (10L)
BRB - stone #11 is heavy (11H)
BRL - stone #10 is heavy (10H)
BRR - stone #9 is light (9L)
LBB - stone #8 is light (8L)
LBL - stone #6 is light (6L)
LBR - stone #7 is light (7L)
LLB - stone #2 is heavy (2H)
LLR - stone #1 is heavy (1H)
LRB - stone #5 is light (5L)
LRL - stone #3 is heavy (3H)
LRR - stone #4 is heavy (4H)
RBB - stone #8 is heavy (8H)
RBL - stone #7 is heavy (7H)
RBR - stone #6 is heavy (6H)
RLB - stone #5 is heavy (5H)
RLL - stone #4 is light (4L)
RLR - stone #3 is light (3L)
RRB - stone #2 is light (2L)
RRL - stone #1 is light (1L)

So in 3 weighings you can cover all possible combinations to figure out which stone has the diamond (1-12) and whether it is heavier or lighter (H/L).

There are more complicated ways to do this with shifting combinations and so forth (see answer #3 above), but my method is a nice algorithmic way to get the answer without having to do a lot of branching, etc.

So the answer is you can find the stone with the diamond (and know whether it is heavier or lighter) in exactly 3 weighings.

Answer 2

Separate the stones into four groups of three, groups A,B,C and D.

Use the first scale to attempt to balance A and B.

If A and B balance the stone is either in Group C or D.

If they do not balance the diamond stone (ds) is on one of the sides, and one side will be lighter or heavier. For the sake of discussion we will assume that A is lighter, and B is heavier. Then weight A with C on the next scale. If A and C are not balanced, we know that the diamond stone is group A, and that it is lighter. If A and C do balance we know the (ds) to be in group B, and that it is heavier, since that group was heavier.

From here we know that the stone is either light or heavy, and we have narrowed it down to three stones. Using the last scale balance two of the stones from the group. If the balance, the third stone is the (ds). If they do not, and they are the group A stones, the lighter is the (ds). If they are from group B the heavier is the (ds).

Now, let us assume that groups A and B balance from the beginning. Now we know that the (ds) is either in Group C or D, and we do not know whether it is heavier or lighter. So, take group C, and weight it against group B, which we know to be a standard group. If they balance the stone is in D, if they do not the stone is in C and you can tell whether the stone is lighter or heavier by how it balances with B.

Then seperate the stones from C and weight two of them...

If C and B balance, you know the stone is in group D, but you do not know its weight. From there you could determine within 2 which stone was the (ds) and know whether it was lighter or heavier, but I don't see a way you could know which of the two it was without bending the rules (i.e. taking stones off of the scale individually).

I see now from reading the above response that the answer to thi problem does indeed lie with an algorithm, simply comparing combinations by forking as I proposed will not solve the problem.

Tiger Striped Dog MD

Answer 3

separate them into 4 groups
weigh the first two groups:

Case 1) they are equal, then the diamond is in the other group.
choose two stones from the third group, and weigh them:
Case 1.1) they are equal, then it is one of the last two rocks. Choose one of them, and weigh it with on of the known rocks. If it is equal, then it is in the last stone. If it is not equal, then it it is in the stone that was weighed.

Case 2)The first two sets are not equal. Label the stones on the heavier side a, b, c, and d, and the stones on the lighter side, e, f, g, and h, and the 4 that weren't weighed call them all x.
Place e, f, g, and d and x, x, x, h on the scale:
2.1) They are equal. Thus we know that it is non of them, so e=f=g=d=h=x, and a+b+c+d=a+b+c+x > e+f+g+h=x+x+x+x thus it is one of a, b, or c, and the diamond is heavier.
Place a and b on the balance:
2.1.1)if equal, then it is c
2.1.2) if not equal, then it is the heavier one.
2.2) e,f,g,d>x,x,x,h. Since e,f,g changed sides and the balance stayed the same, we know it's not them. Since it isn't balanced, we know it's not a, b, or c. Thus it is d or h.
Place d on the scale with an x
2.2.1)equal then it is h.
2.2.2)not equal, then it is d.
2.3) e,f,g,d Place e and f on a balance
2.3.1)equal, then it is g.
2.3.2)not equal, then it is the lighter one.


Bandf is correct, it is always easier to use a chart. But I assume he had to use 27 (24 if you are able to see the 3 impossible cases without computing) case arguments to create that chart . . . it's the prep work to finish work comparison, just depends on what you like more.

Answer 4

I assume you can't see this embedded diamond, otherwise, there would be no question here.

I would place, on one scale, 6 stones on either side, and take the lighter side. My next attempt would weigh 3 and 3, on another scale, and take the lighter. My last step would be to weigh one and one, on the last scale, and if they are different, it's obvious, if not, then the rule of exclusion points to the last stone.

(I'm just assuming the stone with the diamond in it weighs less than the ones without. Of course, this would depend on the type of stone, and the size of the diamond...)

Answer 5

positioned one stone on each fringe of the stability scale. in the experience that they are an same weight, upload one extra stone to each part. back, if the weights are an same, upload yet another to each part, till you positioned on a pair of stones which reason the dimensions to tip. you at present know that one of those 2 is the stone with the diamond, and each of the others don't have a diamond. on your second scale, position between both stones that ought to have the diamond, and yet another stone that you know would not. in the experience that they weigh an same, repeat this with the different stone that ought to have the diamond. Now you may have chanced on the stone with the diamond. once you weigh both achieveable stones adversarial to 2 stones you know don't have diamonds, you know which stones ought to have the diamond, so that you could be in a position to inform if it replaced into heavier or lighters. If I defined it in a confusing way i'm sorry. and that i don't think of the first area ability making use of the first scale more advantageous than once, yet when it does, sorry back.

Answer 6

You weigh four and four. If they weigh the same, the diamond is in the last four. Your then weigh two and two of the last four, The heavier side has the diamonds, You then weigh one and one. In the first try, if one side weighs more, you split to two and two then one and one.

Answer 7

i think you and your boy Friend should get a life
and can give this up

Answer 8

i agree with bandf"s response