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You have 12 coins 11 are real, one is fake but you don't know if it is lighter or heavier. You have a scale that gives you output but you don't understand it (let's say it is written in a foreign language that you don't understand). Using 4 weightings how do you determine who is the fake coin?

For each possible answer (left, right, equal) there is only one output. Let's say you put 2 coins on the scale and the left is heavier, it A will be presented on the scale. But you see only A which you can't derive that it means that left is heavier because it is in some foreign language that you don't understand. But - A will be always if left is heavier, B for if right is heavier and C if it is equal, you'll just won't know it when you'll see the output.

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This was quite a fun twist on weighing puzzles. The fact we don't know what the scale outputs mean requires us to, as part of our weighing procedure, decipher the scale's output. Luckily, we only need identify the fake coin and not whether it's lighter or heavier - I'm pretty sure distinguishing between lighter and heavier is impossible with only the given apparatus.

Let's label the coins as ABCDEFGHIJKL, and the three possible outputs of the scale as X, Y, Z. When weighing one set of coins against another, the order is important - the first set always goes on the left side of the scale (or whatever passes for "left").

Weighing 1

Weigh ABCDEF against GHIJKL. This will always be imbalanced, which is a good thing in this case. Let's label the scale's output as X - we now know that X corresponds to an imbalanced state. This is the only single weighing where we know whether the scales balance or not - otherwise, there are some scenarios where we included the fake coin and some where we omitted it, so we would gain no immediate information about the scale's outputs.

Weighing 2

Weigh AB against GH. There are only two outcomes possible - the scales remain imbalanced in the same way as weighing 1 (case a), or they now balance (case b). In case a, the scale outputs X again; in case b, we see a different output, which we will label as Y. An output of Y thus means the scale is balanced.

Case a - there are 4 candidates for the fake coin (A, B, G, H).

Case b - the fake coin is one of the other 8, but we now know when the scales are balanced, which is a huge advantage as we can now distinguish between all three possible weighing results (as any output we have not yet seen must by elimination be Z, which means the scale is imbalanced in the opposite way to X).

Weighing 3

Case a - Weigh A against G. Again, there are only two possible outcomes. If the scale outputs X, the scales are still imbalanced, so either A or G is fake. Otherwise, A must balance with G, so either B or H is fake. Either way, there only remain two possible fakes.

Case b - This now resolves in the same way as the standard 12-coin problem, except we don't know the difference between lighter and heavier (but this doesn't matter as we merely need to identify the fake coin, we don't need to deduce if it is light or heavy). Weigh CDJ against EFI. There are three sub-cases here:

Subcase i - we get X, so the fake must be either C, D, or I, as those coins stayed on the same side as in weighing 1.

Subcase ii - we get Y, so the fake must be one of the two coins we didn't weigh here - K or L.

Subcase iii - we get Z, so the fake must be either E, F, or J, as those coins are on the opposite side to where they were in weighing 1.

Weighing 4

Case a - we only have two candidates left (either A and G, or B and H). We can thus weigh one candidate against a known normal coin, keeping the candidate on the same side as it has been so far. If we get X - that coin is the fake; if we don't, by elimination the last candidate is the fake.

Case bi - weigh C against D. If we get X, C is the fake. If we get Y, I is the fake. If we get Z, D is the fake.

Case bii - weigh K against L. If we get X, L is the fake. If we get Z, K is the fake. We cannot get Y as one of these two is definitely fake.

Case biii - weigh E against F. If we get X, E is the fake. If we get Y, J is the fake. If we get Z, F is the fake.

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  • $\begingroup$ First of all - thank you very much! nice trick. I just have a question - in subcase i shouldn't it be C, D or J instead of C, D or I? $\endgroup$
    – user13084415
    Commented Dec 4, 2021 at 17:52
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    $\begingroup$ I'm pretty sure distinguishing between lighter and heavier is impossible with only the given apparatus. We can with one additional weighting: put 2 real coins in one scale, and 1 real coin in the other. You will now know the meaning of all readings. $\endgroup$
    – Abigail
    Commented Dec 4, 2021 at 23:08
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    $\begingroup$ Only in a situation where the extra weight or the reduced weight of the coin is trivial. Imagine coins that normally weigh a fraction of a gram... I wouldn't be able to tell by feel whether a particular coin was even 13 times the weight of the others or not. Or vis versa... if sufficiently light, I'd still have to weigh them, but would have absolutely no way of determining whether the odd coin out was lighter or heavier. Or alternatively, coins where all of them are too heavy to be picked up by hand, and are placed on the scales by machine. $\endgroup$
    – Richard Winters
    Commented Dec 6, 2021 at 7:56
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    $\begingroup$ @RichardWinters Tthe point is that you can do this extra weighing after identifying the fake, so can make sure the three coins you use for it all have the same weight. $\endgroup$
    – Especially Lime
    Commented Dec 8, 2021 at 12:09
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    $\begingroup$ The point is that under certain conditions, you can't know whether the fake coin is 13 times heavier (or lighter) than the rest of the coins or not. If the fake coin is 13 times heavier, then even compared to ALL the real coins combined, the fake coin would still be heavier, but you're assuming that all the real coins are heavier. If this is a more realistic scenario though, where you are handling all the coins and so the fake can only be marginally lighter/heavier than the other coins, yes, you can figure out lighter/heavier easily. But the edge case still exists. $\endgroup$
    – Richard Winters
    Commented Dec 9, 2021 at 1:21
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Once you can identify which symbol means "equal", this reduces to the original problem. That is because if you follow a valid strategy assuming that A means "left heavier" and B means "right heavier", and in fact they are the other way around, you get exactly the same results as you would have done if your assumption was correct and the same coin was fake, but its weight differed in the opposite way. In particular, you will still identify the correct fake coin.

So just identify the "equal" symbol by weighing no coins on each side, then follow a strategy to identify the fake in three additional weighings.

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It's well known how to solve the standard three weighing problem (or N weighing problem with $(3^N-3)/2$ coins) where you must pick out all the weighings before you get any results; for example MADO vs LIKE, METO vs FIND, FAKE vs COIN (with coins relabeled). (Mnemonic "Ma, do like me to find fake coin".)

Follow that strategy for the first N weighings. There will be one coin which is different if X means BALANCE, one coin which is different if Y means BALANCE, and one coin which is different if Z means BALANCE. For the last weighing, weigh any other two coins against each other. That will tell you which label means BALANCE, and thus which coin is different.

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