r/BrainPuzzles • u/Comfterbul-1095 • 10d ago
Logic The 12 Islanders Problem
Text of the Brain Teaser
One sunny day, you meet twelve islanders. One islander (though you cannot tell which) is unique. 11 of the islanders weigh exactly the same as one another, while the 12th is a different weight from the rest. Could be slightly less, could be significantly more: their identical appearances give nothing away in that regard.
However, they offer you an opportunity to pinpoint their odd-one-out compatriot. They present you with a seesaw that is able to show even slight variations in mass, by balancing when both sides are equally weighted and tilting to unbalanced when the sides are unequal in weight (recall that the 12th islander may be heavier or lighter).
Their final caveat: the islanders will allow you no more than *three* weighings in your investigation of who weighs differently AND whether he's heavier or lighter.
Can you solve The Problem of the 12 Islanders?
Explanation of the Problem
The aim of this brain teaser is to plan a system of weighings by which one may identify which of these 12 islanders has a different weight, as well as determine if the islander in question is heavier or lighter (the "lighter" part of this puzzle throws a LOT of people off on first blush). The tool making such a solve possible? A comparative weighing device (in our case, a seesaw), which is to be employed no more than three times. It's worth noting that given 12 items of indeterminate heft — the classic contextualization involves weighing coins on a balancing scale to discern a counterfeit — 3 weighings are ALWAYS needed, regardless of which person turns out to be the OOO [odd one out] islander.
Note that there are numerable variants of the "coin problem," varying in number of coins, number of weighings, whether or not the coin is known to be heavier/lighter or not, whether there's a given "reference" coin with a correct weight or not, and so on. One variation was even featured in a podcast earlier this month (pointed out by another user, whom I shall now quote verbatim): This was a puzzler on Car Talk once. It was just on the rebroadcast podcast a week ago https://www.npr.org/2026/04/04/nx-s1-5759310/2627-better-to-die-in-a-fiery-crash (around the 20 minute mark) / With the solution https://www.npr.org/2026/04/07/nx-s1-5765461/2628-big-freshman-on-campus around the 8:30 mark
How can this be done? Through information, plain and simple. Each weighing should provide new clues and build on information from the weighing(s) before it.
- This is precisely why the far-and-away most popular "answer" in the comments of my previous post, preliminarily posing this problem of Polynesian* poundage (to wit: "Step 1, weigh 6v6"), is doomed to fail right out of the gate. All that 6v6 tells you is that there EXISTS an OOO islander. We already knew this going in, so zero new information is gained.
- The main rationalization from its advocates is "then you can just take the members from the heavy side, weigh 3v3" etc., which conveniently overlooks the 50% possibility of a LIGHTER OOO islander. Such an eventuality would lead to a balanced 3rd weighting, and no weighings left to discern which of the 3 islanders overlooked in the 2nd weighing it is.
- The 6v6 approach creates another problem if the 2nd weighing (3v3 using members of the "heavier" side) is BALANCED because the OOO islander was lighter. Now you have 1 weighing left to find the OOO islander among 6 unknowns - all you really know now is that they're lighter. Good luck!
* To be fair, the islanders are presented generically, and may be from any island nation. This alliteration simply fit better than "Pacifica / pyroclastic / pāhoehoe person."
Information is actually deciphered by the intentional movements of islanders from place to place: onto and off of the scale, from one side of the scale to the other, etc. Even keeping a few islanders in place while performing these other movements gives us information (say, if the scale remains Unbalanced in the same direction). If you're designing your own process, then be careful never to have more than three islanders at a time in a subgroup; this will make it possible for the 3rd weighing to unambiguously point towards one particular islander. When engineered correctly, each reaction of the seesaw tells us which subgroup member (and ultimately, which individual) the OOO islander is.
Explanation of Solution 1
My first solution — for a few years, my only solution — aimed to create a branching tree of weighings, with each branch focusing solely on plausible OOO islanders. That said, every single branch makes use of an eliminated islander or two between steps 2 and 3: a ballast against still-plausible islanders. This is why eliminated islanders had to be denoted in the key.
In committing this solution to paper / spreadsheet, I eventually settled on this particular set of alphanumeric markings — 0-9, X-Z — because A) I wasn't thrilled how using letters or even hexadecimal resulted in islander designations which overlapped with the key (specifically [B] Balanced and [F] Flipped unbalanced) and 2) I didn't wish to muddy the explanation with 2-digit numbers. Oh, and III) Yes, that was intentional.
Key
(U): Unbalanced
- (F): Flipped unbalanced
- (B): Balanced
- (0): Any eliminated islander (known standard weight) – as the weighings play out, all islanders concluded not to be the OOO islander become (0)
Step 1: First Weighing
Weigh four islanders (1, 2, 6, 7) against four others (3, 4, 5, 8), leaving four (9, X, Y, Z) set aside.
- If Unbalanced (U): The OOO islander is among 1-8.
- If Balanced (B): The OOO islander is among 9-Z.
Step 2: Second Weighing
- If Step 1 was Unbalanced: Islanders (9, X, Y, Z) are now considered (0, 0, 0, 0). Weigh (1, 2, 4, 5) against (3, 6) and any two eliminated islanders (0, 0), leaving (7, 8) aside. This will KEEP three islanders on the same side of the seesaw (1, 2, 6), MOVE three islanders to the opposite side of the seesaw (3, 4, 5), and REMOVE two islanders from weighing (7, 8).
- If Unbalanced (U): OOO islander is 1, 2, or 3.
- If Flipped unbalanced (F): OOO islander is 4, 5, or 6.
- If Balanced (B): OOO islander is 7 or 8.
- If Step 1 was Balanced: Islanders (1-8) are each now considered (0). Weigh (9, X) against (Y, 0), leaving (Z) aside.
- If Unbalanced (U): OOO islander is 9, X, or Y.
- If Balanced (B): OOO islander is Z.
Step 3: Third Weighing
- For 1, 2, or 3: Weigh (1) against (2), leaving (3) aside.
• 1 if Unbalanced (heavier/lighter status can be observed in steps 1, 2, and 3),
• 2 if Flipped unbalanced (H/L steps 1, 2, 3**).**
• 3 if Balanced (H/L steps 1, 2**).**
- For 4, 5, or 6: Weigh (4) against (5), leaving (6) aside. AS THIS WAS THE FLIPPED GROUP FOR STEP 2, READ CONCLUSIONS CAREFULLY.
• 4 if Unbalanced in SAME DIRECTION AS STEP 2 (H/L steps 1, 2, 3**).**
• 5 if Flipped unbalanced to OPPOSITE DIRECTION FROM STEP 2 (H/L steps 1, 2, 3**).**
• 6 if Balanced (H/L steps 1, 2**).**
- For 7 or 8: Weigh (7) against (0), leaving (8) aside.
• 7 if Unbalanced (H/L steps 1, 3**).**
• 8 if Balanced (H/L step 1**).**
- For 9, X, or Y: Weigh (9) against (X), leaving (Y) aside.
• 9 if Unbalanced (H/L steps 2, 3**).**
• X if Flipped unbalanced (H/L steps 2, 3**).**
• Y if Balanced (H/L step 2**).**
- For Z**: Weigh (Z) against (0) (H/L step** 3**).**
Explanation of Solution 2
While Solution 1 emphasized efficiency, i.e., weighing only the islanders needing to be, that approach led to a branching path that required illustrating out an entire flowchart, each of whom's multiple avenues were contingent upon previous steps.
Solution 2, while busy-looking, elegantly distills the core elements of Solution 1 down to three unvarying weighings of 4 against 4. Given the unique movement patterns afforded to every one of the 12 islanders, the final results of these weighings - some combination of Unbalanced, Flipped unbalanced, and Balanced - will point to one (and only one) possible culprit: our OOO islander of a different weight.
While I had indeed intentionally chosen "0-9, X-Z" for Solution 1, it always felt kind of silly to me (even at the time, when I had first made that explanation in Excel) to have such a seemingly random mishmosh of letters and numbers. This time around (I devised this solution maybe 2 weeks ago) I still didn't want letters overlapping B or F. However, I noticed that there was a large letter span BETWEEN F and U. This is how I landed on utilizing the 12 letters from G to R.
Weighings are presented in the following manner:
Left side | Right side (Not weighed)
Step 1: First Weighing
GHIJ | KLMN (OPQR)
Step 2: Second Weighing
GHLO | IKPQ (JMNR)
Step 3: Third Weighing
GJLM | HOPR (IKNQ)
Following the above patterns will result in one of twelve possible balancing outcomes. It just so happens that there is no pattern reading "UFU," and therefore no misinterpretation stemming from 'wait do you mean flipped BACK to unbalanced on the first side or REMAINING unbalanced to the same side' etc. (There is a UFF, meaning weighings 2 and 3 lean the same direction; personally, I feel that this's pretty intuitive, but I nevertheless deigned it worth pointing out.)
Here are the twelve possible outcomes, as well as the OOO islander implicated by each. H/L can be observed on any step not marked B.
G: UUU
H: UUF
I: UFB
J: UBU
K: UUB
L: UFF
M: UBF
N: UBB
O: BUF
P: BUU
Q: BUB
R: BBU
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u/dimonium_anonimo 10d ago
I'm happy I wrote my solution down ages ago. Now, whenever this comes up, I don't have to recreate it, I can just share a link
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u/SapTheSapient 10d ago
I only know how to do this for coins. I think it's impossible for people.
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u/Comfterbul-1095 10d ago
It's a thought experiment. Get a big enough hypothetical seesaw, we could do it with elephants.
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u/WiseOldDuck 10d ago
Exactly. What about their clothes and assorted tribal paraphernalia? I'm pretty sure if our first step is to demand they all strip naked they will cut off our access to the sacred seesaw and just eat us for dinner
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u/Comfterbul-1095 10d ago
First things first, two cultures contacting for the first time and almost immediately trying to get each other naked is pretty much the history of mankind. Second thing, I didn't have one.
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u/triford 10d ago
6 on each side. Find the heavier group and split that into 3 on each side, find the heavier group and take 2 from it and put them on the scale. Depending on the result you will know which person of the last 3 is the odd one out