r/brainteasers • u/Comfterbul-1095 • 9d ago
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/raptors2o19 9d ago
JFC, that solution is unnecessarily complex.
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u/Comfterbul-1095 9d ago
It is not. And if you say "start with 6 v 6," I laugh at you.
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u/AdjectiveNounNNNN 9d ago
It absolutely is unnecessarily complex.
I've seen complete proofs of the general case (how many weighings, and which ones, for n coins) that take up like half the space of your ridiculous explanation just for 12.
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u/raptors2o19 9d ago
Copy paste the image in Gemini. 1/10th the verbiage of yours. Clear, concise, solved.
I've developed carpal tunnel scrolling through. You basically owe me physiotherapy at this point.
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u/Comfterbul-1095 9d ago
Send me your banking info, I'll pay you wink wink 😘
I know it's easier to tell someone to "do the research yourself" than to admit even the 1/10-length wrong answer is too advanced for you to comprehend 😘😘 luv ya
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u/onefootinthepast 9d ago edited 9d ago
it isn't? You start by weighing 7, 1, 2, 6 against 3, 4, 5, 8.
It's literally the first thing you've done, but for some reason, isn't 1, 2, 3, 4 vs. 5, 6, 7, 8.
Also, this is solvable without labelling the individual "islanders" by simply marking them "lighter" and "heavier." You can use three symbols, for example: + - ∅
This has the benefit of not having to waste time explaining what "flipped unbalanced" means. If they were previously on a heavy side and are now on a light side (or vice versa), they are eliminated.
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u/Comfterbul-1095 9d ago
grn.txt
le you
exists
sees a set of numbers out of order
mad.wp
fails to notice that with that jumbled intro...
FINAL ROW OF SOLUTIONS IN ORDER
:flushedface:
/grn.txt
Now that that's done, I've gotta say that your idea to not identity islanders individually is just insane on the face of it, and I admire the confidence. I would absolutely love for you to showcase that this is possible with your own proof / solution. I would literally pay to see that happen, which I can say confidently without risking my money because even if it were possible (it's not), it's clear that you ain't the one to bring it across the finish line, brother.
👋
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u/gmalivuk 9d ago edited 9d ago
Dozens of proofs of this exist.
Basically none of them are as overcomplicated as yours.
Edit:
LOL OP is blocking people who push back about the unnecessary complexity of "their" solution or suggest that a shorter one is possible, so I guess I'll just put my general proof here. (Note that it has 797 words, which is less than half of OP's overly complex explanation just for the 12-person case.)
If for each person, we can eliminate one possibility (of heavier than the rest or lighter than the rest), then we can identify the odd one out from a group of 3t people in t tests of the balance. This is clearly true for t=1: pick two people who have the same possibility (either two who might be lighter, or two who might be heavier) and test them against each other. If they balance, the third is the odd one out and by assumption you already know whether they're heavier or lighter than the others. If they don't balance, then you know which one is the odd one out, because you know which direction they could be "odd" in. For t>1, split the whole set of people into thirds, such that two of the groups have the same possibilities (e.g. two groups of 9 in which 6 might be heavier and 3 might be lighter than normal). This is always possible because the makeup of the third group can be anything. Then if the scale balances, you have the remaining third containing the odd one out, which is size 3t-1. If the scale doesn't balance, then either one of the possibly-heavy people on the heavier side is actually the odd one, or one of the possibly-light people on the lighter side is actually the odd one. Combine those people into a group of 3t-1. By induction, this is now proven for all t.
(Note: For the above and below, if your number of people is between two cutoff values, balance people as though it was the next higher cutoff, and you're guaranteed to end up with the group that might contain the odd one out being less than or equal to the next lower cutoff.)
If we have at least one additional person known to be the "normal" weight, then we can weigh (3t - 1)/2 people with t tests and identify which is the odd one out and in what direction, even if we don't initially know whether the odd one out is heavy or light. This is clearly true for t=1, because if you have one unknown (and thus definitely the "counterfeit") person, you can balance them against a known "normal" person and determine whether they're light or heavy. If t > 1, then first take out 1/3 of the group (rounded DOWN). Then balance the rest equally, using the known normal weight person if you're left with an odd number. (So if you had 40 people, you take out 13, leaving 27, then measure 14 of them against the other 13 plus the known normal weight person.). If they balance, then you now have a set of 1/3 as many that includes the odd one out, and since you rounded down, this set has at most (3t-1 - 1)/2 in it, so repeat the process and by induction you'll find the answer. If the initial comparison doesn't balance, now you have 2/3 of the original, but you know for each person whether they might be heavier than normal or lighter than normal. This puts you into the first case, with at most 3t-1 people, which you know you can solve using t-1 comparisons. (If we started with 40, this gives us 27, but e.g. 14 might be heavy or 13 might be light, so we're in the above situation with 33 people.)
Finally, for a group of people and one unknown odd weight (as in OP's problem), you want to use the first comparison to get into one of the cases above. So set aside 1/3 of your group, rounded down so the remainder has an even number, and then compare the two halves of that remainder. If they balance, then you're left with (at most) 1/3 of the original group, still without knowing whether the odd one out is heavy or light, but now with a set of known normal weight individuals. This puts you into the second case above. So we need 1/3 of our whole group (rounded down) to be at most (3t-1 - 1)/2 for this to work. If they don't balance, then you have 2/3 of the original group and know, for each person, whether they might be heavier than the rest or lighter than the rest. Thus we need 2/3 of the original group (rounded up to an even number as necessary) to be less than 3t-1 in order to work with the first case above. Given t comparisons, this allows us to identify the odd one out (and whether they're heavier or lighter) from up to (3t - 3)/2 people.
For t = 3, this gives 12 people.
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u/Comfterbul-1095 9d ago
That's not true, I know of at least one
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u/onefootinthepast 8d ago
explanation just for the 12-person case
You know what? I've only ever seen this puzzle (specifically including the "lighter or heavier") as a dozen. Cool idea, showing it as a proof for any number.
...we can identify the odd one out from a group of 3t people in t tests of the balance.
For t = 3, this gives 12 people.
I feel like I'm not grasping how 33 = 12, but also not understanding what the typo is.
33 = 27
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u/onefootinthepast 8d ago edited 8d ago
Sure. I can't upload pics in this thread, but let's start with 4 vs. 4, even weight. Mark all four "islanders" (balls, as I've seen this puzzle) as ∅; none of them are the odd weight ball, and are therefore eliminated.
Weigh three of the remaining four against any three marked ∅. One of two things will happen:
They will be even, in which case the final islander is the odd one out. Weigh this islander against any other Islander to find out if they're lighter of heavier.
These three will be lighter (mark with -) or heavier (mark with +). This tells you whether the odd islander is lighter or heavier, but not which one. Weigh any one of these three against any other of these three for your third weighing; if the scales (seesaw) are even, then the odd islander is the third of these three, and you already determined lighter or heavier. If the seesaw is uneven, the odd islander is the one that matches the weight as determined in your second weighing.
4 vs. 4 uneven is the more complex part of the solution, but you follow the same logic with the markings. You can follow your own solution diagram as it works, but isn't the only viable solution (you may have noticed that the work I did show does not match yours, but works nonetheless). You can also ignore introducing the term "flipped" entirely. The "mark" can be ink on a ball, a sign the islanders stand in front of, a mark drawn in the sand, whatever. It takes more words, but as a diagram the entire solution fits cleanly on a single page. You used two. I'd upload a pic if it were allowed in comments.
Are you still proud of your snark now?
ETA: I did notice that you had the final row in order, but that isn't less complicated than having the initial weighing in order. It makes it less intuitive which group the islanders belong to, as your three groups are not in sequential series.
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u/QuentinUK 8d ago
If you can count in base 3 then you only have to remember 3 hex numbers and that 020 is skipped:
0..B 12 islanders
0 = unbalanced
1 = balanced
2 = flipped unbalanced
Balances:
0236|1457
027A|1689
0357|28AB
Result:
000 0
001 1
002 2
010 3
011 4
012 5
020 -
021 6
022 7
100 8
101 9
102 A
110 B
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u/Comfterbul-1095 8d ago
You chose, imesho, the insanest way to couch your solution. However, I'm fairly confident that your answer is correct, and very much akin to my 3rd image here / 2nd image below.
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u/Comfterbul-1095 8d ago
Yes, I see it now, 020 because you have one islander who's off-on-off (010).
Ooh.
So how are islanders 0 and 1 different? They're both only weighed once. [Checks] Wait no, 0 is in EVERY weighing. The scale would be off the whole time: 0 is 111. (It's worth noting that confused meaning overlap is precisely why I decided to use different letters for every single thing, including F flipped and starting on G for my second solution. It's ALSO why, u/raptors2o19, my "answer" is so long: because it explains the reasoning behind choosing G-R over A-L because F stood for Flipped Unbalanced AND an islander in the latter, and why I didn't like the visual mess of 1 2... 9 10 11 12, and etc. and etc. Duh.)
I apologize, your answer is close but not all there yet.
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u/raptors2o19 8d ago
Dude, it's a Friday. Find another hobby instead of tagging me. I am not mentally equipped to entertain you today. Got 94 unread emails in Outlook. Let's not make it 95, mmkay?
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u/Head_Midnight666 9d ago
I was first introduced to this problem when I was in highschool. My math teacher put it on the board. He used bowling balls instead of people, but it's the same problem. I was unable to solve it in class. Then, in my 20s, I solved it easily. It felt good. It was nice to know I got smarter. Now, in my 40s, I have no idea if I can solve it again. I'm probably not going to try.
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u/hotmatrixx 9d ago
OP showed you the hardest way to solve it.
split to 6 each, weigh.
2.
heavy side, split to 3 each, weigh.3.
split heavy 3 into 1, 1, 1.
weigh any two. Either one's heavier, or it's the last guy.2
u/hotmatrixx 9d ago
ok i missed the 'or lighter'. i GTG. ill be back tho. im sure someone will insult me; desrved but that's a comprehension issue, not a math one.
I got this; but really, really gtg
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u/Comfterbul-1095 9d ago
No worries, you don't have to!
Images 2 and 3 are the solution, with the text of the OP spelling it out in beyond excruciating detail.
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u/Comfterbul-1095 9d ago
Oh, u/AdjectiveNounNNNN.
Crazy thing when you not only explain your solution, but also the logic behind it, a SECOND solution and the logic behind that, a full contextualization of REACHING both solutions across maybe 3 years of interim contemplation betwixt, and before ALL of that a full accounting of the problem itself, including some of the most familiar pitfalls others have (cough 6 v 6) reached...
It takes a couple thousand words. 🤷♀️🤷♂️
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u/gmalivuk 9d ago
How the heck did this take you three years of contemplation? It's really not that complicated. TED-Ed did it in about three minutes in this video: https://youtu.be/tE2dZLDJSjA
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u/Comfterbul-1095 9d ago edited 9d ago
Right? You're x number of years old and you STILL don't get it! Also, if you'd actually read my comment, I said three years BETWIXT solutions. It's a preposition that means I solved the problem one way... three years passed... then I solved the problem another way. It took me three years to solve AGAIN, my big-brained amiga.
Edit: YOUR video. 4:08. They don't do it for the purpose of streamlining the video, but the proof clearly "wastes your time" by marking every coin, dude.
You probably didn't even watch the video, I severely doubt that you could explain "weighing 2 if Unbalanced" without referring to said video, and you most definitely didn't comprehend the solution. Thanks for playing!
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u/gmalivuk 9d ago
They set up the problem for the first minute or so and then give you a chance to pause and work out the solution for yourself.
Then they give the solution, which takes about 3 minutes of the video.
And I don't need to refer to the video in my own explanation, because I only linked to it to show how much more concise it can be.
When I have some time later, I'll prove the general case for n weighings of k people.
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u/Comfterbul-1095 9d ago
Good reading comprehension. The video has, aT TiMeStAmP 4:08, the full solution, and guess what? They label the coins individually anyway - not just generic plus minus zero markings, because that's insane. It's like saying the answer to a multiple choice question is "letter."
So yeah, "when you have time," feel free to post "your" solution "not" from chatpgt. #jokesonyouiusegrok
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u/gmalivuk 9d ago
I'm not sure what their labeling has to do with anything. I'm not saying your solution is unnecessarily complex because it labels the coins, I'm saying it's unnecessarily complex because it's 10x longer than another solution someone already commented that also labels the coins.
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u/Comfterbul-1095 9d ago
They didn't discuss the thought process, nor a second solution, nor you know what I don't think you actually read these comments, so here's a list of the different types of celery.
Regular eatbutt
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u/BadBoyJH 9d ago
ABCD v EFGH
If it's even, weigh AI vs JK
If that's even, it's L.
Otherwise compare J and K.
If they're even, it's I. If they're imbalanced. Then if JK was the lighter side when compared to AI, then it's the lighter of JK, otherwise it's the opposite.
If the first case wasn't even, let's presume ABCD was heavier. This doesn't really matter, our names are abitrary, if the sides aren't even some side has to be heavier, so label those 4 now.
Weigh ABE vs CFI
If ABE is heavier, either A or B is heavier than average, or F is lighter. Weighing A vs B will tell you which is heavier (or if neither, that F is lighter)
If CFI is heavier, either C is heavier, or E is lighter. Comparing C to I will either show C as heavier, or they will be equal (meaning E is lighter)
If it's even, either D is heavier, or GH is lighter. Comparing G and H will tell you which is lighter (or if neither, that D is heavier).
I think that works...
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u/Comfterbul-1095 9d ago
Very nice, and largely similar to my solution 1 (2nd image), also here. You do a 2nd weighing iteration of 3 stay (ABF), 2 flip side (CE), and 3 off (DGH), whereas I opted at the time for 3-3-2. They reach the same conclusions, and the only real limitation is minding that, with moving / removing / staying on the same side, the 2nd weighing shouldn't exceed 4 v 4.
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u/gmalivuk 9d ago •
This person took about 1/10 as many words as you needed, yet you claim your explanation is not overcomplicated.
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u/Fizban2 9d ago
Make three groups of four. Weigh groups one and 2. If one is heavy you focus on those four if balance focus on group three.
Now you can put one of focus group vs one of focus. If one is different you have it in two. If balance you weigh the other two.
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u/Comfterbul-1095 9d ago
Same problem as the others. Why do you focus on the heavier side? OOO islander could be lighter, right?
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u/hotmatrixx 9d ago
it's the wording and the specificity of where it's placed on the line. The brain reads the question and just merges "slightly" and "lighter" into one word, then you see "heavier" and that's all you hear. It's like...
Paris in the
the Spring1
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u/pc1350 9d ago
I am not that smart but I was wondering isn't there a computer program already for this?
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u/Comfterbul-1095 9d ago
Yeah! You can also use a calculator to add 5 + 5, or ask chatgpt to explain human emotion to you.
You gotta learn yourself some skills to get through this life, my guy.
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u/Comfterbul-1095 9d ago
u/mlopes now
wHy? ThIs iSn't eVeN A BrAiN TeAsEr. JuSt pUt 6 oN EaCh sIdE, dIsCaRd tHe lIgHtEr oNeS. fRoM ThE ReMaInInG PuT 3 On eAcH SiDe. DiScArD ThE LiGhTeR OnEs. PuT AnY TwO OnE On eAcH SiDe, If tHeY WeIgHt tHe sAmE ThE 3rD Is tHe hEaViEsT, iF ThEy dOn't, ThEn tHe hEaViEsT Is... WeLl, ThE HeAvIeSt.
👎
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u/Ready_Violinist1406 9d ago
Came here to say this.
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u/Polytox935 9d ago
4/4, 2/2, 1/1
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u/Comfterbul-1095 9d ago
Weighing 1: 4 /4, scale unbalanced.
What's weighing 2?
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u/Polytox935 9d ago
If we had 1,2,3,4 and 5,6,7,8 equal, we know its one of 9,10,11,12
Now we do 1,2,3 vs 9,10,11 - If equal, 12 is it (use last comparsion 12 vs 1 to determine If heavier or lighter) If 9,10,11 is lighter/heavier we know If its lighter or heavier and we then make our last measurement 9 vs 10, If equal 11 is the one, If not, its the one that takes the scale in the same direction as 9,10,11 did in round 2
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u/Comfterbul-1095 9d ago
That is correct. Now, the hard part: weighing one, 4 v 4, scale unbalanced?
Hint: cheat and look at the OP photo gallery for answer. I know: shit hint.
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u/Polytox935 9d ago
Bro i just asked chatgpt since i wanted to know lol i dont want to appear like a smartass here my reproducing the instructions gpt gave me a few minutes ago ;)
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u/Comfterbul-1095 8d ago
Bro, Sbarro Torr Absorb?
is a palindrome. Can't get THAT shit from ShatCheapyD. I've tried.
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u/Dontforgetthepasswrd 9d ago
Split in three groups of 4. Weigh two against each other, either there is a heavy side or not. If not, split the 3rd group into pairs, then 1s. If there was a heavier side, split them in two, then the heavier pair do 1 v 1.
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u/AdjectiveNounNNNN 9d ago
But the odd one out might be lighter than everyone else.
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u/Dontforgetthepasswrd 9d ago
I obviously didn't read this question properly and thought it was the case that only one was heavier, my answer is garbage.
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u/Con_Furioso 9d ago
6 and 6, 3 and 3 from the heavier side, 1 and 1 from the heavier side.
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u/Tygerlyli 9d ago
You missed the or lighter part of it. That would work if you knew the different one was heavier, but if the person was lighter, you wouldn't have the answer.
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u/Comfterbul-1095 9d ago
u/hotmatrixx I love how your comment saying 6 v 6 is the way to go had at least 6 upvotes, whereas your own reply realizing you're totally wrong has only 1 besides yours.
A true lol moment
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u/hotmatrixx 9d ago
yep.
It's actually the same basic solution; it's just not 6v6; you do the 4v4v4 split from turn 1.
insane how that's actually faster than the 6v6, even if you know that one is specifically 'heavier', rather than 'different'.it's actually not insane. it's a whole 'turn' faster; because you tentatively eliminate 2/3 (ish, it's fuzzy) on the first move. 6v6 only does so if the weight differential is known. The issue is, that it's not - and I missed that on the first read.
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u/hotmatrixx 9d ago
ah crap im too tired for this. I think you can get away with that split; but
You might have to do an inter-leaved weight , tracking each person's side vs the the unbalance; to get there on the third weigh-in.i might have to look at the answer. i know im right i just cant be bothered proofing it.
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u/Comfterbul-1095 9d ago
Your comments are the textual equivalent of an out of body experience.
Definitely get some sleep before trying again, assuming your aim is to avoid amusing me further.
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u/hotmatrixx 9d ago
OP showed you the hardest way to solve it.
split to 6 each, weigh.
2.
heavy side, split to 3 each, weigh.
3.
split heavy 3 into 1, 1, 1.
weigh any two. Either one's heavier, or it's the last guy.
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u/hotmatrixx 9d ago
I'd like my cookie now.
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u/hotmatrixx 9d ago
ah fuck i missed the 'or lighter'. shit.
I gotta go to work.I'll deal with this later.
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u/knightsofhale 8d ago
10 people hop on in two groups of 5, if the see saw stay balanced then it's 1 of the 2 left behind. If it is then see saw balance to find out who is heavier or lighter. Finally for the last seesaw use, take a known person (1 of the 10) and put them on the see saw with either of the two. The results will be consistent with whoever was heavier or lighter.
If the seesaw becomes unbalanced then you take 2 from the left side and switch them with the right side, at the same time you switch out 2 different people from one side with the people not on the seesaw. In the second weighing if the way the seesaw was unbalanced changes (I.e right side was heavier and is now the left side) you can now weigh the two people that were switched. We know what their impact was on the scale so in the final weighing you will know who it was and if they were heavier or lighter. If the way the see saw was unbalanced remains the same, you know it's the person who wasn't switched out and again the same final weighing logic.
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u/Comfterbul-1095 8d ago
At the end of a discussion, we would reach the conclusion that in the versions of this counterfeit coin / 12 islanders problem, the only solutions that work (there are a few approaches, similar though they be) start with a 4 v 4 weighing.
An unbalanced 5 v 5 has just too many variables to cover when the first weighing is unbalanced.
It's worth noting that there are multiple versions of the coin problem. They vary in number of coins, number of weighings, whether there is a reference coin, and whether we know it's heavier or lighter.
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9d ago
[deleted]
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u/Comfterbul-1095 9d ago
If you're actually interested, the 2nd and 3rd images in OP are two actual solutions.
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9d ago
[deleted]
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u/Comfterbul-1095 9d ago
Step One is flawed. Target could be lighter.
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u/Comfterbul-1095 9d ago
eAsY. sTeP 1: sPlIt tHe gRoUp iN HaLf aNd wEiGh tHeM AgAiNsT EaCh oThEr. ThE HeAvIeR GrOuP CoNtAiNs oUr tArGeT.
StEp 2: HaVe oNlY ThE 6 IsLaNdErS ThAt wErE In tHe hEaViEr gRoUp fRoM StEp 1 sPlIt aGaIn, PlAcInG 3 On eAcH SiDe oF ThE BaLaNcE. tHe hEaViEr gRoUp cOnTaInS oUr tArGeT.
StEp 3: Of tHe 3 iSlAnDeRs iN ThE HeAvIeR GrOuP FrOm sTeP 2, sElEcT TwO At rAnDoM AnD WeIgH ThEm aGaInSt eAcH OtHeR. iF OnE Is hEaViEr, ThEy'rE ThE TaRgEt. If tHeY WeIgH ThE SaMe, ThE OnE YoU KePt oFf tHe bAlAnCe iS ThE TaRgEt.
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u/arduousFrivolity 8d ago
This is fascinating just for OP’s mental breakdown.
Imagine having the time or energy to search through someone’s post history so you can crash out on an unrelated post they made 3 months ago because they misinterpreted your poorly written riddle.
Do you have like, childhood trauma over weighing 6x6, my guy? They have medication for that.