A quick review of tonight’s **“Brooklyn Nine-Nine” **coming up just as soon as I stop eating the islanders…

Even stripped down to two stories instead of the show’s usual three, “Captain Peralta” bit off more than it could reasonably chew in terms of its title plot. Jake’s abandonment by his father Roger is one of the defining events of his life, so the idea of Roger trying to patch things up is a big deal – even though it turned out, as Charles assumed, that he was just going to let Jake down yet again. The whole thing needed more build-up than it got, whether as part of a multi-episode arc (assuming Bradley Whitford was up for doing more than one), or simply pacing this story differently. it felt like we got dropped in on the idea that the Peralta men were communicating again, and then we got caught up in the case that was the real reason Roger had contacted Jake, and the inevitable disappointment felt quick and perfunctory. There were some funny moments (particularly anything having to do with the Hungarian sausage restaurant, or the reprise of Norman Greenbaum’s “Spirit in the Sky” for Scully’s mad dash to the men’s room), and the emotional payoff of Holt acting like a father to Jake was nice, but on the whole this played out in a more empty and predictable fashion than this story about the main character should have.

As for the brain teaser subplot, I liked the pairing of Terry and Amy as the brains versus the more instinctual team of Gina, Rosa and (to their chagrin) Hitchcock. The show is always very smart about what makes each character combination distinct and funny, and seeing the two allegedly smarter characters get on each other’s nerves was fun, as was having Gina figure out that Holt was using them to help him cheat. I’m told, by the way, that there *is *a solution to the puzzle, but I’ll be damned if I can figure it out, and I’ve been trying since I first saw the episode a few weeks ago.

What did everybody else think? And how would you best use the see-saw?

My answer to the puzzle is that it’s impossible to determine the answer without more information.

It’s really not that hard… The guy below me beat me to posting it, but I solved it in about 2 minutes

not impossible. not even that hard.

I’ll give it a shot

First weighing 1,2,3,4-5,6,7,8.

If they balance:

2nd weighing is 1,2-9,10

If they balance

Weigh 1- 11 If balanced answer is 12 else 11

If 2nd weighing doesn’t balance

Weigh 1-9 If balanced 10 else 9

If first weighing doesn’t balance and 5,6,7,8 is heavier

Weigh 1,5,6-2,7,8

If balanced

weigh 1-3 if balanced answer is 4 else 3

If 2,7,8 is heavy

then 7,8 is heavy or 1 is light so weigh 7-8 and whichever is heavier is the answer. If they balance answer is 1.

If 1,5,6 is heavy

then 5,6 is heavy or 2 is light etc….

The rest of the cases are pretty self explanatory but it can get a little complicated. Hopefully, it made some semblance of sense.

This solution might leave you knowing the answer is 12 but not knowing if he’s light or heavy. I don’t remember whether the puzzle required that or just knowing who the differently weighted person was.

I thought we were just responsible for figuring out which person it was. Either way, your solution seems more efficient.

It only required to find one islanders with different weight, so I believe you have solved it.

Thought so too. If they are asking heavier or lighter you will already know with your method if it’s

Thought so too. If they are also asking for heavier or lighter you will already know the answer with your method if it is one of people 1-8. If 1,2,3,4=5,6,7,8 I would measure 9,10,11against 1.2,3. If heavier measure 9 vs 10. The heavier person is the outlier. If equal 11 is heavy. If 9,10,11 equal 1,2,3, 12 is the outlier. measure 12 against 1 to determine heavy or light.

You can only use the seesaw three times though. That is what makes it a more challenging puzzle. Otherwise it is super simple.

If the puzzle was to figure out which islander weighed differently, why say the one weighed slightly more or less? Why not say he weighs differently? And he also only says “You must find out which” without explicitly saying what.

easier is weigh 6 on each side. toss lighter six. weigh remaining 3 on each side. toss lighter 3 . out of the 3 remaining take any two. if balanced, third guy is heavier. if one side havier it is the heavy guy.

Your answer is right for the most part, the only thing that can be improved upon is your

“If they balance:

2nd weighing is 1,2-9,10

If they balance

Weigh 1- 11 If balanced answer is 12 else 11”

A better way to do this and know whether its lighter or heavier is to use a set of three. 1,2,3-9,10,11 this way if it is balanced you have one more use to check whether 12 is lighter or heavier. If you know that 9,10,11 is lighter or heavier than 1,2,3 you can then compare among themselves to figure out the odd one out.

The problem said “one islander is heavier or lighter” so it’s unsolvable. It seems as though you are all trying to solve assuming one islander is heavier. That wasn’t the problem as stated. If one person is heavier OR lighter, you have no idea where that person is sitting on an unbalanced scale, so you can’t just reorganize the heavier group and weigh again – what if the person is lighter?

Trial 1: 4 against 4

One side is heavier, you have 4 light candidates and 4 heavy candidates (8)

The other side is heavier, you have the same (8)

Neither side is heavier, you have 4 candidates for either light or heavy (8)

Case 1 and 2, trial 2:

You weigh two light candidates and one heavy candidate against one light candidate, one heavy candidate, and one known neutral.

Side 1 is lighter: You now have 2 light candidates (Side 1) and 1 heavy candidate (Side 2) (3)

Side 2 is lighter: You now have 1 heavy candidate (Side 1) and 1 light candidate (Side 2)

(2)

Both sides are even: You now have 2 heavy candidates and 1 light candidate (Those who were in trial 1 and not trial 2). (3)

Case 2, trial 2:

You weigh two of the remaining four candidates against one of the remaining candidates and a known neutral.

Side 1 is heavier: Either one of the two on that side is heavier, or the one on the other side is lighter. (3)

Side 2 is heavier: Either one of the two on side 1 is lighter, or the one on side 2 is heavier. (3)

Same: Either the last remaining unweighed person is lighter, or he is heavier. (2)

For trial three, if you have two of the same weight option you weigh them against each other, if you don’t you weigh one of them against a known netural. Or if you only have one candidate left but don’t know if he’s heavier or lighter, weigh him against anyone.

The trick is, you start with 24 options, and you have to get every branch lower than 9 on the first trial, and every branch lower than 3 on the second.

When he said heavier or lighter he meant either or, so the solution is simply to whey 6 against 6, eliminate the lighter ones. Then whey 3 against 3, again eliminate the lighter ones. Then you’re left with 3, at this point just whey 1 against 1 if they’re even then the third guy is the heavier if they’re not then you have your answer as well

Moe, my understanding and solution were the same as yours, but that just seems way too easy for a brain teaser so I figured I must have misunderstood the premise. I guess we must not only determine who is different, but also whether he is lighter or heavier.

Also, I should note that people from Quebec don’t have accents like people from France. Like, at all. Especially not in Drummondville.

Moe, if you think about it, you can just dismiss the lighter because you don’t know whether the heavy side or light side contains the person with the different weight so this solution doesn’t work.

can’t just dismiss, not can. Sorry. That’s what happens when I can’t edit!

Moe your solution wouldn’t work. The different person could be lighter or they could be heavier. Your solution would only work if they were in fact heavier. You can’t eliminate 6 after the first trial because you don’t know which side includes the different person the heavy side or the light side.

Good to see someone else got it. It took me a little while, but really what it came down to was the realization that a swap (switching one potentially light guy with one potentially heavy) results in a unique group that can be tested in test 3 if need be.

With this it’s pretty easy. Doing a 4 on 4 test means that if they are even you can easily determine the odd man in the remaining 4, and if it’s not balanced, removing 3 creates one easy to test group, swapping 2 creates a second easy to test group, and that leaves a remaining easy to test 3 man group. Once you realize the swap operation allows for the easy creation of a 3rd case in the second test, the rest really falls into place.

@Moe – Let’s assume that the brain teaser DOES specify that you don’t know whether or not the odd man is heavier or lighter. There is still a solution, which is why I think that is what Cap. Holt meant. If you know to start whether the odd man is heavier or lighter there are many solutions that work within 3 steps.

I figured it out, posted the solution here :

[github.com]

The solution that @Chris published here is very similar to mine. I’m double checking to make sure it’s right.

The comment that @thatguy left is accurate, its the swap, or shift operation that makes the solution possible. Definitely took me a second too. Ugh.

Can’t decide which I enjoyed more — Terry’s face covered in donut powder, or the fact that the smuggled ED drug was called Turgidol.

Did anyone else think Hitchcock was going to figure it out after Gina and Rosa kicked him off the team?

Trick is to use some of the islanders you already weeded out.

1st Weigh:

3 versus 3

if equal, then that is the “normal” batch of 6. If not equal, then this is “abnormal” batch of 6.

2nd Weigh

3 of the “abnormal batch” against 3 of the “normal batch”

If equal, you know that the other 3 in the abnormal batch has the culprit. If not equal, you know the culprit was in this abnormal batch of 3. Either way, if you’ve been keeping track of the weighing, you’ll know at this point whether the abnormal islander is heavier or lighter.

3rd Weigh

1 of the abnormal (from the final remaining batch of 3 abnormals) against 1 of the other abnormals from the same batch. If they are equal, you now know it was the non-weighed final one of that batch. If the two being weighed are not equal, you can still figure it out because you know whether the abnormal islander is heavier or lighter.

What if the first two weighings are equal? Which is possible as you only use 9 of the 12 for those first two. Then it becomes impossible to know which of the final three is the outlier.

@Howard The one problem I see with that process is that if your first two weighs both have come up equal you still don’t know if the person is lighter or heavier. So on the final weigh if the 1v1 comes up unbalanced you wont know which on to pick.

@howard – logical flaw in your step 2. In the way you are doing it, weighing groups of 3 against each other, if the first 2 measurements are equal, then you can’t know heavier / lighter.

@howard – logical flaw in your step 2. In the way you are doing it, weighing groups of 3 against each other, if the first 2 measurements are equal, then you can’t know heavier / lighter.

@people who are @howard. howard is right. no flaws.

Omg why are people coming up with these complicated solutions? Whey 6 against 6, eliminate the lighter ones. Then whey 3 against 3, again eliminate the lighter ones. Then you’re left with 3, at this point just whey 1 against 1 if they’re even then the third guy is the heavier if they’re not then you have your answer as well

What if the person is lighter then the others? It could be either heavier or lighter

So… the problem is about making cheese?

Are curds also involved?

If they are, we’re one-third on the way to making poutine! Aww, yeah!

key point is the odd man out can be heavier OR lighter. going 6 v 6 doesn’t give you any answers, could be the guy is on the heavier side because he weighs more, or he could be on the lighter side because he weighs less. 6v6 doesn’t solve. scroll up to GREG, thats the solution.

Firstly, it’s spelled “weigh”

Secondly, you’re not even close to correct. It’s stated that the odd man out could be heavier or lighter, so when you eliminate the 6 lighter ones after your first trial, you could be eliminating the person that is different and your next 3v3 trial could balance out. Then you’re stuck with a group of 6 and only 1 trial left. Your solution is as good as your spelling.

This only works if everybody is the same weight except the heavier one, listen to the riddle closely

I was under the impression that it was either or, because rosa later in the episode says I’ll smash fatty with the seesaw so I assumed you were supposed to pick one and solve it based on that assumption, my bad..

The lack of a father figure seems like it could be an important part of Jake’s character, but as a standalone episode, “emotionally unavailable special guest star dad” is pretty sitcom 101.

I can only do it in four uses of the seesaw…groups of 4…you have to measure twice to figure out which group is a different weight and if its heavy or light…if both groups of four are equal, you know the third group has the target person, but you still need to measure against one of the first two to check if they’re heavy or light…conversely if one of the first two groups sinks, you measure it against the third group. If they’re equal, the other group is light. If they sink again, they’re heavy.

The third use you split the group in half and narrow it down to two people. But I need a fourth use to find the exact person.

So basically this is a useless post, sorry everyone

It’s really easy, the only thing the captain asked was if the person was heavier or lighter (and I re-watched the question to make sure). You only need 2 moves.

So measure 3 versus 3. If neither side move, than those 6 is a control group that all weighs the same. If they do move, than the 6 who are not included is the control group who weighs the same.

Then do, those 6 versus the 6 you didn’t measure. If the control group rises, the person is heavier, if the control group drops, the person is lighter.

Once again, he did not ask who the person is.

he did say “which;” but given the long sentence, he could have meant which man. i’m not going to argue semantics at length, though; i like the solution.

Just rewatched it, this actually blew my mind. Great job man. There is a solution if you have to find out who the person is, but the way he worded it, you have the correct answer.

I only watched through the episode once, but I also don’t recall Holt requiring that the odd person out be identified. The actual riddle, one I’ve heard before, does require you to identify that person. It’s unfortunate that they posed the riddle wrong in the episode. :(

I am just finished the episode right now, however, this is really bugging me. I came up with something that I find plausible but before I’d break out the pencil and paper, I thought I’s try and collaborate with people likely smarter than me on the internet. I bet there is something wrong with my solution and I can’t see it.

Anyway, the crux is since you need to know whether the one-of-these-things-is-not-like-the-other islander is lighter or heavier, you need to have each islander set up in such a way so that none of the islanders have the same relationship to each other across the 3 weigh ins. You could do this by having a unique distribution for each islander across three separate groups for each weigh in: left, right, and NA. So you would need 4 v 4 on each separate weigh-in and 4 set aside.

So, I think example makes the Islanders distinct for each other:

1: L, L, R

2: L, R, L

3: R, L, L

4: L, L, NA

5: L, NA, L

6: NA, L, L

7: R, R, L

8: R, L, R

9: L, R, R

10: R, R, NA

11: R, NA, R

12: NA, R, R

Sorry about that, I was dummy. I realized when I pressed send that I actually had just made the first half of a distribution table rather than a workable distribution. You would need to split up the groups so that you could get these 24 outcomes but the distribution above wouldn’t work:

1: L, L, R

2: L, R, L

3: R, L, L

4: L, L, NA

5: L, NA, L

6: NA, L, L

7: R, R, L

8: R, L, R

9: L, R, R

10: R, R, NA

11: R, NA, R

12: NA, R, R

13. L, R, NA

14. L, NA, R

15. R, L, NA

16. R, NA, L

17. NA, R, L

18. NA, L, R

19. NA, NA, L

20. NA, NA, R

21. NA, L, NA

22. NA, R, NA

23. L, NA, NA

24. R, NA, NA

When I get a distribution that works I’ll update the post.

Your theory is decent, your flaw is in the setup. 12 islanders, 3 postions, and 3 weigh ins. In each weigh in, you need 4L, 4R, and 4NA. Meaning in total you have to end with twelve of each postion. You do not, you only have 6NA total. You placed an extra L and R in each set. Simply count the Ls an Rs in each weigh in on your list and you’ll see.

Second thing, I figured out whether they heavy or light as well as which islander, I missed the part where he didn’t care who it was just what. So I’m not sure your theory would actually work because although each islander has a seperste set of postions, there is am islander who mirrors each. Thus begging the question is one lighter or one heavier. Ex.

1L,R,Na

2R,L,Na

First test L is heavy

Second test R is heavy

Third test balance. So the there’s ten that differ can’t fit the bill. But is 1 heavy or is 2 light?

To figure it out for sure you have to isolate which islander is different and then use that against the other data in the tests to decide if he is light or heavy,

And that is complicated to explain in text, some of these guys are very close to having it with the 4v4 start.

i treated it a bit more like a riddle, as i didn’t see a solution presented otherwise.

#1 weigh all twelve. one side will be heavier. oh well.

remove one from both sides at a time (doesn’t constitute a new use… it’s a seesaw, you can’t just all hop off at once) the last couple to exit before the seesaw evens out has your culprit – continue exiting and once all are away, move on to –

#2 weigh either of those two men against ANYONE other than the one he was paired with (referred to as “control” for next step). result does not matter.

#3 weigh “control” against the other of the two men in the odd group out.

combine results from #s 2 and 3 for solution, including whether odd man was heavier or lighter than the rest of the group.

is this against the spirit of the exercise? who’s to say? i couldn’t come up with a viable solution otherwise, and it just seems so perfectly “outside the box” for a brain teaser/riddle.

comparison 1: 6v6 (take heavier group of 6 and continue to comparison 2)

comparison 2: 3v3 (take heavier group of 3 and continue to comparison 3)

comparison 3: 1v1 (take heavier one from the wieghing, otherwise if balanced, it is the one left out)

you’re dumb

i was trying to find which individual person was lighter/heavier check my other comment to find if they are in fact heavier or lighter.

I misinterpreted it :(

My answer is so simple

First weigh 1,2,3,4,5,6 vs 7,8,9,10,11,12

(Take the side with the unequal weight)

Now weigh 3 people vs 3 people

Take the side of unequal eight and remember if they were heavier or lighter than the other side

Now weigh 2 of the 3 people

If they are equal, the last person has the unequal weight and remember if they were heavier or lighter from the second time you weight ppl (3 vs 3)

If it’s unequal, then you know which person is heavier/lighter

Here is the answer to the puzzle. [puzzles.nigelcoldwell.co.uk]

Thank you Ryan. Rather than having the #nerdfail face for the rest of the night, I gave in to the temptation and clicked on the link you provided. Now that you put my mind at ease, I am thinking at a Scully-esque rate of synapse fires per minute and thinking about Captain Holt holding two puppies.

Weigh them in groups of 4.

Comparison 1: 4v4 (hold 4)

Case 1: Balanced

Use held 4 against other 4 and see if they go up or down (heavier/lighter).

Case 2: Unbalanced

Use held 4 against other 4 and see if they are balanced.

If held 4 go up then man is heavier, if held four go down then man is lighter.

If they are balanced then use held 4 against other 4 from first weighing and compare.

#DoneLikeDinner

~Amtian

How many times can you answer this?

weigh 1,2 &3 vs 4,5&6 if they balance then you then u weigh 7,8&9 vs 10,11&12 and you come up with an answer of e.g. 78&9 weighs more than you know its one of those 3 if you weigh 7 vs 8 and they balance than you know 9 is the heavy one etc.

Simple. The question is:

“There are 12 men on an Island, 11 weigh exactly the same amount but one of the is slightly lighter of heavier, you must figure out which”

Based on the wording you *only* need to figure out if the individual is heavier or lighter, you don’t need to actually find the specific individual. Therefore:

1) Break the men into four groups of three

2) Weigh group A against group B

3A) If they weigh the same, then weigh group A against group C followed by D. One of these groups will be either heavier or lighter than A. Your answer is whichever this is.

3B) If A and B weigh differently, weigh group A against group B. If A and C weigh the same, group B is the group that is heavier or lighter, your answer is whichever it is. If there is a difference between A and C, group A is the group that is either heavier or lighter, your answer is whichever this is.

Split to groups of four, measure two groups. If two groups are not even, note heavier/lighter side and swap two balls across sides. If direction changes take and measure those two to find heavy/lighter. If not measure other two. If first was even, measure remaining four two on each side and then swap one. If direction changes its the one you moved if not its the other.

Greg clearly got it first and has a correct solution (though not the only solution). He wins. Story over.

Greg clearly got it first and has a correct solution (though not the only solution). He wins. Story over.

I know the captain says you can’t have 6 on each side of the seesaw, but funk that.

Put 6 on each side. One side is heavier. Take the 6 from the heavier side and put 3 on each side of the seesaw. You now have three people and one more use of the seesaw. Take 2 of the remaining three and put one on each side of the seesaw. If one is heavier, you have your answer. If they both weigh the same, the one not on the seesaw is the heavier person.

I don’t know why the captain said you couldn’t have 12 people on the seesaw, but my way works. Seesaws can totally handle 12 people.

You have to both identify the person and if heavy or light 6v6 tells you nothing other than 1 on side a is light OR 1 on side b is heavy. Which we already knew. This is a futile waste of time as indicated by holt Ryan has the correct solution as does the individual who previously posted the same one.

simple he wants to know what islander weighs different. weigh4 vs 4 on first weighing this will give you 4 people left if the scale balance and if it dosent you take the 4 that differ and weigh 1vs 1 this will tell you who is different weight on the 2nd 1v1 which is the 3rd weighing

Weigh six and six. Take the six on the heavy/light side and weigh three and three. Take the three and weigh two of them. If one is heavy/lighter, there’s the answer. If they’re even, it’ll be the left out guy of the final three, and it can be determined by if the first two trials were heavier/lighter

It depends on their definition of a use. Riddles usually involve bending the rules. So say you have 6v6, have one from each side walk toward the fulcrum until the sides even out. You know one of those 2 are either lighter or heavier. Then you can weigh them against those you know are of the same weight and figure out which is lighter or heavier.

Couldn’t we just ask everyone what they weigh? It was never stated that no talking was allowed.

Or just look at them.

You can get it down to one of two persons in three moves, but you won’t know which one is lighter or heavier. You can only solve if you know the different person is either lighter or heavier. Bummer…

My answer to the puzzle is to load the seesaw for the first weighing one person at a time. Meaning that load person one on side A and person two on side B. Repeat until the seesaw is not balanced after a pair has been loaded. Then weigh each member of the last pair that made the seesaw not balanced against any one of the other people (called control). One of the final pair added will balance with the control, the other will not, and show that person to be lighter or heavier.

Well, the comments have a lot of discussion about and solutions to the puzzle, but I didn’t catch what the puzzle actually was.

Answer to the Riddle:

[Round 1]:

Weigh 4 vs. 4 (leaving 4 aside)

There are two very different strategies, depending on what happens in round one:

A. If Round 1 balances: then those 8 are known “same” and the “different” one is in the remaining 4 “unknowns”

[Round 2]: weigh 2 “sames” against 2 “unknowns”

-If Round 2 balances, then all from round 2 are “sames” and the “different” islander is in the remaining group of two “unknowns”

[Round 3]: weigh one of the “sames” against one of the remaining “unknowns.” If it balances, then the other “unknown” islander is the different one. If it doesn’t balance, then the “unknown” islander you just tried is the “different” one.

-If Round 2 doesn’t balance then you know that the “different” islander is in the group you just tried against the known “sames.”

[Round 3]: Weigh one of the “sames” against one of the “unknown” islanders from round 2. If it balances, then the other “unknown” islander is the different one. If it doesn’t balance, then the “unknown” islander you just tried is the “different” one.

B. If Round 1 doesn’t balance, then you have 4 known “sames” (the ones you didn’t try), 4 suspected heavies from round 1 (we’ll label these H1s), and 4 suspected lights from round 1 (L1s). Because we don’t know if the “different” islander is heavy or light, we don’t know which group he is in. At this point, it is important to record which side of the seesaw each villager has been on, because this will help you figure out who is different.

[Round 2]: Make two groups of three out of the people from round 1.

Group 1: Two on the heavy side of the scale from round 1 and a light {H1, H1, L1}

Group 2: Two on the light side of the scale from round 1 and one heavy {L1, L1, H1}

Not weighed in Round 2: One from the light side in round 1 and one from the heavy side {L1, H1}

Again, there are two very different strategies for Round 3, depending on what happens in round 2.

a. If round 2 balances, then the “different” islander is in the group you didn’t weigh in round 2. So round 3 is easy:

[Round 3]: Weigh a known “same” from a previous round against one of the remaining two “unkowns.” If round 3 balances, the remaining islander is the “different.” If it doesn’t, the “unknown” you just tried is the “different” one.

b. This is the tricky part. If round 2 doesn’t balance, then you can eliminate the two islanders you didn’t use in round 2, but you still have 6 islanders, all of whom might be the “different” one and only one more round left. This is why you recorded which side of the seesaw they were on, however.

The secret is that the “different” islander will always fall on the same side of the seesaw, because he is the only one influencing its movement. Everyone else is essentially neutral, so they can be on both the heavy and the light sides of the seesaw.

So after round 2 you have six people:

Heavy Side in Round 2:

1: H1, H2

2: H1, H2

3. L1, H2

or

1: H1, H2

2: L1, H2

3: L1, H2

Light side in round 2:

4: L1, L2

5: H1, L2

6: H1, L2

or

4: L1, L2

5: L1, L2

6: H1, L2

You can “eliminate” anyone who was on different sides as not being able to influence the scale and thus being a “same.” This will get you down to 3 possible candidates.

So you will either have this:

1: H1, H2

2: H1, H2 (leave this guy out of round 3)

3: L1, L2

or this:

1: L1, L2

2: L1, L2 (leave this guy out of round 3)

3: H1, H2

depending on which group the “different” islander was in during round 2 and whether he is heavy or light. It doesn’t matter how the split ends up.

[Round 3]: For round three, you will weigh one of the people who was in the light group both rounds and one of the people who was in the heavy group both rounds against two people who are known “sames,” leaving out one of the people (who is either light or heavy, depending on the distribution of round 2).

-If round 3 balances, then the “unknown” you excluded is the “different” islander.

-If round 3 doesn’t balance, take note if the “unknown” group of two is heavier or lighter than the “same” group. If the “unknown” group is heavier, then the H1, H2 islander is the “different” one. If the “unknown” group is lighter then the L1, L2 islander is the “different” one.

1. Split islanders into 2 groups; A group is 1,2,3,4,5,6- B group is 7,8,9,10,11,12.

2. Take A and split in half once more. Weigh the sub groups against each other 1,2,3 vs 4,5,6.

3. If A is balanced eliminate all of A group. If A is off then eliminate all of B group. We will say A was unbalanced (4,5,6 was heavey) and B was eliminated.

4. Now take 3 from each Group A and B and switch them: A=1,2,3,7,8,9 B=4,5,6,10,11,12

5. Repeat step 2 with 1,2,3 vs 7,8,9.

6. We know that from elimination 7,8,9 are all the same. So if the sub groups are even, eliminate 1,2,3 (islander is heavy) if not eliminate 4,5,6 (islander is light). For our puposes, we will pick even, leaving 4,5,6 (the heavy islander).

7. weigh 4vs5. If one is heavier, that’s your islander. If they are even, 6 is your islander.

Divide the people into 4 groups named 123, abc ,xyz and, nop.

first use: 123n vs. abco

if balanced one of wyz or p are the odd one.

if not balanced one of 123n or abco are the odd one.

second use: move 123 to the other side of the seesaw

remove abc and place xyz in 123’s original place.

or xyzn vs. 123o

if balanced twice the odd one is p

if out of balance in the same direction both times the answer is either n or o

if balanced the first time and not the second then one of xyz is the odd one and the odd one is either heavier or lighter depending on the action of the seesaw.

if not balanced the first time and balanced the second time, one of abc is the odd one and is lighter or heavier depending on the action of the seesaw.

third use: let’s review, if unbalanced in the same direction twice weigh either n or o against anyone else if balanced the one not weighed is the odd one,

if out of balance vice versa.

or of the three groups (abc, 123, and xyz) we know two things from the first two weightings: which one has the odd person and whether he is lighter or heavier. Weigh two from that group if balanced the odd one is the one not weighed if unbalanced the one that matches the unbalance of the group is the one.

The End

You all do realize that the writers have trolled you all and set an army of Internet nerds towards solving a problem their first year lit teacher mocks them with every year, right?

I might be missing something because the problem seems simple to me. Take 6 of the islanders and weigh 3 vs. 3. If the seesaw moves, your abnormal islander is in this batch, if it doesn’t, it’s in the other.

Now, weigh the batch with the abnormal islander against the other batch. If the abnormal side goes down, he’s heavier. If not, he’s lighter.

Now, have the islanders get off the see-saw one at a time from each side. When the seesaw balances, the last person who got off from the normal side is your outlier.

*last person who got off from the ABnormal side.

He said you can only use the sea-saw 3 times but he doesn’t say everyone has to get on at the same time.

So…you get the islanders to go on the sea-saw one at a time alternating the side they join. If it balances at each second guy then all those currently on the sea-saw weigh the same. When you get an unbalance you know that the lighter or heavier guy is one of the last two to get on. – That’s your first time using the sea-saw.

So you take someone you know to be the standard weight either one of the first guys to get on or consequently if the first two guys that go on were unevenly weighted then any of the other islanders. You take the standard guy and put him on one side then one of the guys who unbalanced the sea-saw and put him on the other. If it balances then you know it’s the other guy. Then you just think back to which side he was on – was it higher or lower to determine whether he is lighter or heavier? Or weigh him again against one of the others if you want. If it doesn’t balance then it’s the guy on the sea-saw who you’re not using for the weight standard.

Here is the solution.

Weigh 1: 1,2,3,4 vs 5,6,7,8

If 1,2,3,4 goes up, either 1,2,3,4 is light or 5,6,7,8 is heavy. 9,10,11,12 are normal.

Weigh 2: 1L,2L,5H vs 3L,6H,9N.

If 1L,2L,5H goes up, either 1,2 is light or 6 is heavy. Weigh 3: 1L vs 2L. Whichever side goes up is the light one. If balanced, 6 is heavy.

If 1L,2L,5H goes down, then either 5 is heavy or 3 is light. Weigh 3: 5H vs any normal. If 5 goes down, it is heavy. If balanced, 3 is light.

If Weigh 2 is balanced, then either 4 is light or 7,8 is heavy. Weigh 3: 7H vs 8H. Whichever side goes down is heavy. If balanced, 4 is light.

If Weigh 1 is balanced, then 9,10,11,12 is either light or heavy. 1,2,3,4,5,6,7,8 are normal.

Weigh 2: 9LH,10LH,1N vs 11LH,2N,3N.

If 9,10,1 goes up, then either 9,10 is light or 11 is heavy. Weigh 3: 9L vs 10L. Whichever side goes up is light. If balanced, then 11 is heavy.

If 9,10,1 goes down, then either 9,10 is heavy or 11 is light. Weigh 3: 9H vs 10H. Whichever side goes down is heavy. If balanced, then 11 is light.

If Weigh 2 is even, then 12 is either light or heavy. 1-11 are normal. Weigh 3: 12LH vs any normal to determine whether 12 is light or heavy.

By the way, I love Brooklyn Nine-Nine!! Great show.

I believe in the episode Captain Holt said that one islander was heavier or lighter you decide which, everyone went with heavier… So I’ll go with heavier as well

Measurement #1

1-6 on one side and 7-12 on the other (I’ll say 1-6 was heavier)

Measurement #2

1-3 on one side and 4-6 on the other (I’ll say 1-3 was heavier)

Measurement #3

#1 on one side and #2 on the other (whichever side goes down was the heavy one, if they are equal then it was #3 that was the heavy one)

Since when did Jake know his dad? I thought he didn’t even know his identity.

ok got it in 2 tries. and it dosent ask who is the odd weight just if they are heavier or lighter.

(1) 1,2,3 vs 4,5,6

-if equal than 7,8,9,10,11,12 has the odd weight.

(2) 1,2,3,4,5,6 vs 7,8,9,10,11,12

-if 7,8,9,10,11,12 is heavier than that odd person on the island is heaver.

-if 7,8,9,10,11,12 is lighter than that odd person on the island is lighter.

-if unequal than 1,2,3,4,5,6 has the odd weight.

(2) 1,2,3,4,5,6 vs 7,8,9,10,11,12

-if 1,2,3,4,5,6 is heaver than that odd person on the island is heaver.

-if 1,2,3,4,5,6 is lighter than that odd person on the island is lighter.

Solution:

You only have to use the seesaw twice.

The goal is to figure out if the one man out is heavier or lighter than the rest (1 out of 12 men is either heavier or lighter).

Split the 12 men into two groups of 6.

6 on one side, 6 on the other side of the seesaw.

Label the side that goes up as the ‘light’ side and label the side that goes down as the ‘heavy’ side.

Choose only one side of 6, heavy or light, as the other side doesn’t matter. Split that group into two groups of 3 and put them on the seesaw.

Theoretically, only one of these groups of 6 will move the seesaw and the other group will be equal weight and will not move the seesaw.

If you choose the ‘heavy’ group and the seesaw moves then the 1 man is heavier than the rest but if the seesaw doesn’t move then the 1 man is lighter than the rest and vice versa.

Explanation:

All but one man is a different weight than the rest (1 out of 12). Split into groups of 6, one man will make the weight difference to make the seesaw move. That one man stays in that group and if that man stays in the group you chose and you split that group into two groups of 3, the man with a different weight is on one side of the seesaw and will again make the seesaw move. If that one man is not in the group you chose, the two groups of 3 will be equal weight and the seesaw will not move. This tells us that the group that moves the seesaw, heavy or light, is the answer.

Has anyone else noticed that the ratio of comments about the puzzle to those about the episode itself (or even B99 in general) is about 10:1?

I re-watched the episode again, and the Captain never says we have to identify who the person is, but rather whether that person is lighter or heavier than the other 11 islanders. You can identify whether the odd person is lighter or heavier using the see saw only twice.

Split the islanders into three groups of 4.

For the first use of the see-saw, place Group A on one side and Group B on the other.

One of two results can happen. Either Group A and Group B are the same, or one group is heavier/lighter than the other.

Scenario 1- Group A and Group B are the same.

Place Group A on one side of the see saw and Group C on the other. If Group A rises the odd person is heavier than the other 11. If Group A sinks the odd person is lighter than the other 11.

Scenario 2 – Group A and Group B are different.

Take whichever group sank and place it on one side of the see saw and place Group C on the other side. If Group C rises then the odd person is heavier than the other 11. If Group C is the same then the odd person is lighter than the other 11.

This seems to be the most reasonable solution to the brain-teaser. (Can anyone find a flaw?)

[www.mathsisfun.com]

I don’t need the seesaw at all.

Have the 2 guys dig a hole on the beach and have it fill with mostly with water. Mark the level of the water.

Then have each of the 12 men take turns getting in the hole. The water level for the 11 of the same weight will change the water level the exact same.

The person with the different weight will be the only one that has the water level either higher (heavier) or lower (lighter) than the rest.

You now know who is the different weight and if he is lighter or heavier. It’s called water displacement.

nice, I posted mine, then saw yours. I agree

[www.puzzle.dse.nl]

Here’s the answer

I say the see saw is only misdirection, As they are on an island water is bound to be near by. Everyone should go float on their backs and whoever is more or less buoyant is the one of un equal weight.

Divide them into four groups of three. One of the groups will have the islander that weights slightly different so 3 of the groups will weight the same and one will weight different (either less than the other three or more than the other three). Let’s say the 4 groups of 3 are group A B C and D.

First time on the saw:

Group A on one side Group B on the other

Second time on the saw:

Group C on one side Group D on the other

The results on these two weighings are predictable. One weighing will yield equal weights and a see-saw that doesn’t tilt and another weighing will yield a see-saw that tilts in one direction or another.

For the sake of convienience let’s say A and B weigh the same in test one and it tilts towards C in test two. Now for the third test weigh A against C. If it tilts towards C we know that groups A and D weigh the same and it tilts towards C because it contains the heavier islander. It cannot tilt towards A but it could yield a flat see-saw meaning A and C (and also B) weigh the same, this would imply that the result of test two (C weighing more than D) is because group D contains a lighter islander. So all you have to do is weigh 3 vs. 3, then the other 3 vs.3 then take 3 from the first weighing and weigh them against 3 from the last weighing to get an answer.

6 on a side 6 on the other side. Take the heavier side split them 3 on each side of their even then the guy is lighter and on the other 6. If there is a heavier side then take one put on other side take other off. if even then heavy guy is off see saw other wise you will see who is heavier on see saw.

Here’s my solution:

Step 1: Weigh 1, 2, 3, 4 vs 5, 6, 7, 8

If they are even, then the different man has to be among 9, 10, 11 or 12. So…

Step 2: Weigh 9 vs 10. If they are even, then the different man has to be 11 or 12. So…

Step 3: Weigh 9 vs. 11. If they are even, then we now know that 9=10=11, so the different man is #12. If they are uneven, we now know that because 9=10, the different guy has to be #11.

If Step 2 had been uneven, that means the different man is either #9 or #10. So then Step 3 is still to weigh 9 vs. 11. If they’re even, we know that #10 is the different man; else, it’s #9.

Going back to Step 1… If Step 1 is uneven, then take note of which side is heavier or lighter. (1+2+3+4 or 5+6+7+8)

Step 2: weigh 1+2+5 vs. 3+6+7. If they are the same, then the different man is either #4 or #8. So…

Step 3: weigh a control we know is of average weight, say #9, versus #4. If they are the same, our different man is #8. If they are different, then the different man is #4.

If this step #2 is uneven, then again take note of which is heavier. Then switch it up again for

Step #3: weigh 1+6 vs. 2+7.

Evaluate the results of all three steps:

If in Step #1 1+2+3+4 had been heavier,

then if in Step #2 1+2+5 had been heavier

and now in Step #3 if 1+6 is heavier, then the heavier man is #1 (consistently on the heavier team).

else in Step #3 2+7 is now heavier, and so the heavier man is #2 (consistently on the heavier team)

else in Step #2 3+6+7 is heavier, so #3 is the heavier man, consistently on the heavier team (and step 3’s 1+6 will = 2+7)

else in Step #1 5+6+7+8 is heavier. So

if in Step #2 1+2+5 is heavier, then the heavier man is #5 (consistently on the heavier team — and in step #3 1+6 will = 2+7)

else in Step #2 3+6+7 is heavier. So

if in Step #3 1+6 is heavier, then the heavier man is #6 (consistently on the heavier team)

else in Step #3 2+7 is heavier, so the heavier man is #7 (consistently on the heavier team)

Oops, my solution is incorrect. It presumes that the different islander is heavier. But what if 1+2+3+4 > 5+6+7+8 because one islander (say #6) is lighter? Then, all my presumptions in my 2nd half don’t work.

Actually, it looks like Keith Graves Barrett’s answer is the right one — it tells you whether the islander is heavier or lighter, then winnows it down to the right one in 3 steps.

This spreadsheet really clears it all up without the crazy amount of text everyone on here is using and confusing everyone.

[docs.google.com]

enjoy!

post script to previous post. One possibility

was left out of solution: If after the first two weighing

the results are unbalanced in one direction and then unbalanced in the other then: one of 123 is the odd one and the odd one is either heavier or lighter depending on the action of the seesaw.