SOLUTION!! Holts 12 Men on an Island Puzzle Brooklyn Nine Nine | SUPER SIMPLE!!

[@pierrealexguev]

Somebody gives that men a pair of Beyonce tickets.

[@Mitskii1]

BUAHAHAHAHA

[@thelonewolf4689]

Rosa’s solution is still the best

[@salvadorarreolarodriguez6165]

No...

[@adampkalb]

"You use the seesaw to press down on their necks until fatty confesses."

[@Ashborne985]

Incorrect...and disturbing!

[@MildExplosion]

Thanks for satisfying my curiosity in the middle of a Brooklyn 99 binge watch session! Your solution is so elegant.

[@ali94hn]

I have not been able to sleep in peace for the past year and a half! Thanks for clearing this up for me😅

[@LifeProjX]

Its Official! You have all bullied me enough to force me to make another video on this Puzzle haha. I will post a video explaining the puzzle completely (including the algorithm for sorting the islanders into their positions in the chart) Not only that, but I will break a world record for the most number of islanders ever solved for which hopefully will only stand for a short period of time as I hope one of you will break my world record shortly after! Video Coming Soon!

[@owaishere]

Its a riddle, not a maths problem. You put all 12 on the see saw. Make sure they get off one by one, alternating sides. Then wait til it balances and the last pair that jumped off will be the odd one out

[@llain007]

It’s on an island, all 12 men are placed in the water surrounding the island. All 12 men then exhale fully causing them to sink, the one that doesn’t sink at the same rate is the islander of a different weight.

[@aaryan3411]

Holy shit.

[@aryanchauhan9786]

nah wont work. buoyancy depends on shape as well.

[@darrell5703]

Muscle density and fattiness of the men would determine buoyancy as well. This is coming from a man who took swimming lessons in his 30’s and came to find out that I can’t float because I’m too “dense”.

[@lookdawg187]

@@aryanchauhan9786 If the object displaces an amount of water equal to its own weight, the buoyant force acting on it will be equal to gravity-and the object will float. But, if the object weighs more than the water it displaces, the buoyant force acting on it will be less than gravity, and it will sink. A giant ass ship will sink faster when the air is gone, just like a man will sink faster or slower if they are heavier or lighter.

[@treborheminway3814]

You are not measuring displacement, your measuring density. To calculate the weight, you would need to know volume! You could have a lighter man, that's more dense because he's small.

[@adampkalb]

Rosa: One guy is slightly heavier. There's a seesaw. You use the seesaw to press down on their necks until fatty confesses. Islander #4: (screams) It was me! I'm the fatter one! You got me!

[@LifeProjX]

haha, really the best way to solve the problem

[@rudyolfschiermeister8167]

Im surprised the creator of brooklyn 99 didn't comment on this

[@LifeProjX]

I don't know if I could handle knowing that a creator of Brooklyn 99 watched a video that I made. I am clearly not cool enough for something like that.

[@austinheinbuch8642]

I’ve seen this riddle elsewhere, they didn’t create it.

[@lexscapephotography]

The question from the episode is not who but which. As in "which is it": heavier or lighter? Divide the 12 people into 3 groups of 4 people (G1, G2 and G3) and weigh G1 against G2 and G3, which is done by using the seesaw only twice. If G1 is heavier than G2 and G3, then the answer to the question "which" is "heavier": 1 person on the island (in G1) is heavier than the other 11. If G1 is heavier than G2, but weighs the same as G3, then 1 person (in G2) is lighter than the other 11. If G1 and G2 weigh the same, then if G3 is heavier, the 12th person (in G3) is heavier. If G3 is lighter, then the 12th person (in G3) is lighter. Finally, if G1 is lighter than G2, then if G1 is also lighter than G3, then the 12th person (in G1) is lighter. But if G1 and G3 weigh the same, then the 12th person (in G2) is heavier than the other 11.

[@parthkhaladkar9017]

I was looking for this comment! The wording in the show IS sort of unclear.

[@Doomstryker]

The way you explain it won’t work in just three tries. You took two weighing(s) just to determine which group has the lighter or heavier person. Which leaves you with one group of four person to determine the odd man. Which is impossible. There’s no solution that works where you have just one set of three groups. The first part you explained is correct but after the first weighing the grouping must be rearranged.

[@Aimran0198]

@@Doomstryker can you explain further? From what I see its correct because hes using G1 as a control and the other two groups as tests?

[@aknotienenestefito]

This might be me not understanding the premise clearly cause of language barrier, but i though in the way captain holt phrased it, he only asked to know if the person was ligther or heavier, not who the person was. Glad to know the right answer finally.

[@pewpewandrew3171]

he asked which

[@JohnnyRawks]

You are correct. Your understanding of English is better than anyone else's on here by the looks of it.

[@illeatthat]

He did, which is what most of these responses don't acknowledge, they just assume in step two that you can say that 7/8 is heavier- when it could be that it's the other number being lighter

[@giannitata9880]

You put six people on either side. Two at a time jump off(one on each side) When the seesaw goes level you know one of the last two people to jump off is either heavier or lighter. You've only used the seesaw once. Have those to sit on either side of the seesaw. One side will go up the other will go down. Leave one of them on and put someone else on the other side. If one side stays up, you know that's the lighter person. If one side stays down you know that's the heavier person. If they even out you'll know it's the last person to leave the seesaw

[@darkvallon]

I don't know why Holt discarded Santiago's solution as I came to the same: 1-weigh 6 vs 6, then take the group of 6 that tilts up or down 2-weigh 3 vs 3 and do the same. 3- here you have just 3 people, weigh any 2 of them, even If It doesn't tilt you'l know the person outside is the discrepant one. Why is this incorrect?

[@LifeProjX]

Jonatas Sousa it’s because you won’t know if there is a light person on one side or a heavy person on the other side. A 6v6 only works if you know whether the islander is heavy or light before you start

[@hakonskaanes7527]

2-weigh have 2 outcomes: tilt or no tilt. Your sulusion does not factor in a non tilt in weigh nr 2

[@PracticeYourPassion]

Brilliant. Thanks for running through a couple of examples. Also, installing a Whiteboard in your personal studio, so legit. I might need to do that

[@emanuelgabbay180]

Amazing solve! Also you managed to figure out whether they are heavier or lighter 🤯

[@missy3163]

Only need to find out if they r lighter or heavier, so could put 1st set of 4 on with 2nd set of 4 and if its uneven(L) then take one side off and put the 3rd set of 4 on if it stays the same you know the person is heavier, if it's even put 2nd 4 back on with the 3rd 4 and depending on which side it goes to you know whether heavier or lighter

[@squidy7771]

I'm usually not so into puzzles, but this one was really fun to figure out for myself. I actually got something different, but in my answer it isn't the same 3 actions every time. Still a really fun puzzle and a great explanation!

[@LifeProjX]

Thanks! It was a good puzzle. Thanks for watching

[@guidotie6997]

What was your solution?

[@squidy7771]

@@guidotie6997 ​ @Guidotie first test: you weigh 1,2,3,4 against 5,6,7,8. If they're equal then: you weigh 1,9 against 10,11. If they're equal then 12 is the answer. If not, write down which side was heavier. Then you weigh 10 against 11. If they're equal, then 9 is the odd one out. If they're not equal, and 10,11 were heavier than 1,9, then the heavier one is the answer. If 10,11 were lighter than 1,9, the lighter one is the answer. Now: if the first weighing tips to one side, write down which side it tipped to. For the example, let's say 1,2,3,4 are heavier. Then, Put on each side of the scale two people from one side and one from the other. let's say, 1,2,5 against 3,4,6. If it doesn't tip, that means 7 or 8 are the culprit. Because they were on the lighter side in measurement 1, you weigh them against each other and the lighter one is the culprit. Now, if measurement 2 tips to one side, let's say to 1,2,5, that means that either 1 or 2 are making that side heavier, or that 6 is making the other side lighter (based on the fact that 1,2,3,4 tipped down on measurement 1). So you measure 1 against 2, and the heavier one is who you're looking for. If it's equal, the answer is number 6. Ok that's it

[@illeatthat]

@@squidy7771 why is "the lighter one" the culprit, when it could be heavier?

[@squidy7771]

@@illeatthat if the scale tips to one side, then one of them is the culprit. Now, we know this individual was on the lighter side of the scale in the previous measurement, meaning that whoever the culprit is is lighter than the rest. That's why if the scale tips, it's because one of the guys is lighter than the rest, meaning he is on the lighter side of the scale.

[@shadowdante1102]

Another way is to split the 12 people in 2 groups. 1st measure, 6 on one side 6 one the other, one side with be heavier or lighter 2nd time, take the group with the weight disparent person, split that group in half, 3 to 3, reweight them. 3rd time. Now you will be down to 3 people. take 2 of those people, split them into two groups. put them on the scale. Now you have one person on each side of the scale, and one person off to the side. Now, no matter what you know which is weight disparent. Because if the scale comes out equal you know the person off to the side has the off weight, or if they are on the scale, you can easily see which one it is.

[@LifeProjX]

if you do 6v6 you won't know if there is a heavy person on one side or a light person on the other.

[@shadowdante1102]

@@LifeProjX Oh right. I forgot that we didnt know if it was heavier or lighter. Thank you.

[@kensearle4892]

Great solution! Good to think through the various options. My answer is perhaps a bit of Kobishi Maru but I heard this could be asked in an interview where time to solve is highly reduced.... I am assuming that as they get off the sea-saw it does not count as a measurement otherwise that doubles the measurements. Let's take advantage of that! For round 1, I put 6 on one side and 6 on the other. The seesaw will obviously tilt one way but we don't know who it is yet. As Round 1 ends, one person from each side gets off at the same time. We observe when the seesaw is level, we know he is one of the two that just got off. For our 2nd measurement, we put one of the remaining two (A or B) on the seesaw with any of the others. If it tilts, A is the one and we get a bonus knowing he is heavier or lighter. Otherwise, it's B who is off the seesaw and we are done. For bonus, we could use the 3rd measurement to put B on with any others to tell if he is heavier or lighter.

[@bartwisialowski4634]

Great solution! That can use the same 3 uses in all cases doesn't make it the simplest. Other methods require fewer table entries and have more intuitive results on which person is light / heavy.

[@charliewinterss]

How did you manage to get the chart to determine weather the islander is heavy or light? *by that I mean all the rcl =light Cll =heavy rcc =light

[@LifeProjX]

kzbin.info/www/bejne/pqjbYqyqj6xob6M

[@ayesnapsnots]

12 is only used once, if he was heavier it'd be ccr, if lighter it's ccl. The logic is sound, but the final variable of lighter vs heavier is the problem

[@bencheevers6693]

I found an internet answer that was absolutely wrong, I thought this was a tough puzzle, this was great!

[@LifeProjX]

thanks! Ive got a new video coming that really dissects this problem and allows you to solve for any number if islanders.

[@jackprotti]

6 and 6 Pick the side that goes down 3 and 3 If they're even then the person is lighter If they're not even the person is heavier

[@trollking2669]

How bout 6v6, then 3v3 then leave a guy out when doing 1v1. If they balance out, then guy left out has different weight. If it doesnt, ask the rest if the guys to lift the last 3 guys.

[@evannguyen4085]

You have to know if the person is heavier or lighter, if u do 6v6 you get no information with that use, you just know one person is different which you already know

[@guidodenbroeder935]

Then you only have information about two groups instead of three.

[@mikiyastesfaye3202]

6 vs 6, take the heavy 6 and split into two 3 vs 3, take the heavy 3 and do 1 vs 1. If tilt left, guy on left. If tilt right, guy on right. If equal, guy not put on the see saw.

[@Legion343]

But you don't know if they're heavier or lighter, so no matter which way it tilts, it could be either. It could go up because that guy is heavier, but it could also go up because the opposite guy is ligher.

[@SimmoSC]

How do you know who to put on the see-saw for the second and third use? Like the video by the way 👍

[@Bhiraata]

You assign a number from 1 to 12 for each person beforehand. This way you'll always know who to weight at a given time.

[@JasmineGeminiDragon]

@@Bhiraata okay, so I did that and there are some numbers that come up the same with others. So do we have to make sure that each number is not paired up with another number that it was paired up with before? How do we know which group goes to which side and how to mix them up question mark you see, I put it in an order, of one two three four five six seven eight nine 10 11 12, and I had one two three four on the left side of the Seesaw, and I had 5 6 7 8 on the right side of the Seesaw, and I had nine 10 11 12 on the middle. However, when I put 9 7 5 10 on one side of the Seesaw and 1 4 8 12 on the other side of the Seesaw and the rest in the middle, and then of course another combination... Some numbers came out with the same combo. Some numbers had RLC twice. So now I'm even more mixed up or am I overthinking this?

[@pannja123]

@@JasmineGeminiDragon You make the LCR arrangements first, then put the people on or off the seasaw accordingly.

[@missy3163]

Only need to find out if they r lighter or heavier, so could put 1st set of 4 on with 2nd set of 4 and if its uneven(L) then take one side off and put the 3rd set of 4 on if it stays the same you know the person is heavier, if it's even put 2nd 4 back on with the 3rd 4 and depending on which side it goes to you know whether heavier or lighter

[@JasmineGeminiDragon]

What formula did you use to place which guy on which side? Some combinations came up for more than one guy.

[@Bronzescorpion]

Find the 27th permutations first. Group them so you they are paired in reverse (left becomes right). Note that the three permutations CCC, LLL and RRR are not used. Now the only thing you have to make sure of, is that there can only sit four people on one side. Note that a C will always stay at the same position. Start by picking four permutations witch start on the left and assign them to the first four people, make sure out of the 8 possible pairs that you pick ones with different places for the second and third letter as evenly as possible: 1: LLR RRL 2: LRR RLL 3: LCL RCR 4: LCC RCC The next four are given, it is those that begin with R: 5: RRC LLC 6: RCL LCR 7: RLR LRL 8: RLC LRC Now look at you distribution of the three weightings: L C R 1234 5678 178 346 25 36 458 127 Now distribute the last four pairs accordingly: a: CCR CCL b: CRC CLC c: CRL CLR d: CLL CRR L C R 1234 abcd 5678 178d 346a 25bc 36cd 458b 127a Notice how easy it is to switch the last four pairs around so they match up. CCR/CCL can be changed if there are too many on the right side and too few on the left in the third iteration and it doesn't affect any of the others. Lastly see that my solution is different than the guys in the videos, there are more ways to to it, but they all have the same principle.

[@dennischolo909]

yeah but why would you pick the second set of numbers? or the third set? are they just random numbers?

[@LifeProjX]

No I made an algorithm. I've been planning on making a video about it but haven't got around to it. its pretty cool tho. Ill try and make the video soon.

[@Gollum491]

Did it with 2 steps. Measure 6 of the men 3 vs 3. Since half of the men are being weighed there’s a 50% Mr.(+|-) is either being weighed or not weighed. If he is weighed the first time, the scale will tilt, telling us which half of the group he is in. If the scale is balanced on the first move then (+|-) is not being weighed. The second move is the same for both since you now know either way in which half (+|-) resides. If you know weigh the two halves against each other the scales will be unbalanced, but now you know where our man is! Two move guaranteed answer every time! (Seesaw scale, same thing haha, appreciate your time!!)

[@kensearle4892]

I thought about that too, but the catch is we don't know if he is heavier or lighter so with 3 vs 3 being weighed we don't know which side he is on. It would tell us whether he is in the group of 6 on the sea saw or the group of 6 off. With only 2 more measurements, I have not been able to get from 6 to 1 that way.

[@Gollum491]

@@kensearle4892 holt doesn’t ask for who the individual is, he just wants to know if he’s lighter or heavier. That’s the only reason my solution works haha

[@xjustxjustinex]

But he only says you need to figure out “which”. Grammatically that would refer to which is the odd man out, or which (heavier or lighter) it is. Not both.

[@TheVillainOfTheYear]

Holt never defined what constitutes a use. If you define "use" as number of times you put men on the seesaw, then there is a simpler solution that will always yield the correct result: Use 1: start with six on each side. In this configuration the seesaw will be tilted because the weight will be uneven. Have one islander from each side get off. If the seesaw remains tilted, those two were the same. Remove one more from each side. Repeat until the weight equalizes when two men get off. One of those two is the odd man. Label those two A and B. Use 2: Weigh man A against any other man but B. If they are balanced, B is the odd man out. If they are not balanced, A is. Use 3 is a red herring.

[@Bronzescorpion]

Use is the number of times the weight is being tested, where is no reason to think it is only when persons climb on it.

[@chencobb824]

Split the group into sets of 3 (4 people per set) group a and b go first. If one side is heavier then that group has the heavier person, if the seesaw is equal then it must be group c. Take the group that is heavier and split them into two, repeat process, which ever side is heavier take them and split the two again and then you have the answer with the third use. Would this work?

[@sgtpluck8344]

Maybe the other group has a lighter person......

[@kerenhargor]

How did you reach the solution tough?

[@iam_a_sad_khan]

This is actually a really popular Graph Theory problem. Computer Science or Mathematics people should know the solution.

[@LifeProjX]

Not sure what graph theory is, but it sounds cool. might have to spend some time looking into it.

[@iam_a_sad_khan]

@@LifeProjX If you love puzzle solving, you'll definitely love graph theory. You should start with the Konigsberg bridge problem. This problem was the birth of graph theory. Maybe you can make videos about more graph theory problems! Edit: Might wanna start with this video. kzbin.info/www/bejne/pIvahKJqq8uqbMU

[@dennischolo909]

i could see if the first 8 were even then the odd one out is 9 10 11 12 but what is the reason for 1259 34 10 11? how did you come to this set of numbers? or the next set? let me know and thanks.

[@LifeProjX]

Here is my video showing how to get the numbers kzbin.info/www/bejne/pqjbYqyqj6xob6M

[@LifeProjX]

I have a better video coming out showing a more complete solution

[@andrewboryszewski1217]

How did you form the matrix of XXX? Why didn't you put LCX in the left column, for instance?

[@LifeProjX]

check out the other video I made on the problem. it explains it better. kzbin.info/www/bejne/pqjbYqyqj6xob6M

[@beautifulbaldguy]

While I did come up with a way to solve it, it was nowhere near this simple. Very well done!

[@joederrett6619]

I do it in 2 uses of the seesaw. Am I missing something? split the 12 into 2 groups of 6 (group A and group B). split group A into 2 groups of 3 and place onto seesaw. If balanced The odd person is in group B. Measure A v B if B weighs the most the odd person is heavier. If A weighs down odd person is lighter. If unbalanced odd person is in A. Weigh A v B. If A weighs down odd person is heavier. If B weighs down odd person is in lighter. 2 uses. 🤷‍♂️

[@alpozdemir914]

how are you gonna find out who is lighter or heavier ?

[@alpozdemir914]

I mean lets just say number 8 is the odd one how are you gonna figure it out

[@geoffreycalligan1045]

To complicated the first use you should do 6 vs 6 and when the men are getting off have 1 man from each side step off at a time until the seesaw is balanced then take the 2 men who stepped off just before the seesaw balanced and 1 man you now know weighs the same as the others and use him to weigh against the last 2 men to step off one at a time you 1vs1 and 1vs1

[@LifeProjX]

I thought of server "work arounds" while solving the puzzle. I could argue that each time someone stepped off that would be considered a use. Since I solved it I have done a lot of research of the problem and it is quite famous. also known as the 12 marble problem or the 12 coin problem. Understanding the intention of the problem I know its not just a word play game and does have a mathematical solution.

[@TheVillainOfTheYear]

@@LifeProjX do you really think a riddle that can be asked and apparently explained over dinner would require a chalk board?

[@LifeProjX]

TheVillainOfTheYear this riddle probably would.

[@assumptionisthemotherofall2402]

Take all 12 men put 6 on see saw the other 6 on the other end which ever side is lighter or heavier you keep that group kill the other 6. Next weigh divide the 6 into two groups of 3 repeat process kill the 3 that side which is lighter or heavier now using see saw last time you have 1 sit on each side while 1 waits on the end if they weight the same the see saw won’t move and that means guy on side is the one you’re looking for if one side is lighter or heavier you found the guy ….🙌

[@RoverNOut]

Made it sound so easy! How did you decided the order of people?

[@LifeProjX]

I have a sorting algorithm. Ive been planning on making a video about it, but haven't gotten around to it.

[@musawirmehfuz6623]

So If I understood correctly the answer is person 11?

[@julieandtomkenyon1079]

See saws react to torque, not weight. You need a hanging balance, not a see-saw. If two men with identical weight got on opposite sides, it's unlikely the see-saw would balance unless their centers of mass were exactly the same distance from the fulcrum.

[@JohnnyRawks]

The problem, as stated, does not say you have to identify the individual. It asks you to determine whether the exceptional individual is lighter or heavier. That's all. You only need two uses of the see saw to solve this problem. 1. Six on one side, six on the other. Take the heavy six and.... 2. spilt the heavy six in half, three on each side. Balanced = Lighter. Unbalanced = Heavier.

[@mrtiny7707]

I have kinda simpler solution. Just separate it to 3 group 1234, 5678, abcd. 1st situation 1234 and 5678 is balanced, now this one is too easy so i won't talk about it. If you can't do it yourself then just give up. 2nd situation they are not balanced then I will take 3 from the lighter side away (I suppose it is 678), then put 3 from the heavier size to it so we will have 5234. Now I take 3 from the other group and add to the heavier side and we will have 1bcd and 5234. Now it's quite easy but need focusing. 1st case the seesaw doesn't change, it means either 1 is heavier or 5 is lighter. Just try one of them with the regular one. 2nd case the seesaw is balanced then there is a lighter one in 678 (it was in the lighter side). Just weight 2 of them and we'll find out. Last case the seesaw change side, it means one in 234 is heavier than the regular one. Voila.

[@framu3214]

I have come up with a solution in which you do 4 vs 4, 2 vs 2, and 1 vs 1. I would not say that it is simpler but it is curious.

[@framu3214]

Well in the second use it is 2 VS. 2 or 3 VS. 3 depending on what happened in the first use.

[@will-oy1wz]

That one works if we know whether or not the man is lighter *or* heavier, but in this case, we don't know. Ergo, for the first 4v4 weighing, if the see saw tilts, it's impossible to know which of the 4 to pick from to do the 2v2 weighing. Ergo, the solution in the video is necessary.

[@framu3214]

Will Leet No, my solution works even if you don't know if it is heavier or lighter

[@satyanarayanbera7659]

1st Use: sit 5 people on each side. 2 standing beside. Outcome 1 : See saw is balanced, means the overweight guy is one of the two standing, sit them in either side , solved in two attempts. Too easy. So let's look at the alternative. Outcome 2: one side with 5 men is heavier, you have eliminated 5 from the other side and the two standing, so now with 5 men you move on two 2nd use. 2nd use: Sit two on each side, remaining one is standing. Outcome 1: Seesaw is balanced, the standing one is the overweight one. Outcome 2: SeeSaw is not balanced, one side is heavier, like before the 2 in the lighter side and the one standing are eliminated. Only two people left. Use 3: sit the two in either side , the heavier one's side goes down. We have a winner 🎉

[@viniciusresende1188]

SOMEONE PLEASE TELL ME WHAT IS THIS EPISODE AND SERIES ???????

[@LifeProjX]

Brooklyn Nine-Nine season 2 episode 18

[@joshuagonzales3539]

He is taking the long turn

[@richardthompson1125]

"Can you work out which is heavier or lighter by only using the see saw 3 times?" "Sure thing, just let me use it another 72 times first"

[@escadrila9271]

Right,makes no sense?? Just because he chooses number and it works with his algorithm. We cant assume that he is lighter or heavier and then say yes it will go there and there.even if the first try the scale is uneven it still means that there are 8 possible answers,because he can be either heavier or lighter. Wich group to chose than.

[@geiranadal1453]

A bit late here, but I only saw this episode a few days ago, but this is the solution I came up with during the episode. You put 6x6 on the see-saw. It will tilt either right og left. then you take off one person on each side until the see-saw levels. Then you weigh the 2 last persons to leave the see-saw against another person. One will be equal the other will be either lighter or heavier.

[@jylounen]

That's more than 3 uses of the seesaw

[@tj71520]

Thanks for making it easy to understand. I just subbed. I saw the problem on the show as well.

[@LifeProjX]

Thanks for the sub!

[@joshuagonzales3539]

U all never ask can I visualize because I’ll be able to see who is who is

[@amir-ng6jv]

This puzzle broke my brain

[@illeatthat]

Why have you grouped them into 4? The least most common nominal figure of all the numbers involved?

[@aryamannkhare9505]

Why not apply the Monty hall problem solution to this puzzle? I mean divide the 12 islanders into 3 groups of 4, and apply the same logic which is applied in the Monty Hall problem.

[@LifeProjX]

Im going to have to put some thought into that, but on the surface it seems like you are just increasing the odds of finding the group the person is in. maybe elaborate and I can think on it a little more.

[@annabelleyang1881]

Great solution! Could you link the video that explains how you got it please?

[@LifeProjX]

thanks! you can check out the explanation here kzbin.info/www/bejne/pqjbYqyqj6xob6M

[@Zmanj413]

Could you do 5 on left 5 on right, if it doesn’t tilt then 1v1. If it does tilt put 2 and 2 from the side that tilted, if it’s balanced it’s the one left out otherwise put the two that tilted in a 1v1 you’ve found your odd one

[@shrillburger1233]

I have a solution but I need to know what is a utulisation

[@Tio225O]

That’s weirdly really cool

[@harshchahal75]

Anyone reading this please go to 3:29 and pause we don't know if the person is heavy or light lets choose No. 5 heavy just to go through this. Starting with the riddle it could be any one of 1,2,3,4,5,6,7,8,9,10,11 or 12. Step 1- See-saw tilts towards right. (no.5 heavy). Which proves 9,10,11 and 12 are same weight can be eliminated. So the person must be any one of 1,2,3,4,5,6,7 or 8. Step 2- See-saw tilts towards left (no. 5 heavy). which proves 6,7,8 and 12 are the same weight so can be eliminated. we already determined 12 in the fist step so additionally we know 6,7 and 8 are also same weight. So the person must be any one of 1,2,34 or 5. Step 3- See- saw doesn't tilt (no. 5 is not there). which proves the people on the see-saw are same weight so no. 3,7,9,10,1,4,6 and 12 can be eliminated. From all three steps we get to know 1,3,4,6,7,8,9,10,11 and 12 are the same weight. So we are are still left with no. 2 and no.5. which never got eliminated and we will never know which one of them weigh differently. you are absolutely right about the configuration but you will always be left with 2 people not one. I beg you to prove me wrong cuz I want to have a solution.

[@LifeProjX]

It really good thought, but there is more information that you are forgetting. the 2 and 5 swap from being on the same side of the scale to being on opposite sides of the scale. if I take everyone off the scale except the 2 and 5 (for a visualization) for the 2 to be the correct answer it would have to switch from being light in weigh 1 to heavy in weigh 2 (same with everyone still on the scale). I didn't use an elimination method to solve this, but if I did I can eliminate #2 because it can't change weight. great thought tho!

[@avenger_aawin4971]

if u put 3 v 3 on the first move you can eliminate 6 if it's balanced then the odd one is withing the 6 that haven't been weighed if it's not balanced then the odd one is one of the 6 that have been weighed

[@alesioser]

Is the odd one the heavier or the lighter?

[@avenger_aawin4971]

@@alesioser then you can place the 6 normals on one side and the other 6 the ones with the odd fella on the other side if the group with the odd one leans down then the odd one is heavier if it leans up the the odd one is lighter

[@alesioser]

@@avenger_aawin4971 it doesn't make any sense. In just 3 tries you will never get the heavier or lighter guy with your theory

[@avenger_aawin4971]

​@@alesioseryou only need to figure out if he's lighter or heavier 0:43 'one of them is slightly lighter or heavier you must figure out which' you don't need to know who the guy is; you just need to figure out if he's lighter or heavier 12 men A B C D E F G H I J K L 1st try: A B C vs. D E F if it's balance (let's call this outcome V) then we know that the abnormal man is one of (G H I J K L) then we compare (A B C D E F) with (G H I J K L) (after outcome V)2nd try: (A B C D E F) vs. (G H I J K L) if (G H I J K L) goes up then the man with the different weight is lighter if (G H I J K L) was lower then that means that the odd man has a higher weight *** after the first try if the seesaw wasn't balanced (let's call this outcome W) then we know that the odd man is in (A B C D E F) then we compare (A B C D E F) with (G H I J K L) (if first try outcome W) 2nd try: (A B C D E F) vs. (G H I J K L) if (A B C D E F) was lower then that means that the odd man has a higher weight if (A B C D E F) goes up then the man with the different weight is lighter

[@alesioser]

@@avenger_aawin4971 One is lighter or heavier, you must figure out which. So you have to know which person it is. And think about this, would be harder to know who the person is than knowing who's heavier or lighter, so, by logic, the "Creator" of this problem would do it the as harder as he can.

[@viniciusresende1188]

Whay the season and ep.?????????

[@LifeProjX]

Brooklyn Nine-Nine season 2 episode 18

[@egaputera7290]

hi, very cool and understood! can you share the algorithm (or way of thinking) for you figure out the solution here.

[@LifeProjX]

Yea, have been planning to for a while. I'll try soon.

[@loz6176]

At first i was like they stupid it’s 6 vs 6 and then the heaviest you split into 3 vs 3 and then you do 1 vs 1 and if it’s equal it’s the one you left out if there’s one leaning away it the heaviest and then I realised it’s heavier OR lighter and I was like arghhgg 😂😂

[@anandsrivastava2514]

Same bro😂

[@CT-sf2rg]

I mean you could start with 4 vs 4 and leave 4 out and if the seesaw stays the same then it’s the left out group that’s the heaviest or one of the two groups compared. Then you can separate the 4 into a 2 vs 2 and then a 1 vs 1 but idk if that’s right but I can’t see how it isn’t

[@loz6176]

CT 481 because it not heaviest of lightest, it’s just a different weight, that’s where I was confused

[@theinkstainedtalonx3949]

The ink stained talon X I figured out the first teaser You put three people on each side If one side is different you know that they are within those six If not they get of and the other six get on You have one person get of from both sides until the balance changes If the balance changes you know it’s one of those two Then you have one guy sit on the seesaw with a guy from the other group that you know can’t be it If it balances you know it was the other dude and if it tilts you know it’s him

[@bruner999]

The main question is just to find out if the twelfth man is heavier or lighter not who this man is. The solution is quite simple. You divide them into four groups of three. Then you weigh two groups together on the see saw and you spend two weighings, this gives one measurement that shows balance and one that dose not. You have to remember wher did the see saw leaned with the groups thet were not balanced. Then you take on of the groups and sit it with a one of the groups that showed balance. If thay are agen in balance you have to look at the resolt of the previous weighing if they are not you can see if the twelfth man is heavier or lighter as the seesaw will tilt up or down agenst the weight of the calibrated group. After three weighings we determin if the twelfth man is heavier or lighter.

[@theoneandonlyjazztheautobo8597]

Take 3 groups of four people each of the four people split up into two and then get on opposite ends of the seesaw if it balances out then do it again 2 more times

[@nataliefisetm.d.8952]

You have a great mind!😀

[@LifeProjX]

Thank You!

[@andomneraio457]

Wish u had explained the solution with a test run rather than proving you know the outcome of the tilt for every number depending on that number being light or heavy.

[@duzzixx6860]

But how do you know if the person is lighter or heavier, Holt doesent say if hes heavier or lighter

[@sirhoneybadger5875]

You might of figured it out but if you're still curious. Take person 11 for example. The seesaw can either go CRC or CLC. The way the seesaw tilts is dependent on whether or not the person is heavier or lighter, it'll always tilt to the heavier side. So all you would have to do is observe what happens. If the seesaw goes 'CRC' then you know that person 11 is heavier since the seesaw tilted to the right, where person 11 is. This means you can tell who is the odd one out and also whether or not they're heavier or lighter. Hope this helps.

[@pewpewandrew3171]

that’s why he made the solution...

[@Jared-W]

Yes because actual mind twisters are this complex, complex enough that you would need a giant formula to solve it. There is no way that they expected thier standard viewer to figure this out.

[@0123Dayan]

I solved it with a simple method of 5-5-2 and only 3 uses of the scale.

[@jmahoney1979]

This seems easier 1st 6 on each side 2nd 3 on each side (split the heavy group from round 1) 3rd weigh 1 on each side from the last 3 remaining. A. If 3rd round seesaw is flat, the man not on the seesaw is heaviest. If the heaviest usvon the seesaw, the seesaw will show who's heaviest

[@ramonparelladamartin4877]

I thought that too, but it doesn't work, since it can be heavier or lighter.

[@omarofi9750]

Why u dont compare 4 to 4 and than u know which 4 is different than compare 2 to 2 and than 1 to 1

[@LifeProjX]

You don't know whether the person is heavy or light. you won't know if there is a heavy person on the right or a light person on the left

[@andrewbaggs7743]

Ha ha, I figured it out in 6 minutes; with a couple glasses of fine red wine. This is what simple is..1st measure 3 and 3..balanced, set all aside. Unbalanced; you divide into 2 groups. 2nd measure; you take 3 of the balanced group vs 3 of the unbalanced, the unbalanced group will go up or down (heavy/ light!) or stay balanced. If balanced set aside. If unbalanced

[@andrewbaggs7743]

You now have 3 unbalanced people left but you know that 1 is heavier or lighter than the others from the previous measurement now Measure 2 on the see saw and you have your answer .. if balanced the loner is the heavy/lighter one or the one heavier or lighter on the see saw based on the previous measurement is the odd man out! Baaaam!

[@rafaellima381]

the problem with your solution is that there is no rational behind it. it is just brute force im not sure if it is possible to work it rationally as slit and conquer or something... but your solution only works if you have paper and pen and you are able to map everysingle possible outcome before so if you were in an oral evaluation, or if you were in the island itself you would be able to do that [even if you know you answer] you would need to memorize all this 24 outcomes and the 3 seasaw configurations

[@LifeProjX]

I know, unfortunately I have been talked into making a 3rd video about this problem. All the prep work is done and all I need to do is film it. I promise you will have no questions after the next video. I am going to explain how to solve yourself and by the end I will work through a solution to 1093 islanders in 7 weighs.

[@ghostrider9261]

Excellent bro

[@LifeProjX]

Thanks!

[@dericheitor1706]

4 on each side. which one is heavier/lighter/centered. if imbalanced with 4 on each. split the heavier/lighter side into two groups of two. whichever one is imbalanced, then compare those two with one on each side . solved. if the first 8 people were balanced, measure the other four not on there. whichever side goes up or down, split them into one on each side. solved.

[@rumpelstillkills8036]

You deserve those beyonce knowles rock concert tickets.

[@LifeProjX]

Thank you!

[@aergiaflexed3116]

In the show version, you do not need to identify which person is the lighter/heavier person, just whether he is lighter, or heavier. This can be achieved without the need for an algorithm, but rather by simply weighing triads, starting with 1-2-3 vs. 4-5-6, then 1-2-3 vs 7-8-9. Depending on the results of those two, you will know whether it's necessary to weigh 1-2-3 vs 10-11-12.

[@thevoicelesswriter]

He did say "which".

[@deathreaper9692]

But im still confused, cause doesnt that mean you could say islander 9 is heavier or 6 is lighter etc, like is there a fixed anwser who it is and what they are or can it basically be anyone being anything if that makes any sense, im bad at explaning things lol

[@LifeProjX]

It's a layout that answers the question no matter who it is and whether they are heavier or lighter. because the information is laid out symmetrically if person 9 is heavy it will give the symmetrical opposite answer as if he were light and that's what makes the chart work. Im not sure if that answers your question. I wasn't 100% sure what you were asking

[@deathreaper9692]

@@LifeProjX thank you for anwsering. I understand the chart but from what i understand there isnt a fixed anwser. So if i would tell this math problem to a friend there isnt a anwser i could give them as like "its islander 6, and he is heavier". Because from what i understand from the video it can be any of them.

[@LifeProjX]

DeathReaper 969 correct, it could be any islander. That’s what makes the puzzle hard. You need to find a solution that works no matter who the islander is

[@deathreaper9692]

@@LifeProjX oke cool, thanks for helping out man, i really appreciatie it. Keep it up mate✌️

[@BCMSi]

What if the real test was comprehension. Holt doesn't ask you to figure out 'who AND whether they're lighter or heavier' as is on the whiteboard. He simply says. One is heavier or lighter, you must figure out which. As in, you only needed to figure out whether the odd man out is heavier or lighter. Which is easy, but leaves a lot of people focusing on the wrong thing and doing a bunch of extra and unnecessary work because they didn't take the time to understand the ask :)

[@LifeProjX]

possibly for the show. but the problem is a pretty famous math problem that's why I went with the famous math problem solution.

[@marcM98]

how has Holt not just googled it like me before?

[@0123Dayan]

I found a very much easier solution with a 5-5-2 method, and using the scale only 3 times. :)

[@LifeProjX]

Id love to hear it

[@fabooshka]

Is the algorithm video out?

[@LifeProjX]

Fabrobin I keep trying to quit making videos, but everyone keeps wanting shit haha. I’ll try and make it next week.

[@julia-7520]

ok so hear me out- assuming that the odd one out doesn’t know what everyone else weighs then you just ask them one by one to tell them their weight and the odd one reveals himself

[@LifeProjX]

I love the way your brain works. Im not sure thats the idea behind the riddle, but definitely the most creative reply I've had.

[@joshuagonzales3539]

There is

[@MariaFernanda-mm8jn]

Não entendi :(

[@alesioser]

I know the solution: You give a big bunch of food to one of them letting the others starve. No need to use the see-saw, that guy is gonna be the heavier! ha!

[@LifeProjX]

I like the way your brain works.

[@daltonjackson5259]

There is a much much simpler answer. Hint, the see saw is a red hearing.

[@susioeandno1else]

I sort of understand it... so thanks... Can you do the solution for the *BONE* problem? (B99 fans will get the reference)

[@LifeProjX]

Haha, already working on it. I just have to have all my t's crossed and my i's dotted before I disagree with vsauce and everyone else on the internet.

[@jasab93]

6 + 6, right falls, use they 6 for 2nd try. 3 + 3, right falls, use they 3 for last try. 1 left, 1 middle, 1 right. Whatever way if falls that person is heavier

[@LifeProjX]

J T B that is incorrect

[@jasab93]

@@LifeProjX how is that incorrect 🤣🤣🤣 thats literally it. I dont see you explaining why its incorrect

[@luminite1863]

@@jasab93 because the person could be heavier OR lighter. You cant tell which one is the guy

[@LifeProjX]

boyf riends is correct. If you do a 6v6 you won’t know whether there is a heavy person on one side or a light person on the other.

[@renatabisi1952]

Come on! I was having fun! More math!

[@LifeProjX]

I love math!

[@sgtpluck8344]

Clear as mud. Don't understand that at all. Why is 11 (or any other one number) weighed only once. Why he is not heavy? There's a CRC as well as a CLC...

[@kathrynsavage2211]

It’s been a year for you but I’m just now watching this. They can only check the weight 3 times. I don’t know how he got the order but I’m guessing it’s to get the grid to have a different “code” for each scenario. but it’s in a way where you can figure out the odd man out. If it’s 11 then it doesn’t matter who is weighed with him because they are all the same weight. You can “eliminate” all the people in row one for being the same weight (if it’s 11) so some are used again because you know it isn’t them, and you cycle through the last 4 people that weren’t weighed in the first scale, and then since 11 isn’t on the last scale you know he was the odd one out and can figure out if 11 was heavy or light depending on if the scale tipped towards 11 or away :)

[@kathrynsavage2211]

In the case of 10 and 11 both being new, you only want one of the two being weighted the 3rd time to see which of the two it is :)

[@UnluckyGambler]

i worked out this myself in 10 mins. booyah.

[@LifeProjX]

thats better than me!

[@iversia8546]

But you still don’t really know who it is. Your choosing whoever islander you want and assume that he could be light or heavy. This video gives me all the probability of every islander if he is light or heavier, but still I have no idea who is truly the one who different from all other islanders, and whether he was heavier or lighter. I’m still confused 😅.

[@LifeProjX]

the solution gives you the ability to identify the islander no matter who it is. If you picked the islander and whether they are heavy our light and didn't tell me who it was or whether they were heavy or light. I could ask you at each stage which way the scale would tilt and after all 3 stages I would have the answer.

[@michaelsullivan4703]

Ok, but... Split them into 3 groups. Challenge 1234v5678. If one side is heavier weigh 12v34 and then 1v2. If 1234v5678 is even then those 8 are among the 11 normies, and the fatty is in the third group. So you split that group in the same way as before, 12v34, 1v2. It’s not that complicated

[@joshuagonzales3539]

But u never use all 12

[@-_sak_-]

THANKS !! !! !! !! !! !! !!

[@goolmegaming1266]

Not tryna flex, but I solved this riddle in 10 seconds. I was just curious if this was a joke and the tv show wanted the riddle too be easy or if it actually was hard for some people. I am in the top 2 % tho and a member of mensa, so sometimes I dont know if its hard or easy for the average person.

[@LifeProjX]

definitely took me longer than 10 seconds. any secrets on how you came up with the solution?

[@goolmegaming1266]

@@LifeProjX Some riddles just come very easy too me, this one did. I just went into paint and drew somewhat the same of what you showed here.

[@goolmegaming1266]

@@LifeProjX My mind explodes with ideas when comming too riddles like these. First I thought I had too start with 6-6 but then I realised immediatly that it had too be 4-4 because no matter the outcome then, if its center or right or left, you end up with 4 people and from there I already knew the answear too the rest.