The latest episode of **“Brooklyn Nine-Nine” **featured a subplot where Captain Holt gave his underlings the following brain-teaser:

There are twelve men on an island.

Eleven weigh exactly the same amount, but

one of them is slightly lighter or

heavier. You must figure out which.

The island has no scales, but

there is a see-saw — the exciting catch:

you can only use it three times.

The “Brooklyn” detectives couldn't solve the problem, but many of you quickly went to work to figure it out. Now for confirmation, here's an exclusive clip of Andre Braugher as Captain Holt explaining the exciting manner of solving it. Enjoy.

A new episode of “Brooklyn” airs Sunday night at 8:30 Eastern on FOX.

Did they upload the wrong take? He repeats a line in the middle after stumbling… I thought it was part of a joke that it was about to get over complicated and jumbled up, but seems it was just a flubbed line that was forgotten about. Or they just didn’t care and wanted to wrap ;)

They just uploaded it and didn’t care. It wasn’t color corrected either.

Yeah, I suspect they weren’t going to make Braugher do multiple takes on a web extra, especially since he nailed everything before and after.

Ahh the joys of region locked content. Damn you FOX. Give Canada some love!

ZenMate, the chrome extension that gives me access to oh so many US restricted videos.

Hola app works too on Chrome!

Speaking of unavailable content – does anyone know a way to view locked UK content (or Scandinavia) if you’re in the U.S.?

Unlocator. Google it.

I love this show.

VPN BABY! OH YEAH!

Had to choose a server based in Kansas, but could have chosen one in Vegas or Washington or Dallas! Either way I got to see the fun that is Captain Holt doing his thing.

I must admit I could have used Smart DNS!

Great stuff.

Pretty fun!

He has perfected that deadpan so well that several times I actually thought the video had locked up

I think there is a mistake. Started at :30 he takes the case of 1-4 being heavier than 5-8, so either 1, 2, 3, or 4 is heavier than standard, or 5, 6, 7, or 8 is lighter than standard (and 9, 10, 11, & 12 are all standard).

Now weigh 1, 2, & 5 against 3, 6, & 9. If the balance tips the same way as before, you know 1 or 2 is heavier or “3 is lighter”. Yet we know that the lighter one must be in 4-8. So it should have been 1 or 2 is heavier or 6 is lighter because 3 must be either the standard weight or heavier, because if 3 were lighter 1-4 couldn’t be heavier than 5-8.

Or so I think. I might have made a mistake along the way.

If 1/2/5 is heavier than 3/6/9 then 1 or 2 heavy or 6 light. 4 cannot be light or first scale lied. if 7 or 8 is light than 1/2/5 = 3/6/9.

Correct. So am I correct in my assessment that the “3 is lighter” comment was in error? I’m just trying to make sure my understanding is accurate.

You’re correct Scott. it should be 6 is lighter, so just substitute 6 where you hear Holt say 3.

Also, he doesn’t go through all the options, but I did it on paper. If 1-4 is lighter than 5-8, the procedures Holt described are just reversed. However, if 1-4 equals 5-8, then it’s a little bit tricky to narrow down which of Islander’s 9-12 is lighter or heavier.

In that instance, you have to weigh 9 & 10 against 1 & 11. If equal, then 12 is odd man out. you just weight 12 against anyone else to see if he’s lighter or heavier.

If 9 & 10 are heavier than 1 & 11, then you know that either 9 & 10 are heavier, or 11 is lighter. This is the same outcome as Holt’s deduction of 1 & 2 being heavier or 6 being lighter, and you follow the same procedure. If 9 &10 are lighter than 1 & 11, just reverse the procedure.

It WAS fun!

You can also divide them into two groups of six. Take each group of six and divide it into two groups of three.

Weigh group 1-3 against 4-6. Then 7-9 against 10-12. One group of three will be lighter. Lets says it’s 1-3.

Then you weigh 1 and 2. If 1 is lighter it’s him, if 2 is lighter it’s him. If they weigh the same then it’s 3.

Of course it’s the solution. I would have taken 6 vs 6, then 3 vs 3 and finally the same 3rd test as you did. It’s pretty easy actually, Captain Holt is not good at maths ^^

The problem with this answer is that one of them could be lighter or heavier.

Test 1: 1-3 vs 4-6 = 1-3 Lighter

Test 2: 7-9 vs 10-12 = Same

Test 3: 1 vs 2 = Same

With no more possible tests, you don’t know if 3 is lighter or one of 4-6 is heavier

Bill’s right. Kevin’s method wouldn’t work.

I haven’t watched the clip, but it seems pretty obvious to me. You have 12 people. You put 6 on one side of the see-saw and 6 on the other. Whichever side is heavier, you get rid of those 6. Then you have 6 left. You put 3 on one side and 3 on the other. Whichever side is heavier, you get rid of those 3. Then you have 3 left. You put 1 on one side and 1 on the other. If it is balanced, you know they weight the same, and you know the last one weighs less, if it isn’t you know which one weighs less.

The point is you don’t know if the person is lighter or heavier than the rest of them….

I guess I’m not sure what you mean Alex. The see-saw is essentially a scale. If you put a person on each side that weigh the same it would be balanced. If you put two people on each side who weigh the same it would be balanced. And so forth. So you’ll know which side is heavier by which side is on the ground. Thus 6 and 6, then 3 and 3, then 1 and 1.

Daniel, the riddle requires you to state whether the odd man out is either heavier than the other 11 islanders or lighter than the other 11 islanders. Your method narrows down the odd man out, but doesn’t identify whether the man is heavier or lighter than the other 11 islanders. A fourth weighing would do that, but alas, only three are allowed…

I used letters.

First, 12 men – 11 same weight, one different

A-B-C-D-E-F G-H-I-J-K-L

Second, solve for 4 people — seesaw A-B; if equal, then replace A with C, and if unequal then C is diff, else D is diff; if unequal to start, then replace A with C and if equal then A, else B is diff (2 attempts)

Third, solve for 12 people — (i) seesaw ABCD-EFGH; if equal, then IJK or L are diff, and solve per Case 4 (total of 3 attempts); if unequal+, then ABCD are lighter or EFGH are heavier; then (ii) seesaw ABEF-CGJK, and if equal then D is lighter or H is heavier, then (iiia) seesaw DH-AB, and if unequal+ then D, else H; if ABEF-CGJK is unequal+, then AB is lighter or G is heavier, then (iiib) seesaw AG-CE, and if unequal+ then A, else G, and if equal then B; if ABEF-CGJK is unequal-, then EF is heavier or C is lighter, then (iiic) seesaw CE-AB, and if equal then F, if unequal+ then C, else E.

This seems like way too many uses of the seesaw. You can only use it three times total.

this content is currently unavailable in my area. joke.

Huh, that’s way too complicated if even true (I didn’t bother to check.) Here’s the simple solution.

First, define the question. I made this mistake at first, but the question is “Is the 12th man heavier or lighter”. You don’t really care who it is.

Split the 12 in to 3 Groups of 4 men A, B and C. Measure A to B, B to C and C to A. One of the weighings will be even, the other group remaining will either be heavier or lighter in both other weighings telling you if the man is heavier or lighter.

So, here’s an example. Lets say someone in Group A is the 12th guy. So, when B and C are weighed they will be even. When you compared A to B, A will either be heavier or lighter giving you the answer. You’ll get the same answer when comparing A to C as well.

Okay a follow up. The question is intentionally ambiguous. “You must figure out which.” This could mean which man or which weight (heavy or light). You can figure out either, but you can’t figure out both. You can either know the man is heavier or lighter or which man is heavier or lighter, but not both. The person asking the question will always say you’re wrong choosing the opposite interpretation of your solution.

As phrased in the episode, the question is unfortunately ambiguous. However, it was not intentional. The actual riddle is you must determine who the outlier is AND whether he’s lighter or heavier. Having to figure out both using the seesaw just three times is what makes the riddle so challenging.

Yep, you’re right. There is an actual solution to know who and if he is lighter or heavier than the others. This is what I came up with.

SOLUTION

Step 1.

Group 12 men in to 3 groups of 4 and call the groups A,B,C

Each member of group is 1-4, so A1,A2,…

Compare A to B

A > B: A heavy, B light

A

Ax compare C1 to C2.C1 > C2: C1 heavy

C1 < C2: C2 heavy

C1 == C2: C3 is light.

Cx < Ax compare C1 to C2.

C1 C2: C2 light

C1 == C2: C3 heavy.

Cx == Ax compare C4 with A1.

C4 > A1: C4 heavy

C4 Cx: compare A1 to A2

A1 > A2: A1 heavy

A1 < A2: A2 heavy

A1 == A2: B2 light

Ax B2: B2 light

B1 B4: B4 light

B3 Cx: compare A1 to A2

A1 > A2: A2 light

A1 < A2: A1 light

A1 == A2: B2 heavy

Ax B2: B1 heavy

B1 B4: B3 heavy

B3 < B4: B4 heavy

B3 == B4: A4 light

Yikes. Looks like large chunks of solution was removed because of html filter and using less than symbol.

You can see full solution here: [dl.dropboxusercontent.com]

You’ve got a couple outcomes that are impossible. When you compare B1 to B2 in the final compare it isn’t possible to get B2 heavy or B1 light in the final 2 cases respectively. These occur in the 2 scenarios that Ax and Bx are not equal in the first comparison. This is definitely the most complete and comprehensible solutions I’ve seen in this thread.

Are you talking about these lines:

if A heavy, B light

Compare A1+A2+B1 (Ax) to C1+B2+A3 (Cx)

Ax > Cx: compare A1 to A2

A1 == A2: B2 light

This is correct. When Ax > Cx then either A1,A2 is heavy or B2 is light. If A1 == A2 then that leaves B2. If you’re talking about another comparison list the lines of logic as above.

If you look through all the possible outcomes you will notices there are 2 outcomes where B2 is heavy and 2 outcomes where B2 is light. There can only be one outcome of each solution so obviously some lines outcomes are not possible or in error. It turns out you have 1 impossible outcome and quite a few errors in the last block. In your notation this is the line that is not possible.

if A heavy, B light

Compare A1+A2+B1 (Ax) to C1+B2+A3 (Cx)

Ax B2: B2 light.

Note if B2 Cx.

Here are the errors in the last block.

if A light, B heavy

Compare A1+A2+B1 (Ax) to C1+B2+A3 (Cx); remainder is A4+B3+B4

Ax > Cx: compare A1 to A2

A1 > A2: A2 light (Conflicts with Ax > Cx)

A1 Cx)

A1 == A2: B2 heavy (Conflicts with Ax > Cx)

Ax B2: B1 heavy (Conflicts with Ax < Cx)

B1 B4: B3 heavy

B3 Cx: compare B1 to B2

B1 > B2: B1 heavy

B1 < B2: B2 heavy

B1 == B2: A2 light

Bx A2: Not Possible

A1 A4: A4 light

A3 < A4: A3 light

A3 == A4: B4 heavy

And I ran into the problems you were originally having I will spell out the obsolete line when A heavy and B Light

Compare Ax and Cx

Ax is less than Cx

Compare B1 and B2

B1 is Greater than B2 is impossible

The block when A is light and B is heavy has errors as well. To remedy this take the block where A is heavy and B is light. and replace all the A’s with B’s and B’s with A’s. Obviously there will be one impossible outcome in this block.

You’re right, but I fixed it. The duplication was from a condition that wasn’t possible, so it didn’t matter what the answer was because it would never be true. But I did fix the language. I’m not surprised the last part was all screwy because I just flipped symbols and changed heavy to light thinking that was right. This time I just change A to B and B to A and left the methods and logic alone. The solution is still under the same link.

[dl.dropboxusercontent.com]

Ah, I see you figured it out too. I had the window open and didn’t refresh until submitting the answer.

How did I not see it before? The answer is: “This content is not currently available.”

I typed in your symptoms and it says that you could have network connectivity problems

Can this be done in another way?

I did: (Going with an islander is heavier)

6 and 6, which ever group hits the ground has our culprit. Dismiss the ‘other’ 6.

Now weight 3 and 3 and you will be left with the heavier 3.

Pick two of them, if the balance out, the third is the heavier one and if they don’t balance, well, you have your answer.

Did I do good?

You don’t know whether the islander is heavier or lighter so you cannot make that assumption. The episode gave poor phrasing of the riddle; that’s why the actual solution is much more complicated. You must figure out who the outlier is and whether he’s lighter or heavier.

I guessed it right. I would have won the tickets

Pick 2 men (A&B) and set aside. The remaining men are (C,D,E,F,G,H,I,J,K & L)

If A>

<all.I think that solves it in multiple directions and using the 3 times rule.

1st usage:

You put 6 guys on each side and you take 2 of ’em off (one from each side until a seesaw balance.

2nd usage:

You take then the last pair (or pair that made a balance) and weight one of them against the one of normal weight. If it is equal you do the third usage (3rd guy against a guy of a normal weight), if not he is lighter or heavier than a normal.

Sorry for my bad English :)

I have it as 4 sets of 3. You do 3 vs. 3. If it’s even, you take one set off and replace with another. If it’s still even, you know the unused set of 3 has the guy. You than pick 2 of the 3 of the group and go 1 on 1. If it’s even, the unused guy weighs different. If it’s not than you have your answer on the scale.

Just saw the episode and had to try and figure it out, lol. The answer is heavier (since you are restricted to using the seesaw only 3 times). It would take a lot more than 3 times to determine the “odd man out” if he was lighter. (It simply adds too many variables to isolate the odd man out in only three attempts). Since you’re only allowed 3 attempts, he would have to be heavier. Visuals always help. Instead of using numbers and letters to represent the people and groupings, I just visualized 11 Pee Wee Hermans and 1 sumo wrestler on the Island. Break the people up into 3 groups. After the first seesaw (if it’s even… all eight on the seesaw are gone, if it’s not… the heavier group stays and the remaining 8 are gone). Regardless, 8 Pee Wee Hermans are immediately voted off the Island! Haha. Now only 3 Pee Wee Hermans and one sumo are left. Break them up into two groups of two people. After the second seesaw, two more Pee Wee Hermans are immediately voted off the island. Now you are left with just one Pee Wee Herman and the sumo. Put them on the seesaw and use it for the third and final time, and you have your man, and the answer. Run the same scenario with 11 Pee Wee Hermans and a skinny, much lighter midget. You cant determine for certain who the odd man out is in only three attempts. If I hadn’t read the above posts, I wouldn’t have even had a clue of where to begin, lol.

if one or two are the same, 3 is lighter???

not possible, 3 is from the heavy group

I don’t know if this would be considered cheating, but what I did was divide into 3 groups of 4. On each end of the seesaw have two people balanced against each other either by sitting on a crossbeam or just leaning back while counterbalancing each other. If the four are equal they will balance perfectly. The one group of four that doesn’t balance perfectly will have one end that isn’t balanced. If that end is up it is the lighter person on that end (ie whose board is higher or who has to lean back more to balance) who is slightly lighter. if the end is down it is the heavier person on that end ( ie board lower or closer to center to balance.) This would take a MAXimum of 3 uses of seesaw as the group that is off balance would be immediadely apparent and the different individual from that group could be determined in the single group weigh without additional uses. If the individual is in the first group, it would only take one use of the seesaw. Of course this would take some amazing coordination, but we are talking hypotheticals ;)

This is a possible answer, but not the best one. It can be done in just two uses of the see-saw.

1. Split all twelve six and six and weigh

2. Take the side that goes down and split into three and three. a) if it stays even you know that one member of the first group was lighter. b) if one side goes down then a member of that side is heavier.

Or… you could toss out the seesaw and just line everyone up in the sand and see who’s sand pit is deeper or shallower than the rest.

Or, you just ask each man how much he weighs. Problem solved.

abcd > efgh; abe > cdi; a > c; a+

abcd > efgh; abe > cdi; a = c; b+

abcd > efgh; abe d; c+

abcd > efgh; abe < cdi; c efgh; abe efgh; abe = cdi; f > g; g-

abcd > efgh; abe = cdi; f efgh; abe = cdi; f = g; h-

abcd cdi; c > d; d-

abcd cdi; c < d; c-

abcd cdi; c = d; e+

abcd < efgh; abe < cdi; a < c; a-

abcd < efgh; abe < cdi; a = c; b-

abcd g; f+

abcd < efgh; abe = cdi; f < g; g+

abcd jk ; j > k; k-

abcd = efgh; ai > jk ; j jk ; j = k; i+

abcd = efgh; ai k; j+

abcd = efgh; ai < jk ; j < k; k+

abcd = efgh; ai l; l-

abcd = efgh; ai = jk ; a < l; l+

What does one use of the seesaw mean? One loading and unloading? If so, I propose that the simplest answer is to load two men at a time until the seesaw gets unbalanced. One of the last pair of men loaded is the outlier and can be determined by balancing each of them to any other individual with the remaining two seesaw uses.

I just went through the entire thing (and corrected the error that he meant that 6 is lighter when he said that 3 is lighter) and came up with this simplified version of the solution. I even came up with a way to solve it if 1-4 is lighter than 5-8 that works. However, there’s one problem. I have no idea whatsoever how this would work if 1-4 weighed the same as 5-8. Then there’s no telling if 9-12 is heavier or lighter than 1-8 without using the see-saw an additional time.

Anyway, here’s the solution that I came up with.

If 1-4 > 5-8 , then do 1,2,5 v 3,6,9

If 1,2,5 > 3,6,9 , then do 1 v 2

Whichever goes down is heavier

If 1 = 2 , 6 is lighter

If 1,2,5 12 , 5 is lighter

If 5 = 12 , 3 is heavier

If 1,2,5 = 3,6,9 , then do 7 v 8

Whichever goes up is lighter

If 7 = 8 , 4 is heavier

If 1-4 2,7,9 , then do 5 v 6

Whichever goes down is heavier

If 5 = 6 , 2 is lighter

If 1,5,6 12 , 1 is lighter

If 1 = 12 , 7 is heavier

If 1,5,6 = 2,7,9 , then do 3 v 4

Whichever goes up is lighter

If 3 = 4 , 8 is heavier

Sorry, the second part came out a bit wonky. Here it is again:

If 1-4 2,7,9 , then do 5 v 6

Whichever goes down is heavier

If 5 = 6 , 2 is lighter

If 1,5,6 12 , 1 is lighter

If 1 = 12 , 7 is heavier

If 1,5,6 = 2,7,9 , then do 3 v 4

Whichever goes up is lighter

If 3 = 4 , 8 is heavier

I’m not sure the answer but I think it is: as simple as 1,2,3

3 on each side

2 on each side

1 on each side

Anyway, it obviously could be done with 2 on each side 3 times, but I don’t think the answer would be that obvious.

I figured it out. Took a bit of thought.

I figured it out.

Take 6vs6, it will drop to one side. Weigh two more times 3vs3 on each side. One will balance. One will not. If the one on the low side balances the answer is lighter. If the one on the high side balances. The answer is heavier. I figured this out in minutes so everyone stop trying to “math” when it’s a simple answet

Divide them into 3 groups of 4 people.

Put any two groups on each side of the see-saw. (First Use)

Condition 1

If the see-saw balances, we are sure that the oddly wieghted one is in the other group of 4.

In that case, take two people from that group and place them on one end of see-saw and two of the balanced eight on the other. (Second Use)

Condition 1.1

If the see saw balances, remove all but one from the seesaw and put one of the remaining two opposite them. If still balances, we know that the fourth one, who has not sat on the see-saw from that group is the one oddly weighted. (Third Use)

Condition 1.2

If the see saw is not balanced, remove one from each end. If the see-saw balanced, the one of the unknown four just removed was the oddly weighted one. Otherwise the one who stayed is the oddly weighted one.(Third Use)

Condition 2

If the two groups of 4 don’t balance remember which side was lighter, have three get off one end and the remaining person swap places with one of the other four. Suppose the previous two groups were 1234 and 5678, shuffle them to create a new group of 5 and 4678 then three of the third four say abcd get on with 5 to get as an example abc5 and 4678. (Second Use)

Condition 2.1.1

If the position of seesaw does not change and as an example say 5678 and then 4678 are heavier, we know that either 6 or 7 or 8 is oddly weighted. Now put 7 on one end and 8 on the other. If one is heavier they are the odd one otherwise it is 6. (Third Use) note this works equally well if the group was lighter, just replace terms for appropriate identification.

Condition 2.1.2

If the seesaw reverses, ether 4 or 5 is the oddly weighted one. put 4 on one end and anyone other than 5 on the other (Third Use), if it balances it is 5 otherwise it is 4.

Condition 2.1.3

If the seesaw balances we know that either 1 or 2 or 3 is oddly weighted. Say as example 1234 were lighter. Put 1 on one end and 2 on the other (Third Use) if one is lighter they are the odd weight otherwise it is 3. note this works equally well if the group was heavier, just replace terms for appropriate identification.

Done – easy peasy

It is easier than everyone makes it. A seesaw is binary. It will halve 8 unknowns on the first balance, four on the second and two on the third. Set it up so deduction eliminates everything else and your gold. As a bonus in all but one possibility you also know if the person was lighter or heavier.

(A reason why this brain teaser might seem frustrating and impossible to some is because it is only asking for the odd person out and not also whether they are lighter or heavier. It is impossible to know both for sure in only three steps.)