To recap, Captain Holt promised Beyonce tickets to anybody on his staff who could solve this puzzle. I couldn’t find this on the web, so I turn to the brains of SDMB.
Twelve men are on a desert island. They all have identical weights except for one of them, who is either slightly lighter or slightly heavier than the others.
The only other thing on the island is a seesaw. There are no scales or means to measure weight otherwise. Can you determine which man has the different weight? You only get to use the seesaw three times.
Santiago started with “You take six guys…” and the captain said that was incorrect. Diaz said you hold the seesaw board to all twelve throats until one confesses. That wasn’t it either.
By the end, Linetti figured out the captain didn’t know either. He was about to meet his old captain who gave him the puzzle twenty years ago and he still hadn’t determined the answer.
It may be that it’s impossible to answer, and that it’s just something to keep people from bothering the quizzer with their personal problems.
My best guess is that you weigh four vs four. You get two scenarios:
If the seesaw balances, the odd guy is one of the four remaining. Put one of that group on each end. If it balances, he’s one of the two left and you can determine which one on the third try. If it tilts, replace one of them with one of the two remaining. If it tilts, the one is the ground is the heavier one. If it balances, the odd guy is one of the two left, but all your tries are used up.
If the seesaw tilts, replace one four-member squad with another. If that balances, the odd guy is one of the four that was replaced and he’s heavier. If it tilts, the odd guy is one of the four guys in the air and he’s lighter. That just leaves one try left in either case however, and it’s not enough to identify the odd guy.
Not sure if this is CS or GQ, but since it appeared in the show I’ll try it here.
So part of the plot was the precinct captain (played by Andre Braugher but can’t think of his character name right now) was going to meet a former superior who, 20 years before, asked him a puzzle question that the captain was never able to figure out. Every time he gets together with the former superior he asks if he figured it out yet. So he tasks several members of the precinct to figure it out for him with a prize of 2 tickets to a concert.
The puzzle was this: 12 people are on an island. 11 weigh exactly the same, but one is slightly heavier or slightly lighter. The only way to compare their weight is by using a seesaw on the island, but he seesaw can only be used 3 times.
It seemed ambiguous whether you were supposed to know whether the person was either heavier or lighter, or just that you know one person weighs differently, but not which direction.
If it’s the second one (you don’t know heavier or lighter) it does seem pretty unsolvable. But if it’s the first one (you know whether the one person is heavier or lighter) it seems simple:
Divide the 12 into two groups of six. Take one group of 6 and weight 3 vs. 3 on the seesaw. If you get an imbalance, it’s easy, you can skip the next step and will only have to use the seesaw twice.
If the seesaw balances, do the same with the second group of 6.
Once you find the group of 3 with the imbalance, take any 2 of those 3 and test them. Either there’s an imbalance and you know it’s one of those two, or there’s a balance and you know it’s the third one.
On the show they never came up with an answer. But it seems pretty clear-cut to me, depending on how you interpret the heavier/lighter part. Am I correct or am I missing something?
I just posted this exact question. Ya just beat me to it.
It was ambiguous to me whether you were supposed to know whether the one person was heavier or lighter, or whether you just knew there was a weight discrepancy in one person but not whether heavier or lighter.
If it’s the second one, I have no idea. If it’s the first one, it seems pretty easy:
Pick 6 of the 12- weigh 3 vs. 3 on the seesaw.
If it balances, take the second 6 and find the 3 with the imbalance.
Take the 3 with the imbalance, weigh 2 of those. If there’s an imbalance, it’s one of those two. If there’s a balance, it’s the third one.
If you don’t know whether he’s lighter or heavier:
Put 4 people on each side, if they balance it’s one of the remaining 4. Take 2 of the known normal guys and put them on one side and 2 of the unknowns on the other. If it balances it’s one of the 2 remaining, if it doesn’t it’s one of the 2 on the “unknown side.” Use one of the knowns and one of the unknowns. If it tips it’s the unknown guy and if it balances it’s the remaining guy.
If the original scenario tips… I’ll get back to this.
Well, that was part of what I was wondering- it was ambiguous as to whether you were supposed to know whether the one person was either heavier or lighter, or whether you just knew one person weighed differently, but not in which direction. If it’s the former, my method works. If it’s the latter, I have no idea.
First, weigh two of the groups against each other. If they balance, then the imbalanced person is in the third group. To find him, take three of the people and weigh them against three of the folks that we know for sure are real (from the first eight people). If they also balance, then the remaining person is the target, but we still don’t know if he’s lighter or heavier, so weigh him against anyone from the first eight to determine that. But that second weighing didn’t balance, then we know that the target is one of the three, and we also know he is lighter or heavier, because we know which side had the valid people: if the seesaw went down, the target’s on that side and heavier; if it went up, he’s on that side and lighter – since the other side is known good people. So weigh two of the three people against each other: if they balance then the third person is the target. If they don’t balance we know which person is the target, because we already know if he is lighter or heavier.
Suppose the first weighing was not balanced? That means the target person is in one of the groups, but we don’t know which group. Let’s mark the group that was heavier as Group X, and the group that was lighter Group Y. For the second seesaw weighing, weigh two guys from Group X and one guy from Group Y against the other two guys of Group X and one guy of Group Y. If they balance, the target must be one of the remaining two; a final weigh between them will tell which one it is. (They are both from Group Y so we know that it’s lighter). If the second trial didn’t balance, look at the side that is heavier: Either the target is one of the two Group X guys, or it can be the group Y guy on the other side of the seesaw. Take the two Group X guys and weigh them against each other. If they balance, then it is the Group Y guy that is the target, and if they don’t balance, then the heavier one is the target.
Seems like an odd formulation for it, given how difficult it would be to get precision from people sitting on a seesaw. I get that it’s a puzzle and you’re supposed to ignore the details; it just feels like it pushes it a bit. Almost as valid to say you just throw the men in the water and see how they float.
For a moment I was wondering why Beyonce would be interested in having tickets to see the staff, though.
If all you want to do is figure out which guy is the odd man out, and not whether he’s heavier or lighter, you can solve it for 13 guys with only 3 weighings, but 14 is impossible.
I actually started out right (weigh 4 against 4). But didn’t get much farther than that. But I then did what Amy should have and just googled the answer.
I think it’s something like this
suppose the men are labeled A to L
divide into 4 groups;
A-D, E-H, I-L
additional rules, if you compare sides, put imaginary + or – above the letters indicating potential heavier or lighter men, keep track of this, if the men are reasoned to not be the outlier, revert their letter to a 0. For examples A & B > C & D and all other comparisons are equal then E to L are 0 and either A or B is + (heavier) or C or D is – (lighter)
choose two groups to compare
comparison 1: A-D vs E-H
if balanced then check I-L
check three of the I-L group against three dummies 0 (note that you know all the other men way the same)
Comparison 2: I, J, K vs 000
if balanced then check the one you left out against any
if inbalanced you know wether he’s heavier or lighter, as 000 is the reference group
compare two against each other and draw conclusions (if balanced, the one left out is the heavier/lighter, if inbalanced, well you know if it is because one is heavier or the other is lighter from before)
again from comparison one:
if inbalanced, keep track of which sides are heavier/lighter during comparisons and which letters belong to them, put an imaginary + or – above these letters
switch sides of two men for another comparison, this will help you pinpoint the last four potentials, with two +’s and two –‘s
compare a potential +, a potential – and a 0 with two 0’s and either a potential + or – , you’ll see that the outcome will pinpoint the person you’re looking for and if he’s heavier or lighter (marked with + or – )
I double checked and I think this is it, took me long enough
The answer is pretty easy to find on Google. And I’m not saying we shouldn’t have a thread about it.
But don’t they have internet in the Nine-Nine universe? The Captain has been trying to solve this for 20 years and never thought to Google it?
(As the esteemed **N9IWP **pointed out, as well.)
They probably changed it to men instead of coins in the show precisely to make it harder to Google. But “12 same weight heavier lighter” still brings up plenty of correct hits.
Bricker’s answer is kind of amusing. You can tell he learned the riddle with counterfeit coins because he slips and says “real” and “valid” person. I thought it was amusing anyway.
