Brooklyn 99 riddle: Weighing Islanders

  • This one comes from this week's Brooklyn Nine-Nine episode!

    There is an island with 12 islanders. All of the islanders individually weigh exactly the same amount, except for one, who either weighs more or less than the other 11.

    You must use a see-saw to figure out whose weight is different, and you may only use the see-saw 3 times. There are no scales or other weighing device on the island.

    How can you find out which islander is the one that has a different weight?

    Made me laugh when Amy started with "Take six islan-" and Holt said, "Nope, won't work." I originally thought it would take 6 on one side, 6 on the other, and go from there, but then I realized... the islander can weigh more OR less, so you couldn't find out that way. I like Rosa's answer... squeeze until fatty confesses. :p

    haha yeah loved the episode! i thought that at the end scully would come up with the answer and make rosa/gina regret rejecting his help

    I wonder if this is the highest-viewed duplicate question of all time?

    sorry my bad. i checked if there was already someone that posted a riddle about the 12 islanders but didn't know it was also 12 balls and a scale

    @emdee - Here is the link that you posted in the question. Feel free to provide your own comment and I can delete this one. **Brooklyn 99 officially released a video of the answer with Captain Holt!**

    @randal'thor I don't understand that - it's the highest viewed question by 19k views, so it's not even close. How does that happen on an (originally) poor quality dupe?

    @mdc32 Eye-catching title, I guess.

    By the way, a see-saw would take distances into account, wouldn't it? You'd have to ensure that each pair of islanders was sitting exactly the same distance apart on the seesaw.

  • AeJey

    AeJey Correct answer

    6 years ago

    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.)

    Edit: In 11/12 cases you know whether the person is lighter or heavier as the seasaw dictates it. The only case where you don’t is 1.1.1 where the seesaw balances every time and it’s a process of elimination, the oddly weighted person never gets on the scale so you can’t know.

    " It is impossible to know both for sure in only three steps." That's not true. You can figure out both in three steps. Example for condition 1: put three "unknowns" against three "knowns". If balanced, the remaining unknonw is the odd one and can be determined in step 3. If the unknowns go down, you know that one of them is heavy. In step 3 place one unknown on each side. If balanced, the 3rd unknown is heavy. If not balanced, the one going down is heavy. Works the other way around with "light" and "up" instead of "heavy" and "down"

    "It is impossible to know both for sure in only three steps". Wow how did this even get 8 upvotes...mind boggling

    @Hilmar: It's indeed not that easy to know both, but yes it's possible like you said:

    This shouldn't be the accepted answer since it isn't actually the solution... It is a good write up though

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