Brooklyn 99 riddle: Weighing Islanders [duplicate]

Question

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?

Answer 1

Divide them into 3 groups of 4 people.

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

Condition 1

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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

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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.

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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.

Answer 2

OK, I think I have it, now the problem of explaining it, here goes:

We are going to name the islanders 1 2 3 4 5 6 7 8 9 10 11 12

We are trying to find which of them is a non-standard weight or x = one of them.

//are comments during explanation

Use 1:

1 2 3 4 against 5 6 7 8

I) 1 2 3 4 = 5 6 7 8 then Use 2: 9 against 10 //9 10 11 or 12 are x

     A) 9 > 10 or 9 < 10 then Use 3:  9 against 11    //9 or 10 are x

          i)  9 > 11 or 9 < 11 then 9 = x

         ii)  9 = 11  then 10 = x

     B) 9 = 10 then Use 3:  9 against 11      //11 or 12 are x

          i) 9 > 11 or 9 < 11 then 11 = x

         ii) 9 = 11 then 12 = x

//Ok four down, eight to go, that was the easy part

II) 1 2 3 4 > 5 6 7 8 then Use 2: 1 2 3 5 against 4 10 11 12 //10 11 12 are not x now

  A) 1 2 3 5 > 4 10 11 12 then Use 3: 1 against 2  //1 2 or 3 are x now and x is heavier than the rest

         i) 1 > 2 then 1 is x //x is heavier

        ii) 1 < 2 then 2 is x

       iii) 1 = 2 then 3 is x

  B) 1 2 3 5 < 4 10 11 12 then Use 3:  4 against 12 // 4 or 5 is x. The switched 4 and 5 caused a reversal

        i) 4 > 12 or 4 < 12 then 4 is x

       ii) 4 = 12 then 5 is x

 C) 1 2 3 5 = 4 10 11 12 then Use 3:  6 against 7   //6 7 or 8 are x and lighter than the rest 

        i) 6 > 7 then 7 is x

       ii) 6 < 7 then 6 is x

      iii) 6 = 7 then 8 is x

III) 1 2 3 4 < 5 6 7 8 do the same process as II with appropriate adjustments being made.

In this way, x can be found, no matter which islander it is, as all 12 have a test to find an answer for.

Hope that makes sense. :D

Answer 3

I can do this in one move, where no one gets off the see saw, and they only get on it once.

It is a see saw, not a giant scale! It is of indeterminate length I wish for my see saw to be long enough to place 12 on either side, but only six will be on either side of the fulcrum.

Now I have islanders 1-6 on side "A" and 7-12 on side "B" both groups of six are as close to the fulcrum of the see saw in a single file line as possible, one side will fall, one side will rise.
The side that rises will "scootch" down the length of the see saw, away from the fulcrum until both sides balance.
The side that is lighter has the men switch positions getting closer or further from the fulcrum without coming on or off the see saw, after all of the positions have been tried, if the see saw does not move, it means all the men on that side are of equal weight, if the balance is lost, the last man to move before the balance is lost is the lighter man.
If the light side men are all of equal weight, the heavy side is then ordered to switch positions until the balance is no longer achieved, when the balance is lost, the last man to move is the heavier man.

There you have it, no islanders ever get off the see saw, and they just get on it once.