### Brooklyn Nine-Nine Riddle: Weighing Islanders

• A brain teaser posed by the character Captain Ray Holt of 'Brooklyn Nine-Nine' played by Andre Braugher during episode 18 of season 2:

"There are twelve(12) men on an island, eleven(11) weigh exactly the same amount, but one of them is slightly lighter or heavier, [the object is to] figure out [whether he is lighter or heavier]. A standard see-saw must be used and only three times."

(Rephrased for clarity, below is a verbatim transcript of 'Holts' dialogue)

"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."

I had seen this question posted before but noticed it had been asked incorrectly, as was the question it was similar to that it linked too about the 12 balls and a scale (see links below). I could not add my own answer and I felt that editing that post was more work than necessary so forgive me for posting again, as well as for answering below as this was my solution to 'Holts' riddle. Thankyou for reading and understanding. ((http://puzzling.stackexchange.com/questions/9979/brooklyn-99-riddle-weighing-islanders )) ((http://puzzling.stackexchange.com/questions/183/twelve-balls-and-a-scale ))

Please explain your claim that is was not asked properly. I believe it was in http://puzzling.stackexchange.com/questions/183/twelve-balls-and-a-scale

@RoccoRuscitti - Here is a video of Holt's solution. This should help to clarify the intent of his question as well as to explain his answer.

• Corvus Correct answer

6 years ago

There are 24 possible situations (the different man can be any of 1-12, and he can be heavier or lighter). Thus we need to log224 bits of information to solve the puzzle. You can weigh three combinations of men on the see-saw. Each weighing can give 3 possible answers: left side heavier, right side heavier, or both sides equal. Thus in principle we can get log227 bits from the three comparisons. So in principle, we should be able to solve the problem. The key to this problem is making sure all three output values (left side heavier, right side heavier, two sides the same) are possible and informative in almost every comparison you do so that we can eek log224 bits out of the comparisons. Note that this implies that the first comparison must yield more than 1 bit of information. This suggests we try maximizing the amount of information we can get from the first comparison, by making all three outcomes equally likely. Comparing (1,2,3,4) to (5,6,7,8) does exactly this. Similar logic will help us design all further comparisons.

Here is one solution:

Number the men 1,2,3...12. First weigh 1,2,3,4 against 5,6,7,8. One of two things will happen:

1) They are equal. Now we know that the different man is among {9,10,11,12}. Weigh 9,10,11 against 1,2,3. If these are equal, the different man is 12. Weigh 12 against 1 to find out whether 12 is heaver or lighter. If the 9,10,11 differs from 1,2,3, then weigh 9 against 10. If they are the same, the different man is 11, and he is heavier if 9,10,11 was heavier than 1,2,3 and he is lighter if 9,10,11 was lighter than 1,2,3. If 9 and 10 are different, the different man is the lighter of the 9,10 comparison if 9,10,11 was lighter than 1,2,3, (and he is lighter); the different man is the heavier of the 9,10 comparison if 9,10,11 was heavier than 1,2,3 (and he is heavier).

2) They are different. Without loss of generality suppose that 1,2,3,4 is heavier than 5,6,7,8. (We could always relabel the men so that this is true). We know {9,10,11,12} all weigh the same.

Weigh 1,2,5,6,7 against 8,9,10,11,12:

a) If 1,2,5,6,7 is heavier, then either 1 or 2 heavier, or 8 is lighter. Weigh 1 against 2. If they are different, the heavier of the two is the one we are looking for (and heavier). If they are the same, 8 is the one we are looking for (and lighter).

b) If 1,2,5,6,7 is lighter, then one of 5,6,7 is different and lighter. Weigh 5 against 6. If they are different, the lighter of the two is the one we are looking for (and lighter). If they are the same, 7 is different (and lighter).

c) If they are the same, then one of 3,4 is different. Weigh them against each other. The one who is heavier is the different man (and heavier).

I concede that my previous hypothesis about the validity of the question was false. @Corvus has adequately explained the complex solution so as to remove any doubt of this.

License under CC-BY-SA with attribution

Content dated before 6/26/2020 9:53 AM
• {{ error }}