Brooklyn 99 is a great show. I didn't always watch it, but it's solid. One thing I do like is that all of the characters are supposed to be smart and deal with cool shit. The most famous moment is the Monty Hall episode, but I really enjoyed Holt's Islander Riddle.
If you don't want to watch the video, then here's the puzzle.
There are 12 men on an island. 11 weigh exactly the same amount, but one of them is slightly lighter or heaver. 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 stopped the episode and figured it out. It took a while but it was fun.
You get information on what you don't measure and you also get information on light/heaviness. That's three results per measurement.
Divide the 12 into three groups of 4 (A-D, E-H, and I-L), and then compare two of them.
Let's say we compare E-H and I-L. If they're equal then we know then 12th man is in A-D, and A-D is the suspect group.
If they're not equal, one of those two group is the suspect group and the other is a reference. A-D is also a reference.
Example: The E-H side rises and the I-L side drops. Either the man is in E-H and he's light or he's in I-L and heavy.
The second comparison has to get us to the third. The third comparison has to give a solid answer from three men: two men who are either both light or both heavy are compared. If they're equal, the third is the 12th man. If they're expected to be light/heavy, it's the lighter/heavier of the two.
If the first comparison was unequal, then second comparison has to compare between 9 men. All 8 of our suspect men and 1 reference man. Mix the groups and remove 3 suspect men (2 from the light/heavy side and 1 from the heavy/light side). Then compare.
Example: We add A, making an EAI group and an FGJ group. The EAI group drops and the FGJ group rises. Because FG and I yielded the same results as the first comparison, they're our final three.
If it went the other way, we'd be comparing E to J.
If they were equal, we'd compare H, J, and K.
We're down to 2 or 3 men and we know whether they're light or heavy.
If the first comparison was equal, then we look at A-D, compare ABC to a reference like EFG. If it's heavy, then the 12th man is heavy and in ABC. If it's light, then the 12th man is light and in ABC. We are down to three men or fewer and we know if they're light or heavy.
Unless ABC was equal to EFG. Then we know it's D, and we only need to make a third comparison to a reference to determine if he's light or heavy. The actual instructions are unclear: are we finding which man is light/heavy or which man and which of the two--light/heavy--he is. I guess we can find both so it's not important.
Since we have three men and accurately expect each to be either light or heavy, we compare two that are expected to be the same: light/light or heavy/heavy.
Example: We compare FG. If they're equal, I is the 12th man and he is heavy. If F drops and G rises, then we know that one of them is the 12th man and he's light so it must be G. If G drops and F rises, then F is the 12th man by the same logic.
If the 2nd measurement had been the opposite, we'd compare E to a reference. If E is lighter than the reference, then it's the 12th man. If they're equal then J is the 12th man and he's heavy. E will never be heavier than the reference. There is no condition in which E, which was on the lighter side of each measurement could be the 12th man and heavy.
If the 2nd measurement was a tie, then we expect J and K to be heavy and we're at the same logic as comparing FG except heavy instead of light.
If the 1st comparison was equal then we expect ABC to all be heavy/light (determined by the 2nd comparison). Compare any one to the other. If ABC was heavier than the reference, we compare A and B. If A is heavier, it's A. If B is heavier, then it's B. If they're equal then C is the 12th man and he's heavy.
Brooklyn 99 didn't give us an answer, except for this bit. I feel it's not an outtake because it's not funny. It feels like a DVD special feature.
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