How do you solve that? I saw a solution in the comments where it says to start with numbering all the people and butting 1234 and 5678 on the see saw, then it says if they weight the same then continue and that seems to work. But if they dont weigh the same it doesnt work and it doesnt say what to do in that case.

[–] adrian783@lemmy.world 0 points 2 years ago (1 children)

you can do it like you weight 6v6 then 3v3 then for the last weighing you weight the 2 out of 3.

or you weigh 4v4 to find out which grouping of 4 the light weight person is in, then do 2v2 and 1v1.

[–] ChairmanMeow@programming.dev 0 points 2 years ago (1 children)

You don't know if the person is lighter or heavier yet.

[–] Sagifurius@lemm.ee 0 points 2 years ago* (last edited 2 years ago) (1 children)

That's not the question. Either the scales balance, and the third is heavier or lighter, or the scales don't balance and you get both answers, but the question is purposely framed this way

[–] ChairmanMeow@programming.dev 0 points 2 years ago (1 children)

I mean that not knowing it is part of the question, and the proposed solution doesn't work without knowing if the person is heavier or lighter.

If you know if the person is heavier or lighter, the question becomes trivial.

[–] Sagifurius@lemm.ee 0 points 2 years ago (2 children)

The question is to figure out who is different, not how they are different. That takes one more step, half the time.

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[–] Alph4d0g@discuss.tchncs.de 0 points 2 years ago (1 children)

I'm sure it didn't go well. If it was somehow framed in a sycophantic way where the police were led to believe it was their idea, I'm sure it would have gone better. Wait that might not be too difficult to do.

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