this post was submitted on 29 Jul 2023
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[–] mofongo@lemm.ee 158 points 1 year ago (39 children)

Thats actually a really good dilemma if you think about it. Like if everyone doubles it you basically don’t kill anyone. But you’ll always risk that there’s some psycho who likes killing and then you will have killed more. And if these choices continue endlessly you will eventually find someone like this. So killing immediately should be the right thing to do.

[–] ghariksforge@lemmy.world 44 points 1 year ago (16 children)

At some people you will run out of people to tie to the tracks.

[–] Diplomjodler@feddit.de 12 points 1 year ago (12 children)

How many branches is that going to take? Just out of interest.

[–] Magikjak@lemmy.world 46 points 1 year ago (1 children)
[–] alerternate@lemmy.world 26 points 1 year ago (1 children)

math checks out. log2(8 billion) ~= 32.9

[–] ghariksforge@lemmy.world 11 points 1 year ago (2 children)

It's a little more complicated than that. You have to be summing everyone who is still tied to all the previous tracks. It needs to be a geometric sum formula.

[–] RatMaster@sh.itjust.works 11 points 1 year ago (1 children)

You could just move them over whenever someone decides to double it up. That way the person that was going to die alone is doomed to die anyway. 😂

[–] Patawagon@lemmy.world 4 points 1 year ago (1 children)

I wonder if we do stack it till every human is tied to the track whos at the lever?

Does the train/AI decide?

[–] fneu@discuss.tchncs.de 3 points 1 year ago

The current version of GPT is willing to sacrifice billions of robots for humanities sake. However, it proposed to strap alien civilizations to the track next. And in that case it would choose to save the larger group…

[–] sabazius@lemmy.world 10 points 1 year ago* (last edited 1 year ago) (1 children)

It'll just be one fewer junctions. 2^n is always one more than the sum of 2^1+...2^(n-1)

[–] Magikjak@lemmy.world 1 points 1 year ago

I think you have to include 2^0 for that to be true?

e.g 2^0 = 1, 2^1 = 2 2^0 + 2^1 = 1 + 2 = 3, 2^2 = 4 … 7, 8 15,16 31, 32 etc.

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