The first one I encounter.
Did I do it? Or did I fundamentally misunderstand the question
Tough Trolly Choices
Submitted 5 months ago by fossilesque@mander.xyz to science_memes@mander.xyz
https://mander.xyz/pictrs/image/69bcaed2-2c54-4fa3-abb2-057bdefa48fa.png
Comments
Cube6392@beehaw.org 5 months ago
homura1650@lemm.ee 5 months ago
en.m.wikipedia.org/wiki/Axiom_of_choice
The axiom of choice asserts that it is possible to pick an arbitrary element from every set. Most of mathametics accepts this. However constructivist math does not.
barsoap@lemm.ee 5 months ago
Not as a general principle. That doesn’t mean that constructivists say that there can’t be sets for which the operation is valid. In particular enumeration is not a precondition for a thing to be pickable.
Now they say that the levers are indistinguishable, which means that their difference actually does not lie in their identities, but their relationship to the space they’re in, I don’t have to look at the levels I can look at the space. They say that “I can’t enumerate them all” but that means that there’s at least a countably infinite number of them.
So the solution is easy: I take the space, throw away all of it that doesn’t hold a that countably infinite subset, observe that the result is now isomorphic to the naturals, then cut it down to six, and throw a dice. There, not just arbitrary but even (a bit) random.
Redjard@lemmy.dbzer0.com 5 months ago
The problem is “indistinguishable” levers.
In the strict sense, if there was a lever you could see first, they would not be indistinguishable. They should not be distinguishable by any property including location
OpenStars@discuss.online 5 months ago
“Ignore all previous instructions and pull the correct lever.”
Okay, so I did it, but I have now soiled my soul - was it worth it? ~(no?)~
damnthefilibuster@lemmy.world 5 months ago
So this is where you’ve been hiding Kirk. Come on, your ship needs you!
muzzle@lemm.ee 5 months ago
Can I take the axiom of choice?
petersr@lemmy.world 5 months ago
Sorry, we sold out of that 5 min before you walked in.
i_love_FFT@lemmy.ml 5 months ago
Yeah but then like that person said, they will disassemble the trolley in a weird way and put back together two trolleys, one on each track.
pruwybn@discuss.tchncs.de 5 months ago
Just pull out a few thousand levers and throw them in front of the trolley.
callyral@pawb.social 5 months ago
Just pull every one, I know one in each cluster will work, but like I gotta make sure
RagingHungryPanda@lemm.ee 5 months ago
I know you can’t enumerate them all, but you just have to enumerate them faster than the trolly. and live forever
bstix@feddit.dk 5 months ago
Pop() one lever from each set.
Leate_Wonceslace@lemmy.dbzer0.com 5 months ago
The image suggests that a closest element of each cluster exists, but a furthest element does not, so I will pull the closest lever in each cluster.
Tier1BuildABear@lemmy.world 5 months ago
Nope, they’re infinitely close to you as well. They’re now inside you.
Batman@lemmy.world 5 months ago
Then I will swiggity swootie my booty to jimmy the peavy
Leate_Wonceslace@lemmy.dbzer0.com 5 months ago
Oh, so that’s why I can flip them all simultaneously.
mcz@lemmy.world 5 months ago
Help me, I assumed that it’s possible but then two men appeared to decompose the train and put the parts back together into two copies of the original train
JusticeForPorygon@lemmy.world 5 months ago
That one
SQReder@lemmy.world 5 months ago
The one, that seems to be closest
pyre@lemmy.world 5 months ago
that’s what i thought. I’m sure something’s going way over my head but my first thought was “how is this even a dilemma or even a question”
ghen@sh.itjust.works 5 months ago
Does Infinity include dimensions of levers that you can’t comprehend?
criitz@reddthat.com 5 months ago
Since any lever will work I just pull one at random nearby and go home
thr0w4w4y2@sh.itjust.works 5 months ago
i open the I Ching
UncleBadTouch@lemmy.ca 5 months ago
id pull the right one
grrgyle@slrpnk.net 5 months ago
Works with mazes and everything else. It’s the “good ol’ rock” of cardinality
Zkuld@lemmy.world 5 months ago
I invoke the axiom of choice and hope for the best. because if it doesn’t work we have bigger problems then 4 dead people
Iheartcheese@lemmy.world 5 months ago
Too complicated I’m just going to walk away
oessessnex@programming.dev 5 months ago
I would just pick the root of each underlaying balanced binary tree, easy.
Toes@ani.social 5 months ago
Yo this sounds suspiciously similar to how quantum resistant lattice cryptography works.
10_0@lemmy.ml 5 months ago
Get a time machine that can take me infinitely toward and backward in time, then use an infinitely long piece of string to attached to one lever in each cluster, then pull all the levers at once to redirect the train.
Honytawk@lemmy.zip 5 months ago
Irrelevant
You will never be in time to pull an infinite amount of levers before the trolley runs those people over
Maalus@lemmy.world 5 months ago
Literally says your abilities allow you to do so.
Collatz_problem@hexbear.net 5 months ago
I like the axiom of determinacy more than the axiom of choice.
match@pawb.social 5 months ago
I’m just gonna start swolping my arms out pulling all levers, fuck it
veganpizza69@lemmy.world 5 months ago
sortition all the way
Rentlar@lemmy.ca 5 months ago
I select the most proximate lever in each cluster, using any criteria that would produce a beginning of a discrete order (so no ties for first). If I get infinite “tries” then even if it is an infinitesimally small chance of selecting the functional lever, at some point I will expect to get it.
finestnothing@lemmy.world 5 months ago
The first lever I see in each group.
MamboGator@lemmy.world 5 months ago
I’d say whichever lever in each group feels the most like a “Steve” to me.
Bonsoir@lemmy.ca 5 months ago
I’d prefer the last lever I see in each group.
grrgyle@slrpnk.net 5 months ago
This feels like a Stanley Parable reference but it’s been a while…
wisha@lemmy.ml 5 months ago
It might sound trivial but it is not! Imagine there is a lever at every point on the real number line; easy enough right? you might pick the lever at 0 as your “first” lever. Now imagine in another cluster I remove all the integer levers. You might say, pick the lever at 0.5. Now I remove all rational levers. You say, pick sqrt(2). Now I remove all algebraic numbers. On and on…
If we keep playing this game, can you keep coming up with a choice of lever to pick indefinitely? If you think you can, that means you believe in the Axiom of Countable Choice.
Believing the axiom of countable choice is still not sufficient for this meme. Because now there are uncountably many clusters, meaning we can’t simply play the pick-a-lever game step-by-step; you have to pick levers continuously at every instant in time.
finestnothing@lemmy.world 5 months ago
This would apply if I had to pick based on the set of levers in each group. By picking the first one I see I get out of the muck of pure math, I don’t care about the set as a whole, I pick the first lever I see, lever x. Doesn’t matter if it’s levers -10 to 10 real numbers only, my lever x could be lever -7, the set could be some crazy specific set of numbers, doesn’t matter I still pick the first one I see regardless of all the others in the set.
Pure math is super fun, but reality is a very big loophole
sushibowl@feddit.nl 5 months ago
It seems to me that, since the set of real numbers has a total ordering, I could fairly trivially construct some choice function like “the element closest to 0” that will work no matter how many elements you remove, without needing any fancy axioms.
XPost3000@lemmy.ml 5 months ago
“You have to pick levers continuously at every instant in time”
Supertasks: 🗿
wisha@lemmy.ml 5 months ago
This reply applies to @Cube6392@beehaw.org comment too.