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Tough Trolly Choices

⁨219⁩ ⁨likes⁩

Submitted ⁨⁨11⁩ ⁨months⁩ ago⁩ by ⁨fossilesque@mander.xyz⁩ to ⁨science_memes@mander.xyz⁩

https://mander.xyz/pictrs/image/69bcaed2-2c54-4fa3-abb2-057bdefa48fa.png

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Comments

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  • finestnothing@lemmy.world ⁨11⁩ ⁨months⁩ ago

    The first lever I see in each group.

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    • MamboGator@lemmy.world ⁨11⁩ ⁨months⁩ ago

      I’d say whichever lever in each group feels the most like a “Steve” to me.

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      • Bonsoir@lemmy.ca ⁨11⁩ ⁨months⁩ ago

        I’d prefer the last lever I see in each group.

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      • grrgyle@slrpnk.net ⁨11⁩ ⁨months⁩ ago

        This feels like a Stanley Parable reference but it’s been a while…

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    • wisha@lemmy.ml ⁨11⁩ ⁨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.

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      • finestnothing@lemmy.world ⁨11⁩ ⁨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

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      • sushibowl@feddit.nl ⁨11⁩ ⁨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.

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      • XPost3000@lemmy.ml ⁨11⁩ ⁨months⁩ ago

        “You have to pick levers continuously at every instant in time”

        Supertasks: 🗿

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      • wisha@lemmy.ml ⁨11⁩ ⁨months⁩ ago

        This reply applies to @Cube6392@beehaw.org comment too.

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  • Cube6392@beehaw.org ⁨11⁩ ⁨months⁩ ago

    The first one I encounter.

    Did I do it? Or did I fundamentally misunderstand the question

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    • homura1650@lemm.ee ⁨11⁩ ⁨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.

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      • barsoap@lemm.ee ⁨11⁩ ⁨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.

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    • Redjard@lemmy.dbzer0.com ⁨11⁩ ⁨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

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  • OpenStars@discuss.online ⁨11⁩ ⁨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?)~

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    • damnthefilibuster@lemmy.world ⁨11⁩ ⁨months⁩ ago

      So this is where you’ve been hiding Kirk. Come on, your ship needs you!

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  • muzzle@lemm.ee ⁨11⁩ ⁨months⁩ ago

    Can I take the axiom of choice?

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    • petersr@lemmy.world ⁨11⁩ ⁨months⁩ ago

      Sorry, we sold out of that 5 min before you walked in.

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    • i_love_FFT@lemmy.ml ⁨11⁩ ⁨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.

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  • pruwybn@discuss.tchncs.de ⁨11⁩ ⁨months⁩ ago

    Just pull out a few thousand levers and throw them in front of the trolley.

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  • callyral@pawb.social ⁨11⁩ ⁨months⁩ ago

    Just pull every one, I know one in each cluster will work, but like I gotta make sure

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  • RagingHungryPanda@lemm.ee ⁨11⁩ ⁨months⁩ ago

    I know you can’t enumerate them all, but you just have to enumerate them faster than the trolly. and live forever

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  • bstix@feddit.dk ⁨11⁩ ⁨months⁩ ago

    Pop() one lever from each set.

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  • Leate_Wonceslace@lemmy.dbzer0.com ⁨11⁩ ⁨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.

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    • Tier1BuildABear@lemmy.world ⁨11⁩ ⁨months⁩ ago

      Nope, they’re infinitely close to you as well. They’re now inside you.

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      • Batman@lemmy.world ⁨11⁩ ⁨months⁩ ago

        Then I will swiggity swootie my booty to jimmy the peavy

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      • Leate_Wonceslace@lemmy.dbzer0.com ⁨11⁩ ⁨months⁩ ago

        Oh, so that’s why I can flip them all simultaneously.

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  • mcz@lemmy.world ⁨11⁩ ⁨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

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  • JusticeForPorygon@lemmy.world ⁨11⁩ ⁨months⁩ ago

    That one

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  • SQReder@lemmy.world ⁨11⁩ ⁨months⁩ ago

    The one, that seems to be closest

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    • pyre@lemmy.world ⁨11⁩ ⁨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”

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    • ghen@sh.itjust.works ⁨11⁩ ⁨months⁩ ago

      Does Infinity include dimensions of levers that you can’t comprehend?

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  • criitz@reddthat.com ⁨11⁩ ⁨months⁩ ago

    Since any lever will work I just pull one at random nearby and go home

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  • thr0w4w4y2@sh.itjust.works ⁨11⁩ ⁨months⁩ ago

    i open the I Ching

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  • UncleBadTouch@lemmy.ca ⁨11⁩ ⁨months⁩ ago

    id pull the right one

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    • grrgyle@slrpnk.net ⁨11⁩ ⁨months⁩ ago

      Works with mazes and everything else. It’s the “good ol’ rock” of cardinality

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  • Zkuld@lemmy.world ⁨11⁩ ⁨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

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  • Iheartcheese@lemmy.world ⁨11⁩ ⁨months⁩ ago

    Too complicated I’m just going to walk away

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    • spizzat2@lemm.ee ⁨11⁩ ⁨months⁩ ago

      Murderer…

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      • Iheartcheese@lemmy.world ⁨11⁩ ⁨months⁩ ago

        Image

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  • oessessnex@programming.dev ⁨11⁩ ⁨months⁩ ago

    I would just pick the root of each underlaying balanced binary tree, easy.

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  • Toes@ani.social ⁨11⁩ ⁨months⁩ ago

    Yo this sounds suspiciously similar to how quantum resistant lattice cryptography works.

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  • 10_0@lemmy.ml ⁨11⁩ ⁨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.

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  • Honytawk@lemmy.zip ⁨11⁩ ⁨months⁩ ago

    Irrelevant

    You will never be in time to pull an infinite amount of levers before the trolley runs those people over

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    • Maalus@lemmy.world ⁨11⁩ ⁨months⁩ ago

      Literally says your abilities allow you to do so.

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  • Collatz_problem@hexbear.net ⁨11⁩ ⁨months⁩ ago

    I like the axiom of determinacy more than the axiom of choice.

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  • match@pawb.social ⁨11⁩ ⁨months⁩ ago

    I’m just gonna start swolping my arms out pulling all levers, fuck it

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  • veganpizza69@lemmy.world ⁨11⁩ ⁨months⁩ ago

    sortition all the way

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  • Rentlar@lemmy.ca ⁨11⁩ ⁨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.

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