• 3 Posts
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Joined 4 years ago
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Cake day: May 19th, 2021

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  • wisha@lemmy.mltoScience Memes@mander.xyzTough Trolly Choices
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    6 months ago

    They have to get smaller to fit the problem statement- if all levers are the same size or have some nonzero minimum size then the full set of levers would be countable!

    Now we play the game again 🤓. I start by removing the levers in the field/scale of view of your microscope’s default orientation.


  • wisha@lemmy.mltoScience Memes@mander.xyzTough Trolly Choices
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    6 months ago

    But look at the picture: the levers are not all the same size- they get progressively smaller until (I assume from the ellipsis) they become infinitesimally small. If a cluster has this dense side facing you, then you won’t “see” a lever at all. You would only see a uniform sea of gray or whatever color the levers are. You now have to choose where to zoom in to see your first lever.



  • wisha@lemmy.mltoScience Memes@mander.xyzTough Trolly Choices
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    6 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 which lever to pick indefinitely (as long as I haven’t removed all the levers)? 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.



  • wisha@lemmy.mltoLinux@lemmy.ml*Permanently Deleted*
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    6 months ago

    We could take this further and let developers specify exactly the dependencies they need! No more bloated runtimes! App A could specify libfoo>=1.23.45 while app B specify libfoo<1.24 and Flatpak could resolve the compatible version automatically!

    Serious answer: If space saving is the goal, traditional packaging is the way to go. Allowing multiple runtimes is a slippery slope away from the core idea of Flatpak (simplest dependency management possible so developers don’t have to test many configurations).

    (Not that there’s anything wrong with traditional packaging with more complicated dependency management - it’s just not Flatpak’s thing).