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

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  • Yeah, the four color problem becomes obvious to the brain if you try to place five territories on a plane (or a sphere) that are all adjacent to each other.

    I think one of the earliest attempts at the 4 color problem proved exactly that (that C5 graph cannot be planar). Search engines are failing me in finding the source on this though.

    But any way, that result is not sufficient to proof the 4-color theorem. A graph doesn’t need to have a C5 subgraph to make it impossible to 4-color. Think of two C4 graphs. Choose one vertex from each- call them A and B. Connect A and B together. Now make a new vertex called C and connect C to every vertex except A and B. The result should be a C5-free graph that cannot be 4-colored.















  • wisha@lemmy.mltoScience Memes@mander.xyzTough Trolly Choices
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    5 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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    5 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.



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    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 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.