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[Mind_Goblin @FishtankGoblin]

The man she wants.

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@2bucx Help is available.

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@algekalipso @SpiritDirtbag So, the proof is not going to be shared?

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@bIiccy Eyebrow of what

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@Blair_TankWatch @BurtBronx Lol ok, fair play.

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@Blair_TankWatch @BurtBronx I larp as a retard to cover the fact that I'm a (slightly) schizo socially awkward weirdo obsessed with sex. Everyone is worthy of criticism, but what do you gain from calling out Burt like this? We all know he's ND, SD, just D in general

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@t_saar @AbolitionRising https://t.co/ACFzD4npZA

Timestamped, this interaction starts around 32:33

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@shoe0nhead Who is she. I am in love

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@cremieuxrecueil What link does that little 2 lead to?

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@Super_PK_ @aurel_pr Like in K = ⌊(⌊b^ϵ⌋)^c⌋? Consider just d = ⌊b^ϵ⌋. Since ϵ is 2^-75, the quantity b^ϵ is the 2^75th root of a (positive) integer b. This makes b^ϵ a positive real number, not an integer. But d is a count and must be an integer, so the floor function ⌊⌋ rounds it down.

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@Super_PK_ @aurel_pr Lol, just wanted to show my work for that n. Idk if the improvements being made by people reduce it, but basically the numbers being multiplied are so huge this can't actually be done on a computer. It is a "galactic algorithm."

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@Super_PK_ @aurel_pr From sec. 5.2 K ≥ 6, so b^2^-131 ≥ 6 and b ≥ 6^2^131.

Sec. 9.2.1 says b = ⌈log₂n⌉, so n ≥ 2^b ≥ 2^6^2^131 ≈ 2^10^10^39. This n is the bit length of the input integers.

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@Super_PK_ @aurel_pr From original paper section 9.1, c = 2^-56 and ϵ = 2^-75 so ϵ•c = 2^-131.

Sec. 9.2.1 says d = ⌊b^ϵ⌋ and K = ⌊d^c⌋, so K = ⌊(⌊b^ϵ⌋)^c⌋ ≈ ⌊b^(ϵ•c)⌋ = ⌊b^2^-131⌋.

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@lgs_joe @abbas_mehrabian I think that's right. As OP describes, ω can be investigated using tools from algebraic geometry and numerical optimization. I don't think it would be wrong to call the result "nonconstructive" in the technical sense. The openai paper doesn't give a new algorithm.

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@nicotheraped @BurtBronx That's very kind and very true. Vulpa w.

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@kingdoodoo69 It's cute.

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@FiresOfTruth Surely this isn't real, right? It can't be that a small town is considering absorbing 10x its population in migrants? And the local government said they were "not bound by any public objections"?

I see this stuff and it seems fake, not real. Why do this.

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@Blair_TankWatch This is AGI

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@Auralien251166 Hanging chads.