No #bitcoin is NOT O(n^2) in terms of bandwidth. Every node has a fixed set of connections. Bandwidth is O(n).
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@Datavetaren@mmeijeri@Nightwolf42 But no rigorous argument has ever been made that Bitcoin is still secure when m=logn rather than n.
@Datavetaren @mmeijeri @Nightwolf42 The only existing proofs of Bitcoin's security up to 51% attack are for the case m=n => O(mn) = O(n^2).
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@NickSzabo4@mmeijeri@Nightwolf42 Sure, but no existing client is doing that. It's even hard to know n precisely. -
@Datavetaren@mmeijeri@Nightwolf42 "That's the way they're doing it" is in no way a proof or even strong evidence of Bitcoin's security. -
@NickSzabo4@mmeijeri@Nightwolf42 Hey, yes, I'm not arguing against that. I'm just saying how things work in practice.
End of conversation
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