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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Replying to @Datavetaren
@Datavetaren@Nightwolf42 Provably secure Bitcoin would require true broadcast: O(n^2). We don't know how secure lesser messaging is.1 reply 1 retweet 4 likes -
Replying to @NickSzabo4
@NickSzabo4@Nightwolf42 The clients I've looked at maintains a fixed set of open connections. So live graph consists of C*N edges.3 replies 0 retweets 0 likes -
Replying to @Datavetaren
@Datavetaren@Nightwolf42 We don't know how secure that is. Some specific problems with it have already been found (e.g eclipse attacks).1 reply 2 retweets 1 like -
Replying to @NickSzabo4
@Datavetaren@Nightwolf42 Increasing C (i.e. treating as not a constant after all) is a good general strategy for reducing these risks.2 replies 2 retweets 1 like
@Datavetaren @Nightwolf42 As the value of the network grows we should be increasing the "constant" C.
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