@NickSzabo4 @Nightwolf42 The clients I've looked at maintains a fixed set of open connections. So live graph consists of C*N edges.
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Replying to @Datavetaren
@Datavetaren@NickSzabo4@Nightwolf42 Every full node still needs to see all transactions.1 reply 1 retweet 1 like -
Replying to @mmeijeri
@mmeijeri@NickSzabo4@Nightwolf42 Yes and these tx flow on the edges of a graph where the number of edges is O(n) or O(n log n) if C=log n.1 reply 1 retweet 1 like -
Replying to @Datavetaren
@Datavetaren@NickSzabo4@Nightwolf42 And each of these needs to transfer O(n) txs.1 reply 0 retweets 0 likes -
Replying to @mmeijeri
@mmeijeri@NickSzabo4@Nightwolf42 O(n*m) if m num of tx. But n and m are independent. Also, we're talking about bandwidth between nodes.3 replies 0 retweets 0 likes -
Replying to @Datavetaren
@mmeijeri@NickSzabo4@Nightwolf42 Or at least m is not a linear function of n. Number of tx is related to number of users. Perhaps m=log n.3 replies 0 retweets 1 like -
Replying to @Datavetaren
@Datavetaren@mmeijeri@Nightwolf42 But no rigorous argument has ever been made that Bitcoin is still secure when m=logn rather than n.2 replies 3 retweets 4 likes -
Replying to @NickSzabo4
@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).1 reply 2 retweets 5 likes -
Replying to @NickSzabo4
@NickSzabo4@mmeijeri@Nightwolf42 Sure, but no existing client is doing that. It's even hard to know n precisely.1 reply 0 retweets 0 likes -
Replying to @Datavetaren
@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.2 replies 1 retweet 3 likes
@Datavetaren @mmeijeri @Nightwolf42 I encourage you to try to prove or disprove your m=log(n) theory. It would be an important result.
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