4 ms·
> Generally speaking, "the value of a graph is proportional to the square of the number of edges" No, what Metcalfe's law assumes is that the value of the grap
by atleta 3y ago
> Generally speaking, "the value of a graph is proportional to the square of the number of edges"
No, what Metcalfe's law assumes is that the value of the graph is proportional to the number of edges (not their square). And from that assumption and the fact that the graph is fully connected follows that it's proportional to the square of the number of nodes. (Because you can have (n-1)*n/2 edges with n nodes in a fully connected graph.
And hence, the Reiser quote above is similar but it emphasizes something else: it states what Metcalfe's law (I think) uses as a premise (or implicit claim) that the value is in the connections. Because it's not necessarily a fully connected graph.
Edit: originally I've given (n-1)*2/2 as the number of edges instead of (n-1)*n/2.
- dekhn 3y agoSo, the wikipedia first line is wrong, you're saying? "Metcalfe's law states that the financial value or influence of a telecommunications network is proportional to the square of the number of connected users of the system" Later in the article it seems like the original stating is more consistent with what you said, but everything I've ever learned in network theory and practice shows that it scales as a power of number of edges, and the logarithm of the number of nodes.
- gnramires 3y agoThe number of connected users is different from the number of user connections :) The first is the number of nodes, the second edges.
- dekhn 3y agoUhh, are you sure? I believe "connected users" refers to edges. Otherwise it would be stated as "users connected to the network". It could explain my misunderstanding, and also seems consistent with the explanation later in the article, but it's also completely the opposite of what we observe on the internet; for example, the value of the web is definitely not in its in number of pages, but in the value and quality of the connections between the pages.
- zvr 3y agoTwo users in the network: A and B; one connection: AB. Three users in the network: A, B, and C; three connections: AB, AC, BC. Four users in the network: A, B, C, and D; six connections AB, AC, AD, BC, BD, CD. Metcalfe's law says value increases as 1-3-6-... instead of 2-3-4. In graph terms, users are nodes, connections are edges, and in a fully-connected graph edges are in order of the square of nodes.
- dekhn 3y agoYes, I see I had completely misread and misunderstood the original law. But ethernetworks aren't fully connected (they tend to have lots of local connections that then are connected to each other through routing).
- jacquesm 3y agoI think the difference between logical and physical connections is what drives the confusion here. If two nodes can reach each other somehow then for Metcalfe's law they are connected, even if there is no direct connection between them.
- dekhn 3y agoYes, I realized that shortly after reading the replies. Thanks for stating it explicitly. Once again, my brain's inability to parse english caused a multi-decade misunderstanding. Realistically, the only metric that I can think of that makes sense here isn't proportional to |V| or |E| but to the betweenness connectivity of the graph and the average distance between nodes.
- jacquesm 3y agoYou actually have a very valid point: given that there is such a thing as the maximum ttl at some point that 'logically connected' network will become more and more sparse depending on how 'wide' the network really is. I wonder if there are already parts of the V4 net that are so far removed from each other that this is an issue.
- amomchilov 3y agoPromotional to the number of _edges_. Edges are proportional to the square of the number of nodes, so the value of the network overall is proportional to the square of the nodes. Think of it this way, for every new user added to the network: * the new user is enriched proportional to the number of existing users * every existing user is enriched by the 1 new user This double-counting is what gives it the quadratic growth.
- atleta 3y agoYep. That's what I was saying too. The first line of my comment quotes the GP and I was correcting that.