4 ms·
A fully connected graph of n users contains n*(n-1) links if users don't connect to themselves (I would describe this as polynomial growth in the number of chan
by todd8 9y ago
A fully connected graph of n users contains n*(n-1) links if users don't connect to themselves (I would describe this as polynomial growth in the number of channels, a lot better than exponential). A chat broker that acts as a switchboard between users could, I suppose, reduce this to a linear relationship between users and channels.
- adamweston 9y agoLike a bus?
- amelius 9y agoYou have to divide that by 2. It's "half the matrix".
- todd8 9y agoThanks, my mistake. You're right, there are n(n-1)/2 arcs in the fully connected graph (K_n).