3 ms·
There's one huge flaw in the exponential math here: duplicates. There's a good chance that a large number of people your friends know are already included in yo
by incongruity 13y ago
There's one huge flaw in the exponential math here: duplicates. There's a good chance that a large number of people your friends know are already included in your network – the percentages will vary, but it definitely cuts down the pure exponential growth.
- johnsoft 13y agoSo instead of exponential growth, each additional suspect causes the total to asymptotically approach 100% of the vertices in the graph. I can't help but feel that's what they're really going for with three hops.
- olefoo 13y agoSeveral simplifying assumptions were made; in a real world scenario you would have a certain number of duplicates that is shared with any given contact; but the overlap of redundancies between contacts is going to be low. You also need to figure that some of your contacts have an order of magnitude more contacts than your your average contact, and that they are more likely be connected to other super-connectors. So your second order-contacts are likely to be richer in people who have more contacts than the average networker... Even building a decent estimator for this sort of problem is a hard problem; since you need to get some good estimates for a number of distributions ( number of contacts, number of duplicates between pairs of contacts, number super-connector bilateral contacts, etc.). My general sense is that duplication of contacts has little effect on the number of contacts of contacts of contacts; and that the number of super-connectors you know would have a greater impact.