3 ms·
> But after you've got there, you don't have to be that much better than the competition to win. I don't follow, did you mean it's O(log(n))? Otherwise a socia
by taneq 4y ago
> But after you've got there, you don't have to be that much better than the competition to win.
I don't follow, did you mean it's O(log(n))? Otherwise a social network with 2mil users is still more than twice as good as a social network with 1mil users.
- btilly 4y agoIf the total value of the network is O(n log(n)) then the value per user is only O(log(n)). So comparing a 2 million user network with a 1 million user network, if the second one is on a product that is 10% better, users should find switching an improvement.
- taneq 4y agoThanks for explaining! I'd always heard that O(n^2) value for social networks but never really thought about who that value is for. Now that you mention it, the value of a network to an individual user really is only O(n) or even O(log(n)). So while the company still has a massive cash advantage, it's much harder to deepen that moat and secure your customers with it than it seemed.
- coryrc 4y agoIt's often said the value of social networking is O(n^2), which is basically impossible to overcome, versus n lg n, which makes the difference between 2 million and 20 million users a lot smaller (too tired for the math, sorry, maybe 20x?) instead of 100x.