3 ms·

Since O(n^2) is used as a proxy for “something superlinear, but don’t get hung up on how much”, you might also choose O(n^(1+ε)), an upper bound characterised b

• brap 1y ago
  Why not O(nlog*(n)) startups
  • sfpotter 1y ago
    Not that it matters, but O(n log n) is often referred to as "quasilinear", but O(n^(1+eps)) is regarded as "superlinear" (and in fact grows faster than O(n log n) for any eps > 0).
  • sshine 1y ago
    Because, as the article states, > The analogy [between asymptotic growth and company economic growth] breaks down when you force or imply something numerically specific about your growth rate, or introduce functions with different growth rates like logs or exponentials. For now we will (somewhat unprincipledly) stick with two sole classes: O(n) and O(n^2). Perhaps choosing a better two functions could more closely explain the growth dynamics of network effects, which could be more exponential. I think the analogy diminishes in value if you try to directly numerically match it to some growth metric. So the article specifically tries to be unspecific about what superlinearity we're talking about, and also calls it vaguely superlinear. Since O(n^(1+ε)) is arbitrarily superlinear (O(n^1) = O(n), and ε is some arbitrary small amount, making it superlinear by definition, and practically nothing else), it is a good choice when that is all you wish to say. If you went with O(n log n), you'd get the same questions as with O(n^2): Why not O(...something else...): That's not the point! :-D
• ocean_moist 1y ago
  Sure makes sense.