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Oct 15, 2019 at 15:45 comment added Zach Teitler This was the subject of an SMBC (Saturday Morning Breakfast Cereal) webcomic: smbc-comics.com/comic/mathematicians
Jan 18, 2019 at 2:30 comment added Solveit @TomChurch How do we know $O(n^2)$ is impossible?
Sep 27, 2018 at 6:09 comment added Tom Church $O(n^2)$ is impossible; the conjecture is $O(n^{2+\varepsilon})$. But by the same token the proofs here do not actually prove e.g. $O(n^{2.374})$, but rather $O(n^{2.374+\varepsilon})$.
Sep 11, 2018 at 11:14 comment added Simd @FedericoPoloni $O(n^2 \log{n})$ is certainly in $O(n^{2+\epsilon})$ for $\epsilon > 0$ but $O(n^{2+\epsilon})$ is not in $O(n^2 \log{n})$. The question is whether $O(n^2 \log{n})$ would also be shocking.
Sep 9, 2018 at 10:08 comment added Federico Poloni @Mitch A function in $O(n^2\log n)$ is also in $O(n^{2+\epsilon})$ for any $\epsilon>0$; that is just a more general form that includes essentially all "$n^2$ times logarithmic factors" classes.
Sep 9, 2018 at 0:26 comment added Mitch @StevenStadnicki, Christopher: Isn't $O(n^2 \log(n))$ a viable conjecture?
Sep 6, 2018 at 17:11 comment added Steven Stadnicki @Christopher $O(n^2)$ I think would be a shock, but $O(n^{2+\epsilon})$ for any $\epsilon$ is a lot more plausible.
Sep 6, 2018 at 15:59 history made wiki Post Made Community Wiki by S. Carnahan
Sep 6, 2018 at 10:36 comment added Christopher It's probably worth pointing out that [according to the papers, I don't know the subject myself] the strong-but-widely-believed conjecture - analogous to the Riemann hypothesis in the example in the question - is that the asymptotic complexity is $O(n^2)$.
Sep 6, 2018 at 9:02 history edited Federico Poloni CC BY-SA 4.0
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Sep 6, 2018 at 8:55 history edited Federico Poloni CC BY-SA 4.0
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Sep 6, 2018 at 8:50 history answered Federico Poloni CC BY-SA 4.0