Timeline for Faster Algorithm for Computing Norm
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 1, 2013 at 4:16 | vote | accept | Jackson Walters | ||
| Mar 27, 2013 at 23:14 | review | First posts | |||
| Mar 28, 2013 at 8:53 | |||||
| Mar 25, 2013 at 13:50 | history | tweeted | twitter.com/#!/StackCompSci/status/316185234653724672 | ||
| Mar 25, 2013 at 13:33 | answer | added | Yuval Filmus | timeline score: 1 | |
| Mar 25, 2013 at 13:27 | comment | added | Jackson Walters | Of course $O(n^2)$ means $n^2$ is an upper bound, but I think you knew what I meant. @vonbrand The "weird" notation is standard and that is not what that means. It asks for the largest $k$ such that $i | n$ for all $1 \le i \le k$. For instance, $||n||=1$ for $n$ odd. | |
| Mar 25, 2013 at 11:10 | comment | added | vonbrand | If the weird notation means select the largest $k$ that divides $n$, it is just $n$. $O(1)$. | |
| Mar 25, 2013 at 10:34 | history | edited | Raphael | CC BY-SA 3.0 |
added 80 characters in body; edited tags
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| Mar 25, 2013 at 10:32 | comment | added | Raphael | What have you tried? Also, "faster than $O(n^2)$" makes no sense. | |
| Mar 25, 2013 at 6:47 | comment | added | Karolis Juodelė |
Why is it $O(n^2)$? Don't you just need to do for k = 1 to infinity, if n mod k != 0, return k-1? I wonder if I misunderstood your question.
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| Mar 25, 2013 at 5:15 | history | asked | Jackson Walters | CC BY-SA 3.0 |