304 reputation
419
bio website
location Czech Republic
age 24
visits member for 2 years, 2 months
seen yesterday

student of Czech technical university in Prague


2d
awarded  Constituent
Dec
9
awarded  Caucus
Dec
1
awarded  Notable Question
Oct
17
awarded  Popular Question
Sep
28
answered Prove or Disprove? $\log(n^n)\text{ is } \Theta(\log n)$
Sep
28
answered how can i prove that square root of n is space constructible
Sep
7
accepted Time complexity, proof $\log(n + c) \in O(\log(n))$
Sep
7
comment Time complexity, proof $\log(n + c) \in O(\log(n))$
But what if $c < 1$? Will it still hold?
Sep
7
asked Time complexity, proof $\log(n + c) \in O(\log(n))$
Sep
7
awarded  Custodian
Sep
7
reviewed Approve Solving $1/n^{\lg (n)}$
Sep
7
accepted Solving $1/n^{\lg (n)}$
Sep
7
asked Solving $1/n^{\lg (n)}$
Aug
1
awarded  Popular Question
Jul
2
awarded  Curious
Jun
3
accepted $f(n) \in o(g(n))$ and $g(n) \in o(f(n))$
Jun
3
comment $f(n) \in o(g(n))$ and $g(n) \in o(f(n))$
Ok, right, but small o notation requires the second function to grow asymptotically faster than $f(n)$ - equality is not permitted here!
Jun
3
comment $f(n) \in o(g(n))$ and $g(n) \in o(f(n))$
That is right, $f$ is not in $o(f(n))$ because small $o(g(n))$ notation means, that $g(n)$ is bigger (not $\geq$) than $f(n)$
Jun
3
comment $f(n) \in o(g(n))$ and $g(n) \in o(f(n))$
But the second definition is basically big O notation, right? Small o notation is stronger...
Jun
3
asked $f(n) \in o(g(n))$ and $g(n) \in o(f(n))$