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Results tagged with Search options user 94479
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Questions about asymptotic notations and analysis

Hint: use $\log(a+b) = \log a + \log(1+b/a)$ this; Write the $\log(n^n+n)$ as $\log(n^n+n) = \log(n^n) + \log(1+n/n^n)$ $\log(n^n) + \log(1+n/n^n) \in \mathcal{O}(n \log n)$ note $n/n^n \rightarro … answered Oct 11 '18 by kelalaka$T(n)= a T(n/b)+ f(n) $A generalization that usually works; set$n= b^k$. With backward substitution method.$T(n)=4T(n/2)+n^2 log^2 n $take$n = 2^k\begin{align} T(2^k) & = 4 T(2^{k-1}) … answered Oct 11 '18 by kelalaka We can write the for loop as the sums; $$\sum_{i=1}^{n} \sum_{j = 1}^{i} 1 = \sum_{i=1}^{n}i = \frac{n(n+1)}{2} \in\mathcal{O}(n^2) \, .$$ Note: set the starting values fromi = 1$and$j = 1$, and … answered Dec 23 '18 by kelalaka$\log_{100}n \in \mathcal{O}(log(n))$since base change is a constant.$\log n \in \mathcal{O}(n) \log n \in \mathcal{O}(n^2) \log n \in \mathcal{O}(n^3) \log n \in \mathcal{O}(n^{3/2}) $… answered Oct 17 '18 by kelalaka Using David's definition; a function is super-exponential if it grows faster than any exponential function. More formally, this means that it is$\omega(c^n)$for every constant$c$, i.e., if$\lim …
answered Nov 10 '18 by kelalaka
Here a list from Wikipedia, The lower in the table the bigger complexity class;  \begin{array}{|l|l|} \hline Name & \text{Running Time} \\ \hline \text{Constant time} & \mathcal{O}(1) \\ \text{Inver …
answered Oct 6 '18 by kelalaka