Niclas Alexandersson edited b) Algorithm.tex  almost 10 years ago

Commit id: c653cc41eed7ed0079dcb98483f2ce41f438d666

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&< \sum_{i=1}^{2^{n+1}}{\left[ (m+1)^2(i-1) \right]}\\  &= (m+1)^22^{n}\left(2^{n+1} - 1\right)  \end{align*}  \end{subequations}  Looking at these bounds again in terms of $p$ and $k$, we get that:  \begin{subequations}  \begin{align*}  T_{\mathtt{min}}(p, k) &= \floor{\log_2(p)}^22^{\floor{\log_2(k)}-1}\left(2^{\floor{\log_2(k)}} - 1\right)\\  T_{\mathtt{max}}(m, n) &= (\floor{\log_2(p)}+1)^22^{\floor{\log_2(k)}}\left(2^{\floor{\log_2(k)}+1} - 1\right)  \end{align*}  \end{subequations}