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Xavier Holt edited begin_claim_The_algorithm_runs__.tex
about 8 years ago
Commit id: 006d32bd82b2db3cc94c240c2cbd9d5afb03a5ca
deletions | additions
diff --git a/begin_claim_The_algorithm_runs__.tex b/begin_claim_The_algorithm_runs__.tex
index ec542cb..a49e525 100644
--- a/begin_claim_The_algorithm_runs__.tex
+++ b/begin_claim_The_algorithm_runs__.tex
...
\begin{claim}
The algorithm runs in
$O(m\log^2m)$ $O(n\log n)$
\end{claim}
\begin{proof}
Assume.
\begin{align*}
R(n) &= 2R(n/2) + O(\log n) + O(n) \\
&=2\left(2R(n/4) + O(\log n/2) + O(n/2)\right) + O(\log n) + O(n)\\
&= iO(n) + \sum_{j=0}^{i-1} 2^j O(\log \frac{n}{2^i}) + 2^iR(\frac{n}{2^i})\\
&= O(n\log n) + \sum_{j=0}^{\log n-1} 2^j O(\log \frac{n}{2^i}) + 2^{\log n}R(1)\\
&= O(n\log n) + n^{\log 2}O(1) + \sum_{j=0}^{\log n-1} 2^j (\log n - \log{2^j})\\
&= O(n\log n) + O(n) + \left(\log n\sum_{j=0}^{\log n-1} 2^j - \sum_{j=0}^{\log n-1} 2^j\log{2^j} \right)\\
&= O(n\log n) + \left(\log n \times 2^{\log n} - C \right) &\text{with $C\geq0$}\\
&= O(n\log n) + \left(n\log n - C \right) &\text{inner bracket $\geq 0$}\\
&\leq O(n\log n) + O(n\log n)\\
&\in O(n\log n)
\end{align*}
\end{proof}