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| author | Thibaut Horel <thibaut.horel@gmail.com> | 2014-09-11 11:40:46 -0400 |
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| committer | Thibaut Horel <thibaut.horel@gmail.com> | 2014-09-11 11:40:46 -0400 |
| commit | 18cfe48ab7c7c543175fc7bb17e85a3ae7aca917 (patch) | |
| tree | d6bcc47d2ea2dcbcbc1795c78e8f02d6a9469e6c | |
| parent | 2240f5cc3708235143be6efb48e3fb93b2bf1f8a (diff) | |
| download | cs224-18cfe48ab7c7c543175fc7bb17e85a3ae7aca917.tar.gz | |
Remove junk
| -rw-r--r-- | lecture/main.tex | 9 |
1 files changed, 0 insertions, 9 deletions
diff --git a/lecture/main.tex b/lecture/main.tex index bef50ba..8b273ca 100644 --- a/lecture/main.tex +++ b/lecture/main.tex @@ -305,15 +305,6 @@ that this expected number is constant. proof of the lemma. \end{proof} -\begin{proof} - $\E(time to query(x)) \leq \E(number of full intervals containing x)\leq - \sum_{k=1}^\infty\sum_{intervals I, |I|=k, h(x)\in I} \Pr(full)$ - $\sum_{k=1}^\infty k\max_I \Pr(I full)$ - we need to show that the $\max$ is $O(1/k^3)$ so that the sum converges. - $(m=2n)$, $n=number of diff keys active$. - $$\Pr(I full) \leq \Pr\big(|L(I)-\E L(I)|\geq \E L(I)\big)$$ -\end{proof} - \begin{proof}[Proof of Proposition~\ref{prop:unfinished}] By applying the previous lemma: \begin{align*} |
