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authorThibaut Horel <thibaut.horel@gmail.com>2014-09-11 11:40:46 -0400
committerThibaut Horel <thibaut.horel@gmail.com>2014-09-11 11:40:46 -0400
commit18cfe48ab7c7c543175fc7bb17e85a3ae7aca917 (patch)
treed6bcc47d2ea2dcbcbc1795c78e8f02d6a9469e6c /lecture
parent2240f5cc3708235143be6efb48e3fb93b2bf1f8a (diff)
downloadcs224-18cfe48ab7c7c543175fc7bb17e85a3ae7aca917.tar.gz
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@@ -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*}