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authorThibaut Horel <thibaut.horel@gmail.com>2014-11-05 13:35:10 -0500
committerThibaut Horel <thibaut.horel@gmail.com>2014-11-05 13:35:10 -0500
commitfd5452fbb038ed78b01eedfad68357f171479d1d (patch)
treea4e78018e0cad797f406f23e49097961787b1d44 /ps5
parent27ad83b1a951a26a431c5aabbb6a5cd21430a516 (diff)
downloadcs224-fd5452fbb038ed78b01eedfad68357f171479d1d.tar.gz
[ps5] Fix latex mistake
Diffstat (limited to 'ps5')
-rw-r--r--ps5/main.tex2
1 files changed, 1 insertions, 1 deletions
diff --git a/ps5/main.tex b/ps5/main.tex
index 662ffa1..b04f24e 100644
--- a/ps5/main.tex
+++ b/ps5/main.tex
@@ -313,7 +313,7 @@ concludes the proof of the question.
\paragraph{(e)} Using Markov's inequality, let $X$ be the number of
monochromatic edges after applying the previous rounding scheme. We have:
\begin{displaymath}
- \Pr\left[X\geq \frac{n}{4}\left] \leq \frac{n/6}{n/4} = \frac{2}{3}
+ \Pr\left[X\geq \frac{n}{4}\right] \leq \frac{n/6}{n/4} = \frac{2}{3}
\end{displaymath}
Hence, by repeating the previous rounding scheme $O(\log n)$ times, we
know that at least once we will have less that $\frac{n}{4}$ monochromatic edges