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| author | Thibaut Horel <thibaut.horel@gmail.com> | 2014-11-05 13:35:10 -0500 |
|---|---|---|
| committer | Thibaut Horel <thibaut.horel@gmail.com> | 2014-11-05 13:35:10 -0500 |
| commit | fd5452fbb038ed78b01eedfad68357f171479d1d (patch) | |
| tree | a4e78018e0cad797f406f23e49097961787b1d44 /ps5 | |
| parent | 27ad83b1a951a26a431c5aabbb6a5cd21430a516 (diff) | |
| download | cs224-fd5452fbb038ed78b01eedfad68357f171479d1d.tar.gz | |
[ps5] Fix latex mistake
Diffstat (limited to 'ps5')
| -rw-r--r-- | ps5/main.tex | 2 |
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 |
