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authorThibaut Horel <thibaut.horel@gmail.com>2015-04-08 00:24:02 -0400
committerThibaut Horel <thibaut.horel@gmail.com>2015-04-08 00:24:02 -0400
commit9db365927c14737dc5f480c20a7ade9b476f21a0 (patch)
treef95108ef37a087fc586bcf555862f4e4d8ac4b98
parent517260d60f6fb23192f6abee650fa6c8d5a986ef (diff)
downloadcs225-9db365927c14737dc5f480c20a7ade9b476f21a0.tar.gz
[ps4] forgot to commit last minute typo fixes
-rw-r--r--ps4/main.tex12
1 files changed, 6 insertions, 6 deletions
diff --git a/ps4/main.tex b/ps4/main.tex
index 3b13700..46dd97e 100644
--- a/ps4/main.tex
+++ b/ps4/main.tex
@@ -156,9 +156,9 @@ $(1-\eps)(\frac{1}{2}-\frac{\eps}{2}) \geq (\frac{1}{2}
-\eps)$. The resulting list length will be $\sqrt{\frac{m2^m}{d}}l_2
=\Theta(\frac{1}{\eps^5})$.
-The inner Hadamard code as $2^m = \Theta(\frac{n}{\eps^6})$ code-words, so the
-list-decoding can be done in polynomial time by brute-force testing all
-code-words. Finally, since $m = \Theta(\log \frac{n}{\eps^6})$, The outer
+The inner Hadamard code has $2^m = \Theta(\frac{n}{\eps^6})$ code-words, so the
+inner list-decoding can be done in polynomial time by brute-force testing all
+code-words. Finally, since $m = \Theta(\log \frac{n}{\eps^6})$, the outer
RS code can be list-decoded in polynomial time using the algorithm seen in
class. Note that by \textbf{2.} we only need to apply this algorithm $l_2
= \Theta(\frac{1}{\eps^2})$ times, so the overall resulting running time is
@@ -229,10 +229,10 @@ $m_2$ of the code of Problem 5.2.3:
= 1 - d\big(\Enc_1(m_1), \Enc_2(m_2)\big)\frac{1}{2}
\end{displaymath}
where the last equality used the following reasoning (this is basically the
-same thing as \textbf{1.} plus using that Hadamard codes have constant
-distance):
+same thing as part \textbf{1.} of Problem 5.2.3 plus using that Hadamard codes
+have constant distance):
\begin{itemize}
- \item on the positions where $\Enc_1(m_1)$ and $\Enc_1(m_2)$ coincide, then
+ \item on the positions where $\Enc_1(m_1)$ and $\Enc_1(m_2)$ coincide,
the resulting blocks after applying $\Enc_2$ will be identical and do
not contribute anything to the $d\big(\Enc(m_1), \Enc(m_2)\big)$.
\item on the positions where $\Enc_1(m_1)$ and $\Enc_1(m_2)$ differ, the