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| author | Thibaut Horel <thibaut.horel@gmail.com> | 2015-04-08 00:24:02 -0400 |
|---|---|---|
| committer | Thibaut Horel <thibaut.horel@gmail.com> | 2015-04-08 00:24:02 -0400 |
| commit | 9db365927c14737dc5f480c20a7ade9b476f21a0 (patch) | |
| tree | f95108ef37a087fc586bcf555862f4e4d8ac4b98 /ps4/main.tex | |
| parent | 517260d60f6fb23192f6abee650fa6c8d5a986ef (diff) | |
| download | cs225-9db365927c14737dc5f480c20a7ade9b476f21a0.tar.gz | |
[ps4] forgot to commit last minute typo fixes
Diffstat (limited to 'ps4/main.tex')
| -rw-r--r-- | ps4/main.tex | 12 |
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 |
