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Section Week 12

This is an outline of the topics we covered in the twelfth week of class.

Subsection Tuesday 3/31

Upcoming.

Today: How to do GRS decoding
Thursday: Why this worked...

ELP.

\begin{equation*} \Lambda(x) = \prod_{j \in J}(1 - \alpha_{j} x) \end{equation*}

EVP.

\begin{equation*} \Gamma(x) = \sum_{j \in J}e_{j} v_{j} \prod_{m \in J \setminus \{j\}}(1 - \alpha_{m} x) \end{equation*}

Key equation.

\begin{equation*} \Lambda(x) S(x) \equiv \Gamma(x) \pmod{x^{d-1}} \end{equation*}

Extended Euclid Alg:.

\begin{equation*} q_{i}(x), \ r_{i}(x), \ s_{i}(x), \ t_{i}(x) \end{equation*}

Final values:.

\begin{align*} h&= 2\\ t_{h}&= 8x^{2} + 8 \implies \Lambda = x^{2} + 1 \quad (\text{using }t_{h}(0))\\ r_{h}&= 10x + 6 \implies \Gamma = 11x + 4 \end{align*}

Subsection Thursday 4/2

Coding Theorist of the Day.

Elwyn Berlekamp, 1940-2019

History / Timeline for RS Codes.

Timeline depicting major decoding events for RS codes from 1960 until 2002
RS codes invented in 1960; some codes could be decoded with Gorenstein-Ziegler BCH decoder in 1963. The Berlekamp-Massey algorithm with complexity \(O(n^2)\) invented 1969. Sugiyama’s algorithm with the same complexity invented 1975. RS codes were used for the Voyager missions in 1977. In 1986 Berlekamp-Welch was invented with complexity \(O(n^3)\text{.}\) This was the first fast algorithm focused on the polynomial evaluation point of view. Starting in 1996 Sudan and others create list-decoding algorithms. In 2002 Gao created a fast algorithm for polynomial evaluation point of view.
Figure 41. Timeline of Developments in RS Codes

Why Do They Work.

Sugiyama
  1. \(\displaystyle \Lambda(x) S(x) \equiv \Gamma(x) \pmod{x^{d-1}}\)
  2. \(\deg(\Lambda(x)) \le \frac{d-1}{2}\) &\(\deg(\Gamma(x)) < \frac{d-1}{2}\)
  3. \(\gcd(\Lambda(x), \Gamma(x)) = 1\) &\(\Lambda(0) = 1\)
Remark 44.
\(\Lambda(x)\) &\(\Gamma(x) = \text{ELP}\) &\(\text{EEP}\) have 1)–3).
Example 46. Worksheet Problem 1 Setup.
\(n=8\text{,}\) \(k=4\text{,}\) \(e=2\text{,}\) \(e+k-1=5\)
\(y_{i} E(\alpha_{i}) = Q(\alpha_{i})\)
\(E(x) = x^{2} + e_{1} x + e_{0}\)
\(Q(x) = q_{0} + q_{1} x + q_{2} x^{2} + q_{3} x^{3} + q_{4} x^{4} + q_{5} x^{5}\)
  • \(i=1\text{:}\)
    \begin{equation*} 0 = q_{0} + q_{1} + q_{2} + q_{3} + q_{4} + q_{5} \end{equation*}
  • \(i=2\text{:}\)
    \begin{equation*} 0 = q_{0} + 4q_{1} + 3q_{2} + 12q_{3} + 9q_{4} + 10q_{5} = Q(4) = Q(\alpha_{2}) \end{equation*}
  • \(i=7\text{:}\) \(y_{7} E(\alpha_{7}) = Q(\alpha_{7})\)
    \begin{equation*} 3 E(5) = Q(5) \end{equation*}
    \begin{equation*} 3(12 + 5e_{1} + e_{0}) = q_{0} + 5q_{1} + 12q_{2} + 8q_{3} + 9q_{4} + 5q_{5} \end{equation*}
\begin{equation*} \begin{bmatrix}1&1&1&1&1&1&0&0 \\&&&&&&&\\&&&A&&&&\\&&&&&&&\\ 1&5&12&8&9&5&10&11\end{bmatrix} \begin{pmatrix}q_{0} \\ q_{1} \\ q_{2} \\ q_{3} \\ q_{4} \\ q_{5} \\ e_{0} \\ e_{1}\end{pmatrix} = \begin{pmatrix}0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 0 \\ 10 \\ 1\end{pmatrix} \end{equation*}
Worksheet Problem 2 Results
\begin{align*} Q(x)&= 8x + 4x^{2} + 5x^{3} + 10x^{4} + x^{5}\\ E(x)&= 8 + 9x + x^{2} \end{align*}