Posted by: atri | December 5, 2007

## Lecture 41:Parvaresh Vardy Decoder

In today’s lecture, we saw how to decode folded RS codes of rate $R$ up to $1-\sqrt[m+1]{(mR)^m}$ fraction of errors. For suitable choice of parameters, one can correct $1-\epsilon$ fraction of errors with rate $\Omega(\epsilon/\log(1/\epsilon))$ (I think the stated an incorrect bound on the rate in class). We still need to prove the following lemmas

• There exists an irreducible polynomial $E(X)$ of degree $q-1$ such that for any polynomial $f(X)$ of degree at most $q-2$, the following is true: $f(X)^q\equiv f(\gamma X)\mod(E(X))$, where $\gamma$ is the generator of the field $\mathbb{F}_q$.
• Given an polynomial $f(X)$ that needs to be output in the second step of the list decoding algorithm, we have $T(f(X),f(\gamma X))\equiv 0$, where $f(X)$ and $f(\gamma X)$ are thought of as elements of $\mathbb{F}_q[X]/E(X)\equiv \mathbb{F}_{q^{q-1}}$. Recall that $T(Y,Z)$ was defined as $T_0(Y,Z)\mod(E(X))$, where $T_0(Y,Z)$ (with coefficients from $\mathbb{F}_q[X]$) is just $Q_0(X,Y,Z)$. Finally, recall $Q_0(X,Y,Z)$ is the largest factor of $Q(X,Y,Z)$ from step 1 of the algorithm that is not divisible by $E(X)$.

In the next (and last!) lecture, we will quickly prove the above two lemmas. Then we will see how a small change to the algorithm allows us to decode from $1-R-\epsilon$ fraction of errors in polynomial time (for constant $\epsilon>0$). If we have more time, we will either briefly cover list decoding of binary codes and/or topics that we did not cover in any detail in this class.