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 
up to ![1-\sqrt[m+1]{(mR)^m}](http://l.wordpress.com/latex.php?latex=1-%5Csqrt%5Bm%2B1%5D%7B%28mR%29%5Em%7D&bg=ffffff&fg=000000&s=0 "1-\sqrt[m+1]{(mR)^m}")
fraction of errors. For suitable choice of parameters, one can correct 
fraction of errors with rate 
(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
of degree 
such that for any polynomial 
of degree at most 
, the following is true: 
, where 
is the generator of the field 
.
- Given an polynomial that needs to be output in the second step of the list decoding algorithm, we have
, where and 
are thought of as elements of ![\mathbb{F}_q[X]/E(X)\equiv \mathbb{F}_{q^{q-1}}](http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BF%7D_q%5BX%5D%2FE%28X%29%5Cequiv+%5Cmathbb%7BF%7D_%7Bq%5E%7Bq-1%7D%7D&bg=ffffff&fg=000000&s=0 "\mathbb{F}_q[X]/E(X)\equiv \mathbb{F}_{q^{q-1}}")
. Recall that 
was defined as 
, where 
(with coefficients from ![\mathbb{F}_q[X]](http://l.wordpress.com/latex.php?latex=%5Cmathbb%7BF%7D_q%5BX%5D&bg=ffffff&fg=000000&s=0 "\mathbb{F}_q[X]")
) is just 
. Finally, recall is the largest factor of 
from step 1 of the algorithm that is not divisible by .
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 
fraction of errors in polynomial time (for constant 
). 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.
