I'm working on trying to understand how to use Reed-Solomon decoding to make Shamir secret sharing robust to cheaters as mentioned here. I'm following the setup shown at the bottom of page 40 in this paper.
So, here is my simple example. I'm working in $\mathbb{Z}_{11}$. Let $p(x)=8+3x$ be the polynomial I'm using for sharing (so my secret is $8$). I compute 4 shares using $x=1,2,3,4$ and get $0,3,6,9$ as the shares.
From the setup in the second paper, I set $G=\left(\begin{array}{cccc} 1 & 1 & 1 & 1\\1&2&3&4\end{array}\right)$. Thanks to the answer I got yesterday, I know that $H=\left(\begin{array}{cccc} 1 & 9 & 1 & 0\\2&8&0&1\end{array}\right)$.
Say I receive the word $r=\left(\begin{array}{cccc} 0 & 4 & 6 & 9\end{array}\right)$, notice the error in the second position. Thus $rH^T=\left(\begin{array}{cc} 9 & 8\end{array}\right)$, which are my syndromes.
From Wikipedia we have the error locator polynomial $\Lambda(x)=1+\Lambda_1x^1+\dots+\Lambda_vx^v$ where the $\Lambda_i$ can be computed from:
Note that in my case $v=1$ so the system of equations is very simple $S_1\Lambda_1=-S_2$ or $9\Lambda_1=-8=3$ so $\Lambda_1=4$ and $\Lambda(x)=4x+1$.
The root of $\Lambda(x)$ I got as $8$.
Given the roots, I should be able to find the error locators. According to Wikipedia, "the error locators are the reciprocals of those roots", so my error locator is $8^{-1}=7$.
From there, Wikipedia says the error location is the log base $a$ of the locator. Since I did my encoding as $p(1),p(2),p(3),p(4)$ instead of $p(a^1),p(a^2),p(a^3),p(a^4)$ there really isn't an $a$.
Is decoding possible based on the way I did my encoding? Do I need to change my encoding method?
If I don't need to change, where do I go from here to find and correct the error? Have I done things right up to this point?
On a side note, I put my syndromes into the berlekamp-massey algorithm using Sage and got the polynomial $x+4$ in return which has $7$ as its root.