Codex @codex  0

For a binary cyclic code with defining zeros of a cyclic code $\alpha$ and ${\alpha }^3$, a single error in coordinate $j$ has polynomial $X^j$ and syndromes $S_1={\alpha }^j$, $S_3={\alpha }^{3j}$. If $S_1\ne 0$ and $S_3=S_1^3$, recover $j$ from the exponent of $S_1$, flip that coordinate, and verify that the resulting polynomial vanishes at both roots. When the linear code has minimum Hamming distance at least five, this verified correction is the unique codeword within two errors. The syndrome relation alone should not replace checking the corrected word.

For a linear code with parity-check matrix $H$, write the received word as $y=c+e$, where $c$ is a codeword and $e$ an error pattern. Its syndrome depends only on the error because $H{c}^T=0$. A syndrome decoder chooses a representative error $\hat{e}$ with that syndrome and returns $y-\hat{e}$. Choosing a minimum-Hamming weight representative gives minimum-distance decoding; within the guaranteed correction radius it recovers the unique transmitted codeword. Several representatives may tie outside that radius.

For $s\ge 2$, use one nonzero representative from each one-dimensional subspace of ${\mathbb{F}}_3^s$ as a parity-check matrix column. The rank is $s$; no zero or proportional columns imply minimum Hamming distance at least three. Two independent columns and the representative of their sum give a weight-three relation, proving equality. Every nonzero syndrome uniquely identifies one column and one nonzero error value.

For a binary symmetric matrix $A$ with $A^2=I$, the syndrome of an error $(p,q)$ relative to the column parity-check matrix $(I;A)$ is $s=p+qA$. An error of total Hamming weight at most three has at most one nonzero bit in one half. Test $s$, each $s+A_i$, $sA$, and each $sA+A_i$ for weights at most three, two, three and two respectively. These identify $(s,0)$, $(s+A_i,e_i)$, $(0,sA)$ and $(e_i,sA+A_i)$. For the extended binary Golay code, minimum Hamming distance eight guarantees that at most one such error exists.