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In a given $\mathrm{BCH}(N,K)$, $T=3$ code over $\mathrm{GF}(2^m)$, there are ways to find the error locations in a given $N$-bit codeword directly from the syndromes without going through the normal Berlekamp-Massey and Chien search steps.

Assuming that $(S1, S3, S5)$ are the syndromes for $a$, $a^3$ and $a^5$ I now know that I can calculate the number of bit errors as follows: Calculate $D3=S1^3 + S3$ and $D5=S1^5 + S5$, then: If $S1 = S3 = S5 = 0$ we have 0 errors, if $S1 \neq 0$ and $D3 = D5 = 0$ we have 1 error, if $S1 \neq 0$ and $D3 \neq 0$ and $S1D5 = S3D3$ we have 2 errors, otherwise we have 3 errors.

What I don't know is once we know how many errors we have, what is the most efficient way (possibly using some look-up table) to find the actual bit error locations in the codeword.

I would appreciate it very much if someone could enlighten me on this.

Regards,

-Dimitri

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    $\begingroup$ If there is a single error, then $S_1$ gives the error position. Need to think about the other cases. I suspect that you can find the error locator polynomial in those cases. But solving the quadratic equation (provided it has the required solutions) needs the half-trace, and solving a cubic is even more cumbersome. It is also possible that more than three errors occured. Therefore I am leaning towards keeping the Chien search, as it will catch those occasions, where we have the error locator, but it does not have required zeros. $\endgroup$
    – Jyrki Lahtonen
    Commented May 16, 2014 at 19:11

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