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I' m not getting how syndromes are calculated for bch codes so I tried finding examples but still I don't seem to have it

To calculate the first syndrome for the received message polynomial $R(x)=1+X^8+X^{11}+X^{14}$ in the (15,7) minimal polynomial code, we use the minimal polynomial $ϕ_1(x)=1+X^3+X^4$ in the finite field $GF(2^4​)$.

To determine the syndrome $S = (S_1, S_2, S_3, S_4)$ the $R(X)$ is divided by each of the minimal polynomials, for $ϕ_1(X)$ the remainder is $b_1(X) = (1 + X^2 +X^3)$

$S_1 =b_1(\alpha)= \alpha^{11}$

Dividing $R(X)$ by $ϕ_1(X)$ does not give me the same remainder (using elements with the primitive irreducible polynomial $X^4+x^3+1$ ) . I have $1+X^6$

Edit : the full example here (I summarized with some typos above)

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user159729
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    What is $\alpha$? My guess would be that it is a root of $\phi_1(x)$, but do tell. Why do you have both $X$ and $x$ appearing? Are they to mean the same thing? I have never heard of a minimal polynomial code and I have worked with these objects for a few decades :-) – Jyrki Lahtonen Jul 12 '24 at 03:34
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    What do you mean with $1+>X^2+X^3$? – Jyrki Lahtonen Jul 12 '24 at 03:38
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    Anyway, I think that the remainder of $R(x)$ modulo $1+x^3+x^4$ is $x+x^2+x^3$, which is also the remainder of $1+x^6$, so you have probably done some calculation correctly. The remainder of $x^11$, OTOH, is $1+x^2+x^3$, so $\alpha^{11}=1+\alpha^2+\alpha^3$. – Jyrki Lahtonen Jul 12 '24 at 03:44
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    You are probably trying to use the BCH-code with generator polynomial $g(x)=m_1(x)m_3(x)$, where $m_3(x)$ is the minimal polynomial of $\alpha^3$. Because $\alpha^3$ has order five, its minimal polynomial is $m_3(x)=1+x+x^2+x^3+x^4$. – Jyrki Lahtonen Jul 12 '24 at 03:47
  • I can't shake the feeling that you are not comfortable enough with finite field arithmetic. Alas, my prepared example uses the minimal polynomial $1+x+x^4$ to define this field. In other words, the element $\gamma$ in my answer has the role of $\alpha^{-1}$ in your notation. Or, equivalently, your $\alpha$ is my $\gamma^{-1}=\gamma^{14}$. – Jyrki Lahtonen Jul 12 '24 at 03:50
  • it's a typo it's $1+X^2+X^3$ – user159729 Jul 12 '24 at 08:18
  • $\alpha$ is the primitive irreducible polynomial root – user159729 Jul 12 '24 at 08:28

1 Answers1

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The question matches the changes I made to my test code: $GF(2^4):1 + x^3 + x^4$, $m_1 m_3 = 1 + x + x^2 + x^4 + x^8$, $v(x) = x^2 + x^5 + x^8 + x^{11} + x^{14}$

I'm not aware of calculating syndromes from the minimum polynomials. Instead my code calculates syndromes by treating each bit of a message as if it was a 4 bit element of $GF(2^4)$ with values 0 or 1: For i = 1 to 4: $S(i) = R(x) \ \bmod \ (x - \alpha^i) = R(\alpha^i)$, which matches the Wiki article:

https://en.wikipedia.org/wiki/BCH_code#Calculate_the_syndromes

For $R(x)=1+x^8+x^{11}+x^{14}$, $E(x) = 1 + x^2 + x^5$, but BCH(15,7) code can only correct 2 errors (or 1 error + 2 erasures or 4 erasures). Syndromes: $\alpha^8, \alpha^1, \alpha^6, \alpha^2$ = $1110_2, 0010_2, 1111_2, 0100_2$ .

Although not normally used for decoding:

$R(x) \bmod m_1(x) = E(x) \bmod m_1(x) = \alpha^8 = 1110_2$

$R(x) \bmod m_3(x) = E(x) \bmod m_3(x) = \alpha^2 = 0100_2$

rcgldr
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