# Finding BCH code syndromes

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)

• 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 :-) Commented Jul 12 at 3:34
• What do you mean with $1+>X^2+X^3$? Commented Jul 12 at 3:38
• 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 $\alpha^{11}=1+\alpha^2+\alpha^3$. Commented Jul 12 at 3:44
• 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 $m_3(x)=1+x+x^2+x^3+x^4$. Commented Jul 12 at 3:47
• 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 $\gamma^{-1}=\gamma^{14}$. Commented Jul 12 at 3:50

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:

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

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

• I added the complete example Commented Jul 12 at 8:40
• @user159729 - I updated my answer. Commented Jul 12 at 9:13
• +1 for going through all, and making sense of the question. Commented Jul 13 at 3:27
• I think the example I found is simply wrong ,the remainder is wrong . Is it not true Syndromes $S_i=R(\alpha^i)$ as α is the primitive irreducible polynomial root ? Commented Jul 14 at 1:30
• As mentioned in my answer and in the Wiki article, $S_i = R(\alpha^i)$. The irreducible polynomial does not have to be primitive. BCH code will work with $GF(2^4): x^4 + x^3 + x^2 + x + 1$, with $\alpha = x+1 = 0011_2$ . Commented Jul 14 at 4:08