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Consider a Reed-Solomon code over a finite field of $\mathbb{F}_q$. Why is the typical block size chosen to be $q-1$ [1][2][3]? The reasoning I saw around this is that in order to maximize the rate for a fixed minimum distance. But wouldn't choosing $q$ as the code size would be even slightly better? Does it have to do something with how BCH decoders work? Wikipedia lists both $q$ and $q-1$ as a possible good choice for the block size, but doesn't cite any reference.

[1]: Manz: Fehlerkorrigierende Codes 
[2]: Pretzel: Error-correcting Codes and Finite Fields 
[3]: Hoffman: Coding Theory: The Essentials
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  • $\begingroup$ did you see my updated answer? $\endgroup$
    – kodlu
    Dec 19, 2023 at 13:26

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Reed Solomon codes are cyclic. A cyclic code [over whatever Galois field] of length $n$ is defined with respect to reduction modulo $x^n-1$ to get the cyclic property.

Now, the generator polynomial $g$ and the parity check polynomials $h$ are defined by $g(x)h(x)=x^n-1,$ more exactly one is chosen and then the other one is determined using this equation.

The factorization of $x^n-1$ is crucial in determining the properties of the code. The nonzero elements of a finite field $GF(q)$ obey the equation $x^{q-1}-1=0,$ and form the multiplicative group $GF(q)^{\ast}.$

Thus the only choice is to make the length of the code either $q-1$ or a divisor of $q-1.$ This is because, it is known that $x^n-1$ is divisible by $x^m-1$ if and only if $n$ is divisible by $m.$ Usually we take $n=q-1,$ for efficiency since it gives the longest code for a fixed alphabet size.

Remark: For Reed-Solomon which have alphabet $GF(q)$ the factorization is trivial (no cyclotomic cosets need to be computed which are used to determine minimum distance for binary codes). So we take $g$ to have a sequence of consecutive powers of $\alpha$ the primitive element as its roots.

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