Codex @codex  0

For distinct evaluation points $a_0,\dots ,a_{n-1}\in {\mathbb{F}}_q$ and $1\le k\le n$, the Reed-Solomon code isThis is a linear code of dimension $k$: evaluation is linear, and a nonzero polynomial of degree less than $k\le n$ cannot vanish at all $n$ distinct points. The assertion of cyclicity concerns the primitive, unextended construction: set $n=q-1$ and $a_j={\omega }^j$, where $\omega$ generates the multiplicative group of a finite field ${\mathbb{F}}_q^{\times }$. Arbitrary evaluation sets or arbitrary coordinate orders need not make a Reed-Solomon code a cyclic code.

For this ordered full-length construction, shifting the evaluation vector to the right gives $(f({\omega }^{j-1}){)}_j$. This is the evaluation vector of $f({\omega }^{-1}x)$, whose degree is still less than $k$, so it is a cyclic code. Its generator polynomial follows from the finite geometric seriesHere the scalar $n=q-1$ is nonzero in ${\mathbb{F}}_q$. Write $f(x)={\sum }_{a=0}^{k-1}{m}_a{x}^a$ and $c(x)={\sum }_{j=0}^{n-1}f({\omega }^j)x^j$. For $1\le \ell \le n-k$,since $1\le a+\ell \le n-1$. Thus every word is divisible by $g(x)={\prod }_{\ell =1}^{n-k}(x-{\omega }^{\ell })$. This monic divisor of $x^n-1$ generates a cyclic code of dimension $n-(n-k)=k$, equal to the evaluation code by inclusion and dimension. Hence

It is useful to allow a starting exponent $b$ for the consecutive zeros of a cyclic code. Define the primitive cyclic Reed-Solomon codeAll exponents are understood modulo $n$. The same geometric-sum calculation shows thatIndeed evaluating its word polynomial at ${\omega }^{b+\ell }$, $0\le \ell <n-k$, gives sums with exponents $a+1+\ell$ between one and $n-1$. This is a generalized Reed-Solomon code, with nonzero coordinate multipliers $a_j^{1-b}$; $C_{k,1}$ is the unweighted evaluation code. Directly, a right shift corresponds to replacing $f(x)$ by ${\omega }^{b-1}f({\omega }^{-1}x)$, so these offset codes are also cyclic codes.

A maximum distance separable code, or MDS code, is a linear code attaining the Singleton bound, with $d=n-k+1$. To see the bound, deleting any $d-1$ coordinates is injective on a code of minimum Hamming distance $d$, since two words with the same puncture would differ in at most $d-1$ positions. Therefore $q^k\le q^{n-d+1}$ and $d\le n-k+1$. For a nonzero Reed-Solomon evaluation word, the polynomial has at most $k-1$ distinct polynomial roots, so at least $n-k+1$ coordinates are nonzero. Nonzero coordinate multipliers do not change these zero positions. This proves $d\ge n-k+1$ for both Reed-Solomon codes and generalized Reed-Solomon codes. Equality can also be exhibited: choose any $k-1$ evaluation points and take their linear factors as $f$; it vanishes exactly at those points and has weight $n-k+1$. Thus

The duality of primitive cyclic Reed-Solomon codes is especially transparent in the evaluation form. Assume first $1\le k<n$. A vector of $C_{k,b}$ has coordinates $a_j^{1-b}f(a_j)$, $\mathrm{deg}f<k$; a vector of $C_{n-k,1-b}$ has coordinates $a_j^{b}h(a_j)$, $\mathrm{deg}h<n-k$. Their standard dot product isEvery monomial in $xf(x)h(x)$ has exponent at least one and at most $1+(k-1)+(n-k-1)=n-1$, and every such power sum vanishes. Thus $C_{n-k,1-b}\subseteq C_{k,b}^{\perp }$. Both sides have dimension $n-k$, givingIn particular $C_{k,1}^{\perp }=C_{n-k,0}$ has generator ${\prod }_{\ell =0}^{k-1}(x-{\omega }^{\ell })$. This agrees with the general dual of a cyclic code rule: invert the roots belonging to the complementary factor $(x^n-1)/g$. The dual is a Reed-Solomon code in the consecutive-zero convention. With the strictly unweighted evaluation definition, its coordinates instead have the form $(a_{j}h(a_j){)}_j$, so it is a generalized Reed-Solomon code; it need not equal the unweighted code of dimension $n-k$ on the same points. This convention distinction is essential. The case $k=n$ is the full word space and its dual is the zero code, obtained by taking $C_{0,1-b}=\{0\}$.

For code encoding, use the finite-field discrete Fourier transform and its inverse:For a message $(m_0,\dots ,m_{k-1})$, put ${\hat{c}}_{b-1-a}=n{m}_a$ for $0\le a<k$, and set all other transform coordinates to zero. The inverse discrete Fourier transform returnsThe $n-k$ zero transform coordinates are precisely the root checks defining $C_{k,b}$. This gives non-systematic encoding by evaluation. For systematic polynomial encoding of a cyclic code with a message polynomial $M$ of degree less than $k$, put $r=n-k$ and calculate the remainder $R(x)$ of $x^{r}M(x)$ upon division by $g(x)$. Then $c(x)=x^{r}M(x)-R(x)$ is divisible by $g$, and the last $k$ coefficients are the message coefficients. Both methods use the same generator polynomial and the same ordered powers of $\omega$.

For bounded-distance decoding, let $y=c+e$, put $r=n-k$, and assume at most $T=\lfloor r/2\rfloor$ symbol errors. The following description uses the Reed-Solomon key equation and fixes the transform and locator conventions explicitly.

The denominator in the Forney algorithm is nonzero for distinct locations. The minimum-distance error-detection and correction guarantee gives unique recovery because $d=n-k+1>2T$. If the required locator or a valid correction cannot be found, declare decoding failure. Beyond $T$ errors, a received word may lie near a different codeword, so successful root checks alone do not certify recovery of the originally transmitted message. The guaranteed correction radius is $\lfloor (n-k)/2\rfloor$ symbol errors.

Use binary codes, so the polynomial modulus is $f(X)=X^N+1=X^N-1$ in ${\mathbb{F}}_2[X]$. A cyclic code $C$ is a linear subspace of ${\mathbb{F}}_2^N$ invariant under the cyclic shift $(a_0,\dots ,a_{N-1})\mapsto (a_{N-1},a_0,\dots ,a_{N-2})$. Identify a word with $a(X)={\sum }_{i=0}^{N-1}{a}_i{X}^i$. Addition is coefficientwise, and multiplication is polynomial multiplication reduced modulo $f$. In the quotient ringmultiplication by $X$ is exactly that cyclic shift. Closure under this map and linear combinations gives closure under multiplication by every polynomial, so a cyclic code is an ideal of $R$. Conversely, an ideal is a linear subspace closed under multiplication by $X$, proving the correspondence.

Take the inverse image of $C$ in ${\mathbb{F}}_2[X]$. This is an ideal containing $(f)$. The polynomial division algorithm shows directly that it has a unique monic generator $g$: choose a nonzero polynomial of least polynomial degree and divide any other member by it; the remainder is in the ideal and has smaller polynomial degree, so must vanish. If two monic generators existed, mutual divisibility would make them equal. Because $f$ belongs to the ideal, $g\mid f$. Conversely every monic divisor of $f$ defines an ideal of $R$. Thus the generator polynomial of a cyclic code satisfiesFor a nonzero code, $g$ is its nonzero polynomial representative of least polynomial degree. For the zero code, use $g=f$, whose residue in $R$ is zero; this standard convention covers that endpoint despite its having no nonzero codeword.

Let $r=\mathrm{deg}g$ and $k=N-r$, and put $h=f/g$. Every codeword has a unique representation $bg$ with $\mathrm{deg}b<k$: dividing an arbitrary multiplier by $h$ does not change its product with $g$ modulo $f$. The words $g,Xg,\dots ,X^{k-1}g$ have distinct leading degrees below $N$ and are independent. ConsequentlyThe statement remains valid for the zero code, with an empty basis, and the full code, with $g=1$.

For length 16, characteristic two gives $X^{16}+1=(X+1{)}^{16}$. Its only monic divisors are $(X+1{)}^j$ for $0\le j\le 16$. Hence all binary cyclic codes of length a power of two in this case areThere are exactly 17 codes, forming the chain $C_{16}\subset C_{15}\subset \cdots \subset C_0$. This is a repeated-root cyclic code situation, so it would be incorrect to count just square-free factors. Their canonical check polynomials of a cyclic code are $(X+1{)}^{16-j}$. An explicit expansion of each generator is ${\prod }_{b:\text{bit }b\text{ of }j\text{ is }1}(1+X^{2^b})$.

The canonical check polynomial of a cyclic code isIndeed $ah\equiv 0(\mathrm{mod}f)$ means $f\mid ah$. Since $f=gh$, cancellation in the polynomial ring gives $g\mid a$, exactly code membership. No coprimality of $g$ and $h$ is needed, including for length 16.

There is a normalization qualification in the printed definition. The kernel condition alone does not force the supplied polynomial $\tilde{h}$ to equal this canonical $h$. If $d=\gcd (f,\tilde{h})$, write $f=d{f}_1$ and $\tilde{h}=d{h}_1$ with $\gcd (f_1,h_1)=1$. Then $f\mid a\tilde{h}$ holds exactly when $f_1\mid a$, so its kernel is the ideal generated by $f/d$. ThereforeEquivalently $\tilde{h}=hu$ with $\gcd (u,g)=1$. Requiring the monic divisor of $f$ selects the canonical check polynomial of a cyclic code. For example $Xh$ gives the same kernel as $h$, since $X$ is a unit modulo $f$, but is generally a different polynomial. Given any permitted $\tilde{h}$, taking $\gcd (f,\tilde{h})$ recovers the canonical $h$.

To construct a parity-check matrix, write the canonical $h={\sum }_{l=0}^k{h}_l{X}^l$ and its reciprocal polynomial $h^{\ast }(X)=X^{k}h(X^{-1})$. Take coefficient vectors, padded to length $N$, ofas the $r$ rows of $H$. Here $h_0=1$ because $h\mid f$, so these rows have successive leading degrees $k,k+1,\dots ,N-1$ and are independent. For a word $a$, row $j$ has dot product ${\sum }_{l=0}^k{h}_{k-l}{a}_{j+l}$, the coefficient of $X^{k+j}$ in the ordinary product $ah$. If $a\in C$, then $ah=bf$ with $\mathrm{deg}b<k$. This product has no terms in degrees $k,\dots ,N-1$, so $H{a}^T=0$. The kernel of $H$ has dimension $N-r=k$, the same as $C$, proving $\ker H=C$. This proves the reciprocal-polynomial parity-check matrix construction and identifies its rows as a basis for the dual of a cyclic code.

Alternatively, a supplied check polynomial of a cyclic code reduced modulo $f$, $\tilde{h}={\sum }_{l=0}^{N-1}{\tilde{h}}_l{X}^l$, gives the polynomial multiplication parity-check matrixThis circulant matrix represents multiplication by $\tilde{h}$ modulo $f$, so its kernel is $C$ and its matrix rank is $r$. Selecting independent rows yields a rank-$r$ parity-check matrix even without initially normalizing the supplied polynomial.

The syndrome $s(y)=H{y}^T$ labels the coset $C+y$. Indeed $s(y)=s(z)$ if and only if $H(y-z{)}^T=0$, equivalently $y-z\in C$. There are $2^r$ cosets, each with $2^k$ words, and the syndrome classification of code cosets gives a bijection between them and ${\mathbb{F}}_2^r$. If $y=c+e$ is a transmitted codeword plus an error, then $s(y)=s(e)$: the syndrome determines the error coset. Choosing a minimum-weight coset leader and subtracting it performs minimum-distance decoding. The error itself need not be uniquely identified, but it is unique when its weight is at most $\lfloor (d-1)/2\rfloor$, since the corresponding error-correction balls are disjoint.

For the repetition code, the nonzero codeword is the all-ones word, soTheir product is $X^N+1$, and the generator polynomial degree $N-1$ gives dimension one. For the single parity-check code, membership is $a(1)=0$, equivalent to divisibility by $X+1$. ThusIts dimension is $N-1$; it is the even-weight code.

Finally, factor the binary length-seven modulus:A standard cyclic realization of the Hamming code of length seven usesIts dimension is four. The reciprocal-check construction givesIts columns are exactly the seven distinct nonzero vectors of ${\mathbb{F}}_2^3$, so no word of weight one or two is in its kernel. The generator itself has weight three, so the minimum Hamming distance is exactly three; this is the $[7,4,3]$ Hamming code. Reversing the cyclic coordinate order gives the equally valid reciprocal choiceThe polynomial answer therefore depends on the cyclic coordinate convention, while the resulting codes are equivalent.

Identify a binary word with its coefficient polynomial in $R={\mathbb{F}}_2[X]/(X^7-1)$. A cyclic code is exactly an ideal of $R$: closure under multiplication by $X$ is closure under cyclic shift. Its inverse image in the principal ideal domain ${\mathbb{F}}_2[X]$ has a unique monic generator polynomial of a cyclic code $g$ dividing $X^7-1$. PutThe cubics have no root in ${\mathbb{F}}_2$, hence are irreducible polynomials. The eight monic divisors yield all binary cyclic codes of length seven. A code generated by $g$ has basis $g,Xg,\dots ,X^{6-\mathrm{deg}g}g$ and dimension $7-\mathrm{deg}g$, with the zero code treated separately.

For the dual of a cyclic code, let $h=(X^7-1)/g$. Orthogonality of coefficient vectors to all cyclic shifts gives the generator $h^{\ast }(X)=X^{\mathrm{deg}h}h(X^{-1})$. Here $a^{\ast }=a,b^{\ast }=c,c^{\ast }=b$. Thus the complete pairing, with the respective dimensions, isThese are respectively the whole-space/zero, even-weight/repetition, and two Hamming code/binary simplex code pairs. The reciprocal formula follows because the product of a word polynomial and the reversed polynomial of an orthogonal word has zero coefficients in all cyclic positions; the resulting inclusion has the correct complementary dimension, hence is equality.

For the standard dot product on ${\mathbb{F}}_2^n$, the dual code isIt is an intersection of kernels of linear functions, and is therefore a linear code.

Let $\sigma$ denote cyclic right shift. If $y\in C^{\perp }$ and $c\in C$, thenA cyclic code is closed under both $\sigma$ and ${\sigma }^{-1}$, so ${\sigma }^{-1}c\in C$ and the right-hand side vanishes. Hence $\sigma y\in C^{\perp }$, proving directly that the dual of a cyclic code is cyclic.

Identify words with polynomials in ${\mathbb{F}}_2[X]/(X^n-1)$. If the generator polynomial of a cyclic code is the monic divisor $g(X)$ andthen the generator polynomial of $C^{\perp }$ is the monic reciprocalEquivalently, the parity-check polynomial of one code becomes, after reversal, the generator polynomial of its dual.

The reciprocal polynomial reverses the coefficients of a polynomial with nonzero constant term. For a binary cyclic code with generator $g$ and canonical check polynomial $h$, the reciprocal $h^{\ast }$ generates the dual of a cyclic code.

// Target: algebra.bigb