Constructing the binary Golay code
I'm reading up about the binary Golay code of length $23$. I know it's a cyclic code and I also know it's a quadratic residue code. I've read that we can consider the linear code over $mathbb{F}_2$ generated by the $23$ vectors $c_i$ with coordinates $(c_i)_j = 1$ if $(j-i)$ is a quadratic residue mod $23$ and $0$ otherwise.
Why is this true? How does this relate to it being a quadratic residue code? I thought that $x^{23} + 1 = (x+1)(x^{11}+x^{10}+x^6+x^5+x^4+x^2+1)(x^{11} + x^9 + x^7 + x^6 + x^5 + x + 1)$, then we can chose either of the polynomials of degree $11$ to be a generator polynomial. Clearly the construction above would yield a completely different generator matrix than these generator polynomials would give. How is this possible?
abstract-algebra information-theory coding-theory
asked Nov 27 at 0:56
the man
697715
add a comment |
I'm reading up about the binary Golay code of length $23$. I know it's a cyclic code and I also know it's a quadratic residue code. I've read that we can consider the linear code over $mathbb{F}_2$ generated by the $23$ vectors $c_i$ with coordinates $(c_i)_j = 1$ if $(j-i)$ is a quadratic residue mod $23$ and $0$ otherwise.
Why is this true? How does this relate to it being a quadratic residue code? I thought that $x^{23} + 1 = (x+1)(x^{11}+x^{10}+x^6+x^5+x^4+x^2+1)(x^{11} + x^9 + x^7 + x^6 + x^5 + x + 1)$, then we can chose either of the polynomials of degree $11$ to be a generator polynomial. Clearly the construction above would yield a completely different generator matrix than these generator polynomials would give. How is this possible?
abstract-algebra information-theory coding-theory
asked Nov 27 at 0:56
the man
697715
add a comment |
I'm reading up about the binary Golay code of length $23$. I know it's a cyclic code and I also know it's a quadratic residue code. I've read that we can consider the linear code over $mathbb{F}_2$ generated by the $23$ vectors $c_i$ with coordinates $(c_i)_j = 1$ if $(j-i)$ is a quadratic residue mod $23$ and $0$ otherwise.
Why is this true? How does this relate to it being a quadratic residue code? I thought that $x^{23} + 1 = (x+1)(x^{11}+x^{10}+x^6+x^5+x^4+x^2+1)(x^{11} + x^9 + x^7 + x^6 + x^5 + x + 1)$, then we can chose either of the polynomials of degree $11$ to be a generator polynomial. Clearly the construction above would yield a completely different generator matrix than these generator polynomials would give. How is this possible?
abstract-algebra information-theory coding-theory
asked Nov 27 at 0:56
the man
697715
I'm reading up about the binary Golay code of length $23$. I know it's a cyclic code and I also know it's a quadratic residue code. I've read that we can consider the linear code over $mathbb{F}_2$ generated by the $23$ vectors $c_i$ with coordinates $(c_i)_j = 1$ if $(j-i)$ is a quadratic residue mod $23$ and $0$ otherwise.
Why is this true? How does this relate to it being a quadratic residue code? I thought that $x^{23} + 1 = (x+1)(x^{11}+x^{10}+x^6+x^5+x^4+x^2+1)(x^{11} + x^9 + x^7 + x^6 + x^5 + x + 1)$, then we can chose either of the polynomials of degree $11$ to be a generator polynomial. Clearly the construction above would yield a completely different generator matrix than these generator polynomials would give. How is this possible?
abstract-algebra information-theory coding-theory
abstract-algebra information-theory coding-theory
asked Nov 27 at 0:56
the man
697715
asked Nov 27 at 0:56
the man
697715
asked Nov 27 at 0:56
the man
697715
asked Nov 27 at 0:56
the man
697715
asked Nov 27 at 0:56
the man
697715
697715
add a comment |
add a comment |
1 Answer
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The generator matrix of a linear code is not unique.
An $[n,k,d]_{q}$ code is a $k$-dimensional subspace of $mathbb{F}_{q}^{n}$, and so there are
$$prod_{i=0}^{k-1} (q^{k}-q^{i})$$
different ordered bases;
each one of these will give you a different generator matrix.
And that is just the number of generator matrices for a given code,
you have even more if you consider different but equivalent codes.
So choosing an irreducible $11$-degree factor of $x^{23}-1$ will give you a cyclic basis for your code, giving you a cyclic generator matrix. Using the quadratic residue construction gives you a different generator matrix, will most likely even give you a different code than you obtain from the cyclic construction (as in, a different set of codewords), but the codes are equivalent (ie can be obtained from each other by a permutation of coordinates).
answered Dec 2 at 1:43
Morgan Rodgers
9,57021439
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
oldest
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active
oldest
votes
The generator matrix of a linear code is not unique.
An $[n,k,d]_{q}$ code is a $k$-dimensional subspace of $mathbb{F}_{q}^{n}$, and so there are
$$prod_{i=0}^{k-1} (q^{k}-q^{i})$$
different ordered bases;
each one of these will give you a different generator matrix.
And that is just the number of generator matrices for a given code,
you have even more if you consider different but equivalent codes.
So choosing an irreducible $11$-degree factor of $x^{23}-1$ will give you a cyclic basis for your code, giving you a cyclic generator matrix. Using the quadratic residue construction gives you a different generator matrix, will most likely even give you a different code than you obtain from the cyclic construction (as in, a different set of codewords), but the codes are equivalent (ie can be obtained from each other by a permutation of coordinates).
answered Dec 2 at 1:43
Morgan Rodgers
9,57021439
add a comment |
The generator matrix of a linear code is not unique.
An $[n,k,d]_{q}$ code is a $k$-dimensional subspace of $mathbb{F}_{q}^{n}$, and so there are
$$prod_{i=0}^{k-1} (q^{k}-q^{i})$$
different ordered bases;
each one of these will give you a different generator matrix.
And that is just the number of generator matrices for a given code,
you have even more if you consider different but equivalent codes.
So choosing an irreducible $11$-degree factor of $x^{23}-1$ will give you a cyclic basis for your code, giving you a cyclic generator matrix. Using the quadratic residue construction gives you a different generator matrix, will most likely even give you a different code than you obtain from the cyclic construction (as in, a different set of codewords), but the codes are equivalent (ie can be obtained from each other by a permutation of coordinates).
answered Dec 2 at 1:43
Morgan Rodgers
9,57021439
add a comment |
The generator matrix of a linear code is not unique.
An $[n,k,d]_{q}$ code is a $k$-dimensional subspace of $mathbb{F}_{q}^{n}$, and so there are
$$prod_{i=0}^{k-1} (q^{k}-q^{i})$$
different ordered bases;
each one of these will give you a different generator matrix.
And that is just the number of generator matrices for a given code,
you have even more if you consider different but equivalent codes.
So choosing an irreducible $11$-degree factor of $x^{23}-1$ will give you a cyclic basis for your code, giving you a cyclic generator matrix. Using the quadratic residue construction gives you a different generator matrix, will most likely even give you a different code than you obtain from the cyclic construction (as in, a different set of codewords), but the codes are equivalent (ie can be obtained from each other by a permutation of coordinates).
answered Dec 2 at 1:43
Morgan Rodgers
9,57021439
The generator matrix of a linear code is not unique.
An $[n,k,d]_{q}$ code is a $k$-dimensional subspace of $mathbb{F}_{q}^{n}$, and so there are
$$prod_{i=0}^{k-1} (q^{k}-q^{i})$$
different ordered bases;
each one of these will give you a different generator matrix.
And that is just the number of generator matrices for a given code,
you have even more if you consider different but equivalent codes.
So choosing an irreducible $11$-degree factor of $x^{23}-1$ will give you a cyclic basis for your code, giving you a cyclic generator matrix. Using the quadratic residue construction gives you a different generator matrix, will most likely even give you a different code than you obtain from the cyclic construction (as in, a different set of codewords), but the codes are equivalent (ie can be obtained from each other by a permutation of coordinates).
answered Dec 2 at 1:43
Morgan Rodgers
9,57021439
answered Dec 2 at 1:43
Morgan Rodgers
9,57021439
answered Dec 2 at 1:43
Morgan Rodgers
9,57021439
answered Dec 2 at 1:43
Morgan Rodgers
9,57021439
9,57021439
add a comment |
add a comment |
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