Mixing
Dimension of cyclic codes
Clash Royale CLAN TAG#URR8PPP
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According to most texts, this is a trivial matter. If $g(x)$ is a generator polynomial of degree $n-k$ of a length $n$ cyclic code $C$, then $(a_0+ ldots +a_k-1X^k-1)g(x) mod X^n - 1$ will give all possible values of $(b_0+ ldots +b_n-1X^n-1)g(x) mod X^n - 1$. I really don't see how this is trivial at all. Any thoughts?
coding-theory
asked Aug 6 at 14:07
Romanda de Gore
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According to most texts, this is a trivial matter. If $g(x)$ is a generator polynomial of degree $n-k$ of a length $n$ cyclic code $C$, then $(a_0+ ldots +a_k-1X^k-1)g(x) mod X^n - 1$ will give all possible values of $(b_0+ ldots +b_n-1X^n-1)g(x) mod X^n - 1$. I really don't see how this is trivial at all. Any thoughts?
coding-theory
asked Aug 6 at 14:07
Romanda de Gore
588
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
According to most texts, this is a trivial matter. If $g(x)$ is a generator polynomial of degree $n-k$ of a length $n$ cyclic code $C$, then $(a_0+ ldots +a_k-1X^k-1)g(x) mod X^n - 1$ will give all possible values of $(b_0+ ldots +b_n-1X^n-1)g(x) mod X^n - 1$. I really don't see how this is trivial at all. Any thoughts?
coding-theory
asked Aug 6 at 14:07
Romanda de Gore
588
According to most texts, this is a trivial matter. If $g(x)$ is a generator polynomial of degree $n-k$ of a length $n$ cyclic code $C$, then $(a_0+ ldots +a_k-1X^k-1)g(x) mod X^n - 1$ will give all possible values of $(b_0+ ldots +b_n-1X^n-1)g(x) mod X^n - 1$. I really don't see how this is trivial at all. Any thoughts?
coding-theory
asked Aug 6 at 14:07
Romanda de Gore
588
edited Aug 6 at 21:52
asked Aug 6 at 14:07
Romanda de Gore
588
asked Aug 6 at 14:07
Romanda de Gore
588
asked Aug 6 at 14:07
Romanda de Gore
588
588
add a comment |Â
add a comment |Â
1 Answer
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Let $c in C$. Then $c=bg$ where $b in (GF(q))^n$. This means $c(X)=b(X)g(X) mod X^n-1$, where, if $l=(l_0, ldots,l_n-1)$, then $l(X)=l_0+ ldots + l_n-1X^n-1$. So, we have, $$c(X)=b(X)g(X) + q(X)(X^n-1)$$ We know that $g(X)$ divides $X^n -1$, say $g(X)h(X)=X^n-1$, hence $$c(X)=b(X)g(X) + q(X)h(X)g(X)$$ Therefore $c(X)=(b(X) +q(X)h(X))g(X)$. This means that $deg(b(X) +q(X)h(X))$ is at most $k-1$, hence the $b(X) +q(X)h(X)$ is the sought after $a_0 + ldots + a_k-1X^k-1$
answered Aug 6 at 22:10
Romanda de Gore
588
add a comment |Â
1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Let $c in C$. Then $c=bg$ where $b in (GF(q))^n$. This means $c(X)=b(X)g(X) mod X^n-1$, where, if $l=(l_0, ldots,l_n-1)$, then $l(X)=l_0+ ldots + l_n-1X^n-1$. So, we have, $$c(X)=b(X)g(X) + q(X)(X^n-1)$$ We know that $g(X)$ divides $X^n -1$, say $g(X)h(X)=X^n-1$, hence $$c(X)=b(X)g(X) + q(X)h(X)g(X)$$ Therefore $c(X)=(b(X) +q(X)h(X))g(X)$. This means that $deg(b(X) +q(X)h(X))$ is at most $k-1$, hence the $b(X) +q(X)h(X)$ is the sought after $a_0 + ldots + a_k-1X^k-1$
answered Aug 6 at 22:10
Romanda de Gore
588
add a comment |Â
up vote
0
down vote
Let $c in C$. Then $c=bg$ where $b in (GF(q))^n$. This means $c(X)=b(X)g(X) mod X^n-1$, where, if $l=(l_0, ldots,l_n-1)$, then $l(X)=l_0+ ldots + l_n-1X^n-1$. So, we have, $$c(X)=b(X)g(X) + q(X)(X^n-1)$$ We know that $g(X)$ divides $X^n -1$, say $g(X)h(X)=X^n-1$, hence $$c(X)=b(X)g(X) + q(X)h(X)g(X)$$ Therefore $c(X)=(b(X) +q(X)h(X))g(X)$. This means that $deg(b(X) +q(X)h(X))$ is at most $k-1$, hence the $b(X) +q(X)h(X)$ is the sought after $a_0 + ldots + a_k-1X^k-1$
answered Aug 6 at 22:10
Romanda de Gore
588
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Let $c in C$. Then $c=bg$ where $b in (GF(q))^n$. This means $c(X)=b(X)g(X) mod X^n-1$, where, if $l=(l_0, ldots,l_n-1)$, then $l(X)=l_0+ ldots + l_n-1X^n-1$. So, we have, $$c(X)=b(X)g(X) + q(X)(X^n-1)$$ We know that $g(X)$ divides $X^n -1$, say $g(X)h(X)=X^n-1$, hence $$c(X)=b(X)g(X) + q(X)h(X)g(X)$$ Therefore $c(X)=(b(X) +q(X)h(X))g(X)$. This means that $deg(b(X) +q(X)h(X))$ is at most $k-1$, hence the $b(X) +q(X)h(X)$ is the sought after $a_0 + ldots + a_k-1X^k-1$
answered Aug 6 at 22:10
Romanda de Gore
588
Let $c in C$. Then $c=bg$ where $b in (GF(q))^n$. This means $c(X)=b(X)g(X) mod X^n-1$, where, if $l=(l_0, ldots,l_n-1)$, then $l(X)=l_0+ ldots + l_n-1X^n-1$. So, we have, $$c(X)=b(X)g(X) + q(X)(X^n-1)$$ We know that $g(X)$ divides $X^n -1$, say $g(X)h(X)=X^n-1$, hence $$c(X)=b(X)g(X) + q(X)h(X)g(X)$$ Therefore $c(X)=(b(X) +q(X)h(X))g(X)$. This means that $deg(b(X) +q(X)h(X))$ is at most $k-1$, hence the $b(X) +q(X)h(X)$ is the sought after $a_0 + ldots + a_k-1X^k-1$
answered Aug 6 at 22:10
Romanda de Gore
588
answered Aug 6 at 22:10
Romanda de Gore
588
answered Aug 6 at 22:10
Romanda de Gore
588
answered Aug 6 at 22:10
Romanda de Gore
588
588
add a comment |Â
add a comment |Â
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