Mixing
Show that a binary perfect linear $[n,k,d]$ code is generated by its words of minimal weight.
$begingroup$
Question
Show that a binary perfect linear $[n,k,d]$ code is generated by its words of minimal weight.
What I have so far
We can try to solve this by induction on the weight of the words in the code.
Induction base: if a word has weight $d$, then it generates itself and has minimal weight.
Induction step: now suppose that we know that for every word of weight $l>d$ we know that it is generated by words of minimal weight. We now look at the words of weight $l+1$.
This is the hardest step of the proof and I'm stuck here. Any suggestions?
coding-theory
edited Jan 7 at 14:49
amWhy
1
asked Jan 7 at 14:26
xzeoxzeo
388111
$endgroup$
add a comment |
$begingroup$
Question
Show that a binary perfect linear $[n,k,d]$ code is generated by its words of minimal weight.
What I have so far
We can try to solve this by induction on the weight of the words in the code.
Induction base: if a word has weight $d$, then it generates itself and has minimal weight.
Induction step: now suppose that we know that for every word of weight $l>d$ we know that it is generated by words of minimal weight. We now look at the words of weight $l+1$.
This is the hardest step of the proof and I'm stuck here. Any suggestions?
coding-theory
edited Jan 7 at 14:49
amWhy
1
asked Jan 7 at 14:26
xzeoxzeo
388111
$endgroup$
1
$begingroup$
You don't seem to be using the "perfect" property anywhere...
$endgroup$
– leonbloy
Jan 7 at 14:44
$begingroup$
Yes I've noticed. Don't really know how to use it here though..
$endgroup$
– xzeo
Jan 7 at 14:48
add a comment |
$begingroup$
Question
Show that a binary perfect linear $[n,k,d]$ code is generated by its words of minimal weight.
What I have so far
We can try to solve this by induction on the weight of the words in the code.
Induction base: if a word has weight $d$, then it generates itself and has minimal weight.
Induction step: now suppose that we know that for every word of weight $l>d$ we know that it is generated by words of minimal weight. We now look at the words of weight $l+1$.
This is the hardest step of the proof and I'm stuck here. Any suggestions?
coding-theory
edited Jan 7 at 14:49
amWhy
1
asked Jan 7 at 14:26
xzeoxzeo
388111
$endgroup$
Question
Show that a binary perfect linear $[n,k,d]$ code is generated by its words of minimal weight.
What I have so far
We can try to solve this by induction on the weight of the words in the code.
Induction base: if a word has weight $d$, then it generates itself and has minimal weight.
Induction step: now suppose that we know that for every word of weight $l>d$ we know that it is generated by words of minimal weight. We now look at the words of weight $l+1$.
This is the hardest step of the proof and I'm stuck here. Any suggestions?
coding-theory
coding-theory
edited Jan 7 at 14:49
amWhy
1
asked Jan 7 at 14:26
xzeoxzeo
388111
edited Jan 7 at 14:49
amWhy
1
asked Jan 7 at 14:26
xzeoxzeo
388111
edited Jan 7 at 14:49
amWhy
1
edited Jan 7 at 14:49
amWhy
1
edited Jan 7 at 14:49
amWhy
1
1
asked Jan 7 at 14:26
xzeoxzeo
388111
asked Jan 7 at 14:26
xzeoxzeo
388111
asked Jan 7 at 14:26
xzeoxzeo
388111
388111
1
$begingroup$
You don't seem to be using the "perfect" property anywhere...
$endgroup$
– leonbloy
Jan 7 at 14:44
$begingroup$
Yes I've noticed. Don't really know how to use it here though..
$endgroup$
– xzeo
Jan 7 at 14:48
add a comment |
1
$begingroup$
You don't seem to be using the "perfect" property anywhere...
$endgroup$
– leonbloy
Jan 7 at 14:44
$begingroup$
Yes I've noticed. Don't really know how to use it here though..
$endgroup$
– xzeo
Jan 7 at 14:48
1
1
$begingroup$
You don't seem to be using the "perfect" property anywhere...
$endgroup$
– leonbloy
Jan 7 at 14:44
$begingroup$
You don't seem to be using the "perfect" property anywhere...
$endgroup$
– leonbloy
Jan 7 at 14:44
$begingroup$
Yes I've noticed. Don't really know how to use it here though..
$endgroup$
– xzeo
Jan 7 at 14:48
$begingroup$
Yes I've noticed. Don't really know how to use it here though..
$endgroup$
– xzeo
Jan 7 at 14:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Ok, so $d=2t+1$.
A plan of attack for the inductive step. Justify everything.
- Let $x$ be a codeword of weight $ell+1$. Let $y$ be a vector formed by selecting $t+1$ of the $1$s in $x$, and setting the rest of the components to zero. In other words, $y$ has weight $t+1$, and $d(y,x)=ell-t$.
- Perfectness implies that there exists a codeword $z$ such that $d(y,z)le t$. Then $z$ must have the minimum weight $2t+1$.
- It follows that the codeword $x-z$ has weight $le ell$. The induction step follows from this.
answered Jan 7 at 18:54
Jyrki LahtonenJyrki Lahtonen
109k13169371
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add a comment |
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$begingroup$
Ok, so $d=2t+1$.
A plan of attack for the inductive step. Justify everything.
- Let $x$ be a codeword of weight $ell+1$. Let $y$ be a vector formed by selecting $t+1$ of the $1$s in $x$, and setting the rest of the components to zero. In other words, $y$ has weight $t+1$, and $d(y,x)=ell-t$.
- Perfectness implies that there exists a codeword $z$ such that $d(y,z)le t$. Then $z$ must have the minimum weight $2t+1$.
- It follows that the codeword $x-z$ has weight $le ell$. The induction step follows from this.
answered Jan 7 at 18:54
Jyrki LahtonenJyrki Lahtonen
109k13169371
$endgroup$
add a comment |
$begingroup$
Ok, so $d=2t+1$.
A plan of attack for the inductive step. Justify everything.
- Let $x$ be a codeword of weight $ell+1$. Let $y$ be a vector formed by selecting $t+1$ of the $1$s in $x$, and setting the rest of the components to zero. In other words, $y$ has weight $t+1$, and $d(y,x)=ell-t$.
- Perfectness implies that there exists a codeword $z$ such that $d(y,z)le t$. Then $z$ must have the minimum weight $2t+1$.
- It follows that the codeword $x-z$ has weight $le ell$. The induction step follows from this.
answered Jan 7 at 18:54
Jyrki LahtonenJyrki Lahtonen
109k13169371
$endgroup$
add a comment |
$begingroup$
Ok, so $d=2t+1$.
A plan of attack for the inductive step. Justify everything.
- Let $x$ be a codeword of weight $ell+1$. Let $y$ be a vector formed by selecting $t+1$ of the $1$s in $x$, and setting the rest of the components to zero. In other words, $y$ has weight $t+1$, and $d(y,x)=ell-t$.
- Perfectness implies that there exists a codeword $z$ such that $d(y,z)le t$. Then $z$ must have the minimum weight $2t+1$.
- It follows that the codeword $x-z$ has weight $le ell$. The induction step follows from this.
answered Jan 7 at 18:54
Jyrki LahtonenJyrki Lahtonen
109k13169371
$endgroup$
Ok, so $d=2t+1$.
A plan of attack for the inductive step. Justify everything.
- Let $x$ be a codeword of weight $ell+1$. Let $y$ be a vector formed by selecting $t+1$ of the $1$s in $x$, and setting the rest of the components to zero. In other words, $y$ has weight $t+1$, and $d(y,x)=ell-t$.
- Perfectness implies that there exists a codeword $z$ such that $d(y,z)le t$. Then $z$ must have the minimum weight $2t+1$.
- It follows that the codeword $x-z$ has weight $le ell$. The induction step follows from this.
answered Jan 7 at 18:54
Jyrki LahtonenJyrki Lahtonen
109k13169371
answered Jan 7 at 18:54
Jyrki LahtonenJyrki Lahtonen
109k13169371
answered Jan 7 at 18:54
Jyrki LahtonenJyrki Lahtonen
109k13169371
answered Jan 7 at 18:54
Jyrki LahtonenJyrki Lahtonen
109k13169371
109k13169371
add a comment |
add a comment |
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1
$begingroup$
You don't seem to be using the "perfect" property anywhere...
$endgroup$
– leonbloy
Jan 7 at 14:44
$begingroup$
Yes I've noticed. Don't really know how to use it here though..
$endgroup$
– xzeo
Jan 7 at 14:48