Underbound length of a ternary linear code
0
$begingroup$
I have to prove the following theorem:
Let $C$ be a ternary linear $[n,2,d]$-code. Prove that $d+ d/3 leq n$.
Now I don't really see what the best approach for this would be. I was thinking maybe the generator matrix, but I didn't really get anywhere when trying this. Any tips?
coding-theory
asked Jan 8 at 18:41
xzeoxzeo
388111
$endgroup$
add a comment |
0
$begingroup$
I have to prove the following theorem:
Let $C$ be a ternary linear $[n,2,d]$-code. Prove that $d+ d/3 leq n$.
Now I don't really see what the best approach for this would be. I was thinking maybe the generator matrix, but I didn't really get anywhere when trying this. Any tips?
coding-theory
asked Jan 8 at 18:41
xzeoxzeo
388111
$endgroup$
$begingroup$
$d+ d/3 leq n$ looks weird, as it would be simpler to write $ frac43 d leq n$ Are you sure you got it right?
$endgroup$
– leonbloy
Jan 8 at 20:47
$begingroup$
Yes I'm sure...
$endgroup$
– xzeo
Jan 9 at 11:08
$begingroup$
Hint: consider the possible columns of $G$ and how many linear combinatios produce a zero in each position....
$endgroup$
– leonbloy
Jan 9 at 13:43
2
$begingroup$
This is a special case of the Griesmer bound. The 2-dimensional case can, indeed, be proven the way leonbloy indicated. But the general result isn't very difficult either.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 15:25
add a comment |
0
0
0
$begingroup$
I have to prove the following theorem:
Let $C$ be a ternary linear $[n,2,d]$-code. Prove that $d+ d/3 leq n$.
Now I don't really see what the best approach for this would be. I was thinking maybe the generator matrix, but I didn't really get anywhere when trying this. Any tips?
coding-theory
asked Jan 8 at 18:41
xzeoxzeo
388111
$endgroup$
I have to prove the following theorem:
Let $C$ be a ternary linear $[n,2,d]$-code. Prove that $d+ d/3 leq n$.
Now I don't really see what the best approach for this would be. I was thinking maybe the generator matrix, but I didn't really get anywhere when trying this. Any tips?
coding-theory
coding-theory
asked Jan 8 at 18:41
xzeoxzeo
388111
asked Jan 8 at 18:41
xzeoxzeo
388111
asked Jan 8 at 18:41
xzeoxzeo
388111
asked Jan 8 at 18:41
xzeoxzeo
388111
asked Jan 8 at 18:41
xzeoxzeo
388111
388111
$begingroup$
$d+ d/3 leq n$ looks weird, as it would be simpler to write $ frac43 d leq n$ Are you sure you got it right?
$endgroup$
– leonbloy
Jan 8 at 20:47
$begingroup$
Yes I'm sure...
$endgroup$
– xzeo
Jan 9 at 11:08
$begingroup$
Hint: consider the possible columns of $G$ and how many linear combinatios produce a zero in each position....
$endgroup$
– leonbloy
Jan 9 at 13:43
2
$begingroup$
This is a special case of the Griesmer bound. The 2-dimensional case can, indeed, be proven the way leonbloy indicated. But the general result isn't very difficult either.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 15:25
add a comment |
$begingroup$
$d+ d/3 leq n$ looks weird, as it would be simpler to write $ frac43 d leq n$ Are you sure you got it right?
$endgroup$
– leonbloy
Jan 8 at 20:47
$begingroup$
Yes I'm sure...
$endgroup$
– xzeo
Jan 9 at 11:08
$begingroup$
Hint: consider the possible columns of $G$ and how many linear combinatios produce a zero in each position....
$endgroup$
– leonbloy
Jan 9 at 13:43
2
$begingroup$
This is a special case of the Griesmer bound. The 2-dimensional case can, indeed, be proven the way leonbloy indicated. But the general result isn't very difficult either.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 15:25
$begingroup$
$d+ d/3 leq n$ looks weird, as it would be simpler to write $ frac43 d leq n$ Are you sure you got it right?
$endgroup$
– leonbloy
Jan 8 at 20:47
$begingroup$
$d+ d/3 leq n$ looks weird, as it would be simpler to write $ frac43 d leq n$ Are you sure you got it right?
$endgroup$
– leonbloy
Jan 8 at 20:47
$begingroup$
Yes I'm sure...
$endgroup$
– xzeo
Jan 9 at 11:08
$begingroup$
Yes I'm sure...
$endgroup$
– xzeo
Jan 9 at 11:08
$begingroup$
Hint: consider the possible columns of $G$ and how many linear combinatios produce a zero in each position....
$endgroup$
– leonbloy
Jan 9 at 13:43
$begingroup$
Hint: consider the possible columns of $G$ and how many linear combinatios produce a zero in each position....
$endgroup$
– leonbloy
Jan 9 at 13:43
2
2
$begingroup$
This is a special case of the Griesmer bound. The 2-dimensional case can, indeed, be proven the way leonbloy indicated. But the general result isn't very difficult either.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 15:25
$begingroup$
This is a special case of the Griesmer bound. The 2-dimensional case can, indeed, be proven the way leonbloy indicated. But the general result isn't very difficult either.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 15:25
add a comment |
0
active
oldest
votes
Your Answer
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066564%2funderbound-length-of-a-ternary-linear-code%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
draft saved
draft discarded
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3066564%2funderbound-length-of-a-ternary-linear-code%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
$d+ d/3 leq n$ looks weird, as it would be simpler to write $ frac43 d leq n$ Are you sure you got it right?
$endgroup$
– leonbloy
Jan 8 at 20:47
$begingroup$
Yes I'm sure...
$endgroup$
– xzeo
Jan 9 at 11:08
$begingroup$
Hint: consider the possible columns of $G$ and how many linear combinatios produce a zero in each position....
$endgroup$
– leonbloy
Jan 9 at 13:43
2
$begingroup$
This is a special case of the Griesmer bound. The 2-dimensional case can, indeed, be proven the way leonbloy indicated. But the general result isn't very difficult either.
$endgroup$
– Jyrki Lahtonen
Jan 9 at 15:25