Asymptotic for binary linear code

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Let $C=[n,k,d]$ is a binary linear code of length $n$, dimension $k$, and minimum distance $d$. Let us consider all possible binary linear codes with $k=d$ and $nin mathbb N$. Is it true that
$$A=mathoplim limits_kto infty frackn=0 ?$$
I look codetables.de and this assumption looks plausible. Are suitable boundaries or hypotheses known?







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    I assume you want the number of codes, not just $frac kn$, in your limit. For fixed $n$, once $k gt frac n2$ there are no codes
    – Ross Millikan
    yesterday











  • @RossMillikan $n=n(k)$ and $n→infty $ as $k→infty $, of course. From your comments it follows that $A=mathoplim sup _k→infty k/n le 1/2$. Сan there exist codes $[10k,k,k]$ for arbitrarily large $k$? If such codes exist only a finite number, no more, then $A le 1/10$ etc.
    – grizzly
    yesterday















up vote
1
down vote

favorite












Let $C=[n,k,d]$ is a binary linear code of length $n$, dimension $k$, and minimum distance $d$. Let us consider all possible binary linear codes with $k=d$ and $nin mathbb N$. Is it true that
$$A=mathoplim limits_kto infty frackn=0 ?$$
I look codetables.de and this assumption looks plausible. Are suitable boundaries or hypotheses known?







share|cite|improve this question

















  • 1




    I assume you want the number of codes, not just $frac kn$, in your limit. For fixed $n$, once $k gt frac n2$ there are no codes
    – Ross Millikan
    yesterday











  • @RossMillikan $n=n(k)$ and $n→infty $ as $k→infty $, of course. From your comments it follows that $A=mathoplim sup _k→infty k/n le 1/2$. Сan there exist codes $[10k,k,k]$ for arbitrarily large $k$? If such codes exist only a finite number, no more, then $A le 1/10$ etc.
    – grizzly
    yesterday













up vote
1
down vote

favorite









up vote
1
down vote

favorite











Let $C=[n,k,d]$ is a binary linear code of length $n$, dimension $k$, and minimum distance $d$. Let us consider all possible binary linear codes with $k=d$ and $nin mathbb N$. Is it true that
$$A=mathoplim limits_kto infty frackn=0 ?$$
I look codetables.de and this assumption looks plausible. Are suitable boundaries or hypotheses known?







share|cite|improve this question













Let $C=[n,k,d]$ is a binary linear code of length $n$, dimension $k$, and minimum distance $d$. Let us consider all possible binary linear codes with $k=d$ and $nin mathbb N$. Is it true that
$$A=mathoplim limits_kto infty frackn=0 ?$$
I look codetables.de and this assumption looks plausible. Are suitable boundaries or hypotheses known?









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited yesterday
























asked yesterday









grizzly

25829




25829







  • 1




    I assume you want the number of codes, not just $frac kn$, in your limit. For fixed $n$, once $k gt frac n2$ there are no codes
    – Ross Millikan
    yesterday











  • @RossMillikan $n=n(k)$ and $n→infty $ as $k→infty $, of course. From your comments it follows that $A=mathoplim sup _k→infty k/n le 1/2$. Сan there exist codes $[10k,k,k]$ for arbitrarily large $k$? If such codes exist only a finite number, no more, then $A le 1/10$ etc.
    – grizzly
    yesterday













  • 1




    I assume you want the number of codes, not just $frac kn$, in your limit. For fixed $n$, once $k gt frac n2$ there are no codes
    – Ross Millikan
    yesterday











  • @RossMillikan $n=n(k)$ and $n→infty $ as $k→infty $, of course. From your comments it follows that $A=mathoplim sup _k→infty k/n le 1/2$. Сan there exist codes $[10k,k,k]$ for arbitrarily large $k$? If such codes exist only a finite number, no more, then $A le 1/10$ etc.
    – grizzly
    yesterday








1




1




I assume you want the number of codes, not just $frac kn$, in your limit. For fixed $n$, once $k gt frac n2$ there are no codes
– Ross Millikan
yesterday





I assume you want the number of codes, not just $frac kn$, in your limit. For fixed $n$, once $k gt frac n2$ there are no codes
– Ross Millikan
yesterday













@RossMillikan $n=n(k)$ and $n→infty $ as $k→infty $, of course. From your comments it follows that $A=mathoplim sup _k→infty k/n le 1/2$. Сan there exist codes $[10k,k,k]$ for arbitrarily large $k$? If such codes exist only a finite number, no more, then $A le 1/10$ etc.
– grizzly
yesterday





@RossMillikan $n=n(k)$ and $n→infty $ as $k→infty $, of course. From your comments it follows that $A=mathoplim sup _k→infty k/n le 1/2$. Сan there exist codes $[10k,k,k]$ for arbitrarily large $k$? If such codes exist only a finite number, no more, then $A le 1/10$ etc.
– grizzly
yesterday











1 Answer
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1
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accepted










The Gilbert Varshamov bound shows the existence of a binary linear code of dimension $$kgeq n-lfloor log_2 V(n,d-1)rfloor,$$
where $V(n,r)$ is the volume of the hamming sphere in the $n$ dimensional hypercube.



Let $d=k$, and find the maximal $k$ satisfying this equation, for each $n$. Along a subsequence of integers, something like the existence of codes with parameters
$$
[Ak_i,k_i,k_i]
$$
with $A=5,$ should hold.



More generally, let $nrightarrow infty$ and use the fact that for each $n$ the binomial coefficients $binomnk,$ are superincreasing in $k$ up to about $k<n/3,$ and the approximation
$$
V(n,k-1)approx binomnk approx 2^n (h(k/n)+o(1)),
$$
where $h$ is binary Shannon entropy (see here) to conclude that the solution $theta=0.227cdots$ to
$$
1-theta=h(theta)
$$
where $theta=k/n,$ means that asymptotically there is a sequence of codes with parameters $[Ak,k,k]$ and $A=1/theta.$






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  • This is very useful for me. Many thanks!
    – grizzly
    7 hours ago










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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
1
down vote



accepted










The Gilbert Varshamov bound shows the existence of a binary linear code of dimension $$kgeq n-lfloor log_2 V(n,d-1)rfloor,$$
where $V(n,r)$ is the volume of the hamming sphere in the $n$ dimensional hypercube.



Let $d=k$, and find the maximal $k$ satisfying this equation, for each $n$. Along a subsequence of integers, something like the existence of codes with parameters
$$
[Ak_i,k_i,k_i]
$$
with $A=5,$ should hold.



More generally, let $nrightarrow infty$ and use the fact that for each $n$ the binomial coefficients $binomnk,$ are superincreasing in $k$ up to about $k<n/3,$ and the approximation
$$
V(n,k-1)approx binomnk approx 2^n (h(k/n)+o(1)),
$$
where $h$ is binary Shannon entropy (see here) to conclude that the solution $theta=0.227cdots$ to
$$
1-theta=h(theta)
$$
where $theta=k/n,$ means that asymptotically there is a sequence of codes with parameters $[Ak,k,k]$ and $A=1/theta.$






share|cite|improve this answer





















  • This is very useful for me. Many thanks!
    – grizzly
    7 hours ago














up vote
1
down vote



accepted










The Gilbert Varshamov bound shows the existence of a binary linear code of dimension $$kgeq n-lfloor log_2 V(n,d-1)rfloor,$$
where $V(n,r)$ is the volume of the hamming sphere in the $n$ dimensional hypercube.



Let $d=k$, and find the maximal $k$ satisfying this equation, for each $n$. Along a subsequence of integers, something like the existence of codes with parameters
$$
[Ak_i,k_i,k_i]
$$
with $A=5,$ should hold.



More generally, let $nrightarrow infty$ and use the fact that for each $n$ the binomial coefficients $binomnk,$ are superincreasing in $k$ up to about $k<n/3,$ and the approximation
$$
V(n,k-1)approx binomnk approx 2^n (h(k/n)+o(1)),
$$
where $h$ is binary Shannon entropy (see here) to conclude that the solution $theta=0.227cdots$ to
$$
1-theta=h(theta)
$$
where $theta=k/n,$ means that asymptotically there is a sequence of codes with parameters $[Ak,k,k]$ and $A=1/theta.$






share|cite|improve this answer





















  • This is very useful for me. Many thanks!
    – grizzly
    7 hours ago












up vote
1
down vote



accepted







up vote
1
down vote



accepted






The Gilbert Varshamov bound shows the existence of a binary linear code of dimension $$kgeq n-lfloor log_2 V(n,d-1)rfloor,$$
where $V(n,r)$ is the volume of the hamming sphere in the $n$ dimensional hypercube.



Let $d=k$, and find the maximal $k$ satisfying this equation, for each $n$. Along a subsequence of integers, something like the existence of codes with parameters
$$
[Ak_i,k_i,k_i]
$$
with $A=5,$ should hold.



More generally, let $nrightarrow infty$ and use the fact that for each $n$ the binomial coefficients $binomnk,$ are superincreasing in $k$ up to about $k<n/3,$ and the approximation
$$
V(n,k-1)approx binomnk approx 2^n (h(k/n)+o(1)),
$$
where $h$ is binary Shannon entropy (see here) to conclude that the solution $theta=0.227cdots$ to
$$
1-theta=h(theta)
$$
where $theta=k/n,$ means that asymptotically there is a sequence of codes with parameters $[Ak,k,k]$ and $A=1/theta.$






share|cite|improve this answer













The Gilbert Varshamov bound shows the existence of a binary linear code of dimension $$kgeq n-lfloor log_2 V(n,d-1)rfloor,$$
where $V(n,r)$ is the volume of the hamming sphere in the $n$ dimensional hypercube.



Let $d=k$, and find the maximal $k$ satisfying this equation, for each $n$. Along a subsequence of integers, something like the existence of codes with parameters
$$
[Ak_i,k_i,k_i]
$$
with $A=5,$ should hold.



More generally, let $nrightarrow infty$ and use the fact that for each $n$ the binomial coefficients $binomnk,$ are superincreasing in $k$ up to about $k<n/3,$ and the approximation
$$
V(n,k-1)approx binomnk approx 2^n (h(k/n)+o(1)),
$$
where $h$ is binary Shannon entropy (see here) to conclude that the solution $theta=0.227cdots$ to
$$
1-theta=h(theta)
$$
where $theta=k/n,$ means that asymptotically there is a sequence of codes with parameters $[Ak,k,k]$ and $A=1/theta.$







share|cite|improve this answer













share|cite|improve this answer



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answered 14 hours ago









kodlu

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  • This is very useful for me. Many thanks!
    – grizzly
    7 hours ago
















  • This is very useful for me. Many thanks!
    – grizzly
    7 hours ago















This is very useful for me. Many thanks!
– grizzly
7 hours ago




This is very useful for me. Many thanks!
– grizzly
7 hours ago












 

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