Show that weight of code is equal to the minimal number of rows in the control matrix Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)How to show that the extended code of cyclic binary hamming code can fix single and two adjacent positions errors?Show by arithmetical arguments that there cannot be a perfect binary code of length $n$ which corrects $e$ errors.Binary Hamming code - number of words weight $i$Proving a formula for the weight distributions of a linear codePlotkin bound for the minimal distance of linear code over $mathbbF_q$Coding theory - Linear applicationSeeing weight of a code from generator matrixTo show the code is not linearHow to show that in a self-dual code all words have weight divisible by 4 or half do it and the anther half don't.Prove that $A_3(4,3)=9$ and construct the correspoding code.
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Show that weight of code is equal to the minimal number of rows in the control matrix
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)How to show that the extended code of cyclic binary hamming code can fix single and two adjacent positions errors?Show by arithmetical arguments that there cannot be a perfect binary code of length $n$ which corrects $e$ errors.Binary Hamming code - number of words weight $i$Proving a formula for the weight distributions of a linear codePlotkin bound for the minimal distance of linear code over $mathbbF_q$Coding theory - Linear applicationSeeing weight of a code from generator matrixTo show the code is not linearHow to show that in a self-dual code all words have weight divisible by 4 or half do it and the anther half don't.Prove that $A_3(4,3)=9$ and construct the correspoding code.
$begingroup$
Show that weight of code is equal to the minimal number of rows in the control matrix.
My solution:
Let $a$ is vector, $C^*$ is control matrix, $w(a)$ - weight of code.
We want to show $a*C^* = 0$
$0 = aC^* = [sum_ia_ic_i1, sum_ia_ic_i2,, dots,, sum_ia_ic_in] = sum_i^na_ic_i = sumc_i:a_i=1$,
where $c_i:a_i = 1 = w(a)$ - weight of code.
So, $sum_j=1^lc_ij = 0 rightarrow l geq min(w(a))$.
How to show it in another side, that $l leq min(w(a))$?
abstract-algebra cryptography coding-theory
$endgroup$
add a comment |
$begingroup$
Show that weight of code is equal to the minimal number of rows in the control matrix.
My solution:
Let $a$ is vector, $C^*$ is control matrix, $w(a)$ - weight of code.
We want to show $a*C^* = 0$
$0 = aC^* = [sum_ia_ic_i1, sum_ia_ic_i2,, dots,, sum_ia_ic_in] = sum_i^na_ic_i = sumc_i:a_i=1$,
where $c_i:a_i = 1 = w(a)$ - weight of code.
So, $sum_j=1^lc_ij = 0 rightarrow l geq min(w(a))$.
How to show it in another side, that $l leq min(w(a))$?
abstract-algebra cryptography coding-theory
$endgroup$
add a comment |
$begingroup$
Show that weight of code is equal to the minimal number of rows in the control matrix.
My solution:
Let $a$ is vector, $C^*$ is control matrix, $w(a)$ - weight of code.
We want to show $a*C^* = 0$
$0 = aC^* = [sum_ia_ic_i1, sum_ia_ic_i2,, dots,, sum_ia_ic_in] = sum_i^na_ic_i = sumc_i:a_i=1$,
where $c_i:a_i = 1 = w(a)$ - weight of code.
So, $sum_j=1^lc_ij = 0 rightarrow l geq min(w(a))$.
How to show it in another side, that $l leq min(w(a))$?
abstract-algebra cryptography coding-theory
$endgroup$
Show that weight of code is equal to the minimal number of rows in the control matrix.
My solution:
Let $a$ is vector, $C^*$ is control matrix, $w(a)$ - weight of code.
We want to show $a*C^* = 0$
$0 = aC^* = [sum_ia_ic_i1, sum_ia_ic_i2,, dots,, sum_ia_ic_in] = sum_i^na_ic_i = sumc_i:a_i=1$,
where $c_i:a_i = 1 = w(a)$ - weight of code.
So, $sum_j=1^lc_ij = 0 rightarrow l geq min(w(a))$.
How to show it in another side, that $l leq min(w(a))$?
abstract-algebra cryptography coding-theory
abstract-algebra cryptography coding-theory
asked Mar 27 at 20:49
Aleksandra Aleksandra
546
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