Gilbert = バルシャモフ limit

With the Gilbert = バルシャモフ limit (British: Gilbert-Varshamov bound), I point to the limit of the parameter of the mark (it may not be a linear mark). With "Gilbert = Shannon = バルシャモフ limit" (GSV limit).

Theorem

Mark of q 進数 ${\displaystyle \ C}$  But, it is the head ${\displaystyle n}$  The で's smallest Hamming distance ${\displaystyle d}$  であるとき, the possible maximum size (the total number of mark word) ${\displaystyle \ A_{q}(n,d)}$  とする. In addition, the mark of q 進数,${\displaystyle q}$  Body of the element of the unit ${\displaystyle \mathbb {F} _{q}^{n}}$  It is an upper linear mark.

Then next is managed.

${\displaystyle {\begin{matrix}\\A_{q}(n,d)\geq {\frac {q^{n}}{\sum _{j=0}^{d-1}{\binom {n}{j}}(q-1)^{j}}}\\\\\end{matrix}}}$ 

Using maximum integer k where the next ceremony is managed at this limit in the case of 素数冪 q ${\displaystyle A_{q}(n,d)\geq q^{k}}$  I can become と briefness.

${\displaystyle A_{q}(n,d)\geq {\frac {q^{n}}{\sum _{j=0}^{d-2}{\binom {n-1}{j}}(q-1)^{j}}}}$ 

Proof

Mark ${\displaystyle C}$  Length of the の mark word ${\displaystyle n}$ , smallest Hamming distance ${\displaystyle d}$ , biggest mark number of words

${\displaystyle |C|=A_{q}(n,d)}$ 

とする. Then of all ${\displaystyle x\in \mathbb {F} _{q}^{n}}$  At least one mark word that it is ${\displaystyle c_{x}\in C}$  But, I exist,${\displaystyle x}$  と ${\displaystyle c_{x}}$  Hamming distance between の ${\displaystyle d(x,c_{x})}$  The next that it is is managed.

${\displaystyle d(x,c_{x})\leq d-1}$ 

Otherwise,${\displaystyle x}$  Even if add it as を mark word; the smallest Hamming distance of the mark ${\displaystyle d}$  は does not change${\displaystyle |C|}$  But) which contradicts it in a premise to be biggest.

That is why,${\displaystyle \mathbb {F} _{q}^{n}}$  The whole ${\displaystyle c\in C}$  The を center and radius to do ${\displaystyle d-1}$  It is included in the sum of sets of the ball of all の.

${\displaystyle \mathbb {F} _{q}^{n}=\cup _{c\in C}B(c,d-1)}$ 

Here the size of each ball

${\displaystyle {\begin{matrix}\sum _{j=0}^{d-1}{\binom {n}{j}}(q-1)^{j}\\\end{matrix}}}$ 

となる. This changes maximum d-1 figure among mark word of the n figure (from a supporting place value of the mark word which is the center of the ball),${\displaystyle (q-1)}$ I can assume it the different value of the kind (the mark is q 進数${\displaystyle \mathbb {F} _{q}^{n}}$  ) which can take the value of the kind. Therefore, the following reasoning is managed.

${\displaystyle {\begin{matrix}|\mathbb {F} _{q}^{n}|&=&|\cup _{c\in C}B(c,d-1)|\\\\&\leq &\cup _{c\in C}|B(c,d-1)|\\\\&=&|C|\sum _{j=0}^{d-1}{\binom {n}{j}}(q-1)^{j}\\\\\end{matrix}}}$ 

In other words:

${\displaystyle {\begin{matrix}A_{q}(n,d)&\geq &{\frac {q^{n}}{\sum _{j=0}^{d-1}{\binom {n}{j}}(q-1)^{j}}}\\\end{matrix}}}$ 

となる${\displaystyle |\mathbb {F} _{q}^{n}|=q^{n}}$  であるため).

Allied item

• Singleton limit
• Humming limit
• Johnson limit
• Proto Kyn limit

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