User:ECCclass/Gilbert-Varshamov

In coding theory, the Gilbert–Varshamov bound is a bound on the parameters of a (not necessarily linear) code. It is occasionally known as the Gilbert–Shannon–Varshamov bound (or the GSV bound), but the Gilbert–Varshamov bound is by far the most popular name.

Statement of the bound
Let


 * $$A_q(n,d)$$

denote the maximum possible size of a q-ary code $$C$$ with length n and minimum Hamming weight d (a q-ary code is a code over the field $$\mathbb{F}_q$$ of q elements).

Then:


 * $$A_q(n,d) \geq \frac{q^n}{\sum_{j=0}^{d-1} \binom{n}{j}(q-1)^j}.$$

Proof
Let $$C$$ be a code of length $$n$$ and minimum Hamming distance $$d$$ having maximal size:


 * $$|C|=A_q(n,d).\,$$

Then for all $$x\in\mathbb{F}_q^n$$ there exists at least one codeword $$c_x \in C$$ such that the Hamming distance $$d(x,c_x)$$ between $$x$$ and $$c_x$$ satisfies


 * $$d(x,c_x)\leq d-1$$

since otherwise we could add x to the code whilst maintaining the code's minimum Hamming distance d – a contradiction on the maximality of $$|C|$$.

Hence the whole of $$\mathbb{F}_q^n$$ is contained in the union of all balls of radius d &minus; 1 having their centre at some $$c \in C$$ :


 * $$\mathbb{F}_q^n =\cup_{c \in C} B(c,d-1).\, $$

Now each ball has size



\sum_{j=0}^{d-1} \binom{n}{j}(q-1)^j $$

since we may allow (or choose) up to $$d-1$$ of the $$n$$ components of a codeword to deviate (from the value of the corresponding component of the ball's centre) to one of $$(q-1)$$ possible other values (recall: the code is q-ary: it takes values in $$\mathbb{F}_q^n$$). Hence we deduce



\begin{align} \\ & \leq \sum_{c \in C} |B(c,d-1)| \\ \\ & = |C|\sum_{j=0}^{d-1} \binom{n}{j}(q-1)^j \\ \\ \end{align} $$
 * \mathbb{F}_q^n| & = |\cup_{c \in C} B(c,d-1)| \\

That is:



A_q(n,d) \geq \frac{q^n}{\sum_{j=0}^{d-1} \binom{n}{j}(q-1)^j} $$

(using the fact: $$|\mathbb{F}_q^n|=q^n$$).

An improvement in the prime power case
For q a prime power, one can improve the bound to $$A_q(n,d)\ge q^k$$ where k is the greatest integer for which


 * $$q^k < \frac{q^{n-1}}{\sum_{j=0}^{d-2} \binom{n-1}{j}(q-1)^j}.$$