Griesmer bound

In the mathematics of coding theory, the Griesmer bound, named after James Hugo Griesmer, is a bound on the length of linear binary codes of dimension k and minimum distance d. There is also a very similar version for non-binary codes.

Statement of the bound
For a binary linear code, the Griesmer bound is:


 * $$n\geqslant \sum_{i=0}^{k-1} \left\lceil\frac{d}{2^i}\right\rceil.$$

Proof
Let $$N(k,d)$$ denote the minimum length of a binary code of dimension k and distance d. Let C be such a code. We want to show that


 * $$ N(k,d)\geqslant \sum_{i=0}^{k-1} \left\lceil\frac{d}{2^i}\right\rceil.$$

Let G be a generator matrix of C. We can always suppose that the first row of G is of the form r = (1, ..., 1, 0, ..., 0) with weight d.
 * $$G= \begin{bmatrix}

1 & \dots & 1 & 0 & \dots & 0 \\ \ast & \ast & \ast &   & G' &   \\ \end{bmatrix}$$

The matrix $$G'$$ generates a code $$C'$$, which is called the residual code of $$C.$$ $$C'$$ obviously has dimension $$k'=k-1$$ and length $$n'=N(k,d)-d.$$ $$C'$$ has a distance $$d',$$ but we don't know it. Let $$u\in C'$$ be such that $$w(u)=d'$$. There exists a vector $$v\in \mathbb{F}_2^d$$ such that the concatenation $$(v|u)\in C.$$ Then $$w(v)+w(u)=w(v|u)\geqslant d.$$ On the other hand, also $$(v|u)+r\in C,$$ since $$r\in C$$ and $$C$$ is linear: $$w((v|u)+r)\geqslant d.$$ But


 * $$w((v|u)+r)=w(((1,\cdots,1)+v)|u)=d-w(v)+w(u),$$

so this becomes $$d-w(v)+w(u)\geqslant d$$. By summing this with $$w(v)+w(u)\geqslant d,$$ we obtain $$d+2w(u)\geqslant 2d$$. But $$w(u)=d',$$ so we get $$2d'\geqslant d.$$ As $$d'$$ is integral, we get $$d'\geqslant \left\lceil \tfrac{d}{2} \right\rceil.$$ This implies


 * $$n'\geqslant N\left (k-1,\left\lceil \tfrac{d}{2} \right\rceil \right ),$$

so that


 * $$N(k,d)\geqslant d + N\left (k-1, \left\lceil \tfrac{d}{2} \right\rceil \right ).$$

By induction over k we will eventually get


 * $$ N(k,d)\geqslant \sum_{i=0}^{k-1} \left\lceil\frac{d}{2^i}\right\rceil.$$

Note that at any step the dimension decreases by 1 and the distance is halved, and we use the identity


 * $$\left\lceil\frac{\left\lceil a/2^{k-1}\right\rceil}{2}\right\rceil = \left\lceil\frac{a}{2^k}\right\rceil$$

for any integer a and positive integer k.

Bound for the general case
For a linear code over $$\mathbb{F}_q$$, the Griesmer bound becomes:


 * $$n\geqslant \sum_{i=0}^{k-1} \left\lceil\frac{d}{q^i}\right\rceil.$$

The proof is similar to the binary case and so it is omitted.