Griesmer bound

For a binary linear code, the Griesmer bound is:

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

Let ${\displaystyle N(k,d)}$ 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

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

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.

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

The matrix ${\displaystyle G'}$ generates a code ${\displaystyle C'}$, which is called the residual code of ${\displaystyle C.}$ ${\displaystyle C'}$ obviously has dimension ${\displaystyle k'=k-1}$ and length ${\displaystyle n'=N(k,d)-d.}$ ${\displaystyle C'}$ has a distance ${\displaystyle d',}$ but we don't know it. Let ${\displaystyle u\in C'}$ be such that ${\displaystyle w(u)=d'}$. There exists a vector ${\displaystyle v\in \mathbb {F} _{2}^{d}}$ such that the concatenation ${\displaystyle (v|u)\in C.}$ Then ${\displaystyle w(v)+w(u)=w(v|u)\geqslant d.}$ On the other hand, also ${\displaystyle (v|u)+r\in C,}$ since ${\displaystyle r\in C}$ and ${\displaystyle C}$ is linear: ${\displaystyle w((v|u)+r)\geqslant d.}$ But

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

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

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

so that

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

By induction over k we will eventually get

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

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

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

for any integer a and positive integer k.

For a linear code over ${\displaystyle \mathbb {F} _{q}}$, the Griesmer bound becomes:

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

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

• Singleton bound
• Hamming bound
• Gilbert-Varshamov bound
• Johnson bound
• Plotkin bound
• Elias Bassalygo bound

• J. H. Griesmer, "A bound for error-correcting codes," IBM Journal of Res. and Dev., vol. 4, no. 5, pp. 532-542, 1960.