Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am looking for all $k$-dimensional subspaces of $(\mathbb{Z}/2\mathbb{Z})^n$ up to permutational equivalence.

Is there a database of all $[n,k]$-codes up to equivalence for reasonable values of $(n,k)$? For instance, $1 \leq k \leq 6$ and $6 \leq n \leq 12$ (giving around 35,000 codes).

When $k=1$, there are $n$ such subspaces, each uniquely determined by the Hamming weight of the nonzero element.

When $k=2$, there are 0, 1, 3, 6, 10, 16, 23, 32 … (OEIS:A034198) such subspaces.

When $k=3$, there are 0, 0, 1, 4, 10, 22, 43, 77, … (OEIS:A034357) such subspaces.

When $k=4$, there are 0, 0, 0, 1, 5, 16, 43, 106, … (OEIS:A034358) such subspaces.

For each $k\geq 2$, I have trouble finding representatives of the codes for $n \geq 8$, but surely someone has recorded them somewhere. For $n=8$, there are just 8, 32, 77, 106 such codes for $k=1,2,3,4$, so it seems reasonable to me that someone wrote them down.

I have found several databases of "good" codes, but I actually want the bad codes too.

share|improve this question
For $k=2$ the code is determined up to equivalence by the Hamming weights $n\ge w_1\ge w_2\ge w_3>0$ of the non-zero words subject to the obvious constraints: $2\mid s\le 2n$, where $s=w_1+w_2+w_3$, and $w_2+w_3\ge w_1$. The sequence $(w_i)_{i=1}^3$ of numbers is an invariant of the code. It is also easy see that all codes sharing the same sequence are equivalent by permuting the coordinates in such a way that the 1s get stacked into the beginning. The weight enumerator is always an invariant, but there are known examples of non-equivalent codes with the same weight enumerator. –  Jyrki Lahtonen Jul 1 '11 at 12:16
Rats. The last sentence was meant to say that there are examples of non-equivalent codes sharing the same weight enumerator, but only when $k$ is larger. –  Jyrki Lahtonen Jul 3 '11 at 9:32

1 Answer 1

A more comprehensive list of those numbers can be found here. The numbers are growing extremely fast, so that in all but the smallest cases, it is hopeless to store a complete list of representatives.

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.