Punctured codes' dimension computation

[David Lucas]

For now, computation of the dimension of a punctured codes forces the computation of its generator matrix. It does : dimension = generator_matrix.rank().

As it's not really satisfying, another solution should be found to do that.

[Daniel Augot]

I think there is not a lot of choice. In any case, you need to perform row reduction to get the actual dimension of the punctured code, unless it has been previously done for the mother code.

[Daniel Augot]

Oups I am wrong.

I checked that in Pless' book. Here is the Theorem (btw Sloane's book is wrong)

Theorem 1.5.1 Let C be an [n, k, d] code over F q , and let C* be the code C punctured on the ith coordinate.

(i) If d > 1, C∗ is an [n − 1, k, d∗ ] code where d∗ = d − 1 if C has a minimum weight codeword with a nonzero ith coordinate and d∗ = d otherwise.

(ii) When d = 1, C∗ is an [n − 1, k, 1] code if C has no codeword of weight 1 whose nonzero entry is in coordinate i; otherwise, if k > 1, C∗ is an [n − 1, k − 1, d* ] code with d∗ ≥ 1.

Concerning the dimension, the conditions are easy to check.

For the minimum distance, one has to known about minimum weight codewords, and it is mich harder.