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A parity check matrix for a binary linear code is a matrix $H$ (with linearly independent rows) such that the (right) kernel of $H$ is the codespace. Clearly we can replace any row $r_i$ by $r_i + \sum_{j \in J} r_j$ if $i \not \in J$ to produce a matrix with the same kernel. It is a natural question to ask if there is some matrix $H'$ with the same kernel as $H$ such that all rows of $H'$ have weight at most $k$, where the weight of a row is just the number of nonzero entries in the row. Is there a way to produce such an $H'$ (or even to tell if one exists) without checking all matrices with the same nullspace?

Follow up: is there a way to solve the same problem given a parity check matrix over $GF(4)$?

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  • $\begingroup$ Thank you for your comment @Kevin. To see if I understand - if we could solve my problem then we would know if there at at least $r$ (number of rows of $H$ independent vectors in the dual codespace of weight at most $k$. To find the minimum weight codeword we could try all $k$ from $0$ to the number of columns of $H$. If we ever get a YES answer (that $H'$ exists) then we would know the minimum distance of the dual code is at most $k$. However, if we get a NO answer, doesn't this just say that there are at most $r - 1$ codewords of weight $k$? Does this let us find the minimum weight codeword? $\endgroup$ Commented Jul 19, 2022 at 17:40
  • $\begingroup$ Sounds like a job for integer linear programming. Do you have example data? $\endgroup$
    – RobPratt
    Commented Jul 19, 2022 at 17:54
  • $\begingroup$ @itsabijection Yes, I deleted my comment after realizing it's not obvious how that helps you. Perhaps the hardness proof for finding a min weight codeword can be adapted to your situation. $\endgroup$
    – Kevin
    Commented Jul 19, 2022 at 18:49
  • $\begingroup$ @RobPratt I'm interested to see how integer linear programming can be applied here. I think the problem can re-stated as finding linear combinations of the rows of $H$ with the least maximum weight...you'd have to add that the rows are independent which will complicate things...I can ask a post a separate question is this is different from OP's $\endgroup$
    – unknown
    Commented Jul 19, 2022 at 21:00

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