In Theorem 7 of the paper Quantum Error Correction Via Codes Over GF(4) the authors come up with a sort of reduction algorithm. Right after they claim the $[[85,77,3]]$ Hamming code implies the existence of a $[[9,1,3]]$ linear quantum code.

To be sure, linear here means $GF(4)$-linear, i.e., if we take the elements in the stabilizer of the code and consider them as elements in $GF(4) = \{0,1,\omega,\omega^2\}$ then not only are they closed under addition they are also closed under $\omega$ and $\omega^2$.

I was wondering what the stabilizer generators for this linear $[[9,1,3]]$ code are.

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    $\begingroup$ the results in that paper are hard to verify (and possibly wrong). Take the (classical) hamming code over GF(4) : $[85,81,3]_4$; the example on page 19 claims that its dual is self-orthogonal and leads to a $[[85,77,3]]$ quantum code. According to GAP/GUAVA IsSelfOrthogonalCode(DualCode(HammingCode(4,GF(4))) is false; so I'm not sure how they get the $[[85,77,3]]$. In principle if you have that code you can construct the $[[9,1,3]]$ from it according to the recipe they give. $\endgroup$
    – unknown
    Commented May 4, 2023 at 18:14

1 Answer 1


(partial answer; too long to fit as a comment)

The paper and GUAVA use different definitions for GF(4) orthogonality. With the paper's definition, Hamming(4,GF(4)) is self orthogonal. This gives a $4 \times 85$ matrix over GF(4), $H$, which corresponds to a $[[85,81,3]]$ code; taking $H$ and $\omega H$ gives an $8 \times 85$ matrix $H'$ that corresponds to the $[[85,77,3]]$ code in the paper. Note that the GF(4) rank of $H'$ is still 4, but it defines a different code than $H$. I can get these matrices and verify the weight enumerator of $C=(85,2^4)$ and its dual $C^\perp$ (page 15). So the "answer" to your question is then : take this $8 \times 85$ matrix and delete certain 76 of the 85 columns to get an $8 \times 9$ matrix which defines the $[[9,1,3]]$ code. These 76 columns are the support of all codewords in $C^\perp$ of weight 76. I don't beleive that the problem of finding these columns is tractable for codes this size but I could be wrong.

[update] Instead of calculation which 76 columns to delete I did a random search of which 9 to keep. This found solutions fairly quickly (which means these are not that rate). Here's one :


$[0,1,2,3] \leftrightarrow [0,1,\omega,\omega^2]$

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    $\begingroup$ I can confirm what you wrote down is both $GF(4)$-linear and corresponds to a $[[9,1,3]]$ code. $\endgroup$ Commented May 7, 2023 at 4:27

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