2
$\begingroup$

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.

$\endgroup$
1
  • 1
    $\begingroup$ 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
    May 4, 2023 at 18:14

1 Answer 1

3
$\begingroup$

(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 :

[[1,3,2,1,1,1,0,0,0],
 [0,2,1,2,2,0,2,1,0],
 [2,0,1,2,0,3,0,0,0],
 [3,0,3,0,2,3,3,0,1],
 [2,1,3,2,2,2,0,0,0],
 [0,3,2,3,3,0,3,2,0],
 [3,0,2,3,0,1,0,0,0],
 [1,0,1,0,3,1,1,0,2]]

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

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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