The binary polycyclic-code isoduality characterization

Let $C$ be a polycyclic code over $\mathbb{F}_2$ generated by $g(x)$ dividing $x^n-ax^i-b$ . Binary polycyclic-code isoduality conjecture. The code $C$ is iso-dual if and only if $g^2(x)=x^n-ax^i-b$ . The claim proposes a complete generator-polynomial characterization of iso-dual binary polycyclic codes; the supplied text gives no resolution evidence.

Sources & referencesView supporting material

Primary source

Nuh Aydin, Peihan Liu and Bryan Yoshino, “A Database of Quantum Codes”, arXiv:2108.03567 (2021).

Additional references

2 papers in this index state this conjecture (2021). The statement above is taken from the most recent of them; the others are arXiv:2106.12065.

coding-theory polycyclic-codes

1 week ago

Progress summary

Refreshed 5 days ago

Partially solved

A later paper supports the construction direction but reports that the proposed converse is false, so the complete characterization does not stand.

The conjecture asserts that a binary polycyclic code generated by $g(x)$ is isodual exactly when $g^2(x)=x^n-ax^i-b$ . A 2022 paper records this as an earlier conjecture and studies related isoduality conditions.

Known results

Theorem 4.4 establishes sufficiency: $g^2(x)=x^n-ax^i-b$ implies that the generated code is isodual (2022 paper).
The same condition is described as a construction method, not a necessary one (2022 paper).

May 2026 challenge to the converse

A May 2026 paper states that the relevant necessity direction is incorrect and gives an explicit binary counterexample, while retaining the sufficiency result. Its numbering refers to an earlier Conjecture 4.2, so the source does not completely clarify whether its counterexample matches the exact formulation stated here.

Current status (as of August 2026): The sufficiency direction is supported, while the necessity direction is reported false; the exact correspondence of the published counterexample to this formulation remains to be checked.

Sources

No solutions have been posted yet.