The correspondence conjecture for trinomial-associated polycyclic codes

Let $t_1(x)=x^n-ax^i-b$ and $t_2(x)=x^n-a'x^{n-i}-b'$ be polynomials in $\bbF_q[x]$ with equal order, and let $C_1$ and $C_2$ denote the sets of length- $n$ polycyclic codes over $\bbF_q$ associated with $t_1(x)$ and $t_2(x)$ , respectively. Correspondence conjecture. If

\operatorname{Ord}(t_1(x))=\operatorname{Ord}(t_2(x)),

then $C_1$ and $C_2$ are in a one-to-one correspondence, with corresponding codes equivalent to each other. This proposes a trinomial analogue of known correspondence results for constacyclic codes; the source gives computational evidence but no proof or resolution.

Sources & referencesView supporting material

Primary source

Nuh Aydin, Peihan Liu and Bryan Yoshino, “Polycyclic Codes Associated with Trinomials: Good Codes and Open Questions”, arXiv:2106.12065 (2021).

coding-theory polycyclic-codes

1 week ago

Progress summary

Refreshed 5 days ago

Solved

The proposed rule is false: a concrete counterexample shows that equal-order trinomial polynomials need not produce corresponding equivalent code families.

The 2021 paper formulates the equal-order correspondence as Conjecture $3.5$ , based on computation rather than proof. It concerns trinomials $t_1(x)=x^n-ax^i-b$ and $t_2(x)=x^n-a'x^{n-i}-b'$ over $\mathbb{F}_q$ .

Known results

The original paper proves equal orders in the mutually reciprocal special case, including $a'=a(-b)^{-1}$ and $b'=b^{-1}$ (2021).
Under the stricter condition $t_2(x)=t_1(0)^{-1}t_1^*(x)$ , a later paper proves the required one-to-one correspondence and equivalence (2022).

2022 counterexample

The later paper disproves Conjecture $3.5$ : over $\mathbb{F}_3$ , $t_1(x)=x^{10}+x^8+1$ and $t_2(x)=x^{10}+x^2+2$ both have order $156$ , but their code sets contain $18$ and $12$ linear codes, respectively. Thus the proposed correspondence cannot exist.

Current status (as of August 2026): the original equal-order correspondence conjecture is settled false by the counterexample, while narrower reciprocal and special-family equivalence results remain valid.

Sources

No solutions have been posted yet.