Construction and enumeration for self-dual cyclic codes over ${\mathbb {Z}}_4$ of oddly even length

Abstract

For any positive odd integer n, a precise representation for cyclic codes over ${\mathbb {Z}}_4$ of length 2n is given in terms of the Chinese Remainder Theorem. Using this representation, an efficient encoder for each of these codes is described. Then the dual codes are determined precisely and this is used to study codes which are self-dual. In particular, the number of self-dual cyclic codes over ${\mathbb {Z}}_{4}$ of length 2n can be calculated from 2-cyclotomic cosets modulo n directly. Moreover, mistakes in Blackford (Discret Appl Math 128:27–46, 2003) and Dougherty and Ling (Des Codes Cryptogr 39:127–153, 2006) are corrected. As an application, all 315 self-dual cyclic codes over ${\mathbb {Z}}_4$ of length 30 are listed. Among these codes, there are some new cyclic self-dual ${\mathbb {Z}}_4$-codes ${\mathcal {C}}$ with parameters $(30,|{\mathcal {C}}|=2^{30},d_H=6,d_L=12)$ and $(30,|{\mathcal {C}}|=2^{30},d_H=5,d_L=10)$. From these codes and applying the Gray map from ${\mathbb {Z}}_4$ onto ${\mathbb {F}}_2^2$, formally self-dual and 2-quasicyclic binary codes with basic parameters [60, 30, 12] and [60, 30, 10] are derived respectively.

Appendix: Generator matrices for all 315 self-dual cyclic codes over \({\mathbb {Z}}_4\) of length 30

We identify each polynomial $a(x)=\sum _{j=0}^{29}a_jx^j\in \frac{{\mathbb {Z}}_4[x]}{\langle x^{30}-1\rangle }$ with the vector $(a_0,a_1,\ldots ,a_{29})\in {\mathbb {Z}}_4^{30}$. Then set

• (i)
  $$\begin{aligned} G_1=\left( \begin{array}{c} 2\theta _1(x) \\ 2x\theta _1(x)\end{array}\right) \in \mathrm{M}_{2\times 2n}({\mathbb {Z}}_4), k_{0,1}=0 \text { and } k_{1,1}=2. \end{aligned}$$
• (ii-1)
  $$\begin{aligned} G_{2,1}=\left( \begin{array}{c} 2\theta _2(x) \\ 2x\theta _2(x)\\ 2x^2\theta _2(x)\\ 2x^3\theta _2(x)\end{array}\right) \in \mathrm{M}_{4\times 2n}({\mathbb {Z}}_4), k_{0,2}=0 \text { and } k_{1,2}=4. \end{aligned}$$
• (ii-2)
  $$\begin{aligned} G_{2,2}= & {} \left( \begin{array}{c} (f_2(x)+2h(x))\theta _2(x) \\ x(f_2(x)+2h(x))\theta _2(x)\end{array}\right) \in \mathrm{M}_{2\times 2n}({\mathbb {Z}}_4), k_{0,2}=2 \text { and } \\&\quad k_{1,2}=0, \text { where } h(x)\in {\mathcal {W}}_2=\{1,1+x\}. \end{aligned}$$
• (iii-1)
  $$\begin{aligned} G_{3,1}=\left( \begin{array}{c} 2\theta _3(x) \\ 2x\theta _3(x)\\ \ldots \\ 2x^7\theta _3(x)\end{array}\right) \in \mathrm{M}_{8\times 2n}({\mathbb {Z}}_4), k_{0,2}=0 \text { and } k_{1,2}=8. \end{aligned}$$
• (iii-2)
  $$\begin{aligned} G_{3,2}= & {} \left( \begin{array}{c} (f_3(x)+2h(x))\theta _3(x) \\ x(f_3(x)+2h(x))\theta _3(x) \\ x^2(f_3(x)+2h(x))\theta _3(x) \\ x^3(f_3(x)+2h(x))\theta _3(x) \end{array}\right) \in \mathrm{M}_{4\times 2n}({\mathbb {Z}}_4), k_{0,3}=4 \text { and } \\&\quad k_{1,3}=0, \text { where } h(x)\in {\mathcal {W}}_3=\{x, x+x^2, x^2+x^3, x^3\}. \end{aligned}$$
• (iv-1)
  $$\begin{aligned} G_{4,1}= & {} 0, k_{0,4}=k_{1,4}=0;\\ G_{5,1}= & {} \left( \begin{array}{c} \theta _5(x) \\ x\theta _5(x) \\ \ldots \\ x^7\theta _5(x)\end{array}\right) \in \mathrm{M}_{8\times 2n}({\mathbb {Z}}_4), k_{0,5}=8 \text { and } k_{1,5}=0. \end{aligned}$$
• (iv-2)
  $$\begin{aligned} G_{4,2}= & {} \left( \begin{array}{c} \theta _4(x) \\ x\theta _4(x)\\ \ldots \\ x^7 \theta _4(x)\end{array}\right) \in \mathrm{M}_{8\times 2n}({\mathbb {Z}}_4), k_{0,4}=8 \text { and } k_{1,4}=0;\\ G_{5,2}= & {} 0, k_{0,5}=k_{1,5}=0. \end{aligned}$$
• (iv-3)
  $$\begin{aligned} G_{4,3}= & {} \left( \begin{array}{c} 2\theta _4(x) \\ 2x\theta _4(x)\\ \ldots \\ 2x^7\theta _4(x)\end{array}\right) \in \mathrm{M}_{8\times 2n}({\mathbb {Z}}_4), k_{0,4}=0 \text { and } k_{1,4}=8;\\ G_{5,3}= & {} \left( \begin{array}{c} 2\theta _5(x) \\ 2x\theta _5(x)\\ \ldots \\ 2x^7\theta _5(x)\end{array}\right) \in \mathrm{M}_{8\times 2n}({\mathbb {Z}}_4), k_{0,5}=0 \text { and } k_{1,5}=8. \end{aligned}$$
• (iv-4)
  $$\begin{aligned} G_{4,4}= & {} \left( \begin{array}{c} 2{\overline{f}}_4(x)\theta _4(x)\\ 2x{\overline{f}}_4(x)\theta _4(x)\\ 2x^2{\overline{f}}_4(x)\theta _4(x)\\ 2x^3{\overline{f}}_4(x)\theta _4(x)\end{array}\right) \in \mathrm{M}_{4\times 2n}({\mathbb {Z}}_4), k_{0,4}=0 \text { and } k_{1,4}=4;\\ G_{5,4}= & {} \left( \begin{array}{c} f_5(x)\theta _5(x)\\ xf_5(x)\theta _5(x)\\ x^2f_5(x)\theta _5(x)\\ x^3f_5(x)\theta _5(x)\\ 2\theta _5(x)\\ 2x\theta _5(x)\\ 2x^2\theta _5(x)\\ 2x^3\theta _5(x)\end{array}\right) \in \mathrm{M}_{8\times 2n}({\mathbb {Z}}_4), k_{0,5}=4 \text { and } k_{1,5}=4. \end{aligned}$$
• (iv-5)
  $$\begin{aligned} G_{4,5}= & {} \left( \begin{array}{c} f_4(x)\theta _4(x)\\ xf_4(x)\theta _4(x)\\ x^2f_4(x)\theta _4(x)\\ x^3f_4(x)\theta _4(x)\\ 2\theta _4(x)\\ 2x\theta _4(x)\\ 2x^2\theta _4(x)\\ 2x^3\theta _4(x)\end{array}\right) \in \mathrm{M}_{8\times 2n}({\mathbb {Z}}_4), k_{0,4}=4 \text { and } k_{1,4}=4;\\ G_{5,5}= & {} \left( \begin{array}{c} 2{\overline{f}}_5(x)\theta _5(x)\\ 2x{\overline{f}}_5(x)\theta _5(x)\\ 2x^2{\overline{f}}_5(x)\theta _5(x)\\ 2x^3{\overline{f}}_5(x)\theta _5(x)\end{array}\right) \in \mathrm{M}_{4\times 2n}({\mathbb {Z}}_4), k_{0,5}=0 \text { and } k_{1,5}=4. \end{aligned}$$

• (iv-6) Let $a,b,c,d\in \{0,1\}$.

  $$\begin{aligned} G_{4,6}= & {} \left( \begin{array}{c} (f_4(x)+2(a+bx+cx^2+dx^3)\theta _4(x) \\ x(f_4(x)+2(a+bx+cx^2+dx^3)\theta _4(x) \\ x^2(f_4(x)+2(a+bx+cx^2+dx^3)\theta _4(x) \\ x^3(f_4(x)+2(a+bx+cx^2+dx^3)\theta _4(x) \end{array}\right) \\&\quad \in \mathrm{M}_{4\times 2n}({\mathbb {Z}}_4), k_{0,4}=4 \text { and } k_{1,4}=0;\\ G_{5,6}= & {} \left( \begin{array}{c} (f_5(x)+2(a+dx+cx^2+(1+a+b)x^3)\theta _5(x) \\ x(f_5(x)+2(a+dx+cx^2+(1+a+b)x^3)\theta _5(x) \\ x^2(f_5(x)+2(a+dx+cx^2+(1+a+b)x^3)\theta _5(x) \\ x^3(f_5(x)+2(a+dx+cx^2+(1+a+b)x^3)\theta _5(x)\end{array}\right) \\&\quad \in \mathrm{M}_{4\times 2n}({\mathbb {Z}}_4), k_{0,5}=4 \text { and } k_{1,5}=0, \end{aligned}$$

Then, by Theorem 1, all of the 315 self-dual codes over ${\mathbb {Z}}_4$ of length 30 are generated by one of the following 315 matrices:

$$\begin{aligned} G_{(i,j,l)}=\left( \begin{array}{c}G_1\\ G_{2,i}\\ G_{3,j}\\ G_{4,l} \\ G_{5,l}\end{array}\right) , \ 1\le i,j\le 2, \ 1\le l\le 6. \end{aligned}$$

Precisely, the self-dual codes over ${\mathbb {Z}}_4$ of length 30 with generator matrix $G_{(i,j,l)}$ are of type $4^{k_{2,0}+k_{3,0}+k_{4,0}}2^{2+k_{2,1}+k_{3,1}+k_{4,1}}$.