Nearly optimal codebooks from generalized Boolean bent functions over $ \mathbb{Z}_{4} $

Junchao Zhou; Yunge Xu; Lisha Wang; Nian Li

doi:10.3934/amc.2020121

Article Contents

2022, Volume 16, Issue 3: 485-501. Doi: 10.3934/amc.2020121

This issue Previous Article Three classes of partitioned difference families and their optimal constant composition codes Next Article Codes over

$ \frak m $

-adic completion rings

Nearly optimal codebooks from generalized Boolean bent functions over $ \mathbb{Z}_{4} $

1.
Hubei Key Laboratory of Applied Mathematics, Faculty of Mathematics and Statistics, Hubei University, Wuhan, 430062, China
2.
Faculty of Mathematics and Statistics, Hubei Engineering University, Xiaogan, 432000, China

^* Corresponding author: Lisha Wang
^* Corresponding author: Lisha Wang

Received: February 2020

Revised: September 2020

Early access: December 2020

Published: August 2022

This work was supported by the National Natural Science Foundation of China (Nos. 61702166, 61761166010) and Science Research Project of Hubei Provincial Department of Education (No. B2020150).

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Abstract

In this paper, based on the theory of $ \mathbb{Z}_{4} $-valued quadratic forms we propose several classes of generalized Boolean bent functions over $ \mathbb{Z}_{4} $, and new families of codebooks are constructed from these functions. The codebooks constructed in this paper are nearly optimal with respect to the Welch bound, and their parameters are new. Furthermore, some Boolean bent functions are also derived.

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Mathematics Subject Classification: Primary: 11T71; Secondary: 11T23.

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Table 1. The parameters of codebooks nearly achieving the Welch bound

Parameters $ (N, K) $	Constraints	$ I_{max} $	Ref.
$ \big(p^n, K=\frac{p-1}{2p}(p^n+p^{n/2})+1\big) $	$ p $ is an odd prime	$ \frac{(p+1)p^{n/2}}{2pK} $	[13]
$ \big(q^2, \frac{(q-1)^2}{2}\big) $	$ q $ is a power of an odd prime	$ \frac{q+1}{(q-1)^2} $	[36]
$ \big(q(q+4), \frac{(q+3)(q+1)}{2}\big) $	$ q $ is a prime power	$ \frac{1}{q+1} $	[16]
$ (q, \frac{q-1}{2}) $	$ q $ is a prime power	$ \frac{\sqrt{q}+1}{q-1} $	[16]
$ (p^n-1, \frac{p^n-1}{2}) $	$ p $ is an odd prime	$ \frac{\sqrt{p^n}+1}{p^n-1} $	[35]
$ (q^l+q^{l-1}-1, q^{l-1}) $	$ l>2 $	$ \frac{1}{\sqrt{q^{l-1}}} $	[39]
$ \big((q-1)^k+q^{k-1}, q^{k-1}\big) $	$ k>2, q\geq4 $	$ \frac{\sqrt{q^{k+1}}}{(q-1)^{k}+(-1)^{k+1}} $	[11]
$ \big((q-1)^k+K, K\big) $	$ k>2, K=\frac{(q-1)^{k}+(-1)^{k+1}}{q} $	$ \frac{\sqrt{q^{k-1}}}{K} $	[11]
$ \big((q^s-1)^n+K, K\big) $	$ s>1, n>1 $ and	$ \frac{\sqrt{q^{sn+1}}}{(q^s-1)^{n}+(-1)^{n+1}} $	[20]
	$ K=\frac{(q^s-1)^{n}+(-1)^{n+1}}{q} $
$ \big((q^s-1)^n+q^{sn-1}, q^{sn-1}\big) $	$ s>1, n>1 $	$ \frac{\sqrt{q^{sn+1}}}{(q^s-1)^{n}+(-1)^{n+1}} $	[20]
$ \big(q-1, \frac{q(r-1)}{2r}\big) $	$ r=p^t, q=r^s, p \nmid s $	$ \frac{\sqrt{r}}{\sqrt{q}(\sqrt{r}-1) K} $	[31]
$ \big(q^2, \frac{q(q+1)(r-1)}{2r}\big) $	$ r=p^t, q=r^s $	$ \frac{(r+1)q}{2rK} $	[31]
$ (q^3, q^2) $, $ (q^3+q^2, q^2) $	$ q $ is a prime power	$ \frac{1}{q} $	[19]
$ \big((q-1)q^2, (q-1)q\big) $,	$ q $ is a prime power	$ \frac{1}{q-1} $	[19]
$ \big((q^2-1)q, (q-1)q\big) $
$ \big(q^3-q^2-q+1, (q-1)^2\big) $,	$ q $ is a prime power	$ \frac{q}{(q-1)^2} $	[19]
$ \big(q^3-2q+1, (q-1)^2\big) $,
$ \big((q-1)^2q, (q-1)^2\big) $,
$ \big((q-1)q^2, (q-1)^2\big) $
$ \big((q-1)^2q, (q-1)(q-2)\big) $,	$ q $ is a prime power	$ \frac{q}{(q-1)(q-2)} $	[19]
$ \big(q^3-q^2-2q+2, (q-1)(q-2)\big) $
$ \big((q-1)^3, (q-2)^2\big) $,	$ q $ is a prime power	$ \frac{q}{(q-2)^2} $	[19]
$ \big(q^3-2q^2-q+3, (q-2)^2\big) $
$ (2^{\frac{n}{r}+n}+2^n, 2^n) $	$ 1<r<n, r\mid n $	$ \frac{1}{\sqrt{2^n}} $	This paper

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