DAFNet: A dual attention-guided fuzzy network for cardiac MRI segmentation

Yuxin Luo; Yu Fang; Guofei Zeng; Yibin Lu; Li Du; Lisha Nie; Pu-Yeh Wu; Dechuan Zhang; Longling Fan; Yuxin Luo; Yu Fang; Guofei Zeng; Yibin Lu; Li Du; Lisha Nie; Pu-Yeh Wu; Dechuan Zhang; Longling Fan

doi:10.3934/math.2024429

AIMS Mathematics

2024, Volume 9, Issue 4: 8814-8833. doi: 10.3934/math.2024429

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Research article Special Issues

DAFNet: A dual attention-guided fuzzy network for cardiac MRI segmentation

1.
Faculty of Science, Kunming University of Science and Technology, Kunming, 650500, China
2.
Key Laboratory of Applied Statistics and Data Analysis, Department of Education of Yunnan Province, Kunming, 650500, China
3.
Department of Radiology, Chongqing Hospital of Traditional Chinese Medicine, Chongqing, 400021, China
4.
GE Healthcare, MR Research China, Beijing, 100176, China
† These two authors contributed equally.

Received: 05 December 2023 Revised: 07 February 2024 Accepted: 26 February 2024 Published: 01 March 2024
MSC : 68T07, 60A86

Background

In clinical diagnostics, magnetic resonance imaging (MRI) technology plays a crucial role in the recognition of cardiac regions, serving as a pivotal tool to assist physicians in diagnosing cardiac diseases. Despite the notable success of convolutional neural networks (CNNs) in cardiac MRI segmentation, it remains a challenge to use existing CNNs-based methods to deal with fuzzy information in cardiac MRI. Therefore, we proposed a novel network architecture named DAFNet to comprehensively address these challenges.

Methods

The proposed method was used to design a fuzzy convolutional module, which could improve the feature extraction performance of the network by utilizing fuzzy information that was easily ignored in medical images while retaining the advantage of attention mechanism. Then, a multi-scale feature refinement structure was designed in the decoder portion to solve the problem that the decoder structure of the existing network had poor results in obtaining the final segmentation mask. This structure further improved the performance of the network by aggregating segmentation results from multi-scale feature maps. Additionally, we introduced the dynamic convolution theory, which could further increase the pixel segmentation accuracy of the network.

Result

The effectiveness of DAFNet was extensively validated for three datasets. The results demonstrated that the proposed method achieved DSC metrics of 0.942 and 0.885, and HD metricd of 2.50mm and 3.79mm on the first and second dataset, respectively. The recognition accuracy of left ventricular end-diastolic diameter recognition on the third dataset was 98.42%.

Conclusion

Compared with the existing CNNs-based methods, the DAFNet achieved state-of-the-art segmentation performance and verified its effectiveness in clinical diagnosis.

Keywords:

Citation: Yuxin Luo, Yu Fang, Guofei Zeng, Yibin Lu, Li Du, Lisha Nie, Pu-Yeh Wu, Dechuan Zhang, Longling Fan. DAFNet: A dual attention-guided fuzzy network for cardiac MRI segmentation[J]. AIMS Mathematics, 2024, 9(4): 8814-8833. doi: 10.3934/math.2024429

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Abstract

Background

Methods

Result

Conclusion

Compared with the existing CNNs-based methods, the DAFNet achieved state-of-the-art segmentation performance and verified its effectiveness in clinical diagnosis.

1. Introduction

Throughout this paper, let $p$ be an odd prime and $q = p^n$ for some positive integer $n$ . Codebooks (also known as signal sets) with low coherence are typically used to distinguish signals of different users in code division multiple access (CDMA) systems. An $(N, K)$ codebook $\mathcal{C}$ is a finite set $\{\mathbf{c}_0, \mathbf{c}_1, \ldots, \mathbf{c}_{N-1}\}$ , where the codeword $\mathbf{c}_i$ , $0\leq i \leq N-1$ , is a unit norm $1\times K$ complex vector over an alphabet $\mathcal{A}$ . The maximum inner-product correlation $I_{\max}(\mathcal{C})$ of $\mathcal{C}$ is defined by

$\begin{align*} I_{\max}(\mathcal{C}) = \max\limits_{0\leq i\neq j\leq N-1}|\mathbf{c}_i\mathbf{c}_j^{H}|, \end{align*}$

where $\mathbf{c}_j^{H}$ denotes the conjugate transpose of $\mathbf{c}_j$ . The maximal cross-correlation amplitude $I_{\max}(\mathcal{C})$ of $\mathcal{C}$ is an important index of $\mathcal{C}$ , as it can approximately optimize many performance metrics such as outage probability and average signal-to-noise ratio. For a fixed $K$ , researchers are highly interested in designing a codebook $\mathcal{C}$ with the parameter $N$ being as large as possible and $I_{\max}(\mathcal{C})$ being as small as possible simultaneously. Unfortunately, there exists a bound between the parameters $N$ , $K$ and $I_{\max}(\mathcal{C})$ .

Lemma 1. (^[1]) For any $(N, K)$ codebook $\mathcal{C}$ with $N\geq K$ ,

$\begin{align} I_{\max}(\mathcal{C})\geq \sqrt{\frac{N-K}{(N-1)K}}. \end{align}$

(1.1)

The bound in (1.1) is called the Welch bound of $\mathcal{C}$ and is denoted by $I_{w}(\mathcal{C})$ . If the codebook $\mathcal{C}$ achieves $I_{w}(\mathcal{C})$ , then $\mathcal{C}$ is said to be optimal with respect to the Welch bound. However, constructing codebooks achieving the Welch bound is extremely difficult. Hence, many researchers have focused their main energy on constructing asymptotically optimal codebooks, i.e., $I_{\max}(\mathcal{C})$ asymptotically meets the Welch bound $I_{w}(\mathcal{C})$ for sufficiently large $N$ ^{[2,3,4,5,6,7]}.

The objective of this paper is to construct a class of complex codebooks and investigate their maximum inner-product correlation. Results show that these constructed complex codebooks are nearly optimal with respect to the Welch bound, i.e., the ratio of their maximal cross-correlation amplitude to the Welch bound approaches $1$ . These codebooks may have applications in strongly regular graphs ^[8], combinatorial designs ^[9,10], and compressed sensing ^[11,12].

This paper is organized as follows. In Section 2, we review some essential mathematical concepts regarding characters and Gauss sums over finite fields. In Section 3, we present a class of asymptotically optimal codebooks using the trace functions and multiplicative characters over finite fields. Finally, we make a conclusion in Section 4.

2. Preliminaries

In this section, we review some essential mathematical concepts regarding characters and Gauss sums over finite fields. These concepts will play significant roles in proving the main results of this paper.

Let $n$ be a positive integer and $p$ an odd prime. Denote the finite field with $p^n$ elements by $\mathbb{F}_{p^n}$ . The trace function ${\operatorname{Tr}}_n$ from $\mathbb{F}_{p^n}$ to $\mathbb{F}_p$ is defined by

$\begin{align*} {\operatorname{Tr}}_n(x) = \sum\limits_{i = 0}^{n-1}x^{p^i}. \end{align*}$

Let $\zeta_p$ denote a primitive $p$ -th root of complex unity and ${\operatorname{Tr}}_n$ denote the trace function from $\mathbb{F}_{p^n}$ to $\mathbb{F}_p$ . For $x\in \mathbb{F}_{p^n}$ , it can be checked that $\chi_n$ given by $\chi_n(x) = \zeta_p^{{\operatorname{Tr}}_n(x)}$ is an additive character of $\mathbb{F}_{p^n}$ , and $\chi_n$ is called the canonical additive character of $\mathbb{F}_{p^n}$ . Assume $a\in \mathbb{F}_{p^n}$ , then every additive character of $\mathbb{F}_{p^n}$ can be obtained by $\mu_a(x) = \chi_n(ax)$ where $\ x\in \mathbb{F}_{p^n}$ . The orthogonality relation of $\mu_a$ is given by

$\begin{align} \sum\limits_{x\in \mathbb{F}_{p^n}}\mu_a(x) = \left\{\begin{array}{ll} p^n, & \text{if}\ a = 0, \\ 0, & \text{otherwise}. \end{array} \right. \end{align}$

(2.1)

Let $q = p^n$ and $\alpha$ be a primitive element of $\mathbb{F}_{q}^{*}$ , then all multiplicative characters of $\mathbb{F}_{q}$ are given by $\varphi_j(\alpha^i) = \zeta_{q-1}^{ij}$ , where $\zeta_{q-1}$ denotes a primitive $(q-1)$ -th root of unity and $0 \leq i, j\leq q-2$ . The quadratic character of $\mathbb{F}_{q}$ is the character $\varphi_{(q-1)/2}$ , which will be denoted by $\eta_n$ in the sequel, and $\eta_n$ is extended by setting $\eta_n(0) = 0$ . For $\varphi_j$ , its orthogonality relation is given by

$\begin{align*} \sum\limits_{x\in \mathbb{F}_{p^n}}\varphi_j(x) = \left\{\begin{array}{ll} q-1, & \text{if}\ j = 0, \\ 0, & \text{otherwise}. \end{array} \right. \end{align*}$

The Gauss sum $G(\eta_n)$ over $\mathbb{F}_{p^n}$ is defined by

$G(\eta_n) = \sum\limits_{x\in \mathbb{F}_{p^n}^{*}} \eta_n(x)\chi_n(x).$

The explicit value of $G(\eta_n)$ is given in the following lemma.

Lemma 2 (^[13], Theorem 5.15). With symbols and notations above, we have

$\begin{align*} G(\eta_n) = (-1)^{n-1}(-1)^{\frac{(p-1)n}{4}}q^{\frac{1}{2}}. \end{align*}$

The following results on exponential sums will play an important role in proving the main results of this paper.

Lemma 3 (^[13], p.195). With symbols and notations above, we have

$\begin{align*} \eta_1(x) = \frac{1}{p}\sum\limits_{a\in\mathbb{F}_{p}}G(\eta_1)\eta_1(-a)\chi_1(ax), \end{align*}$

where $\eta_1$ denotes the quadratic character and $\chi_1$ the canonical additive character of $\mathbb{F}_p$ .

Lemma 4 (^[13], Theorem 5.33). If $f(x) = a_2x^2+a_1x+a_0\in \mathbb{F}_{p^n}[x]$ with $a_2\neq0$ , then

$\begin{align*} \sum\limits_{x\in \mathbb{F}_{p^n}}\zeta_p^{{\operatorname{Tr}}_n(f(x))} = \eta_n(a_2)G(\eta_n)\zeta_p^{{\operatorname{Tr}}_n\left(a_0-a_1^2(4a_2)^{-1}\right)}. \end{align*}$

Lemma 5 (^[14], Theorem 2). Let $n = 2m$ be an even integer and $z\in \mathbb{F}_{p}^{*}$ , then

$\begin{align*} \sum\limits_{x\in \mathbb{F}_{p^n}}\zeta_p^{z{\operatorname{Tr}}_n\left(x^{p^m+1}\right)} = -p^m. \end{align*}$

Lemma 6 (^[15]). Let $n = 2m$ be an even integer, $a\in \mathbb{F}_{p^m}^{*}$ , and $b\in \mathbb{F}_{p^n}$ , then

$\begin{align*} \sum\limits_{x\in \mathbb{F}_{p^{n}}}\zeta_p^{{\operatorname{Tr}}_m\left(ax^{p^m+1}\right)+{\operatorname{Tr}}_n(bx)} = -p^m\zeta_p^{-{\operatorname{Tr}}_m\left(\frac{b^{p^m+1}}{a}\right)}. \end{align*}$

Lemma 7 (^[16], Lemma 3.12). If $A$ and $B$ are finite abelian groups, then there is an isomorphism

$\begin{align*} \widehat{A\times B}\cong \widehat{A} \times \widehat{B}, \end{align*}$

where $\widehat{A}$ consists of all characters of $A$ .

By this lemma, we know that

$\widehat{\mathbb{F}_{p^n}^{+}\times \mathbb{F}_{p^n}^{+}} = \{\mu_{a, b}: a, b\in \mathbb{F}_{p^n}\}$

where

$\mu_{a, b}(x, y) = \zeta_p^{{\operatorname{Tr}}_n(ax+by)}$

for $x, y\in \mathbb{F}_{p^n}$ .

3. Proofs and main results

In this section, we always suppose that $n = 2m$ is an even integer and $p$ is an odd prime. The set $D$ is defined as follows:

$\begin{align*} D = \left\{(x, y)\in \mathbb{F}_{p^n}\times \mathbb{F}_{p^n}:\eta_1\left({\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)\right) = 1\right\}, \end{align*}$

where $\eta_1$ is the quadratic character of $\mathbb{F}_p$ . A codebook $\mathcal{C}$ is constructed by

$\begin{align} \mathcal{C} = \left\{\mathbf{c}_{a, b}:a, b\in \mathbb{F}_{p^n}\right\}, \end{align}$

(3.1)

where $\mathbf{c}_{a, b} = \frac{1}{\sqrt{|D|}}\left(\mu_{a, b}(x, y)\right)_{(x, y)\in D}$ , $\mu_{a, b}(x, y) = \zeta_p^{{\operatorname{Tr}}_n(ax+by)}$ for $(x, y)\in D$ and $|D|$ denotes the cardinality of the set $D$ .

Lemma 8. With symbols and notations as above, we have

$\begin{align*} |D| = \frac{p-1}{2}\left(p^{2n-1}-(-1)^{\frac{n(p-1)}{4}}p^{n-1}\right). \end{align*}$

Proof. Let

$\begin{align*} A_1 = \sum\limits_{x, y\in \mathbb{F}_{p^n} \atop {\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right) = 0}1, \ A_2 = \sum\limits_{x, y\in \mathbb{F}_{p^n}\atop {\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)\neq0}\eta_1\left({\operatorname{Tr}}_n(x^2+y^{p^m+1})\right). \end{align*}$

Note that

$\begin{align*} \sum\limits_{x, y\in \mathbb{F}_{p^n} \atop {\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right) = 0}1+\sum\limits_{x, y\in \mathbb{F}_{p^n}\atop {\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)\neq0}1 = p^{2n}. \end{align*}$

Together with the definition of $D$ , we have

$\begin{align} |D|& = \sum\limits_{x, y\in\mathbb{F}_{p^n}\atop {\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)\neq0}\frac{\eta_1\left(\mathrm{Tr}_n\left(x^2+y^{p^m+1}\right)\right)+1}2 \\ & = \frac12\sum\limits_{x, y\in\mathbb{F}_{p^n}\atop{\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)\neq0}\eta_1\left(\mathrm{Tr}_n\left(x^2+y^{p^m+1}\right)\right) +\frac12\sum\limits_{x, y\in\mathbb{F}_{p^n}\atop{\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)\neq0}1 \\ & = \frac{A_{2}}{2}+\frac{p^{2n}}{2}-\frac{1}{2}\sum\limits_{x, y\in\mathbb{F}_{p^n}\atop{\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right) = 0}1 \\ & = \frac12\left(p^{2n}-A_1+A_2\right). \end{align}$

(3.2)

By definition, we have

$\begin{align} A_1& = \frac{1}{p}\sum\limits_{x, y\in \mathbb{F}_{p^n}}\sum\limits_{z\in \mathbb{F}_p}\zeta_p^{z{\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)} \\ & = p^{2n-1}+\frac{1}{p}\sum\limits_{z\in \mathbb{F}_{p}^{*}}\sum\limits_{x, y\in \mathbb{F}_{p^n}}\zeta_p^{{\operatorname{Tr}}_n\left(zx^2\right)+{\operatorname{Tr}}_n\left(zy^{p^m+1}\right)} \\ & = p^{2n-1}+(-1)^{\frac{n(p-1)}{4}}p^{n-1}(p-1). \end{align}$

(3.3)

where the last equality follows from Lemmas 2, 4, and 5. Note that $\eta_n(z) = 1$ for $z\in \mathbb{F}_{p}^{*}$ if $n$ is even. By Lemma 3, we obtain

$\begin{align*} A_2 = \frac{G(\eta_n)}{p}\sum\limits_{a\in \mathbb{F}_{p}^{*}}\eta_1(-a)\sum\limits_{x\in \mathbb{F}_{p^n}}\zeta_p^{{\operatorname{Tr}}_n\left(ax^2\right)}\sum\limits_{y\in \mathbb{F}_{p^n}}\zeta_p^{{\operatorname{Tr}}_n\left(ay^{p^m+1}\right)} \end{align*}$

Using Lemmas 4 and 5, we get

$A_2 = -p^{m-1}G^2(\eta_n)\sum\limits_{a\in \mathbb{F}_{p}^{*}}\eta_1(-a) = 0.$

The desired conclusion follows from (3.2) and (3.3). □

Example 1. Let $p = 5$ and $n = 2$ . By the Magma program, we know that $|D| = 240$ , which is consistent with Lemma 8.

Theorem 9. Let symbols and notations be the same as before, then the codebook $\mathcal{C}$ defined in (3.1) has parameters $[p^{2n}, K]$ ,

$\begin{align*} K = \frac{p-1}{2}\left(p^{2n-1}-(-1)^{\frac{n(p-1)}{4}}p^{n-1}\right), \end{align*}$

and

$I_{\max}(\mathcal{C}) = (p+1)p^{n-1}/(2K).$

Proof. By the definition of the set $\mathcal{C}$ and Lemma 8, we deduce that $\mathcal{C}$ is a $[p^{2n}, K]$ codebook. If $a, b\in \mathbb{F}_{p^n}$ and $(a, b)\neq (0, 0)$ , then we have

$\begin{align*} \sum\limits_{x, y\in \mathbb{F}_{p^n} }\zeta_p^{{\operatorname{Tr}}_n(ax+by)} = \sum\limits_{x, y\in \mathbb{F}_{p^n} \atop {\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right) = 0}\zeta_p^{{\operatorname{Tr}}_n(ax+by)}+\sum\limits_{x, y\in \mathbb{F}_{p^n} \atop {\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)\neq0}\zeta_p^{{\operatorname{Tr}}_n(ax+by)} = 0 \end{align*}$

This implies that

$\begin{align*} \sum\limits_{x, y\in \mathbb{F}_{p^n} \atop {\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right) = 0}\zeta_p^{{\operatorname{Tr}}_n(ax+by)} = -\sum\limits_{x, y\in \mathbb{F}_{p^n} \atop {\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)\neq0}\zeta_p^{{\operatorname{Tr}}_n(ax+by)}. \end{align*}$

For $a, b\in \mathbb{F}_{p^n}$ and $(a, b)\neq (0, 0)$ , we have that

$\begin{align} \sum\limits_{(x, y)\in D}\mu_{a, b}(x, y) = &\sum\limits_{x, y\in \mathbb{F}_{p^n} \atop {\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)\neq0}\zeta_p^{{\operatorname{Tr}}_n(ax+by)}\frac{\eta_1\left({\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)\right)+1}{2} \\ = &-\frac{1}{2}\sum\limits_{x, y\in \mathbb{F}_{p^n} \atop {\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right) = 0}\zeta_p^{{\operatorname{Tr}}_n(ax+by)}+\frac{1}{2}\sum\limits_{x, y\in \mathbb{F}_{p^n} \atop {\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)\neq0}\zeta_p^{{\operatorname{Tr}}_n(ax+by)}\eta_1\left({\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)\right) \\ = &\frac{1}{2}(-B_1+B_2), \end{align}$

(3.4)

where

$\begin{align*} B_1 = \sum\limits_{x, y\in \mathbb{F}_{p^n} \atop {\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right) = 0}\zeta_p^{{\operatorname{Tr}}_n(ax+by)}, \ B_2 = \sum\limits_{x, y\in \mathbb{F}_{p^n} }\zeta_p^{{\operatorname{Tr}}_n(ax+by)}\eta_1\left({\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)\right). \end{align*}$

By (2.1), we derive that

$\begin{align*} \sum\limits_{z\in \mathbb{F}_p}\zeta_p^{z{\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)} = \left\{\begin{array}{ll} p, & \text{if}\ {\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right) = 0, \\ 0, & \text{otherwise}. \end{array} \right. \end{align*}$

Combining Lemmas 4 and 6, we get that

$\begin{align} B_1 = &\frac{1}{p}\sum\limits_{x, y\in \mathbb{F}_{p^n} }\sum\limits_{z\in \mathbb{F}_p} \zeta_p^{{\operatorname{Tr}}_n(ax+by)} \zeta_p^{z{\operatorname{Tr}}_n\left(x^2+y^{p^m+1}\right)}\\ = &\frac{1}{p}\sum\limits_{z\in \mathbb{F}_p^{*}}\sum\limits_{x, y\in \mathbb{F}_{p^n} }\zeta_p^{{\operatorname{Tr}}_n\left(zx^2+ax\right)}\zeta_p^{{\operatorname{Tr}}_n\left(zy^{p^m+1}+by\right)}\\ = &-p^{m-1}G(\eta_n)\sum\limits_{z\in \mathbb{F}_{p}^{*}}\eta_n(z)\zeta_p^{{\operatorname{Tr}}_n\left(a^2+b^{p^m+1}\right)z} \\ = &\left\{\begin{array}{ll} -p^{m-1}(p-1)G(\eta_n), & \text{if}\ {\operatorname{Tr}}_n\left(a^2+b^{p^m+1}\right) = 0, \\ p^{m-1}G(\eta_n), & \text{if}\ {\operatorname{Tr}}_n\left(a^2+b^{p^m+1}\right)\neq0, \end{array} \right. \end{align}$

(3.5)

where $G(\eta_n)$ is given in Lemma 2. By Lemma 3, we have that

$\begin{align*} B_2 = \frac{G(\eta_1)}{p}\sum\limits_{z\in \mathbb{F}_{p}^{*}}\eta_1(-z)\sum\limits_{x\in \mathbb{F}_{p^n}}\zeta_p^{{\operatorname{Tr}}_n\left(zx^2+ax\right)}\sum\limits_{y\in \mathbb{F}_{p^n}}\zeta_p^{{\operatorname{Tr}}_n\left(zy^{p^m+1}+by\right)}. \end{align*}$

Moreover, by Lemmas 4 and 6, we obtain

$\begin{align} B_2& = -p^{m-1}G(\eta_1)G(\eta_n)\sum\limits_{z\in \mathbb{F}_{p}^{*}}\eta_1(z)\zeta_p^{z{\operatorname{Tr}}_n\left(a^2+b^{p^m+1}\right)} \\ & = \left\{\begin{array}{ll} 0, & \text{if}\ {\operatorname{Tr}}_n\left(a^2+b^{p^m+1}\right) = 0, \\ -p^{m}\left(\frac{-1}{p}\right)G(\eta_n)\eta_1\left({\operatorname{Tr}}_n\left(a^2+b^{p^m+1}\right)\right), & \text{if}\ {\operatorname{Tr}}_n\left(a^2+b^{p^m+1}\right)\neq0. \end{array} \right. \end{align}$

(3.6)

It follows from Lemma 2 and (3.4) that

$\begin{align} \sum\limits_{(x, y)\in D}\mu_{a, b}(x, y)\in \left\{\frac{1}{2}(p-1)p^{m-1}G(\eta_n), -\frac{1}{2}(p+1)p^{m-1}G(\eta_n)\right\}. \end{align}$

(3.7)

For any two distinct codewords $\mathbf{c}_{z_1, z_2}$ , $\mathbf{c}_{z_1^{'}, z_2^{'}}\in \mathcal{C}$ , i.e., $(z_1, z_2)\neq (z_1^{'}, z_2^{'})$ , it is easy to check that

$\begin{align} \left|\mathbf{c}_{z_1, z_2}\mathbf{c}_{z_1^{'}, z_2^{'}}^{H}\right| = \frac{1}{K}\left|\sum\limits_{(x, y)\in D}\mu_{z_1-z_1^{'}, z_2-z_2^{'}}(x, y)\right|. \end{align}$

(3.8)

Combining (3.7) and (3.8), we get that $I_{\max}(\mathcal{C}) = (p+1)p^{n-1}/(2K).$ □

Example 2. Let $f(x)$ be an irreducible polynomial over the field $\mathbb{F}_3$ and $f(x) = x^2+x+2$ in $\mathbb{F}_3[x]$ . Suppose that $p = 3$ , $n = 2$ , and $\alpha$ is a root of $f(x)$ over $\mathbb{F}_3$ , then $m = 1$ , $q = 3^{2}$ , and $\mathbb{F}_{9} = \mathbb{F}_{3}(\alpha)$ . It can be verified that the set $D$ consists of the following $30$ elements:

$\begin{align*} D& = \left\{(x, y)\in\mathbb{F}_{9}\times\mathbb{F}_{9}:\mathrm{Tr}_{2}\left(x^{2}+y^{4}\right) = 1\right\} \\ & = \left\{ \begin{array}{lllll} (1+2\alpha, 0), &(2+\alpha, 0), &(1, 1), &(1, 2), &(1, 1+2\alpha), \\ (1, 2+\alpha), &(2, 1), &(2, 2), &(2, 1+2\alpha), &(2, 2+\alpha), \\ (\alpha, \alpha), &(\alpha, 2\alpha), &(\alpha, 1+\alpha), &(\alpha, 2+2\alpha), &(2\alpha, \alpha), \\ (2\alpha, 2\alpha), &(2\alpha, 1+\alpha), &(2\alpha, 2+2\alpha), &(0, \alpha), &(0, 2\alpha), \\ (0, 1+\alpha), &(0, 2+2\alpha), &(1+\alpha, \alpha), &(1+\alpha, 2\alpha), &(1+\alpha, 1+\alpha), \\ (1+\alpha, 2+2\alpha), &(2+2\alpha, \alpha), &(2+2\alpha, 2\alpha), &(2+2\alpha, 1+\alpha), &(2+2\alpha, 2+2\alpha) \end{array} \right\}. \end{align*}$

The corresponding codebook $\mathcal{C}$ is given by

$\begin{align*} \mathcal{C} = \left\{\frac{1}{\sqrt{30}}\left(\zeta_3^{{\operatorname{Tr}}_2(ax+by)}\right)_{(x, y)\in D}:a, b\in \mathbb{F}_9\right\}, \end{align*}$

where $\zeta_3 = e^{\frac{2\pi\sqrt{-1}}{3}}$ and ${\operatorname{Tr}}_2$ denotes the trace function from $\mathbb{F}_9$ to $\mathbb{F}_3$ .

Corollary 10. The codebook $\mathcal{C}$ constructed in (3.1) is asymptotically optimal with respect to the Welch bound.

Proof. The corresponding Welch bound is

$\begin{align*} I_{w}(\mathcal{C}) = \sqrt{\frac{p^{2n-1}(p+1)+(-1)^{\frac{n(p-1)}{4}}p^{n-1}(p-1)}{\left(p^{2n}-1\right)(p-1)\left(p^{2n-1}-(-1)^{\frac{n(p-1)}{4}}p^{n-1}\right)}} . \end{align*}$

We deduce that

$\begin{align*} \lim\limits_{p^n\rightarrow +\infty} \frac{I_{w}(\mathcal{C})}{I_{\max}(\mathcal{C})} = \lim\limits_{p^n\rightarrow +\infty}\sqrt{\frac{4K\left(p^{2n}-K\right)}{\left(p^{2n}-1\right)(p+1)^2p^{2n-2}}} = 1, \end{align*}$

which implies that $\mathcal{C}$ asymptotically meets the Welch bound. □

In , we show some parameters of some specific codebooks defined in (3.1). From this table, we conclude that $I_{\max}(\mathcal{C})$ is very close to $I_w(\mathcal{C})$ for largely enough $p$ , which ensures the correctness of Theorem 9 and Corollary 1.

Table 1. The parameters of the codebook

$\mathcal{C}$ in (3.1) for

$n = 4$ .

$p$	$N$	$K$	$I_{\max}(\mathcal{C})$	$I_w(\mathcal{C})$	$I_{\max}(\mathcal{C})/I_w(\mathcal{C})$
3	$3^8$	$2160$	1/40	$1.762 \times 10^{-2}$	1.4185
7	$7^8$	$2469600$	1/1800	$4.811\times 10^{-4}$	1.1548
11	$11^8$	$97429200$	1/12200	$7.483 \times 10^{-5}$	1.0955
13	$13^8$	$376477920$	1/24480	$3.782 \times 10^{-5}$	1.0801
17	$17^8$	$3282670080$	1/74240	$1.2699 \times 10^{-5}$	1.0607

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4. Concluding remarks

This paper presented a family of codebooks by the combination of additive characters and multiplicative characters over finite fields. Results show that the constructed codebooks are asymptotically optimal in the sense that the maximum cross correlation amplitude of the codebooks asymptotically achieves the Welch bound. As a comparison, parameters of some known nearly optimal codebooks and the constructed ones are listed in . From this table, we can conclude that the parameters of $\mathcal{C}$ are not covered by those in ^{[2,3,4,5,6,17,18,19]}. This means the presented codebooks have new parameters.

Table 2. The parameters of codebooks asymptotically meeting the Welch bound.

Ref.	Parameters $(N, K)$	Constraints
^[2]	$\left((q-1)^\ell+M, M \right)$ , $M=\frac{(q-1)^\ell+(-1)^{\ell+1}}{q}$ .	$q$ is a prime power, $\ell>2$ .
^[3]	$\left(2K+(-1)^{ln}, K\right)$ , $K=\frac{(q_1-1)^n\cdots (q_l-1)^n-(-1)^{ln}}{2}$ .	$1\leq i\leq l$ , $s_i>1$ , $q_i=2^{s_i},$ $l>1$ and $n>1$ .
^[4]	$\left((q^s-1)^m+q^{sm-1}, q^{sm-1}\right)$	$s>1$ , $m>1$ , $q$ is a prime power.
^[5]	$\left((p_{{\min}}+1)Q^2, Q^2\right)$	$Q>1$ is an integer, $p_{{\min}}$ is the smallest prime factor of $Q$ .
^[6]	$\left(p_{\min}N_1N_2, N_1N_2\right)$	$N_1\geq1$ , $N_2=N_1+o(N_1)$ , $p_{{\min}}$ is the smallest prime factor of $N_2$ .
^[6]	$\left(p_{\min}N_1N_2, N_1N_2\right)$	$N_1\geq1$ , $N_2=N_1+o(N_1)$ , $p_{{\min}}$ is the smallest prime factor of $N_2$ .
^[17]	$\left(q_1q_2\cdots q_l, \left(q_1q_2\cdots q_l-1\right)/2\right)$	$1\leq i\leq l$ , $q_i$ is a prime power, $q_i\equiv3\pmod{4}$
^[18]	$\left(q, \frac{q+1}{2}\right)$	$q$ is a prime power.
^[19]	$\left(q^3+q^2, q^2\right)$ or $\left(q^3+q^2-q, q^2-q\right)$	$q$ is a prime power.
Thm. 3.1	$\left(p^{2n}, \frac{p-1}{2}\left(p^{2n-1}-(-1)^{\frac{n(p-1)}{4}}p^{n-1}\right)\right)$	$p$ is an odd prime, $n=2m$ , $m$ is a positive integer.

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Use of AI tools declaration

The authors declare they have not used Artificial Intelligence (AI) tools in the creation of this article.

Acknowledgments

This work was supported by the Innovation Project of Engineering Research Center of Integration and Application of Digital Learning Technology (No.1221003), Humanities and Social Sciences Youth Foundation of Ministry of Education of China (No. 22YJC870018), the Science and Technology Development Fund of Tianjin Education Commission for Higher Education (No. 2020KJ112, 2022KJ075, KYQD1817), the National Natural Science Foundation of China (Grant No. 12301670), the Natural Science Foundation of Tianjin (Grant No. 23JCQNJC00050), Haihe Lab. of Information Technology Application Innovation (No. 22HHXCJC00002), Fundamental Research Funds for the Central Universities, China (Grant No. ZY2301, BH2316), the Open Project of Tianjin Key Laboratory of Autonomous Intelligence Technology and Systems (No. TJKL-AITS-20241004, No. TJKL-AITS-20241006).

Conflict of interest

The authors declare no conflicts of interest.

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